@KirkDBorne: Introduction to Graph Theory: https://arxiv.org/abs/2308.04512 (454-page PDF) — edges, vertices, trees, networks, paths…

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Summary

Darij Grinberg's 454-page graduate-level textbook 'Introduction to Graph Theory' (Spring 2025 edition, Drexel University Math 530) covers simple graphs, trees, Eulerian/Hamiltonian cycles, colorings, matchings, and network flows, with around a hundred exercises.

Introduction to Graph Theory: https://t.co/wP0nfijOxN (454-page PDF) — edges, vertices, trees, networks, paths, and graphs (and theorems!) ————— #Mathematics #NetworkScience https://t.co/5sNMHObIh7

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An introduction to graph theory(Text for Math 530 in Spring 2022 at Drexel University)

Source: https://arxiv.org/html/2308.04512 Darij Grinberg††thanks:Drexel University, Korman Center, 15 S 33rd Street, Office #263, Philadelphia, PA 19104 (USA). //[email protected]//http://www.cip.ifi.lmu.de/˜grinberg/

Spring 2025 edition, June 7, 2025

Abstract

Abstract.This is a graduate-level introduction to graph theory, corresponding to a quarter-long course. It covers simple graphs, multigraphs as well as their directed analogues, and more restrictive classes such as tournaments, trees and arborescences. Among the features discussed are Eulerian circuits, Hamiltonian cycles, spanning trees, the matrix-tree and BEST theorems, proper colorings, Turan’s theorem, bipartite matching and the Menger and Gallai–Milgram theorems. The basics of network flows are introduced in order to prove Hall’s marriage theorem.

Around a hundred exercises are included (without solutions).

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1Preface

1.1What is this?

This is a course ongraphs– a rather elementary concept (actually a cluster of closely related concepts) that can be seen all over mathematics. We will discuss several kinds of graphs (simple graphs, multigraphs, directed graphs, etc.) and study their features and properties. In particular, we will encounter walks on graphs, matchings of graphs, flows on networks (networks are graphs with extra data), and take a closer look at certain types of graphs such as trees and tournaments.

The theory of graphs goes back at least to Leonhard Euler, who ina 1736 paper[Euler36](see[Euler53]for anEnglish translation) solved a puzzle about an optimal tour of the town of Königsberg. It saw some more developments in the 19th century and straight-up exploded in the 20th; now it is one of the most active fields of mathematics. There are now dozens (if not hundreds) textbooks available on the subject, such as

•the comprehensive works[BonMur08],[Berge91],[Ore74],[Bollob98], [Dieste17],[ChLeZh16],[Jungni13]
•or the more introductory[Ore96],[BenWil06, Chapters 5–6],[Bollob71],[Griffi21],[Galvin21],[Guicha16, Chapter 5],[Harary69],[Harju14], [HaHiMo08, Chapter 1],[Wilson10],[Tait21],[LeLeMe18, Chapters 10–13],[Merris00],[Bona23, Part III],[Ruohon13],[KelTro17],[LoPeVe03],[West01],[Verstr21],[HarRin03],[Bickle24].

These texts are written at different levels of sophistication, rigor and detail, are tailored to different audiences, and (beyond the absolute basics) often cover different ground (for instance,[Dieste17]distinguishes itself by treating infinite and random graphs, whereas[Even12]focusses on algorithms, and[Griffi21]is strong on applications).

The present notes are self-contained and do not follow any existing book. Nevertheless, I recommend skimming the texts cited above to gain a wider perspective on graph theory (far beyond what we can cover in an introductory course), and perhaps marking some for later reading. Our focus in these notes is on the more discrete and algebraic sides of graph theory (finite graphs of various kinds, existential results, counting formulas), and they are limited both by the time constraints (being written for a quarter-long course) and the limits of my own knowledge.

1.1.1Remarks

Prerequisites.These notes target a graduate-level (or advanced undergraduate) reader. A certain mathematical sophistication and willingness to think along (as well as invent one’s own examples) is expected. Beyond that, the main prerequisites are the basic properties of determinants, polynomials and finite sums. Rings and fields are occasionally mentioned, but the reader can make do with just the most basic examples thereof (ℚ\mathbb{Q},ℝ\mathbb{R}, polynomial rings and matrix rings; also the finite field𝔽2\mathbb{F}_{2}in a few places). No analysis (or even calculus) is required anywhere in this text.

Course websites.These notes were written for my Math 530 course at Drexel University in Spring 2022. The website of this course can be found at

An older, but similarly structured course is my Spring 2017 course at the University of Minnesota. Its website is available at

and contains some additional materials (such as solutions to some selected exercises, a few more detailed topics, and astub of a text[17s]that covers parts of our Chapter2in more depth). If you are reading the present notes on the arXiv, then said additional materials can also be found as ancillary files to this arXiv submission.

Exercises.These notes include exercises of varying difficulty and significance. Almost all of the exercises are optional (i.e., they are not used anywhere in the text, except perhaps in other exercises), but they often provide practice, context and additional inspiration. Naturally, one person’s inspiration is another’s distraction, so I do not recommend assigning too much importance to any specific exercise; it is usually better to read on than to dwell for hours. However, a dozen minutes of thought per exercise will likely not be a waste of time.

Acknowledgments.I have learned a lot from conversations with Joel Brewster Lewis, Lukas Katthän and Victor Reiner. Chiara Libera Carnevale, Amanda Johnson, Victor Michel and Eric Müller corrected errors in previous versions of these notes. I am indebted to all of the above, and would appreciate any further input – please [email protected] any corrections (however small) and suggestions.

1.2Notations

The following notations will be used throughout these notes:

•We letℕ={0,1,2,…}\mathbb{N}=\left\{0,1,2,\ldots\right\}. Thus,0∈ℕ0\in\mathbb{N}.
•The size (i.e., cardinality) of a finite setSSis denoted by|S|\left|S\right|.
•IfSSis a set, then thepowersetofSSmeans the set of all subsets ofSS. This powerset will be denoted by𝒫⁡(S)\mathcal{P}\left(S\right). Moreover, ifSSis a set, andkkis an integer, then𝒫k(S)\mathcal{P}_{k}\left(S\right)will mean the set of allkk-element subsets ofSS. For instance, 𝒫2({1,2,3})={{1,2},{1,3},{2,3}}.\mathcal{P}_{2}\left(\left\{1,2,3\right\}\right)=\left\{\left\{1,2\right\},\ \left\{1,3\right\},\ \left\{2,3\right\}\right\}.
•For any numbernnand anyk∈ℕk\in\mathbb{N}, we define thebinomial coefficient(nk)\dbinom{n}{k}to be the number n(n−1)(n−2)⋯(n−k+1)k!=∏i=0k−1(n−i)k!.\dfrac{n\left(n-1\right)\left(n-2\right)\cdots\left(n-k+1\right)}{k!}=\dfrac{\prod\limits_{i=0}^{k-1}\left(n-i\right)}{k!}. These binomial coefficients have many interesting properties, which can often be found in textbooks on enumerative combinatorics (e.g.,[19fco, Chapter 2]). Some of the most important ones are the following: - –The factorial formula: Ifn,k∈ℕn,k\in\mathbb{N}andn≥kn\geq k, then(nk)=n!k!⋅(n−k)!\dbinom{n}{k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}. - –The combinatorial interpretation: Ifn,k∈ℕn,k\in\mathbb{N}, and ifSSis annn-element set, then(nk)\dbinom{n}{k}is the number of allkk-element subsets ofSS(in other words,|𝒫k(S)|=(nk)\left|\mathcal{P}_{k}\left(S\right)\right|=\dbinom{n}{k}). - –Pascal’s recursion: For any numbernnand any positive integerkk, we have (nk)=(n−1k−1)+(n−1k).\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}.

2Simple graphs

2.1Definitions

The first type of graphs that we will consider are the “simple graphs”, named so because of their very simple definition:

Definition 2.1.

Asimple graphis a pair(V,E)\left(V,E\right), whereVVis a finite set, and whereEEis a subset of𝒫2(V)\mathcal{P}_{2}\left(V\right).

To remind,𝒫2(V)\mathcal{P}_{2}\left(V\right)is the set of all22-element subsets ofVV. Thus, a simple graph is a pair(V,E)\left(V,E\right), whereVVis a finite set, andEEis a set consisting of22-element subsets ofVV. We will abbreviate the word “simple graph” as “graph” in this chapter, but later (in Chapter3) we will learn some more advanced and general notions of “graphs”.

Example 2.2.

Here is a simple graph:

({1,2,3,4},{{1,3},{1,4},{3,4}}).\left(\left\{1,2,3,4\right\},\ \ \left\{\left\{1,3\right\},\ \ \left\{1,4\right\},\ \ \left\{3,4\right\}\right\}\right).

Example 2.3.

For anyn∈ℕn\in\mathbb{N}, we can define a simple graphCopn\operatorname*{Cop}\nolimits_{n}to be the pair(V,E)\left(V,E\right), whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}and

E={{u,v}∈𝒫2(V)∣gcd⁡(u,v)=1}.E=\left\{\left\{u,v\right\}\in\mathcal{P}_{2}\left(V\right)\ \mid\ \gcd\left(u,v\right)=1\right\}.We call this thenn-th coprimality graph.

(Some authors do not requireVVto be finite in Definition2.1; this leads toinfinite graphs. But I shall leave this can of worms closed for this quarter.)

The purpose of simple graphs is to encode relations on a finite set – specifically the kind of relations that are binary (i.e., relate pairs of elements), symmetric (i.e., mutual) and irreflexive (i.e., an element cannot be related to itself). For example, the graphCopn\operatorname*{Cop}\nolimits_{n}in Example2.3encodes the coprimality (aka coprimeness) relation on the set{1,2,…,n}\left\{1,2,\ldots,n\right\}, except that the latter relation is not irreflexive (11is coprime to11, but{1,1}\left\{1,1\right\}is not inEE; thus, the graphCopn\operatorname*{Cop}\nolimits_{n}“forgets” that11is coprime to11). For another example, ifVVis a set of people, andEEis the set of{u,v}∈𝒫2(V)\left\{u,v\right\}\in\mathcal{P}_{2}\left(V\right)such thatuuhas been married tovvat some point, then(V,E)\left(V,E\right)is a simple graph. Even in 2022, marriage to oneself is not a thing, so all marriages can be encoded as22-element subsets.111The more standard example for a social graph would be a “friendship graph”; here,VVis again a set of people, butEEis now the set of{u,v}∈𝒫2(V)\left\{u,v\right\}\in\mathcal{P}_{2}\left(V\right)such thatuuandvvare friends. Of course, this only works if you think of friendship as being automatically mutual (true for facebook friendship, questionable for the actual thing).

The following notations provide a quick way to reference the elements ofVVandEEwhen given a graph(V,E)\left(V,E\right):

Definition 2.4.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

(a)The setVVis called thevertex setofGG; it is denoted byV⁡(G)\operatorname{V}\left(G\right). (Notice that the letter “V\operatorname{V}” in “V⁡(G)\operatorname{V}\left(G\right)” is upright, as opposed to the letter “VV” in “(V,E)\left(V,E\right)”, which is italic. These are two different symbols, and have different meanings: The letterVVstands for the specific setVVwhich is the first component of the pairGG, whereas the letterV\operatorname{V}is part of the notationV⁡(G)\operatorname{V}\left(G\right)for the vertex set of any graph. Thus, ifH=(W,F)H=\left(W,F\right)is another graph, thenV⁡(H)\operatorname{V}\left(H\right)isWW, notVV.) The elements ofVVare called thevertices(or thenodes) ofGG.
(b)The setEEis called theedge setofGG; it is denoted byE⁡(G)\operatorname{E}\left(G\right). (Again, the letter “E\operatorname{E}” in “E⁡(G)\operatorname{E}\left(G\right)” is upright, and stands for a different thing than the “EE”.) The elements ofEEare called theedgesofGG. Whenuuandvvare two elements ofVV, we shall often use the notationuvuvfor{u,v}\left\{u,v\right\}; thus, each edge ofGGhas the formuvuvfor two distinct elementsuuandvvofVV. Of course, we always haveuv=vuuv=vu. Notice that each simple graphGGsatisfiesG=(V⁡(G),E⁡(G))G=\left(\operatorname{V}\left(G\right),\operatorname{E}\left(G\right)\right).
(c)Two verticesuuandvvofGGare said to beadjacent(to each other) ifuv∈Euv\in E(that is, ifuvuvis an edge ofGG). In this case, the edgeuvuvis said tojoinuuwithvv(orconnectuuandvv); the verticesuuandvvare called theendpointsof this edge. When the graphGGis not obvious from the context, we shall often say “adjacent inGG” instead of just “adjacent”. Two verticesuuandvvofGGare said to benon-adjacent(to each other) if they are not adjacent (i.e., ifuv∉Euv\notin E).
(d)Letvvbe a vertex ofGG(that is,v∈Vv\in V). Then, theneighborsofvv(inGG) are the verticesuuofGGthat satisfyvu∈Evu\in E. In other words, theneighborsofvvare the vertices ofGGthat are adjacent tovv.

Example 2.5.

LetGGbe the simple graph

({1,2,3,4},{{1,3},{1,4},{3,4}})\left(\left\{1,2,3,4\right\},\ \ \left\{\left\{1,3\right\},\ \ \left\{1,4\right\},\ \ \left\{3,4\right\}\right\}\right)from Example2.2. Then, its vertex set and its edge set are

V(G)={1,2,3,4}andE(G)={{1,3},{1,4},{3,4}}={13,14,34}\operatorname*{V}\left(G\right)=\left\{1,2,3,4\right\}\ \ \ \ \ \text{and}\ \ \ \ \ \operatorname*{E}\left(G\right)=\left\{\left\{1,3\right\},\ \ \left\{1,4\right\},\ \ \left\{3,4\right\}\right\}=\left\{13,\ 14,\ 34\right\}(using our notationuvuvfor{u,v}\left\{u,v\right\}). The vertices11and33are adjacent (since13∈E(G)13\in\operatorname*{E}\left(G\right)), but the vertices11and22are not (since12∉E(G)12\notin\operatorname*{E}\left(G\right)). The neighbors of11are33and44. The endpoints of the edge3434are33and44.

2.2Drawing graphs

There is a common method to represent graphs visually: Namely, a graph can be drawn as a set of points in the plane and a set of curves connecting some of these points with each other.

More precisely:

Definition 2.6.

A simple graphGGcan be visually represented bydrawingit on the plane. To do so, we represent each vertex ofGGby a point (at which we put the name of the vertex), and then, for each edgeuvuvofGG, we draw a curve that connects the point representinguuwith the point representingvv. The positions of the points and the shapes of the curves can be chosen freely, as long as they allow the reader to unambiguously reconstruct the graphGGfrom the picture. (Thus, for example, the curves should not pass through any points other than the ones they mean to connect.)

Example 2.7.

Let us draw some simple graphs.

(a)The simple graph({1,2,3},{12,23})\left(\left\{1,2,3\right\},\ \left\{12,23\right\}\right)(where we are again using the shorthand notationuvuvfor{u,v}\left\{u,v\right\}) can be drawn as follows:

123.\hbox to132.18pt{\vbox to18.37pt{\pgfpicture\makeatletter\hbox{\hskip 9.18588pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.64 0 C 169.64 6.71 164.19 12.16 157.48 12.16 C 150.77 12.16 145.32 6.71 145.32 0 C 145.32 -6.71 150.77 -12.16 157.48 -12.16 C 164.19 -12.16 169.64 -6.71 169.64 0 Z M 157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{111.31104pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 154.02 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.71 0 L 66.03 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 91.45 0 L 144.77 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This is (in a sense) the simplest way to draw this graph: The edges are represented by straight lines. But we can draw it in several other ways as well – e.g., as follows:

123.\hbox to132.18pt{\vbox to40.69pt{\pgfpicture\makeatletter\hbox{\hskip 9.18588pt\lower-31.50575pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.64 0 C 169.64 6.71 164.19 12.16 157.48 12.16 C 150.77 12.16 145.32 6.71 145.32 0 C 145.32 -6.71 150.77 -12.16 157.48 -12.16 C 164.19 -12.16 169.64 -6.71 169.64 0 Z M 157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{111.31104pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 154.02 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 9.74 -8.17 C 50.96 -42.76 106.52 -42.76 147.74 -8.17}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 144.77 0 L 91.45 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Here, we have placed the points representing the vertices1,2,31,2,3differently. As a consequence, we were not able to draw the edge1212as a straight line, because it would then have overlapped with the vertex33, which would make the graph ambiguous (the edge1212could be mistaken for two edges1313and3232).

Here are three further drawings of the same graph({1,2,3},{12,23})\left(\left\{1,2,3\right\},\ \left\{12,23\right\}\right):

123123123.\begin{tabular}[c]{|ccc|}\hline\cr\vrule\hfil\hfil\lx@intercol\vrule\lx@intercol &\hfil\hfil\lx@intercol\vrule\lx@intercol &\\ \vrule\lx@intercol\hfil\ \ \ $\hbox to132.18pt{\vbox to44.69pt{\pgfpicture\makeatletter\hbox{\hskip 9.18588pt\lower-31.50575pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.64 0 C 169.64 6.71 164.19 12.16 157.48 12.16 C 150.77 12.16 145.32 6.71 145.32 0 C 145.32 -6.71 150.77 -12.16 157.48 -12.16 C 164.19 -12.16 169.64 -6.71 169.64 0 Z M 157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{111.31104pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 154.02 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 9.74 -8.17 C 50.96 -42.76 106.52 -42.76 147.74 -8.17}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 146.47 6.36 C 127.32 17.42 108.91 17.42 89.75 6.36}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ \ $\hfil\lx@intercol\vrule\lx@intercol &\lx@intercol\hfil\ \ \ $\hbox to82.39pt{\vbox to61.05pt{\pgfpicture\makeatletter\hbox{\hskip 9.18588pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 56.45 59.06 C 56.45 65.77 51.01 71.21 44.29 71.21 C 37.58 71.21 32.13 65.77 32.13 59.06 C 32.13 52.34 37.58 46.9 44.29 46.9 C 51.01 46.9 56.45 52.34 56.45 59.06 Z M 44.29 59.06}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{29.50935pt}{39.45692pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 40.83 54.6)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 100.74 0 C 100.74 6.71 95.3 12.16 88.58 12.16 C 81.87 12.16 76.43 6.71 76.43 0 C 76.43 -6.71 81.87 -12.16 88.58 -12.16 C 95.3 -12.16 100.74 -6.71 100.74 0 Z M 88.58 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{61.5187pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 85.12 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 7.63 10.17 L 36.67 48.89}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 51.92 48.89 L 80.96 10.17}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ \ $\hfil\lx@intercol\vrule\lx@intercol &\ \ \ $\hbox to82.39pt{\vbox to61.05pt{\pgfpicture\makeatletter\hbox{\hskip 9.18588pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} 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{}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.69 0.69 C 35.14 1.92 52.29 24.78 47.18 46.68}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 41.4 46.68 C 36.3 24.78 53.44 1.92 75.89 0.69}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ \ $\\ \vrule\hfil\hfil\lx@intercol\vrule\lx@intercol &\hfil\hfil\lx@intercol\vrule\lx@intercol &\\ \hline\cr\end{tabular}\ \ \ \ . (b)Consider the55-th coprimality graphCop5\operatorname*{Cop}\nolimits_{5}defined in Example2.3. Here is one way to draw it:

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36.49 68.17 36.49 74.89 Z M 24.33 74.89}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{15.08499pt}{50.89864pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 20.87 70.43)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -51.55 46.28 C -51.55 53 -56.99 58.44 -63.7 58.44 C -70.42 58.44 -75.86 53 -75.86 46.28 C -75.86 39.57 -70.42 34.13 -63.7 34.13 C -56.99 34.13 -51.55 39.57 -51.55 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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.65 3.93 L -51.61 42.35}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.65 -3.93 L -51.61 -42.35}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} 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-62.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -63.7 33.57 L -63.7 -33.57}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -56.23 36 L 16.86 -64.6}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -51.61 -50.21 L 12.24 -70.96}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Here is another way to draw the same graphCop5\operatorname*{Cop}\nolimits_{5}, with fewer intersections between edges:

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-51.55 -46.28 Z M -63.7 -46.28}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-48.53775pt}{-36.67035pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -67.16 -50.74)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 36.49 -74.89 C 36.49 -68.17 31.05 -62.73 24.33 -62.73 C 17.62 -62.73 12.18 -68.17 12.18 -74.89 C 12.18 -81.6 17.62 -87.04 24.33 -87.04 C 31.05 -87.04 36.49 -81.6 36.49 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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -56.23 36 L 16.86 -64.6}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -51.61 -50.21 L 12.24 -70.96}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .By appropriately repositioning the points corresponding to the five vertices ofCop5\operatorname*{Cop}\nolimits_{5}, we can actually get rid of all intersections and make all the edges straight (as opposed to curved). Can you find out how?

(c)Let us draw one further graph: the simple graph({1,2,3,4,5},𝒫2({1,2,3,4,5}))\left(\left\{1,2,3,4,5\right\},\ \mathcal{P}_{2}\left({\left\{1,2,3,4,5\right\}}\right)\right). This is the simple graph whose vertices are1,2,3,4,51,2,3,4,5, and whose edges are all possible two-element sets consisting of its vertices (i.e., each pair of two distinct vertices is adjacent). We shall later call this graph the “complete graphK5K_{5}”. Here is a simple way to draw this graph:

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useful for many purposes; for example, it makes the abstract symmetry of this graph (i.e., the fact that, roughly speaking, its vertices1,2,3,4,51,2,3,4,5are “equal in rights”) obvious. But sometimes, you might want to draw it differently, to minimize the number of intersecting curves. Here is a drawing with fewer intersections:

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{\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.9 77.53 C -41.52 88.88 -97.37 12.02 -70.06 -35.27}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 35.34 68.53 C 81.63 41.81 81.63 -41.81 35.34 -68.53}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -37.15 11.59 L -58.4 -34.73}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -25.53 12.11 L 18.01 -63.86}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -51.61 -50.21 L 12.24 -70.96}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .In this drawing, we have only one intersection between two curves left. Can we get rid of all intersections?

This is a question of topology, not of combinatorics, since it really is about curves in the plane rather than about finite sets and graphs. The answer is “no”. (That is, no matter how you draw this graph in the plane, you will always have at least one pair of curves intersect.) This is a classical result (one of the first theorems in the theory ofplanar graphs), and proofs of it can be found in various textbooks (e.g.,[FriFri98, Theorem 4.1.2], which is generally a good introduction to planar graph theory even if it uses terminology somewhat different from ours). Note that any proof must use some analysis or topology, since the result relies on the notion of a (continuous) curve in the plane (if curves were allowed to be non-continuous, then they could “jump over” one another, so they could easily avoid intersecting!).

2.3A first fact: The Ramsey numberR⁡(3,3)=6R\left(3,3\right)=6

Enough definitions; let’s state a first result:

Proposition 2.8.

LetGGbe a simple graph with|V⁡(G)|≥6\left|\operatorname{V}\left(G\right)\right|\geq 6(that is,GGhas at least66vertices). Then, at least one of the following two statements holds:

•Statement 1:There exist three distinct verticesaa,bbandccofGGsuch thatabab,bcbcandcacaare edges ofGG.
•Statement 2:There exist three distinct verticesaa,bbandccofGGsuch that none ofabab,bcbcandcacais an edge ofGG.

In other words, Proposition2.8says that if a graphGGhas at least66vertices, then we can either find three distinct vertices that are mutually adjacent222by which we mean (of course) that any twodistinctones among these three vertices are adjacentor find three distinct vertices that are mutually non-adjacent (i.e., no two of them are adjacent), or both. Often, this is restated as follows: “In any group of at least six people, you can always find three that are (pairwise) friends to each other, or three no two of whom are friends” (provided that friendship is a symmetric relation).

We will give some examples in a moment, but first let us introduce some convenient terminology:

Definition 2.9.

LetGGbe a simple graph.

(a)A set{a,b,c}\left\{a,b,c\right\}of three distinct vertices ofGGis said to be atriangle(ofGG) if every two distinct vertices in this set are adjacent (i.e., ifabab,bcbcandcacaare edges ofGG).
(b)A set{a,b,c}\left\{a,b,c\right\}of three distinct vertices ofGGis said to be ananti-triangle(ofGG) if no two distinct vertices in this set are adjacent (i.e., if none ofabab,bcbcandcacais an edge ofGG).

Thus, Proposition2.8says that every simple graph with at least66vertices contains a triangle or an anti-triangle (or both).

Example 2.10.

Let us show two examples of graphsGGto which Proposition2.8applies, as well as an example to which it does not:

(a)LetGGbe the graph(V,E)\left(V,E\right), where V\displaystyle V={1,2,3,4,5,6}and\displaystyle=\left\{1,2,3,4,5,6\right\}\qquad\text{and}E\displaystyle E={{1,2},{2,3},{3,4},{4,5},{5,6},{6,1}}.\displaystyle=\left\{\left\{1,2\right\},\left\{2,3\right\},\left\{3,4\right\},\left\{4,5\right\},\left\{5,6\right\},\left\{6,1\right\}\right\}.(This graph can be drawn in such a way as to look like a hexagon: 123456.\hbox to132.18pt{\vbox to116.94pt{\pgfpicture\makeatletter\hbox{\hskip 66.0914pt\lower-58.46764pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} 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\lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -27.21 68.19 C -27.21 74.91 -32.66 80.35 -39.37 80.35 C -46.08 80.35 -51.53 74.91 -51.53 68.19 C -51.53 61.48 -46.08 56.03 -39.37 56.03 C -32.66 56.03 -27.21 61.48 -27.21 68.19 Z M -39.37 68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.95276pt}{46.05954pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.83 63.73)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 0 C -66.58 6.71 -72.03 12.16 -78.74 12.16 C -85.45 12.16 -90.9 6.71 -90.9 0 C -90.9 -6.71 -85.45 -12.16 -78.74 -12.16 C -72.03 -12.16 -66.58 -6.71 -66.58 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -27.21 -68.19 C -27.21 -61.48 -32.66 -56.03 -39.37 -56.03 C -46.08 -56.03 -51.53 -61.48 -51.53 -68.19 C -51.53 -74.91 -46.08 -80.35 -39.37 -80.35 C -32.66 -80.35 -27.21 -74.91 -27.21 -68.19 Z M -39.37 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.95276pt}{-52.50397pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.83 -72.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 51.53 -68.19 C 51.53 -61.48 46.08 -56.03 39.37 -56.03 C 32.66 -56.03 27.21 -61.48 27.21 -68.19 C 27.21 -74.91 32.66 -80.35 39.37 -80.35 C 46.08 -80.35 51.53 -74.91 51.53 -68.19 Z M 39.37 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{25.95276pt}{-52.50397pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 35.91 -72.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}{{}}{}{{}}{}{{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{\hbox{\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 72.38 11.01 L 45.73 57.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 26.66 68.19 L -26.66 68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -45.73 57.18 L -72.38 11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -72.38 -11.01 L -45.73 -57.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -26.66 -68.19 L 26.66 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 45.73 -57.18 L 72.38 -11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .) This graph satisfies Proposition2.8, since{1,3,5}\left\{1,3,5\right\}is an anti-triangle (or since{2,4,6}\left\{2,4,6\right\}is an anti-triangle).
(b)LetGGbe the graph(V,E)\left(V,E\right), where V\displaystyle V={1,2,3,4,5,6}and\displaystyle=\left\{1,2,3,4,5,6\right\}\qquad\text{and}E\displaystyle E={{1,2},{2,3},{3,4},{4,5},{5,6},{6,1},{1,3},{4,6}}.\displaystyle=\left\{\left\{1,2\right\},\left\{2,3\right\},\left\{3,4\right\},\left\{4,5\right\},\left\{5,6\right\},\left\{6,1\right\},\left\{1,3\right\},\left\{4,6\right\}\right\}.(This graph can be drawn in such a way as to look like a hexagon with two extra diagonals: 123456.\hbox to132.18pt{\vbox to116.94pt{\pgfpicture\makeatletter\hbox{\hskip 66.0914pt\lower-58.46764pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} 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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -27.21 68.19 C -27.21 74.91 -32.66 80.35 -39.37 80.35 C -46.08 80.35 -51.53 74.91 -51.53 68.19 C -51.53 61.48 -46.08 56.03 -39.37 56.03 C -32.66 56.03 -27.21 61.48 -27.21 68.19 Z M -39.37 68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.95276pt}{46.05954pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.83 63.73)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 0 C -66.58 6.71 -72.03 12.16 -78.74 12.16 C -85.45 12.16 -90.9 6.71 -90.9 0 C -90.9 -6.71 -85.45 -12.16 -78.74 -12.16 C -72.03 -12.16 -66.58 -6.71 -66.58 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} 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{{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 51.53 -68.19 C 51.53 -61.48 46.08 -56.03 39.37 -56.03 C 32.66 -56.03 27.21 -61.48 27.21 -68.19 C 27.21 -74.91 32.66 -80.35 39.37 -80.35 C 46.08 -80.35 51.53 -74.91 51.53 -68.19 Z M 39.37 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{25.95276pt}{-52.50397pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 35.91 -72.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} 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\lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -26.66 -68.19 L 26.66 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 45.73 -57.18 L 72.38 -11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 67.73 6.36 L -28.36 61.84}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 28.36 -61.84 L -67.73 -6.36}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .) This graph satisfies Proposition2.8, since{1,2,3}\left\{1,2,3\right\}is a triangle.
(c)LetGGbe the graph(V,E)\left(V,E\right), where V\displaystyle V={1,2,3,4,5}and\displaystyle=\left\{1,2,3,4,5\right\}\qquad\text{and}E\displaystyle E={{1,2},{2,3},{3,4},{4,5},{5,1}}.\displaystyle=\left\{\left\{1,2\right\},\left\{2,3\right\},\left\{3,4\right\},\left\{4,5\right\},\left\{5,1\right\}\right\}.(This graph can be drawn to look like a pentagon: 12345.\hbox to121.32pt{\vbox to126.61pt{\pgfpicture\makeatletter\hbox{\hskip 55.22363pt\lower-63.30673pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} 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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -63.7 33.57 L -63.7 -33.57}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -51.61 -50.21 L 12.24 -70.96}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 31.8 -64.6 L 71.27 -10.28}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .) Proposition2.8says nothing about this graph, since this graph does not satisfy the assumption of Proposition2.8(in fact, its number of vertices|V⁡(G)|\left|\operatorname{V}\left(G\right)\right|fails to be≥6\geq 6). By itself, this does not yield that the claim of Proposition2.8is false for this graph. However, it is easy to check that the claim actuallyisfalse for this graph: It has neither a triangle nor an anti-triangle.

Proof 2.11(Proof of Proposition2.8.).

We need to prove thatGGhas a triangle or an anti-triangle (or both).

Choose any vertexu∈V⁡(G)u\in\operatorname{V}\left(G\right). (This is clearly possible, since|V⁡(G)|≥6≥1\left|\operatorname{V}\left(G\right)\right|\geq 6\geq 1.) Then, there are at least55vertices distinct fromuu(sinceGGhas at least66vertices). We are in one of the following two cases:

Case 1:The vertexuuhas at least33neighbors.

Case 2:The vertexuuhas at most22neighbors.

Let us consider Case 1 first. In this case, the vertexuuhas at least33neighbors. Hence, we can find three distinct neighborspp,qqandrrofuu. Consider thesepp,qqandrr. If one (or more) ofpqpq,qrqrandrprpis an edge ofGG, thenGGhas a triangle (for example, ifpqpqis an edge ofGG, then{u,p,q}\left\{u,p,q\right\}is a triangle). If not, thenGGhas an anti-triangle (namely,{p,q,r}\left\{p,q,r\right\}). Thus, in either case, our proof is complete in Case 1.

Let us now consider Case 2. In this case, the vertexuuhas at most22neighbors. Hence, the vertexuuhas at least33non-neighbors333The word “non-neighbor” shall here mean a vertex that is not adjacent touuanddistinct fromuu. Thus,uudoes not count as a non-neighbor ofuu.(since there are at least55vertices distinct fromuuin total). Thus, we can find three distinct non-neighborspp,qqandrrofuu. Consider thesepp,qqandrr. If all ofpqpq,qrqrandrprpare edges ofGG, thenGGhas a triangle (namely,{p,q,r}\left\{p,q,r\right\}). If not, thenGGhas an anti-triangle (for example, ifpqpqis not an edge ofGG, then{u,p,q}\left\{u,p,q\right\}is an anti-triangle). In either case, we are thus done with the proof in Case 2. Thus, both cases are resolved, and the proof is complete.

Notice the symmetry between Case 1 and Case 2 in our above proof: the arguments used were almost the same, except that neighbors and non-neighbors swapped roles.

Proposition2.8is the first result in a field of graph theory known asRamsey theory. I shall not dwell on this field in this course, but let me make a few more remarks. The first step beyond Proposition2.8is the following generalization (see, e.g.,[Tomesc85, problem 14.11]for a proof):

Proposition 2.13.

Letrrandssbe two positive integers. LetGGbe a simple graph with|V⁡(G)|≥(r+s−2r−1)\left|\operatorname{V}\left(G\right)\right|\geq\dbinom{r+s-2}{r-1}. Then, at least one of the following two statements holds:

•Statement 1:There existrrdistinct vertices ofGGthat are mutually adjacent (i.e., each two distinct ones among theserrvertices are adjacent).
•Statement 2:There existssdistinct vertices ofGGthat are mutually non-adjacent (i.e., no two distinct ones among thesessvertices are adjacent).

Applying Proposition2.13tor=3r=3ands=3s=3, we can recover Proposition2.8.

One might wonder whether the number(r+s−2r−1)\dbinom{r+s-2}{r-1}in Proposition2.13can be improved – i.e., whether we can replace it by a smaller number without making Proposition2.13false. In the case ofr=3r=3ands=3s=3, this is impossible, because the number66in Proposition2.8cannot be made smaller444Indeed, we saw in Example2.10(c)that55vertices would not suffice.. However, for some other values ofrrandss, the value(r+s−2r−1)\dbinom{r+s-2}{r-1}can be improved. (For example, forr=4r=4ands=4s=4, the best possible value is1818rather than(4+4−24−1)=20\dbinom{4+4-2}{4-1}=20.) The smallest possible value that could stand in place of(r+s−2r−1)\dbinom{r+s-2}{r-1}in Proposition2.13is called theRamsey numberR⁡(r,s)R\left(r,s\right); thus, we have just showed thatR⁡(3,3)=6R\left(3,3\right)=6. FindingR⁡(r,s)R\left(r,s\right)for higher values ofrrandssis a hard computational challenge; here are some values that have been found with the help of computers:

R⁡(3,4)\displaystyle R\left(3,4\right)=9;R⁡(3,5)=14;R⁡(3,6)=18;R⁡(3,7)=23;\displaystyle=9;\qquad R\left(3,5\right)=14;\qquad R\left(3,6\right)=18;\qquad R\left(3,7\right)=23;R⁡(3,8)\displaystyle R\left(3,8\right)=28;R⁡(3,9)=36;R⁡(4,4)=18;R⁡(4,5)=25.\displaystyle=28;\qquad R\left(3,9\right)=36;\qquad R\left(4,4\right)=18;\qquad R\left(4,5\right)=25.(We are only considering the casesr≤sr\leq s, since it is easy to see thatR⁡(r,s)=R⁡(s,r)R\left(r,s\right)=R\left(s,r\right)for allrrandss. Also, the trivial valuesR⁡(1,s)=1R\left(1,s\right)=1andR⁡(2,s)=s+1R\left(2,s\right)=s+1fors≥2s\geq 2are omitted.) The Ramsey numberR⁡(5,5)R\left(5,5\right)is still unknown (although it is known that43≤R⁡(5,5)≤4643\leq R\left(5,5\right)\leq 46; see[AngMcK24]for details).

Proposition2.13can be further generalized to a result calledRamsey’s theorem. The idea behind the generalization is to slightly change the point of view, and replace the simple graphGGby a complete graph (i.e., a simple graph in which every two distinct vertices are adjacent) whose edges are colored in two colors (say, blue and red). This is a completely equivalent concept, because the concepts of “adjacent” and “non-adjacent” inGGcan be identified with the concepts of “adjacent through a blue edge” (i.e., the edge connecting them is colored blue) and “adjacent through a red edge”, respectively. Statements 1 and 2 then turn into “there existrrdistinct vertices that are mutually adjacent through blue edges” and “there existssdistinct vertices that are mutually adjacent through red edges”, respectively. From this point of view, it is only logical to generalize Proposition2.13further to the case when the edges of a complete graph are colored inkk(rather than two) colors. The corresponding generalization is known as Ramsey’s theorem. We refer to the well-written Wikipedia pagehttps://en.wikipedia.org/wiki/Ramsey’s_theoremfor a treatment of this generalization with proof, as well as a table of known Ramsey numbersR⁡(r,s)R\left(r,s\right)and a self-contained (if somewhat terse) proof of Proposition2.13. Ramsey’s theorem can be generalized and varied further; this usually goes under the name “Ramsey theory”. For elementary introductions, see the Cut-the-knot pagehttp://www.cut-the-knot.org/Curriculum/Combinatorics/ThreeOrThree.shtml, the above-mentioned Wikipedia article, as well as the texts by Harju[Harju14], Bollobas[Bollob98]and West[West01]. Several other texts on combinatorics have a chapter on Ramsey theory, e.g.,[Tomesc85, Chapter 14].

There is one more direction in which Proposition2.8can be improved a bit: A graphGGwith at least66vertices has not only one triangle or anti-triangle, but at least two of them (this can include having one triangle and one anti-triangle). Proving this makes for a nice exercise:

Exercise 2.14.

LetGGbe a simple graph. Atriangle-or-anti-triangleinGGmeans a set that is either a triangle or an anti-triangle.

(a)Assume that|V⁡(G)|≥6\left|\operatorname{V}\left(G\right)\right|\geq 6. Prove thatGGhas at least two triangle-or-anti-triangles. (For comparison: Proposition2.8shows thatGGhas at least one triangle-or-anti-triangle.)
(b)Assume that|V⁡(G)|=m+6\left|\operatorname{V}\left(G\right)\right|=m+6for somem∈ℕm\in\mathbb{N}. Prove thatGGhas at leastm+1m+1triangle-or-anti-triangles.

[Solution:This is Exercise 1 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

2.4Degrees

2.4.1Definition and basic properties

Thedegreeof a vertex in a simple graph just counts how many edges contain this vertex:

Definition 2.15.

LetG=(V,E)G=\left(V,E\right)be a simple graph. Letv∈Vv\in Vbe a vertex. Then, thedegreeofvv(with respect toGG) is defined to be

deg⁡v:=\displaystyle\deg v:=(the number of edgese∈Ethat containv)\displaystyle\ \left(\text{the number of edges }e\in E\text{ that contain }v\right)=\displaystyle=(the number of neighbors ofv)\displaystyle\ \left(\text{the number of neighbors of }v\right)=\displaystyle=|{u∈V∣uv∈E}|\displaystyle\ \left|\left\{u\in V\ \mid\ uv\in E\right\}\right|=\displaystyle=|{e∈E∣v∈e}|.\displaystyle\ \left|\left\{e\in E\ \mid\ v\in e\right\}\right|.

(These equalities are pretty easy to check: Each edgee∈Ee\in Ethat containsvvcontains exactly one neighbor ofvv, and conversely, each neighbor ofvvbelongs to exactly one edge that containsvv. However, these equalities are specific to simple graphs, and won’t hold any more once we move on to multigraphs.)

For example, in the graph

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deg⁡1=3,deg⁡2=2,deg⁡3=3,deg⁡4=2,deg⁡5=0.\deg 1=3,\ \ \ \ \ \ \ \ \ \ \deg 2=2,\ \ \ \ \ \ \ \ \ \ \deg 3=3,\ \ \ \ \ \ \ \ \ \ \deg 4=2,\ \ \ \ \ \ \ \ \ \ \deg 5=0. Here are some basic properties of degrees in simple graphs:

Proposition 2.16.

LetGGbe a simple graph withnnvertices. Letvvbe a vertex ofGG. Then,

deg⁡v∈{0,1,…,n−1}.\deg v\in\left\{0,1,\ldots,n-1\right\}.

Proof 2.17.

All neighbors ofvvbelong to the(n−1)\left(n-1\right)-element setV(G)∖{v}\operatorname*{V}\left(G\right)\setminus\left\{v\right\}. Thus, their number is≤n−1\leq n-1.

Proposition 2.18(Euler 1736).

LetGGbe a simple graph. Then, the sum of the degrees of all vertices ofGGequals twice the number of edges ofGG. In other words,

∑v∈V(G)deg⁡v=2⋅|E(G)|.\sum\limits_{v\in\operatorname*{V}\left(G\right)}\deg v=2\cdot\left|\operatorname*{E}\left(G\right)\right|.

Proof 2.19.

Write the simple graphGGasG=(V,E)G=\left(V,E\right); thus,V(G)=V\operatorname*{V}\left(G\right)=VandE(G)=E\operatorname*{E}\left(G\right)=E.

Now, letNNbe the number of all pairs(v,e)∈V×E\left(v,e\right)\in V\times Esuch thatv∈ev\in e. We computeNNin two different ways (this is called “double-counting”):

1.We can obtainNNby computing, for eachv∈Vv\in V, the number of alle∈Ee\in Ethat satisfyv∈ev\in e, and then summing these numbers over allvv. Since these numbers are just the degreesdeg⁡v\deg v, the result will be∑v∈Vdeg⁡v\sum\limits\limits_{v\in V}\deg v.
2.On the other hand, we can obtainNNby computing, for eache∈Ee\in E, the number of allv∈Vv\in Vthat satisfyv∈ev\in e, and summing these numbers over allee. Since eache∈Ee\in Econtains exactly22verticesv∈Vv\in V, this result will be∑e∈E2=|E|⋅2=2⋅|E|\sum\limits\limits_{e\in E}2=\left|E\right|\cdot 2=2\cdot\left|E\right|.

Since these two results must be equal (because they both equalNN), we thus see that∑v∈Vdeg⁡v=2⋅|E|\sum\limits\limits_{v\in V}\deg v=2\cdot\left|E\right|. But this is the claim of Proposition2.18.

Corollary 2.20(handshake lemma).

LetGGbe a simple graph. Then, the number of verticesvvofGGwhose degreedeg⁡v\deg vis odd is even.

Proof 2.21.

Proposition2.18yields that∑v∈V(G)deg⁡v=2⋅|E(G)|\sum\limits\limits_{v\in\operatorname*{V}\left(G\right)}\deg v=2\cdot\left|\operatorname*{E}\left(G\right)\right|. Hence,∑v∈V(G)deg⁡v\sum\limits\limits_{v\in\operatorname*{V}\left(G\right)}\deg vis even. However, if a sum of integers is even, then it must have an even number of odd addends. Thus, the sum∑v∈V(G)deg⁡v\sum\limits\limits_{v\in\operatorname*{V}\left(G\right)}\deg vmust have an even number of odd addends. In other words, the number of verticesvvofGGwhose degreedeg⁡v\deg vis odd is even.

Corollary2.20is often stated as follows: In a group of people, the number of persons with an odd number of friends (in the group) is even. It is also known as thehandshake lemma.

Here is another property of degrees in a simple graph:

Proposition 2.22.

LetGGbe a simple graph with at least two vertices. Then, there exist two distinct verticesvvandwwofGGthat have the same degree.

Proof 2.23.

Assume the contrary. So the degrees of allnnvertices ofGGare distinct, wheren=|V(G)|n=\left|\operatorname*{V}\left(G\right)\right|.

In other words, the map

deg:V(G)\displaystyle\deg:\operatorname*{V}\left(G\right)→{0,1,…,n−1},\displaystyle\rightarrow\left\{0,1,\ldots,n-1\right\},v\displaystyle v↦deg⁡v\displaystyle\mapsto\deg vis injective. But this is a map between two finite sets of the same size (nn). When such a map is injective, it has to be bijective (by the pigeonhole principle). Therefore, in particular, it takes both00andn−1n-1as values.

In other words, there are a vertexuuwith degree00and a vertexvvwith degreen−1n-1. Are these two vertices adjacent or not? Yes because ofdeg⁡v=n−1\deg v=n-1; no because ofdeg⁡u=0\deg u=0. Contradiction!

(Fine print: The two verticesuuandvvmust be distinct, since0≠n−10\neq n-1. It is here that we are using the “at least two vertices” assumption!)

2.4.2Mantel’s theorem

Here is an application of counting neighbors to proving a fact about graphs. This is known asMantel’s theorem:

Theorem 2.24(Mantel’s theorem).

LetGGbe a simple graph withnnvertices andeeedges. Assume thate>n2/4e>n^{2}/4. Then,GGhas a triangle (i.e., three distinct vertices that are pairwise adjacent).

Example 2.25.

LetGGbe the graph(V,E)\left(V,E\right), where

V\displaystyle V={1,2,3,4,5,6};\displaystyle=\left\{1,2,3,4,5,6\right\};E\displaystyle E={12,23,34,45,56,61,14,25,36}.\displaystyle=\left\{12,\ 23,\ 34,\ 45,\ 56,\ 61,\ 14,\ 25,\ 36\right\}.Here is a drawing:

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\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -33.01 57.18 L 33.01 -57.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}.This graph has no triangle (which, by the way, is easy to verify without checking all possibilities: just observe that every edge ofGGjoins two vertices of different parity, but a triangle would necessarily have two vertices of equal parity). Thus, by the contrapositive of Mantel’s theorem, it satisfiese≤n2/4e\leq n^{2}/4withn=6n=6ande=9e=9. This is indeed true because9=62/49=6^{2}/4. But this also entails that if we add any further edge toGG, then we obtain a triangle.

Proof 2.26(Proof of Mantel’s theorem.).

We will prove the theorem by strong induction onnn. Thus, we assume (as the induction hypothesis) that the theorem holds for all graphs with fewer thannnvertices. We must now prove it for our graphGGwithnnvertices. LetV=V(G)V=\operatorname*{V}\left(G\right)andE=E(G)E=\operatorname*{E}\left(G\right), so thatG=(V,E)G=\left(V,E\right).

We must prove thatGGhas a triangle. Assume the contrary. Thus,GGhas no triangle.

Frome>n2/4≥0e>n^{2}/4\geq 0, we see thatGGhas an edge. Pick any such edge, and call itvwvw. Thus,v≠wv\neq w.

Let us now color each edge ofGGwith one of three colors, as follows:

•The edgevwvwis colored black.
•Each edge that contains exactly one ofvvandwwis colored red.
•All other edges are colored blue.

The following picture shows an example of this coloring:

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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ . We now count the edges of each color:

•There is exactly11black edge – namely,vwvw.
•How many red edges can there be? I claim that there are at mostn−2n-2. Indeed, each vertex other thanvvandwwis connected to at most one ofvvandwwby a red edge, since otherwise it would form a triangle withvvandww.
•How many blue edges can there be? The vertices other thanvvandww, along with the blue edges that join them, form a graph withn−2n-2vertices; this graph has no triangles (sinceGGhas no triangles). By the induction hypothesis, however, if this graph had more than(n−2)2/4\left(n-2\right)^{2}/4edges, then it would have a triangle. Thus, it has≤(n−2)2/4\leq\left(n-2\right)^{2}/4edges. In other words, there are≤(n−2)2/4\leq\left(n-2\right)^{2}/4blue edges.

In total, the number of edges is therefore

≤1+(n−2)+(n−2)2/4=n2/4.\leq 1+\left(n-2\right)+\left(n-2\right)^{2}/4=n^{2}/4.In other words,e≤n2/4e\leq n^{2}/4. This contradictse>n2/4e>n^{2}/4. This is the contradiction we were looking for, so the induction is complete.

Quick question: What about equality? Can a graph withnnvertices and exactlyn2/4n^{2}/4edges have no triangles? Yes (for evennn). Indeed, for any evennn, we can take the graph

({1,2,…,n},{ij∣i≢jmod⁡2})\left(\left\{1,2,\ldots,n\right\},\ \left\{ij\ \mid\ i\not\equiv j\operatorname{mod}2\right\}\right)(keep in mind thatijijmeans the22-element set{i,j}\left\{i,j\right\}here, not the producti⋅ji\cdot j). We can also do this for oddnn, and obtain a graph with(n2−1)/4\left(n^{2}-1\right)/4edges (which is as close ton2/4n^{2}/4as we can get whennnis odd – after all, the number of edges has to be an integer). So the bound in Mantel’s theorem is optimal (as far as integers are concerned).

The following exercise can be regarded as a “mirror version” of Mantel’s theorem:

Exercise 2.27.

LetGGbe a simple graph withnnvertices andeeedges. Assume thate<n⁡(n−2)/4e<n\left(n-2\right)/4. Prove thatGGhas an anti-triangle (i.e., three distinct vertices that are pairwise non-adjacent).

[Solution:This is Exercise 2 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

Mantel’s theorem can be generalized:

Theorem 2.28(Turan’s theorem).

Letrrbe a positive integer. LetGGbe a simple graph withnnvertices andeeedges. Assume that

e>r−1r⋅n22.e>\dfrac{r-1}{r}\cdot\dfrac{n^{2}}{2}.Then, there existr+1r+1distinct vertices ofGGthat are mutually adjacent.

Mantel’s theorem is the particular case forr=2r=2. We will see a proof of Turan’s theorem later (Theorem7.16). Mantel’s and Turan’s theorems are two of the simplest results ofextremal graph theory– the study of how inequalities between some graph parameters (in our case: the numbers of vertices and edges) imply the existence of certain substructures (in our case: of a triangle or ofr+1r+1mutually adjacent vertices). Deeper introductions to this subject can be found in[Zhao23, Chapters 1 and 5]and[Jukna11].

Exercise 2.29.

LetG=(V,E)G=\left(V,E\right)be a simple graph. Setn=|V|n=\left|V\right|. Prove that we can find some edgese1,e2,…,eke_{1},e_{2},\ldots,e_{k}ofGGand some trianglest1,t2,…,tℓt_{1},t_{2},\ldots,t_{\ell}ofGGsuch thatk+ℓ≤n2/4k+\ell\leq n^{2}/4and such that each edgee∈E∖{e1,e2,…,ek}e\in E\setminus\left\{e_{1},e_{2},\ldots,e_{k}\right\}is a subset of (at least) one of the trianglest1,t2,…,tℓt_{1},t_{2},\ldots,t_{\ell}.

[Remark:In other words, this exercise is claiming that all edges ofGGcan be covered by at mostn2/4n^{2}/4edge-or-triangles. Here, anedge-or-trianglemeans either an edge or a triangle ofGG, and the word “covers” means that each edge ofGGis a subset of the chosen edge-or-triangles.]

[Hint:Imitate the above proof of Mantel’s theorem.]

Exercise 2.31.

LetGGbe a simple graph withnnvertices andkkedges, wheren>0n>0. Prove thatGGhas at leastk3n(4k−n2)\dfrac{k}{3n}\left(4k-n^{2}\right)triangles.

[Hint:First argue that for any edgevwvwofGG, the total number of triangles that containvvandwwis at leastdeg⁡v+deg⁡w−n\deg v+\deg w-n. Then, use the inequalityn⁡(a12+a22+⋯+an2)≥(a1+a2+⋯+an)2n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)\geq\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}, which holds for anynnreal numbersa1,a2,…,ana_{1},a_{2},\ldots,a_{n}. (This is a particular case of the Cauchy–Schwarz inequality or the Chebyshev inequality or the Jensen inequality – pick your favorite!)]

Exercise 2.33.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

An edgee={u,v}e=\left\{u,v\right\}ofGGwill be calledoddif the numberdeg⁡u+deg⁡v\deg u+\deg vis odd.

Prove that the number of odd edges ofGGis even.

[Hint:There are several solutions. One usesmodular arithmeticand (in particular) the congruencem2≡mmod2m^{2}\equiv m\mod 2for every integermm. Other solutions use nothing but common sense.]

Exercise 2.34.

LetG=(V,E)G=\left(V,E\right)be a simple graph. LetSSbe a subset ofVV, and letk=|S|k=\left|S\right|. Prove that

∑v∈Sdeg⁡v≤k⁡(k−1)+∑v∈V∖Smin⁡{deg⁡v,k}.\sum\limits_{v\in S}\deg v\leq k\left(k-1\right)+\sum\limits_{v\in V\setminus S}\min\left\{\deg v,k\right\}.

2.5Graph isomorphism

Two graphs can be distinct and yet “the same up to the names of their vertices”: for instance,

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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Let us formalize this:

Definition 2.36.

LetGGandHHbe two simple graphs.

(a)Agraph isomorphism(orisomorphism) fromGGtoHHmeans a bijectionϕ:V(G)→V(H)\phi:\operatorname*{V}\left(G\right)\rightarrow\operatorname*{V}\left(H\right)that “preserves edges”, i.e., that has the following property: For any two verticesuuandvvofGG, we have (uv∈E(G))⟺(ϕ⁡(u)ϕ(v)∈E(H)).\left(uv\in\operatorname*{E}\left(G\right)\right)\Longleftrightarrow\left(\phi\left(u\right)\phi\left(v\right)\in\operatorname*{E}\left(H\right)\right).
(b)We say thatGGandHHareisomorphic(this is writtenG≅HG\cong H) if there exists a graph isomorphism fromGGtoHH.

Here are two examples:

•The two graphs are isomorphic, because the bijection between their vertex sets that sends1,2,31,2,3to1,3,21,3,2is an isomorphism. Another isomorphism between the same two graphs sends1,2,31,2,3to2,3,12,3,1.
•The two graphs 112233445566andAABBCC112233are isomorphic, because the bijection between their vertex sets that sends1,2,3,4,5,61,2,3,4,5,6to1,B,3,A,2,C1,B,3,A,2,Cis an isomorphism.

Here are some basic properties of isomorphisms (the proofs are straightforward):

Proposition 2.37.

LetGGandHHbe two graphs. The inverse of a graph isomorphismϕ\phifromGGtoHHis a graph isomorphism fromHHtoGG.

Proposition 2.38.

LetGG,HHandIIbe three graphs. Ifϕ\phiis a graph isomorphism fromGGtoHH, andψ\psiis a graph isomorphism fromHHtoII, thenψ∘ϕ\psi\circ\phiis a graph isomorphism fromGGtoII.

As a consequence of these two propositions, it is easy to see that the relation≅\cong(on the class of all graphs) is an equivalence relation.

Graph isomorphisms preserve all “intrinsic” properties of a graph. For example:

Proposition 2.39.

LetGGandHHbe two simple graphs, andϕ\phia graph isomorphism fromGGtoHH. Then:

(a)For everyv∈V(G)v\in\operatorname*{V}\left(G\right), we havedegG⁡v=degH⁡(ϕ⁡(v))\deg_{G}v=\deg_{H}\left(\phi\left(v\right)\right). Here,degG⁡v\deg_{G}vmeans the degree ofvvas a vertex ofGG, whereasdegH⁡(ϕ⁡(v))\deg_{H}\left(\phi\left(v\right)\right)means the degree ofϕ⁡(v)\phi\left(v\right)as a vertex ofHH.
(b)We have|E(H)|=|E(G)|\left|\operatorname*{E}\left(H\right)\right|=\left|\operatorname*{E}\left(G\right)\right|.
(c)We have|V(H)|=|V(G)|\left|\operatorname*{V}\left(H\right)\right|=\left|\operatorname*{V}\left(G\right)\right|.

One use of graph isomorphisms is to relabel the vertices of a graph. For example, we can relabel the vertices of annn-vertex graph as1,2,…,n1,2,\ldots,n, or as any othernndistinct objects:

Proposition 2.40.

LetGGbe a simple graph. LetSSbe a finite set such that|S|=|V(G)|\left|S\right|=\left|\operatorname*{V}\left(G\right)\right|. Then, there exists a simple graphHHthat is isomorphic toGGand has vertex setV(H)=S\operatorname*{V}\left(H\right)=S.

Proof 2.41.

Straightforward.

2.6Some families of graphs

We will now define some particularly significant families of graphs.

2.6.1Complete and empty graphs

The simplest families of graphs are the complete graphs and the empty graphs:

Definition 2.42.

LetVVbe a finite set.

(a)Thecomplete graphonVVmeans the simple graph(V,𝒫2(V))\left(V,\ \mathcal{P}_{2}\left(V\right)\right). It is the simple graph with vertex setVVin which every two distinct vertices are adjacent. IfV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}, then the complete graph onVVis denotedKnK_{n}.
(b)Theempty graphonVVmeans the simple graph(V,∅)\left(V,\ \varnothing\right). It is the simple graph with vertex setVVand no edges.

The following pictures show the complete graph and the empty graph on the set{1,2,3,4,5}\left\{1,2,3,4,5\right\}:

complete graphempty graph11223344551122334455The complete one is calledK5K_{5}.

Here are the complete graphsK0,K1,K2,K3,K4K_{0},K_{1},K_{2},K_{3},K_{4}:

K0K_{0}K1K_{1}K2K_{2}K3K_{3}K4K_{4}11112211223311223344 Note that a simple graphGGis isomorphic to the complete graphKnK_{n}if and only if it hasnnvertices and is a complete graph (i.e., every two distinct vertices are adjacent).

Question:Given two finite setsVVandWW, what are the isomorphisms from the complete graph onVVto the complete graph onWW?

2.6.2Path and cycle graphs

Next come two families of graphs with fairly simple shapes:

Definition 2.43.

For eachn∈ℕn\in\mathbb{N}, we define thenn-th path graphPnP_{n}to be the simple graph

({1,2,…,n},{{i,i+1}∣1≤i<n})\displaystyle\left(\left\{1,2,\ldots,n\right\},\ \ \left\{\left\{i,i+1\right\}\ \mid\ 1\leq i<n\right\}\right)=({1,2,…,n},{12,23,34,…,(n−1)n}).\displaystyle=\left(\left\{1,2,\ldots,n\right\},\ \ \left\{12,\ 23,\ 34,\ \ldots,\ \left(n-1\right)n\right\}\right).This graph hasnnvertices andn−1n-1edges (unlessn=0n=0, in which case it has00edges).

Definition 2.44.

For eachn>1n>1, we define thenn-th cycle graphCnC_{n}to be the simple graph

({1,2,…,n},{{i,i+1}∣1≤i<n}∪{{n,1}})\displaystyle\left(\left\{1,2,\ldots,n\right\},\ \ \left\{\left\{i,i+1\right\}\ \mid\ 1\leq i<n\right\}\cup\left\{\left\{n,1\right\}\right\}\right)=({1,2,…,n},{12,23,34,…,(n−1)n,n1}).\displaystyle=\left(\left\{1,2,\ldots,n\right\},\ \ \left\{12,\ 23,\ 34,\ \ldots,\ \left(n-1\right)n,\ n1\right\}\right).This graph hasnnvertices andnnedges (unlessn=2n=2, in which case it has11edge only). (We will later modify the definition of the22-nd cycle graphC2C_{2}somewhat, in order to force it to have22edges. But we cannot do this yet, since a simple graph with22vertices cannot have22edges.)

The following pictures show the path graphP5P_{5}and the cycle graphC5C_{5}:

path graphcycle graph11223344551122334455Of course, it is more common to draw the path graph stretched out horizontally:

Note that the cycle graphC3C_{3}is identical with the complete graphK3K_{3}.

Question:What are the graph isomorphisms fromPnP_{n}to itself?

Answer:One such isomorphism is the identity mapid:{1,2,…,n}→{1,2,…,n}\operatorname*{id}:\left\{1,2,\ldots,n\right\}\rightarrow\left\{1,2,\ldots,n\right\}. Another is the “reversal” map

{1,2,…,n}\displaystyle\left\{1,2,\ldots,n\right\}→{1,2,…,n},\displaystyle\rightarrow\left\{1,2,\ldots,n\right\},i\displaystyle i↦n+1−i.\displaystyle\mapsto n+1-i.There are no others.

Question:What are the graph isomorphisms fromCnC_{n}to itself?

Answer:For anyk∈ℤk\in\mathbb{Z}, we can define a “rotation bykkvertices”, which is the map

{1,2,…,n}\displaystyle\left\{1,2,\ldots,n\right\}→{1,2,…,n},\displaystyle\rightarrow\left\{1,2,\ldots,n\right\},i\displaystyle i↦(i+kreduced modulonto an element of{1,2,…,n}).\displaystyle\mapsto\left(i+k\text{ reduced modulo }n\text{ to an element of }\left\{1,2,\ldots,n\right\}\right).Thus we getnnrotations (one for eachk∈{1,2,…,n}k\in\left\{1,2,\ldots,n\right\}); all of them are graph isomorphisms.

There are also the reflections, which are the maps

{1,2,…,n}\displaystyle\left\{1,2,\ldots,n\right\}→{1,2,…,n},\displaystyle\rightarrow\left\{1,2,\ldots,n\right\},i\displaystyle i↦(k−ireduced modulonto an element of{1,2,…,n})\displaystyle\mapsto\left(k-i\text{ reduced modulo }n\text{ to an element of }\left\{1,2,\ldots,n\right\}\right)fork∈ℤk\in\mathbb{Z}. There arennof them, too, and they are isomorphisms as well.

Altogether we obtain2n2nisomorphisms (forn>2n>2), and there are no others. (The group they form is thenn-th dihedral group.)

2.6.3Kneser graphs

Here is a more exotic family of graphs:

Example 2.45.

IfSSis a finite set, and ifk∈ℕk\in\mathbb{N}, then we define thekk-th Kneser graph ofSSto be the simple graph

KS,k:=(𝒫k(S),{IJ∣I,J∈𝒫k(S)andI∩J=∅}).K_{S,k}:=\left(\mathcal{P}_{k}\left(S\right),\ \ \left\{IJ\ \mid\ I,J\in\mathcal{P}_{k}\left(S\right)\text{ and }I\cap J=\varnothing\right\}\right).The vertices ofKS,kK_{S,k}are thekk-element subsets ofSS, and two such subsets are adjacent if they are disjoint.

The graphK{1,2,…,5},2K_{\left\{1,2,\ldots,5\right\},2}is called thePetersen graph; here is how it looks like:

{1,2}\left\{1,2\right\}{2,3}\left\{2,3\right\}{3,4}\left\{3,4\right\}{4,5}\left\{4,5\right\}{1,5}\left\{1,5\right\}{3,5}\left\{3,5\right\}{1,4}\left\{1,4\right\}{2,5}\left\{2,5\right\}{1,3}\left\{1,3\right\}{2,4}\left\{2,4\right\}

2.7Subgraphs

Definition 2.46.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

(a)AsubgraphofGGmeans a simple graph of the formH=(W,F)H=\left(W,F\right), whereW⊆VW\subseteq VandF⊆EF\subseteq E. In other words, a subgraph ofGGmeans a simple graph whose vertices are vertices ofGGand whose edges are edges ofGG.
(b)LetSSbe a subset ofVV. Theinduced subgraph ofGGon the setSSdenotes the subgraph (S,E∩𝒫2(S))\left(S,\ \ E\cap\mathcal{P}_{2}\left(S\right)\right)ofGG. In other words, it denotes the subgraph ofGGwhose vertices are the elements ofSS, and whose edges are precisely those edges ofGGwhose both endpoints belong toSS. We denote this induced subgraph byG⁡[S]G\left[S\right].
(c)Aninduced subgraphofGGmeans a subgraph ofGGthat is the induced subgraph ofGGonSSfor someS⊆VS\subseteq V.

Thus, a subgraph of a graphGGis obtained by throwing away some vertices and some edges ofGG(in such a way, of course, that no edges remain “dangling” – i.e., if you throw away a vertex, then you must throw away all edges that contain this vertex). Such a subgraph is an induced subgraph if no edges are removed without need – i.e., if you removed only those edges that lost some of their endpoints. Thus, induced subgraphs can be characterized as follows:

Proposition 2.47.

LetHHbe a subgraph of a simple graphGG. Then,HHis an induced subgraph ofGGif and only if each edgeuvuvofGGwhose endpointsuuandvvbelong toV(H)\operatorname*{V}\left(H\right)is an edge ofHH.

Proof 2.48.

This is a matter of understanding the definition.

Example 2.49.

The following table shows a graphGGand two of its subgraphs: one that is induced and one that is not.

graphGGinduced subgraphsubgraph that ison the set{1,2,3}\left\{1,2,3\right\}not induced11223344112233112233

Example 2.50.

Letn>1n>1be an integer.

(a)The path graphPnP_{n}is a subgraph of the cycle graphCnC_{n}. It is not an induced subgraph (forn>2n>2), because it contains the two verticesnnand11ofCnC_{n}but does not contain the edgen1n1.
(b)The path graphPn−1P_{n-1}is an induced subgraph ofPnP_{n}. (Namely, it is the induced subgraph ofPnP_{n}on the set{1,2,…,n−1}\left\{1,2,\ldots,n-1\right\}.)
(c)Assume thatn>3n>3. IsCn−1C_{n-1}a subgraph ofCnC_{n}? No, because the edge(n−1)1\left(n-1\right)1belongs toCn−1C_{n-1}but not toCnC_{n}.

The following is easy:

Proposition 2.51.

LetGGbe a simple graph, and letHHbe a subgraph ofGG. Assume thatHHis a complete graph. Then,HHis automatically an induced subgraph ofGG.

Proof 2.52.

This follows from Proposition2.47, since the completeness ofHHmeans that each22-element subset{u,v}\left\{u,v\right\}of the vertex set ofHHis an edge ofHH.

We note that triangles in a graph can be characterized in terms of complete subgraphs. Namely, a triangle “is” the same as a complete subgraph (or, equivalently, induced complete subgraph) with three vertices:

Thus, instead of saying “triangle ofGG”, one often says “aK3K_{3}inGG” or “aC3C_{3}inGG”. Generally, “anHHinGG” (whereHHandGGare two graphs) means a subgraph ofGGthat is isomorphic toHH. (In the case whenH=K3=C3H=K_{3}=C_{3}, it does not matter whether we require it to be a subgraph or an induced subgraph, since a complete subgraph has to be induced automatically.)

Example 2.54.

LetGGbe the following simple graph:

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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 111.13 0 L 184.14 0 M 187.86 8.99 L 107.41 89.44 M 111.13 98.42 L 184.14 98.42 M 187.86 89.44 L 107.41 8.99}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 196.85 12.71 L 196.85 85.71 M 209.56 98.42 L 282.56 98.42 M 286.28 89.44 L 205.83 8.99 M 209.56 0 L 282.56 0 M 295.27 12.71 L 295.27 85.71}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .It has several subgraphs isomorphic toC4C_{4}, including the three “little squares” (note that the middle square produces three different subgraphs isomorphic toC4C_{4}: you can either keep the edges23,37,76,6223,37,76,62or keep the edges23,36,67,7223,36,67,72or keep the edges26,63,37,7226,63,37,72) as well as some less obvious subgraphs (e.g., the one with vertices3,6,7,83,6,7,8and edges36,67,78,8336,67,78,83). But only one of these subgraphs is an induced subgraph (namely, the leftmost “little square”, which isG⁡[{1,2,5,6}]G\left[\left\{1,2,5,6\right\}\right]), since all the others have extraneous edges. The middle square yields an induced subgraph isomorphic toK4K_{4}, namelyG⁡[{2,3,6,7}]G\left[\left\{2,3,6,7\right\}\right].

Exercise 2.55.

Letnnbe a positive integer. LetSSbe a simple graph with2n2nvertices. Prove thatSShas two distinct vertices that have an even number of common neighbors.

Exercise 2.56.

Letn≥2n\geq 2be an integer. LetGGbe a simple graph withnnvertices.

(a)DescribeGGif the degrees of the vertices ofGGare1,1,…,1,n−11,1,\ldots,1,n-1.
(b)Letaaandbbbe two positive integers such thata+b=na+b=n. DescribeGGif the degrees of the vertices ofGGare1,1,…,1,a,b1,1,\ldots,1,a,b.

Here, to “describe”GGmeans to explicitly determine (with proof) a graph that is isomorphic toGG.

2.8Disjoint unions

Another way of constructing new graphs from old is the disjoint union. The idea is simple: Taking the disjoint unionG1⊔G2⊔⋯⊔GkG_{1}\sqcup G_{2}\sqcup\cdots\sqcup G_{k}of several simple graphsG1,G2,…,GkG_{1},G_{2},\ldots,G_{k}means putting the graphs alongside each other and treating the result as one big graph. To make this formally watertight, we have to relabel each vertexvvof each graphGiG_{i}as the pair(i,v)\left(i,v\right), so that vertices coming from different graphs appear as different even if they were equal. For example, the disjoint unionC3⊔C4C_{3}\sqcup C_{4}of the two cycle graphsC3C_{3}andC4C_{4}should not be

(which makes no sense, because there are two points labelled11in this picture, but a graph can have only one vertex11), but rather should be

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So here is the formal definition:

Definition 2.58.

LetG1,G2,…,GkG_{1},G_{2},\ldots,G_{k}be simple graphs, whereGi=(Vi,Ei)G_{i}=\left(V_{i},E_{i}\right)for eachi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}. Thedisjoint unionof thesekkgraphsG1,G2,…,GkG_{1},G_{2},\ldots,G_{k}is defined to be the simple graph(V,E)\left(V,E\right), where

V\displaystyle V={(i,v)∣i∈{1,2,…,k}andv∈Vi}and\displaystyle=\left\{\left(i,v\right)\ \mid\ i\in\left\{1,2,\ldots,k\right\}\text{ and }v\in V_{i}\right\}\ \ \ \ \ \ \ \ \ \ \text{and}E\displaystyle E={{(i,v1),(i,v2)}∣i∈{1,2,…,k}and{v1,v2}∈Ei}.\displaystyle=\left\{\left\{\left(i,v_{1}\right),\left(i,v_{2}\right)\right\}\ \mid\ i\in\left\{1,2,\ldots,k\right\}\text{ and }\left\{v_{1},v_{2}\right\}\in E_{i}\right\}.This disjoint union is denoted byG1⊔G2⊔⋯⊔GkG_{1}\sqcup G_{2}\sqcup\cdots\sqcup G_{k}.

Note: IfGGandHHare two graphs, then the two graphsG⊔HG\sqcup HandH⊔GH\sqcup Gare isomorphic, but not the same graph (unlessG=HG=H). For example,C3⊔C4C_{3}\sqcup C_{4}has a vertex(2,4)\left(2,4\right), butC4⊔C3C_{4}\sqcup C_{3}does not.

2.9Walks and paths

We now come to the definitions of walks and paths – two of the most fundamental features that graphs can have. In particular, Euler’s 1736 paper, where graphs were first studied, is about certain kinds of walks.

2.9.1Definitions

Imagine a graph as a road network, where each vertex is a town and each edge is a (bidirectional) road. By successively walking along several edges, you can often get from a town to another even if they are not adjacent. This is made formal in the concept of a “walk”:

Definition 2.59.

LetGGbe a simple graph. Then:

(a)Awalk(inGG) means a finite sequence(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)of vertices ofGG(withk≥0k\geq 0) such that all ofv0v1,v1v2,v2v3,…,vk−1vkv_{0}v_{1},\ v_{1}v_{2},\ v_{2}v_{3},\ \ldots,\ v_{k-1}v_{k}are edges ofGG. (The latter condition is vacuously true ifk=0k=0.)
(b)If𝐰=(v0,v1,…,vk)\mathbf{w}=\left(v_{0},v_{1},\ldots,v_{k}\right)is a walk inGG, then: - •Theverticesof𝐰\mathbf{w}are defined to bev0,v1,…,vkv_{0},v_{1},\ldots,v_{k}. - •Theedgesof𝐰\mathbf{w}are defined to bev0v1,v1v2,v2v3,…,vk−1vkv_{0}v_{1},\ v_{1}v_{2},\ v_{2}v_{3},\ \ldots,\ v_{k-1}v_{k}. - •The nonnegative integerkkis called thelengthof𝐰\mathbf{w}. (This is the number of all edges of𝐰\mathbf{w}, counted with multiplicity. It is11smaller than the number of all vertices of𝐰\mathbf{w}, counted with multiplicity.) - •The vertexv0v_{0}is called thestarting pointof𝐰\mathbf{w}. We say that𝐰\mathbf{w}starts(orbegins) atv0v_{0}. - •The vertexvkv_{k}is called theending pointof𝐰\mathbf{w}. We say that𝐰\mathbf{w}endsatvkv_{k}.
(c)Apath(inGG) means a walk (inGG) whose vertices are distinct. In other words, a path means a walk(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)such thatv0,v1,…,vkv_{0},v_{1},\ldots,v_{k}are distinct.
(d)Letppandqqbe two vertices ofGG. Awalk frompptoqqmeans a walk that starts atppand ends atqq. Apath frompptoqqmeans a path that starts atppand ends atqq.
(e)We often say “walk ofGG” and “path ofGG” instead of “walk inGG” and “path inGG”, respectively.

Example 2.60.

LetGGbe the graph

({1,2,3,4,5,6},{12,23,34,45,56,61,13}).\left(\left\{1,2,3,4,5,6\right\},\ \ \left\{12,\ 23,\ 34,\ 45,\ 56,\ 61,\ 13\right\}\right).This graph looks as follows:

Then:

•The sequence(1,3,4,5,6,1,3,2)\left(1,3,4,5,6,1,3,2\right)of vertices ofGGis a walk inGG. This walk is a walk from11to22. It is not a path. The length of this walk is77.
•The sequence(1,2,4,3)\left(1,2,4,3\right)of vertices ofGGis not a walk, since2424is not an edge ofGG. Hence, it is not a path either.
•The sequence(1,3,2,1)\left(1,3,2,1\right)is a walk from11to11. It has length33. It is not a path.
•The sequence(1,2,1)\left(1,2,1\right)is a walk from11to11. It has length22. It is not a path.
•The sequence(5)\left(5\right)is a walk from55to55. It has length00. It is a path. More generally, each vertexvvofGGproduces a length-00path(v)\left(v\right).
•The sequence(5,4)\left(5,4\right)is a walk from55to44. It has length11. It is a path. More generally, each edgeuvuvofGGproduces a length-11path(u,v)\left(u,v\right).

Intuitively, we can think of walks and paths as follows:

•Awalkof a graph is a way of walking from one vertex to another (or to the same vertex) by following a sequence of edges.
•Apathis a walk whose vertices are distinct (i.e., each vertex appears at most once in the walk).

Exercise 2.61.

LetGGbe a simple graph. Let𝐰\mathbf{w}be a path inGG. Prove that the edges of𝐰\mathbf{w}are distinct. (This may look obvious when you can point to a picture; but we ask you to give a rigorous proof!)

[Solution:This is Exercise 3 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

2.9.2Composing/concatenating and reversing walks

Here are some simple things we can do with walks and paths.

First, we can “splice” two walks together if the ending point of the first is the starting point of the second:

Proposition 2.62.

LetGGbe a simple graph. Letuu,vvandwwbe three vertices ofGG. Let𝐚=(a0,a1,…,ak)\mathbf{a}=\left(a_{0},a_{1},\ldots,a_{k}\right)be a walk fromuutovv. Let𝐛=(b0,b1,…,bℓ)\mathbf{b}=\left(b_{0},b_{1},\ldots,b_{\ell}\right)be a walk fromvvtoww. Then,

(a0,a1,…,ak,b1,b2,…,bℓ)\displaystyle\left(a_{0},a_{1},\ldots,a_{k},b_{1},b_{2},\ldots,b_{\ell}\right)=(a0,a1,…,ak−1,b0,b1,…,bℓ)\displaystyle=\left(a_{0},a_{1},\ldots,a_{k-1},b_{0},b_{1},\ldots,b_{\ell}\right)=(a0,a1,…,ak−1,v,b1,b2,…,bℓ)\displaystyle=\left(a_{0},a_{1},\ldots,a_{k-1},v,b_{1},b_{2},\ldots,b_{\ell}\right)is a walk fromuutoww. This walk shall be denoted𝐚∗𝐛\mathbf{a}\ast\mathbf{b}.

Proof 2.63.

Intuitively clear and straightforward to verify.

Note that “splicing” does not preserve path-ness: If𝐚\mathbf{a}and𝐛\mathbf{b}are two paths, then the walk𝐚∗𝐛\mathbf{a}\ast\mathbf{b}is not necessarily a path.

Another thing that can be done with walks (and this one does preserve path-ness) is walking them backwards:

Proposition 2.64.

LetGGbe a simple graph. Letuuandvvbe two vertices ofGG. Let𝐚=(a0,a1,…,ak)\mathbf{a}=\left(a_{0},a_{1},\ldots,a_{k}\right)be a walk fromuutovv. Then:

(a)The list(ak,ak−1,…,a0)\left(a_{k},a_{k-1},\ldots,a_{0}\right)is a walk fromvvtouu. We denote this walk byrev𝐚\operatorname*{rev}\mathbf{a}and call it thereversalof𝐚\mathbf{a}.
(b)If𝐚\mathbf{a}is a path, thenrev𝐚\operatorname*{rev}\mathbf{a}is a path again.

Proof 2.65.

Intuitively clear and straightforward to verify.

2.9.3Reducing walks to paths

A path is just a walk without repeated vertices. If you have a walk, you can turn it into a path by removing “loops” (or “digressions”):

Proposition 2.66.

LetGGbe a simple graph. Letuuandvvbe two vertices ofGG. Let𝐚=(a0,a1,…,ak)\mathbf{a}=\left(a_{0},a_{1},\ldots,a_{k}\right)be a walk fromuutovv. Assume that𝐚\mathbf{a}is not a path. Then, there exists a walk fromuutovvwhose length is smaller thankk.

Proof 2.67.

Since𝐚\mathbf{a}is not a path, two of its vertices are equal. In other words, there existi<ji<jsuch thatai=aja_{i}=a_{j}. Consider theseiiandjj. Now, consider the tuple

(a0,a1,…,ai⏟the firsti+1vertices of𝐚,aj+1,aj+2,…,ak⏟the lastk−jvertices of𝐚)\left(\underbrace{a_{0},a_{1},\ldots,a_{i}}_{\text{the first }i+1\text{ vertices of }\mathbf{a}},\underbrace{a_{j+1},a_{j+2},\ldots,a_{k}}_{\text{the last }k-j\text{ vertices of }\mathbf{a}}\right)(this is just𝐚\mathbf{a}with the part betweenaia_{i}andaja_{j}cut out). This tuple is a walk fromuutovv, and its length isi⏟<j+(k−j)<j+(k−j)=k\underbrace{i}_{<j}+\left(k-j\right)<j+\left(k-j\right)=k. So we have found a walk fromuutovvwhose length is smaller thankk. This proves the proposition.

Example 2.68.

Consider the walk(1,3,4,5,6,1,3,2)\left(1,3,4,5,6,1,3,2\right)from Example2.60. Then, Proposition2.66tells us that there is a walk from11to22that has smaller length. You can find this walk by removing the part between the two33’s. You get the walk(1,3,2)\left(1,3,2\right). This is actually a path.

Corollary 2.69(When there is a walk, there is a path).

LetGGbe a simple graph. Letuuandvvbe two vertices ofGG. Assume that there is a walk fromuutovvof lengthkkfor somek∈ℕk\in\mathbb{N}. Then, there is a path fromuutovvof length≤k\leq k.

Proof 2.70.

Proposition2.66says that if there is a walk fromuutovvthat is not a path, then there is a walk fromuutovvhaving shorter length. Apply this repeatedly, until you get a path. (You will eventually get a path, because the length cannot keep decreasing forever.)

2.9.4Remark on algorithms

We take a little break from proving structural theorems in order to address some important computational questions. As always in these notes, we will only scratch the surface and content ourselves with simple but not quite optimal algorithms.

Given a simple graphGGand two verticesuuandvvofGG, we can ask ourselves the following questions:

Question 1:DoesGGhave a walk fromuutovv? Question 2:DoesGGhave a path fromuutovv? Question 3:Find a shortest path fromuutovv(that is, a path fromuutovvhaving the smallest possible length), or determine that no such path exists. Question 4:Given a numberk∈ℕk\in\mathbb{N}, find a walk fromuutovvhaving lengthkk, or determine that no such walk exists. Question 5:Given a numberk∈ℕk\in\mathbb{N}, find a path fromuutovvhaving lengthkk, or determine that no such path exists.

Corollary2.69reveals that Questions 1 and 2 are equivalent (indeed, the existence of a walk fromuutovventails the existence of a path fromuutovvby Corollary2.69, whereas the converse is obvious). Question 3 is clearly a stronger version of Question 2 (in the sense that any answer to Question 3 will automatically answer Question 2 as well).

With a bit more thought, it is easily seen that Question 4 is a stronger version of Question 3. Indeed, Corollary2.69shows that a shortest walk fromuutovv(if it exists) must also be a shortest path fromuutovv. However, any path fromuutovvmust have length≤n−1\leq n-1, wherennis the number of vertices ofGG(since a path of lengthkkhask+1k+1distinct vertices, butGGhas onlynnvertices to spare). Hence, if there is no walk of length≤n−1\leq n-1fromuutovv, then there is no path fromuutovvwhatsoever. Thus, if we answer Question 4 for all valuesk∈{0,1,…,n−1}k\in\left\{0,1,\ldots,n-1\right\}, then we obtain either a shortest path fromuutovv(by taking the smallestkkfor which the answer is positive, and then picking the resulting walk, which must be a shortest path by what we previously said), or proof positive that no path fromuutovvexists (if the answer for eachk∈{0,1,…,n−1}k\in\left\{0,1,\ldots,n-1\right\}is negative).

Thus, answering Question 4 will yield answers to Questions 1, 2 and 3.

Let us now outline a way how Question 4 can be answered using a recursive algorithm. Specifically, we recurse onkk. The base case (k=0k=0) is easy: A walk fromuutovvhaving length00exists ifu=vu=vand does not exist otherwise. The interesting part is the recursion step: Assume that the integerkkis positive, and that we already know how to answer Question 4 fork−1k-1instead ofkk. Now, let us answer it forkk. To do so, we observe that any walk fromuutovvhaving lengthkkmust have the form(u,…,w,v)\left(u,\ldots,w,v\right), where the penultimate vertexwwis some neighbor ofvv. Moreover, if we remove the last vertexvvfrom our walk(u,…,w,v)\left(u,\ldots,w,v\right), then we obtain a walk(u,…,w)\left(u,\ldots,w\right)of lengthk−1k-1. Hence, we can find a walk fromuutovvhaving lengthkkas follows:

•We make a list of all neighbors ofvv. We go through this list in some arbitrary order.
•For each neighborwwin this list, we try to find a walk fromuutowwhaving lengthk−1k-1(this is a matter of answering Question 4 fork−1k-1instead ofkk, so we supposedly already know how to do this). If such a walk exists, then we simply insertvvat its end, and thus obtain a walk fromuutovvhaving lengthkk. Thus we obtain a positive answer to our question.
•If we have gone through our whole list of neighbors ofvvwithout finding a walk fromuutovvhaving lengthkk, then no such walk exists, and thus we have found a negative answer.

This recursive algorithm answers Question 4, and is fast enough to be practically viable if implemented well. (In the language of complexity theory, it is apolynomial time algorithm666To be specific: Its running time can be bounded in a polynomial ofnnandkk, wherennis the number of vertices ofGG..) Much more efficient algorithms exist, however. In applications, a generalized version of Question 3 often appears, asking for a path that is shortest not in the sense of smallest length, but in the sense of smallest “weighted length” (i.e., different edges contribute differently to this “length”). This generalized question is one of the most fundamental algorithmic problems in computer science, known as theshortest path problem, and various algorithms can be found onits Wikipedia pageand in algorithm-focussed texts such as[Griffi21, §3.5],[KelTro17, §12.3],[Even12, §1.5],[Schrij17, Chapter 1]or (for a royal treatment)[Schrij03, Chapters 6–8].

Question 5 looks superficially similar to Question 4, yet it differs in the most important way: There is no efficient algorithm known for answering it! In the language of complexity theory, it is anNP-hard problem, which means that a polynomial-time algorithm for it is not expected to exist (although this is the kind of negative that appears near-impossible to prove at the current stage of the discipline). It is still technically a finite problem (there are only finitely many possible paths inGG, and thus one can theoretically try them all), and there is even a polynomial-time algorithm for any fixed value ofkk(again, a trivial one: check all thenk+1n^{k+1}possible(k+1)\left(k+1\right)-tuples of vertices ofGGfor whether they are paths fromuutovv), but the complexity of this algorithm grows exponentially inkk, which makes it useless in practice.

2.9.5The equivalence relation “path-connected”

We can use the concepts of walks and paths to define a certain equivalence relation on the vertex setV(G)\operatorname*{V}\left(G\right)of any graphGG:

Definition 2.71.

LetGGbe a simple graph. We define a binary relation≃G\simeq_{G}on the setV(G)\operatorname*{V}\left(G\right)as follows: For two verticesuuandvvofGG, we shall haveu≃Gvu\simeq_{G}vif and only if there exists a walk fromuutovvinGG.

This binary relation≃G\simeq_{G}is called “path-connectedness” or just “connectedness”. When two verticesuuandvvsatisfyu≃Gvu\simeq_{G}v, we say that “uuandvvarepath-connected”.

Proposition 2.72.

LetGGbe a simple graph. Then, the relation≃G\simeq_{G}is an equivalence relation.

Proof 2.73.

We need to show that≃G\simeq_{G}is symmetric, reflexive and transitive.

•Symmetry:Ifu≃Gvu\simeq_{G}v, thenv≃Guv\simeq_{G}u, because we can take a walk fromuutovvand reverse it.
•Reflexivity:We always haveu≃Guu\simeq_{G}u, since the trivial walk(u)\left(u\right)is a walk fromuutouu.
•Transitivity:Ifu≃Gvu\simeq_{G}vandv≃Gwv\simeq_{G}w, thenu≃Gwu\simeq_{G}w, because (as we know from Proposition2.62) we can take a walk𝐚\mathbf{a}fromuutovvand a walk𝐛\mathbf{b}fromvvtowwand combine them to form the walk𝐚∗𝐛\mathbf{a}\ast\mathbf{b}defined in Proposition2.62.

Proposition 2.74.

LetGGbe a simple graph. Letuuandvvbe two vertices ofGG. Then,u≃Gvu\simeq_{G}vif and only if there exists a path fromuutovv.

Proof 2.75.

⟸:\Longleftarrow:Clear, since any path is a walk.

⟹:\Longrightarrow:This is just saying that if there is a walk fromuutovv, then there is a path fromuutovv. But this follows from Corollary2.69.

2.9.6Connected components and connectedness

The equivalence relation≃G\simeq_{G}introduced in Definition2.71allows us to define two important concepts:

Definition 2.76.

LetGGbe a simple graph. The equivalence classes of the equivalence relation≃G\simeq_{G}are called theconnected components(or, for short,components) ofGG.

Definition 2.77.

LetGGbe a simple graph. We say thatGGisconnectedifGGhas exactly one component.

Thus, a simple graphGGis connected if and only if it has at least one component (i.e., it has at least one vertex) and it has at most one component (i.e., each two of its vertices are path-connected).

Example 2.78.

LetGGbe the graph with vertex set{1,2,…,9}\left\{1,2,\ldots,9\right\}and such that two verticesiiandjjare adjacent if and only if|i−j|=3\left|i-j\right|=3. What are the components ofGG?

The graphGGlooks like this:

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-48.41}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This looks like a jumbled mess, so you might think that all vertices are mutually path-connected. But this is not the case, because edges that cross in a drawing do not necessarily have endpoints in common. Walks can only move from one edge to another at a common endpoint. Thus, there are much fewer walks than the picture might suggest. We have1≃G4≃G71\simeq_{G}4\simeq_{G}7and2≃G5≃G82\simeq_{G}5\simeq_{G}8and3≃G6≃G93\simeq_{G}6\simeq_{G}9, but there are no further≃G\simeq_{G}-relations. In fact, two vertices ofGGare adjacent only if they are congruent modulo33(as numbers), and therefore you cannot move from one modulo-33congruence class to another by walking along edges ofGG. So the components ofGGare{1,4,7}\left\{1,4,7\right\}and{2,5,8}\left\{2,5,8\right\}and{3,6,9}\left\{3,6,9\right\}. The graphGGis not connected.

Example 2.79.

LetGGbe the graph with vertex set{1,2,…,9}\left\{1,2,\ldots,9\right\}and such that two verticesiiandjjare adjacent if and only if|i−j|=6\left|i-j\right|=6. This graph looks like this:

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38.46 72.47 43.9 72.47 50.61 Z M 60.32 50.61}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{41.09172pt}{33.35617pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 56.86 46.15)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 25.83 77.54 C 25.83 84.26 20.39 89.7 13.67 89.7 C 6.96 89.7 1.52 84.26 1.52 77.54 C 1.52 70.83 6.96 65.39 13.67 65.39 C 20.39 65.39 25.83 70.83 25.83 77.54 Z M 13.67 77.54}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{7.38135pt}{52.81934pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 10.21 73.09)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -27.21 68.19 C -27.21 74.91 -32.66 80.35 -39.37 80.35 C -46.08 80.35 -51.53 74.91 -51.53 68.19 C -51.53 61.48 -46.08 56.03 -39.37 56.03 C -32.66 56.03 -27.21 61.48 -27.21 68.19 Z M -39.37 68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.95276pt}{46.05954pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.83 63.73)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -61.83 26.93 C -61.83 33.65 -67.28 39.09 -73.99 39.09 C -80.71 39.09 -86.15 33.65 -86.15 26.93 C -86.15 20.22 -80.71 14.77 -73.99 14.77 C -67.28 14.77 -61.83 20.22 -61.83 26.93 Z M -73.99 26.93}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-55.97395pt}{16.24094pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -77.45 22.47)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -61.83 -26.93 C -61.83 -20.22 -67.28 -14.77 -73.99 -14.77 C -80.71 -14.77 -86.15 -20.22 -86.15 -26.93 C -86.15 -33.65 -80.71 -39.09 -73.99 -39.09 C -67.28 -39.09 -61.83 -33.65 -61.83 -26.93 Z M -73.99 -26.93}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-55.97395pt}{-22.68536pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -77.45 -31.39)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -27.21 -68.19 C -27.21 -61.48 -32.66 -56.03 -39.37 -56.03 C -46.08 -56.03 -51.53 -61.48 -51.53 -68.19 C -51.53 -74.91 -46.08 -80.35 -39.37 -80.35 C -32.66 -80.35 -27.21 -74.91 -27.21 -68.19 Z M -39.37 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.95276pt}{-52.50397pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.83 -72.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 25.83 -77.54 C 25.83 -70.83 20.39 -65.39 13.67 -65.39 C 6.96 -65.39 1.52 -70.83 1.52 -77.54 C 1.52 -84.26 6.96 -89.7 13.67 -89.7 C 20.39 -89.7 25.83 -84.26 25.83 -77.54 Z M 13.67 -77.54}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{7.38135pt}{-59.26376pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 10.21 -82)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 72.47 -50.61 C 72.47 -43.9 67.03 -38.46 60.32 -38.46 C 53.6 -38.46 48.16 -43.9 48.16 -50.61 C 48.16 -57.33 53.6 -62.77 60.32 -62.77 C 67.03 -62.77 72.47 -57.33 72.47 -50.61 Z M 60.32 -50.61}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{41.09172pt}{-39.8006pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 56.86 -55.07)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 67.73 -6.36 L -28.36 -61.84}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 55.97 38.67 L 18.02 -65.6}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 18.02 65.6 L 55.97 -38.67}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .What are the components ofGG? They are{1,7}\left\{1,7\right\}and{2,8}\left\{2,8\right\}and{3,9}\left\{3,9\right\}and{4}\left\{4\right\}and{5}\left\{5\right\}and{6}\left\{6\right\}. Note that three of these six components are singleton sets. The graphGGis not connected.

Example 2.80.

LetGGbe the graph with vertex set{1,2,…,9}\left\{1,2,\ldots,9\right\}and such that two verticesiiandjjare adjacent if and only if|i−j|=3\left|i-j\right|=3or|i−j|=4\left|i-j\right|=4. This graph looks like this:

123456789.\hbox to128.75pt{\vbox to130.46pt{\pgfpicture\makeatletter\hbox{\hskip 62.65984pt\lower-65.22743pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 72.47 50.61 C 72.47 57.33 67.03 62.77 60.32 62.77 C 53.6 62.77 48.16 57.33 48.16 50.61 C 48.16 43.9 53.6 38.46 60.32 38.46 C 67.03 38.46 72.47 43.9 72.47 50.61 Z M 60.32 50.61}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{41.09172pt}{33.35617pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 56.86 46.15)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 25.83 77.54 C 25.83 84.26 20.39 89.7 13.67 89.7 C 6.96 89.7 1.52 84.26 1.52 77.54 C 1.52 70.83 6.96 65.39 13.67 65.39 C 20.39 65.39 25.83 70.83 25.83 77.54 Z 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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -62.98 20.58 L 49.31 -44.26}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .We can take a long walk throughGG:

(1,4,7,3,6,9,5,2,5,8).\left(1,4,7,3,6,9,5,2,5,8\right).This walk traverses every vertex ofGG; thus, any two vertices ofGGare path-connected. Hence,GGhas only one component, namely{1,2,…,9}\left\{1,2,\ldots,9\right\}. Thus,GGis connected.

Example 2.81.

The complete graph on a nonempty set is connected. The complete graph on the empty set is not connected, since it has00(not11) components.

Example 2.82.

The empty graph on a finite setVVhas|V|\left|V\right|many components (those are the singleton sets{v}\left\{v\right\}forv∈Vv\in V). Thus, it is connected if and only if|V|=1\left|V\right|=1.

Exercise 2.83.

Letk∈ℕk\in\mathbb{N}. LetSSbe a finite set.

Recall that theKneser graphKS,kK_{S,k}is the simple graph whose vertices are thekk-element subsets ofSS, and whose edges are the unordered pairs{A,B}\left\{A,B\right\}consisting of two such subsetsAAandBBthat satisfyA∩B=∅A\cap B=\varnothing.

Prove that this Kneser graphKS,kK_{S,k}is connected if|S|≥2k+1\left|S\right|\geq 2k+1.

[Remark:Can the “if” here be replaced by an “if and only if”? Not quite, because the graphKS,kK_{S,k}is also connected if|S|=2\left|S\right|=2andk=1k=1(in which case it has two vertices and one edge), or if|S|=k\left|S\right|=k(in which case it has only one vertex), or ifk=0k=0(in which case it has only one vertex). But these are the only “exceptions”.]

2.9.7Induced subgraphs on components

Recall that ifSSis a subset of the vertex set of a simple graphGG, thenG⁡[S]G\left[S\right]denotes the induced subgraph ofGGonSS(see Definition2.46(b)). The following is not hard to see:

Proposition 2.84.

LetGGbe a simple graph. LetCCbe a component ofGG.

Then, the graphG⁡[C]G\left[C\right](that is, the induced subgraph ofGGon the setCC) is connected.

Proof 2.85.

We need to show thatG⁡[C]G\left[C\right]is connected. In other words, we need to show thatG⁡[C]G\left[C\right]has exactly11component.

Clearly,G⁡[C]G\left[C\right]has at least one vertex (sinceCCis a component, i.e., an equivalence class of≃G\simeq_{G}, but equivalence classes are always nonempty), thus has at least11component. So we only need to show thatG⁡[C]G\left[C\right]has no more than11component. In other words, we need to show that any two vertices ofG⁡[C]G\left[C\right]are path-connected inG⁡[C]G\left[C\right].

So letuuandvvbe two vertices ofG⁡[C]G\left[C\right]. Then,u,v∈Cu,v\in C, and thereforeu≃Gvu\simeq_{G}v(sinceCCis a component ofGG). In other words, there exists a walk𝐰=(w0,w1,…,wk)\mathbf{w}=\left(w_{0},w_{1},\ldots,w_{k}\right)fromuutovvinGG. We shall now prove that this walk𝐰\mathbf{w}is actually a walk ofG⁡[C]G\left[C\right]. In other words, we shall prove that all vertices of𝐰\mathbf{w}belong toCC.

But this is easy: Ifwiw_{i}is a vertex of𝐰\mathbf{w}, then(w0,w1,…,wi)\left(w_{0},w_{1},\ldots,w_{i}\right)is a walk fromuutowiw_{i}inGG, and therefore we haveu≃Gwiu\simeq_{G}w_{i}, so thatwiw_{i}belongs to the same component ofGGasuu; but that component isCC. Thus, we have shown that each vertexwiw_{i}of𝐰\mathbf{w}belongs toCC. Therefore,𝐰\mathbf{w}is a walk of the graphG⁡[C]G\left[C\right]. Consequently, it shows thatu≃G⁡[C]vu\simeq_{G\left[C\right]}v.

We have now proved thatu≃G⁡[C]vu\simeq_{G\left[C\right]}vfor any two verticesuuandvvofG⁡[C]G\left[C\right]. Hence, the relation≃G⁡[C]\simeq_{G\left[C\right]}has no more than11equivalence class. In other words, the graphG⁡[C]G\left[C\right]has no more than11component. This completes our proof.

Note that the converse of Proposition2.84is not true: IfSSis a subset of the vertex set of a simple graphGGsuch thatG⁡[S]G\left[S\right]is connected, then it does not follow thatSSis a component ofGG.

Proposition 2.86.

LetGGbe a simple graph. LetC1,C2,…,CkC_{1},C_{2},\ldots,C_{k}be all components ofGG(listed without repetition).

Thus,GGis isomorphic to the disjoint unionG⁡[C1]⊔G⁡[C2]⊔⋯⊔G⁡[Ck]G\left[C_{1}\right]\sqcup G\left[C_{2}\right]\sqcup\cdots\sqcup G\left[C_{k}\right].

Proof 2.87.

Consider the bijection fromV(G⁡[C1]⊔G⁡[C2]⊔⋯⊔G⁡[Ck])\operatorname*{V}\left(G\left[C_{1}\right]\sqcup G\left[C_{2}\right]\sqcup\cdots\sqcup G\left[C_{k}\right]\right)toV(G)\operatorname*{V}\left(G\right)that sends each vertex(i,v)\left(i,v\right)ofG⁡[C1]⊔G⁡[C2]⊔⋯⊔G⁡[Ck]G\left[C_{1}\right]\sqcup G\left[C_{2}\right]\sqcup\cdots\sqcup G\left[C_{k}\right]to the vertexvvofGG. We claim that this bijection is a graph isomorphism. In order to prove this, we need to check that there are no edges ofGGthat join vertices in different components. But this is easy: If two vertices in different components ofGGwere adjacent, then they would be path-connected, and thus would actually belong to the same component.

The upshot of these results is that every simple graph can be decomposed into a disjoint union of its components (or, more precisely, of the induced subgraphs on its components). Each of these components is a connected graph. Moreover, this is easily seen to be the only way to decompose the graph into a disjoint union of connected graphs.

2.9.8Some exercises on connectedness

Exercise 2.88.

LetGGbe a simple graph withV⁡(G)≠∅\operatorname{V}\left(G\right)\neq\varnothing. Show that the following two statements are equivalent:

•Statement 1:The graphGGis connected.
•Statement 2:For every two nonempty subsetsAAandBBofV⁡(G)\operatorname{V}\left(G\right)satisfyingA∩B=∅A\cap B=\varnothingandA∪B=V⁡(G)A\cup B=\operatorname{V}\left(G\right), there exista∈Aa\in Aandb∈Bb\in Bsuch thatab∈E⁡(G)ab\in\operatorname{E}\left(G\right). (In other words: Whenever we subdivide the vertex setV⁡(G)\operatorname{V}\left(G\right)ofGGinto two nonempty subsets, there will be at least one edge ofGGconnecting a vertex in one subset to a vertex in another.)

[Solution:This is Exercise 7 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 2.89.

LetVVbe a nonempty finite set. LetGGandHHbe two simple graphs such thatV⁡(G)=V⁡(H)=V\operatorname{V}\left(G\right)=\operatorname{V}\left(H\right)=V. Assume that for eachu∈Vu\in Vandv∈Vv\in V, there exists a path fromuutovvinGGor a path fromuutovvinHH. Prove that at least one of the graphsGGandHHis connected.

[Solution:This is Exercise 8 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 2.90.

LetG=(V,E)G=\left(V,E\right)be a simple graph. Thecomplement graphG¯\overline{G}ofGGis defined to be the simple graph(V,𝒫2(V)∖E)\left(V,\ \mathcal{P}_{2}\left(V\right)\setminus E\right). (Thus, two distinct verticesuuandvvinVVare adjacent inG¯\overline{G}if and only if they are not adjacent inGG.)

Prove that at least one of the following two statements holds:

•Statement 1:For eachu∈Vu\in Vandv∈Vv\in V, there exists a path fromuutovvinGGof length≤3\leq 3.
•Statement 2:For eachu∈Vu\in Vandv∈Vv\in V, there exists a path fromuutovvinG¯\overline{G}of length≤2\leq 2.

[Solution:This is Exercise 9 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 2.91.

Letn≥2n\geq 2be an integer. LetGGbe a connected simple graph withnnvertices.

(a)DescribeGGif the degrees of the vertices ofGGare1,1,2,2,…,21,1,2,2,\ldots,2(exactly two11’s andn−2n-2many22’s).
(b)DescribeGGif the degrees of the vertices ofGGare1,1,…,1,n−11,1,\ldots,1,n-1.
(c)DescribeGGif the degrees of the vertices ofGGare2,2,…,22,2,\ldots,2.

Here, to “describe”GGmeans to explicitly determine (with proof) a graph that is isomorphic toGG.

The following exercise is not explicitly concerned with connectedness and components, but it might help to think about components to solve it (although there are solutions that do not use them):

Exercise 2.92.

LetGGbe a simple graph withnnvertices. Assume that each vertex ofGGhas at least one neighbor.

AmatchingofGGshall mean a setFFof edges ofGGsuch that no two edges inFFhave a vertex in common. Letmmbe the largest size of a matching ofGG.

Anedge coverofGGshall mean a setFFof edges ofGGsuch that each vertex ofGGis contained in at least one edgee∈Fe\in F. Letccbe the smallest size of an edge cover ofGG.

Prove thatc+m=nc+m=n.

2.10Closed walks and cycles

Here are two further kinds of walks:

Definition 2.94.

LetGGbe a simple graph.

(a)Aclosed walkofGGmeans a walk whose first vertex is identical with its last vertex. In other words, it means a walk(w0,w1,…,wk)\left(w_{0},w_{1},\ldots,w_{k}\right)withw0=wkw_{0}=w_{k}. Sometimes, closed walks are also known ascircuits(but many authors use this latter word for something slightly different).
(b)AcycleofGGmeans a closed walk(w0,w1,…,wk)\left(w_{0},w_{1},\ldots,w_{k}\right)such thatk≥3k\geq 3and such that the verticesw0,w1,…,wk−1w_{0},w_{1},\ldots,w_{k-1}are distinct.

Example 2.95.

LetGGbe the simple graph

({1,2,3,4,5,6},{12,23,34,45,56,61,13}).\left(\left\{1,2,3,4,5,6\right\},\ \ \left\{12,\ 23,\ 34,\ 45,\ 56,\ 61,\ 13\right\}\right).This graph looks as follows (we have already seen it in Example2.60):

Then:

•The sequence(1,3,2,1,6,5,6,1)\left(1,3,2,1,6,5,6,1\right)is a closed walk ofGG. But it is very much not a cycle.
•The sequences(1,2,3,1)\left(1,2,3,1\right)and(1,3,4,5,6,1)\left(1,3,4,5,6,1\right)and(1,2,3,4,5,6,1)\left(1,2,3,4,5,6,1\right)are cycles ofGG. You can get further cycles by rotating these sequences (in a proper sense of this word – e.g., rotating(1,2,3,1)\left(1,2,3,1\right)gives(2,3,1,2)\left(2,3,1,2\right)and(3,1,2,3)\left(3,1,2,3\right)) and by reversing them. Every cycle ofGGcan be obtained in this way.
•The sequences(1)\left(1\right)and(1,2,1)\left(1,2,1\right)are closed walks, but not cycles ofGG(since they fail thek≥3k\geq 3condition).
•The sequence(1,2,3)\left(1,2,3\right)is a walk, but not a closed walk, since1≠31\neq 3.

Authors have different opinions about whether(1,2,3,1)\left(1,2,3,1\right)and(1,3,2,1)\left(1,3,2,1\right)count as different cycles. Fortunately, this matters only if you want to count cycles, but not for the existence or non-existence of cycles.

We have now defined paths (in an arbitrary graph) and also path graphsPnP_{n}; we have also defined cycles (in an arbitrary graph) and also cycle graphsCnC_{n}. Besides their similar names, are they related? The answer is “yes”:

Proposition 2.96.

LetGGbe a simple graph.

(a)If(p0,p1,…,pk)\left(p_{0},p_{1},\ldots,p_{k}\right)is a path ofGG, then there is a subgraph ofGGisomorphic to the path graphPk+1P_{k+1}, namely the subgraph({p0,p1,…,pk},{pipi+1∣0≤i<k})\left(\left\{p_{0},p_{1},\ldots,p_{k}\right\},\ \left\{p_{i}p_{i+1}\ \mid\ 0\leq i<k\right\}\right). (If this subgraph is actually an induced subgraph ofGG, then the path(p0,p1,…,pk)\left(p_{0},p_{1},\ldots,p_{k}\right)is called an “induced path”.) Conversely, any subgraph ofGGisomorphic toPk+1P_{k+1}gives a path ofGG.
(b)Now, assume thatk≥3k\geq 3. If(c0,c1,…,ck)\left(c_{0},c_{1},\ldots,c_{k}\right)is a cycle ofGG, then there is a subgraph ofGGisomorphic to the cycle graphCkC_{k}, namely the subgraph({c0,c1,…,ck},{cici+1∣0≤i<k})\left(\left\{c_{0},c_{1},\ldots,c_{k}\right\},\ \left\{c_{i}c_{i+1}\ \mid\ 0\leq i<k\right\}\right). (If this subgraph is actually an induced subgraph ofGG, then the cycle(c0,c1,…,ck)\left(c_{0},c_{1},\ldots,c_{k}\right)is called an “induced cycle”.) Conversely, any subgraph ofGGisomorphic toCkC_{k}gives a cycle ofGG.

Proof 2.97.

Straightforward.

Certain graphs contain cycles; other graphs don’t. For instance, the complete graphKnK_{n}contains a lot of cycles (whenn≥3n\geq 3), whereas the path graphPnP_{n}contains none. Let us try to find some criteria for when a graph can and when it cannot have cycles777Mantel’s theorem already gives such a criterion for cycles of length33(because a cycle of length33is the same as a triangle).:

Proposition 2.98.

LetGGbe a simple graph. Let𝐰\mathbf{w}be a walk ofGGsuch that no two consecutive edges of𝐰\mathbf{w}are identical. (By “consecutive edges”, we mean edges of the formwi−1wiw_{i-1}w_{i}andwiwi+1w_{i}w_{i+1}, wherewi−1,wi,wi+1w_{i-1},w_{i},w_{i+1}are three consecutive vertices of𝐰\mathbf{w}.)

Then,𝐰\mathbf{w}either is a path or contains a cycle (i.e., there exists a cycle ofGGwhose edges are edges of𝐰\mathbf{w}).

Example 2.99.

LetGGbe as in Example2.95. Then,(2,1,3,2,1,6)\left(2,1,3,2,1,6\right)is a walk𝐰\mathbf{w}ofGGsuch that no two consecutive edges of𝐰\mathbf{w}are identical (even though the edge2121appears twice in this walk). On the other hand,(2,1,3,1,6)\left(2,1,3,1,6\right)is not such a walk (since its two consecutive edges1313and3131are identical).

Proof 2.100(Proof of Proposition2.98.).

We assume that𝐰\mathbf{w}is not a path. We must then show that𝐰\mathbf{w}contains a cycle.

Write𝐰\mathbf{w}as𝐰=(w0,w1,…,wk)\mathbf{w}=\left(w_{0},w_{1},\ldots,w_{k}\right). Since𝐰\mathbf{w}is not a path, two of the verticesw0,w1,…,wkw_{0},w_{1},\ldots,w_{k}must be equal. In other words, there exists a pair(i,j)\left(i,j\right)of integersiiandjjwithi<ji<jandwi=wjw_{i}=w_{j}. Among all such pairs, we pick one withminimumdifferencej−ij-i. We shall show that the walk(wi,wi+1,…,wj)\left(w_{i},w_{i+1},\ldots,w_{j}\right)is a cycle.

First, this walk is clearly a closed walk (sincewi=wjw_{i}=w_{j}). It thus remains to show thatj−i≥3j-i\geq 3and that the verticeswi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct. The distinctness ofwi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}follows from the minimality ofj−ij-i. To show thatj−i≥3j-i\geq 3, we assume the contrary. Thus,j−ij-iis either11or22(sincei<ji<j). Butj−ij-icannot be11, since the endpoints of an edge cannot be equal (since our graph is a simple graph). Soj−ij-imust be22. Thus,wi=wi+2w_{i}=w_{i+2}. Therefore, the two edgeswiwi+1w_{i}w_{i+1}andwi+1wi+2w_{i+1}w_{i+2}are identical. But this contradicts the fact that no two consecutive edges of𝐰\mathbf{w}are identical. Contradiction, qed.

Corollary 2.101.

LetGGbe a simple graph. Assume thatGGhas a closed walk𝐰\mathbf{w}of length>0>0such that no two consecutive edges of𝐰\mathbf{w}are identical. Then,GGhas a cycle.

Proof 2.102.

This follows from Proposition2.98, since𝐰\mathbf{w}is not a path.

Theorem 2.103.

LetGGbe a simple graph. Letuuandvvbe two vertices inGG. Assume that there are two distinct paths fromuutovv. Then,GGhas a cycle.

Proof 2.104.

More generally, we shall prove this theorem with the word “path” replaced by “backtrack-free walk”, where a “backtrack-free walk” means a walk𝐰\mathbf{w}such that no two consecutive edges of𝐰\mathbf{w}are identical. This is a generalization of the theorem, since every path is a backtrack-free walk (why?).

So we claim the following:

Claim 1:Let𝐩\mathbf{p}and𝐪\mathbf{q}be two distinct backtrack-free walks that start at the same vertex and end at the same vertex. Then,GGhas a cycle.

We shall prove Claim 1 by induction on the length of𝐩\mathbf{p}:

Base case:As the base case, we note that Claim 1 is vacuously true when the length of𝐩\mathbf{p}is−1-1(since there are no walks of length−1-1).

Induction step:We fix an integerNN, and we assume that Claim 1 is proved in the case when the length of𝐩\mathbf{p}isN−1N-1. We must now show that it is also true when the length of𝐩\mathbf{p}isNN.

So let𝐩=(p0,p1,…,pa)\mathbf{p}=\left(p_{0},p_{1},\ldots,p_{a}\right)and𝐪=(q0,q1,…,qb)\mathbf{q}=\left(q_{0},q_{1},\ldots,q_{b}\right)be two distinct backtrack-free walks that start at the same vertex and end at the same vertex and satisfya=Na=N. We must find a cycle.

The walks𝐩\mathbf{p}and𝐪\mathbf{q}are distinct but start at the same vertex, so they cannot both be trivial888We say that a walk istrivialif it has length00.. If one of them is trivial, then the other is a closed walk (because a trivial walk is a closed walk), and then our goal follows from Corollary2.101in this case (because we have a nontrivial closed backtrack-free walk). Hence, from now on, we WLOG assume that neither of the two walks𝐩\mathbf{p}and𝐪\mathbf{q}is trivial. Thus, each of these two walks has a last edge. The last edge of𝐩\mathbf{p}ispa−1pap_{a-1}p_{a}, whereas the last edge of𝐪\mathbf{q}isqb−1qbq_{b-1}q_{b}.

Two cases are possible:

Case 1:We havepa−1pa=qb−1qbp_{a-1}p_{a}=q_{b-1}q_{b}.

Case 2:We havepa−1pa≠qb−1qbp_{a-1}p_{a}\neq q_{b-1}q_{b}.

Let us consider Case 1 first. In this case, the last edgespa−1pap_{a-1}p_{a}andqb−1qbq_{b-1}q_{b}of the two walks𝐩\mathbf{p}and𝐪\mathbf{q}are identical, so the second-to-last vertices of these two walks must also be identical. Thus, if we remove these last edges from both walks, then we obtain two shorter backtrack-free walks(p0,p1,…,pa−1)\left(p_{0},p_{1},\ldots,p_{a-1}\right)and(q0,q1,…,qb−1)\left(q_{0},q_{1},\ldots,q_{b-1}\right)that again start at the same vertex and end at the same vertex, but the length of the first of them isa−1=N−1a-1=N-1. Hence, by the induction hypothesis, we can apply Claim 1 to these two shorter walks (instead of𝐩\mathbf{p}and𝐪\mathbf{q}), and we conclude thatGGhas a cycle. So we are done in Case 1.

Let us now consider Case 2. In this case, we combine the two walks𝐩\mathbf{p}and𝐪\mathbf{q}(more precisely,𝐩\mathbf{p}and the reversal of𝐪\mathbf{q}) to obtain the closed walk

(p0,p1,…,pa−1,pa=qb,qb−1,…,q0).\left(p_{0},p_{1},\ldots,p_{a-1},p_{a}=q_{b},q_{b-1},\ldots,q_{0}\right).This closed walk is backtrack-free (since(p0,p1,…,pa)\left(p_{0},p_{1},\ldots,p_{a}\right)and(q0,q1,…,qb)\left(q_{0},q_{1},\ldots,q_{b}\right)are backtrack-free, and sincepa−1pa≠qb−1qbp_{a-1}p_{a}\neq q_{b-1}q_{b}) and has length>0>0(since it contains at least the edgepa−1pap_{a-1}p_{a}). Hence, Corollary2.101entails thatGGhas a cycle.

We have thus found a cycle in both Cases 1 and 2. This completes the induction step. Thus, we have proved Claim 1. As we said, Theorem2.103follows from it.

Exercise 2.105.

LetGGbe a simple graph.

(a)Prove that ifGGhas a closed walk of odd length, thenGGhas a cycle of odd length.
(b)Is it true that ifGGhas a closed walk of length not divisible by33, thenGGhas a cycle of length not divisible by33?
(c)Does the answer to part(b)change if we replace “walk” by “non-backtracking walk”? (A walk𝐰\mathbf{w}with edgese1,e2,…,eke_{1},e_{2},\ldots,e_{k}(in this order) is said to benon-backtrackingif eachi∈{1,2,…,k−1}i\in\left\{1,2,\ldots,k-1\right\}satisfiesei≠ei+1e_{i}\neq e_{i+1}.)
(d)Atrail(in a graph) means a walk whose edges are distinct (but whose vertices are not necessarily distinct). Does the answer to part(b)change if we replace “walk” by “trail”?

(Proofs and counterexamples should be given.)

2.11The longest path trick

Here is another proposition that guarantees the existence of cycles in a graph under certain circumstances. More importantly, its proof illustrates a useful tactic in dealing with graphs:

Proposition 2.106.

LetGGbe a simple graph with at least one vertex. Letd>1d>1be an integer. Assume that each vertex ofGGhas degree≥d\geq d. Then,GGhas a cycle of length≥d+1\geq d+1.

Proof 2.107.

Let𝐩=(v0,v1,…,vm)\mathbf{p}=\left(v_{0},v_{1},\ldots,v_{m}\right)be alongestpath ofGG. (Why doesGGhave a longest path? Let’s see: Any path ofGGhas length≤|V|−1\leq\left|V\right|-1, since its vertices have to be distinct. Moreover,GGhas at least one vertex and thus has at least one path. A finite nonempty set of integers has a largest element. Thus,GGhas a longest path.)

The vertexv0v_{0}has degree≥d\geq d(by assumption), and thus has≥d\geq dneighbors (since the degree of a vertex is the number of its neighbors).

If all neighbors ofv0v_{0}belonged to the set{v1,v2,…,vd−1}\left\{v_{1},v_{2},\ldots,v_{d-1}\right\}999Ifd−1>md-1>m, then this set should be understood to mean{v1,v2,…,vm}\left\{v_{1},v_{2},\ldots,v_{m}\right\}., then the number of neighbors ofv0v_{0}would be at mostd−1d-1, which would contradict the previous sentence. Thus, there exists at least one neighboruuofv0v_{0}that doesnotbelong to this set{v1,v2,…,vd−1}\left\{v_{1},v_{2},\ldots,v_{d-1}\right\}. Consider thisuu. Then,u≠v0u\neq v_{0}(since a vertex cannot be its own neighbor).

Attaching the vertexuuto the front of the path𝐩\mathbf{p}, we obtain a walk

𝐩′:=(u,v0,v1,…,vm).\mathbf{p}^{\prime}:=\left(u,v_{0},v_{1},\ldots,v_{m}\right).If we hadu∉{v0,v1,…,vm}u\notin\left\{v_{0},v_{1},\ldots,v_{m}\right\}, then this walk𝐩′\mathbf{p}^{\prime}would be a path; but this would contradict the fact that𝐩\mathbf{p}is alongestpath ofGG. Thus, we must haveu∈{v0,v1,…,vm}u\in\left\{v_{0},v_{1},\ldots,v_{m}\right\}. In other words,u=viu=v_{i}for somei∈{0,1,…,m}i\in\left\{0,1,\ldots,m\right\}. Consider thisii. Sinceu≠v0u\neq v_{0}andu∉{v1,v2,…,vd−1}u\notin\left\{v_{1},v_{2},\ldots,v_{d-1}\right\}, we thus havei≥di\geq d. Here is a picture:

v0v_{0}v1v_{1}v2v_{2}vi=uv_{i}=u⋯\cdots⋯\cdots Now, consider the walk

𝐜:=(u,v0,v1,…,vi).\mathbf{c}:=\left(u,v_{0},v_{1},\ldots,v_{i}\right).This is a closed walk (sinceu=viu=v_{i}) and has lengthi+1≥d+1i+1\geq d+1(sincei≥di\geq d). If we can show that𝐜\mathbf{c}is a cycle, then we have thus found a cycle of length≥d+1\geq d+1, so we will be done.

It thus remains to prove that𝐜\mathbf{c}is a cycle. Let us do this. We need to check that the verticesu,v0,v1,…,vi−1u,v_{0},v_{1},\ldots,v_{i-1}are distinct, and that the length of𝐜\mathbf{c}is≥3\geq 3. The latter claim is clear: The length of𝐜\mathbf{c}isi+1≥d+1≥3i+1\geq d+1\geq 3(sinced>1d>1andd∈ℤd\in\mathbb{Z}). The former claim is not much harder: Sinceu=viu=v_{i}, the verticesu,v0,v1,…,vi−1u,v_{0},v_{1},\ldots,v_{i-1}are just the verticesvi,v0,v1,…,vi−1v_{i},v_{0},v_{1},\ldots,v_{i-1}, and thus are distinct because they are distinct vertices of the path𝐩\mathbf{p}. The proof of Proposition2.106is thus complete.

2.12Bridges

One question that will later prove crucial is: What happens to a graph if we remove a single edge from it? Let us first define a notation for this:

Definition 2.108.

LetG=(V,E)G=\left(V,E\right)be a simple graph. Leteebe an edge ofGG. Then,G∖eG\setminus ewill mean the graph obtained fromGGby removing this edgeee. In other words,

G∖e:=(V,E∖{e}).G\setminus e:=\left(V,\ E\setminus\left\{e\right\}\right).

Some authors writeG−eG-eforG∖eG\setminus e.

Theorem 2.109.

LetGGbe a simple graph. Leteebe an edge ofGG. Then:

(a)Ifeeis an edge of some cycle ofGG, then the components ofG∖eG\setminus eare precisely the components ofGG. (Keep in mind that the components are sets of vertices. It is these sets that we are talking about here, not the induced subgraphs on these sets.)
(b)Ifeeappears in no cycle ofGG(in other words, there exists no cycle ofGGsuch thateeis an edge of this cycle), then the graphG∖eG\setminus ehas one more component thanGG.

Example 2.110.

LetGGbe the graph shown in the following picture:

(where we have labeled the edgesaaandbbfor further reference). This graph has44components. The edgeaais an edge of a cycle ofGG, whereas the edgebbappears in no cycle ofGG. Thus, if we sete=ae=a, then Theorem2.109(a)shows that the components ofG∖eG\setminus eare precisely the components ofGG. This graphG∖eG\setminus efore=ae=alooks as follows:

and visibly has44components. On the other hand, if we sete=be=b, then Theorem2.109(b)shows that the graphG∖eG\setminus ehas one more component thanGG. This graphG∖eG\setminus efore=be=blooks as follows:

and visibly has55components.

Proof 2.111(Proof of Theorem2.109.).

We will only sketch the proof. For details, see[21f6, §6.7].

Letuuandvvbe the endpoints ofee, so thate=uve=uv. Note that(u,v)\left(u,v\right)is a path ofGG, and thus we haveu≃Gvu\simeq_{G}v.

(a)Assume thateeis an edge of some cycle ofGG. Then, if you removeeefrom this cycle, then you still have a path fromuutovvleft (as the remaining edges of the cycle function as a detour), and this path is a path ofG∖eG\setminus e. Thus,u≃G∖evu\simeq_{G\setminus e}v.

Now, we must show that the components ofG∖eG\setminus eare precisely the components ofGG. This will clearly follow if we can show that the relation≃G∖e\simeq_{G\setminus e}is precisely the relation≃G\simeq_{G}(because the components of a graph are the equivalence classes of its≃\simeqrelation). So let us prove the latter fact.

We must show that two verticesxxandyyofGGsatisfyx≃G∖eyx\simeq_{G\setminus e}yif and only if they satisfyx≃Gyx\simeq_{G}y. The “only if” part is obvious (since a walk ofG∖eG\setminus eis always a walk ofGG). It thus remains to prove the “if” part. So we assume thatxxandyyare two vertices ofGGsatisfyingx≃Gyx\simeq_{G}y, and we want to show thatx≃G∖eyx\simeq_{G\setminus e}y.

Fromx≃Gyx\simeq_{G}y, we conclude thatGGhas a path fromxxtoyy(by Proposition2.74). If this path does not use101010We say that a walk𝐰\mathbf{w}usesan edgeffifffis an edge of𝐰\mathbf{w}.the edgeee, then it is a path fromxxtoyyinG∖eG\setminus e, and thus we havex≃G∖eyx\simeq_{G\setminus e}y, which is what we wanted to prove. So we WLOG assume that this path does use the edgeee. Thus, this path contains the endpointsuuandvvof this edgeee. We WLOG assume thatuuappears beforevvon this path (otherwise, just swapuuwithvv). Thus, this path looks as follows:

(x,…,u,v,…,y).\left(x,\ldots,u,v,\ldots,y\right).If we remove the edgee=uve=uv, then this path breaks into two smaller paths

(x,…,u)and(v,…,y)\left(x,\ldots,u\right)\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left(v,\ldots,y\right)(since the edges of a path are distinct, soeeappears only once in it). Both of these two smaller paths are paths ofG∖eG\setminus e. Thus,x≃G∖eux\simeq_{G\setminus e}uandv≃G∖eyv\simeq_{G\setminus e}y. Now, recalling that≃G∖e\simeq_{G\setminus e}is an equivalence relation, we combine these results to obtain

x≃G∖eu≃G∖ev≃G∖ey.x\simeq_{G\setminus e}u\simeq_{G\setminus e}v\simeq_{G\setminus e}y.Hence,x≃G∖eyx\simeq_{G\setminus e}y. This completes the proof of Theorem2.109(a).

(b)Assume thateeappears in no cycle ofGG. We must prove that the graphG∖eG\setminus ehas one more component thanGG. To do so, it suffices to show the following:

Claim 1:The component ofGGthat containsuuandvv(this component exists, sinceu≃Gvu\simeq_{G}v) breaks into two components ofG∖eG\setminus ewhen the edgeeeis removed.

Claim 2:All other components ofGGremain components ofG∖eG\setminus e.

Claim 2 is pretty clear: The components ofGGthat don’t containuuandvvdo not change at all wheneeis removed (since they contain neither endpoint ofee). Thus, they remain components ofG∖eG\setminus e. (Formalizing this is a nice exercise in formalization; see[21f6, §6.7].)

It remains to prove Claim 1. We introduce some notations:

•LetCCbe the component ofGGthat containsuuandvv.
•LetAAbe the component ofG∖eG\setminus ethat containsuu.
•LetBBbe the component ofG∖eG\setminus ethat containsvv.

Then, we must show thatA∪B=CA\cup B=CandA∩B=∅A\cap B=\varnothing.

To see thatA∩B=∅A\cap B=\varnothing, we need to show thatu≃G∖evu\simeq_{G\setminus e}vdoesnothold (sinceAAandBBare the equivalence classes ofuuandvvwith respect to the relation≃G∖e\simeq_{G\setminus e}). So let us do this. Assume the contrary. Thus,u≃G∖evu\simeq_{G\setminus e}v. Hence, there exists a path fromuutovvinG∖eG\setminus e. Sincee=uve=uv, we can “close” this path by appending the vertexuuto its end; the result is a cycle of the graphGGthat contains the edgeee. But this contradicts our assumption that no cycle ofGGcontainsee. This contradiction shows that our assumption was wrong. Thus, we conclude thatu≃G∖evu\simeq_{G\setminus e}vdoesnothold. Hence, as we said,A∩B=∅A\cap B=\varnothing.

It remains to show thatA∪B=CA\cup B=C. SinceAAandBBare clearly subsets ofCC(because each walk ofG∖eG\setminus eis a walk ofGG, and thus each component ofG∖eG\setminus eis a subset of a component ofGG), we haveA∪B⊆CA\cup B\subseteq C, and therefore we only need to show thatC⊆A∪BC\subseteq A\cup B. In other words, we need to show that eachc∈Cc\in Cbelongs toA∪BA\cup B.

Let us show this. Letc∈Cc\in Cbe a vertex. Then,c≃Guc\simeq_{G}u(sinceCCis the component ofGGcontaininguu). Therefore,GGhas a path𝐩\mathbf{p}fromcctouu. Consider this path𝐩\mathbf{p}. Two cases are possible:

•Case 1:This path𝐩\mathbf{p}does not use the edgeee. In this case,𝐩\mathbf{p}is a path ofG∖eG\setminus e, and thus we obtainc≃G∖euc\simeq_{G\setminus e}u. In other words,c∈Ac\in A(sinceAAis the component ofG∖eG\setminus econtaininguu).
•Case 2:This path𝐩\mathbf{p}does use the edgeee. In this case, the edgeeemust be the last edge of𝐩\mathbf{p}(since the path𝐩\mathbf{p}would otherwise contain the vertexuutwice111111Indeed, the path𝐩\mathbf{p}already ends inuu. If it would containeeanywhere other than at the very end, then it would thus contain the vertexuutwice (sinceuuis an endpoint ofee).; but a path cannot contain a vertex twice), and the last two vertices of𝐩\mathbf{p}must bevvanduuin this order. Thus, by removing the last vertex from𝐩\mathbf{p}, we obtain a path fromcctovv, and this latter path is a path ofG∖eG\setminus e(since it no longer containsuuand therefore does not useee). This yieldsc≃G∖evc\simeq_{G\setminus e}v. In other words,c∈Bc\in B(sinceBBis the component ofG∖eG\setminus econtainingvv).

In either of these two cases, we have shown thatccbelongs to one ofAAandBB. In other words,c∈A∪Bc\in A\cup B. This is precisely what we wanted to show. This completes the proof of Theorem2.109(b).

We introduce some fairly standard terminology:

Definition 2.112.

Leteebe an edge of a simple graphGG.

(a)We say thateeis abridge(ofGG) ifeeappears in no cycle ofGG.
(b)We say thateeis acut-edge(ofGG) if the graphG∖eG\setminus ehas more components thanGG.

Corollary 2.113.

Leteebe an edge of a simple graphGG. Then,eeis a bridge if and only ifeeis a cut-edge.

Proof 2.114.

Follows from Theorem2.109.

We can also define “cut-vertices”: A vertexvvof a graphGGis said to be acut-vertexif the graphG∖vG\setminus v(that is, the graphGGwith the vertexvvremoved121212When we remove a vertex, we must of course also remove all edges that contain this vertex.) has more components thanGG. Unfortunately, there doesn’t seem to be an analogue of Corollary2.113for cut-vertices. Note also that removing a vertex (unlike removing an edge) can add more than one component to the graph (or it can also subtract11component if this vertex had degree00). For example, removing the vertex00from the graph

results in an empty graph on the set{1,2,3,4}\left\{1,2,3,4\right\}, so the number of components has increased from11to44.

2.13Dominating sets

2.13.1Definition and basic facts

Here is another concept we can define for a graph:

Definition 2.115.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

A subsetUUofVVis said to bedominating(forGG) if it has the following property: Each vertexv∈V∖Uv\in V\setminus Uhas at least one neighbor inUU.

Adominating setforGG(ordominating setofGG) will mean a subset ofVVthat is dominating.

Example 2.116.

Consider the cycle graph

C5=({1,2,3,4,5},{12,23,34,45,51})=12345.C_{5}=\left(\left\{1,2,3,4,5\right\},\ \ \left\{12,\ 23,\ 34,\ 45,\ 51\right\}\right)=\hbox to121.32pt{\vbox to126.61pt{\pgfpicture\makeatletter\hbox{\hskip 55.22363pt\lower-63.30673pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} 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set{1,3}\left\{1,3\right\}is a dominating set forC5C_{5}, since all three vertices2,4,52,4,5that don’t belong to{1,3}\left\{1,3\right\}have neighbors in{1,3}\left\{1,3\right\}. The set{1,5}\left\{1,5\right\}is not a dominating set forC5C_{5}, since the vertex33has no neighbor in{1,5}\left\{1,5\right\}. There is no dominating set forC5C_{5}that has size00or11, but there are several of size22, and every subset of size≥3\geq 3is dominating.

Here are some more examples:

•IfG=(V,E)G=\left(V,E\right)is a simple graph, then the whole vertex setVVis always dominating, whereas the empty set∅\varnothingis dominating only whenV=∅V=\varnothing.
•IfG=(V,E)G=\left(V,E\right)is a complete graph, then any nonempty subset ofVVis dominating.
•IfG=(V,E)G=\left(V,E\right)is an empty graph, then onlyVVis dominating.

Clearly, the “denser” a graph is (i.e., the more edges it has), the “easier” it is for a set to be dominating. Often, a graph is given, and one is interested in finding a dominating set of the smallest possible size131313Supposedly, this has applications in mobile networking: For example, you might want to choose a set of routers in a given network so that each node is either a router or directly connected (i.e., adjacent) to one.. As the case of an empty graph reveals, sometimes the only choice is the whole vertex set. However, in many cases, we can do better. Namely, we need to require that the graph has no isolated vertices:

Definition 2.117.

LetGGbe a simple graph. A vertexvvofGGis said to beisolatedif it has no neighbors (i.e., ifdeg⁡v=0\deg v=0).

An isolated vertex has to belong to every dominating set (since otherwise, it would need a neighbor in that set, but it has no neighbors). Thus, isolated vertices do not contribute much to the study of dominating sets, other than inflating their size. Therefore, when we look for dominating sets, we can restrict ourselves to graphs with no isolated vertices. There, we have the following result:

Proposition 2.118.

LetG=(V,E)G=\left(V,E\right)be a simple graph that has no isolated vertices. Then:

(a)There exists a dominating subset ofVVthat has size≤|V|/2\leq\left|V\right|/2.
(b)There exist two disjoint dominating subsetsAAandBBofVVsuch thatA∪B=VA\cup B=V.

One proof of this proposition will be given in Exercise2.121below (homework set #2 exercise 4). Another appears in[17s, §3.6].

For specific graphs, the bound|V|/2\left|V\right|/2in Proposition2.118(a)can often be improved. Here is an example:

Exercise 2.119.

Letn≥3n\geq 3be an integer. Find a formula for the smallest size of a dominating set of the cycle graphCnC_{n}. You can use theceiling functionx↦⌈x⌉x\mapsto\left\lceil x\right\rceil, which sends a real numberxxto the smallest integer that is≥x\geq x.

Exercise 2.120.

Letnnandkkbe positive integers such thatn≥k⁡(k+1)n\geq k\left(k+1\right)andk>1k>1. Recall (from Subsection2.6.3) the Kneser graphKGn,kKG_{n,k}, whose vertices are thekk-element subsets of{1,2,…,n}\left\{1,2,\ldots,n\right\}, and whose edges are the unordered pairs{A,B}\left\{A,B\right\}of such subsets withA∩B=∅A\cap B=\varnothing.

Prove that the minimum size of a dominating set ofKGn,kKG_{n,k}isk+1k+1.

Exercise 2.121.

LetG=(V,E)G=\left(V,E\right)be a connected simple graph with at least two vertices.

Thedistanced⁡(v,w)d\left(v,w\right)between two verticesvvandwwofGGis defined to be the smallest length of a path fromvvtoww. (In particular,d⁡(v,v)=0d\left(v,v\right)=0for eachv∈Vv\in V.)

Fix a vertexv∈Vv\in V. Define two subsets

A={w∈V∣d⁡(v,w)is even}andB={w∈V∣d⁡(v,w)is odd}A=\left\{w\in V\mid d\left(v,w\right)\text{ is even}\right\}\qquad\text{ and }\qquad B=\left\{w\in V\mid d\left(v,w\right)\text{ is odd}\right\}ofVV.

(a)Prove thatAAis dominating.
(b)Prove thatBBis dominating.
(c)Prove that there exists a dominating set ofGGthat has size≤|V|/2\leq\left|V\right|/2.
(d)Prove that the claim of part(c)holds even if we don’t assume thatGGis connected, as long as we assume that each vertex ofGGhas at least one neighbor. (In other words, prove Proposition2.118(a).)

2.13.2The number of dominating sets

Next, we state a rather surprising recent result about the number of dominating sets of a graph:

Theorem 2.122(Brouwer’s dominating set theorem).

LetGGbe a simple graph. Then, the number of dominating sets ofGGis odd.

Three proofs of this theorem are given in Brouwer’s note[Brouwe09].141414Other proofs can be found in the AoPS threadhttps://artofproblemsolving.com/community/c6h358772p1960068. (This thread is concerned with a superficially different contest problem, but the latter problem is quickly revealed to be Theorem2.122in a number-theoretical disguise.)Let me show the one I like the most. We first need a notation:

Definition 2.123.

LetG=(V,E)G=\left(V,E\right)be a simple graph. Adetached pairwill mean a pair(A,B)\left(A,B\right)of two disjoint subsetsAAandBBofVVsuch that there exists no edgeab∈Eab\in Ewitha∈Aa\in Aandb∈Bb\in B.

Example 2.124.

Consider the cycle graph

C6=({1,2,3,4,5,6},{12,23,34,45,56,61})=123456.C_{6}=\left(\left\{1,2,3,4,5,6\right\},\ \left\{12,\ 23,\ 34,\ 45,\ 56,\ 61\right\}\right)=\hbox to132.18pt{\vbox to116.94pt{\pgfpicture\makeatletter\hbox{\hskip 66.0914pt\lower-58.46764pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} 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\lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -72.38 -11.01 L -45.73 -57.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -26.66 -68.19 L 26.66 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 45.73 -57.18 L 72.38 -11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}.Then,({1,2},{4,5})\left(\left\{1,2\right\},\left\{4,5\right\}\right)is a detached pair, whereas({1,2},{3,4})\left(\left\{1,2\right\},\left\{3,4\right\}\right)is not (since2323is an edge). Of course, there are many other detached pairs; in particular, any pair of the form(∅,B)\left(\varnothing,B\right)or(A,∅)\left(A,\varnothing\right)is detached.

Let me stress that the word “pair” always means “ordered pair” unless I say otherwise. So, if(A,B)\left(A,B\right)is a detached pair, then(B,A)\left(B,A\right)is a different detached pair, unlessA=B=∅A=B=\varnothing.

Here is an attempt at a proof of Theorem2.122. It is a nice example of how to apply known results to new graphs to obtain new results. The only problem is, it shows a result that is a bit at odds with the claim of the theorem…

Proof 2.125(Proof of Theorem2.122, attempt 1.).

Write the graphGGas(V,E)\left(V,E\right).

Recall that𝒫⁡(V)\mathcal{P}\left(V\right)denotes the set of all subsets ofVV.

Construct a new graphHHwith the vertex set𝒫⁡(V)\mathcal{P}\left(V\right)as follows: Two subsetsAAandBBofVVare adjacent as vertices ofHHif and only if(A,B)\left(A,B\right)is a detached pair. (Note that if the original graphGGhasnnvertices, then this graphHHhas2n2^{n}vertices. It is huge!)

I claim that the vertices ofHHthat have odd degree are precisely the subsets ofVVthat are dominating. In other words:

Claim 1:LetAAbe a subset ofVV. Then, the vertexAAofHHhas odd degree if and only ifAAis a dominating set ofGG.

[Proof of Claim 1:We letN⁡(A)N\left(A\right)denote the set of all vertices ofGGthat have a neighbor inAA. (This may or may not be disjoint fromAA.)

The neighbors ofAA(as a vertex inHH) are precisely the subsetsBBofVVsuch that(A,B)\left(A,B\right)is a detached pair (by the definition ofHH). In other words, they are the subsetsBBofVVthat are disjoint fromAAand also have no neighbors inAA(by the definition of a “detached pair”). In other words, they are the subsetsBBofVVthat are disjoint fromAAand also disjoint fromN⁡(A)N\left(A\right). In other words, they are the subsets of the setV∖(A∪N⁡(A))V\setminus\left(A\cup N\left(A\right)\right). Hence, the number of such subsetsBBis2|V∖(A∪N⁡(A))|2^{\left|V\setminus\left(A\cup N\left(A\right)\right)\right|}.

The degree ofAA(as a vertex ofHH) is the number of neighbors ofAAinHH. Thus, this degree is2|V∖(A∪N⁡(A))|2^{\left|V\setminus\left(A\cup N\left(A\right)\right)\right|}(because we have just shown that the number of neighbors ofAAis2|V∖(A∪N⁡(A))|2^{\left|V\setminus\left(A\cup N\left(A\right)\right)\right|}). But2k2^{k}is odd if and only ifk=0k=0. Thus, we conclude that the degree ofAA(as a vertex ofHH) is odd if and only if|V∖(A∪N⁡(A))|=0\left|V\setminus\left(A\cup N\left(A\right)\right)\right|=0. The condition|V∖(A∪N⁡(A))|=0\left|V\setminus\left(A\cup N\left(A\right)\right)\right|=0can be rewritten as follows:

(|V∖(A∪N⁡(A))|=0)\displaystyle\ \left(\left|V\setminus\left(A\cup N\left(A\right)\right)\right|=0\right)⟺(V∖(A∪N⁡(A))=∅)\displaystyle\Longleftrightarrow\ \left(V\setminus\left(A\cup N\left(A\right)\right)=\varnothing\right)⟺(V⊆A∪N⁡(A))\displaystyle\Longleftrightarrow\ \left(V\subseteq A\cup N\left(A\right)\right)⟺(V∖A⊆N⁡(A))\displaystyle\Longleftrightarrow\ \left(V\setminus A\subseteq N\left(A\right)\right)⟺(each vertexv∈V∖Abelongs toN(A))\displaystyle\Longleftrightarrow\ \left(\text{each vertex }v\in V\setminus A\text{ belongs to }N\left(A\right)\right)⟺(each vertexv∈V∖Ahas a neighbor inA)\displaystyle\Longleftrightarrow\ \left(\text{each vertex }v\in V\setminus A\text{ has a neighbor in }A\right)⟺(Ais dominating)(by the definition of “dominating”).\displaystyle\Longleftrightarrow\ \left(A\text{ is dominating}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of \textquotedblleft dominating\textquotedblright}\right).Thus, what we have just shown is that the degree ofAA(as a vertex ofHH) is odd if and only ifAAis dominating. This proves Claim 1.]

Claim 1 shows that the vertices ofHHthat have odd degree are precisely the dominating sets ofGG. But the handshake lemma (Corollary2.20) tells us that any simple graph has an even number of vertices of odd degree. Applying this toHH, we conclude that there is an even number of dominating sets ofGG.

Huh? We want to show that there is anoddnumber of dominating sets ofGG, not an even number! Why did we just get the opposite result?

Puzzle: Find the mistake in our above reasoning! The answer will be revealed on the next page.

So what was the mistake in our reasoning?

The mistake is that our definition ofHHrequires the vertex∅\varnothingofHHto be adjacent to itself (since(∅,∅)\left(\varnothing,\varnothing\right)is a detached pair); but a vertex of a simple graph cannot be adjacent to itself. So we need to tweak the definition ofHHsomewhat:

Proof 2.126(Correction of the above proof of Theorem2.122.).

Define the graphHHas above, but do not try to have∅\varnothingadjacent to itself. (This is the only vertex that creates any trouble, because a detached pair(A,B)\left(A,B\right)cannot satisfyA=BA=Bunless bothAAandBBare∅\varnothing.)

We WLOG assume thatV≠∅V\neq\varnothing(otherwise, the claim is obvious). Thus, the empty set∅\varnothingis not dominating.

Our Claim 1 needs to be modified as follows:

Claim 1’:LetAAbe a subset ofVV. Then, the vertexAAofHHhas odd degree if and only ifAAis empty or a dominating set ofGG.

This can be proved in the same way as we “proved” Claim 1 above; we just need to treat theA=∅A=\varnothingcase separately now (but this case is easy:∅\varnothingis adjacent to all other vertices ofHH, and thus has degree2|V|−12^{\left|V\right|}-1, which is odd).

So we conclude (using the handshake lemma) that the number of empty or dominating sets is even. Subtracting11for the empty set, we conclude that the number of dominating sets is odd (since the empty set is not dominating). This proves Brouwer’s theorem (Theorem2.122).

There are other ways to prove Brouwer’s theorem as well. A particularly nice one was found by Irene Heinrich and Peter Tittmann in 2017; they gave an “explicit” formula for the number of dominating sets that shows that this number is odd ([HeiTit17, Theorem 8], restated using the language of detached pairs):

Theorem 2.127(Heinrich–Tittmann formula).

LetG=(V,E)G=\left(V,E\right)be a simple graph withnnvertices. Assume thatn>0n>0.

Then:

(a)The numbersα\alphaandβ\betaare even.
(b)The number of dominating sets ofGGis2n−1+α−β2^{n}-1+\alpha-\beta.

Part(a)of this theorem is obvious (recall that if(A,B)\left(A,B\right)is a detached pair, then so is(B,A)\left(B,A\right)). Part(b)is the interesting part. In[17s, §3.3–§3.4], I give a long but elementary proof.

More recently ([HeiTit18]), Heinrich and Tittmann have refined their formula to allow counting dominating sets of a given size. Their main result is the following formula (exercise 5 on homework set #2):

Exercise 2.128.

LetG=(V,E)G=\left(V,E\right)be a simple graph with at least one vertex. Letn=|V|n=\left|V\right|. Adetached pairmeans a pair(A,B)\left(A,B\right)of two disjoint subsetsAAandBBofVVsuch that there exists no edgeab∈Eab\in Ewitha∈Aa\in Aandb∈Bb\in B.

Prove the following generalization of the Heinrich–Tittmann formula:

∑Sis a dominatingset ofGx|S|=(1+x)n−1+∑(A,B)is a detached pair;A≠∅;B≠∅(−1)|A|x|B|.\sum\limits_{\begin{subarray}{c}S\text{ is a dominating}\\ \text{set of }G\end{subarray}}x^{\left|S\right|}=\left(1+x\right)^{n}-1+\sum\limits_{\begin{subarray}{c}\left(A,B\right)\text{ is a detached pair;}\\ A\neq\varnothing;\ B\neq\varnothing\end{subarray}}\left(-1\right)^{\left|A\right|}x^{\left|B\right|}.(Here, both sides are polynomials in a single indeterminatexxwith coefficients inℤ\mathbb{Z}.)

[Hint:This is a generalization of the Heinrich–Tittmann formula for the number of dominating sets. (The latter formula can be obtained fairly easily by substitutingx=1x=1into the above and subsequently cancelling the addends with|A|≢|B|mod⁡2\left|A\right|\not\equiv\left|B\right|\operatorname{mod}2against each other.) You are free to copy arguments from[17s]and change whatever needs to be changed. (Some lemmas can even be used without any changes – they can then be cited without proof.)]

The following exercise gives a generalization of Theorem2.122(to recover Theorem2.122from it, setk=1k=1):

Exercise 2.129.

Letkkbe a positive integer. LetG=(V,E)G=\left(V,E\right)be a simple graph. A subsetUUofVVwill be calledkk-path-dominatingif for everyv∈Vv\in V, there exists a path of length≤k\leq kfromvvto some element ofUU.

Prove that the number of allkk-path-dominating subsets ofVVis odd.

[Hint:This is not as substantial a generalization as it may look. The shortest proof is very short.]

[Solution:This is Exercise 6 on homework set #1 from my Spring 2017 course; seethe course pagefor solutions.]

2.14Hamiltonian paths and cycles

2.14.1Basics

Now to something different. Here is a quick question: Given a simple graphGG, when is there a closedwalkthat contains each vertex ofGG?

The answer is easy: WhenGGis connected. Indeed, if a simple graphGGis connected, then we can label its vertices byv1,v2,…,vnv_{1},v_{2},\ldots,v_{n}arbitrarily, and we then get a closed walk by composing a walk fromv1v_{1}tov2v_{2}with a walk fromv2v_{2}tov3v_{3}with a walk fromv3v_{3}tov4v_{4}and so on, ending with a walk fromvnv_{n}tov1v_{1}. This closed walk will certainly contain each vertex. Conversely, such a walk cannot exist ifGGis not connected.

The question becomes a lot more interesting if we replace “closed walk” by “path” or “cycle”. The resulting objects have a name:

Definition 2.130.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

(a)AHamiltonian pathinGGmeans a walk ofGGthat contains each vertex ofGGexactly once. Obviously, it is a path.
(b)AHamiltonian cycleinGGmeans a cycle(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)ofGGsuch that each vertex ofGGappears exactly once amongv0,v1,…,vk−1v_{0},v_{1},\ldots,v_{k-1}.

Some graphs have Hamiltonian paths; some don’t. Having a Hamiltonian cycle is even stronger than having a Hamiltonian path, because if(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)is a Hamiltonian cycle ofGG, then(v0,v1,…,vk−1)\left(v_{0},v_{1},\ldots,v_{k-1}\right)is a Hamiltonian path ofGG.

Convention 2.14.1.

In the following, we will abbreviate:

•“Hamiltonian path” as “hamp”;
•“Hamiltonian cycle” as “hamc”.

Example 2.131.

Which of the following eight graphs have hamps? Which have hamcs?

112233445566A=A=112233445566B=B=11223300C=C=22113344D=D=11223344556600E=E=1122334455667788F=F=11223344551′1^{\prime}2′2^{\prime}3′3^{\prime}4′4^{\prime}5′5^{\prime}G=G=11223344551′1^{\prime}2′2^{\prime}3′3^{\prime}4′4^{\prime}5′5^{\prime}H=H= Answers:

•The graphAAhas a hamc(1,2,3,4,5,6,1)\left(1,2,3,4,5,6,1\right), and thus a hamp(1,2,3,4,5,6)\left(1,2,3,4,5,6\right). (Recall that a graph that has a hamc always has a hamp, since we can simply remove the last vertex from a hamc to obtain a hamp.)
•The graphBBhas a hamp(2,3,1,4,5,6)\left(2,3,1,4,5,6\right), but no hamc. The easiest way to see thatBBhas no hamc is the following: The edge1414is a cut-edge (i.e., removing it renders the graph disconnected), thus a bridge (i.e., an edge that appears in no cycle); therefore, any cycle must stay entirely “on one side” of this edge.
•The graphCChas a hamp(0,1,2,3)\left(0,1,2,3\right), but no hamc. The argument for the non-existence of a hamc is the same as forBB: The edge0101is a bridge.
•The graphDDhas neither a hamp nor a hamc, because it is not connected. Only a connected graph can have a hamp.
•The graphEEhas a hamp(0,3,2,1,6,5,4)\left(0,3,2,1,6,5,4\right), but no hamc (checking this requires some work, though).
•The graphFFhas a hamc(1,2,3,4,8,7,6,5,1)\left(1,2,3,4,8,7,6,5,1\right), thus also a hamp.
•The graphGGhas a hamc(1,2,3,4,5,5′,4′,3′,2′,1′,1)\left(1,2,3,4,5,5^{\prime},4^{\prime},3^{\prime},2^{\prime},1^{\prime},1\right), thus also a hamp.
•The graphHH(which, by the way, is isomorphic to the Petersen graph from Subsection2.6.3) has a hamp(1,3,5,2,4,4′,3′,2′,1′,5′)\left(1,3,5,2,4,4^{\prime},3^{\prime},2^{\prime},1^{\prime},5^{\prime}\right), but no hamc (but this is not obvious! seethe Wikipedia articlefor an argument).

In general, finding a hamp or a hamc, or proving that none exists, is a hard problem. It can always be solved by brute force (i.e., by trying all lists of distinct vertices and checking if there is a hamp among them, and likewise for hamcs), but this quickly becomes forbiddingly laborious as the size of the graph increases. Some faster algorithms exist (in particular, there is one of running timeO⁡(n22n)O\left(n^{2}2^{n}\right), wherennis the number of vertices), but no polynomial-time algorithm is known. The problem (both in its hamp version and in its hamc version) is known to be NP-hard (in the language of complexity theory). In practice, hamps and hamcs can often be found with some wit and perseverance; proofs of their non-existence can often be obtained with some logic and case analysis (see the above example for some sample arguments). Seethe Wikipedia page for “Hamiltonian path problem” for more information.

The problem of finding hamps is related to the so-called “traveling salesman problem” (TSP), which asks for a hamp with “minimum weight” in a weighted graph (each edge has a number assigned to it, which is called its “weight”, and the weight of a hamp is the sum of the weights of the edges it uses). There is a lot of computer-science literature about this problem.

2.14.2Sufficient criteria: Ore and Dirac

We shall now show some necessary criteria and some sufficient criteria (but no necessary-and-sufficient criteria) for the existence of hamps and hamcs. Here is the most famous sufficient criterion, known asOre’s theorem:

Theorem 2.132(Ore’s theorem).

LetG=(V,E)G=\left(V,E\right)be a simple graph withnnvertices, wheren≥3n\geq 3.

Assume thatdeg⁡x+deg⁡y≥n\deg x+\deg y\geq nfor any two non-adjacent distinct verticesxxandyy.

Then,GGhas a hamc.

There are various proofs of this theorem scattered around; see[Harju14, Theorem 3.6]or[Guicha16, Theorem 5.3.2]. We shall give a proof that is essentially Ore’s original proof[Ore60](somewhat restated as in the “Algorithm” section onthe Wikipedia page for “Ore’s theorem”):

Proof 2.133(Proof of Theorem2.132.).

Alisting(ofVV) shall mean a list of elements ofVVthat contains each element exactly once. It must clearly be annn-tuple.

Thehamnessof a listing(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)will mean the number of alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}such thatvivi+1∈Ev_{i}v_{i+1}\in E. Here, we setvn+1=v1v_{n+1}=v_{1}. (Visually, it is best to represent a listing(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)by drawing the verticesv1,v2,…,vnv_{1},v_{2},\ldots,v_{n}on a circle in this order. Its hamness then counts how often two successive vertices on the circle are adjacent in the graphGG.) Note that the hamness of a listing(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)does not change if we cyclically rotate the listing (i.e., transform it into(v2,v3,…,vn,v1)\left(v_{2},v_{3},\ldots,v_{n},v_{1}\right)).

Clearly, if we can find a listing(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)of hamness≥n\geq n, then all ofv1v2,v2v3,…,vnv1v_{1}v_{2},\ v_{2}v_{3},\ \ldots,\ v_{n}v_{1}are edges ofGG, and thus(v1,v2,…,vn,v1)\left(v_{1},v_{2},\ldots,v_{n},v_{1}\right)is a hamc ofGG. Thus, we need to find a listing of hamness≥n\geq n.

To do so, I will show that if you have a listing of hamness<n<n, then you can slightly modify it to get a listing of larger hamness. In other words, I will show the following:

Claim 1:Let(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)be a listing of hamnessk<nk<n. Then, there exists a listing of hamness larger thankk.

[Proof of Claim 1:Since the listing(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)has hamnessk<nk<n, there exists somei∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}such thatvivi+1∉Ev_{i}v_{i+1}\notin E. Pick such anii. Thus, the verticesviv_{i}andvi+1v_{i+1}ofGGare non-adjacent (and distinct, of course). The “deg⁡x+deg⁡y≥n\deg x+\deg y\geq n” assumption of the theorem thus yieldsdeg⁡(vi)+deg⁡(vi+1)≥n\deg\left(v_{i}\right)+\deg\left(v_{i+1}\right)\geq n.

However,

deg⁡(vi)\displaystyle\deg\left(v_{i}\right)=|{w∈V∣viw∈E}|\displaystyle=\left|\left\{w\in V\ \mid\ v_{i}w\in E\right\}\right|=|{j∈{1,2,…,n}∣vivj∈E}|\displaystyle=\left|\left\{j\in\left\{1,2,\ldots,n\right\}\ \mid\ v_{i}v_{j}\in E\right\}\right|=|{j∈{1,2,…,n}∖{i}∣vivj∈E}|\displaystyle=\left|\left\{j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}\ \mid\ v_{i}v_{j}\in E\right\}\right|(becausej=ij=icould not satisfyvivj∈Ev_{i}v_{j}\in Eanyway) and

deg⁡(vi+1)\displaystyle\deg\left(v_{i+1}\right)=|{w∈V∣vi+1w∈E}|\displaystyle=\left|\left\{w\in V\ \mid\ v_{i+1}w\in E\right\}\right|=|{j∈{1,2,…,n}∣vi+1vj+1∈E}|\displaystyle=\left|\left\{j\in\left\{1,2,\ldots,n\right\}\ \mid\ v_{i+1}v_{j+1}\in E\right\}\right|(since(v2,v3,…,vn+1)is a listing ofV(becausevn+1=v1))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since }\left(v_{2},v_{3},\ldots,v_{n+1}\right)\text{ is a listing of }V\\ \text{(because }v_{n+1}=v_{1}\text{)}\end{array}\right)=|{j∈{1,2,…,n}∖{i}∣vi+1vj+1∈E}|\displaystyle=\left|\left\{j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}\ \mid\ v_{i+1}v_{j+1}\in E\right\}\right|(becausej=ij=icould not satisfyvi+1vj+1∈Ev_{i+1}v_{j+1}\in Eanyway). In light of these two equalities, we can rewrite the inequalitydeg⁡(vi)+deg⁡(vi+1)≥n\deg\left(v_{i}\right)+\deg\left(v_{i+1}\right)\geq nas

|{j∈{1,2,…,n}∖{i}∣vivj∈E}|\displaystyle\left|\left\{j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}\ \mid\ v_{i}v_{j}\in E\right\}\right|+|{j∈{1,2,…,n}∖{i}∣vi+1vj+1∈E}|≥n.\displaystyle\ \ \ \ \ \ \ \ \ \ +\left|\left\{j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}\ \mid\ v_{i+1}v_{j+1}\in E\right\}\right|\geq n.Thus, the two subsets{j∈{1,2,…,n}∖{i}∣vivj∈E}\left\{j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}\ \mid\ v_{i}v_{j}\in E\right\}and {j∈{1,2,…,n}∖{i}∣vi+1vj+1∈E}\left\{j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}\ \mid\ v_{i+1}v_{j+1}\in E\right\}of the(n−1)\left(n-1\right)-element set{1,2,…,n}∖{i}\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}have total size≥n\geq n(that is, the sum of their sizes is≥n\geq n). Hence, these two subsets must overlap (i.e., have an element in common). In other words, there exists aj∈{1,2,…,n}∖{i}j\in\left\{1,2,\ldots,n\right\}\setminus\left\{i\right\}that satisfies bothvivj∈Ev_{i}v_{j}\in Eandvi+1vj+1∈Ev_{i+1}v_{j+1}\in E. Pick such ajj.

Now, consider a new listing obtained from the old listing(v1,v2,…,vn)\left(v_{1},v_{2},\ldots,v_{n}\right)as follows:

•First, cyclically rotate the old listing so that it begins withvi+1v_{i+1}. Thus, you get the listing(vi+1,vi+2,…,vn,v1,v2,…,vi)\left(v_{i+1},v_{i+2},\ldots,v_{n},v_{1},v_{2},\ldots,v_{i}\right).
•Then, reverse the part of the listing starting atvi+1v_{i+1}and ending atvjv_{j}. Thus, you get the new listing (vj,vj−1,…,vi+1⏟This is the reversed part;it may or may not “wrap around”(i.e., contain…,v1,vn,…somewhere).,vj+1,vj+2,…,vi⏟This is the part thatwas not reversed.).\left(\underbrace{v_{j},v_{j-1},\ldots,v_{i+1}}_{\begin{subarray}{c}\text{This is the reversed part;}\\ \text{it may or may not \textquotedblleft wrap around\textquotedblright}\\ \text{(i.e., contain }\ldots,v_{1},v_{n},\ldots\text{ somewhere).}\end{subarray}},\underbrace{v_{j+1},v_{j+2},\ldots,v_{i}}_{\begin{subarray}{c}\text{This is the part that}\\ \text{was not reversed.}\end{subarray}}\right).This is the new listing we want.

I claim that this new listing has hamness larger thankk. Indeed, rotating the old listing clearly did not change its hamness. But reversing the part fromvi+1v_{i+1}tovjv_{j}clearly did: After the reversal, the edgesvivi+1v_{i}v_{i+1}andvjvj+1v_{j}v_{j+1}no longer count towards the hamness (if they were edges to begin with), but the edgesvivjv_{i}v_{j}andvi+1vj+1v_{i+1}v_{j+1}started counting towards the hamness. This is a good bargain, because it means that the hamness gained+2+2from the newly-counted edgesvivjv_{i}v_{j}andvi+1vj+1v_{i+1}v_{j+1}(which, as we know, both exist), while only losing00or11(since the edgevivi+1v_{i}v_{i+1}did not exist, whereas the edgevjvj+1v_{j}v_{j+1}may or may not have been lost). Thus, the hamness of the new listing is larger than the hamness of the old listing either by11or22. In other words, it is larger thanmmby at least11or22. This proves Claim 1.]

Now, we can start withanylisting ofVVand keep modifying it using Claim 1, increasing its hamness each time, until its hamness becomes≥n\geq n. But once its hamness is≥n\geq n, we have found a hamc (as explained above). Theorem2.132is thus proven.

As a particular case, we get a less general but more memorable criterion calledDirac’s criterion:

Corollary 2.134(Dirac’s theorem).

LetG=(V,E)G=\left(V,E\right)be a simple graph withnnvertices, wheren≥3n\geq 3.

Assume thatdeg⁡x≥n2\deg x\geq\dfrac{n}{2}for each vertexx∈Vx\in V.

Then,GGhas a hamc.

Proof 2.135.

Follows from Ore’s theorem, since any two verticesxxandyyofGGsatisfydeg⁡x⏟≥n2+deg⁡y⏟≥n2≥n2+n2=n\underbrace{\deg x}_{\geq\dfrac{n}{2}}+\underbrace{\deg y}_{\geq\dfrac{n}{2}}\geq\dfrac{n}{2}+\dfrac{n}{2}=n.

Exercise 2.136.

(a)LetG=(V,E)G=\left(V,E\right)be a simple graph, and letuuandvvbe two distinct vertices ofGGthat are not adjacent. Letn=|V|n=\left|V\right|. Assume thatdeg⁡u+deg⁡v≥n\deg u+\deg v\geq n. LetG′=(V,E∪{uv})G^{\prime}=\left(V,E\cup\left\{uv\right\}\right)be the simple graph obtained fromGGby adding a new edgeuvuv. Assume thatG′G^{\prime}has a hamc. Prove thatGGhas a hamc.
(b)Does this remain true if we replace “hamc” by “hamp”?

2.14.3A necessary criterion

So much for sufficient criteria. What about necessary criteria?

Proposition 2.137.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

For each subsetSSofVV, we letG∖SG\setminus Sbe the induced subgraph ofGGon the setV∖SV\setminus S. (In other words, this is the graph obtained fromGGby removing all vertices inSSand removing all edges that have at least one endpoint inSS.)

(For example, if11223344556600G=G=andS={3,6}S=\left\{3,6\right\}, then1122445500G∖S=G\setminus S=.)

Also, we letconn(H)\operatorname*{conn}\left(H\right)denote the number of connected components of a simple graphHH.

(a)IfGGhas a hamc, then every nonemptyS⊆VS\subseteq Vsatisfiesconn(G∖S)≤|S|\operatorname*{conn}\left(G\setminus S\right)\leq\left|S\right|.
(b)IfGGhas a hamp, then everyS⊆VS\subseteq Vsatisfiesconn(G∖S)≤|S|+1\operatorname*{conn}\left(G\setminus S\right)\leq\left|S\right|+1.

For example, part(a)of this proposition shows that the graphEEfrom Example2.131has no hamc, because if we takeSSto be{3,6}\left\{3,6\right\}, thenconn(G∖S)=3\operatorname*{conn}\left(G\setminus S\right)=3whereas|S|=2\left|S\right|=2. Thus, the proposition can be used to rule out the existence of hamps and hamcs in some cases.

Proof 2.138(Proof of Proposition2.137.).

†\dagger†\dagger†\dagger†\dagger→remove the verticesmarked with daggers\overset{\begin{subarray}{c}\text{remove the vertices}\\ \text{marked with daggers}\end{subarray}}{\to}Of course, our graphGGitself may not be a cycle, but if it has a hamc, then the removal of the vertices inSSwill split the hamc into at most|S|\left|S\right|paths (according to the preceding sentence), and thus the graphG∖SG\setminus Swill have≤|S|\leq\left|S\right|many components (just using the surviving edges of the hamc alone). Taking into account all the other edges ofGGcan only decrease the number of components.

(b)This is analogous to part(a).

This proposition often (but not always) gives a quick way of convincing yourself that a graph has no hamc or hamp. Alas, its converse is false. Case in point: The Petersen graph (defined in Subsection2.6.3) has no hamc, but it does satisfy the “every nonemptyS⊆VS\subseteq Vsatisfiesconn(G∖S)≤|S|\operatorname*{conn}\left(G\setminus S\right)\leq\left|S\right|” condition of Proposition2.137(a).

2.14.4Hypercubes

Now, let us move on to a concrete example of a graph that has a hamc.

Definition 2.139.

Letn∈ℕn\in\mathbb{N}. Thenn-hypercubeQnQ_{n}(more precisely, thenn-th hypercube graph) is the simple graph with vertex set

{0,1}n={(a1,a2,…,an)∣eachaibelongs to{0,1}}\left\{0,1\right\}^{n}=\left\{\left(a_{1},a_{2},\ldots,a_{n}\right)\ \mid\ \text{each }a_{i}\text{ belongs to }\left\{0,1\right\}\right\}and edge set defined as follows: A vertex(a1,a2,…,an)∈{0,1}n\left(a_{1},a_{2},\ldots,a_{n}\right)\in\left\{0,1\right\}^{n}is adjacent to a vertex(b1,b2,…,bn)∈{0,1}n\left(b_{1},b_{2},\ldots,b_{n}\right)\in\left\{0,1\right\}^{n}if and only if there existsexactlyonei∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}such thatai≠bia_{i}\neq b_{i}. (For example, inQ4Q_{4}, the vertex(0,1,1,0)\left(0,1,1,0\right)is adjacent to(0,1,0,0)\left(0,1,0,0\right).)

The elements of{0,1}n\left\{0,1\right\}^{n}are often calledbitstrings(orbinary words), and their entries are called theirbits(orletters). So two bitstrings are adjacent inQnQ_{n}if and only if they differ in exactly one bit.

We often write a bitstring(a1,a2,…,an)\left(a_{1},a_{2},\ldots,a_{n}\right)asa1a2⋯ana_{1}a_{2}\cdots a_{n}. (For example, we write(0,1,1,0)\left(0,1,1,0\right)as01100110.)

Example 2.140.

Here is how thenn-hypercubesQnQ_{n}look like forn=1,2,3n=1,2,3:

This should explain the name “hypercube”. The00-hypercubeQ0Q_{0}is a graph with just one vertex (namely, the empty bitstring()\left({}\right)).

Theorem 2.141(Gray).

Letn≥2n\geq 2. Then, the graphQnQ_{n}has a hamc.

Such hamcs are known asGray codes. They are circular lists of bitstrings of lengthnnsuch that two consecutive bitstrings in the list always differ in exactly one bit. Seethe Wikipedia article on “Gray codes”for applications.

Example 2.142.

Here is one hamp ofQ3Q_{3}(depicted by drawing its edges while omitting all other edges of the graph):

000000001001010010011011100100101101110110111111

Proof 2.143(Proof of Theorem2.141.).

We will show something stronger:

Claim 1:For eachn≥1n\geq 1, thenn-hypercubeQnQ_{n}has a hamp from00⋯000\cdots 0to100⋯0100\cdots 0. (Keep in mind that00⋯000\cdots 0and100⋯0100\cdots 0are bitstrings, not numbers: 00⋯0=(0,0,…,0⏟nzeroes);100⋯0=(1,0,0,…,0⏟n−1zeroes).00\cdots 0=\left(\underbrace{0,0,\ldots,0}_{n\text{ zeroes}}\right);\ \ \ \ \ \ \ \ \ \ 100\cdots 0=\left(1,\underbrace{0,0,\ldots,0}_{n-1\text{ zeroes}}\right).)

[Proof of Claim 1:We induct onnn.

Induction base:A look atQ1Q_{1}reveals a hamp from00to11.

Induction step:FixN≥2N\geq 2. We assume that Claim 1 holds forn=N−1n=N-1. In other words,QN−1Q_{N-1}has a hamp from00⋯0⏟N−1zeroes\underbrace{00\cdots 0}_{N-1\text{ zeroes}}to100⋯0⏟N−2zeroes1\underbrace{00\cdots 0}_{N-2\text{ zeroes}}. Let𝐩\mathbf{p}be such a hamp.

By attaching a00to the front of each bitstring (= vertex) in𝐩\mathbf{p}, we obtain a path𝐪\mathbf{q}from00⋯0⏟Nzeroes\underbrace{00\cdots 0}_{N\text{ zeroes}}to0100⋯0⏟N−2zeroes01\underbrace{00\cdots 0}_{N-2\text{ zeroes}}inQNQ_{N}.

By attaching a11to the front of each bitstring (= vertex) in𝐩\mathbf{p}, we obtain a path𝐫\mathbf{r}from100⋯0⏟N−1zeroes1\underbrace{00\cdots 0}_{N-1\text{ zeroes}}to1100⋯0⏟N−2zeroes11\underbrace{00\cdots 0}_{N-2\text{ zeroes}}inQNQ_{N}.

Now, we assemble a hamp from00⋯0⏟Nzeroes\underbrace{00\cdots 0}_{N\text{ zeroes}}to100⋯0⏟N−1zeroes1\underbrace{00\cdots 0}_{N-1\text{ zeroes}}inQNQ_{N}as follows:

•Start at00⋯0⏟Nzeroes\underbrace{00\cdots 0}_{N\text{ zeroes}}, and follow the path𝐪\mathbf{q}to its end (i.e., to0100⋯0⏟N−2zeroes01\underbrace{00\cdots 0}_{N-2\text{ zeroes}}).
•Then, move to the adjacent vertex1100⋯0⏟N−2zeroes11\underbrace{00\cdots 0}_{N-2\text{ zeroes}}.
•Then, follow the path𝐫\mathbf{r}backwards, ending up at100⋯0⏟N−1zeroes1\underbrace{00\cdots 0}_{N-1\text{ zeroes}}.

This shows that Claim 1 holds forn=Nn=N, too.]

Claim 1 tells us that thenn-hypercubeQnQ_{n}has a hamp from00⋯000\cdots 0to100⋯0100\cdots 0. Since its starting point00⋯000\cdots 0and its ending point100⋯0100\cdots 0are adjacent, we can turn this hamp into a hamc by appending the starting point00⋯000\cdots 0again at the end. This proves Theorem2.141.

2.14.5Cartesian products

Theorem2.141can in fact be generalized. To state the generalization, we define theCartesian productof two graphs:

Definition 2.144.

LetG=(V,E)G=\left(V,E\right)andH=(W,F)H=\left(W,F\right)be two simple graphs. TheCartesian productG×HG\times Hof these two graphs is defined to be the simple graph(V×W,E′∪F′)\left(V\times W,\ E^{\prime}\cup F^{\prime}\right), where

E′\displaystyle E^{\prime}:={(v1,w)(v2,w)∣v1v2∈Eandw∈W}and\displaystyle:=\left\{\left(v_{1},w\right)\left(v_{2},w\right)\ \mid\ v_{1}v_{2}\in E\text{ and }w\in W\right\}\ \ \ \ \ \ \ \ \ \ \text{and}F′\displaystyle F^{\prime}:={(v,w1)(v,w2)∣w1w2∈Fandv∈V}.\displaystyle:=\left\{\left(v,w_{1}\right)\left(v,w_{2}\right)\ \mid\ w_{1}w_{2}\in F\text{ and }v\in V\right\}.In other words, it is the graph whose vertices are pairs(v,w)∈V×W\left(v,w\right)\in V\times Wconsisting of a vertex ofGGand a vertex ofHH, and whose edges are of the forms

(v1,w)(v2,w)wherev1v2∈Eandw∈W\left(v_{1},w\right)\left(v_{2},w\right)\ \ \ \ \ \ \ \ \ \ \text{where }v_{1}v_{2}\in E\text{ and }w\in Wand

(v,w1)(v,w2)wherew1w2∈Fandv∈V.\left(v,w_{1}\right)\left(v,w_{2}\right)\ \ \ \ \ \ \ \ \ \ \text{where }w_{1}w_{2}\in F\text{ and }v\in V.

For example, the Cartesian productG×P2G\times P_{2}of a simple graphGGwith the22-path graphP2P_{2}can be constructed by overlaying two copies ofGGand additionally joining each vertex of the first copy to the corresponding vertex of the second copy by an edge. (The vertices of the first copy are the(v,1)\left(v,1\right), whereas the vertices of the second copy are the(v,2)\left(v,2\right).) For a specific example, here is the55-cycle graphC5C_{5}and the Cartesian productC5×P2C_{5}\times P_{2}:

C5C_{5}C5×P2C_{5}\times P_{2}1122334455(1,1)\left(1,1\right)(2,1)\left(2,1\right)(3,1)\left(3,1\right)(4,1)\left(4,1\right)(5,1)\left(5,1\right)(1,2)\left(1,2\right)(2,2)\left(2,2\right)(3,2)\left(3,2\right)(4,2)\left(4,2\right)(5,2)\left(5,2\right)As another instance of the above description ofG×P2G\times P_{2}, it is easy to see the following:

Proposition 2.145.

We haveQn≅Qn−1×P2Q_{n}\cong Q_{n-1}\times P_{2}for eachn≥1n\geq 1. (See Definition2.139for the definitions ofQnQ_{n}andQn−1Q_{n-1}.)

Proof 2.146.

This is Exercise 1(a)on homework set #2 from my Spring 2017 course; seethe course pagefor solutions.

Now, we claim the following:

Theorem 2.147.

LetGGandHHbe two simple graphs. Assume that each of the two graphsGGandHHhas a hamp. Then:

(a)The Cartesian productG×HG\times Hhas a hamp.
(b)Now assume furthermore that at least one of the two numbers|V(G)|\left|\operatorname*{V}\left(G\right)\right|and|V(H)|\left|\operatorname*{V}\left(H\right)\right|is even, and that both numbers|V(G)|\left|\operatorname*{V}\left(G\right)\right|and|V(H)|\left|\operatorname*{V}\left(H\right)\right|are larger than11. Then, the Cartesian productG×HG\times Hhas a hamc.

Proof 2.148.

Now, Theorem2.141can be reproved (again by inducting onnn) using Theorem2.147(b)and Proposition2.145, sinceP2P_{2}has a hamp and since|V(P2)|=2\left|\operatorname*{V}\left(P_{2}\right)\right|=2is even. (Convince yourself that this works!)

2.14.6Subset graphs

Thenn-hypercubeQnQ_{n}can be reinterpreted in terms of subsets of{1,2,…,n}\left\{1,2,\ldots,n\right\}. Namely: Letn∈ℕn\in\mathbb{N}. LetGnG_{n}be the simple graph whose vertex set is the powerset𝒫⁡({1,2,…,n})\mathcal{P}\left(\left\{1,2,\ldots,n\right\}\right)of{1,2,…,n}\left\{1,2,\ldots,n\right\}(that is, the vertices are all2n2^{n}subsets of{1,2,…,n}\left\{1,2,\ldots,n\right\}), and whose edges are determined as follows: Two verticesSSandTTare adjacent if and only if one of the two setsSSandTTis obtained from the other by inserting an extra element (i.e., we have eitherS=T∪{s}S=T\cup\left\{s\right\}for somes∉Ts\notin T, orT=S∪{t}T=S\cup\left\{t\right\}for somet∉St\notin S). Then,Gn≅QnG_{n}\cong Q_{n}, since the map

{0,1}n\displaystyle\left\{0,1\right\}^{n}→𝒫⁡({1,2,…,n}),\displaystyle\rightarrow\mathcal{P}\left(\left\{1,2,\ldots,n\right\}\right),(a1,a2,…,an)\displaystyle\left(a_{1},a_{2},\ldots,a_{n}\right)↦{i∈{1,2,…,n}∣ai=1}\displaystyle\mapsto\left\{i\in\left\{1,2,\ldots,n\right\}\ \mid\ a_{i}=1\right\}is a graph isomorphism fromQnQ_{n}toGnG_{n}.

Thus, Theorem2.141shows that for eachn≥2n\geq 2, the graphGnG_{n}has a hamc. In other words, for eachn≥2n\geq 2, we can list all subsets of{1,2,…,n}\left\{1,2,\ldots,n\right\}in a circular list in such a way that each subset on this list is obtained from the previous one by inserting or removing a single element. For example, forn=3n=3, here is such a list:

∅,{1},{1,2},{2},{2,3},{1,2,3},{1,3},{3}.\varnothing,\ \ \left\{1\right\},\ \ \left\{1,2\right\},\ \ \left\{2\right\},\ \ \left\{2,3\right\},\ \ \left\{1,2,3\right\},\ \ \left\{1,3\right\},\ \ \left\{3\right\}. A long-standing question only resolved a few years ago asked whether the same can be done with the subsets of{1,2,…,n}\left\{1,2,\ldots,n\right\}having sizen±12\dfrac{n\pm 1}{2}whennnis odd. For example, forn=3n=3, we can do it as follows:

{1},{1,2},{2},{2,3},{3},{1,3}.\left\{1\right\},\ \ \left\{1,2\right\},\ \ \left\{2\right\},\ \ \left\{2,3\right\},\ \ \left\{3\right\},\ \ \left\{1,3\right\}.In other words, ifn≥3n\geq 3is odd, and ifGn′G_{n}^{\prime}is the induced subgraph ofGnG_{n}on the set of all subsetsJJof{1,2,…,n}\left\{1,2,\ldots,n\right\}that satisfy|J|∈{n−12,n+12}\left|J\right|\in\left\{\dfrac{n-1}{2},\dfrac{n+1}{2}\right\}, then doesGn′G_{n}^{\prime}have a hamc?

SinceGn≅QnG_{n}\cong Q_{n}, we can restate this question equivalently as follows: Ifn≥3n\geq 3is odd, and ifQn′Q_{n}^{\prime}is the induced subgraph ofQnQ_{n}on the set

{a1a2⋯an∈{0,1}n∣∑i=1nai∈{n−12,n+12}},\left\{a_{1}a_{2}\cdots a_{n}\in\left\{0,1\right\}^{n}\ \mid\ \sum\limits_{i=1}^{n}a_{i}\in\left\{\dfrac{n-1}{2},\dfrac{n+1}{2}\right\}\right\},then doesQn′Q_{n}^{\prime}have a hamc?

In 2014, Torsten Mütze proved that the answer is “yes”. See[Mutze14]for his truly nontrivial proof,[MeMuNa22]for an analogue for Kneser graphs, and[Mutze22]for a recent survey of similar questions. (Cf. alsochange ringing.)

The following exercise provides another generalization of Theorem2.141:

Exercise 2.149.

Letnnandkkbe two integers such thatn>k>0n>k>0. Define the simple graphQn,kQ_{n,k}as follows: Its vertices are the bitstrings(a1,a2,…,an)∈{0,1}n\left(a_{1},a_{2},\ldots,a_{n}\right)\in\left\{0,1\right\}^{n}; two such bitstrings are adjacent if and only if they differ in exactlykkbits (in other words: two vertices(a1,a2,…,an)\left(a_{1},a_{2},\ldots,a_{n}\right)and(b1,b2,…,bn)\left(b_{1},b_{2},\ldots,b_{n}\right)are adjacent if and only if the number ofi∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}satisfyingai≠bia_{i}\neq b_{i}equalskk). (Thus,Qn,1Q_{n,1}is thenn-hypercube graphQnQ_{n}.)

(a)DoesQn,kQ_{n,k}have a hamc whenkkis even? (Recall that “hamc” is short for “Hamiltonian cycle”.)
(b)DoesQn,kQ_{n,k}have a hamc whenkkis odd?

[Hint:One way to approach part(b)is by identifying the set{0,1}\left\{0,1\right\}with the field𝔽2\mathbb{F}_{2}with two elements. The bitstrings(a1,a2,…,an)∈{0,1}n\left(a_{1},a_{2},\ldots,a_{n}\right)\in\left\{0,1\right\}^{n}thus become the size-nnrow vectors in the𝔽2\mathbb{F}_{2}-vector space𝔽2n\mathbb{F}_{2}^{n}. Lete1,e2,…,ene_{1},e_{2},\ldots,e_{n}be the standard basis vectors of𝔽2n\mathbb{F}_{2}^{n}(so thateie_{i}has a11in itsii-th position and zeroes everywhere else). Then, two vectors are adjacent in thenn-hypercube graphQnQ_{n}(resp. in the graphQn,kQ_{n,k}) if and only if their difference is one of the standard basis vectors (resp., a sum ofkkdistinct standard basis vectors). Try to use this to find a graph isomorphism fromQnQ_{n}to a subgraph ofQn,kQ_{n,k}.]

The next exercise extends the idea of our proof of Theorem2.141:

Exercise 2.150.

Letn≥1n\geq 1. LetQnQ_{n}be thenn-hypercube graph, as in Definition2.139. Recall that “hamp” is short for “Hamiltonian path”.

At what vertices can a hamp ofQnQ_{n}end if it starts at the vertex00⋯000\cdots 0? (Find all possibilities, and prove that they are possible and all other vertices are impossible.)

3Multigraphs

3.1Definitions

So far, we have been working with simple graphs. We shall now introduce several other kinds of graphs, starting with themultigraphs.

Definition 3.1.

LetVVbe a set. Then,𝒫1,2(V)\mathcal{P}_{1,2}\left(V\right)shall mean the set of all11-element or22-element subsets ofVV. In other words,

𝒫1,2(V):=\displaystyle\mathcal{P}_{1,2}\left(V\right):={S⊆V∣|S|∈{1,2}}\displaystyle\left\{S\subseteq V\ \mid\ \left|S\right|\in\left\{1,2\right\}\right\}=\displaystyle={{u,v}∣u,v∈Vnot necessarily distinct}.\displaystyle\left\{\left\{u,v\right\}\ \mid\ u,v\in V\text{ not necessarily distinct}\right\}.

For instance,

𝒫1,2({1,2,3})={{1},{2},{3},{1,2},{1,3},{2,3}}.\mathcal{P}_{1,2}\left(\left\{1,2,3\right\}\right)=\left\{\left\{1\right\},\ \left\{2\right\},\ \left\{3\right\},\ \left\{1,2\right\},\ \left\{1,3\right\},\ \left\{2,3\right\}\right\}. We can now define multigraphs:

Definition 3.2.

Amultigraphis a triple(V,E,φ)\left(V,E,\varphi\right), whereVVandEEare two finite sets, andφ:E→𝒫1,2(V)\varphi:E\rightarrow\mathcal{P}_{1,2}\left(V\right)is a map.

Example 3.3.

Here is a multigraph:

1122334455λ\lambdaα\alphaβ\betaγ\gammaδ\deltaε\varepsilonκ\kappaFormally speaking, this multigraph is the triple(V,E,φ)\left(V,E,\varphi\right), where

V={1,2,3,4,5},E={α,β,γ,δ,ε,κ,λ},V=\left\{1,2,3,4,5\right\},\ \ \ \ \ \ \ \ \ \ E=\left\{\alpha,\beta,\gamma,\delta,\varepsilon,\kappa,\lambda\right\},and whereφ:E→𝒫1,2(V)\varphi:E\rightarrow\mathcal{P}_{1,2}\left(V\right)is the map that sendsα,β,γ,δ,ε,κ,λ\alpha,\beta,\gamma,\delta,\varepsilon,\kappa,\lambdato{1,2},{2,3},{2,3},{4,5},{4,5},{4,5},{1}\left\{1,2\right\},\left\{2,3\right\},\left\{2,3\right\},\left\{4,5\right\},\left\{4,5\right\},\left\{4,5\right\},\left\{1\right\}, respectively. (Of course, you can write{1}\left\{1\right\}as{1,1}\left\{1,1\right\}.)

This suggests the following terminology (most of which is a calque of our previously defined terminology for simple graphs):

Definition 3.4.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Then:

(a)The elements ofVVare called theverticesofGG. The setVVis called thevertex setofGG, and is denotedV(G)\operatorname*{V}\left(G\right).
(b)The elements ofEEare called theedgesofGG. The setEEis called theedge setofGG, and is denotedE(G)\operatorname*{E}\left(G\right).
(c)Ifeeis an edge ofGG, then the elements ofφ⁡(e)\varphi\left(e\right)are called theendpointsofee.
(d)We say that an edgeeecontainsa vertexvvifv∈φ⁡(e)v\in\varphi\left(e\right)(in other words, ifvvis an endpoint ofee).
(e)Two verticesuuandvvare said to beadjacentif there exists an edgee∈Ee\in Ewhose endpoints areuuandvv.
(f)Two edgeseeandffare said to beparallelifφ⁡(e)=φ⁡(f)\varphi\left(e\right)=\varphi\left(f\right). (In the above example, any two of the edgesδ,ε,κ\delta,\varepsilon,\kappaare parallel.)
(g)We say thatGGhasno parallel edgesifGGhas no two distinct edges that are parallel.
(h)An edgeeeis called aloop(orself-loop) ifφ⁡(e)\varphi\left(e\right)is a11-element set (i.e., ifeehas only one endpoint). (In Example3.3, the edgeλ\lambdais a loop.)
(i)We say thatGGislooplessifGGhas no loops (among its edges).
(j)Thedegreedeg⁡v\deg v(also writtendegG⁡v\deg_{G}v) of a vertexvvofGGis defined to be the number of edges that containvv, where loops are counted twice. In other words, deg⁡v=\displaystyle\deg v=degG⁡v\displaystyle\deg_{G}v:⁣=\displaystyle:=|{e∈E∣v∈φ⁡(e)}|⏟this counts all edgesthat containv+|{e∈E∣φ⁡(e)={v}}|⏟this counts all loopsthat containvonce again.\displaystyle\underbrace{\left|\left\{e\in E\ \mid\ v\in\varphi\left(e\right)\right\}\right|}_{\begin{subarray}{c}\text{this counts all edges}\\ \text{that contain }v\end{subarray}}+\underbrace{\left|\left\{e\in E\ \mid\ \varphi\left(e\right)=\left\{v\right\}\right\}\right|}_{\begin{subarray}{c}\text{this counts all loops}\\ \text{that contain }v\text{ once again}\end{subarray}}.(Note that, unlike in the case of a simple graph,deg⁡v\deg visnotthe number of neighbors ofvv, unless it happens thatvvis not contained in any loops or parallel edges.) (For example, in Example3.3, we havedeg⁡1=3\deg 1=3anddeg⁡2=3\deg 2=3anddeg⁡3=2\deg 3=2anddeg⁡4=3\deg 4=3anddeg⁡5=3\deg 5=3.)
(k)AwalkinGGmeans a list of the form (v0,e1,v1,e2,v2,…,ek,vk)(withk≥0),\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{with }k\geq 0\right),wherev0,v1,…,vkv_{0},v_{1},\ldots,v_{k}are vertices ofGG, wheree1,e2,…,eke_{1},e_{2},\ldots,e_{k}are edges ofGG, and where eachi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}satisfies φ⁡(ei)={vi−1,vi}\varphi\left(e_{i}\right)=\left\{v_{i-1},v_{i}\right\}(that is, the endpoints of each edgeeie_{i}arevi−1v_{i-1}andviv_{i}). Note that we have to record both the verticesandthe edges in our walk, since we want the walk to “know” which edges it traverses. (For instance, in Example3.3, the two walks(1,α,2,β,3)\left(1,\alpha,2,\beta,3\right)and(1,α,2,γ,3)\left(1,\alpha,2,\gamma,3\right)are distinct.) Theverticesof a walk(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)arev0,v1,…,vkv_{0},v_{1},\ldots,v_{k}; theedgesof this walk aree1,e2,…,eke_{1},e_{2},\ldots,e_{k}. This walk is said tostartatv0v_{0}andendatvkv_{k}; it is also said to be awalk fromv0v_{0}tovkv_{k}. Itsstarting pointisv0v_{0}, and itsending pointisvkv_{k}. Itslengthiskk.
(l)Apathmeans a walk whose vertices are distinct.
(m)The notions of “path-connected” and “connected” and “component” are defined exactly as for simple graphs. The symbol≃G\simeq_{G}still means “path-connected”.
(n)Aclosed walk(orcircuit) means a walk(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)withvk=v0v_{k}=v_{0}.
(o)Acyclemeans a closed walk(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)such that - •the verticesv0,v1,…,vk−1v_{0},v_{1},\ldots,v_{k-1}are distinct; - •the edgese1,e2,…,eke_{1},e_{2},\ldots,e_{k}are distinct; - •we havek≥1k\geq 1. (Note that we are not requiringk≥3k\geq 3any more, as we did for simple graphs. Thus, in Example3.3, both(2,β,3,γ,2)\left(2,\beta,3,\gamma,2\right)and(1,λ,1)\left(1,\lambda,1\right)are cycles, but(2,β,3,β,2)\left(2,\beta,3,\beta,2\right)is not. The purpose of the “k≥3k\geq 3” requirement for cycles in simple graphs was to disallow closed walks such as(2,β,3,β,2)\left(2,\beta,3,\beta,2\right)from being cycles; but they are now excluded by the “the edgese1,e2,…,eke_{1},e_{2},\ldots,e_{k}are distinct” condition.)
(p)Hamiltonian paths and cycles are defined as for simple graphs.
(q)We draw a multigraph by drawing each vertex as a point, each edge as a curve, and labeling both the vertices and the edges (or not, if we don’t care about what they are). An example of such a drawing appeared in Example3.3.

So there are two differences between simple graphs and multigraphs:

1.A multigraph can have loops, whereas a simple graph cannot.
2.In a simple graph, an edgeeeisa set of two vertices, whereas in a multigraph, an edgeeehasa set of two vertices (possibly two equal ones, ifeeis a loop) assigned to it by the mapφ\varphi. This not only allows for parallel edges, but also lets us store some information in the “identities” of the edges.

Nevertheless, the two notions have much in common; thus, they are both called “graphs”:

Convention 3.1.1.

The word “graph” means either “simple graph” or “multigraph”. The precise meaning should usually be understood from the context. (I will try not to use it when it could cause confusion.)

Fortunately, simple graphs and multigraphs have many properties in common, and often it is not hard to derive a result about multigraphs from the analogous result about simple graphs or vice versa. We will soon explore how some of the properties we have seen in the previous chapter can be adapted to multigraphs. First, however, let us explain how to convert multigraphs into simple graphs and vice versa.

3.2Conversions

We can turn each multigraph into a simple graph, but at a cost of losing some information:

Definition 3.5.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Then, theunderlying simple graphGsimpG^{\operatorname*{simp}}ofGGmeans the simple graph

(V,{φ(e)∣e∈Eis not a loop}).\left(V,\ \left\{\varphi\left(e\right)\ \mid\ e\in E\text{ is not a loop}\right\}\right).In other words, it is the simple graph with vertex setVVin which two distinct verticesuuandvvare adjacent if and only ifuuandvvare adjacent inGG. Thus,GsimpG^{\operatorname*{simp}}is obtained fromGGby removing loops and “collapsing” parallel edges to a single edge.

For example, the underlying simple graph of the multigraphGGin Example3.3is

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Conversely, each simple graph can be viewed as a multigraph:

Definition 3.6.

LetG=(V,E)G=\left(V,E\right)be a simple graph. Then, thecorresponding multigraphGmultG^{\operatorname*{mult}}is defined to be the multigraph

(V,E,ι),\left(V,E,\iota\right),whereι:E→𝒫1,2(V)\iota:E\rightarrow\mathcal{P}_{1,2}\left(V\right)is the map sending eache∈Ee\in Etoeeitself.

Example 3.7.

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As we said, the “underlying simple graph” constructionG↦GsimpG\mapsto G^{\operatorname*{simp}}destroys information, so it is irreversible. This being said, the two constructionsG↦GsimpG\mapsto G^{\operatorname*{simp}}andG↦GmultG\mapsto G^{\operatorname*{mult}}come fairly close to undoing one another:151515In the following proposition, we will use the notion of an “isomorphism of multigraphs”. A rigorous definition of this notion is given in Definition3.17further below (but it is more or less what you would expect: it is a way to relabel the vertices and the edges of one multigraph to obtain those of another).

Proposition 3.8.

(a)IfGGis a simple graph, then(Gmult)simp=G\left(G^{\operatorname*{mult}}\right)^{\operatorname*{simp}}=G.
(b)IfGGis a loopless multigraph that has no parallel edges, then(Gsimp)mult≅G\left(G^{\operatorname*{simp}}\right)^{\operatorname*{mult}}\cong G. (This is just an isomorphism, not an equality, since the “identities” of the edges ofGGhave been forgotten inGsimpG^{\operatorname*{simp}}and cannot be recovered.)
(c)IfGGis a multigraph that has loops or (distinct) parallel edges, then the multigraph(Gsimp)mult\left(G^{\operatorname*{simp}}\right)^{\operatorname*{mult}}has fewer edges thanGGand thus is not isomorphic toGG.

Proof 3.9.

A matter of understanding the definitions.

We will often identify a simple graphGGwith the corresponding multigraphGmultG^{\operatorname*{mult}}. This may be dangerous, because we have defined notions such as adjacency, walks, paths, cycles, etc. both for simple graphs and for multigraphs; thus, when we identify a simple graphGGwith the multigraphGmultG^{\operatorname*{mult}}, we are potentially inviting ambiguity (for example, does “cycle ofGG” mean a cycle of the simple graphGGor of the multigraphGmultG^{\operatorname*{mult}}?). Fortunately, this ambiguity is harmless, because wheneverGGis a simple graph, any of the notions we defined forGGis equivalent to the corresponding notion for the multigraphGmultG^{\operatorname*{mult}}. For example, for the notions of a cycle, we have the following:

Proposition 3.10.

LetGGbe a simple graph. Then:

(a)If(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)is a cycle of the multigraphGmultG^{\operatorname*{mult}}, then(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)is a cycle of the simple graphGG.
(b)Conversely, if(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)is a cycle of the simple graphGG, then(v0,{v0,v1},v1,{v1,v2},v2,…,vk−1,{vk−1,vk},vk)\left(v_{0},\left\{v_{0},v_{1}\right\},v_{1},\left\{v_{1},v_{2}\right\},v_{2},\ldots,v_{k-1},\left\{v_{k-1},v_{k}\right\},v_{k}\right)is a cycle of the multigraphGmultG^{\operatorname*{mult}}.

Proof 3.11.

This is not completely obvious, since our definitions of a cycle of a simple graph and of a cycle of a multigraph were somewhat different. The proof boils down to checking the following two statements:

1.If(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)is a cycle of the simple graphGG, then its edges {v0,v1},{v1,v2},…,{vk−1,vk}\left\{v_{0},v_{1}\right\},\left\{v_{1},v_{2}\right\},\ldots,\left\{v_{k-1},v_{k}\right\}are distinct.
2.If(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)is a cycle of the multigraphGmultG^{\operatorname*{mult}}, thenk≥3k\geq 3.

Checking statement 2 is easy (we cannot havek=1k=1sinceGmultG^{\operatorname*{mult}}has no loops, and we cannot havek=2k=2since this would lead toe1=e2e_{1}=e_{2}). Statement 1 is also clear, since the distinctness of thekkverticesv0,v1,…,vk−1v_{0},v_{1},\ldots,v_{k-1}forces the22-element sets formed from thesekkvertices to also be distinct (and since the edges{v0,v1},{v1,v2},…,{vk−1,vk}={vk−1,v0}\left\{v_{0},v_{1}\right\},\left\{v_{1},v_{2}\right\},\ldots,\left\{v_{k-1},v_{k}\right\}=\left\{v_{k-1},v_{0}\right\}are such22-element sets).

For all other notions discussed above, it is even more obvious that there is no ambiguity.

3.3Generalizing from simple graphs to multigraphs

Now, as promised, we shall revisit the results of Chapter2, and see which of them also hold for multigraphs instead of simple graphs.

3.3.1The Ramsey numberR⁡(3,3)R\left(3,3\right)

One of the first properties of simple graphs that we proved is the following (Proposition2.8):

Proposition 3.12.

LetGGbe a simple graph with|V⁡(G)|≥6\left|\operatorname{V}\left(G\right)\right|\geq 6(that is,GGhas at least66vertices). Then, at least one of the following two statements holds:

•Statement 1:There exist three distinct verticesaa,bbandccofGGsuch thatabab,bcbcandcacaare edges ofGG.
•Statement 2:There exist three distinct verticesaa,bbandccofGGsuch that none ofabab,bcbcandcacais an edge ofGG.

This is still true for multigraphs161616Of course, we should understand it appropriately: i.e., we should read “ababis an edge” as “there is an edge with endpointsaaandbb”., because replacing a multigraphGGby the underlying simple graphGsimpG^{\operatorname*{simp}}does not change the meaning of the statement.

3.3.2Degrees

In Definition2.15, we defined the degree of a vertexvvin a simple graphG=(V,E)G=\left(V,E\right)by

deg⁡v:=\displaystyle\deg v:=(the number of edgese∈Ethat containv)\displaystyle\left(\text{the number of edges }e\in E\text{ that contain }v\right)=\displaystyle=(the number of neighbors ofv)\displaystyle\left(\text{the number of neighbors of }v\right)=\displaystyle=|{u∈V∣uv∈E}|\displaystyle\left|\left\{u\in V\ \mid\ uv\in E\right\}\right|=\displaystyle=|{e∈E∣v∈e}|.\displaystyle\left|\left\{e\in E\ \mid\ v\in e\right\}\right|.These equalitiesno longer holdwhenGGis a multigraph. Parallel edges correspond to the same neighbor, so the number of neighbors ofvvis only a lower bound ondeg⁡v\deg v. (For instance, in Example3.3, the vertex22has degree33but only22neighbors.)

Proposition2.16(which says that ifGGis a simple graph withnnvertices, then any vertexvvofGGhas degreedeg⁡v∈{0,1,…,n−1}\deg v\in\left\{0,1,\ldots,n-1\right\}) alsono longer holdsfor multigraphs, because you can have arbitrarily many edges in a multigraph with just11or22vertices. (You can even have parallel loops!)

Is Proposition2.18true for multigraphs? Yes, because we have said that loops should count twice in the definition of the degree. The proof needs some tweaking, though. Let me give a slightly different proof; but first, let me state the claim for multigraphs as a proposition of its own:

Proposition 3.13(Euler 1736 for multigraphs).

LetGGbe a multigraph. Then, the sum of the degrees of all vertices ofGGequals twice the number of edges ofGG. In other words,

∑v∈V(G)deg⁡v=2⋅|E(G)|.\sum\limits_{v\in\operatorname*{V}\left(G\right)}\deg v=2\cdot\left|\operatorname*{E}\left(G\right)\right|.

Proof 3.14.

WriteGGasG=(V,E,φ)G=\left(V,E,\varphi\right); thus,V(G)=V\operatorname*{V}\left(G\right)=VandE(G)=E\operatorname*{E}\left(G\right)=E.

For each edgeee, let us (arbitrarily) choose one endpoint ofeeand denote it byα⁡(e)\alpha\left(e\right). The other endpoint will be calledβ⁡(e)\beta\left(e\right). Ifeeis a loop, then we setβ⁡(e)=α⁡(e)\beta\left(e\right)=\alpha\left(e\right). Then, for each vertexvv, we have

deg⁡v\displaystyle\deg v=(the number ofe∈Esuch thatv=α⁡(e))\displaystyle=\left(\text{the number of }e\in E\text{ such that }v=\alpha\left(e\right)\right)+(the number ofe∈Esuch thatv=β(e))\displaystyle\ \ \ \ \ \ \ \ \ \ +\left(\text{the number of }e\in E\text{ such that }v=\beta\left(e\right)\right)(note how loops get counted twice on the right hand side, because ife∈Ee\in Eis a loop, thenvvis bothα⁡(e)\alpha\left(e\right)andβ⁡(e)\beta\left(e\right)at the same time). Summing up this equality over allv∈Vv\in V, we obtain

∑v∈Vdeg⁡v\displaystyle\sum\limits_{v\in V}\deg v=∑v∈V(the number ofe∈Esuch thatv=α⁡(e))\displaystyle=\sum\limits_{v\in V}\left(\text{the number of }e\in E\text{ such that }v=\alpha\left(e\right)\right)+∑v∈V(the number ofe∈Esuch thatv=β(e)).\displaystyle\ \ \ \ \ \ \ \ \ \ +\sum\limits_{v\in V}\left(\text{the number of }e\in E\text{ such that }v=\beta\left(e\right)\right).However,

∑v∈V(the number ofe∈Esuch thatv=α⁡(e))=|E|,\sum\limits_{v\in V}\left(\text{the number of }e\in E\text{ such that }v=\alpha\left(e\right)\right)=\left|E\right|,since each edgee∈Ee\in Eis counted in exactly one addend of this sum. Similarly,

∑v∈V(the number ofe∈Esuch thatv=β⁡(e))=|E|.\sum\limits_{v\in V}\left(\text{the number of }e\in E\text{ such that }v=\beta\left(e\right)\right)=\left|E\right|.Thus, the above equality becomes

∑v∈Vdeg⁡v\displaystyle\sum\limits_{v\in V}\deg v=∑v∈V(the number ofe∈Esuch thatv=α(e))⏟=|E|\displaystyle=\underbrace{\sum\limits_{v\in V}\left(\text{the number of }e\in E\text{ such that }v=\alpha\left(e\right)\right)}_{=\left|E\right|}+∑v∈V(the number ofe∈Esuch thatv=β(e))⏟=|E|\displaystyle\ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{v\in V}\left(\text{the number of }e\in E\text{ such that }v=\beta\left(e\right)\right)}_{=\left|E\right|}=|E|+|E|=2⋅|E|.\displaystyle=\left|E\right|+\left|E\right|=2\cdot\left|E\right|.This proves Proposition3.13.

This is a good motivation for counting loops twice in the definition of a degree.

The handshake lemma (Corollary2.20)still holds for multigraphs. In other words, we have the following:

Corollary 3.15(handshake lemma).

LetGGbe a multigraph. Then, the number of verticesvvofGGwhose degreedeg⁡v\deg vis odd is even.

Proof 3.16.

This follows from Proposition3.13in the same way as for simple graphs.

Proposition2.22fails for multigraphs. For example, the multigraph 112233has three vertices with degrees1,2,31,2,3. Fortunately, Proposition2.22was more of a curiosity than a useful fact.

Mantel’s theorem (Theorem2.24) alsofails for multigraphs, because we can join two vertices with a lot of parallel edges and thus satisfye>n2/4e>n^{2}/4for stupid reasons without ever creating a triangle. Thus, Turan’s theorem (Theorem2.28) also fails for multigraphs.

3.3.3Graph isomorphisms

Graph isomorphy (and isomorphisms) can still be defined for multigraphs, but the definition is not the same as for simple graphs. Graph isomorphisms can no longer be defined merely as bijections between the vertex sets, since we also need to specify what they do to the edges. Instead, we define them as follows:

Definition 3.17.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)andH=(W,F,ψ)H=\left(W,F,\psi\right)be two multigraphs.

(a)Agraph isomorphism(orisomorphism) fromGGtoHHmeans apair(α,β)\left(\alpha,\beta\right)of bijections α:V→Wandβ:E→F\alpha:V\rightarrow W\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \beta:E\rightarrow Fwith the property that ife∈Ee\in E, then the endpoints ofβ⁡(e)\beta\left(e\right)are the images underα\alphaof the endpoints ofee. (This property can also be restated as a commutative diagram EβφFψ𝒫1,2(V)𝒫⁡(α)𝒫1,2(W),\lx@xy@svg{\hbox{\raise 2.5pt\hbox{\kern 19.68755pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\halign{\entry@#!@&&\entry@@#!@\cr&\cr&\crcr}}}\ignorespaces{\hbox{\kern-6.97916pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise-2.5pt\hbox{ $\\textstyle\{E\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\}$ }}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 39.11479pt\raise 6.1111pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-1.75pt\hbox{ $\\scriptstyle\{\\beta\}$ }}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 81.85425pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern-11.25975pt\raise-15.83333pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.8264pt\hbox{ $\\scriptstyle\{\\varphi\}$ }}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 0.0pt\raise-23.66666pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 81.85425pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise-2.5pt\hbox{ $\\textstyle\{F\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\}$ }}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 88.76398pt\raise-15.83333pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-1.75pt\hbox{ $\\scriptstyle\{\\psi\}$ }}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 88.76398pt\raise-23.66666pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern-19.68755pt\raise-31.66666pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise-2.5pt\hbox{ $\\textstyle\{\\mathcal\{P\}\_\{1,2\}\\left$V\\right$\\ignorespaces\\ignorespaces\\ignorespaces\\ignorespaces\}$ }}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern 32.9701pt\raise-38.16666pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern 0.0pt\raise-1.75pt\hbox{ $\\scriptstyle\{\\mathcal\{P\}\\left$\\alpha\\right$\}$ }}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 67.68755pt\raise-31.66666pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern 67.68755pt\raise-31.66666pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise-2.5pt\hbox{ $\\textstyle\{\\mathcal\{P\}\_\{1,2\}\\left$W\\right$\}$ }}}}}}}\ignorespaces}}}}\ignorespaces\ \ ,where𝒫⁡(α)\mathcal{P}\left(\alpha\right)is the map from𝒫1,2(V)\mathcal{P}_{1,2}\left(V\right)to𝒫1,2(W)\mathcal{P}_{1,2}\left(W\right)that sends each subset{u,v}∈𝒫1,2(V)\left\{u,v\right\}\in\mathcal{P}_{1,2}\left(V\right)to{α⁡(u),α⁡(v)}∈𝒫1,2(W)\left\{\alpha\left(u\right),\alpha\left(v\right)\right\}\in\mathcal{P}_{1,2}\left(W\right). If you are used to category theory, this restatement may look more natural to you.)
(b)We say thatGGandHHareisomorphic(this is writtenG≅HG\cong H) if there exists a graph isomorphism fromGGtoHH.

Again, isomorphy of multigraphs is an equivalence relation.

3.3.4Complete graphs, paths, cycles

In Definition2.42, Definition2.43and Definition2.44, we defined the complete graphsKnK_{n}, the path graphsPnP_{n}and the cycle graphsCnC_{n}as simple graphs. Thus, all of them can be viewed as multigraphs if one so desires (since each simple graphGGgives rise to a multigraphGmultG^{\operatorname*{mult}}).

However, using multigraphs, we can extend our definition ofnn-th cycle graphsCnC_{n}to the casen=1n=1and also tweak it in the casen=2n=2to make it more natural. We do this as follows:

Definition 3.18.

We modify the definition of cycle graphs (Definition2.44) as follows:

(a)Weredefinethe22-nd cycle graphC2C_{2}to be the multigraph with two vertices11and22and two parallel edges with endpoints11and22. (We don’t care what the edges are, only that there are two of them and each has endpoints11and22.) Thus, it looks as follows:1122.
(b)We define the11-st cycle graphC1C_{1}to be the multigraph with one vertex11and one edge (which is necessarily a loop). Thus, it looks as follows:11.

This has the effect that thenn-th cycle graphCnC_{n}has exactlynnedges for eachn≥1n\geq 1(rather than having11edge forn=2n=2, as it did back when it was a simple graph).

3.3.5Induced submultigraphs

In Definition2.46, we defined subgraphs and induced subgraphs of a simple graph. The corresponding notions for multigraphs are defined as follows:

Definition 3.19.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph.

(a)AsubmultigraphofGGmeans a multigraph of the formH=(W,F,ψ)H=\left(W,F,\psi\right), whereW⊆VW\subseteq VandF⊆EF\subseteq Eandψ=φ∣F\psi=\varphi\mid_{F}. In other words, a submultigraph ofGGmeans a multigraphHHwhose vertices are vertices ofGGand whose edges are edges ofGGand whose edges have the same endpoints inHHas they do inGG. We often abbreviate “submultigraph” as “subgraph”.
(b)LetSSbe a subset ofVV. Theinduced submultigraph ofGGon the setSSdenotes the submultigraph (S,E′,φ∣E′)\left(S,\ \ E^{\prime},\ \ \varphi\mid_{E^{\prime}}\right)ofGG, where E′:={e∈E∣all endpoints ofebelong toS}.E^{\prime}:=\left\{e\in E\ \mid\ \text{all endpoints of }e\text{ belong to }S\right\}.In other words, it denotes the submultigraph ofGGwhose vertices are the elements ofSS, and whose edges are precisely those edges ofGGwhose both endpoints belong toSS. We denote this induced submultigraph byG⁡[S]G\left[S\right].
(c)Aninduced submultigraphofGGmeans a submultigraph ofGGthat is the induced submultigraph ofGGonSSfor someS⊆VS\subseteq V.

The infix “multi” is often omitted. So we often speak of “subgraphs” instead of “submultigraphs”.

With these definitions, we can now identify cycles in a multigraph with subgraphs isomorphic to a cycle graph: A cycle of lengthnnin a multigraphGGis “the same as” a submultigraph ofGGisomorphic toCnC_{n}. (We leave the details to the reader.)

3.3.6Disjoint unions

In Section2.8, we defined the disjoint union of two or more simple graphs. The analogous definition for multigraphs is straightforward and left to the reader.

3.3.7Walks

We already defined walks, paths, closed walks and cycles for multigraphs back in Section3.1. Thelengthof a walk is still defined to be its number of edges. Now, let’s see which of their basic properties (seen in Section2.9) still hold for multigraphs.

First of all, the edges of a path are still always distinct. This is just as easy to prove as for simple graphs.

Next, let us see how two walks can be “spliced” together:

Proposition 3.20.

LetGGbe a multigraph. Letuu,vvandwwbe three vertices ofGG. Let𝐚=(a0,e1,a1,…,ek,ak)\mathbf{a}=\left(a_{0},e_{1},a_{1},\ldots,e_{k},a_{k}\right)be a walk fromuutovv. Let𝐛=(b0,f1,b1,…,fℓ,bℓ)\mathbf{b}=\left(b_{0},f_{1},b_{1},\ldots,f_{\ell},b_{\ell}\right)be a walk fromvvtoww. Then,

(a0,e1,a1,…,ek,ak,f1,b1,f2,b2,…,fℓ,bℓ)\displaystyle\left(a_{0},e_{1},a_{1},\ldots,e_{k},a_{k},f_{1},b_{1},f_{2},b_{2},\ldots,f_{\ell},b_{\ell}\right)=(a0,e1,a1,…,ak−1,ek,b0,f1,b1,…,fℓ,bℓ)\displaystyle=\left(a_{0},e_{1},a_{1},\ldots,a_{k-1},e_{k},b_{0},f_{1},b_{1},\ldots,f_{\ell},b_{\ell}\right)=(a0,e1,a1,…,ak−1,ek,v,f1,b1,…,fℓ,bℓ)\displaystyle=\left(a_{0},e_{1},a_{1},\ldots,a_{k-1},e_{k},v,f_{1},b_{1},\ldots,f_{\ell},b_{\ell}\right)is a walk fromuutoww. This walk shall be denoted𝐚∗𝐛\mathbf{a}\ast\mathbf{b}.

Walks can be reversed (i.e., walked in backwards direction):

Proposition 3.21.

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Let𝐚=(a0,e1,a1,…,ek,ak)\mathbf{a}=\left(a_{0},e_{1},a_{1},\ldots,e_{k},a_{k}\right)be a walk fromuutovv. Then:

(a)The list(ak,ek,ak−1,ek−1,…,e1,a0)\left(a_{k},e_{k},a_{k-1},e_{k-1},\ldots,e_{1},a_{0}\right)is a walk fromvvtouu. We denote this walk byrev𝐚\operatorname*{rev}\mathbf{a}and call it thereversalof𝐚\mathbf{a}.
(b)If𝐚\mathbf{a}is a path, thenrev𝐚\operatorname*{rev}\mathbf{a}is a path again.

Walks that are not paths contain smaller walks between the same vertices:

Proposition 3.22.

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Let𝐚=(a0,e1,a1,…,ek,ak)\mathbf{a}=\left(a_{0},e_{1},a_{1},\ldots,e_{k},a_{k}\right)be a walk fromuutovv. Assume that𝐚\mathbf{a}is not a path. Then, there exists a walk fromuutovvwhose length is smaller thankk.

Corollary 3.23(When there is a walk, there is a path).

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Assume that there is a walk fromuutovvof lengthkkfor somek∈ℕk\in\mathbb{N}. Then, there is a path fromuutovvof length≤k\leq k.

All these results can be proved in the same way as their counterparts for simple graphs (Proposition2.62, Proposition2.64, Proposition2.66and Corollary2.69); the only change needed is to record the edges in the walk.

Corollary3.23can be strengthened a bit:

Corollary 3.24(When there is a walk, there is a path).

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Assume that there is a walk𝐰\mathbf{w}fromuutovvof lengthkkfor somek∈ℕk\in\mathbb{N}. Then, there is a path𝐩\mathbf{p}fromuutovvof length≤k\leq kthat is contained in𝐰\mathbf{w}(meaning that each edge of𝐩\mathbf{p}is an edge of𝐰\mathbf{w}, and each vertex of𝐩\mathbf{p}is a vertex of𝐰\mathbf{w}).

Proof 3.25.

Replay the proof of Corollary3.23(and the proofs of the underlying results) and observe that each time we shorten our walk by “removing a loop”, we obtain a walk that is contained in𝐰\mathbf{w}.

Given a multigraphGGand two verticesuuandvvofGG, we can ask ourselves the same five Questions 1, 2, 3, 4 and 5 that we asked for a simple graphGGin Subsection2.9.4. The answers we gave in that subsection still apply without requiring substantial changes; the only necessary modification is that we now have to keep track of the edges in a path or walk. (The reader can easily fill in the details here.)

3.3.8Path-connectedness

The relation “path-connected” is defined for multigraphs just as it is for simple graphs (Definition2.71), and is still denoted≃G\simeq_{G}. It is still an equivalence relation (and the proof is the same as for simple graphs). The following also holds (with the same proof as for simple graphs):

Proposition 3.26.

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Then,u≃Gvu\simeq_{G}vif and only if there exists a path fromuutovv.

The definitions of “components” and “connected” for multigraphs are the same as for simple graphs (Definition2.76and Definition2.77). The following propositions can be proved in the same way as we proved their analogues for simple graphs (Proposition2.84and Proposition2.86):

Proposition 3.27.

LetGGbe a multigraph. LetCCbe a component ofGG.

Then, the multigraphG⁡[C]G\left[C\right](that is, the induced submultigraph ofGGon the setCC) is connected.

Proposition 3.28.

LetGGbe a multigraph. LetC1,C2,…,CkC_{1},C_{2},\ldots,C_{k}be all components ofGG(listed without repetition).

Thus,GGis isomorphic to the disjoint unionG⁡[C1]⊔G⁡[C2]⊔⋯⊔G⁡[Ck]G\left[C_{1}\right]\sqcup G\left[C_{2}\right]\sqcup\cdots\sqcup G\left[C_{k}\right].

The following proposition is an analogue of Proposition2.98for multigraphs:

Proposition 3.29.

LetGGbe a multigraph. Let𝐰\mathbf{w}be a walk ofGGsuch that no two consecutive edges of𝐰\mathbf{w}are identical. (By “consecutive edges”, we mean edges of the formei−1e_{i-1}andeie_{i}, wheree1,e2,…,eke_{1},e_{2},\ldots,e_{k}are the edges of𝐰\mathbf{w}from first to last.)

Then,𝐰\mathbf{w}either is a path or contains a cycle (i.e., there exists a cycle ofGGwhose edges are edges of𝐰\mathbf{w}).

Proof 3.30.

The proof of this proposition for multigraphs is more or less the same as it was for simple graphs (i.e., as the proof of Proposition2.98), with a mild difference in how we prove that the walk(wi,wi+1,…,wj)\left(w_{i},w_{i+1},\ldots,w_{j}\right)is a cycle (of course, this walk is no longer(wi,wi+1,…,wj)\left(w_{i},w_{i+1},\ldots,w_{j}\right)now, but rather(wi,ei+1,wi+1,…,ej,wj)\left(w_{i},e_{i+1},w_{i+1},\ldots,e_{j},w_{j}\right), because the edges need to be included).171717Here are some details: We assume that𝐰\mathbf{w}is not a path, and we write the walk𝐰\mathbf{w}as(w0,e1,w1,e2,w2,…,ek,wk)\left(w_{0},e_{1},w_{1},e_{2},w_{2},\ldots,e_{k},w_{k}\right). Then, there exists a pair(i,j)\left(i,j\right)of integersiiandjjwithi<ji<jandwi=wjw_{i}=w_{j}. Among all such pairs, we pick one withminimumdifferencej−ij-i. Then,(wi,ei+1,wi+1,…,ej,wj)\left(w_{i},e_{i+1},w_{i+1},\ldots,e_{j},w_{j}\right)is a closed walk. We claim that this closed walk is a cycle. To do so, we need to show that1.the verticeswi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct;2.the edgesei+1,ei+2,…,eje_{i+1},e_{i+2},\ldots,e_{j}are distinct;3.we havej−i≥1j-i\geq 1.The first of these claims follows from the minimality ofj−ij-i. The third follows fromi<ji<j. It remains to prove the second claim. In other words, it remains to prove that the edgesei+1,ei+2,…,eje_{i+1},e_{i+2},\ldots,e_{j}are distinct, i.e., that we haveea≠ebe_{a}\neq e_{b}for any two integersaaandbbsatisfyingi<a<b≤ji<a<b\leq j. Let us do this. Letaaandbbbe two integers satisfyingi<a<b≤ji<a<b\leq j. We must show thatea≠ebe_{a}\neq e_{b}. We distinguish two cases: the casea=b−1a=b-1and the casea≠b−1a\neq b-1.•Ifa=b−1a=b-1, theneae_{a}andebe_{b}are two consecutive edges of𝐰\mathbf{w}and thus distinct (since we assumed that no two consecutive edges of𝐰\mathbf{w}are identical). Thus,ea≠ebe_{a}\neq e_{b}is proved in the case whena=b−1a=b-1.•Now, consider the case whena≠b−1a\neq b-1. In this case, we must havea<b−1a<b-1(sincea<ba<bentailsa≤b−1a\leq b-1). Also,i≤a−1i\leq a-1(sincei<ai<a). Hence,i≤a−1<a<b−1≤j−1i\leq a-1<a<b-1\leq j-1(sinceb≤jb\leq j). Therefore,a−1a-1,aaandb−1b-1are three distinct elements of the set{i,i+1,…,j−1}\left\{i,i+1,\ldots,j-1\right\}. Consequently,wa−1,wa,wb−1w_{a-1},w_{a},w_{b-1}are three distinct vertices (since the verticeswi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct). Therefore,wb−1∉{wa−1,wa}=φ⁡(ea)w_{b-1}\notin\left\{w_{a-1},w_{a}\right\}=\varphi\left(e_{a}\right)(since𝐰\mathbf{w}is a walk, so that the edgeeae_{a}has endpointswa−1w_{a-1}andwaw_{a}). However,φ⁡(eb)={wb−1,wb}\varphi\left(e_{b}\right)=\left\{w_{b-1},w_{b}\right\}(since𝐰\mathbf{w}is a walk, so that the edgeebe_{b}has endpointswb−1w_{b-1}andwbw_{b}). Now, comparingwb−1∈{wb−1,wb}=φ⁡(eb)w_{b-1}\in\left\{w_{b-1},w_{b}\right\}=\varphi\left(e_{b}\right)withwb−1∉φ⁡(ea)w_{b-1}\notin\varphi\left(e_{a}\right), we see that the setsφ⁡(eb)\varphi\left(e_{b}\right)andφ⁡(ea)\varphi\left(e_{a}\right)must be distinct (sinceφ⁡(eb)\varphi\left(e_{b}\right)containswb−1w_{b-1}butφ⁡(ea)\varphi\left(e_{a}\right)does not). In other words,φ⁡(eb)≠φ⁡(ea)\varphi\left(e_{b}\right)\neq\varphi\left(e_{a}\right). Hence,eb≠eae_{b}\neq e_{a}. In other words,ea≠ebe_{a}\neq e_{b}. Thus,ea≠ebe_{a}\neq e_{b}is proved in the case whena≠b−1a\neq b-1.We have now provedea≠ebe_{a}\neq e_{b}in both cases, so we are done.

Just as for simple graphs, we get the following corollary:

Corollary 3.31.

LetGGbe a multigraph. Assume thatGGhas a closed walk𝐰\mathbf{w}of length>0>0such that no two consecutive edges of𝐰\mathbf{w}are identical. Then,GGhas a cycle.

The analogue of Theorem2.103for multigraphs is true as well:

Theorem 3.32.

LetGGbe a multigraph. Letuuandvvbe two vertices inGG. Assume that there are two distinct paths fromuutovv. Then,GGhas a cycle.

Proof 3.33.

For simple graphs, this was proved as Theorem2.103above. The same proof applies to multigraphs, once the obvious changes are made (e.g., instead ofpa−1pap_{a-1}p_{a}andqb−1qbq_{b-1}q_{b}, we need to take the last edges of the two walks𝐩\mathbf{p}and𝐪\mathbf{q}).

In contrast, Proposition2.106isfalsefor multigraphs. In fact, we can take a multigraph with a single vertex and lots of loops around it. In that case, its degree can be very large, but it has no cycles of length>1>1.

3.3.9G∖eG\setminus e, bridges and cut-edges

Next, we extend the definition ofG∖eG\setminus e(Definition2.108) to multigraphs:

Definition 3.34.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Leteebe an edge ofGG. Then,G∖eG\setminus ewill mean the graph obtained fromGGby removing this edgeee. In other words,

G∖e:=(V,E∖{e},φ∣E∖{e}).G\setminus e:=\left(V,\ E\setminus\left\{e\right\},\ \varphi\mid_{E\setminus\left\{e\right\}}\right).

Some authors writeG−eG-eforG∖eG\setminus e.

The analogue of Theorem2.109for multigraphs holds (and can be proved in the same way as Theorem2.109):

Theorem 3.35.

LetGGbe a multigraph. Leteebe an edge ofGG. Then:

(a)Ifeeis an edge of some cycle ofGG, then the components ofG∖eG\setminus eare precisely the components ofGG. (Keep in mind that the components are sets of vertices. It is these sets that we are talking about here, not the induced subgraphs on these sets.)
(b)Ifeeappears in no cycle ofGG(in other words, there exists no cycle ofGGsuch thateeis an edge of this cycle), then the graphG∖eG\setminus ehas one more component thanGG.

Note that an edgeeethat is a loop always is an edge of a cycle (indeed, it creates a cycle of length11), and can never appear on any path; thus, removing such an edgeeeobviously does not change the path-connectedness relation.

Defining cut-edges and bridges just as we did for simple graphs (Definition2.112), we equally recover the following corollary:

Corollary 3.36.

Leteebe an edge of a multigraphGG. Then,eeis a bridge if and only ifeeis a cut-edge.

Proof 3.37.

Just like the proof of Corollary2.113.

3.3.10Dominating sets

We defined and studied dominating sets in Section2.13. We could define dominating sets for multigraphs in the same way as for simple graphs, but we would not get anything new this way. Indeed, ifGGis a multigraph, then the dominating sets ofGGare precisely the dominating sets ofGsimpG^{\operatorname*{simp}}. Thus, we can reduce any claims about dominating sets of multigraphs to analogous claims about simple graphs.

3.3.11Hamiltonian paths and cycles

Hamiltonian paths and Hamiltonian cycles were defined for multigraphs in Definition3.4. It is easy to see that a multigraphGGhas a Hamiltonian path or Hamiltonian cycle if and only if the corresponding simple graphGsimpG^{\operatorname*{simp}}has one, with only one exception: A multigraphGGwith a single vertex has a Hamiltonian cycle if it has a loop, but the corresponding simple graphGsimpG^{\operatorname*{simp}}does not have a Hamiltonian cycle.

This does not mean, however, that everything we proved about Hamiltonian paths still applies to multigraphs. For instance, neither Ore’s theorem (Theorem2.132) nor Dirac’s theorem (Corollary2.134) holds for multigraphs, because we could duplicate edges to make degrees arbitrarily large, without necessarily creating a hamc.

Proposition2.137still holds for multigraphs, but this is clear because it can be derived from the corresponding property ofGsimpG^{\operatorname*{simp}}.

3.3.12Exercises

Exercise 3.38.

Which of the Exercises2.29,2.31,2.55,2.91,2.92,2.33and2.56remain true if “simple graph” is replaced by “multigraph”?

(For each exercise that becomes false, provide a counterexample. For each exercise that remains true, either provide a new solution that works for multigraphs, or argue that the solution we have seen applies verbatim to multigraphs, or derive the multigraph case from the simple graph case.)

Exercise 3.39.

LetGGbe a multigraph with at least one edge. Assume that each vertex ofGGhas even degree. Prove thatGGhas a cycle.

[Solution:This is Exercise 4 on midterm #1 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 3.40.

LetGGbe a multigraph with at least one vertex. Letd>2d>2be an integer. Assume thatdeg⁡v>2\deg v>2for each vertexvvofGG. Prove thatGGhas a cycle whose length is not divisible bydd.

Exercise 3.41.

LetGGbe a multigraph. Assume thatGGhas exactly two vertices of odd degree. Prove that these two vertices are path-connected.

Exercise 3.42.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph that has no loops.

Ife∈Ee\in Eis an edge that contains a vertexv∈Vv\in V, then we lete/ve/vdenote the endpoint ofeedistinct fromvv.

For eachv∈Vv\in V, we define a rational numberqvq_{v}by

qv=∑e∈E;v∈φ⁡(e)deg⁡(e/v)deg⁡v.q_{v}=\sum\limits_{\begin{subarray}{c}e\in E;\\ v\in\varphi\left(e\right)\end{subarray}}\dfrac{\deg\left(e/v\right)}{\deg v}.(Note that the denominatordeg⁡v\deg von the right hand side is nonzero whenever the sum is nonempty!)

(Thus,qvq_{v}is the average degree of the neighbors ofvv, weighted with the number of edges that joinvvto the respective neighbors. Ifvvhas no neighbors, thenqv=0q_{v}=0.)

Prove that

∑v∈Vqv≥∑v∈Vdeg⁡v.\sum\limits_{v\in V}q_{v}\geq\sum\limits_{v\in V}\deg v. (In other words, in a social network, your average friend has, on average, more friends than you do!)

[Hint:Any positive realsxxandyysatisfyxy+yx≥2\dfrac{x}{y}+\dfrac{y}{x}\geq 2. Why, and how does this help?]

Exercise 3.43.

LetFFbe any field. (For instance,FFcan beℚ\mathbb{Q}orℝ\mathbb{R}orℂ\mathbb{C}.)

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}.

For each edgee∈Ee\in E, we construct a column vectorχe∈Fn\chi_{e}\in F^{n}(that is, a column vector withnnentries) as follows:

•Ifeeis a loop, then we letχe\chi_{e}be the zero vector.
•Otherwise, we letuuandvvbe the two endpoints ofee, and we letχe\chi_{e}be the column vector that has a11in itsuu-th position, a−1-1in itsvv-th position, and00s in all other positions. (This depends on which endpoint we calluuand which endpoint we callvv, but we just make some choice and stick with it. The result will be true no matter how we choose.)

LetMMbe then×|E|n\times\left|E\right|-matrix overFFwhose columns are the column vectorsχe\chi_{e}for alle∈Ee\in E(we order them in some way; the exact order doesn’t matter). Prove that

rankM=|V|−connG,\operatorname*{rank}M=\left|V\right|-\operatorname*{conn}G,whereconnG\operatorname*{conn}Gdenotes the number of components ofGG.

[Example:Here is an example: LetGGbe the multigraph

1122334455aaccbbeeddffgghh(so thatn=5n=5). Then, if we choose the endpoints ofbbto be22and55in this order, then we haveχb=(0100−1)\chi_{b}=\begin{pmatrix}0\\ 1\\ 0\\ 0\\ -1\end{pmatrix}. (Choosing them to be55and22instead, we would obtainχb=(0−1001)\chi_{b}=\begin{pmatrix}0\\ -1\\ 0\\ 0\\ 1\end{pmatrix}.) If we do the same for all edges ofGG(that is, we choose the smaller endpoint asuuand the larger endpoint asvv), and if we order the columns so that they correspond to the edgesa,b,c,d,e,f,g,ha,b,c,d,e,f,g,hfrom left to right, then the matrixMMcomes out as follows:

M=(10100000−110010000001−1010000001−100−1−1−10−100).M=\begin{pmatrix}1&0&1&0&0&0&0&0\\ -1&1&0&0&1&0&0&0\\ 0&0&0&1&-1&0&1&0\\ 0&0&0&0&0&1&-1&0\\ 0&-1&-1&-1&0&-1&0&0\end{pmatrix}.It is easy to see thatrankM=4\operatorname*{rank}M=4, which is precisely|V|−connG\left|V\right|-\operatorname*{conn}G.]

[Remark:The claim of the exercise can be restated as follows: The span of the vectorsχe\chi_{e}for alle∈Ee\in Ehas dimension|V|−connG\left|V\right|-\operatorname*{conn}G.

Topologists will recognize the matrixMMas (a matrix that represents) the boundary operator∂:C1(G)→C0(G)\partial:C_{1}\left(G\right)\rightarrow C_{0}\left(G\right), whereGGis viewed as a CW-complex.]

Exercise 3.44.

IfGGis a multigraph, thenconn⁡G\operatorname{conn}Gshall denote the number of connected components ofGG. (Note that this is00whenGGhas no vertices, and11ifGGis connected.)

Let(V,H,φ)\left(V,H,\varphi\right)be a multigraph. LetEEandFFbe two subsets ofHH.

(a)Prove that conn⁡(V,E,φ∣E)+conn⁡(V,F,φ∣F)\displaystyle\operatorname{conn}\left(V,\ E,\ \varphi\mid_{E}\right)+\operatorname{conn}\left(V,\ F,\ \varphi\mid_{F}\right)≤conn⁡(V,E∪F,φ∣E∪F)+conn⁡(V,E∩F,φ∣E∩F).\displaystyle\leq\operatorname{conn}\left(V,\ E\cup F,\ \varphi\mid_{E\cup F}\right)+\operatorname{conn}\left(V,\ E\cap F,\ \varphi\mid_{E\cap F}\right).(2) [Hint:Feel free to restrict yourself to the case of a simple graph; in this case,EEandFFare two subsets of𝒫2(V)\mathcal{P}_{2}\left(V\right), and you have to show that conn⁡(V,E)+conn⁡(V,F)≤conn⁡(V,E∪F)+conn⁡(V,E∩F).\operatorname{conn}\left(V,\ E\right)+\operatorname{conn}\left(V,\ F\right)\leq\operatorname{conn}\left(V,\ E\cup F\right)+\operatorname{conn}\left(V,\ E\cap F\right).This isn’t any easier than the general case, but saves you the hassle of carrying the mapφ\varphiaround.]
(b)Give an example where the inequality (2) doesnotbecome an equality.

[Solution:This is Exercise 3 on homework set #3 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 3.45.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a connected multigraph with2m2medges, wherem∈ℕm\in\mathbb{N}. A set{e,f}\left\{e,f\right\}of two distinct edges will be called afriendly coupleifeeandffhave at least one endpoint in common. Prove that the edge set ofGGcan be decomposed intommdisjoint friendly couples (i.e., there existmmdisjoint friendly couples{e1,f1},{e2,f2},…,{em,fm}\left\{e_{1},f_{1}\right\},\left\{e_{2},f_{2}\right\},\ldots,\left\{e_{m},f_{m}\right\}such thatE={e1,f1,e2,f2,…,em,fm}E=\left\{e_{1},f_{1},e_{2},f_{2},\ldots,e_{m},f_{m}\right\}). (“Disjoint” means “disjoint as sets” – i.e., having no edges in common.)

[Example:Here is a graph with an even number of edges:

One possible decomposition of its edge set into disjoint friendly couples is{a,y},{b,z},{c,x}\left\{a,y\right\},\left\{b,z\right\},\left\{c,x\right\}.]

[Hint:Induct on|E|\left|E\right|. Pick a vertexvvof degree>1>1and consider the components ofG∖vG\setminus v.]

Exercise 3.46.

Letn≥0n\geq 0. Letd1,d2,…,dnd_{1},d_{2},\ldots,d_{n}bennnonnegative integers such thatd1+d2+⋯+dnd_{1}+d_{2}+\cdots+d_{n}is even.

(a)Prove that there exists a multigraphGGwith vertex set{1,2,…,n}\left\{1,2,\ldots,n\right\}such that alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}satisfydeg⁡i=di\deg i=d_{i}.
(b)Prove that there exists a loopless multigraphGGwith vertex set{1,2,…,n}\left\{1,2,\ldots,n\right\}such that alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}satisfydeg⁡i=di\deg i=d_{i}if and only if eachi∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}satisfies the inequality ∑j∈{1,2,…,n};j≠idj≥di.\sum\limits_{\begin{subarray}{c}j\in\left\{1,2,\ldots,n\right\};\\ j\neq i\end{subarray}}d_{j}\geq d_{i}.(3)

[Remark:The inequality (3) is the “nn-gon inequality”: It is equivalent to the existence of a (possibly degenerate)nn-gon with sidelengthsd1,d2,…,dnd_{1},d_{2},\ldots,d_{n}.]

Exercise 3.47.

LetGGbe a loopless multigraph. Recall that atrail(inGG) means a walk whose edges are distinct (but whose vertices are not necessarily distinct). Letuuandvvbe two vertices ofGG. As usual, “trail fromuutovv” means “trail that starts atuuand ends atvv”. Prove that

(the number of trails fromutovinG)\displaystyle\left(\text{the number of trails from }u\text{ to }v\text{ in }G\right)≡(the number of paths fromutovinG)mod2.\displaystyle\equiv\left(\text{the number of paths from }u\text{ to }v\text{ in }G\right)\operatorname{mod}2. [Hint:Try to pair up the non-path trails into pairs. Make sure to prove that this pairing is well-defined (i.e., each non-path trail𝐭\mathbf{t}has exactly one partner, which is not itself, and that𝐭\mathbf{t}is the designated partner of its partner!).]

Exercise 3.48.

LetGGbe a multigraph such that every vertex ofGGhas even degree. Letuuandvvbe two distinct vertices ofGG. Prove that the number of paths fromuutovvis even.

[Hint:When you add an edge joininguutovv, the graphGGbecomes a graph with exactly two odd-degree verticesuuandvv, and the claim becomes “the number of paths fromuutovvis odd” (why?). In this form, the claim turns out to be easier to prove. Indeed, any path must start with some edge…

Keep in mind that paths can be replaced by trails, by Exercise3.47.]

Exercise 3.49.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph such that|E|>|V|\left|E\right|>\left|V\right|. Prove thatGGhas a cycle of length≤2n+23\leq\dfrac{2n+2}{3}, wheren=|V|n=\left|V\right|.

[Solution:This is Exercise 8 on midterm #3 from my Spring 2017 course (except that the simple graph was replaced by a multigraph); seethe course pagefor solutions.]

3.4Eulerian circuits and walks

3.4.1Definitions

Let us now move on to a new feature of multigraphs, one that we have not yet studied (even for simple graphs).

Recall that a Hamiltonian path or cycle is a path or cycle that contains all vertices of the graph. Being a path or cycle, it has to contain each of them exactly once (except, in the case of a cycle, of its starting point).

What about a walk or closed walk that contains alledgesexactly once instead? These are called “Eulerian” walks or circuits; here is the formal definition:

Definition 3.50.

LetGGbe a multigraph.

(a)A walk ofGGis said to beEulerianif each edge ofGGappears exactly once in this walk. (In other words: A walk(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)ofGGis said to beEulerianif for each edgeeeofGG, there exists exactly onei∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}such thate=eie=e_{i}.)
(b)AnEulerian circuitofGGmeans a circuit (i.e., closed walk) ofGGthat is Eulerian. (Strictly speaking, the preceding sentence is redundant, but we still said it to stress the notion of an Eulerian circuit.)

Unlike for Hamiltonian paths and cycles, an Eulerian walk or circuit is usually not a path or cycle. Also, finding an Eulerian walk in a multigraphGGis not the same as finding an Eulerian walk in the simple graphGsimpG^{\operatorname*{simp}}. (Nevertheless, some authors call Eulerian walks “Eulerian paths” and call Eulerian circuits “Eulerian cycles”. This is rather confusing.)

Example 3.51.

Consider the following multigraphs:

1122334455A=A=aaccbbeeddffgg11223344B=B=aabbccdd11223344C=C=ggccddeeffaabb11223344D=D=cceeaa11223344E=E=aabbccddeef\qquad fgg11223344F=F=aabbccdd1122334455G=G=aabbccdde\quad eff\quadgghh11H=H=

•The multigraphAAhas an Eulerian walk(3,d,5,b,2,e,3,g,4,f,5,c,1,a,2)\left(3,d,5,b,2,e,3,g,4,f,5,c,1,a,2\right). ButAAhas no Eulerian circuit. The easiest way to see this is by observing thatAAhas a vertex of odd degree (e.g., the vertex22). If an Eulerian circuit were to exist, then it would have to enter this vertex as often as it exited it; but this would mean that the degree of this vertex would be even (because each edge containing this vertex would be used exactly once either to enter or to exit it, except for loops, which would be used twice). So, more generally, any multigraph that has a vertex of odd degree cannot have an Eulerian circuit.
•The multigraphBBhas an Eulerian circuit(1,a,2,b,3,c,4,d,1)\left(1,a,2,b,3,c,4,d,1\right), and thus of course an Eulerian walk (since any Eulerian circuit is an Eulerian walk).
•The multigraphCChas an Eulerian circuit(1,g,1,b,2,c,3,d,2,e,4,f,2,a,1)\left(1,g,1,b,2,c,3,d,2,e,4,f,2,a,1\right).
•The multigraphDDhas no Eulerian walk. Indeed, it has four vertices of odd degree. Ifvvis a vertex of odd degree, then any Eulerian walk has to either start or end atvv(since otherwise, the walk would enter and leavevvequally often, but then the degree ofvvwould be even). But a walk can only have one starting point and one ending point. This allows for two vertices of odd degree, but not more than two. So, more generally, any multigraph that has more than two vertices of odd degree cannot have an Eulerian walk.
•The multigraphEEhas no Eulerian walk. The reason is the same as forDD. Note thatEEis the famous multigraph of bridges in Königsberg, as studied by Euler in 1736 (seethe Wikipedia page for “Seven bridges of Königsberg”for the backstory).
•The multigraphFFhas no Eulerian walk, since it has two components, each containing at least one edge. (An Eulerian walk would have to contain both edgesbbandcc, but there is no way to walk between them, since they belong to different components.)
•The multigraphGGhas an Eulerian walk, namely(3,b,2,h,5,g,1,a,2,f,4,d,1,e,3,c,4)\left(3,b,2,h,5,g,1,a,2,f,4,d,1,e,3,c,4\right). It has no Eulerian circuit, since it has two vertices of odd degree.
•The multigraphHHhas an Eulerian circuit, namely(1)\left(1\right).

Exercise 3.53.

Letnnbe a positive integer. Recall from Definition2.42(a)thatKnK_{n}denotes the complete graph onnnvertices. This is the graph with vertex setV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}and edge set𝒫2(V)\mathcal{P}_{2}\left(V\right)(so each two distinct vertices are adjacent).

Find Eulerian circuits for the graphsK3K_{3},K5K_{5}, andK7K_{7}.

[Solution:This is Exercise 2 on homework set #2 from my Spring 2017 course; seethe course pagefor solutions.]

3.4.2The Euler–Hierholzer theorem

How hard is it to find an Eulerian walk or circuit in a multigraph, or to check if there is any? Surprisingly, this is a lot easier than the same questions for Hamiltonian paths or cycles. The second question in particular is answered (for connected multigraphs) by theEuler–Hierholzer theorem:

Theorem 3.54(Euler, Hierholzer).

LetGGbe a connected multigraph. Then:

(a)The multigraphGGhas an Eulerian circuit if and only if each vertex ofGGhas even degree.
(b)The multigraphGGhas an Eulerian walk if and only if all but at most two vertices ofGGhave even degree.

We already proved the “⟹\Longrightarrow” directions of both parts(a)and(b)in Example3.51. It remains to prove the “⟸\Longleftarrow” directions. Euler did not actually prove them in his 1736 paper181818although he pretended to do so, in a beautifully transparent example of “proof by trivialization”[Euler36, §21], but Hierholzer did in his 1873 paper[Hierho73]. The “standard” proof can be found in many texts, such as[Guicha16, Theorem 5.2.2 and Theorem 5.2.3]. I will sketch a different proof, which I learnt from[LeLeMe18, Problem 12.35]and which is actually rather close to Hierholzer’s original one[Hierho73]. We begin with the following definition:

Definition 3.55.

LetGGbe a multigraph. AtrailofGGmeans a walk ofGGwhose edges are distinct.

So a trail can repeat vertices, but cannot repeat edges.

Thus, an Eulerian walk has to be a trail. A trail cannot be longer than an Eulerian walk. Hence, a reasonable way to try constructing an Eulerian walk is to start with some trail, and make it progressively longer until it becomes Eulerian (hopefully).

This suggests the following approach to proving the “⟸\Longleftarrow” directions of Theorem3.54: We pick the longest trail ofGGand argue that (under the right assumptions) it has to be Eulerian, since otherwise there would be a way to make it longer. Of course, we need to find such a way. Here is the first step:

Lemma 3.56.

LetGGbe a multigraph with at least one vertex. Then,GGhas a longest trail.

Proof 3.57.

Clearly,GGhas at least one trail (e.g., a length-00trail from a vertex to itself). Moreover,GGhas only finitely many trails (since each edge ofGGcan only be used once in a trail, and there are only finitely many edges). Hence, the maximum principle proves the lemma.

Our goal now is to show that under appropriate conditions, such a longest trail will be Eulerian. This will require two further lemmas.

First, one more piece of notation: We say that an edgeeeof a multigraphGGintersectsa walk𝐰\mathbf{w}if at least one endpoint ofeeis a vertex of𝐰\mathbf{w}. Here is how this can look like:

𝐰\mathbf{w}𝐰\mathbf{w}𝐰\mathbf{w}𝐰\mathbf{w}ee(here, the edges of𝐰\mathbf{w}are marked with a “𝐰\mathbf{w}” underneath them) or

𝐰\mathbf{w}𝐰\mathbf{w}𝐰\mathbf{w}𝐰\mathbf{w}ee(here, the endpoint ofeethat is a vertex of𝐰\mathbf{w}happens to be the starting point of𝐰\mathbf{w}) or

𝐰\mathbf{w}𝐰\mathbf{w}𝐰\mathbf{w}𝐰\mathbf{w}ee(here, both endpoints ofeehappen to be vertices of𝐰\mathbf{w}). Be careful with such pictures, though: A walk doesn’t have to be a path; it can visit a vertex any number of times!

Lemma 3.58.

LetGGbe a connected multigraph. Let𝐰\mathbf{w}be a walk ofGG. Assume that there exists an edge ofGGthat is not an edge of𝐰\mathbf{w}.

Then, there exists an edge ofGGthat is not an edge of𝐰\mathbf{w}but intersects𝐰\mathbf{w}.

Proof 3.59.

We assumed that there exists an edge ofGGthat is not an edge of𝐰\mathbf{w}. Pick such an edge, and call itff.

A “𝐰\mathbf{w}-ff-path” will mean a path from a vertex of𝐰\mathbf{w}to an endpoint offf. Such a path clearly exists, sinceGGis connected. Thus, we can pick ashortestsuch path. If this shortest path has length00, then we are done (sinceffintersects𝐰\mathbf{w}in this case). If not, we consider the first edge of this path. This first edge cannot be an edge of𝐰\mathbf{w}, because otherwise we could remove it from the path and get an even shorter𝐰\mathbf{w}-ff-path. But it clearly intersects𝐰\mathbf{w}. So we have found an edge ofGGthat is not an edge of𝐰\mathbf{w}but intersects𝐰\mathbf{w}. This proves the lemma.

Lemma 3.60.

LetGGbe a multigraph such that each vertex ofGGhas even degree. Let𝐰\mathbf{w}be a longest trail ofGG. Then,𝐰\mathbf{w}is a closed walk.

Proof 3.61.

Assume the contrary. Letuube the starting point andvvthe ending point of𝐰\mathbf{w}. Since we assumed that𝐰\mathbf{w}is not a closed walk, we thus haveu≠vu\neq v.

Consider the edges of𝐰\mathbf{w}that containvv. Such edges are of two kinds: those by which𝐰\mathbf{w}entersvv(this means thatvvcomes immediately after this edge in𝐰\mathbf{w}), and those by which𝐰\mathbf{w}leavesvv(this means thatvvcomes immediately before this edge in𝐰\mathbf{w}).191919Loops whose only endpoint isvvcount as both.Except for the very last edge of𝐰\mathbf{w}, each edge of the former kind is immediately followed by an edge of the latter kind; conversely, each edge of the latter kind is immediately preceded by an edge of the former kind (since𝐰\mathbf{w}starts at the vertexuu, which is distinct fromvv). Hence, the walk𝐰\mathbf{w}has exactly one more edge enteringvvthan it has edges leavingvv. Thus, the number of edges of𝐰\mathbf{w}that containvv(with loops counting twice) is odd. However, the total number of edges ofGGthat containvv(with loops counting twice) is even (because it is the degree ofvv, but we assumed that each vertex ofGGhas even degree). So these two numbers are distinct. Thus, there is at least one edge ofGGthat containsvvbut is not an edge of𝐰\mathbf{w}.

Fix such an edge and call itff. Now, appendffto the trail𝐰\mathbf{w}at the end. The result will be a trail (sinceffis not an edge of𝐰\mathbf{w}) that is longer than𝐰\mathbf{w}. But this contradicts the fact that𝐰\mathbf{w}is a longest trail. Thus, the lemma is proved.

We can now finish the proof of the Euler–Hierholzer theorem:

Proof 3.62(Proof of Theorem3.54.).

(a)⟹:\Longrightarrow:We proved this back in Example3.51.

⟸:\Longleftarrow:Assume that each vertex ofGGhas even degree.

By Lemma3.56, we know thatGGhas a longest trail. Fix such a longest trail, and call it𝐰\mathbf{w}. Then, Lemma3.60shows that𝐰\mathbf{w}is a closed walk.

We claim that𝐰\mathbf{w}is Eulerian. Indeed, assume the contrary. Then, there exists an edge ofGGthat is not an edge of𝐰\mathbf{w}. Hence, Lemma3.58shows that there exists an edge ofGGthat is not an edge of𝐰\mathbf{w}but intersects𝐰\mathbf{w}. Fix such an edge, and call itff.

Sinceffintersects𝐰\mathbf{w}, there exists an endpointvvofffthat is a vertex of𝐰\mathbf{w}. Consider thisvv. Since𝐰\mathbf{w}is aclosedtrail, we can WLOG assume that𝐰\mathbf{w}starts and ends atvv(since we can otherwise achieve this by rotating202020Rotatinga closed walk(w0,e1,w1,e2,w2,…,ek,wk)\left(w_{0},e_{1},w_{1},e_{2},w_{2},\ldots,e_{k},w_{k}\right)means moving its first vertex and its first edge to the end, i.e., replacing the walk by(w1,e2,w2,e3,w3,…,ek,wk,e1,w1)\left(w_{1},e_{2},w_{2},e_{3},w_{3},\ldots,e_{k},w_{k},e_{1},w_{1}\right). This always results in a closed walk again. For example, if(1,a,2,b,3,c,1)\left(1,a,2,b,3,c,1\right)is a closed walk, then we can rotate it to obtain(2,b,3,c,1,a,2)\left(2,b,3,c,1,a,2\right); then, rotating it one more time, we obtain(3,c,1,a,2,b,3)\left(3,c,1,a,2,b,3\right). Clearly, by rotating a closed walk several times, we can make it start at any of its vertices. Moreover, if we rotate a closed trail, then we obtain a closed trail.𝐰\mathbf{w}). Then, we can append the edgeffto the trail𝐰\mathbf{w}. This results in a new trail (sinceffis not an edge of𝐰\mathbf{w}) that is longer than𝐰\mathbf{w}. And this contradicts the fact that𝐰\mathbf{w}is a longest trail ofGG.

This contradiction proves that𝐰\mathbf{w}is Eulerian. Hence,𝐰\mathbf{w}is an Eulerian circuit (since𝐰\mathbf{w}is a closed walk). Thus, the “⟸\Longleftarrow” direction of Theorem3.54(a)is proven.

(b)⟹:\Longrightarrow:Already proved in Example3.51.

⟸:\Longleftarrow:Assume that all but at most two vertices ofGGhave even degree. We must prove thatGGhas an Eulerian walk.

If each vertex ofGGhas even degree, then this follows from Theorem3.54(a), since every Eulerian circuit is an Eulerian walk. Thus, we WLOG assume that not each vertex ofGGhas even degree. In other words, the number of vertices ofGGhaving odd degree is positive.

The handshake lemma for multigraphs (i.e., Corollary3.15) shows that the number of vertices ofGGhaving odd degree is even. Furthermore, this number is at most22(since all but at most two vertices ofGGhave even degree). So this number is even, positive and at most22. Thus, this number is22. In other words, the multigraphGGhas exactly two vertices having odd degree. Letuuandvvbe these two vertices.

Add a new edgeeethat has endpointsuuandvvto the multigraphGG(do this even if there already is such an edge!212121This is a time to be grateful for the notion of a multigraph. We could not do this with simple graphs!). LetG′G^{\prime}denote the resulting multigraph. Then, inG′G^{\prime}, each vertex has even degree (since the newly added edgeeehas increased the degrees ofuuandvvby11, thus turning them from odd to even). Moreover,G′G^{\prime}is still connected (sinceGGwas connected, and the newly added edgeeecan hardly take that away). Thus, we can apply Theorem3.54(a)toG′G^{\prime}instead ofGG. As a result, we conclude thatG′G^{\prime}has an Eulerian circuit. Cutting the newly added edgeeeout of this Eulerian circuit222222More precisely: We rotate this circuit untileebecomes its last edge, and then we remove this last edge to obtain a walk., we obtain an Eulerian walk ofGG. Hence,GGhas an Eulerian walk. Thus, the “⟸\Longleftarrow” direction of Theorem3.54(b)is proven.

Note:If you look closely at the above proof, you will see hidden in it an algorithm forfindingEulerian circuits and walks.232323You might be skeptical about this. After all, in order to apply Lemma3.60, we need a longest trail, so you might wonder how we can find a longest trail to begin with. Fortunately, we don’t need to take Lemma3.60this literally. Our above proof of Lemma3.60can be used even if𝐰\mathbf{w}isnota longest trail. In this case, however, instead of showing that𝐰\mathbf{w}is a closed walk, this proof may show us a way how to make𝐰\mathbf{w}longer. In other words, by following this proof, we may discover a trail longer than𝐰\mathbf{w}. In this case, we can replace𝐰\mathbf{w}by this longer trail, and then apply Lemma3.60again. We can repeat this over and over again, until we do end up with a closed walk. (This will eventually happen, since we know that a trail cannot be longer than the total number of edges ofGG.)

Exercise 3.63.

LetGGbe a connected multigraph. Letmmbe the number of vertices ofGGthat have odd degree. Prove that we can addm/2m/2new edges toGGin such a way that the resulting multigraph will have an Eulerian circuit. (It is allowed to add an edge even if there is already an edge between the same two vertices.)

[Solution:This exercise is Exercise 6 on midterm #1 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 3.64.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Theline graphL⁡(G)L\left(G\right)is defined as the simple graph(E,F)\left(E,F\right), where

F={{e1,e2}∈𝒫2(E)∣φ⁡(e1)∩φ⁡(e2)≠∅}.F=\left\{\left\{e_{1},e_{2}\right\}\in\mathcal{P}_{2}\left({E}\right)\ \mid\ \varphi\left(e_{1}\right)\cap\varphi\left(e_{2}\right)\neq\varnothing\right\}.(In other words,L⁡(G)L\left(G\right)is the graph whoseverticesare theedgesofGG, and in which two verticese1e_{1}ande2e_{2}are adjacent if and only if the edgese1e_{1}ande2e_{2}ofGGshare a common endpoint.)

[Example:Here is a multigraphGGalong with its line graphL⁡(G)L\left(G\right):

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-61.64}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ $\\ \hline\cr\end{tabular}\ \ \ .Note thatL⁡(G)L\left(G\right)does not always determineGGuniquely.]

Assume that|V|>1\left|V\right|>1. Prove the following:

(a)IfGGhas a Hamiltonian path, thenL⁡(G)L\left(G\right)has a Hamiltonian path.
(b)IfGGhas an Eulerian walk, thenL⁡(G)L\left(G\right)has a Hamiltonian path.

[Solution:This exercise is Exercise 2 on midterm #1 from my Spring 2017 course (generalized from simple graphs to multigraphs); seethe course pagefor solutions.]

4Digraphs and multidigraphs

4.1Definitions

We have so far seen two concepts of graphs: simple graphs and multigraphs.

For all their differences, these two concepts have one thing in common: The two endpoints of an edge are equal in rights. Thus, when defining walks, each edge serves as a “two-way road”. Hence, such graphs are good at modelling symmetric relations between things.

We shall now introduce two analogous versions of “graphs” in which the edges have directions. These versions are known asdirected graphs(short:digraphs). In such directed graphs, each edge will have a specified starting point (its “source”) and a specified ending point (its “target”). Correspondingly, we will draw these edges as arrows, and we will only allow using them in one direction (viz., from source to target) when we walk down the graph. Here are the definitions in detail:

Definition 4.1.

Asimple digraphis a pair(V,A)\left(V,A\right), whereVVis a finite set, and whereAAis a subset ofV×VV\times V.

Definition 4.2.

LetD=(V,A)D=\left(V,A\right)be a simple digraph.

(a)The setVVis called thevertex setofDD; it is denoted byV(D)\operatorname*{V}\left(D\right). Its elements are called thevertices(ornodes) ofDD.
(b)The setAAis called thearc setofDD; it is denoted byA(D)\operatorname*{A}\left(D\right). Its elements are called thearcs(ordirected edges) ofDD. Whenuuandvvare two elements ofVV, we will occasionally useuvuvas a shorthand for the pair(u,v)\left(u,v\right). Note that this means an ordered pair now!
(c)If(u,v)\left(u,v\right)is an arc ofDD(or, more generally, a pair inV×VV\times V), thenuuis called thesourceof this arc, andvvis called thetargetof this arc.
(d)We drawDDas follows: We represent each vertex ofDDby a point, and each arcuvuvby an arrow that goes from the point representinguuto the point representingvv.
(e)An arc(u,v)\left(u,v\right)is called aloop(orself-loop) ifu=vu=v. (In other words, an arc is a loop if and only if its source is its target.)

Example 4.3.

For eachn∈ℕn\in\mathbb{N}, we define thedivisibility digraph on{1,2,…,n}\left\{1,2,\ldots,n\right\}to be the simple digraph(V,A)\left(V,A\right), whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}and

A={(i,j)∈V×V∣idividesj}.A=\left\{\left(i,j\right)\in V\times V\ \mid\ i\text{ divides }j\right\}. For example, forn=6n=6, this digraph looks as follows:

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\lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(4)Note that the divisibility digraph on{1,2,…,n}\left\{1,2,\ldots,n\right\}hasnnloops(1,1),(2,2),…,(n,n)\left(1,1\right),\ \left(2,2\right),\ \ldots,\ \left(n,n\right).

Note that simple digraphs (unlike simple graphs) are allowed to have loops (i.e., arcs of the form(v,v)\left(v,v\right)).

Definition 4.4.

Amultidigraphis a triple(V,A,ψ)\left(V,A,\psi\right), whereVVandAAare two finite sets, andψ:A→V×V\psi:A\rightarrow V\times Vis a map.

Definition 4.5.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph.

(a)The setVVis called thevertex setofDD; it is denoted byV(D)\operatorname*{V}\left(D\right). Its elements are called thevertices(ornodes) ofDD.
(b)The setAAis called thearc setofDD; it is denoted byA(D)\operatorname*{A}\left(D\right). Its elements are called thearcs(ordirected edges) ofDD.
(c)Ifaais an arc ofDD, and ifψ⁡(a)=(u,v)\psi\left(a\right)=\left(u,v\right), then the vertexuuis called thesourceofaa, and the vertexvvis called thetargetofaa.
(d)We drawDDas follows: We represent each vertex ofDDby a point, and each arcaaby an arrow that goes from the point representinguuto the point representingvv, where(u,v)=ψ⁡(a)\left(u,v\right)=\psi\left(a\right).
(e)An arcaaofDDis called aloop(orself-loop) if its source is its target (i.e., if it satisfiesu=vu=v, where(u,v)=ψ⁡(a)\left(u,v\right)=\psi\left(a\right)).

Example 4.6.

Here is a multidigraph:

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-90.9 6.71 -90.9 0 C -90.9 -6.71 -85.45 -12.16 -78.74 -12.16 C -72.03 -12.16 -66.58 -6.71 -66.58 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 39.37 C 90.9 46.08 85.45 51.53 78.74 51.53 C 72.03 51.53 66.58 46.08 66.58 39.37 C 66.58 32.66 72.03 27.21 78.74 27.21 C 85.45 27.21 90.9 32.66 90.9 39.37 Z M 78.74 39.37}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{25.23055pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 34.91)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 -39.37 C 90.9 -32.66 85.45 -27.21 78.74 -27.21 C 72.03 -27.21 66.58 -32.66 66.58 -39.37 C 66.58 -46.08 72.03 -51.53 78.74 -51.53 C 85.45 -51.53 90.9 -46.08 90.9 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\lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-31.09564pt}{23.97519pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -43.03 33.17)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} 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\lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.59853pt}{-30.91962pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.34 -42.78)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.37 5.68 L 65.89 32.94}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.89442}{0.44722}{-0.44722}{0.89442}{48.1528pt}{24.07646pt}\lxSVG@begingroup@{transform=matrix(0.89442 0.44722 -0.44722 0.89442 66.63 33.31)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{26.28899pt}{18.15936pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 36.38 25.13)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 78.74 26.66 L 78.74 -25}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} 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{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 67.37 -33.69 L 12.85 -6.43}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{-0.89442}{0.44722}{-0.44722}{-0.89442}{8.75272pt}{-4.37627pt}\lxSVG@begingroup@{transform=matrix(-0.89442 0.44722 -0.44722 -0.89442 12.11 -6.06)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{26.12463pt}{-22.46494pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 36.15 -31.08)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(5)Formally speaking, this multidigraph is the triple(V,A,ψ)\left(V,A,\psi\right), whereV={1,2,3,4}V=\left\{1,2,3,4\right\}andA={a,b,c,d,e}A=\left\{a,b,c,d,e\right\}andψ⁡(a)=(1,2)\psi\left(a\right)=\left(1,2\right)andψ⁡(b)=(1,2)\psi\left(b\right)=\left(1,2\right)andψ⁡(c)=(2,3)\psi\left(c\right)=\left(2,3\right)andψ⁡(d)=(3,4)\psi\left(d\right)=\left(3,4\right)andψ⁡(e)=(4,2)\psi\left(e\right)=\left(4,2\right).

Example 4.7.

Here is another multidigraph:

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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.0}{-1.0}{1.0}{0.0}{56.90552pt}{-18.66689pt}\lxSVG@begingroup@{transform=matrix(0.0 -1.0 1.0 0.0 78.74 -25.83)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{60.83853pt}{-1.18056pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 84.18 -1.63)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(6)Formally speaking, this multidigraph is the triple(V,A,ψ)\left(V,A,\psi\right), whereV={1,2,3,4,5}V=\left\{1,2,3,4,5\right\}andA={a,b,c,d,e,f,g,h}A=\left\{a,b,c,d,e,f,g,h\right\}andψ⁡(a)=(1,2)\psi\left(a\right)=\left(1,2\right)andψ⁡(b)=(2,5)\psi\left(b\right)=\left(2,5\right)and so on.

Thus, simple digraphs and multidigraphs are analogues of simple graphs and multigraphs, respectively, in which the edges have been replaced by arcs (“edges endowed with a direction”). The analogy is perfect but for the fact that simple graphs forbid loops but simple digraphs allow loops (but different authors have different opinions on this).

Convention 4.1.1.

The word “digraph” means either “simple digraph” or “multidigraph”, depending on the context.

The word “digraph” was originally a shorthand for “directed graph”, but by now it is a technical term that is perfectly understood by everyone in the subject. (It is also understood by linguists, but in a rather different way.)

Definition 4.8.

A digraphDDis said to belooplessif it has no loops.

4.2Outdegrees and indegrees

What can we do with digraphs? Many of the things we have done with graphs can be modified to work with digraphs (although not all their properties will still hold). For example, the notion of the degree of a vertex in a graph has the following two counterpart notions for digraphs:

Definition 4.9.

LetDDbe a digraph with vertex setVV. (This can be either a simple digraph or a multidigraph.) Letv∈Vv\in Vbe any vertex. Then:

(a)Theoutdegreeofvvdenotes the number of arcs ofDDwhose source isvv. This outdegree is denoteddeg+⁡v\deg^{+}v.
(b)Theindegreeofvvdenotes the number of arcs ofDDwhose target isvv. This indegree is denoteddeg−⁡v\deg^{-}v.

Example 4.10.

In the divisibility digraph on{1,2,3,4,5,6}\left\{1,2,3,4,5,6\right\}(see (4) for a drawing), we have

deg+⁡1\displaystyle\deg^{+}1=6,deg−⁡1=1,deg+⁡2=3,deg−⁡2=2,\displaystyle=6,\ \ \ \ \ \ \ \ \ \ \deg^{-}1=1,\ \ \ \ \ \ \ \ \ \ \deg^{+}2=3,\ \ \ \ \ \ \ \ \ \ \deg^{-}2=2,deg+⁡3\displaystyle\deg^{+}3=2,deg−⁡3=2,deg+⁡4=1,deg−⁡4=3,\displaystyle=2,\ \ \ \ \ \ \ \ \ \ \deg^{-}3=2,\ \ \ \ \ \ \ \ \ \ \deg^{+}4=1,\ \ \ \ \ \ \ \ \ \ \deg^{-}4=3,deg+⁡5\displaystyle\deg^{+}5=1,deg−⁡5=2,deg+⁡6=1,deg−⁡6=4.\displaystyle=1,\ \ \ \ \ \ \ \ \ \ \deg^{-}5=2,\ \ \ \ \ \ \ \ \ \ \deg^{+}6=1,\ \ \ \ \ \ \ \ \ \ \deg^{-}6=4.

Recall Euler’s result (Proposition3.13) saying that in a graph, the sum of all degrees is twice the number of edges. Here is an analogue of this result for digraphs:

Proposition 4.11(diEuler).

LetDDbe a digraph with vertex setVVand arc setAA. Then,

∑v∈Vdeg+⁡v=∑v∈Vdeg−⁡v=|A|.\sum\limits_{v\in V}\deg^{+}v=\sum\limits_{v\in V}\deg^{-}v=\left|A\right|.

Proof 4.12.

By the definition of an outdegree, we have

deg+v=(the number of arcs ofDwhose source isv)\deg^{+}v=\left(\text{the number of arcs of }D\text{ whose source is }v\right)for eachv∈Vv\in V. Thus,

∑v∈Vdeg+⁡v\displaystyle\sum\limits_{v\in V}\deg^{+}v=∑v∈V(the number of arcs ofDwhose source isv)\displaystyle=\sum\limits_{v\in V}\left(\text{the number of arcs of }D\text{ whose source is }v\right)=(the number of all arcs ofD)\displaystyle=\left(\text{the number of all arcs of }D\right)(since each arc ofDhas exactly one source,and thus is counted exactly once in the sum)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since each arc of }D\text{ has exactly one source,}\\ \text{and thus is counted exactly once in the sum}\end{array}\right)=|A|.\displaystyle=\left|A\right|.Similarly,∑v∈Vdeg−⁡v=|A|\sum\limits_{v\in V}\deg^{-}v=\left|A\right|.

(“diEuler” is not a real mathematician; I just gave that moniker to Proposition4.11in order to stress its analogy with Euler’s 1736 result.)

4.3Subdigraphs

Just as we defined subgraphs of a multigraph, we can define subdigraphs (or “submultidigraphs”, to be very precise) of a digraph:

Definition 4.13.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph.

(a)Asubmultidigraph(or, for short,subdigraph) ofDDmeans a multidigraph of the formE=(W,B,χ)E=\left(W,B,\chi\right), whereW⊆VW\subseteq VandB⊆AB\subseteq Aandχ=ψ∣B\chi=\psi\mid_{B}. In other words, a submultidigraph ofDDmeans a multidigraphEEwhose vertices are vertices ofDDand whose arcs are arcs ofDDand whose arcs have the same sources and targets inEEas they have inDD.
(b)LetSSbe a subset ofVV. Theinduced subdigraph ofDDon the setSSdenotes the subdigraph (S,A′,ψ∣A′)\left(S,\ \ A^{\prime},\ \ \psi\mid_{A^{\prime}}\right)ofDD, where A′:={a∈A∣both the source and the target ofabelong toS}.A^{\prime}:=\left\{a\in A\ \mid\ \text{both the source and the target of }a\text{ belong to }S\right\}.In other words, it denotes the subdigraph ofDDwhose vertices are the elements ofSS, and whose arcs are precisely those arcs ofDDwhose sources and targets both belong toSS. We denote this induced subdigraph byD⁡[S]D\left[S\right].
(c)Aninduced subdigraphofDDmeans a subdigraph ofDDthat is the induced subdigraph ofDDonSSfor someS⊆VS\subseteq V.

4.4Conversions

4.4.1Multidigraphs to multigraphs

Any multidigraphDDcan be turned into an (undirected) multigraphGGby “removing the arrowheads” (aka “forgetting the directions of the arcs”):

Definition 4.14.

LetDDbe a multidigraph. Then,DundD^{\operatorname*{und}}will denote the multigraph obtained fromDDby replacing each arc with an edge whose endpoints are the source and the target of this arc. Formally, this is defined as follows: IfD=(V,A,ψ)D=\left(V,A,\psi\right), thenDund=(V,A,φ)D^{\operatorname*{und}}=\left(V,A,\varphi\right), where the mapφ:A→𝒫1,2(V)\varphi:A\rightarrow\mathcal{P}_{1,2}\left(V\right)sends each arca∈Aa\in Ato the set of the entries ofψ⁡(a)\psi\left(a\right)(that is, to the set consisting of the source ofaaand the target ofaa).

We callDundD^{\operatorname*{und}}theunderlying undirected graphofDD.

For example, ifDDis the multidigraph from (6), thenDundD^{\operatorname*{und}}is the following multigraph:

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{}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 6.36 50.38 C 19.23 72.68 59.51 72.68 72.38 50.38}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{25.57191pt}{52.42795pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 35.38 72.54)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 69.75 30.38 L 8.99 -30.38}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{{}{}}}{{}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{19.31488pt}{-3.47221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 26.73 -4.8)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.03 -39.37 L 12.71 -39.37}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{25.46663pt}{-39.33022pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 35.24 -54.42)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 78.74 26.66 L 78.74 -26.66}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{60.83853pt}{-1.18056pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 84.18 -1.63)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .

4.4.2Multigraphs to multidigraphs

We have just seen how to turn any multidigraphDDinto a multigraphDundD^{\operatorname*{und}}by forgetting the directions of the arcs.

Conversely, we can turn a multigraphGGinto a multidigraphGbidirG^{\operatorname*{bidir}}by “duplicating” each edge (more precisely: turning each edge into two arcs with opposite orientations). Here is a formal definition:

Definition 4.15.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. For each edgee∈Ee\in E, let us choose one of the endpoints ofeeand call itses_{e}; the other endpoint will then be calledtet_{e}. (Ifeeis a loop, then we understandtet_{e}to meanses_{e}.)

We then defineGbidirG^{\operatorname*{bidir}}to be the multidigraph(V,E×{1,2},ψ)\left(V,\ \ E\times\left\{1,2\right\},\ \ \psi\right), where the mapψ:E×{1,2}→V×V\psi:E\times\left\{1,2\right\}\rightarrow V\times Vis defined as follows: For each edgee∈Ee\in E, we set

ψ⁡(e,1)=(se,te)andψ⁡(e,2)=(te,se).\psi\left(e,1\right)=\left(s_{e},t_{e}\right)\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \psi\left(e,2\right)=\left(t_{e},s_{e}\right).We callGbidirG^{\operatorname*{bidir}}thebidirectionalized multidigraphofGG.

Note that the mapψ\psidepends on our choice ofses_{e}’s (that is, it depends on which endpoint of an edgeeewe choose to beses_{e}). This makes the definition ofGbidirG^{\operatorname*{bidir}}non-canonical242424If you want to eradicate this dependency, you can rename the arcs(e,1)\left(e,1\right)and(e,2)\left(e,2\right)ofGbidirG^{\operatorname*{bidir}}as(e,se)\left(e,s_{e}\right)and(e,te)\left(e,t_{e}\right), except in the case wheneeis a loop, in which case they remain(e,1)\left(e,1\right)and(e,2)\left(e,2\right). This way,ψ\psino longer depends on any choices. But the independence might not be worth the awkwardness of having loops and non-loops treated differently.. Fortunately, all choices ofses_{e}’s will lead to mutually isomorphic multidigraphsGbidirG^{\operatorname*{bidir}}. (The notion ofisomorphismfor multidigraphs is exactly the one that you expect.)

Example 4.16.

1234G=gabc,\hbox to220.9pt{\vbox to132.18pt{\pgfpicture\makeatletter\hbox{\hskip 154.8058pt\lower-66.0914pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -82.33 0 C -82.33 6.71 -87.77 12.16 -94.49 12.16 C -101.2 12.16 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{\lx@inpgf@ignorespaces{}{}{{}}{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{}{}\lx@inpgf@ignorespaces}{{}} {\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{}{{}}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{{{{{{}}{}{}{}{}{{}}}}}{}{}{}{}}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -106.77 -3.29 C -144.79 -13.48 -144.79 13.48 -106.77 3.29}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{{}{}}}{{}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-106.83319pt}{-1.18056pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -147.83 -1.63)} \pgfsys@hbox{62}\lxSVG@closescope }}} 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{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 8.99 -8.99 L 69.75 -69.75}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{26.30693pt}{-24.51974pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 36.4 -33.93)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -12.71 0 L -81.78 0}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-36.30699pt}{-8.23856pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -50.24 -11.4)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,then

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-5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-44.91808pt}{-19.6788pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -62.15 -27.23)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -13.5 4.92 C -37.82 13.76 -56.67 13.76 -82.54 4.35}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.9397}{-0.34203}{0.34203}{0.9397}{-9.19574pt}{3.347pt}\lxSVG@begingroup@{transform=matrix(0.9397 -0.34203 0.34203 0.9397 -12.72 4.63)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-44.91808pt}{14.6788pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -62.15 20.31)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(Here, for example, we have chosensas_{a}to be22, so thatta=3t_{a}=3andψ⁡(a,1)=(2,3)\psi\left(a,1\right)=\left(2,3\right)andψ⁡(a,2)=(3,2)\psi\left(a,2\right)=\left(3,2\right).) Yes, even the loops ofGGare duplicated inGbidirG^{\operatorname*{bidir}}!

The operation that assigns a multidigraphGbidirG^{\operatorname*{bidir}}to a multigraphGGis injective – i.e., the original graphGGcan be uniquely reconstructed fromGbidirG^{\operatorname*{bidir}}. This is in stark difference to the operationD↦DundD\mapsto D^{\operatorname*{und}}, which destroys information (the directions of the arcs). Note that the multigraph(Gbidir)und\left(G^{\operatorname*{bidir}}\right)^{\operatorname*{und}}is not isomorphic toGG, since each edge ofGGis doubled in(Gbidir)und\left(G^{\operatorname*{bidir}}\right)^{\operatorname*{und}}.

4.4.3Simple digraphs to multidigraphs

Next, we introduce another operation: one that turns simple digraphs into multidigraphs. This is very similar to the operationG↦GmultG\mapsto G^{\operatorname*{mult}}that turns simple graphs into multigraphs, so we will even use the same notation for it. Its definition is as follows:

Definition 4.17.

LetD=(V,A)D=\left(V,A\right)be a simple digraph. Then, thecorresponding multidigraphDmultD^{\operatorname*{mult}}is defined to be the multidigraph

(V,A,ι),\left(V,A,\iota\right),whereι:A→V×V\iota:A\rightarrow V\times Vis the map sending eacha∈Aa\in Atoaaitself.

Example 4.18.

1234D=,\hbox to278.15pt{\vbox to75.28pt{\pgfpicture\makeatletter\hbox{\hskip 98.24751pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 0 C -66.58 6.71 -72.03 12.16 -78.74 12.16 C -85.45 12.16 -90.9 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\lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,then

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4.4.4Multidigraphs to simple digraphs

There is also an operationD↦DsimpD\mapsto D^{\operatorname*{simp}}that turns multidigraphs into simple digraphs:252525I will use a notation that I probably should have introduced before: Ifuuandvvare two vertices of a digraph, then an “arc fromuutovv” means an arc with sourceuuand targetvv.

Definition 4.19.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Then, theunderlying simple digraphDsimpD^{\operatorname*{simp}}ofDDmeans the simple digraph

(V,{ψ⁡(a)∣a∈A}).\left(V,\ \left\{\psi\left(a\right)\ \mid\ a\in A\right\}\right).In other words, it is the simple digraph with vertex setVVin which there is an arc fromuutovvif there exists an arc fromuutovvinDD. Thus,DsimpD^{\operatorname*{simp}}is obtained fromDDby “collapsing” parallel arcs (i.e., arcs having the same source and the same target) to a single arc.

Example 4.20.

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C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Note that the arcsccandddhave not been “collapsed” into one arc, since they do not have the same source and the same target. Likewise, the loopgghas been preserved (unlike for undirected graphs).

4.4.5Multidigraphs as a big tent

We have now established (in this section and also in Section3.2) several operations for converting between different types of graphs. Let us bring them together in a single picture:262626The enthusiastic reader can also define conversions between simple graphs and simple digraphs, which will complete this picture to a “square”; but we shall not have a use for such conversions.

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A takeaway from this all is that multidigraphs are the “most general” notion of graphs we have introduced so far. Indeed, using the operations we have seen so far, we can convert every notion of graphs into a multidigraph:

•Each simple graph becomes a multigraph via theG↦GmultG\mapsto G^{\operatorname*{mult}}operation.
•Each multigraph, in turn, becomes a multidigraph via theG↦GbidirG\mapsto G^{\operatorname*{bidir}}operation.
•Each simple digraph becomes a multidigraph via theD↦DmultD\mapsto D^{\operatorname*{mult}}operation.

Since all three of these operations are injective (i.e., lose no information), we thus can encode each of our four notions of graphs as a multidigraph. Consequently, any theorem about multidigraphs can be specialized to the other three types of graphs. This doesn’t mean that any theorem on any other type of graphs can be generalized to multidigraphs, though (e.g., Mantel’s theorem holds only for simple graphs) – but when it can, we will try to state it at the most general level possible, to avoid doing the same work twice.

4.5Walks, paths, closed walks, cycles

4.5.1Definitions

Let us now define various kinds of walks for simple digraphs and for multidigraphs.

For simple digraphs, we imitate the definitions from Sections2.9and2.10as best as we can, making sure to require all arcs to be traversed in the correct direction:

Definition 4.21.

LetDDbe a simple digraph. Then:

(a)Awalk(inDD) means a finite sequence(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)of vertices ofDD(withk≥0k\geq 0) such that all of the pairsv0v1,v1v2,v2v3,…,vk−1vkv_{0}v_{1},\ v_{1}v_{2},\ v_{2}v_{3},\ \ldots,\ v_{k-1}v_{k}are arcs ofDD. (The latter condition is vacuously true ifk=0k=0.)
(b)If𝐰=(v0,v1,…,vk)\mathbf{w}=\left(v_{0},v_{1},\ldots,v_{k}\right)is a walk inDD, then: - •Theverticesof𝐰\mathbf{w}are defined to bev0,v1,…,vkv_{0},v_{1},\ldots,v_{k}. - •Thearcsof𝐰\mathbf{w}are defined to be the pairsv0v1,v1v2,v2v3,…,vk−1vkv_{0}v_{1},\ v_{1}v_{2},\ v_{2}v_{3},\ \ldots,\ v_{k-1}v_{k}. - •The nonnegative integerkkis called thelengthof𝐰\mathbf{w}. (This is the number of all arcs of𝐰\mathbf{w}, counted with multiplicity. It is11smaller than the number of all vertices of𝐰\mathbf{w}, counted with multiplicity.) - •The vertexv0v_{0}is called thestarting pointof𝐰\mathbf{w}. We say that𝐰\mathbf{w}starts(orbegins) atv0v_{0}. - •The vertexvkv_{k}is called theending pointof𝐰\mathbf{w}. We say that𝐰\mathbf{w}endsatvkv_{k}.
(c)Apath(inDD) means a walk (inDD) whose vertices are distinct. In other words, a path means a walk(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)such thatv0,v1,…,vkv_{0},v_{1},\ldots,v_{k}are distinct.
(d)Letppandqqbe two vertices ofDD. Awalk frompptoqqmeans a walk that starts atppand ends atqq. Apath frompptoqqmeans a path that starts atppand ends atqq.
(e)Aclosed walkofDDmeans a walk whose first vertex is identical with its last vertex. In other words, it means a walk(w0,w1,…,wk)\left(w_{0},w_{1},\ldots,w_{k}\right)withw0=wkw_{0}=w_{k}. Sometimes, closed walks are also known ascircuits(but many authors use this latter word for something slightly different).
(f)AcycleofDDmeans a closed walk(w0,w1,…,wk)\left(w_{0},w_{1},\ldots,w_{k}\right)such thatk≥1k\geq 1and such that the verticesw0,w1,…,wk−1w_{0},w_{1},\ldots,w_{k-1}are distinct.

Note that we replaced the conditionk≥3k\geq 3byk≥1k\geq 1in the definition of a cycle, since simple digraphs can have loops. Fortunately, with the arcs being directed, we no longer have to worry about the same arc being traversed back and forth, so we need no extra condition to rule this out.

Example 4.22.

Consider the simple digraph

1234D=.\hbox to278.15pt{\vbox to75.28pt{\pgfpicture\makeatletter\hbox{\hskip 98.24751pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 0 C -66.58 6.71 -72.03 12.16 -78.74 12.16 C -85.45 12.16 -90.9 6.71 -90.9 0 C -90.9 -6.71 -85.45 -12.16 -78.74 -12.16 C -72.03 -12.16 -66.58 -6.71 -66.58 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 78.74 C 12.16 85.45 6.71 90.9 0 90.9 C -6.71 90.9 -12.16 85.45 -12.16 78.74 C -12.16 72.03 -6.71 66.58 0 66.58 C 6.71 66.58 12.16 72.03 12.16 78.74 Z M 0 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 248.38 0 C 248.38 6.71 242.93 12.16 236.22 12.16 C 229.51 12.16 224.06 6.71 224.06 0 C 224.06 -6.71 229.51 -12.16 236.22 -12.16 C 242.93 -12.16 248.38 -6.71 248.38 0 Z M 236.22 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{168.21655pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 232.76 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-94.9145pt}{-3.41666pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -131.33 -4.73)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -69.75 8.99 L -10.16 68.58}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.7071}{0.7071}{-0.7071}{0.7071}{-6.91965pt}{49.98587pt}\lxSVG@begingroup@{transform=matrix(0.7071 0.7071 -0.7071 0.7071 -9.57 69.17)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 8.99 69.75 L 68.58 10.16}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.7071}{-0.7071}{0.7071}{0.7071}{49.98587pt}{6.91965pt}\lxSVG@begingroup@{transform=matrix(0.7071 -0.7071 0.7071 0.7071 69.17 9.57)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.03 0 L 64.37 0}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{47.11964pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 65.2 0)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 91.45 0 L 221.85 0}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{160.93068pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 222.68 0)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then,(1,2,3,4)\left(1,2,3,4\right)and(1,3,4)\left(1,3,4\right)are two walks ofDD, and these walks are paths. But(2,3,1)\left(2,3,1\right)is not a walk (since you cannot use the arc1313to get from33to11). This digraphDDhas no cycles, and its only closed walks have length00.

Example 4.23.

Consider the simple digraph

1234D=.\hbox to305.92pt{\vbox to82.71pt{\pgfpicture\makeatletter\hbox{\hskip 98.24751pt\lower-16.61922pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 0 C -66.58 6.71 -72.03 12.16 -78.74 12.16 C -85.45 12.16 -90.9 6.71 -90.9 0 C -90.9 -6.71 -85.45 -12.16 -78.74 -12.16 C -72.03 -12.16 -66.58 -6.71 -66.58 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 78.74 C 12.16 85.45 6.71 90.9 0 90.9 C -6.71 90.9 -12.16 85.45 -12.16 78.74 C -12.16 72.03 -6.71 66.58 0 66.58 C 6.71 66.58 12.16 72.03 12.16 78.74 Z M 0 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 248.38 0 C 248.38 6.71 242.93 12.16 236.22 12.16 C 229.51 12.16 224.06 6.71 224.06 0 C 224.06 -6.71 229.51 -12.16 236.22 -12.16 C 242.93 -12.16 248.38 -6.71 248.38 0 Z M 236.22 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{168.21655pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 232.76 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-94.9145pt}{-3.41666pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -131.33 -4.73)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -69.75 8.99 L -10.16 68.58}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.7071}{0.7071}{-0.7071}{0.7071}{-6.91965pt}{49.98587pt}\lxSVG@begingroup@{transform=matrix(0.7071 0.7071 -0.7071 0.7071 -9.57 69.17)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 8.99 69.75 L 68.58 10.16}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.7071}{-0.7071}{0.7071}{0.7071}{49.98587pt}{6.91965pt}\lxSVG@begingroup@{transform=matrix(0.7071 -0.7071 0.7071 0.7071 69.17 9.57)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 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\lxSVG@transformcm{-0.96593}{0.25882}{-0.25882}{-0.96593}{180.16898pt}{-2.53276pt}\lxSVG@begingroup@{transform=matrix(-0.96593 0.25882 -0.25882 -0.96593 249.3 -3.5)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then,(1,2,3,1)\left(1,2,3,1\right)and(3,4,3)\left(3,4,3\right)and(4,4)\left(4,4\right)are cycles ofDD. Moreover,(1,2,3,4,3,1)\left(1,2,3,4,3,1\right)is a closed walk but not a cycle.

Now let’s define the same concepts for multidigraphs, by modifying the analogous definitions for multigraphs we saw in Definition3.4:

Definition 4.24.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Then:

(a)AwalkinDDmeans a list of the form (v0,a1,v1,a2,v2,…,ak,vk)(withk≥0),\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{with }k\geq 0\right),wherev0,v1,…,vkv_{0},v_{1},\ldots,v_{k}are vertices ofDD, wherea1,a2,…,aka_{1},a_{2},\ldots,a_{k}are arcs ofDD, and where eachi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}satisfies ψ⁡(ai)=(vi−1,vi)\psi\left(a_{i}\right)=\left(v_{i-1},v_{i}\right)(that is, each arcaia_{i}has sourcevi−1v_{i-1}and targetviv_{i}). Note that we have to record both the verticesandthe arcs in our walk, since we want the walk to “know” which arcs it traverses. Theverticesof a walk(v0,a1,v1,a2,v2,…,ak,vk)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)arev0,v1,…,vkv_{0},v_{1},\ldots,v_{k}; thearcsof this walk area1,a2,…,aka_{1},a_{2},\ldots,a_{k}. This walk is said tostartatv0v_{0}andendatvkv_{k}; it is also said to be awalk fromv0v_{0}tovkv_{k}. Itsstarting pointisv0v_{0}, and itsending pointisvkv_{k}. Itslengthiskk.
(b)Apathmeans a walk whose vertices are distinct.
(c)Aclosed walk(orcircuit) means a walk(v0,a1,v1,a2,v2,…,ak,vk)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)withvk=v0v_{k}=v_{0}.
(d)Acyclemeans a closed walk(v0,a1,v1,a2,v2,…,ak,vk)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)such that - •the verticesv0,v1,…,vk−1v_{0},v_{1},\ldots,v_{k-1}are distinct; - •we havek≥1k\geq 1. (This automatically implies that the arcsa1,a2,…,aka_{1},a_{2},\ldots,a_{k}are distinct, since each arcaia_{i}has sourcevi−1v_{i-1}.)

Example 4.25.

Consider the multidigraph

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{}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{197.63762pt}{-8.23856pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 273.47 -11.4)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then,(1,a,2,b,3,d,1)\left(1,a,2,b,3,d,1\right)and(3,d,1,c,3)\left(3,d,1,c,3\right)and(4,g,4)\left(4,g,4\right)are three cycles ofDD, whereas(3,d,1,a,2,b,3,d,1,c,3)\left(3,d,1,a,2,b,3,d,1,c,3\right)is a circuit but not a cycle.

4.5.2Basic properties

Now, let us see which properties of walks, paths, closed walks and cycles remain valid for digraphs.

In Proposition2.62, we saw how two walks in a simple graph could be combined (“spliced together”) if the ending point of the first is the starting point of the second. In Proposition3.20, we generalized this to multigraphs. The same holds for multidigraphs:

Proposition 4.26.

LetDDbe a multidigraph. Letuu,vvandwwbe three vertices ofDD. Let𝐚=(a0,e1,a1,…,ek,ak)\mathbf{a}=\left(a_{0},e_{1},a_{1},\ldots,e_{k},a_{k}\right)be a walk fromuutovv. Let𝐛=(b0,f1,b1,…,fℓ,bℓ)\mathbf{b}=\left(b_{0},f_{1},b_{1},\ldots,f_{\ell},b_{\ell}\right)be a walk fromvvtoww. Then,

Proof 4.27.

The same (trivial) argument as for undirected graphs works here.

However, unlike for undirected graphs, we can no longer reverse walks or paths in digraphs. Thus, it often happens that there is a walk fromuutovv, but no walk fromvvtouu.

Reducing a walk to a path (as we did in Proposition2.66for simple graphs and in Proposition3.22for multigraphs) still works for multidigraphs:

Proposition 4.28.

LetDDbe a multidigraph. Letuuandvvbe two vertices ofDD. Let𝐚\mathbf{a}be a walk fromuutovv. Letkkbe the length of𝐚\mathbf{a}. Assume that𝐚\mathbf{a}is not a path. Then, there exists a walk fromuutovvwhose length is smaller thankk.

Corollary 4.29(When there is a walk, there is a path).

LetDDbe a multidigraph. Letuuandvvbe two vertices ofDD. Assume that there is a walk fromuutovvof lengthkkfor somek∈ℕk\in\mathbb{N}. Then, there is a path fromuutovvof length≤k\leq k.

The proofs of these facts are the same as for multigraphs (see Proposition3.22and Corollary3.23, respectively).

The following proposition is an analogue of Proposition2.98for multidigraphs:

Proposition 4.30.

LetDDbe a multidigraph. Let𝐰\mathbf{w}be a walk ofDD. Then,𝐰\mathbf{w}either is a path or contains a cycle (i.e., there exists a cycle ofDDwhose arcs are arcs of𝐰\mathbf{w}).

Proof 4.31.

This follows by the same argument as Proposition2.98.

4.5.3Remark on algorithms

Given a multidigraphDDand two verticesuuandvvofDD, we can pose the same five algorithmic questions (Questions 1, 2, 3, 4 and 5) that we posed for a simple graphGGin Subsection2.9.4. As with multigraphs, the same answers that we gave back then are still valid in our new setting, as long as we replace “neighbors ofvv” by “in-neighbors ofvv” (that is, verticeswwsuch thatDDhas an arc fromwwtovv), and as long as we keep track of the arcs in our paths or walks.

4.5.4Exercises

Exercise 4.32.

LetDDbe a multidigraph with at least one vertex. Prove the following:

(a)If each vertexvvofDDsatisfiesdeg+⁡v>0\deg^{+}v>0, thenDDhas a cycle.
(b)If each vertexvvofDDsatisfiesdeg+⁡v=deg−⁡v=1\deg^{+}v=\deg^{-}v=1, then each vertex ofDDbelongs to exactly one cycle ofDD. Here, two cycles are considered to be identical if one can be obtained from the other by cyclic rotation.

Exercise 4.33.

Letppbe a prime number. Let(a1,a2,a3,…)\left(a_{1},a_{2},a_{3},\ldots\right)be a sequence of integers that is periodic with periodpp(that is, that satisfiesai=ai+pa_{i}=a_{i+p}for eachi>0i>0). Assume thata1+a2+⋯+apa_{1}+a_{2}+\cdots+a_{p}is not divisible bypp. Prove that there exists ani∈{1,2,…,p}i\in\left\{1,2,\ldots,p\right\}such that none of theppnumbers

ai,ai+ai+1,ai+ai+1+ai+2,…,ai+ai+1+⋯+ai+p−1a_{i},\ a_{i}+a_{i+1},\ a_{i}+a_{i+1}+a_{i+2},\ \ldots,\ a_{i}+a_{i+1}+\cdots+a_{i+p-1}(that is, of theppsumsai+ai+1+⋯+aja_{i}+a_{i+1}+\cdots+a_{j}fori≤j<i+pi\leq j<i+p) is divisible bypp.

[Remark:This would be false ifppwas not prime. For instance, forp=4p=4, the sequence(0,2,2,2,0,2,2,2,…)\left(0,2,2,2,0,2,2,2,\ldots\right)would be a counterexample.]

[Hint:Use Exercise4.32(a). What is the digraph, and why does it have a cycle?]

Exercise 4.34.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph.

For two verticesuuandvvofDD, we shall writeu→∗vu\overset{\ast}{\rightarrow}vif there exists a path fromuutovv.

ArootofDDmeans a vertexu∈Vu\in Vsuch that each vertexv∈Vv\in Vsatisfiesu→∗vu\overset{\ast}{\rightarrow}v.

Acommon ancestorof two verticesuuandvvmeans a vertexw∈Vw\in Vsuch thatw→∗uw\overset{\ast}{\rightarrow}uandw→∗vw\overset{\ast}{\rightarrow}v.

Assume thatDDhas at least one vertex. Prove thatDDhas a root if and only if every two vertices inDDhave a common ancestor.

The following exercise is both a directed analogue and a generalization of Mantel’s theorem (Theorem2.24):

Exercise 4.35.

LetDDbe a simple digraph withnnvertices andaaarcs. Assume thatDDhas no loops, and that we havea>n2/2a>n^{2}/2. Prove the following:

(a)The digraphDDhas a cycle of length33.
(b)We define anenhanced33-cycleto be a triple(u,v,w)\left(u,v,w\right)of distinct vertices ofDDsuch that all four pairs(u,v)\left(u,v\right),(v,w)\left(v,w\right),(w,u)\left(w,u\right)and(u,w)\left(u,w\right)are arcs ofDD. Then, the digraphDDhas an enhanced33-cycle.

Exercise 4.36.

LetD=(V,A)D=\left(V,A\right)be a simple digraph that has no cycles.

If𝐯=(v1,v2,…,vn)\mathbf{v}=\left(v_{1},v_{2},\ldots,v_{n}\right)is a list of vertices ofDD(not necessarily a walk!), then aback-cutof𝐯\mathbf{v}shall mean an arca∈Aa\in Awhose source isviv_{i}and whose target isvjv_{j}for somei,j∈{1,2,…,n}i,j\in\left\{1,2,\ldots,n\right\}satisfyingi>ji>j. (Colloquially speaking, a back-cut of𝐯\mathbf{v}is an arc ofDDthat leads from some vertex of𝐯\mathbf{v}to some earlier vertex of𝐯\mathbf{v}.)

A list𝐯=(v1,v2,…,vn)\mathbf{v}=\left(v_{1},v_{2},\ldots,v_{n}\right)of vertices ofDDis said to be atoposort272727This is short for “topological sorting”. I don’t know where this name comes from.ofDDif it contains each vertex ofDDexactly once and has no back-cuts.

Prove the following:

(a)The digraphDDhas at least one toposort.
(b)IfDDhas only one toposort, then this toposort is a Hamiltonian path ofDD. Here, aHamiltonian pathinDDmeans a walk ofDDthat contains each vertex ofDDexactly once.

[Example:For example, the digraph

has two toposorts:(3,2,1,4)\left(3,2,1,4\right)and(3,2,4,1)\left(3,2,4,1\right).]

Exercise 4.37.

Letnnbe a positive integer. LetDDbe a digraph that has no cycles of length≤2\leq 2. Assume thatDDhas at least2n−12^{n-1}vertices. Prove thatDDhas an induced subdigraph that hasnnvertices and has no cycles.

4.5.5The adjacency matrix

A simple way to find the number of walks from a given vertex to a given vertex in a multidigraph is provided by matrix algebra:

Theorem 4.38.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}.

IfMMis any matrix, and ifiiandjjare two positive integers, thenMi,jM_{i,j}shall denote the(i,j)\left(i,j\right)-th entry ofMM(that is, the entry ofMMin theii-th row and thejj-th column).

LetCCbe then×nn\times n-matrix (with real entries) defined by

Ci,j\displaystyle C_{i,j}=(the number of all arcsa∈Awith sourceiand targetj)\displaystyle=\left(\text{the number of all arcs $a\\in A$ with source $i$ and target $j$ }\right)for alli,j∈Vi,j\in V. Letk∈ℕk\in\mathbb{N}, and leti,j∈Vi,j\in V. Then,(Ck)i,j\left(C^{k}\right)_{i,j}equals the number of all walks ofDDhaving starting pointii, ending pointjjand lengthkk.

Proof 4.40(Proof of Theorem4.38.).

Forget that we fixedii,jjandkk. We want to prove the following claim:

Claim 1:Leti∈Vi\in Vandj∈Vj\in Vandk∈ℕk\in\mathbb{N}. Then, (Ck)i,j=(the number of walks fromitojthat have lengthk).\left(C^{k}\right)_{i,j}=\left(\text{the number of walks from }i\text{ to }j\text{ that have length }k\right).

Before we prove this claim, let us recall thatCCis the adjacency matrix ofDD. Thus, for eachi∈Vi\in Vandj∈Vj\in V, we have

Ci,j=(the number of all arcsa∈Awith sourceiand targetj)C_{i,j}=\left(\text{the number of all arcs $a\\in A$ with source $i$ and target $j$ }\right)(by the definition of the adjacency matrix). In other words, for eachi∈Vi\in Vandj∈Vj\in V, we have

Ci,j=(the number of arcs fromitoj),C_{i,j}=\left(\text{the number of arcs from }i\text{ to }j\right),where we agree that an “arc fromiitojj” means an arca∈Aa\in Awith sourceiiand targetjj.

Renamingiiaswwin this statement, we obtain the following: For eachw∈Vw\in Vandj∈Vj\in V, we have

Cw,j=(the number of arcs fromwtoj).C_{w,j}=\left(\text{the number of arcs from }w\text{ to }j\right).(7) Let us also recall that any twon×nn\times n-matricesMMandNNsatisfy

(MN)i,j=∑w=1nMi,wNw,j\left(MN\right)_{i,j}=\sum\limits_{w=1}^{n}M_{i,w}N_{w,j}(8)for anyi∈Vi\in Vandj∈Vj\in V. (Indeed, this is just the rule for how matrices are multiplied.)

We can now prove Claim 1:

[Proof of Claim 1:We shall prove Claim 1 by induction onkk:

Induction base:We shall first prove Claim 1 fork=0k=0.

Indeed, leti∈Vi\in Vandj∈Vj\in V. The00-th power of anyn×nn\times n-matrix is defined to be then×nn\times nidentity matrixInI_{n}; thus,C0=InC^{0}=I_{n}. Hence,

(C0)i,j=(In)i,j={1,ifi=j;0,ifi≠j\left(C^{0}\right)_{i,j}=\left(I_{n}\right)_{i,j}=\begin{cases}1,&\text{if }i=j;\\ 0,&\text{if }i\neq j\end{cases}(9)(by the definition of the identity matrix).

On the other hand, how many walks fromiitojjhave length00? A walk that has length00must consist of a single vertex, which is simultaneously the starting point and the ending point of this walk. Thus, a walk fromiitojjthat has length00exists only wheni=ji=j, and in this case there is exactly one such walk (namely, the walk(i)\left(i\right)). Hence,

(the number of walks fromitojthat have length0)={1,ifi=j;0,ifi≠j.\left(\text{the number of walks from }i\text{ to }j\text{ that have length }0\right)=\begin{cases}1,&\text{if }i=j;\\ 0,&\text{if }i\neq j.\end{cases}Comparing this with (9), we conclude that

(C0)i,j=(the number of walks fromitojthat have length0).\left(C^{0}\right)_{i,j}=\left(\text{the number of walks from }i\text{ to }j\text{ that have length }0\right).(10) Now, forget that we fixediiandjj. We thus have proven (10) for anyi∈Vi\in Vandj∈Vj\in V. In other words, Claim 1 holds fork=0k=0. Thus, the induction base is complete.

Induction step:Letggbe a positive integer. Assume that Claim 1 holds fork=g−1k=g-1. We must show that Claim 1 holds fork=gk=gas well.

We have assumed that Claim 1 holds fork=g−1k=g-1. In other words, for anyi∈Vi\in Vandj∈Vj\in V, we have

(Cg−1)i,j=(the number of walks fromitojthat have lengthg−1).\left(C^{g-1}\right)_{i,j}=\left(\text{the number of walks from }i\text{ to }j\text{ that have length }g-1\right).Renamingjjaswwin this statement, we obtain the following: For anyi∈Vi\in Vandw∈Vw\in V, we have

(Cg−1)i,w=(the number of walks fromitowthat have lengthg−1).\left(C^{g-1}\right)_{i,w}=\left(\text{the number of walks from }i\text{ to }w\text{ that have length }g-1\right).(11) Each walk fromiitojjthat has lengthgghas the form

𝐰=(v0,a1,v1,a2,v2,…,ag−1,vg−1,ag,vg)\mathbf{w}=\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{g-1},v_{g-1},a_{g},v_{g}\right)for some verticesv0,v1,…,vgv_{0},v_{1},\ldots,v_{g}ofDDand some arcsa1,a2,…,aga_{1},a_{2},\ldots,a_{g}ofDDsatisfyingv0=iv_{0}=iandvg=jv_{g}=jand(ψ⁡(ah)=(vh−1,vh)for allh∈{1,2,…,g})\left(\psi\left(a_{h}\right)=\left(v_{h-1},v_{h}\right)\text{ for all }h\in\left\{1,2,\ldots,g\right\}\right). Thus, each such walk𝐰\mathbf{w}can be constructed by the following algorithm:

•First, we choose a vertexwwofDDto serve as the vertexvg−1v_{g-1}(that is, as the penultimate vertex of the walk𝐰\mathbf{w}). This vertexwwmust belong toVV.
•Now, we choose the verticesv0,v1,…,vg−1v_{0},v_{1},\ldots,v_{g-1}(that is, all vertices of our walk except for the last one) and the arcsa1,a2,…,ag−1a_{1},a_{2},\ldots,a_{g-1}(that is, all arcs of our walk except for the last one) in such a way thatvg−1=wv_{g-1}=w. This is tantamount to choosing a walk(v0,a1,v1,a2,v2,…,ag−1,vg−1)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{g-1},v_{g-1}\right)fromiitowwthat has lengthg−1g-1. This choice can be made in(Cg−1)i,w\left(C^{g-1}\right)_{i,w}many ways (because (11) shows that the number of walks fromiitowwthat have lengthg−1g-1is(Cg−1)i,w\left(C^{g-1}\right)_{i,w}).
•We have now determined all but the last vertex and all but the last arc of our walk𝐰\mathbf{w}. We set the last vertexvgv_{g}of our walk to bejj. (This is the only possible option, since our walk𝐰\mathbf{w}has to be a walk fromiitojj.)
•We choose the last arcaga_{g}of our walk𝐰\mathbf{w}. This arcaga_{g}must have sourcevg−1v_{g-1}and targetvgv_{g}; in other words, it must have sourcewwand targetjj(sincevg−1=wv_{g-1}=wandvg=jv_{g}=j). In other words, it must be an arc fromwwtojj. Thus, it can be chosen inCw,jC_{w,j}many ways (because (7) shows that the number of arcs fromwwtojjisCw,jC_{w,j}).

Conversely, of course, this algorithm always constructs a walk fromiitojjthat has lengthgg, and different choices in the algorithm lead to distinct walks. Thus, the total number of walks fromiitojjthat have lengthggequals the total number of choices in the algorithm. But the latter number is∑w∈V(Cg−1)i,wCw,j\sum\limits_{w\in V}\left(C^{g-1}\right)_{i,w}C_{w,j}(since the algorithm first chooses aw∈Vw\in V, then involves a step with(Cg−1)i,w\left(C^{g-1}\right)_{i,w}choices, and then involves a step withCw,jC_{w,j}choices). Hence, the total number of walks fromiitojjthat have lengthggis∑w∈V(Cg−1)i,wCw,j\sum\limits_{w\in V}\left(C^{g-1}\right)_{i,w}C_{w,j}. In other words,

(the number of walks fromitojthat have lengthg)=∑w∈V(Cg−1)i,wCw,j.\left(\text{the number of walks from }i\text{ to }j\text{ that have length }g\right)=\sum\limits_{w\in V}\left(C^{g-1}\right)_{i,w}C_{w,j}.Comparing this with

(Cg⏟=Cg−1C)i,j\displaystyle\left(\underbrace{C^{g}}_{=C^{g-1}C}\right)_{i,j}=(Cg−1C)i,j=∑w=1n(Cg−1)i,wCw,j\displaystyle=\left(C^{g-1}C\right)_{i,j}=\sum\limits_{w=1}^{n}\left(C^{g-1}\right)_{i,w}C_{w,j}(by (8) (applied toM=Cg−1andN=C))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{sol.graph.count-walks.mmult}) (applied to }M=C^{g-1}\text{ and }N=C\text{)}\right)=∑w∈V(Cg−1)i,wCw,j(since{1,2,…,n}=V),\displaystyle=\sum\limits_{w\in V}\left(C^{g-1}\right)_{i,w}C_{w,j}\ \ \ \ \ \ \ \ \ \ \left(\text{since }\left\{1,2,\ldots,n\right\}=V\right),we obtain

(Cg)i,j=(the number of walks fromitojthat have lengthg).\left(C^{g}\right)_{i,j}=\left(\text{the number of walks from }i\text{ to }j\text{ that have length }g\right).(12) Now, forget that we fixediiandjj. We thus have proven (12) for anyi∈Vi\in Vandj∈Vj\in V. In other words, Claim 1 holds fork=gk=g. Thus, the induction step is complete. Hence, Claim 1 is proven by induction.]

Theorem4.38follows immediately from Claim 1.

Exercise 4.41.

LetEEbe the following multidigraph:

Letk∈ℕk\in\mathbb{N}. Compute the number of walks from11to11having lengthkk.

4.5.6Walks and bidirectionalization

Let us take a look at what bidirectionalization (i.e., the operationG↦GbidirG\mapsto G^{\operatorname*{bidir}}that sends a multigraphGGto the multidigraphGbidirG^{\operatorname*{bidir}}) does to walks, paths, closed walks and cycles:

Proposition 4.42.

LetGGbe a multigraph. Then:

(a)The walks ofGGare “more or less the same as” the walks of the multidigraphGbidirG^{\operatorname*{bidir}}. More precisely, each walk ofGGgives rise to a walk ofGbidirG^{\operatorname*{bidir}}(with the same starting point and the same ending point), and conversely, each walk ofGbidirG^{\operatorname*{bidir}}gives rise to a walk ofGG. IfGGhas no loops, then this is a one-to-one correspondence (i.e., a bijection) between the walks ofGGand the walks ofGbidirG^{\operatorname*{bidir}}.
(b)The paths ofGGare “more or less the same as” the paths of the multidigraphGbidirG^{\operatorname*{bidir}}. This is always a one-to-one correspondence, since paths cannot contain loops.
(c)The closed walks ofGGare “more or less the same as” the closed walks of the multidigraphGbidirG^{\operatorname*{bidir}}.
(d)The cycles ofGGare not quite the same as the cycles ofGbidirG^{\operatorname*{bidir}}. In fact, ifeeis an edge ofGGwith two distinct endpointsuuandvv, then(u,e,v,e,u)\left(u,e,v,e,u\right)is not a cycle ofGG, but either(u,(e,1),v,(e,2),u)\left(u,\left(e,1\right),v,\left(e,2\right),u\right)or(u,(e,2),v,(e,1),u)\left(u,\left(e,2\right),v,\left(e,1\right),u\right)is a cycle ofGbidirG^{\operatorname*{bidir}}(this is best seen on a picture:GGhas the edgeuuvveewhereasGbidirG^{\operatorname*{bidir}}has the arc-pairuuvv(e,1)\left(e,1\right)(e,2)\left(e,2\right)), soGbidirG^{\operatorname*{bidir}}usually has more cycles thanGGhas. But it is true that each cycle ofGGgives rise to a cycle ofGbidirG^{\operatorname*{bidir}}.

Proof 4.43.

This is all straightforward.

(a)To transform a walk𝐰\mathbf{w}ofGGinto a walk𝐰′\mathbf{w}^{\prime}ofGbidirG^{\operatorname*{bidir}}(with the same vertices), you must replace each edgeeeof𝐰\mathbf{w}by one of the two arcs(e,1)\left(e,1\right)or(e,2)\left(e,2\right)ofGbidirG^{\operatorname*{bidir}}. (Unlesseeis a loop, only one of these two arcs will do the job, since its direction must match the direction in which𝐰\mathbf{w}uses the edgeee. Wheneeis a loop, however, you get to choose between(e,1)\left(e,1\right)and(e,2)\left(e,2\right).)

Conversely, to transform a walk ofGbidirG^{\operatorname*{bidir}}into a walk ofGG, you must replace each arc(e,i)\left(e,i\right)(withi∈{1,2}i\in\left\{1,2\right\}) by the corresponding edgeeeofGG.

Parts(b)and(c)easily follow from part(a)(indeed, the correspondences between the paths and between the closed walks ofGGand ofGbidirG^{\operatorname*{bidir}}are just restrictions of the correspondence between the walks explained in part(a)). Part(d)is just as easy.

4.6Connectedness strong and weak

4.6.1Strong connectivity and strong components

We defined the “path-connected” relation for undirected graphs using the existence of walks (see Definition2.71). For a digraph, however, the relations “there is a walk fromuutovv” and “there is a walk fromvvtouu” are (in general) distinct and non-symmetric, so I prefer not to give them a symmetric-looking symbol such as≃D\simeq_{D}. Instead, we definestrong path-connectednessto mean the existence ofbothwalks:

Definition 4.44.

LetDDbe a multidigraph. We define a binary relation≃D\simeq_{D}on the setV(D)\operatorname*{V}\left(D\right)as follows: For two verticesuuandvvofDD, we shall haveu≃Dvu\simeq_{D}vif and only if there exists a walk fromuutovvinDDand there exists a walk fromvvtouuinDD.

This binary relation≃D\simeq_{D}is called “strong path-connectedness”. When two verticesuuandvvsatisfyu≃Dvu\simeq_{D}v, we say that “uuandvvarestrongly path-connected”.

Example 4.45.

LetDDbe as in Example4.25. Then,1≃D21\simeq_{D}2, because there exists a walk from11to22inDD(for instance,(1,a,2)\left(1,a,2\right)) and there also exists a walk from22to11inDD(for instance,(2,b,3,d,1)\left(2,b,3,d,1\right)). However, we don’t have3≃D43\simeq_{D}4. Indeed, while there exists a walk from33to44inDD, there exists no walk from44to33inDD.

Proposition 4.46.

LetDDbe a multidigraph. Then, the relation≃D\simeq_{D}is an equivalence relation.

Proof 4.47.

Easy, like for simple graphs.

Again, we can replace “walk” by “path” in the definition of the relation≃D\simeq_{D}:

Proposition 4.48.

LetDDbe a multidigraph. Letuuandvvbe two vertices ofDD. Then,u≃Dvu\simeq_{D}vif and only if there exist a path fromuutovvand a path fromvvtouu.

Proof 4.49.

Easy, like for simple graphs.

Definition 4.50.

LetDDbe a multidigraph. The equivalence classes of the equivalence relation≃D\simeq_{D}are called thestrong componentsofDD.

Definition 4.51.

LetDDbe a multidigraph. We say thatDDisstrongly connectedifDDhas exactly one strong component.

Thus, a multidigraphDDis strongly connected if and only if it has at least one vertex and there is a path from any vertex to any vertex.

Example 4.52.

IfDDis as in Example4.25, then the strong components ofDDare{1,2,3}\left\{1,2,3\right\}and{4}\left\{4\right\}.

In analogy to Proposition2.84, strong components have the following property:

Proposition 4.53.

LetDDbe a multidigraph. LetCCbe a strong component ofDD.

Then, the multidigraphD⁡[C]D\left[C\right](that is, the induced subdigraph ofDDon the setCC) is strongly connected.

Proof 4.54.

Just as in the above proof of Proposition2.84, we can see thatD⁡[C]D\left[C\right]has at least11strong component. It remains to show thatD⁡[C]D\left[C\right]has no more than11component. In other words, it remains to show that any two vertices ofD⁡[C]D\left[C\right]are strongly path-connected inD⁡[C]D\left[C\right].

So letuuandvvbe two vertices ofD⁡[C]D\left[C\right]. Then,u,v∈Cu,v\in C, and thereforeu≃Dvu\simeq_{D}v(sinceCCis a component ofDD). In other words, there exists a walk𝐮=(u0,u1,…,uk)\mathbf{u}=\left(u_{0},u_{1},\ldots,u_{k}\right)fromuutovvinDD, and there exists a walk𝐯=(v0,v1,…,vℓ)\mathbf{v}=\left(v_{0},v_{1},\ldots,v_{\ell}\right)fromvvtouuinDD. We shall now prove that these walks𝐮\mathbf{u}and𝐯\mathbf{v}actually are walks ofD⁡[C]D\left[C\right].

In fact: Ifuiu_{i}is any vertex of𝐮\mathbf{u}, then

•there is a walk fromuutouiu_{i}inDD(namely, the first part(u0,u1,…,ui)\left(u_{0},u_{1},\ldots,u_{i}\right)of𝐮\mathbf{u}), and
•there is a walk fromuiu_{i}touuinDD(namely, the second part(ui,ui+1,…,uk)\left(u_{i},u_{i+1},\ldots,u_{k}\right)of𝐮\mathbf{u}, spliced with the walk𝐯\mathbf{v}),

and therefore we haveu≃Duiu\simeq_{D}u_{i}, so thatuiu_{i}belongs to the same strong component ofDDasuu; but that strong component isCC. Thus, we have shown that each vertexuiu_{i}of𝐮\mathbf{u}belongs toCC. Therefore,𝐮\mathbf{u}is a walk of the digraphD⁡[C]D\left[C\right]. Likewise,𝐯\mathbf{v}is a walk ofD⁡[C]D\left[C\right]as well. Hence, the digraphD⁡[C]D\left[C\right]has a walk fromuutovv(namely,𝐮\mathbf{u}) and a walk fromvvtouu(namely,𝐯\mathbf{v}). In other words,u≃D⁡[C]vu\simeq_{D\left[C\right]}v.

We have now proved thatu≃D⁡[C]vu\simeq_{D\left[C\right]}vfor any two verticesuuandvvofD⁡[C]D\left[C\right]. Hence, the relation≃D⁡[C]\simeq_{D\left[C\right]}has no more than11equivalence class. In other words, the digraphD⁡[C]D\left[C\right]has no more than11strong component. This completes our proof.

4.6.2Weak connectivity and weak components

In comparison, here is a weaker notion of connected components and connectedness:

Definition 4.55.

LetDDbe a multidigraph. Consider its underlying undirected multigraphDundD^{\operatorname*{und}}. The components of this undirected multigraphDundD^{\operatorname*{und}}(that is, the equivalence classes of the equivalence relation≃Dund\simeq_{D^{\operatorname*{und}}}) are called theweak componentsofDD. We say thatDDisweakly connectedifDDhas exactly one weak component (i.e., ifDundD^{\operatorname*{und}}is connected).

Example 4.56.

LetDDbe the following simple digraph:

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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.0}{1.0}{-1.0}{0.0}{79.66812pt}{30.04802pt}\lxSVG@begingroup@{transform=matrix(0.0 1.0 -1.0 0.0 110.24 41.58)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .We treatDDas a multidigraph (namely,DmultD^{\operatorname*{mult}}).

The weak components ofDDare{1,2,3,4,5}\left\{1,2,3,4,5\right\}and{6,7}\left\{6,7\right\}.

The strong components ofDDare{1}\left\{1\right\},{2}\left\{2\right\},{3,4,5}\left\{3,4,5\right\},{6}\left\{6\right\}and{7}\left\{7\right\}. (Indeed, for example, we have1≄D2≄D31\not\simeq_{D}2\not\simeq_{D}3but3≃D4≃D53\simeq_{D}4\simeq_{D}5.)

SoDDis neither strongly nor weakly connected, but has more strong than weak components.

Example 4.57.

The digraph from Example4.22is weakly connected, but not at all strongly connected (indeed, each of its strong components has size11). The digraph from Example4.23, on the other hand, is strongly connected.

Proposition 4.58.

Any strongly connected digraph is weakly connected.

Proof 4.59.

LetDDbe a multidigraph. Then, any walk ofDDis (or, more precisely, gives rise to) a walk ofDundD^{\operatorname*{und}}. Hence, if two verticesuuandvvofDDare strongly path-connected inDD, then they are path-connected inDundD^{\operatorname*{und}}. Therefore, ifDDis strongly connected, thenDundD^{\operatorname*{und}}is connected, but this means thatDDis weakly connected.

Exercise 4.60.

LetDDbe a multidigraph. Prove that the strong components ofDDare the weak components ofDDif and only if each arc ofDDis contained in at least one cycle.

4.6.3Sink components

Let us come back to strong components for a little sidenote. Not all strong components are alike; some have special properties:

Definition 4.61.

LetDDbe a multidigraph. A strong componentCCofDDis said to be asink componentif it has the following property: Ifaais an arc ofDDwhose source lies inCC, then the target ofaaalso lies inCC.

Informally, a sink component is a strong component “with no way out” (i.e., no arcs leading out of it).

Example 4.62.

IfDDis the digraph from Example4.56, then the sink components ofDDare{3,4,5}\left\{3,4,5\right\}and{7}\left\{7\right\}.

Example 4.63.

LetDDbe the following simple digraph:

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\hss}}\lxSVG@closescope\endpgfpicture}}\ \ .We treatDDas a multidigraph (namely,DmultD^{\operatorname*{mult}}). Its strong components are{1,3},{2},{4,8},{5,6,7},{9}\left\{1,3\right\},\ \left\{2\right\},\ \left\{4,8\right\},\ \left\{5,6,7\right\},\ \left\{9\right\}. Among these five strong components, the only sink components are{5,6,7}\left\{5,6,7\right\}and{9}\left\{9\right\}. For example,{4,8}\left\{4,8\right\}is not a sink component, since the arc8989has source in{4,8}\left\{4,8\right\}but target not in{4,8}\left\{4,8\right\}.

Theorem 4.64.

LetDDbe a multidigraph with at least one vertex. Then,DDhas at least one sink component.

Proof 4.65.

Write the multidigraphDDasD=(V,A,ψ)D=\left(V,A,\psi\right). For each vertexv∈Vv\in V, defineDescv\operatorname*{Desc}vto be the set of all verticesw∈Vw\in Vsuch thatDDhas a walk fromvvtoww. (These vertices are sometimes called the “descendants” ofvv; thus the notationDescv\operatorname*{Desc}v.) Note that the existence of a walk fromvvtowwis easily testable algorithmically (see Subsection4.5.3).

The following is easy to see:

Claim 1:Letp,q∈Vp,q\in Vbe two vertices such thatq∈Descpq\in\operatorname*{Desc}p. Then,Descq⊆Descp\operatorname*{Desc}q\subseteq\operatorname*{Desc}p.

Proof 4.66(Proof of Claim 1.).

Letw∈Descqw\in\operatorname*{Desc}q. We must show thatw∈Descpw\in\operatorname*{Desc}p.

We haveq∈Descpq\in\operatorname*{Desc}p; in other words,DDhas a walk frompptoqq(by the definition ofDescp\operatorname*{Desc}p). Likewise, fromw∈Descqw\in\operatorname*{Desc}q, we conclude thatDDhas a walk fromqqtoww. Let us denote these two walks by𝐚\mathbf{a}and𝐛\mathbf{b}. Splicing these two walks together (using Proposition4.26, applied tou=pu=pandv=qv=q), we obtain a walk𝐚∗𝐛\mathbf{a}\ast\mathbf{b}frompptoww. Hence,DDhas a walk frompptoww. In other words, we havew∈Descpw\in\operatorname*{Desc}p(by the definition ofDescp\operatorname*{Desc}p). This proves Claim 1.

Now, recall that the setVVis nonempty (sinceDDhas at least one vertex) and finite. Among all the verticesv∈Vv\in V, we pick one vertexvvfor which the size|Descv|\left|\operatorname*{Desc}v\right|is minimal. Thus,

|Descv|≤|Descw|for eachw∈V.\left|\operatorname*{Desc}v\right|\leq\left|\operatorname*{Desc}w\right|\ \ \ \ \ \ \ \ \ \ \text{for each }w\in V.(13) Now, letCCbe the strong component ofDDthat containsvv. We claim thatCCis a sink component.

To prove this, we consider an arbitrary arcaawhose source lies inCC. Letppbe this source, and letqqbe the target ofaa. Thus,(p,a,q)\left(p,a,q\right)is a walk frompptoqq. Hence,DDhas a walk frompptoqq; in other words,q∈Descpq\in\operatorname*{Desc}p(by the definition ofDescp\operatorname*{Desc}p). Thus, Claim 1 yields

Descq⊆Descp.\operatorname*{Desc}q\subseteq\operatorname*{Desc}p. On the other hand,ppis the source ofaaand thus lies inCC(since we said that the source ofaalies inCC). In other words,pplies in the same strong component ofDDasvv(sinceCCis the strong component ofDDthat containsvv). In other words,p≃Dvp\simeq_{D}v(since the strong components ofDDare the equivalence classes of≃D\simeq_{D}). In other words, there exists a walk frompptovvinDDand there exists a walk fromvvtoppinDD(by the definition of≃D\simeq_{D}). In particular,DDhas a walk fromvvtopp; in other words,p∈Descvp\in\operatorname*{Desc}v. Thus, Claim 1 (applied tovvandppinstead ofppandqq) yieldsDescp⊆Descv\operatorname*{Desc}p\subseteq\operatorname*{Desc}v. Hence,

|Descp|≤|Descv|≤|Descq|(by (13), applied tow=q).\left|\operatorname*{Desc}p\right|\leq\left|\operatorname*{Desc}v\right|\leq\left|\operatorname*{Desc}q\right|\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.mdg.sinkcomp.min}), applied to }w=q\right). However, if two finite setsXXandYYsatisfyX⊆YX\subseteq Yand|Y|≤|X|\left|Y\right|\leq\left|X\right|, thenX=YX=Y(because otherwise,XXwould be a proper subset ofYYand thus satisfy|X|<|Y|\left|X\right|<\left|Y\right|). Applying this fundamental fact toX=DescqX=\operatorname*{Desc}qandY=DescpY=\operatorname*{Desc}p, we conclude thatDescq=Descp\operatorname*{Desc}q=\operatorname*{Desc}p(sinceDescq⊆Descp\operatorname*{Desc}q\subseteq\operatorname*{Desc}pand|Descp|≤|Descq|\left|\operatorname*{Desc}p\right|\leq\left|\operatorname*{Desc}q\right|).

ButDDhas a walk frompptopp(namely, the trivial walk(p)\left(p\right)); in other words,p∈Descpp\in\operatorname*{Desc}p. In other words,p∈Descqp\in\operatorname*{Desc}q(sinceDescq=Descp\operatorname*{Desc}q=\operatorname*{Desc}p). Hence,DDhas a walk fromqqtopp. SinceDDalso has a walk frompptoqq, we thus conclude that there exist both a walk frompptoqqand a walk fromqqtoppinDD. In other words,p≃Dqp\simeq_{D}q(by the definition of≃D\simeq_{D}). Hence,ppandqqlie in the same strong component ofDD(since the strong components ofDDare the equivalence classes of≃D\simeq_{D}). Thus, fromp∈Cp\in C, we obtainq∈Cq\in C(sinceCCis a strong component). In other words, the target ofaalies inCC(sinceqqis the target ofaa).

Forget that we fixedaa. We thus have shown that ifaais an arc ofDDwhose source lies inCC, then the target ofaaalso lies inCC. In other words,CCis a sink component ofDD. Hence,DDhas a sink component, qed.

4.6.4Exercises

Exercise 4.67.

LetD=(V,E,ψ)D=\left(V,E,\psi\right)be a multidigraph.

LetAA,BBandCCbe three subsets ofVVsuch that the induced subdigraphsD⁡[A]D\left[A\right],D⁡[B]D\left[B\right]andD⁡[C]D\left[C\right]are strongly connected.

A cycle ofDDwill be calledeclecticif it contains at least one arc ofD⁡[A]D\left[A\right], at least one arc ofD⁡[B]D\left[B\right]and at least one arc ofD⁡[C]D\left[C\right](although these three arcs are not required to be distinct).

Prove the following:

(a)If the setsB∩CB\cap C,C∩AC\cap AandA∩BA\cap Bare nonempty, butA∩B∩CA\cap B\cap Cis empty, thenDDhas an eclectic cycle.
(b)If the induced subdigraphsD⁡[B∩C]D\left[B\cap C\right],D⁡[C∩A]D\left[C\cap A\right]andD⁡[A∩B]D\left[A\cap B\right]are strongly connected, but the induced subdigraphD⁡[A∩B∩C]D\left[A\cap B\cap C\right]is not strongly connected, thenDDhas an eclectic cycle.

[Note:Keep in mind that the multidigraph with00vertices does not count as strongly connected.]

[Solution:This is a generalization of Exercise 7 on midterm #2 from my Spring 2017 course; seethe course pagefor solutions.]

4.7Eulerian walks and circuits

We have studied Eulerian walks and circuits for (undirected) multigraphs in Section3.4. Let us now define analogous concepts for multidigraphs:

Definition 4.68.

LetDDbe a multidigraph.

(a)A walk ofDDis said to beEulerianif each arc ofDDappears exactly once in this walk. (In other words: A walk(v0,a1,v1,a2,v2,…,ak,vk)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)ofDDis said to beEulerianif for each arcaaofDD, there exists exactly onei∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}such thata=aia=a_{i}.)
(b)AnEulerian circuitofDDmeans a circuit (i.e., closed walk) ofDDthat is Eulerian.

The Euler–Hierholzer theorem gives a necessary and sufficient criterion for a multigraph to have an Eulerian circuit or walk. For multidigraphs, there is an analogous result:

Theorem 4.69(diEuler, diHierholzer).

LetDDbe a weakly connected multidigraph. Then:

(a)The multidigraphDDhas an Eulerian circuit if and only if each vertexvvofDDsatisfiesdeg+⁡v=deg−⁡v\deg^{+}v=\deg^{-}v.
(b)The multidigraphDDhas an Eulerian walk if and only if all but two verticesvvofDDsatisfydeg+⁡v=deg−⁡v\deg^{+}v=\deg^{-}v, and the remaining two verticesvvsatisfy|deg+⁡v−deg−⁡v|≤1\left|\deg^{+}v-\deg^{-}v\right|\leq 1.

Exercise 4.70.

Prove Theorem4.69.

Incidentally, the “each vertexvvofDDsatisfiesdeg+⁡v=deg−⁡v\deg^{+}v=\deg^{-}v” condition has a name:

Definition 4.71.

A multidigraphDDis said to bebalancedif each vertexvvofDDsatisfiesdeg+⁡v=deg−⁡v\deg^{+}v=\deg^{-}v.

So balancedness is necessary and sufficient for the existence of an Eulerian circuit in a weakly connected multidigraph.

The following proposition is obvious:

Proposition 4.72.

LetGGbe a multigraph. Then, the multidigraphGbidirG^{\operatorname*{bidir}}is balanced.

Proof 4.73.

The definition ofGbidirG^{\operatorname*{bidir}}yields that each vertexvvofGbidirG^{\operatorname*{bidir}}satisfiesdeg+⁡v=deg⁡v\deg^{+}v=\deg vanddeg−⁡v=deg⁡v\deg^{-}v=\deg v, wheredeg⁡v\deg vdenotes the degree ofvvas a vertex ofGG. Hence, each vertexvvofGbidirG^{\operatorname*{bidir}}satisfiesdeg+⁡v=deg⁡v=deg−⁡v\deg^{+}v=\deg v=\deg^{-}v. In other words,GbidirG^{\operatorname*{bidir}}is balanced.

Combining this proposition with Theorem4.69(a), we can obtain a curious fact about undirected(!) multigraphs:

Theorem 4.74.

LetGGbe a connected multigraph. Then, the multidigraphGbidirG^{\operatorname*{bidir}}has an Eulerian circuit. In other words, there is a circuit ofGGthat contains each edgeexactly twice, and uses it once in each direction.

Proof 4.75.

The multidigraphGbidirG^{\operatorname*{bidir}}is balanced (by Proposition4.72) and weakly connected (this follows easily from the connectedness ofGG). Hence, Theorem4.69(a)can be applied toD=GbidirD=G^{\operatorname*{bidir}}. Thus,GbidirG^{\operatorname*{bidir}}has an Eulerian circuit. Reinterpreting this circuit as a circuit ofGG, we obtain a circuit ofGGthat contains each edgeexactly twice, and uses it once in each direction. This proves Theorem4.74.

Example 4.76.

LetGGbe the multigraph

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36.94 44.81)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -50.28 0 L 50.28 0}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.16377pt}{-8.23856pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -2.99 -11.4)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 75.7 0 L 184.14 0}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{91.26337pt}{5.87744pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 126.28 8.13)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This graphGGhas no Eulerian circuit, nor even an Eulerian walk (since the vertices0,1,3,40,1,3,4have odd degrees). But Theorem4.74shows that it has a circuit that contains each edgeexactly twice, and uses it once in each direction. And indeed, for example,

(0,x,1,a,2,b,3,y,4,y,3,b,2,a,1,c,3,c,1,x,0)\left(0,x,1,a,2,b,3,y,4,y,3,b,2,a,1,c,3,c,1,x,0\right)is such a circuit.

The following exercise provides yet another analogue of the Euler–Hierholzer theorem (Theorem3.54(a)to be specific), this time not for multidigraphs but for multigraphs with colored edges:

Exercise 4.77.

LetGGbe a connected multigraph. Assume that each edge ofGGhas been colored either red or blue. For any vertexvvofGG, we letdegr⁡v\deg^{r}vdenote the number of red edges containingvv(where red loops are counted twice), and we letdegb⁡v\deg^{b}vdenote the number of blue edges containingvv(where blue loops are counted twice). A walk𝐰\mathbf{w}ofGGis said to bealternatingif no two consecutive edges of𝐰\mathbf{w}have the same color (i.e., if its edges alternate between red and blue, starting with either red or blue). An alternating walk𝐰\mathbf{w}ofGGis said to befully alternatingif the first and the last edges of𝐰\mathbf{w}have different colors (or else if𝐰\mathbf{w}has no edges).

Prove that the multigraphGGhas a fully alternating Eulerian circuit if and only if each vertexvvofGGsatisfiesdegr⁡v=degb⁡v\deg^{r}v=\deg^{b}v.

An example of the situation described in this exercise is shown in the following picture:

1122334455aabbccddeeffgghh(where the edgesa,c,e,ga,c,e,gare colored blue while the edgesb,d,f,hb,d,f,hare colored red). In this example,(1,a,2,b,3,g,4,f,5,e,2,h,2,c,3,d,1)\left(1,a,2,b,3,g,4,f,5,e,2,h,2,c,3,d,1\right)is a fully alternating Eulerian circuit.

4.8Hamiltonian cycles and paths

We can define Hamiltonian paths and cycles for simple digraphs in the same way as we defined them for simple graphs:

Definition 4.78.

LetD=(V,A)D=\left(V,A\right)be a simple digraph.

(a)AHamiltonian pathinDDmeans a walk ofDDthat contains each vertex ofDDexactly once. Obviously, it is a path.
(b)AHamiltonian cycleinDDmeans a cycle(v0,v1,…,vk)\left(v_{0},v_{1},\ldots,v_{k}\right)ofDDsuch that each vertex ofDDappears exactly once amongv0,v1,…,vk−1v_{0},v_{1},\ldots,v_{k-1}.

Convention 4.8.1.

In the following, we will abbreviate:

•“Hamiltonian path” as “hamp”;
•“Hamiltonian cycle” as “hamc”.

Hamps and hamcs for digraphs have some properties analogous to those for simple graphs. In particular, there is an analogue of Ore’s theorem:

Theorem 4.79(Meyniel).

LetD=(V,A)D=\left(V,A\right)be a strongly connected loopless simple digraph withnnvertices. Assume that for each pair(u,v)∈V×V\left(u,v\right)\in V\times Vof two verticesuuandvvsatisfyingu≠vu\neq vand(u,v)∉A\left(u,v\right)\notin Aand(v,u)∉A\left(v,u\right)\notin A, we havedeg⁡u+deg⁡v≥2n−1\deg u+\deg v\geq 2n-1. Here,deg⁡w\deg wmeansdeg+⁡w+deg−⁡w\deg^{+}w+\deg^{-}w. Then,DDhas a hamc.

This is known asMeyniel’s theorem. For its proof (which is far more complicated than the proof of Ore’s theorem), see[BonTho77]or[Berge91, §10.3, Theorem 7]or[BerLiu98](which shows an even more general result). Note that the “strongly connected” condition is needed.

4.9The reverse and complement digraphs

We take a break from studying hamps (Hamiltonian paths) in order to introduce two more operations on simple digraphs.

Definition 4.80.

LetD=(V,A)D=\left(V,A\right)be a simple digraph. Then:

(a)The elements of(V×V)∖A\left(V\times V\right)\setminus Awill be called thenon-arcsofDD.
(b)Thereversalof a pair(i,j)∈V×V\left(i,j\right)\in V\times Vmeans the pair(j,i)\left(j,i\right).
(c)We defineDrevD^{\operatorname*{rev}}as the simple digraph(V,Arev)\left(V,A^{\operatorname*{rev}}\right), where Arev={(j,i)∣(i,j)∈A}.A^{\operatorname*{rev}}=\left\{\left(j,i\right)\ \mid\ \left(i,j\right)\in A\right\}.Thus,DrevD^{\operatorname*{rev}}is the digraph obtained fromDDby reversing each arc (i.e., swapping its source and its target). This is called thereversalofDD.
(d)We defineD¯\overline{D}as the simple digraph(V,(V×V)∖A)\left(V,\ \ \left(V\times V\right)\setminus A\right). This is the digraph that has the same vertices asDD, but whose arcs are precisely the non-arcs ofDD. This digraphD¯\overline{D}is called thecomplementofDD.

Example 4.81.

Let

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Then,

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\lxSVG@transformcm{0.96593}{-0.25882}{0.25882}{0.96593}{-10.41837pt}{2.79158pt}\lxSVG@begingroup@{transform=matrix(0.96593 -0.25882 0.25882 0.96593 -14.42 3.86)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .

Convention 4.9.1.

In the following, the symbol # means “number”. For example,

(# of subsets of{1,2,3})=8.\left(\text{\# of subsets of }\left\{1,2,3\right\}\right)=8.

We shall now try to count hamps in simple digraphs282828See[17s-lec7]for a more detailed treatment of this topic.. As a warmup, here is a particularly simple case:

Proposition 4.82.

LetDDbe the simple digraph(V,A)\left(V,A\right), where

V={1,2,…,n}for somen∈ℕ,V=\left\{1,2,\ldots,n\right\}\ \ \ \ \ \ \ \ \ \ \text{for some }n\in\mathbb{N},and where

A={(i,j)∣i<j}.A=\left\{\left(i,j\right)\ \mid\ i<j\right\}.Then,(# of hamps ofD)=1\left(\text{\# of hamps of }D\right)=1.

Proof 4.83.

It is easy to see that the only hamp ofDDis(1,2,…,n)\left(1,2,\ldots,n\right).

The following is easy, too:

Proposition 4.84.

LetDDbe a simple digraph. Then,

(# of hamps ofDrev)=(# of hamps ofD).\left(\text{\# of hamps of }D^{\operatorname*{rev}}\right)=\left(\text{\# of hamps of }D\right).

Proof 4.85.

The hamps ofDrevD^{\operatorname*{rev}}are obtained from the hamps ofDDby walking backwards.

So far, so boring. What about this:

Theorem 4.86(Berge’s theorem).

LetDDbe a simple digraph. Then,

(# of hamps ofD¯)≡(# of hamps ofD)mod2.\left(\text{\# of hamps of }\overline{D}\right)\equiv\left(\text{\# of hamps of }D\right)\operatorname{mod}2.

This is much less obvious or even expected. We first give an example:

Example 4.87.

LetDDbe the following digraph:

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90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.64 0 C 169.64 6.71 164.19 12.16 157.48 12.16 C 150.77 12.16 145.32 6.71 145.32 0 C 145.32 -6.71 150.77 -12.16 157.48 -12.16 C 164.19 -12.16 169.64 -6.71 169.64 0 Z M 157.48 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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.94 4.35 C 32.04 11.66 46.7 11.66 65.24 4.92}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.9397}{-0.34203}{0.34203}{0.9397}{47.70976pt}{3.34702pt}\lxSVG@begingroup@{transform=matrix(0.9397 -0.34203 0.34203 0.9397 66.02 4.63)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.8 -4.35 C 46.7 -11.66 32.04 -11.66 13.5 -4.92}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} 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\lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This digraph has33hamps:(1,2,3)\left(1,2,3\right)and(2,3,1)\left(2,3,1\right)and(3,1,2)\left(3,1,2\right).

Its complementD¯\overline{D}looks as follows:

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-12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.64 0 C 169.64 6.71 164.19 12.16 157.48 12.16 C 150.77 12.16 145.32 6.71 145.32 0 C 145.32 -6.71 150.77 -12.16 157.48 -12.16 C 164.19 -12.16 169.64 -6.71 169.64 0 Z M 157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{111.31104pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 154.02 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{} {\lx@inpgf@ignorespaces{}{}{{}}{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{}{}\lx@inpgf@ignorespaces}{{}} {\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{}{{}}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{{{{{{}}{}{}{}{}{{}}}}}{}{}{}{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -12.28 -3.29 C -32.1 -8.6 -32.1 8.6 -15.22 4.08}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.96593}{-0.25882}{0.25882}{0.96593}{-10.41837pt}{2.79158pt}\lxSVG@begingroup@{transform=matrix(0.96593 -0.25882 0.25882 0.96593 -14.42 3.86)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{} {\lx@inpgf@ignorespaces{}{}{{}}{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{}{}\lx@inpgf@ignorespaces}{{}} {\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{}{{}}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{{{{{{}}{}{}{}{}{{}}}}}{}{}{}{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.46 -3.29 C 46.64 -8.6 46.64 8.6 63.52 4.08}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.96593}{-0.25882}{0.25882}{0.96593}{46.48715pt}{2.79158pt}\lxSVG@begingroup@{transform=matrix(0.96593 -0.25882 0.25882 0.96593 64.32 3.86)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{} {\lx@inpgf@ignorespaces{}{}{{}}{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{}{}\lx@inpgf@ignorespaces}{{}} {\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{}{{}}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{{{{{{}}{}{}{}{}{{}}}}}{}{}{}{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.76 3.29 C 189.58 8.6 189.58 -8.6 172.7 -4.08}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{-0.96593}{0.25882}{-0.25882}{-0.96593}{124.2294pt}{-2.79158pt}\lxSVG@begingroup@{transform=matrix(-0.96593 0.25882 -0.25882 -0.96593 171.9 -3.86)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 144.77 0 L 93.11 0}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{-1.0}{0.0}{0.0}{-1.0}{66.69139pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(-1.0 0.0 0.0 -1.0 92.28 0)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.01 6.36 C 56.76 32.77 100.72 32.77 145.03 7.19}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.86603}{-0.5}{0.5}{0.86603}{105.3362pt}{4.89293pt}\lxSVG@begingroup@{transform=matrix(0.86603 -0.5 0.5 0.86603 145.75 6.77)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .It has only11hamp:(1,3,2)\left(1,3,2\right).

Thus, in this case, Theorem4.86says that1≡3mod⁡21\equiv 3\operatorname{mod}2.

Proof 4.88(Proof of Theorem4.86.).

(This is an outline; see[17s-lec7, proof of Theorem 1.3.6]for more details.)

Write the simple digraphDDasD=(V,A)D=\left(V,A\right), and assume WLOG thatV≠∅V\neq\varnothing. Setn=|V|n=\left|V\right|.

AVV-listingwill mean a list of elements ofVVthat contains each element ofVVexactly once. (Thus, eachVV-listing is annn-tuple, and there aren!n!manyVV-listings.) Note that aVV-listing is the same as a hamp of the “complete” digraph(V,V×V)\left(V,V\times V\right). Any hamp ofDDor ofD¯\overline{D}is therefore aVV-listing, but not everyVV-listing is a hamp ofDDorD¯\overline{D}.

Ifσ=(σ1,σ2,…,σn)\sigma=\left(\sigma_{1},\sigma_{2},\ldots,\sigma_{n}\right)is aVV-listing, then we define a set

P⁡(σ):={σ1σ2,σ2σ3,…,σn−1σn}.P\left(\sigma\right):=\left\{\sigma_{1}\sigma_{2},\ \sigma_{2}\sigma_{3},\ \ldots,\ \sigma_{n-1}\sigma_{n}\right\}.We call this setP⁡(σ)P\left(\sigma\right)thearc setofσ\sigma. When we regardσ\sigmaas a hamp of(V,V×V)\left(V,V\times V\right), this setP⁡(σ)P\left(\sigma\right)is just the set of all arcs ofσ\sigma. Note that this is an(n−1)\left(n-1\right)-element set. We make a few easy observations (prove them!):

Observation 1:We can reconstruct aVV-listingσ\sigmafrom its arc setP⁡(σ)P\left(\sigma\right). In other words, the mapσ↦P⁡(σ)\sigma\mapsto P\left(\sigma\right)is injective.

Observation 2:Letσ\sigmabe aVV-listing. Then,σ\sigmais a hamp ofDDif and only ifP⁡(σ)⊆AP\left(\sigma\right)\subseteq A.

Observation 3:Letσ\sigmabe aVV-listing. Then,σ\sigmais a hamp ofD¯\overline{D}if and only ifP⁡(σ)⊆(V×V)∖AP\left(\sigma\right)\subseteq\left(V\times V\right)\setminus A.

Now, letNNbe the # of pairs(σ,B)\left(\sigma,B\right), whereσ\sigmais aVV-listing andBBis a subset ofAAsatisfyingB⊆P⁡(σ)B\subseteq P\left(\sigma\right). Thus,

N=∑σis aV-listingNσ,N=\sum\limits_{\sigma\text{ is a }V\text{-listing}}N_{\sigma},where

Nσ=(# of subsetsBofAsatisfyingB⊆P(σ)).N_{\sigma}=\left(\text{\# of subsets }B\text{ of }A\text{ satisfying }B\subseteq P\left(\sigma\right)\right).But we also have

N=∑Bis a subset ofANB,N=\sum\limits_{B\text{ is a subset of }A}N^{B},where

NB=(# ofV-listingsσsatisfyingB⊆P(σ)).N^{B}=\left(\text{\# of }V\text{-listings }\sigma\text{ satisfying }B\subseteq P\left(\sigma\right)\right). Let us now relate these two sums to hamps. We begin with∑σis aV-listingNσ\sum\limits\limits_{\sigma\text{ is a }V\text{-listing}}N_{\sigma}.

We shall use theIverson bracket notation: i.e., the notation[𝒜]\left[\mathcal{A}\right]for the truth value of a statement𝒜\mathcal{A}. This truth value is defined to be the number11if𝒜\mathcal{A}is true, and00if𝒜\mathcal{A}is false. For instance,

[2+2=4]=1and[2+2=5]=0.\left[2+2=4\right]=1\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left[2+2=5\right]=0. For anyVV-listingσ\sigma, we have

Nσ\displaystyle N_{\sigma}=(# of subsetsBofAsatisfyingB⊆P(σ))\displaystyle=\left(\text{\# of subsets }B\text{ of }A\text{ satisfying }B\subseteq P\left(\sigma\right)\right)=(# of subsetsBofA∩P⁡(σ))\displaystyle=\left(\text{\# of subsets }B\text{ of }A\cap P\left(\sigma\right)\right)=2|A∩P⁡(σ)|\displaystyle=2^{\left|A\cap P\left(\sigma\right)\right|}≡[|A∩P(σ)|=0](since2m≡[m=0]mod2for eachm∈ℕ)\displaystyle\equiv\left[\left|A\cap P\left(\sigma\right)\right|=0\right]\ \ \ \ \ \ \ \ \ \ \left(\text{since }2^{m}\equiv\left[m=0\right]\operatorname{mod}2\text{ for each }m\in\mathbb{N}\right)=[A∩P(σ)=∅](since equivalent statements have thesame truth value)\displaystyle=\left[A\cap P\left(\sigma\right)=\varnothing\right]\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since equivalent statements have the}\\ \text{same truth value}\end{array}\right)=[P(σ)⊆(V×V)∖A](sinceP(σ)is always a subset ofV×V)\displaystyle=\left[P\left(\sigma\right)\subseteq\left(V\times V\right)\setminus A\right]\ \ \ \ \ \ \ \ \ \ \left(\text{since }P\left(\sigma\right)\text{ is always a subset of }V\times V\right)=[σis a hamp ofD¯]mod2(by Observation 3).\displaystyle=\left[\sigma\text{ is a hamp of }\overline{D}\right]\operatorname{mod}2\ \ \ \ \ \ \ \ \ \ \left(\text{by Observation 3}\right).So

N\displaystyle N=∑σis aV-listingNσ⏟≡[σis a hamp ofD¯]mod⁡2\displaystyle=\sum\limits_{\sigma\text{ is a }V\text{-listing}}\underbrace{N_{\sigma}}_{\equiv\left[\sigma\text{ is a hamp of }\overline{D}\right]\operatorname{mod}2}≡∑σis aV-listing[σis a hamp ofD¯]\displaystyle\equiv\sum\limits_{\sigma\text{ is a }V\text{-listing}}\left[\sigma\text{ is a hamp of }\overline{D}\right]=(# ofV-listingsσthat are hamps ofD¯)\displaystyle=\left(\text{\# of }V\text{-listings }\sigma\text{ that are hamps of }\overline{D}\right)(because∑σis aV-listing[σis a hamp ofD¯]is a sumof several1’s and several0’s, and the1’s in thissum correspond precisely totheV-listingsσthat are hamps ofD¯)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{because }\sum\limits_{\sigma\text{ is a }V\text{-listing}}\left[\sigma\text{ is a hamp of }\overline{D}\right]\text{ is a sum}\\ \text{of several }1\text{’s and several }0\text{’s, and the }1\text{’s in this}\\ \text{sum correspond precisely to}\\ \text{the }V\text{-listings }\sigma\text{ that are hamps of }\overline{D}\end{array}\right)=(# of hamps ofD¯)mod⁡2.\displaystyle=\left(\text{\# of hamps of }\overline{D}\right)\operatorname{mod}2. What about the other expression forNN? Recall that

N=∑Bis a subset ofANB,N=\sum\limits_{B\text{ is a subset of }A}N^{B},where

NB=(# ofV-listingsσsatisfyingB⊆P(σ)).N^{B}=\left(\text{\# of }V\text{-listings }\sigma\text{ satisfying }B\subseteq P\left(\sigma\right)\right).We want to prove that this sum equals(# of hamps ofD)\left(\text{\# of hamps of }D\right), at least modulo22.

So letBBbe a subset ofAA. We want to knowNBmod⁡2N^{B}\operatorname{mod}2. In other words, we want to know whenNBN^{B}is odd.

Let us first assume thatNBN^{B}is odd, and see what follows from this.

SinceNBN^{B}is odd, we haveNB>0N^{B}>0. Thus, there existsat least oneVV-listingσ\sigmasatisfyingB⊆P⁡(σ)B\subseteq P\left(\sigma\right). We shall now draw some conclusions from this.

First, a definition: Apath coverofVVmeans a set of paths in the “complete” digraph(V,V×V)\left(V,V\times V\right)such that each vertexv∈Vv\in Vis contained in exactly one of these paths. Theset of arcsof such a path cover is simply the set of all arcs of all its paths. For example, ifV={1,2,3,4,5,6,7}V=\left\{1,2,3,4,5,6,7\right\}, then

{(1,3,5),(2),(6),(7,4)}\left\{\left(1,3,5\right),\ \ \left(2\right),\ \ \left(6\right),\ \ \left(7,4\right)\right\}is a path cover ofVV, and its set of arcs is{13,35,74}\left\{13,\ 35,\ 74\right\}.

Now, ponder the following: If we remove an arcvivi+1v_{i}v_{i+1}from a path(v1,v2,…,vk)\left(v_{1},v_{2},\ldots,v_{k}\right), then this path breaks up into two paths(v1,v2,…,vi)\left(v_{1},v_{2},\ldots,v_{i}\right)and(vi+1,vi+2,…,vk)\left(v_{i+1},v_{i+2},\ldots,v_{k}\right). Thus, if we remove some arcs from the arc setP⁡(σ)P\left(\sigma\right)of aVV-listingσ\sigma, then we obtain the set of arcs of a path cover ofVV. (For instance, removing the arcs5252,2626and6767from the arc setP⁡(σ)P\left(\sigma\right)of theVV-listingσ=(1,3,5,2,6,7,4)\sigma=\left(1,3,5,2,6,7,4\right)yields precisely the path cover{(1,3,5),(2),(6),(7,4)}\left\{\left(1,3,5\right),\ \ \left(2\right),\ \ \left(6\right),\ \ \left(7,4\right)\right\}that we just showed as an example.)

Now, recall that there existsat least oneVV-listingσ\sigmasatisfyingB⊆P⁡(σ)B\subseteq P\left(\sigma\right). Hence,BBis obtained by removing some arcs from the arc setP⁡(σ)P\left(\sigma\right)of thisVV-listingσ\sigma. Therefore,BBis the set of arcs of a path cover ofVV(by the claim of the preceding paragraph). Let us say that this path cover consists of exactlyrrpaths. Then,

(# ofV-listingsσsatisfyingB⊆P⁡(σ))=r!,\left(\text{\# of }V\text{-listings }\sigma\text{ satisfying }B\subseteq P\left(\sigma\right)\right)=r!,because any suchVV-listingσ\sigmacan be constructed by concatenating therrpaths in our path cover in some order (and there arer!r!possible orders).

Thus,NB=(# ofV-listingsσsatisfyingB⊆P(σ))=r!N^{B}=\left(\text{\# of }V\text{-listings }\sigma\text{ satisfying }B\subseteq P\left(\sigma\right)\right)=r!. But we have assumed thatNBN^{B}is odd. Sor!r!is odd. Sincerris positive (becauseV≠∅V\neq\varnothing, so our path cover must contain at least one path), this entails thatr=1r=1. So our path cover is just a single path; this path is a path ofDD(since its set of arcsBBis a subset ofAA) and therefore is a hamp ofDD(since it constitutes a path cover ofVVall by itself). If we denote it byσ\sigma, then we haveB=P⁡(σ)B=P\left(\sigma\right)(sinceBBis the set of arcs of the path cover that consists ofσ\sigmaalone).

Forget our assumption thatNBN^{B}is odd. We have thus shown that ifNBN^{B}is odd, thenB=P⁡(σ)B=P\left(\sigma\right)for some hampσ\sigmaofDD.

Conversely, it is easy to see that ifB=P⁡(σ)B=P\left(\sigma\right)for some hampσ\sigmaofDD, thenNBN^{B}is odd (and actually equals11).

Combining these two results, we see thatNBN^{B}is oddif and only ifB=P⁡(σ)B=P\left(\sigma\right)for some hampσ\sigmaofDD. Therefore,

[NBis odd]=[B=P(σ)for some hampσofD].\left[N^{B}\text{ is odd}\right]=\left[B=P\left(\sigma\right)\text{ for some hamp }\sigma\text{ of }D\right].However,

NB\displaystyle N^{B}≡[NBis odd](sincem≡[mis odd]mod2for anym∈ℤ)\displaystyle\equiv\left[N^{B}\text{ is odd}\right]\ \ \ \ \ \ \ \ \ \ \left(\text{since }m\equiv\left[m\text{ is odd}\right]\operatorname{mod}2\text{ for any }m\in\mathbb{Z}\right)=[B=P(σ)for some hampσofD]mod2.\displaystyle=\left[B=P\left(\sigma\right)\text{ for some hamp }\sigma\text{ of }D\right]\operatorname{mod}2. We have proved this congruence for every subsetBBofAA. Thus,

N\displaystyle N=∑Bis a subset ofANB⏟≡[B=P(σ)for some hampσofD]mod2\displaystyle=\sum\limits_{B\text{ is a subset of }A}\underbrace{N^{B}}_{\equiv\left[B=P\left(\sigma\right)\text{ for some hamp }\sigma\text{ of }D\right]\operatorname{mod}2}≡∑Bis a subset ofA[B=P(σ)for some hampσofD]\displaystyle\equiv\sum\limits_{B\text{ is a subset of }A}\left[B=P\left(\sigma\right)\text{ for some hamp }\sigma\text{ of }D\right]=(# of subsetsBofAsuch thatB=P⁡(σ)for some hampσofD)\displaystyle=\left(\text{\# of subsets }B\text{ of }A\text{ such that }B=P\left(\sigma\right)\text{ for some hamp }\sigma\text{ of }D\right)=(# of sets of the formP(σ)for some hampσofD)\displaystyle=\left(\text{\# of sets of the form }P\left(\sigma\right)\text{ for some hamp }\sigma\text{ of }D\right)(because each set of the formP(σ)for somehampσofDis a subset ofA(by Observation 2))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{because each set of the form }P\left(\sigma\right)\text{ for some}\\ \text{hamp }\sigma\text{ of }D\text{ is a subset of }A\text{ (by Observation 2)}\end{array}\right)=(# of hamps ofD)mod⁡2\displaystyle=\left(\text{\# of hamps of }D\right)\operatorname{mod}2(indeed, Observation 1 shows that different hampsσ\sigmahave different setsP⁡(σ)P\left(\sigma\right), so counting the setsP⁡(σ)P\left(\sigma\right)for all hampsσ\sigmais equivalent to counting the hampsσ\sigmathemselves).

Now we have proved thatN≡(# of hamps ofD¯)mod⁡2N\equiv\left(\text{\# of hamps of }\overline{D}\right)\operatorname{mod}2and N≡(# of hamps ofD)mod2N\equiv\left(\text{\# of hamps of }D\right)\operatorname{mod}2. Comparing these two congruences, we obtain

(# of hamps ofD¯)≡(# of hamps ofD)mod2.\left(\text{\# of hamps of }\overline{D}\right)\equiv\left(\text{\# of hamps of }D\right)\operatorname{mod}2.This proves Berge’s theorem.

4.10Tournaments

4.10.1Definition

We now introduce a special class of simple digraphs.

Definition 4.89.

Atournamentis defined to be a loopless simple digraphDDthat satisfies the

•Tournament axiom:For any two distinct verticesuuandvvofDD,exactlyone of(u,v)\left(u,v\right)and(v,u)\left(v,u\right)is an arc ofDD.

Example 4.90.

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\lxSVG@transformcm{-0.96593}{0.25882}{-0.25882}{-0.96593}{53.0975pt}{-2.79158pt}\lxSVG@begingroup@{transform=matrix(-0.96593 0.25882 -0.25882 -0.96593 73.47 -3.86)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ $\\ \hline\cr\end{tabular}\ \ ,because - •the first of these three digraphs violates the tournament axiom foru=1u=1andv=3v=3; - •the second of these three digraphs violates the tournament axiom foru=1u=1andv=2v=2; - •the third of these three digraphs is not a tournament, since it is not loopless.
(c)The digraphDDin Proposition4.82always is a tournament.

Example 4.91.

Here is a tournament with55vertices:

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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .

A tournament can also be viewed as a complete graph, whose each edge has been given a direction.

Using Definition4.80, we can restate the definition of a tournament as follows:

Proposition 4.92.

LetD=(V,A)D=\left(V,A\right)be a loopless simple digraph. Then,DDis a tournament if and only if the non-loop arcs ofD¯\overline{D}are precisely the arcs ofDrevD^{\operatorname*{rev}}.

Proof 4.93.

Easy consequence of definitions.

Exercise 4.94.

LetDDbe a tournament with at least one vertex.

We say that a vertexuuofDDdirectly ownsa vertexwwofDDif(u,w)\left(u,w\right)is an arc ofDD.

We say that a vertexuuofDDindirectly ownsa vertexwwofDDif there exists a vertexvvofDDsuch that both(u,v)\left(u,v\right)and(v,w)\left(v,w\right)are arcs ofDD.

Prove thatDDhas a vertex that (directly or indirectly) owns all other vertices.

[Solution:This exercise appears in[20f, Exercise 6.3.1](restated in the language of players and matches) and in[Maurer80, Theorem 1](restated in the language of chickens and pecking orders). It originates in a study of pecking orders by Landau[Landau53].]

4.10.2The Rédei theorems

Which tournaments have hamps? The answer is surprisingly simple:292929Here we agree to consider the empty list()\left({}\right)to be a hamp of the digraph(∅,∅)\left(\varnothing,\varnothing\right).

Theorem 4.95(Easy Rédei theorem).

A tournament always has at least one hamp.

Even better, and perhaps even more surprisingly:

Theorem 4.96(Hard Rédei theorem).

LetDDbe a tournament. Then,

(# of hamps ofD)is odd.\left(\text{\# of hamps of }D\right)\text{ is odd.}

Our goal now is to prove these two theorems. Clearly, the Easy Rédei Theorem follows from the Hard one, since an odd number cannot be00. Thus, it will suffice to prove the Hard one.

The proof of the hard Rédei theorem will rely on the following crucial lemma:

Lemma 4.97.

LetD=(V,A)D=\left(V,A\right)be a tournament, and letvw∈Avw\in Abe an arc ofDD.

LetD′D^{\prime}be the digraph obtained fromDDby reversing the arcvwvw. In other words, let

D′:=(V,(A∖{vw})∪{wv}).D^{\prime}:=\left(V,\ \ \left(A\setminus\left\{vw\right\}\right)\cup\left\{wv\right\}\right).Then,D′D^{\prime}is again a tournament, and satisfies

(# of hamps ofD)≡(# of hamps ofD′)mod2.\left(\text{\# of hamps of }D\right)\equiv\left(\text{\# of hamps of }D^{\prime}\right)\operatorname{mod}2.

Here is a visualization of the setup of Lemma4.97:

vwD:;vwD′:.\hbox to115.87pt{\vbox to91.45pt{\pgfpicture\makeatletter\hbox{\hskip 41.68742pt\lower-45.72458pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {}{{}}{}{{{}}{\lx@inpgf@ignorespaces}{}{}{}{}{}{}{}}{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 39.37 0 M 102.36 0 C 102.36 34.79 74.16 62.99 39.37 62.99 C 4.58 62.99 -23.62 34.79 -23.62 0 C -23.62 -34.79 4.58 -62.99 39.37 -62.99 C 74.16 -62.99 102.36 -34.79 102.36 0 Z M 39.37 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.18 0 C 11.18 6.18 6.18 11.18 0 11.18 C -6.18 11.18 -11.18 6.18 -11.18 0 C -11.18 -6.18 -6.18 -11.18 0 -11.18 C 6.18 -11.18 11.18 -6.18 11.18 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.603pt}{-2.15277pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.6 -2.98)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 91.1 0 C 91.1 6.83 85.57 12.36 78.74 12.36 C 71.92 12.36 66.38 6.83 66.38 0 C 66.38 -6.83 71.92 -12.36 78.74 -12.36 C 85.57 -12.36 91.1 -6.83 91.1 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{53.19139pt}{-2.15277pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 73.6 -2.98)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} 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{{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.18 0 C 11.18 6.18 6.18 11.18 0 11.18 C -6.18 11.18 -11.18 6.18 -11.18 0 C -11.18 -6.18 -6.18 -11.18 0 -11.18 C 6.18 -11.18 11.18 -6.18 11.18 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.603pt}{-2.15277pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.6 -2.98)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} 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\lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-38.36835pt}{-3.75891pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -53.09 -5.2)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(Here, we are only showing the arcs joiningvvwithww, sinceDDandD′D^{\prime}agree in all other arcs.)

Proof 4.98(Proof of Lemma4.97.).

(This is an outline; see[17s-lec7, proof of Lemma 1.6.2]for more details.)

First of all,D′D^{\prime}is clearly a tournament. It remains to prove the congruence.

We introduce two more digraphs: Let

D0\displaystyle D_{0}:=(the digraphDwith the arcvwremoved)and\displaystyle:=\left(\text{the digraph }D\text{ with the arc }vw\text{ removed}\right)\ \ \ \ \ \ \ \ \ \ \text{and}D2\displaystyle D_{2}:=(the digraphDwith the arcwvadded).\displaystyle:=\left(\text{the digraph }D\text{ with the arc }wv\text{ added}\right).Note that these are not tournaments any more. Here is a comparative illustration of all four digraphsDD,D′D^{\prime},D0D_{0}andD2D_{2}(again showing only the arcs joiningvvwithww, since there are no differences in the other arcs):

vwD:;vwD′:;\displaystyle\hbox to115.87pt{\vbox to91.45pt{\pgfpicture\makeatletter\hbox{\hskip 41.68742pt\lower-45.72458pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {}{{}}{}{{{}}{\lx@inpgf@ignorespaces}{}{}{}{}{}{}{}}{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 39.37 0 M 102.36 0 C 102.36 34.79 74.16 62.99 39.37 62.99 C 4.58 62.99 -23.62 34.79 -23.62 0 C -23.62 -34.79 4.58 -62.99 39.37 -62.99 C 74.16 -62.99 102.36 -34.79 102.36 0 Z M 39.37 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.18 0 C 11.18 6.18 6.18 11.18 0 11.18 C -6.18 11.18 -11.18 6.18 -11.18 0 C -11.18 -6.18 -6.18 -11.18 0 -11.18 C 6.18 -11.18 11.18 -6.18 11.18 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.603pt}{-2.15277pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.6 -2.98)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 91.1 0 C 91.1 6.83 85.57 12.36 78.74 12.36 C 71.92 12.36 66.38 6.83 66.38 0 C 66.38 -6.83 71.92 -12.36 78.74 -12.36 C 85.57 -12.36 91.1 -6.83 91.1 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{53.19139pt}{-2.15277pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 73.6 -2.98)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} 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The digraphD0D_{0}isD′D^{\prime}with the arcwvwvremoved. Therefore, a hamp ofD0D_{0}is the same as a hamp ofD′D^{\prime}that does not use the arcwvwv. Hence,

(# of hamps ofD0)\displaystyle\left(\text{\# of hamps of }D_{0}\right)=(# of hamps ofD′that do not use the arcwv)\displaystyle=\left(\text{\# of hamps of }D^{\prime}\text{ that do not use the arc }wv\right)=(# of hamps ofD′)−(# of hamps ofD′that use the arcwv).\displaystyle=\left(\text{\# of hamps of }D^{\prime}\right)-\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right).Similarly, sinceDDisD2D_{2}with the arcwvwvremoved, we have

(# of hamps ofD)\displaystyle\left(\text{\# of hamps of }D\right)=(# of hamps ofD2)−(# of hamps ofD2that use the arcwv)\displaystyle=\left(\text{\# of hamps of }D_{2}\right)-\left(\text{\# of hamps of }D_{2}\text{ that use the arc }wv\right)=(# of hamps ofD2)−(# of hamps ofD′that use the arcwv)\displaystyle=\left(\text{\# of hamps of }D_{2}\right)-\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right)(the last equality is because a hamp ofD2D_{2}that uses the arcwvwvcannot use the arcvwvw, and therefore is automatically a hamp ofD′D^{\prime}as well, and of course the converse is obviously true).

However, from the previously proved equality

(# of hamps ofD0)\displaystyle\left(\text{\# of hamps of }D_{0}\right)=(# of hamps ofD′)−(# of hamps ofD′that use the arcwv),\displaystyle=\left(\text{\# of hamps of }D^{\prime}\right)-\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right),we obtain

(# of hamps ofD′)\displaystyle\left(\text{\# of hamps of }D^{\prime}\right)=(# of hamps ofD0)+(# of hamps ofD′that use the arcwv)\displaystyle=\left(\text{\# of hamps of }D_{0}\right)+\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right)≡(# of hamps ofD0)−(# of hamps ofD′that use the arcwv)mod2\displaystyle\equiv\left(\text{\# of hamps of }D_{0}\right)-\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right)\operatorname{mod}2(sincex+y≡x−ymod⁡2x+y\equiv x-y\operatorname{mod}2for any integersxxandyy). Thus, if we can show that

(# of hamps ofD2)≡(# of hamps ofD0)mod2,\left(\text{\# of hamps of }D_{2}\right)\equiv\left(\text{\# of hamps of }D_{0}\right)\operatorname{mod}2,then we will be able to conclude that

(# of hamps ofD)\displaystyle\left(\text{\# of hamps of }D\right)=(# of hamps ofD2)⏟≡(# of hamps ofD0)mod⁡2−(# of hamps ofD′that use the arcwv)\displaystyle=\underbrace{\left(\text{\# of hamps of }D_{2}\right)}_{\equiv\left(\text{\# of hamps of }D_{0}\right)\operatorname{mod}2}-\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right)≡(# of hamps ofD0)−(# of hamps ofD′that use the arcwv)\displaystyle\equiv\left(\text{\# of hamps of }D_{0}\right)-\left(\text{\# of hamps of }D^{\prime}\text{ that use the arc }wv\right)≡(# of hamps ofD′)mod⁡2,\displaystyle\equiv\left(\text{\# of hamps of }D^{\prime}\right)\operatorname{mod}2,and the proof of the lemma will be complete.

So let us show this. Recall thatDDis a tournament. Thus, the non-loop arcs ofD¯\overline{D}are precisely the arcs ofDrevD^{\operatorname*{rev}}(by Proposition4.92). Hence, the non-loop arcs ofD0¯\overline{D_{0}}are precisely the arcs ofD2revD_{2}^{\operatorname*{rev}}(sinceD0¯\overline{D_{0}}is justD¯\overline{D}with the extra arcvwvwadded, and sinceD2revD_{2}^{\operatorname*{rev}}is justDrevD^{\operatorname*{rev}}with the extra arcvwvwadded). Therefore, the digraphsD0¯\overline{D_{0}}andD2revD_{2}^{\operatorname*{rev}}are equal “up to loops” (i.e., they have the same vertices and the same non-loop arcs). Since loops don’t matter for hamps, these two digraphs thus have the same of hamps. Hence,

(# of hamps inD0¯)=(# of hamps inD2rev)=(# of hamps inD2)\left(\text{\# of hamps in }\overline{D_{0}}\right)=\left(\text{\# of hamps in }D_{2}^{\operatorname*{rev}}\right)=\left(\text{\# of hamps in }D_{2}\right)(by Proposition4.84), and therefore

(# of hamps inD2)=(# of hamps inD0¯)≡(# of hamps inD0)mod2\left(\text{\# of hamps in }D_{2}\right)=\left(\text{\# of hamps in }\overline{D_{0}}\right)\equiv\left(\text{\# of hamps in }D_{0}\right)\operatorname{mod}2(by Theorem4.86). As explained above, this completes the proof of Lemma4.97.

Now, the Hard Rédei theorem has become easy:

Proof 4.99(Proof of Theorem4.96.).

(This is an outline; see[17s-lec7, proof of Theorem 1.6.1]for more details.)

We need to prove that the # of hamps ofDDis odd. Lemma4.97tells us that the parity of this # does not change when we reverse a single arc ofDD. Thus, of course, if we reverse several arcs ofDD, then this parity does not change either. However, we can WLOG assume that the vertices ofDDare1,2,…,n1,2,\ldots,nfor somen∈ℕn\in\mathbb{N}, and then, by reversing the appropriate arcs, we can ensure that the arcs ofDDare

12,13,14,…,1n,\displaystyle 12,\ 13,\ 14,\ \ldots,\ 1n,23,24,…,2n,\displaystyle\ \ \ \ \ \ \text{\ }23,\ 24,\ \ldots,\ 2n,⋯,\displaystyle\ \ \ \ \ \ \ \ \ \ \cdots,(n−1)n\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{\ \ \ \ \ }\left(n-1\right)n(i.e., each arc ofDDhas the formijijwithi<ji<j). But at this point, the tournamentDDhas only one hamp: namely,(1,2,…,n)\left(1,2,\ldots,n\right). So(# of hamps ofD)=1\left(\text{\# of hamps of }D\right)=1is odd at this point. Since the parity of the # of hamps ofDDhas not changed as we reversed our arcs, we thus conclude that it has always been odd. This proves the Hard Rédei theorem (Theorem4.96).

As we already mentioned, the Easy Rédei theorem follows from the Hard Rédei theorem. But it also has a short self-contained proof ([17s-lec7, Theorem 1.4.9]).

4.10.3Hamiltonian cycles in tournaments

By the Easy Rédei theorem, every tournament has a hamp. But of course, not every tournament has a hamc303030Recall that “hamc” is our shorthand for “Hamiltonian cycle”.. One obstruction is clear:

Proposition 4.101.

If a digraphDDhas a hamc, thenDDis strongly connected.

In general, this is only a necessary criterion for a hamc, not a sufficient one. Not every strongly connected digraph has a hamc. However, it turns out that for tournaments, it is also sufficient, as long as the tournament has enough vertices:

Theorem 4.102(Camion’s theorem).

If a tournamentDDis strongly connected and has at least two vertices, thenDDhas a hamc.

Proof 4.103(Proof sketch.).

A detailed proof can be found in[17s-lec7, Theorem 1.5.5]; here is just a very rough sketch.

LetD=(V,A)D=\left(V,A\right)be a strongly connected tournament with at least two vertices.313131By the way, a tournament with exactly two vertices cannot be strongly connected (as it has only11arc). Thus, by requiringDDto have at least two vertices, we have actually guaranteed thatDDhas at least three vertices.We must show thatDDhas a hamc.

It is easy to see thatDDhas a cycle. Let𝐜=(v1,v2,…,vk,v1)\mathbf{c}=\left(v_{1},v_{2},\ldots,v_{k},v_{1}\right)be a cycle of maximum length. We shall show that𝐜\mathbf{c}is a hamc.

LetCCbe the set{v1,v2,…,vk}\left\{v_{1},v_{2},\ldots,v_{k}\right\}of all vertices of this cycle𝐜\mathbf{c}.

A vertexw∈V∖Cw\in V\setminus Cwill be called ato-vertexif there exists an arc from someviv_{i}toww.

A vertexw∈V∖Cw\in V\setminus Cwill be called afrom-vertexif there exists an arc fromwwto someviv_{i}.

SinceDDis a tournament, each vertex inV∖CV\setminus Cis a to-vertex or a from-vertex. In theory, a vertex could be both (having an arc from someviv_{i}and also an arc to some othervjv_{j}). However, this does not actually happen. To see why, argue as follows:

•If a to-vertexwwhas an arc from someviv_{i}, then it must also have an arc fromvi+1v_{i+1}323232Here, indices are periodic modulokk, so thatvk+1v_{k+1}meansv1v_{1}.(because otherwise there would be an arc fromwwtovi+1v_{i+1}, and then we could make our cycle𝐜\mathbf{c}longer by interjectingwwbetweenviv_{i}andvi+1v_{i+1}; but this would contradict the fact that𝐜\mathbf{c}is a cycle of maximum length).
•Iterating this argument, we see that if a to-vertexwwhas an arc from someviv_{i}, then it must also have an arc fromvi+1v_{i+1}, an arc fromvi+2v_{i+2}, an arc fromvi+3v_{i+3}, and so on; i.e., it must have an arc from each vertex of𝐜\mathbf{c}. Consequently,wwcannot be a from-vertex. This shows that a to-vertex cannot be a from-vertex.

LetFFbe the set of all from-vertices, and letTTbe the set of all to-vertices. Then, as we have just shown,FFandTTare disjoint. Moreover,F∪T=V∖CF\cup T=V\setminus C. Since a to-vertex cannot be a from-vertex, we furthermore conclude that any to-vertex has an arc from each vertex of𝐜\mathbf{c}(otherwise, it would be a from-vertex), and that any from-vertex has an arc to each vertex of𝐜\mathbf{c}(otherwise, it would be a to-vertex).

Next, we argue that there cannot be an arc from a to-vertexttto a from-vertexff. Indeed, if there was such an arc, then we could make the cycle𝐜\mathbf{c}longer by interjectingttandffbetween (say)v1v_{1}andv2v_{2}.

In total, we now know that every vertex ofDDbelongs to one of the three disjoint setsCC,FFandTT, and furthermore there is no arc fromTTtoFF, no arc fromTTtoCC, and no arc fromCCtoFF. Thus, there exists no walk from a vertex inTTto a vertex inCC(because there is no way out ofTT). This would contradict the fact thatDDis strongly connected, unless the setTTis empty. Hence,TTmust be empty. Similarly,FFmust be empty. SinceF∪T=V∖CF\cup T=V\setminus C, this entails thatV∖CV\setminus Cis empty, so thatV=CV=C. In other words, each vertex ofDDis on our cycle𝐜\mathbf{c}. Therefore,𝐜\mathbf{c}is a hamc. This proves Camion’s theorem.

4.10.4Application of tournaments to the Vandermonde determinant

To wrap up the topic of tournaments, let me briefly discuss a curious application of their theory: a combinatorial proof of the Vandermonde determinant formula. See[17s-lec8]for the many details I’ll be omitting.

Recall the Vandermonde determinant formula:

Theorem 4.104(Vandermonde determinant formula).

Letx1,x2,…,xnx_{1},x_{2},\ldots,x_{n}bennnumbers (or, more generally, elements of a commutative ring). Consider then×nn\times n-matrix

V:=(111⋯1x1x2x3⋯xnx12x22x32⋯xn2⋮⋮⋮⋱⋮x1n−1x2n−1x3n−1⋯xnn−1)=(xji−1)1≤i≤n,1≤j≤n.V:=\left(\begin{array}[c]{ccccc}1&1&1&\cdots&1\\ x_{1}&x_{2}&x_{3}&\cdots&x_{n}\\ x_{1}^{2}&x_{2}^{2}&x_{3}^{2}&\cdots&x_{n}^{2}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ x_{1}^{n-1}&x_{2}^{n-1}&x_{3}^{n-1}&\cdots&x_{n}^{n-1}\end{array}\right)=\left(x_{j}^{i-1}\right)_{1\leq i\leq n,\ 1\leq j\leq n}.Then, its determinant is

detV=∏1≤i<j≤n(xj−xi).\det V=\prod\limits_{1\leq i<j\leq n}\left(x_{j}-x_{i}\right).

There are many simple proofs of this theorem (e.g., a few onits ProofWiki page, which works with the transpose matrix). I will now outline a combinatorial one, using tournaments. This proof goes back to Ira Gessel’s 1979 paper[Gessel79].

First, how dodetV\det Vand∏1≤i<j≤n(xi−xj)\prod\limits\limits_{1\leq i<j\leq n}\left(x_{i}-x_{j}\right)relate to tournaments?

As a warmup, let’s assume that we have some numbery(i,j)y_{\left(i,j\right)}given for each pair(i,j)\left(i,j\right)of integers, and let’s expand the product

(y(1,2)+y(2,1))(y(1,3)+y(3,1))(y(2,3)+y(3,2)).\left(y_{\left(1,2\right)}+y_{\left(2,1\right)}\right)\left(y_{\left(1,3\right)}+y_{\left(3,1\right)}\right)\left(y_{\left(2,3\right)}+y_{\left(3,2\right)}\right).The result is a sum of88products, one for each way to pluck an addend out of each of the three little sums:

(y(1,2)+y(2,1))(y(1,3)+y(3,1))(y(2,3)+y(3,2))\displaystyle\left(y_{\left(1,2\right)}+y_{\left(2,1\right)}\right)\left(y_{\left(1,3\right)}+y_{\left(3,1\right)}\right)\left(y_{\left(2,3\right)}+y_{\left(3,2\right)}\right)=y(1,2)y(1,3)y(2,3)+y(1,2)y(1,3)y(3,2)+y(1,2)y(3,1)y(2,3)+y(1,2)y(3,1)y(3,2)\displaystyle=y_{\left(1,2\right)}y_{\left(1,3\right)}y_{\left(2,3\right)}+y_{\left(1,2\right)}y_{\left(1,3\right)}y_{\left(3,2\right)}+y_{\left(1,2\right)}y_{\left(3,1\right)}y_{\left(2,3\right)}+y_{\left(1,2\right)}y_{\left(3,1\right)}y_{\left(3,2\right)}+y(2,1)y(1,3)y(2,3)+y(2,1)y(1,3)y(3,2)+y(2,1)y(3,1)y(2,3)+y(2,1)y(3,1)y(3,2).\displaystyle\ \ \ \ \ \ \ \ \ \ +y_{\left(2,1\right)}y_{\left(1,3\right)}y_{\left(2,3\right)}+y_{\left(2,1\right)}y_{\left(1,3\right)}y_{\left(3,2\right)}+y_{\left(2,1\right)}y_{\left(3,1\right)}y_{\left(2,3\right)}+y_{\left(2,1\right)}y_{\left(3,1\right)}y_{\left(3,2\right)}.Note that each of the88products obtained has the formyaybycy_{a}y_{b}y_{c}, where

•aais one of the pairs(1,2)\left(1,2\right)and(2,1)\left(2,1\right),
•bbis one of the pairs(1,3)\left(1,3\right)and(3,1)\left(3,1\right), and
•ccis one of the pairs(2,3)\left(2,3\right)and(3,2)\left(3,2\right).

We can view these pairsaa,bbandccas the arcs of a tournament with vertex set{1,2,3}\left\{1,2,3\right\}. Thus, our above expansion can be rewritten more compactly as follows:

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\}\}\{\{\\lx@inpgf@ignorespaces\}\}\}\}\\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ &\\ \cline{1-2}\cr\end{tabular}\ \ \ .123123123Here, for convenience, we are drawing an arcijijin blue ifi<ji<jand in red otherwise.

This expansion can be generalized: We have

∏1≤i<j≤n(y(i,j)+y(j,i))=∑Dis a tournamentwith vertex set{1,2,…,n}∏(i,j)is an arc ofDy(i,j).\prod\limits_{1\leq i<j\leq n}\left(y_{\left(i,j\right)}+y_{\left(j,i\right)}\right)=\sum\limits_{\begin{subarray}{c}D\text{ is a tournament}\\ \text{with vertex set }\left\{1,2,\ldots,n\right\}\end{subarray}}\ \ \prod\limits_{\left(i,j\right)\text{ is an arc of }D}y_{\left(i,j\right)}.Substitutingy(i,j)={xj,ifi<j;−xj,ifi≥jy_{\left(i,j\right)}=\begin{cases}x_{j},&\text{if }i<j;\\ -x_{j},&\text{if }i\geq j\end{cases}in this equality, we obtain

∏1≤i<j≤n(xj−xi)\displaystyle\prod\limits_{1\leq i<j\leq n}\left(x_{j}-x_{i}\right)=∑Dis a tournamentwith vertex set{1,2,…,n}∏(i,j)is an arc ofD{xj,ifi<j;−xj,ifi≥j⏟=(−1)(# of red arcs ofD)∏j=1nxjdeg−⁡j(wheredeg−jmeans the indegree ofjinD,and where the “red arcs” are the arcsijwithi>j)\displaystyle=\sum\limits_{\begin{subarray}{c}D\text{ is a tournament}\\ \text{with vertex set }\left\{1,2,\ldots,n\right\}\end{subarray}}\ \ \underbrace{\prod\limits_{\left(i,j\right)\text{ is an arc of }D}\begin{cases}x_{j},&\text{if }i<j;\\ -x_{j},&\text{if }i\geq j\end{cases}}_{\begin{subarray}{c}=\left(-1\right)^{\left(\text{\# of red arcs of }D\right)}\prod\limits\limits_{j=1}^{n}x_{j}^{\deg^{-}j}\\ \text{(where }\deg^{-}j\text{ means the indegree of }j\text{ in }D\text{,}\\ \text{and where the \textquotedblleft red arcs\textquotedblright\ are the arcs }ij\text{ with }i>j\text{)}\end{subarray}}=∑Dis a tournamentwith vertex set{1,2,…,n}(−1)(# of red arcs ofD)∏j=1nxjdeg−⁡j.\displaystyle=\sum\limits_{\begin{subarray}{c}D\text{ is a tournament}\\ \text{with vertex set }\left\{1,2,\ldots,n\right\}\end{subarray}}\left(-1\right)^{\left(\text{\# of red arcs of }D\right)}\prod\limits\limits_{j=1}^{n}x_{j}^{\deg^{-}j}.We shall refer to this sum as the “big sum”.

On the other hand, if we letSnS_{n}be the group of permutations of{1,2,…,n}\left\{1,2,\ldots,n\right\}, and if we denote the sign of a permutationσ\sigmabysignσ\operatorname*{sign}\sigma, then we have

detV=det(VT)=∑σ∈Snsignσ⋅∏j=1nxjσ⁡(j)−1\det V=\det\left(V^{T}\right)=\sum\limits_{\sigma\in S_{n}}\operatorname*{sign}\sigma\cdot\prod\limits_{j=1}^{n}x_{j}^{\sigma\left(j\right)-1}(by the definition of a determinant). We shall refer to this sum as the “small sum”.

Our goal is to prove that the big sum equals the small sum. To prove this, we must verify the following:

1.Each addend of the small sum is an addend of the big sum. Indeed, for each permutationσ∈Sn\sigma\in S_{n}, there is a certain tournamentTσT_{\sigma}that has (−1)(# of red arcs ofTσ)∏j=1nxjdeg−⁡j=signσ⋅∏j=1nxjσ⁡(j)−1.\left(-1\right)^{\left(\text{\# of red arcs of }T_{\sigma}\right)}\prod\limits\limits_{j=1}^{n}x_{j}^{\deg^{-}j}=\operatorname*{sign}\sigma\cdot\prod\limits_{j=1}^{n}x_{j}^{\sigma\left(j\right)-1}.Can you find thisTσT_{\sigma}?
2.All the addends of the big sum that arenotaddends of the small sum cancel each other out. Why? The basic idea is to argue that if a tournamentDDappears in the big sum but not in the small sum, thenDDhas a33-cycle (i.e., a cycle of length33). When we reverse such a33-cycle (i.e., we reverse each of its arcs), the indegrees of all vertices are preserved, but the sign(−1)(# of red arcs ofD)\left(-1\right)^{\left(\text{\# of red arcs of }D\right)}is flipped (since three arcs change their orientation). This suffices to show that for each addend that appears in the big sum but not in the small sum, there is another addend with the same magnitude but with opposite sign. Unfortunately, this in itself does not suffice to ensure that all these addends cancel out; for example, the sum1+1+1+(−1)1+1+1+\left(-1\right)has the same property but does not equal00. We need to show that the # of addends with positive sign (i.e., with(−1)(# of red arcs ofD)=1\left(-1\right)^{\left(\text{\# of red arcs of }D\right)}=1) and a given magnitude equals the # of addends with negative sign (i.e., with(−1)(# of red arcs ofD)=−1\left(-1\right)^{\left(\text{\# of red arcs of }D\right)}=-1) and the same magnitude. One way to achieve this would be by constructing a bijection (aka “perfect matching”) between the “positive” and the “negative” addends. This is tricky here: We would have to decidewhich33-cycle to reverse (as there are usually many of them), and this has to be done in a bijective way (so that two “positive” addends don’t get assigned the same “negative” partner). A less direct, but easier way is the following: Fix a positive integerkk, and consider only the tournaments with exactlykkmany33-cycles. For each such tournament, we can reverse any of itskkmany33-cycles. It can be shown (nice exercise!) that reversing the arcs of a33-cycle does not change the # of all33-cycles; thus, we don’t accidentally change ourkkin the process. Thus, we find a “kk-to-kk” correspondence between the “positive” addends of a given magnitude and the “negative” addends of the same magnitude. As one can easily see, this entails that the former and the latter are equinumerous, and thus really cancel out. The addends that remain are exactly those in the small sum.

As already mentioned, this is only a rough summary of the proof; the details can be found in[17s-lec8].

4.11Exercises on tournaments

There is, of course, much more to say about tournaments. See[Moon13]for a selection of topics. Let us merely hint at some possible directions by giving a few exercises.

The next three exercises use the notion of a “33-cycle”:

Definition 4.105.

A33-cyclein a tournamentD=(V,A)D=\left(V,A\right)means a triple(u,v,w)\left(u,v,w\right)of vertices inVVsuch that all three pairs(u,v)\left(u,v\right),(v,w)\left(v,w\right)and(w,u)\left(w,u\right)belong toAA.

For example, the tournament shown in Example4.91has the nine different33-cycles

(1,4,3),(1,5,3),(2,5,3),(3,1,4),(3,1,5),\displaystyle\left(1,4,3\right),\ \ \ \ \ \ \ \ \ \ \left(1,5,3\right),\ \ \ \ \ \ \ \ \ \ \left(2,5,3\right),\ \ \ \ \ \ \ \ \ \ \left(3,1,4\right),\ \ \ \ \ \ \ \ \ \ \left(3,1,5\right),(3,2,5),(4,3,1),(5,3,1),(5,3,2).\displaystyle\left(3,2,5\right),\ \ \ \ \ \ \ \ \ \ \left(4,3,1\right),\ \ \ \ \ \ \ \ \ \ \left(5,3,1\right),\ \ \ \ \ \ \ \ \ \ \left(5,3,2\right).(Yes, we are counting a33-cycle(u,v,w)\left(u,v,w\right)as being distinct from(v,w,u)\left(v,w,u\right)and(w,u,v)\left(w,u,v\right).)

Exercise 4.106.

LetD=(V,A)D=\left(V,A\right)be a tournament. Setn=|V|n=\left|V\right|andm=∑v∈V(deg−⁡(v)2)m=\sum\limits_{v\in V}\dbinom{\deg^{-}\left(v\right)}{2}.

(a)Show thatm=∑v∈V(deg+⁡(v)2)m=\sum\limits_{v\in V}\dbinom{\deg^{+}\left(v\right)}{2}.
(b)Show that the number of33-cycles inDDis3((n3)−m)3\left(\dbinom{n}{3}-m\right).

[Solution:This is Exercise 5 on homework set #2 from my Spring 2017 course; seethe course pagefor solutions.]

The next exercise uses the notationdegD−v\deg_{D}^{-}vfor the indegree of a vertexvvin a digraphDD. (We usually denote this bydeg−⁡v\deg^{-}v, but sometimes it is important to stress the dependence onDD, sincevvcan be a vertex of two different digraphs.)

Exercise 4.107.

If a tournamentDDhas a33-cycle(u,v,w)\left(u,v,w\right), then we can define a new tournamentDu,v,w′D_{u,v,w}^{\prime}as follows: The vertices ofDu,v,w′D_{u,v,w}^{\prime}shall be the same as those ofDD. The arcs ofDu,v,w′D_{u,v,w}^{\prime}shall be the same as those ofDD, except that the three arcs(u,v)\left(u,v\right),(v,w)\left(v,w\right)and(w,u)\left(w,u\right)are replaced by the three new arcs(v,u)\left(v,u\right),(w,v)\left(w,v\right)and(u,w)\left(u,w\right). (Visually speaking,Du,v,w′D_{u,v,w}^{\prime}is obtained fromDDby turning the arrows on the arcs(u,v)\left(u,v\right),(v,w)\left(v,w\right)and(w,u)\left(w,u\right)around.) We say that the new tournamentDu,v,w′D_{u,v,w}^{\prime}is obtained from the old tournamentDDby a33-cycle reversal operation.

Now, letVVbe a finite set, and letEEandFFbe two tournaments with vertex setVV. Prove thatFFcan be obtained fromEEby a sequence of33-cycle reversal operations if and only if eachv∈Vv\in VsatisfiesdegE−⁡(v)=degF−⁡(v)\deg_{E}^{-}\left(v\right)=\deg_{F}^{-}\left(v\right). (Note that a sequence may be empty, which allows handling the caseE=FE=Feven ifEEhas no33-cycles to reverse.)

[Solution:This is Exercise 6 on homework set #2 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 4.108.

A tournamentD=(V,A)D=\left(V,A\right)is calledtransitiveif it has no33-cycles.

If a tournamentD=(V,A)D=\left(V,A\right)has three distinct verticesuu,vvandwwsatisfying(u,v)∈A\left(u,v\right)\in Aand(v,w)∈A\left(v,w\right)\in A, then we can define a new tournamentDu,v,w′′D_{u,v,w}^{\prime\prime}as follows: The vertices ofDu,v,w′′D_{u,v,w}^{\prime\prime}shall be the same as those ofDD. The arcs ofDu,v,w′′D_{u,v,w}^{\prime\prime}shall be the same as those ofDD, except that the two arcs(u,v)\left(u,v\right)and(v,w)\left(v,w\right)are replaced by the two new arcs(v,u)\left(v,u\right)and(w,v)\left(w,v\right). We say that the new tournamentDu,v,w′′D_{u,v,w}^{\prime\prime}is obtained from the old tournamentDDby a22-path reversal operation.

LetDDbe any tournament. Prove that there is a sequence of22-path reversal operations that transformsDDinto a transitive tournament.

[Solution:This is Exercise 7 on homework set #2 from my Spring 2017 course; seethe course pagefor solutions.]

5Trees and arborescences

Trees are particularly nice graphs. Among other things, they can be characterized as

•the minimal connected graphs on a given set of vertices, or
•the maximal acyclic (= having no cycles) graphs on a given set of vertices, or
•in many other ways.

Arborescences are their closest analogue for digraphs.

In this chapter, we will discuss the theory of trees and some of their applications. Further applications are usually covered in courses in theoretical computer science, but their notion of a tree is somewhat different from ours.

5.1Some general properties of components and cycles

5.1.1Backtrack-free walks revisited

Before we start with trees, let us recall and prove some more facts about general multigraphs. Recall the notion of a “backtrack-free walk” that already had a brief appearance in the proof of Theorem2.103:

Definition 5.1.

LetGGbe a multigraph. Abacktrack-free walkofGGmeans a walk𝐰\mathbf{w}such that no two consecutive edges of𝐰\mathbf{w}are identical.

Here are a few properties of this notion:

Proposition 5.2.

LetGGbe a multigraph. Let𝐰\mathbf{w}be a backtrack-free walk ofGG. Then,𝐰\mathbf{w}either is a path or contains a cycle.

Proof 5.3.

We have already proved this for simple graphs (in Proposition2.98). More or less the same argument works for multigraphs. (“More or less” because the definition of a cycle in a multigraph is slightly different from that in a simple graph; but the proof is easy to adapt.)

Theorem 5.4.

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Assume that there are two distinct backtrack-free walks fromuutovvinGG. Then,GGhas a cycle.

Proof 5.5.

We have already proved this for simple graphs (Claim 1 in the proof of Theorem2.103). More or less the same argument works for multigraphs.

5.1.2Counting components

Next, we shall derive a few properties of the number of components of a graph. Again, we have already done most of the hard work, and we can now derive corollaries. First, we give this number a name:

Definition 5.6.

LetGGbe a multigraph. Then,connG\operatorname*{conn}Gmeans the number of components ofGG. (Some authors also call this numberb0(G)b_{0}\left(G\right). This notation comes from algebraic topology, where it stands for the00-th Betti number. This makes sense, because we can regard a multigraphGGas a topological space. But we won’t need this.)

So a multigraphGGsatisfiesconnG=1\operatorname*{conn}G=1if and only ifGGis connected. Moreover,connG=0\operatorname*{conn}G=0if and only ifGGhas no vertices.

Let us next recall Definition3.34and Theorem3.35(which is an analogue of Theorem2.109and can be proved in more or less the same way). As a consequence of the latter theorem, we obtain the following:

Corollary 5.7.

LetGGbe a multigraph. Leteebe an edge ofGG. Then:

(a)Ifeeis an edge of some cycle ofGG, thenconn(G∖e)=connG\operatorname*{conn}\left(G\setminus e\right)=\operatorname*{conn}G.
(b)Ifeeappears in no cycle ofGG, thenconn(G∖e)=connG+1\operatorname*{conn}\left(G\setminus e\right)=\operatorname*{conn}G+1.
(c)In either case, we haveconn(G∖e)≤connG+1\operatorname*{conn}\left(G\setminus e\right)\leq\operatorname*{conn}G+1.

Proof 5.8.

Part(a)follows from Theorem3.35(a). Part(b)follows from Theorem3.35(b). Part(c)follows by combining parts(a)and(b).

Corollary 5.9.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Then,connG≥|V|−|E|\operatorname*{conn}G\geq\left|V\right|-\left|E\right|.

Proof 5.10.

We induct on|E|\left|E\right|:

Base case:If|E|=0\left|E\right|=0, thenconnG=|V|\operatorname*{conn}G=\left|V\right|(since|E|=0\left|E\right|=0means that the graphGGhas no edges, and thus no two distinct vertices are path-connected); but this rewrites asconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|(since|E|=0\left|E\right|=0). Thus, Corollary5.9is proved for|E|=0\left|E\right|=0.

Induction step:Letk∈ℕk\in\mathbb{N}. Assume (as the induction hypothesis) that Corollary5.9holds for|E|=k\left|E\right|=k. We must now show that it also holds for|E|=k+1\left|E\right|=k+1.

So let us consider a multigraphG=(V,E,φ)G=\left(V,E,\varphi\right)with|E|=k+1\left|E\right|=k+1. Thus,|E|−1=k\left|E\right|-1=k. Pick any edgee∈Ee\in E(such an edge exists, since|E|=k+1≥1>0\left|E\right|=k+1\geq 1>0). Then, the multigraphG∖eG\setminus ehas edge setE∖{e}E\setminus\left\{e\right\}and therefore has|E∖{e}|=|E|−1=k\left|E\setminus\left\{e\right\}\right|=\left|E\right|-1=kmany edges. Hence, by the induction hypothesis, we have

conn(G∖e)≥|V|−|E∖{e}|\operatorname*{conn}\left(G\setminus e\right)\geq\left|V\right|-\left|E\setminus\left\{e\right\}\right|(sinceG∖eG\setminus eis a multigraph with vertex setVVand edge setE∖{e}E\setminus\left\{e\right\}). However, Corollary5.7(c)yieldsconn(G∖e)≤connG+1\operatorname*{conn}\left(G\setminus e\right)\leq\operatorname*{conn}G+1. Thus,

connG\displaystyle\operatorname*{conn}G≥conn(G∖e)⏟≥|V|−|E∖{e}|−1≥|V|−|E∖{e}|⏟=|E|−1−1\displaystyle\geq\underbrace{\operatorname*{conn}\left(G\setminus e\right)}_{\geq\left|V\right|-\left|E\setminus\left\{e\right\}\right|}-\,1\geq\left|V\right|-\underbrace{\left|E\setminus\left\{e\right\}\right|}_{=\left|E\right|-1}-\,1=|V|−(|E|−1)−1=|V|−|E|.\displaystyle=\left|V\right|-\left(\left|E\right|-1\right)-1=\left|V\right|-\left|E\right|.This completes the induction step. Thus, Corollary5.9is proven.

Corollary 5.11.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph that has no cycles. Then,connG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|.

Proof 5.12.

Replay the proof of Corollary5.9, with just a few changes: Instead of applying Corollary5.7(c), apply Corollary5.7(b)(this is allowed becauseGGhas no cycles and thuseeappears in no cycle ofGG). The induction hypothesis can be used because whenGGhas no cycles,G∖eG\setminus ehas no cycles either. All≤\leqand≥\geqsigns in the above proof now can be replaced by==signs (since Corollary5.7(b)claims an equality, not an inequality). The result is thereforeconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|.

Corollary 5.13.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph that has at least one cycle. Then,connG≥|V|−|E|+1\operatorname*{conn}G\geq\left|V\right|-\left|E\right|+1.

Proof 5.14.

Pick an edgee∈Ee\in Ethat belongs to some cycle (such an edge exists, sinceGGhas at least one cycle). Then, Corollary5.7(a)yieldsconn(G∖e)=connG\operatorname*{conn}\left(G\setminus e\right)=\operatorname*{conn}G. However, Corollary5.9(applied toG∖eG\setminus eandE∖{e}E\setminus\left\{e\right\}instead ofGGandEE) yields

conn(G∖e)≥|V|−|E∖{e}|⏟=|E|−1=|V|−(|E|−1)=|V|−|E|+1.\operatorname*{conn}\left(G\setminus e\right)\geq\left|V\right|-\underbrace{\left|E\setminus\left\{e\right\}\right|}_{=\left|E\right|-1}=\left|V\right|-\left(\left|E\right|-1\right)=\left|V\right|-\left|E\right|+1.Sinceconn(G∖e)=connG\operatorname*{conn}\left(G\setminus e\right)=\operatorname*{conn}G, this rewrites asconnG≥|V|−|E|+1\operatorname*{conn}G\geq\left|V\right|-\left|E\right|+1.

We summarize what we have proved into one convenient theorem:

Theorem 5.15.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Then:

(a)We always haveconnG≥|V|−|E|\operatorname*{conn}G\geq\left|V\right|-\left|E\right|.
(b)We haveconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|if and only ifGGhas no cycles.

Proof 5.16.

(a)This is Corollary5.9.

(b)⟸:\Longleftarrow:This is Corollary5.11.

⟹:\Longrightarrow:Assume thatconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|. IfGGhad any cycles, then Corollary5.13would yieldconnG≥|V|−|E|+1>|V|−|E|\operatorname*{conn}G\geq\left|V\right|-\left|E\right|+1>\left|V\right|-\left|E\right|, which would contradictconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|. SoGGhas no cycles. This proves the “⟹\Longrightarrow” direction of Theorem5.15.

5.2Forests and trees

5.2.1Definitions

We now introduce two of the heroes of this chapter:

Definition 5.18.

Aforestis a multigraph with no cycles.

(In particular, a forest therefore cannot contain two distinct parallel edges. It also cannot contain loops.)

Definition 5.19.

Atreeis a connected forest.

Example 5.20.

Consider the following multigraphs:

12345C=cbg1234D=cea1234E=ac1234F=abcdG=1H=.\begin{tabular}[c]{|c|c|}\hline\cr&\\ \hline\cr& $\\hbox to175\.72pt\{\\vbox to139\.87pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 109\.62843pt\\lower\-66\.0914pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-66\.58 0 C \-66\.58 6\.71 \-72\.03 12\.16 \-78\.74 12\.16 C \-85\.45 12\.16 \-90\.9 6\.71 \-90\.9 0 C \-90\.9 \-6\.71 \-85\.45 \-12\.16 \-78\.74 \-12\.16 C \-72\.03 \-12\.16 \-66\.58 \-6\.71 \-66\.58 0 Z M \-78\.74 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-59\.40552pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-82\.2 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 0 C 12\.16 6\.71 6\.71 12\.16 0 12\.16 C \-6\.71 12\.16 \-12\.16 6\.71 \-12\.16 0 C \-12\.16 \-6\.71 \-6\.71 \-12\.16 0 \-12\.16 C 6\.71 \-12\.16 12\.16 \-6\.71 12\.16 0 Z M 0 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 90\.9 78\.74 C 90\.9 85\.45 85\.45 90\.9 78\.74 90\.9 C 72\.03 90\.9 66\.58 85\.45 66\.58 78\.74 C 66\.58 72\.03 72\.03 66\.58 78\.74 66\.58 C 85\.45 66\.58 90\.9 72\.03 90\.9 78\.74 Z M 78\.74 78\.74\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} 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\{\{\}\{\}\{\{ \{\}\{\}\}\}\{ \{\}\{\}\} \{\{\}\{\{\\lx@inpgf@ignorespaces\}\}\}\{\{\}\{\\lx@inpgf@ignorespaces\}\}\{\}\{\{\}\{\\lx@inpgf@ignorespaces\}\} \{\\lx@inpgf@ignorespaces \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-49\.67392pt\}\{\-3\.41666pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-68\.73 \-4\.73$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ \\ \hline\cr\end{tabular}\ \ \ \ \ \ .12345A=acbedfg12345B=cbdg(Yes,GGis an empty graph with no vertices.) Which of them are forests, and which are trees?

•The graphAAis not a forest, since it has a cycle (actually, several cycles). Thus,AAis not a tree either.
•The graphBBis a tree.
•The graphCCis a forest, but not a tree, since it is not connected.
•The graphDDis a tree.
•The graphEEis a forest, but not a tree.
•The graphFFis not a forest, since it has cycles.
•The graphGG(which has no vertices and no edges) is a forest, but not a tree, since it is not connected (recall: a graph is connected if it has11component; butGGhas00components).
•The graphHHis a tree.

5.2.2The tree equivalence theorem

Trees can be described in many ways:

Theorem 5.21(The tree equivalence theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Then, the following eight statements are equivalent:

•Statement T1:The multigraphGGis a tree.
•Statement T2:The multigraphGGhas no loops, and we haveV≠∅V\neq\varnothing, and for eachu∈Vu\in Vandv∈Vv\in V, there is auniquepath fromuutovv.
•Statement T3:We haveV≠∅V\neq\varnothing, and for eachu∈Vu\in Vandv∈Vv\in V, there is auniquebacktrack-free walk fromuutovv.
•Statement T4:The multigraphGGis connected, and we have|E|=|V|−1\left|E\right|=\left|V\right|-1.
•Statement T5:The multigraphGGis connected, and we have|E|<|V|\left|E\right|<\left|V\right|.
•Statement T6:We haveV≠∅V\neq\varnothing, and the graphGGis a forest, but adding any new edge toGGcreates a cycle.
•Statement T7:The multigraphGGis connected, but removing any edge fromGGyields a disconnected (i.e., non-connected) graph.
•Statement T8:The multigraphGGis a forest, and we have|E|≥|V|−1\left|E\right|\geq\left|V\right|-1andV≠∅V\neq\varnothing.

Proof 5.22.

We shall prove the following implications:

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Ti⟹i\LongrightarrowTjj. Since this digraph is strongly connected (i.e., you can travel from Statement Tiito Statement Tjjalong its arcs for anyi,ji,j), this will prove the theorem. So let us prove the implications.

Proof of T1⟹\LongrightarrowT3:Assume that Statement T1 holds. Thus,GGis a tree. Therefore,GGis connected, so thatV≠∅V\neq\varnothing. We must prove that for eachu∈Vu\in Vandv∈Vv\in V, there is auniquebacktrack-free walk fromuutovv. The existence of such a walk is clear (sinceGGis connected, so there is a path fromuutovv). Thus, we only need to show that it is unique. But this is easy: If there were two distinct backtrack-free walks fromuutovv(for someu∈Vu\in Vandv∈Vv\in V), then Theorem5.4would show thatGGhas a cycle, and thusGGcould not be a forest, let alone a tree. Thus, the backtrack-free walk fromuutovvis unique. So we have proved Statement T3. The implication T1⟹\LongrightarrowT3 is thus proved.

Proof of T3⟹\LongrightarrowT2:Assume that Statement T3 holds. We must prove that Statement T2 holds. First,GGhas no loops, because if there was a loopeewith endpointuu, then the two walks(u)\left(u\right)and(u,e,u)\left(u,e,u\right)would be two distinct backtrack-free walks fromuutouu. It remains to prove that for each eachu∈Vu\in Vandv∈Vv\in V, there is auniquepath fromuutovv. However, the existence of a walk fromuutovvalways implies the existence of a path fromuutovv(by Corollary3.23). Moreover, the uniqueness of a backtrack-free walk fromuutovvimplies the uniqueness of a path fromuutovv(since any path is a backtrack-free walk). Thus, Statement T2 follows from Statement T3.

Proof of T2⟹\LongrightarrowT7:Assume that Statement T2 holds. Then,GGis connected. Now, let us remove any edgeeefromGG. Letuuandvvbe the endpoints ofee. Then,u≠vu\neq v(sinceGGhas no loops). There cannot be a path fromuutovvin the graphG∖eG\setminus e(because if there was such a path, then it would also be a path fromuutovvin the graphGG, and this path would be distinct from the path(u,e,v)\left(u,e,v\right); thus, the graphGGwould have at least two paths fromuutovv; but this would contradict the uniqueness part of Statement T2). Hence, the graphG∖eG\setminus eis disconnected. So we have shown thatGGis connected, but removing any edge fromGGyields a disconnected graph. In other words, Statement T7 holds.

Proof of T7⟹\LongrightarrowT1:Assume that Statement T7 holds. We must show thatGGis a tree. SinceGGis connected (by Statement T7), it suffices to show thatGGis a forest, i.e., thatGGhas no cycles. However, ifGGhad any cycle, then we could pick any edgeeeof this cycle, and then we would know thatG∖eG\setminus eis still connected (since Corollary5.7(a)would yieldconn(G∖e)=connG=1\operatorname*{conn}\left(G\setminus e\right)=\operatorname*{conn}G=1), and this would contradict Statement T7. Thus,GGhas no cycles, hence is a forest. This proves Statement T1.

Proof of T1⟹\LongrightarrowT6:Assume that Statement T1 holds. Thus,GGis a tree. We must show that adding any new edge toGGcreates a cycle (since all other parts of Statement T6 are clear).

Indeed, let us add a new edgefftoGG. Letuuandvvbe the endpoints offf. The graphGGis connected, so there is already a path fromuutovvinGG. Combining this path with the edgeff, we obtain a cycle. Thus, the graph obtained fromGGby adding the new edgeffhas a cycle. This completes our proof that Statement T6 holds.

Proof of T6⟹\LongrightarrowT1:Assume that Statement T6 holds. Thus,GGis a forest. We must only show thatGGis connected.

Assume the contrary. Thus, there exist two verticesuuandvvofGGthat are not path-connected inGG. Hence, adding a new edgeffwith endpointsuuandvvto the graphGGcannot create a new cycle (because any such cycle would have to containff(otherwise, it would already be a cycle ofGG, butGGhas no cycles), and then we could removefffrom it to obtain a path fromuutovvinGG; but such a path cannot exist, sinceuuandvvare not path-connected inGG). This contradicts Statement T6.

So we have shown thatGGis connected, and thusGGis a tree. This proves Statement T1.

Proof of T1⟹\LongrightarrowT8:Assume that Statement T1 holds. SoGGis a tree. Clearly,GGis then a forest. We must show that|E|≥|V|−1\left|E\right|\geq\left|V\right|-1.

Theorem5.15(a)yieldsconnG≥|V|−|E|\operatorname*{conn}G\geq\left|V\right|-\left|E\right|. But we haveconnG=1\operatorname*{conn}G=1becauseGGis connected. Thus,1=connG≥|V|−|E|1=\operatorname*{conn}G\geq\left|V\right|-\left|E\right|. In other words,|E|≥|V|−1\left|E\right|\geq\left|V\right|-1. This proves Statement T8.

Proof of T8⟹\LongrightarrowT1:Assume that Statement T8 holds. Thus,GGis a forest. We must only show thatGGis connected. However,GGis a forest, and thus has no cycles. Hence, Theorem5.15(b)yieldsconnG=|V|−|E|≤1\operatorname*{conn}G=\left|V\right|-\left|E\right|\leq 1(since Statement 8 yields|E|≥|V|−1\left|E\right|\geq\left|V\right|-1). On the other hand,connG≥1\operatorname*{conn}G\geq 1(sinceV≠∅V\neq\varnothing). Combining these two inequalities, we obtainconnG=1\operatorname*{conn}G=1. In other words,GGis connected. This yields Statement T1 (sinceGGis a forest).

Proof of T1⟹\LongrightarrowT4:Assume that Statement T1 holds. Then,GGis a tree, hence a connected forest. Therefore,GGhas no cycles (by the definition of a forest). Theorem5.15(b)therefore yieldsconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|. Thus,|V|−|E|=connG=1\left|V\right|-\left|E\right|=\operatorname*{conn}G=1(sinceGGis connected), so that|E|=|V|−1\left|E\right|=\left|V\right|-1. Thus, Statement T4 is proved.

Proof of T4⟹\LongrightarrowT5:The implication T4⟹\LongrightarrowT5 is obvious.

Proof of T5⟹\LongrightarrowT1:Assume that Statement T5 holds. Thus, the multigraphGGis connected, and we have|E|<|V|\left|E\right|<\left|V\right|. Thus,|E|≤|V|−1\left|E\right|\leq\left|V\right|-1. In other words,1≤|V|−|E|1\leq\left|V\right|-\left|E\right|. SinceGGis connected, we haveconnG=1≤|V|−|E|\operatorname*{conn}G=1\leq\left|V\right|-\left|E\right|. However, Theorem5.15(a)yieldsconnG≥|V|−|E|\operatorname*{conn}G\geq\left|V\right|-\left|E\right|. Combining these two inequalities, we obtainconnG=|V|−|E|\operatorname*{conn}G=\left|V\right|-\left|E\right|. Thus, Theorem5.15(b)shows thatGGhas no cycles. In other words,GGis a forest. Hence,GGis a tree (sinceGGis connected). This proves Statement T1.

We have now proved all necessary implications to conclude that all eight statements T1, T2,…\ldots, T8 are equivalent. Theorem5.21is thus proved.

We also observe the following connection between trees and forests:

Proposition 5.23.

LetGGbe a multigraph, and letC1,C2,…,CkC_{1},C_{2},\ldots,C_{k}be its components. Then,GGis a forest if and only if all the induced subgraphsG⁡[C1],G⁡[C2],…,G⁡[Ck]G\left[C_{1}\right],G\left[C_{2}\right],\ldots,G\left[C_{k}\right]are trees.

Proof 5.24.

⟹:\Longrightarrow:Assume thatGGis a forest. Thus,GGhas no cycles. Hence, the induced subgraphsG⁡[C1],G⁡[C2],…,G⁡[Ck]G\left[C_{1}\right],G\left[C_{2}\right],\ldots,G\left[C_{k}\right]have no cycles either (since a cycle in any of them would be a cycle ofGG); in other words, they are forests. But they are furthermore connected (since the induced subgraph on a component is always connected333333This is Proposition3.27.). Hence, they are connected forests, i.e., trees.

⟸:\Longleftarrow:Assume that the induced subgraphsG⁡[C1],G⁡[C2],…,G⁡[Ck]G\left[C_{1}\right],G\left[C_{2}\right],\ldots,G\left[C_{k}\right]are trees. Hence, none of them has a cycle. Thus,GGhas no cycles either (since a cycle ofGGwould have to be fully contained in one of these induced subgraphs343434Indeed, if it wasn’t, then it would contain vertices from different components. But this is impossible, since there are no walks between vertices in different components.). In other words,GGis a forest.

5.2.3Summary

Let us briefly summarize some properties of trees:

IfT=(V,E,φ)T=\left(V,E,\varphi\right)is a tree, then…

•TTis a connected forest. (This is how trees were defined.) Thus,TThas no cycles. (This is how forests were defined.)
•we have|E|=|V|−1\left|E\right|=\left|V\right|-1. (This follows from the implication T1⟹\LongrightarrowT4 in Theorem5.21.)
•adding any new edge toTTcreates a cycle. (This follows from the implication T1⟹\LongrightarrowT6 in Theorem5.21.)
•removing any edge fromTTyields a disconnected (i.e., non-connected) graph. (This follows from the implication T1⟹\LongrightarrowT7 in Theorem5.21.)
•for eachu∈Vu\in Vandv∈Vv\in V, there is auniquebacktrack-free walk fromuutovv. (This follows from the implication T1⟹\LongrightarrowT3 in Theorem5.21.) Moreover, this backtrack-free walk is a path (since any walk fromuutovvcontains a path fromuutovv).

Exercise 5.26.

LetGGbe a multigraph that has no loops. Assume that there exists a vertexuuofGGsuch that

for each vertexvvofGG, there is auniquepath fromuutovvinGG.Prove thatGGis a tree.

[Remark:Pay attention to the quantifiers used here:∃u∀v\exists u\forall v. This differs from the∀u∀v\forall u\forall vin Statement T2 of the tree equivalence theorem (Theorem5.21).]

5.3Leaves

Continuing with our faux-botanical terminology, we define leaves in a tree:

Definition 5.27.

LetTTbe a tree. A vertex ofTTis said to be aleafif its degree is11.

For example, the tree

has three leaves:11,22and44.

How to find a tree with as many leaves as possible (for a given number of vertices)? For anyn≥3n\geq 3, the simple graph

({0,1,…,n−1},{0i∣i>0})\left(\left\{0,1,\ldots,n-1\right\},\ \left\{0i\ \mid\ i>0\right\}\right)is a tree (when considered as a multigraph), and hasn−1n-1leaves (namely, all of1,2,…,n−11,2,\ldots,n-1). This tree is called annn-star graph, as it looks as follows:

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\lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 7.92 -9.94 L 41.17 -51.62}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ \ \ \ \ \ \ \ \ \left(\text{for }n=8\right).It is easy to see that no tree withn≥3n\geq 3vertices can have more thann−1n-1leaves, so thenn-star graph is optimal in this sense. Note that forn=2n=2, thenn-star graph has22leaves, not11.

How to find a tree with as few leaves as possible? For anyn≥2n\geq 2, thenn-path graph

is a tree with only22leaves (viz., the vertices11andnn). Can we find a tree with fewer leaves? Forn=1n=1, yes, because the11-path graphP1P_{1}(this is simply the graph with11vertex and no edges) has no leaves at all. However, forn≥2n\geq 2, thenn-path graph is the best we can do:

Theorem 5.28.

LetTTbe a tree with at least22vertices. Then:

(a)The treeTThas at least22leaves.
(b)Letvvbe a vertex ofTT. Then, there exist two distinct leavesppandqqofTTsuch thatvvlies on the path frompptoqq.

Note that I’m saying “the path” rather than “a path” here. This is allowed, because in a tree, for any two verticesppandqq, there is auniquepath frompptoqq. This follows from Statement T2 in the tree equivalence theorem (Theorem5.21).

Proof 5.29(Proof of Theorem5.28.).

(b)We apply a variant of the “longest path trick”: Among all paths that contain the vertexvv, let𝐰\mathbf{w}be a longest one. Letppbe the starting point of𝐰\mathbf{w}, and letqqbe the ending point of𝐰\mathbf{w}. We shall show thatppandqqare two distinct leaves.

[Here is a picture of𝐰\mathbf{w}, for what it’s worth:

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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 246.3 0 L 302.49 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Of course, the treeTTcan have other edges as well, not just those of𝐰\mathbf{w}.]

First, we observe thatTTis connected (sinceTTis a tree), and has at least one vertexuudistinct fromvv(sinceTThas at least22vertices). Hence,TThas a path𝐫\mathbf{r}that connectsvvtouu. This path𝐫\mathbf{r}must contain at least one edge (sinceu≠vu\neq v). Thus, we have found a path𝐫\mathbf{r}ofTTthat containsvvand contains at least one edge. Hence, the path𝐰\mathbf{w}must contain at least one edge as well (since𝐰\mathbf{w}is a longest path that containsvv, and thus cannot be shorter than𝐫\mathbf{r}). Since𝐰\mathbf{w}is a path frompptoqq, we thus conclude thatp≠qp\neq q(because if a path contains at least one edge, then its starting point is distinct from its ending point).

Now, assume (for the sake of contradiction) thatppis not a leaf. Then,deg⁡p≠1\deg p\neq 1. The path𝐰\mathbf{w}already contains one edge that containspp(namely, the first edge of𝐰\mathbf{w}). Sincedeg⁡p≠1\deg p\neq 1, there must be another edgeffofTTthat contains𝐰\mathbf{w}. Consider thisff. Letp′p^{\prime}be its endpoint distinct frompp(ifffis a loop, then we setp′=pp^{\prime}=p). Appending this edgeff(and its endpoint) to the beginning of the path𝐰\mathbf{w}, we obtain a backtrack-free walk

(p′,f,p,…,v,…,q⏟This is𝐰)\left(p^{\prime},f,\underbrace{p,\ldots,v,\ldots,q}_{\text{This is }\mathbf{w}}\right)(this is backtrack-free sinceffis not the first edge of𝐰\mathbf{w}). According to Proposition5.2, this backtrack-free walk either is a path or contains a cycle. SinceTThas no cycle (becauseTTis a forest), we thus conclude that this backtrack-free walk is a path. It is furthermore a path that containsvvand is longer than𝐰\mathbf{w}(longer by11, in fact). But this contradicts the fact that𝐰\mathbf{w}is a longest path that containsvv. This contradiction shows that our assumption (thatppis not a leaf) was wrong.

Hence,ppis a leaf. A similar argument shows thatqqis a leaf (here, we need to append the new edge at the end of𝐰\mathbf{w}rather than at the beginning). Thus,ppandqqare two distinct leaves ofTT(distinct becausep≠qp\neq q) such thatvvlies on the path frompptoqq(sincevvlies on the path𝐰\mathbf{w}, which is a path frompptoqq). This proves Theorem5.28(b).

(a)Pick any vertexvvofTT. Then, Theorem5.28(b)shows that there exist two distinct leavesppandqqofTTsuch thatvvlies on the path frompptoqq. Thus, in particular, there exist two distinct leavesppandqqofTT. In other words,TThas at least two leaves. This proves Theorem5.28(a).

[Remark:Another way to prove part(a)is to write the treeTTasT=(V,E,φ)T=\left(V,E,\varphi\right), and recall the handshake lemma, which yields

∑v∈Vdeg⁡v\displaystyle\sum\limits_{v\in V}\deg v=2⋅|E|=2⋅(|V|−1)(since|E|=|V|−1in a tree)\displaystyle=2\cdot\left|E\right|=2\cdot\left(\left|V\right|-1\right)\ \ \ \ \ \ \ \ \ \ \left(\text{since }\left|E\right|=\left|V\right|-1\text{ in a tree}\right)=2⋅|V|−2.\displaystyle=2\cdot\left|V\right|-2.Since eachv∈Vv\in Vsatisfiesdeg⁡v≥1\deg v\geq 1(why?), this equality entails that at least two verticesv∈Vv\in Vmust satisfydeg⁡v≤1\deg v\leq 1(since otherwise, the sum∑v∈Vdeg⁡v\sum\limits_{v\in V}\deg vwould be≥2⋅|V|−1\geq 2\cdot\left|V\right|-1), and therefore these two vertices are leaves.]

Leaves are particularly helpful for performing induction on trees. The formal reason for this is the following theorem:

Theorem 5.30(induction principle for trees).

LetTTbe a tree with at least22vertices. Letvvbe a leaf ofTT. LetT∖vT\setminus vbe the multigraph obtained fromTTby removingvvand all edges that containvv(note that there is only one such edge, sincevvis a leaf). Then,T∖vT\setminus vis again a tree.

Here is an example of a treeTTand of the smaller treeT∖vT\setminus vobtained by removing a leafvv(namely,v=3v=3):

112233445566778811224455667788TTT∖vT\setminus v

Proof 5.31(Proof of Theorem5.30.).

WriteTTasT=(V,E,φ)T=\left(V,E,\varphi\right). Thus,T∖vT\setminus vis the induced subgraphT⁡[V∖{v}]T\left[V\setminus\left\{v\right\}\right].

The graphTTis a tree, thus a forest; hence, it has no cycles. Thus, the graphT∖vT\setminus vhas no cycles either. Hence, it is a forest.

Furthermore, this forestT∖vT\setminus vhas at least11vertex (sinceTThas at least22vertices).

We shall now show that any two verticesppandqqofT∖vT\setminus vare path-connected inT∖vT\setminus v.

Indeed, letppandqqbe two vertices ofT∖vT\setminus v. Then,ppandqqare path-connected inTT(sinceTTis connected). Hence, there exists a path𝐰\mathbf{w}frompptoqqinTT. Consider this path𝐰\mathbf{w}. Note thatvvis neither the starting point nor the ending point of this path𝐰\mathbf{w}(sinceppandqqare vertices ofT∖vT\setminus v, and thus distinct fromvv). Hence, ifvvwas a vertex of𝐰\mathbf{w}, then𝐰\mathbf{w}would containtwo distinctedges that containvv(namely, the edge just beforevvand the edge just aftervv). But this is impossible, since there is only one edge available that containsvv(becausevvis a leaf). Thus,vvcannot be a vertex of𝐰\mathbf{w}. Hence, the path𝐰\mathbf{w}does not use the vertexvv, and thus is a path in the graphT∖vT\setminus vas well. So the verticesppandqqare path-connected inT∖vT\setminus v.

We have now shown that any two verticesppandqqofT∖vT\setminus vare path-connected inT∖vT\setminus v. This shows thatT∖vT\setminus vis connected (sinceT∖vT\setminus vhas at least11vertex). Hence,T∖vT\setminus vis a tree (sinceT∖vT\setminus vis a forest).

Theorem5.30has a converse as well:

Theorem 5.32.

LetGGbe a multigraph. Letvvbe a vertex ofGGsuch thatdeg⁡v=1\deg v=1and such thatG∖vG\setminus vis a tree. (Here,G∖vG\setminus vmeans the multigraph obtained fromGGby removing the vertexvvand all edges that containvv.) Then,GGis a tree.

Proof 5.33.

Left to the reader. (The main step is to show that a cycle ofGGcannot containvv.)

Theorem5.30helps prove many properties of trees by induction on the number of vertices. In the induction step, remove a leafvvand apply the induction hypothesis toT∖vT\setminus v.

The following exercise is essentially a generalization of Theorem5.28(a):

Exercise 5.34.

LetTTbe a tree. Letwwbe any vertex ofTT. Prove thatTThas at leastdeg⁡w\deg wmany leaves.

Exercise 5.35.

LetTTbe a tree that has at least two vertices. Abranch vertexofTTmeans a vertexvvofTTsuch thatdeg⁡v≥3\deg v\geq 3.

(a)Prove that ifTThaskkleaves (for somek∈ℕk\in\mathbb{N}), thenTThas at mostk−2k-2branch vertices.
(b)Prove thatTThas two distinct leavesuuandvvsuch that the path fromuutovvcontains at most one branch vertex.

Exercise 5.36.

A dominating set of a multigraphGGis defined to be a dominating set of its underlying simple graphGsimpG^{\operatorname{simp}}.

LetGGbe a forest. Prove that

∑Dis a dominating set ofG(−1)|D|=±1.\sum\limits_{D\text{ is a dominating set of }G}\left(-1\right)^{\left|D\right|}=\pm 1.

Exercise 5.37.

LetTTbe a tree having more than11vertex. LetLLbe the set of leaves ofTT. Prove that it is possible to add|L|−1\left|L\right|-1new edges toTTin such a way that the resulting multigraph has a Hamiltonian cycle.

[Solution:This is Exercise 4 on homework set #3 from my Spring 2017 course; seethe course pagefor solutions.]

5.4Spanning trees

5.4.1Spanning subgraphs

We now proceed to a crucial application of trees. First we define a concept that makes sense for any multigraphs:

Definition 5.38.

Aspanning subgraphof a multigraphG=(V,E,φ)G=\left(V,E,\varphi\right)means a multigraph of the form(V,F,φ∣F)\left(V,F,\varphi\mid_{F}\right), whereFFis a subset ofEE.

In other words, it means a submultigraph ofGGwith the same vertex set asGG.

In other words, it means a multigraph obtained fromGGby removing some edges, but leaving all vertices undisturbed.

Compare this to the notion of an induced subgraph:

•To build aninducedsubgraph, we throw away some vertices but keep all the edges that we can keep. (As usual in mathematics, the words “some vertices” include “no vertices” and “all vertices”.)
•In contrast, to build aspanningsubgraph, we keep all vertices but throw away some edges.

5.4.2Spanning trees

Spanning subgraphs are particularly useful when they are trees:

Definition 5.39.

Aspanning treeof a multigraphGGmeans a spanning subgraph ofGGthat is a tree.

Example 5.40.

LetGGbe the following multigraph:

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\lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -68.5 31.84 C -31.25 4.47 -13.95 -21.48 -3.01 -66.39}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-14.77582pt}{-9.88318pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -20.45 -13.68)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 68.5 31.84 C 31.25 4.47 13.95 -21.48 3.01 -66.39}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{16.48662pt}{-3.7974pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 22.81 -5.25)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.37 -73.06 L 146.11 -5.68}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.43561pt}{-36.69131pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.32 -50.77)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Here is another:

12345εβδν.\hbox to246pt{\vbox to103.73pt{\pgfpicture\makeatletter\hbox{\hskip 122.99692pt\lower-66.0914pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -145.32 0 C -145.32 6.71 -150.77 12.16 -157.48 12.16 C -164.19 12.16 -169.64 6.71 -169.64 0 C -169.64 -6.71 -164.19 -12.16 -157.48 -12.16 C -150.77 -12.16 -145.32 -6.71 -145.32 0 Z M -157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-116.31104pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -160.94 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 39.37 C -66.58 46.08 -72.03 51.53 -78.74 51.53 C -85.45 51.53 -90.9 46.08 -90.9 39.37 C -90.9 32.66 -85.45 27.21 -78.74 27.21 C -72.03 27.21 -66.58 32.66 -66.58 39.37 Z M -78.74 39.37}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{25.23055pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 34.91)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 39.37 C 90.9 46.08 85.45 51.53 78.74 51.53 C 72.03 51.53 66.58 46.08 66.58 39.37 C 66.58 32.66 72.03 27.21 78.74 27.21 C 85.45 27.21 90.9 32.66 90.9 39.37 Z M 78.74 39.37}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{25.23055pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 34.91)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.64 0 C 169.64 6.71 164.19 12.16 157.48 12.16 C 150.77 12.16 145.32 6.71 145.32 0 C 145.32 -6.71 150.77 -12.16 157.48 -12.16 C 164.19 -12.16 169.64 -6.71 169.64 0 Z M 157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{111.31104pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 154.02 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 -78.74 C 12.16 -72.03 6.71 -66.58 0 -66.58 C -6.71 -66.58 -12.16 -72.03 -12.16 -78.74 C -12.16 -85.45 -6.71 -90.9 0 -90.9 C 6.71 -90.9 12.16 -85.45 12.16 -78.74 Z M 0 -78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-60.12773pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -83.2)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -146.11 5.68 L -90.11 33.69}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-87.68987pt}{18.15936pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -121.34 25.13)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -73.63 27.73 C -55.24 -14.19 -40.4 -36.45 -8.78 -69.55}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{{}{}}}{{}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-42.95364pt}{-20.00085pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -59.43 -27.68)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 68.5 31.84 C 31.25 4.47 13.95 -21.48 3.01 -66.39}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{16.48662pt}{-3.7974pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 22.81 -5.25)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.37 -73.06 L 146.11 -5.68}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.43561pt}{-36.69131pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.32 -50.77)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(Yes, this is a different one, becauseα≠β\alpha\neq\beta.) And here is yet another spanning tree ofGG:

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Example 5.41.

Letnnbe a positive integer. Consider the cycle graphCnC_{n}. (We defined this graphCnC_{n}in Definition2.44for alln≥2n\geq 2, but we later redefinedC2C_{2}and definedC1C_{1}in Definition3.18. Here, we are using the latter modified definition.)

The graphCnC_{n}has exactlynnspanning trees. Indeed, any graph obtained fromCnC_{n}by removing a single edge is a spanning tree ofCnC_{n}.

Proof 5.42.

A tree withnnvertices must have exactlyn−1n-1edges (by the implication T1⟹\LongrightarrowT4 in Theorem5.21). Thus, a spanning subgraph ofCnC_{n}can be a tree only if it hasn−1n-1edges, i.e., only if it is obtained fromCnC_{n}by removing a single edge (sinceCnC_{n}hasnnedges in total). Thus,CnC_{n}has at mostnnspanning trees (sinceCnC_{n}hasnnedges that can be removed). It remains to check that any subgraph obtained fromCnC_{n}by removing a single edge is indeed a spanning tree. But this is easy, since all such subgraphs are isomorphic to the path graphPnP_{n}. This proves Example5.41.

Exercise 5.43.

Fixm≥1m\geq 1. LetGGbe the simple graph with3m+23m+2vertices

a,b,x1,x2,…,xm,y1,y2,…,ym,z1,z2,…,zma,b,x_{1},x_{2},\ldots,x_{m},y_{1},y_{2},\ldots,y_{m},z_{1},z_{2},\ldots,z_{m}and the following3m+33m+3edges:

ax1,ay1,az1,\displaystyle ax_{1},\ \ ay_{1},\ \ az_{1},xixi+1,yiyi+1,zizi+1for alli∈{1,2,…,m−1},\displaystyle x_{i}x_{i+1},\ \ y_{i}y_{i+1},\ \ z_{i}z_{i+1}\qquad\text{ for all }i\in\left\{1,2,\ldots,m-1\right\},xmb,ymb,zmb.\displaystyle x_{m}b,\ \ y_{m}b,\ \ z_{m}b.(Thus, the graph consists of two verticesaaandbbconnected by three paths, each of lengthm+1m+1, with no overlaps between the paths except for their starting and ending points. Here is a picture form=3m=3:

aay1y_{1}y2y_{2}y3y_{3}x1x_{1}x2x_{2}x3x_{3}z1z_{1}z2z_{2}z3z_{3}bb) Compute the number of spanning trees ofGG.

[To argue why your number is correct, a sketch of the argument in 1-2 sentences should be enough; a fully rigorous proof is not required.]

[Solution:This is Exercise 2(c)on homework set #3 from my Spring 2017 course; seethe course pagefor solutions.]

5.4.3Spanning forests

A spanning tree of a graphGGcan be regarded as a minimum “backbone” ofGG– that is, a way to keepGGconnected using as few edges as possible. Of course, ifGGis not connected, then this is not possible at all, soGGhas no spanning trees in this case. The best one can hope for is a spanning subgraph that keeps each component ofGGconnected using as few edges as possible. This is known as a “spanning forest”:

Definition 5.44.

Aspanning forestof a multigraphGGmeans a spanning subgraphHHofGGthat is a forest and satisfiesconnH=connG\operatorname*{conn}H=\operatorname*{conn}G.

WhenGGis a connected multigraph, a spanning forest ofGGmeans the same as a spanning tree ofGG.

Exercise 5.45.

Here is an example of a graph and one of its spanning forests:

graphGGspanning forest ofGG11223344556677881122334455667788

5.4.4Existence and construction of a spanning tree

The following theorem is crucial, which is why we will outline four different proofs:

Theorem 5.46.

Each connected multigraphGGhas at least one spanning tree.

Proof 5.47(First proof.).

LetGGbe a connected multigraph. We want to construct a spanning tree ofGG. We try to achieve this by removing edges fromGGone by one, untilGGbecomes a tree. When doing so, we must be careful not to disconnect the graph (i.e., not to destroy its connectedness). According to Theorem3.35, this can be achieved by making sure that we never remove a bridge (i.e., an edge that appears in no cycle). Thus, we keep removing non-bridges (i.e., edges that are not bridges) as long as we can (i.e., until we end up with a graph in which every edge is a bridge).

So here is the algorithm: We start withGG, and we successively remove non-bridges one by one until we no longer have any non-bridges left353535Warning:We cannot remove several non-bridges at once! We have to remove them one by one. Indeed, ifeeandffare two non-bridges ofGG, then there is no guarantee thatffremains a non-bridge inG∖eG\setminus e. So we cannot remove botheeandffsimultaneously; we have to remove one of them and check whether the other is still a non-bridge.. This procedure cannot go on forever, sinceGGhas only finitely many edges. Thus, after finitely many steps, we will end up with a graph that has no non-bridges any more. This resulting graph therefore has no cycles (since any cycle would have at least one edge, and this edge would be a non-bridge), but is still connected (sinceGGwas connected, and we never lost connectedness as we removed only non-bridges). Thus, this resulting graph is a tree. Since it is also a spanning subgraph ofGG(by construction), it is therefore a spanning tree ofGG. This proves Theorem5.46.

Proof 5.48(Second proof (sketched).).

In the above first proof, we constructed a spanning tree ofGGby starting withGGand successively removing edges until we got a tree. Now let us take the opposite strategy: Start with an empty graph on the same vertex set asGG, and successively add edges (fromGG) until we get a connected graph.

Here are some details: We start with a graphLLthat has the same vertex set asGG, but has no edges. Now, we inspect all edgeseeofGGone by one (in some order). For each such edgeee, we add it toLL, but only if it does not create a cycle inLL; otherwise, we discard this edge. Notice that adding an edgeeewith endpointsuuandvvtoLLcreates a cycle if and only ifuuandvvlie in the same component ofLL(before we addee). Thus, we only add an edge toLLif its endpoints lie in different components ofLL; otherwise, we discard it. This way, at the end of the procedure, our graphLLwill still have no cycles (since we never create any cycles). In other words, it will be a forest.

Let me denote this forest byHH. (Thus,HHis theLLat the end of the procedure.) I claim that this forestHHis a spanning tree ofGG. Why? Since we know thatHHis a forest, we only need to show thatHHis connected. Assume the contrary. Thus, there is at least one edgeeeofGGwhose endpoints lie in different components ofHH(why?). This edgeeeis therefore not an edge ofHH. Therefore, at some point during our construction ofHH, we must have discarded this edgeee(instead of adding it toLL). As we know, this means that the endpoints ofeeused to lie in the same component ofLLat the point at which we discardedee. But this entails that these two endpoints lie in the same component ofLLat the end of the procedure as well (because the graphLLnever loses any edges during the procedure, so that any two vertices that used to lie in the same component ofLLat some point will still lie in the same component ofLLever after). In other words, the endpoints ofeelie in the same component ofHH. This contradicts our assumption that the endpoints ofeelie in different components ofHH. This contradiction completes our proof thatHHis connected. Hence,HHis a spanning tree ofGG, and we have proved Theorem5.46again.

Proof 5.49(Third proof.).

This proof takes yet another approach to constructing a spanning tree ofGG: We choose an arbitrary vertexrrofGG, and then progressively “spread a rumor” fromrr. The rumor starts at vertexrr. On day00, onlyrrhas heard the rumor. Every day, every vertex that knows the rumor spreads it to all its neighbors (i.e., all vertices adjacent to it). SinceGGis connected, the rumor will eventually spread to every vertex ofGG. Now, each vertexvv(other thanrr) remembers which other vertexv′v^{\prime}it has first heard the rumor from (if it heard it from several vertices at the same time, it just picks one of them), and picks some edgeeve_{v}that has endpointsvvandv′v^{\prime}(such an edge must exist, sincevvmust have heard the rumor from a neighbor). The edgeseve_{v}for allv∈V∖{r}v\in V\setminus\left\{r\right\}(whereVVis the vertex set ofGG) then form a spanning tree ofGG(that is, the graph with vertex setVVand edge set{ev∣v∈V∖{r}}\left\{e_{v}\ \mid\ v\in V\setminus\left\{r\right\}\right\}is a spanning tree). Why?

Intuitively, this is quite convincing: This graph cannot have cycles (because that would require a time loop) and must be connected (because for any vertexvv, we can trace back the path of the rumor fromrrtovvby following the edgeseve_{v}backwards). To obtain a rigorous proof, we formalize this construction mathematically:

WriteGGasG=(V,E,φ)G=\left(V,E,\varphi\right). Choose any vertexrrofGG.

We shall recursively construct a sequence of subgraphs

(V0,E0,φ0),(V1,E1,φ1),(V2,E2,φ2),…\left(V_{0},E_{0},\varphi_{0}\right),\ \ \ \ \left(V_{1},E_{1},\varphi_{1}\right),\ \ \ \ \left(V_{2},E_{2},\varphi_{2}\right),\ \ \ \ \ldotsofGG. The idea behind these subgraphs is that for eachi∈ℕi\in\mathbb{N}, the setViV_{i}will consist of all verticesvvthat have heard the rumor by dayii, and the setEiE_{i}will consist of the corresponding edgeseve_{v}. The mapφi\varphi_{i}will be the restriction ofφ\varphitoEiE_{i}, of course.

Here is the exact construction of this sequence of subgraphs:

•Recursion base:SetV0:={r}V_{0}:=\left\{r\right\}andE0:=∅E_{0}:=\varnothing. Letφ0\varphi_{0}be the restriction ofφ\varphito the (empty) setE0E_{0}.
•Recursion step:Leti∈ℕi\in\mathbb{N}. Assume that the subgraph(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)ofGGhas already been defined. Now, we set Vi+1:=Vi∪{v∈V∣vis adjacent to some vertex inVi}.V_{i+1}:=V_{i}\cup\left\{v\in V\ \mid\ v\text{ is adjacent to some vertex in }V_{i}\right\}.For eachv∈Vi+1∖Viv\in V_{i+1}\setminus V_{i}, we chooseoneedgeeve_{v}that joins363636We say that an edgejoinsa vertexppto a vertexqqif the endpoints of this edge areppandqq.vvto a vertex inViV_{i}(such an edge exists, sincev∈Vi+1v\in V_{i+1}; if there are several, we just choose a random one). Set Ei+1:=Ei∪{ev∣v∈Vi+1∖Vi}.E_{i+1}:=E_{i}\cup\left\{e_{v}\ \mid\ v\in V_{i+1}\setminus V_{i}\right\}.Finally, we letφi+1\varphi_{i+1}be the restriction of the mapφ\varphito the setEi+1E_{i+1}. This is a map fromEi+1E_{i+1}to𝒫1,2(Vi+1)\mathcal{P}_{1,2}\left(V_{i+1}\right)(because any edgeeve_{v}withv∈Vi+1∖Viv\in V_{i+1}\setminus V_{i}has one endpointvvinVi+1∖Vi⊆Vi+1V_{i+1}\setminus V_{i}\subseteq V_{i+1}and the other endpoint inVi⊆Vi+1V_{i}\subseteq V_{i+1}). Thus,(Vi+1,Ei+1,φi+1)\left(V_{i+1},E_{i+1},\varphi_{i+1}\right)is a well-defined subgraph ofGG.

This construction yields that(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)is a subgraph of(Vi+1,Ei+1,φi+1)\left(V_{i+1},E_{i+1},\varphi_{i+1}\right)for eachi∈ℕi\in\mathbb{N}. Hence,V0⊆V1⊆V2⊆⋯V_{0}\subseteq V_{1}\subseteq V_{2}\subseteq\cdots, so that|V0|≤|V1|≤|V2|≤⋯\left|V_{0}\right|\leq\left|V_{1}\right|\leq\left|V_{2}\right|\leq\cdots. Since a sequence of integers bounded from above cannot keep increasing forever (and the sizes|Vi|\left|V_{i}\right|are bounded from above by|V|\left|V\right|, since eachViV_{i}is a subset ofVV), we thus see that there exists somei∈ℕi\in\mathbb{N}such that|Vi|=|Vi+1|\left|V_{i}\right|=\left|V_{i+1}\right|. Consider thisii. From|Vi|=|Vi+1|\left|V_{i}\right|=\left|V_{i+1}\right|, we obtainVi=Vi+1V_{i}=V_{i+1}(sinceVi⊆Vi+1V_{i}\subseteq V_{i+1}).

In our colloquial model above,Vi=Vi+1V_{i}=V_{i+1}means that no new vertices learn the rumor on dayi+1i+1; it is reasonable to expect that at this point, every vertex has heard the rumor. In other words, we claim thatVi=VV_{i}=V. A rigorous proof of this can be easily given using the fact thatGGis connected373737Here is theproofin detail: We must show thatVi=VV_{i}=V. Assume the contrary. Thus, there exists a vertexu∈V∖Viu\in V\setminus V_{i}. Consider thisuu. The path fromrrtouustarts at a vertex inViV_{i}(sincer∈V0⊆Vir\in V_{0}\subseteq V_{i}) and ends at a vertex inV∖ViV\setminus V_{i}(sinceu∈V∖Viu\in V\setminus V_{i}). Thus, it must cross over fromViV_{i}intoV∖ViV\setminus V_{i}at some point. Therefore, there exists an edge with one endpoint inViV_{i}and the other endpoint inV∖ViV\setminus V_{i}. Letvvandwwbe these two endpoints, so thatv∈Viv\in V_{i}andw∈V∖Viw\in V\setminus V_{i}. Then,wwis adjacent to some vertex inViV_{i}(namely, tovv), and therefore belongs toVi+1V_{i+1}(by the definition ofVi+1V_{i+1}). Hence,w∈Vi+1=Viw\in V_{i+1}=V_{i}. But this contradictsw∉V∖Viw\notin V\setminus V_{i}. This contradiction shows that our assumption was wrong, qed..

Now, we claim that the subgraph(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)is a spanning tree ofGG. To see this, we must show that this subgraph is a forest and is connected (sinceVi=VV_{i}=Valready shows that it is a spanning subgraph). Before we do this, let us give an example:

Example 5.50.

LetGGbe the following multigraph:

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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 169.21 73.85 L 240.24 44.26}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Setr=3r=3. Then, the above construction yields

V0\displaystyle V_{0}={3},\displaystyle=\left\{3\right\},V1\displaystyle V_{1}={3,1,4},\displaystyle=\left\{3,1,4\right\},V2\displaystyle V_{2}={3,1,4,2,5,6,10},\displaystyle=\left\{3,1,4,2,5,6,10\right\},V3\displaystyle V_{3}={3,1,4,2,5,6,10,8,9,7}=V,\displaystyle=\left\{3,1,4,2,5,6,10,8,9,7\right\}=V,so thatVk=VV_{k}=Vfor allk≥3k\geq 3. Thus, we can takei=3i=3. Here is an image of theVkV_{k}as progressively growing circles:

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240.24 44.26}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(The dark-red inner circle isV0V_{0}; the red circle isV1V_{1}; the orange circle isV2V_{2}; the yellow circle isV3=V4=V5=⋯=VV_{3}=V_{4}=V_{5}=\cdots=V.) Finally, the edgeseve_{v}can be chosen to be the following (we are painting them red for clarity):

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The other options would have been equally fine.)

We now return to the general proof. Let us first show the following:

Claim 1:Letj∈ℕj\in\mathbb{N}. Each vertex of the graph(Vj,Ej,φj)\left(V_{j},E_{j},\varphi_{j}\right)is path-connected torrin this graph.

[Proof of Claim 1:We induct onjj:

Base case:Forj=0j=0, Claim 1 is obvious, sinceV0={r}V_{0}=\left\{r\right\}(so the only vertex of the graph in question isrritself).

Induction step:Fix some positive integerkk. Assume (as the induction hypothesis) that Claim 1 holds forj=k−1j=k-1. That is, each vertex of the graph(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right)is path-connected torrin this graph.

Now, letvvbe a vertex of the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right). We must show thatvvis path-connected torrin this graph. Ifv∈Vk−1v\in V_{k-1}, then this follows from the induction hypothesis (since(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right)is a subgraph of(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)). Thus, we WLOG assume thatv∉Vk−1v\notin V_{k-1}from now on. Hence,v∈Vk∖Vk−1v\in V_{k}\setminus V_{k-1}. According to the recursive definition ofEkE_{k}, this entails that there is an edgeev∈Eke_{v}\in E_{k}that joinsvvto some vertexu∈Vk−1u\in V_{k-1}. Consider this latter vertexuu. Then,vvis path-connected touuin the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)(since the edgeeve_{v}provides a length-11path fromvvtouu). However,uuis path-connected torrin the graph(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right)(by the induction hypothesis, sinceu∈Vk−1u\in V_{k-1}), hence also in the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)(since(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right)is a subgraph of(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)). Since the relation “path-connected” is transitive, we conclude from the previous two sentences thatvvis path-connected torrin the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right).

So we have shown that each vertexvvof the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)is path-connected torrin the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right). In other words, Claim 1 holds forj=kj=k. This completes the induction step, and Claim 1 is proved.]

Claim 1 (applied toj=ij=i) shows that each vertex of the graph(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)is path-connected torrin this graph. Since the relation “path-connected” is an equivalence relation, this entails that any two vertices of this graph are path-connected. Thus, the graph(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)is connected (since it has at least one vertex). It remains to prove that this graph(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)is a forest.

Again, we do this using an auxiliary claim:

Claim 2:Letj∈ℕj\in\mathbb{N}. Then, the graph(Vj,Ej,φj)\left(V_{j},E_{j},\varphi_{j}\right)has no cycles.

[Proof of Claim 2:We induct onjj:

Base case:The graph(V0,E0,φ0)\left(V_{0},E_{0},\varphi_{0}\right)has no edges (becauseE0=∅E_{0}=\varnothing) and thus no cycles. Thus, Claim 2 holds forj=0j=0.

Induction step:Fix some positive integerkk. Assume (as the induction hypothesis) that Claim 2 holds forj=k−1j=k-1. That is, the graph(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right)has no cycles.

Now, let𝐜\mathbf{c}be a cycle of the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right). Then,𝐜\mathbf{c}must use at least one edge fromEk∖Ek−1E_{k}\setminus E_{k-1}(since otherwise,𝐜\mathbf{c}would be a cycle of the graph(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right), but this is impossible, since(Vk−1,Ek−1,φk−1)\left(V_{k-1},E_{k-1},\varphi_{k-1}\right)has no cycles). However, each edge fromEk∖Ek−1E_{k}\setminus E_{k-1}has the formeve_{v}for somev∈Vk∖Vk−1v\in V_{k}\setminus V_{k-1}(because of howEkE_{k}was defined). Thus,𝐜\mathbf{c}must have an edge of this form. Consider the corresponding vertexv∈Vk∖Vk−1v\in V_{k}\setminus V_{k-1}. The cycle𝐜\mathbf{c}contains the edgeeve_{v}and therefore also contains its endpointvv. However, (again by the definition ofEkE_{k}) the edgeeve_{v}is theonlyedge inEkE_{k}that contains the vertexvv. Since the edgeeve_{v}is not a loop (because it joins the vertexv∈Vk∖Vk−1v\in V_{k}\setminus V_{k-1}with a vertex inVk−1V_{k-1}), we thus conclude that the vertexvvhas degree11in the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right). Thus, the vertexvvcannot be contained in any cycle of(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)(because a cycle cannot contain a vertex of degree11). This contradicts the fact that the cycle𝐜\mathbf{c}containsvv.

Forget that we fixed𝐜\mathbf{c}. We thus have obtained a contradiction for each cycle𝐜\mathbf{c}of the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right). Hence, the graph(Vk,Ek,φk)\left(V_{k},E_{k},\varphi_{k}\right)has no cycles. In other words, Claim 2 holds forj=kj=k. This completes the induction step, and Claim 2 is proved.]

Applying Claim 2 toj=ij=i, we see that the graph(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)has no cycles. In other words, this graph is a forest. Since it is connected, it is therefore a tree. Since it is a spanning subgraph ofGG, we thus conclude that it is a spanning tree ofGG. Hence, we have constructed a spanning tree ofGG.

We note an important property of this construction:

Claim 3:For eachk∈ℕk\in\mathbb{N}, we have Vk={v∈V∣d⁡(r,v)≤k},V_{k}=\left\{v\in V\ \mid\ d\left(r,v\right)\leq k\right\},whered⁡(r,v)d\left(r,v\right)means the length of a shortest path fromrrtovv.

This is easily proved by induction onkk. Thus, the spanning tree(Vi,Ei,φi)\left(V_{i},E_{i},\varphi_{i}\right)we have constructed has the following property: For eachv∈Vv\in V, the path fromrrtovvin this spanning tree is a shortest path fromrrtovvinGG. For this reason, this spanning tree is called abreadth-first search (“BFS”) tree. Note that the choice of rootrris important here: It is usually not true that the path from an arbitrary vertexuuto an arbitrary vertexvvalong our spanning tree is a shortest path inGG. No spanning tree ofGGhas this property, unlessGGitself is “more or less a tree” (more precisely, unlessGsimpG^{\operatorname*{simp}}is a tree)!

Proof 5.51(Fourth proof of Theorem5.46(sketched).).

We imagine a snake that slithers along the edges ofGG, trying to eventually bite each vertex. It starts at some vertexrr, which it immediately bites. Any time the snake enters a vertexvv, it makes the following step:

•If some neighbor ofvvhas not been bitten yet, then the snake picks such a neighborwwas well as some edgeffthat joinswwwithvv; the snake then moves towwalong the edgeff, bites the vertexwwand marks the edgeff.
•If not, then the snake marks the vertexvvas fully digested and backtracks (along the marked edges) to the last vertex it has visited but not fully digested yet. (To “backtrack” along a marked edge means to use it in the direction opposite to how it was originally used. In other words, if a marked edgeffis initially used to go fromvvtoww, then backtracking along it means walking it fromwwtovv.)

Once backtracking is no longer possible (because there are no more vertices left that are not fully digested), the procedure is finished. I claim that the marked edges at that moment are the edges of a spanning tree ofGG.

I won’t prove this claim in detail, but I will give some hints. First, however, an example:

Example 5.52.

LetGGbe the following connected multigraph:

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\lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 382.93 -10.78 L 325.74 -67.96}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Let our snake start its journey atr=3r=3. It bites this vertex. Then, let’s say that it picks the vertex11as its next victim (it could just as well go to44or77; the snake has many choices, but we follow one possible trip). Thus, it next arrives at vertex11, bites it and marks the edge that brought it to this vertex. As its next destination, it necessarily picks the vertex22(since vertex33has already been bitten). It moves to vertex22, bites it and marks the edge. Next, let’s say that it picks the vertex44(the other option would be88). It thus moves to44, bites it and marks the edge. Proceeding likewise, it then moves to55(the other options are66and1010; the vertices22and33do not qualify since they are already bitten), bites55and marks an edge. From there, let’s say it moves to88, bites88and marks an edge. Now, there is no longer an unbitten neighbor of88to move to. Thus, the snake marks the vertex88as fully digested and backtracks to the last vertex not fully digested – which, at this point, is55. From this vertex55, it moves on to99(this is the only option, since44and88have already been bitten). And so on. Here is one possible outcome of this journey (there are a few more decisions that the snake can make here, so you may get a different one):

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}}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Here, the marked edges are drawn in bold red ink, and endowed with an arrow that represents the direction in which they were first used (e.g., the edge joining22with44has an arrow towards44because it was first used to get from22to44).

Now, as promised, let me outline a proof of the above claim (that the marked edges form a spanning tree ofGG). To wit, argue the following four observations (ideally in this order):

1.After each step, the marked edges are precisely the edges along which the snake has moved so far.
2.After each step, the network of bitten vertices and marked edges is a tree.
3.After enough steps, each bitten vertex is fully digested.
4.At that point, the network of bitten vertices and marked edges is a spanning tree (since each neighbor of a fully digested vertex is bitten, thus fully digested by observation 3).

Details are left to the reader.

The result is that Theorem5.46is proved once again. However, more comes out of the above construction if you know where to look. The spanning treeTTofGGwhose edges are the edges marked by the snake is called adepth-first search (“DFS”) tree. It has the following extra property: Ifuuandvvare two adjacent vertices ofGG, then eitheruulies on the path fromrrtovvinTT, orvvlies on the path fromrrtouuinTT. (This called a “lineal spanning tree”. See[BenWil06, §6.1]for details.)

5.4.5Applications

Spanning trees have lots of applications:

•A spanning tree of a graph can be viewed as a kind of “backbone” of the graph, which in particular provides “canonical” paths between any two vertices. This is useful, e.g., for networking applications where having a choice between different paths would be problematic (see, e.g.,the Spanning Tree Protocol).
•Aww-minimum spanning tree (see Exercise5.59= Homework set #5 exercise 6) solves a global version of the cheapest-path problem. It can also be used for detecting clusters.
•Depth-first search (the algorithm used in our fourth proof of Theorem5.46) can also be used as a way to traverse all vertices of a given graph and return back to the starting point. In particular, this provides an algorithmic way to solve mazes (since a maze can be modeled as a graph, where the vertices correspond to “rooms” and the edges correspond to “doors”). This appears to have been the original motivation for Trémaux to invent depth-first search back in the 19th century.

Here is a more theoretical application of spanning trees:

Definition 5.53.

A vertexvvof a connected multigraphGGis said to be acut-vertexif the graphG∖vG\setminus vis disconnected. (Recall thatG∖vG\setminus vis the multigraph obtained fromGGby removing the vertexvvand all edges that containvv.)

Proposition 5.54.

LetGGbe a connected multigraph with≥2\geq 2vertices. Then, there are at least22vertices ofGGthat arenotcut-vertices.

Proof 5.55.

Pick a spanning treeTTofGG(we know from Theorem5.46that such a spanning tree exists). Then,TThas at least22leaves (by Theorem5.28(a)). But each leaf ofTTis a non-cut-vertex ofGG(why?).

5.4.6Exercises

Exercise 5.57.

LetGGbe a connected multigraph. LetT1T_{1}andT2T_{2}be two spanning trees ofGG.

Prove the following:383838Recall thatE(H)\operatorname*{E}\left(H\right)denotes the edge set of any graphHH.

(a)For anye∈E(T1)∖E(T2)e\in\operatorname*{E}\left(T_{1}\right)\setminus\operatorname*{E}\left(T_{2}\right), there exists anf∈E(T2)∖E(T1)f\in\operatorname*{E}\left(T_{2}\right)\setminus\operatorname*{E}\left(T_{1}\right)with the property that replacingeebyffinT1T_{1}(that is, removing the edgeeefromT1T_{1}and adding the edgeff) results in a spanning tree ofGG.
(b)For anyf∈E(T2)∖E(T1)f\in\operatorname*{E}\left(T_{2}\right)\setminus\operatorname*{E}\left(T_{1}\right), there exists ane∈E(T1)∖E(T2)e\in\operatorname*{E}\left(T_{1}\right)\setminus\operatorname*{E}\left(T_{2}\right)with the property that replacingeebyffinT1T_{1}(that is, removing the edgeeefromT1T_{1}and adding the edgeff) results in a spanning tree ofGG.

[Hint:The two parts look very similar, but (to my knowledge) their proofs are not.]

Exercise 5.58.

LetGGbe a connected multigraph. Let𝒮\mathcal{S}be the simple graph whose vertices are the spanning trees ofGG, and whose edges are defined as follows: Two spanning treesT1T_{1}andT2T_{2}ofGGare adjacent (as vertices of𝒮\mathcal{S}) if and only ifT2T_{2}can be obtained fromT1T_{1}by removing an edge and adding another (i.e., if and only if there exist an edgee1e_{1}ofT1T_{1}and an edgee2e_{2}ofT2T_{2}such thate2≠e1e_{2}\neq e_{1}andT2∖e2=T1∖e1T_{2}\setminus e_{2}=T_{1}\setminus e_{1}).

Prove that the simple graph𝒮\mathcal{S}is itself connected. (In simpler language: Prove that any spanning tree ofGGcan be transformed into any other spanning tree ofGGby a sequence of legal “remove an edge and add another” operations, where such an operation is calledlegalif its result is a spanning tree ofGG.)

[Example:IfGGis the multigraph

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{}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -26.85 -21.89 C -6.23 -25.53 6.23 -25.53 26.85 -21.89}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.16377pt}{-26.03119pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -2.99 -36.02)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 29.13 -12.16 C 14.27 -1.24 7.37 9.11 3.01 27.02}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{{}{}}}{{}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces 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20.93 10.24 31.84}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{23.6171pt}{7.27945pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 32.68 10.07)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}},then the graph𝒮\mathcal{S}looks as follows:

112233112233112233112233112233aaaaaabbbbccccccdddd]

Exercise 5.59.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a connected multigraph. Letw:E→ℝw:E\rightarrow\mathbb{R}be a map that assigns a real numberw⁡(e)w\left(e\right)to each edgeee. We shall call this real numberw⁡(e)w\left(e\right)theweightof the edgeee.

IfH=(W,F,φ∣F)H=\left(W,F,\varphi\mid_{F}\right)is a subgraph ofGG, then theweightw⁡(H)w\left(H\right)ofHHis defined to be∑f∈Fw⁡(f)\sum\limits_{f\in F}w\left(f\right)(that is, the sum of the weights of all edges ofHH).

Aww-minimum spanning treeofGGmeans a spanning tree ofGGthat has the smallest weight among all spanning trees ofGG.

In our first proof of Theorem5.46, we have seen a way to construct a spanning tree ofGGby successively removing non-bridges until only bridges remain. (Anon-bridgemeans an edge that is not a bridge.)

Now, let us perform this algorithm, but taking care to choose a non-bridge of largest weight (among all non-bridges) at each step. Prove that the result will be aww-minimum spanning tree.

Exercise 5.60.

LetGGbe a connected multigraph with an even number of vertices. Prove that there exists a spanning subgraphHHofGGsuch that each vertex ofHHhas odd degree (inHH).

[Hint:One way to solve this begins by reducing the problem to the case whenGGis a tree.]

5.4.7Existence and construction of a spanning forest

So we have learnt that connected graphs have spanning trees. What do disconnected graphs have?

Corollary 5.61.

Each multigraph has a spanning forest.

Proof 5.62.

Apply Theorem5.46to each component of the multigraph. Then, combine the resulting spanning trees into a spanning forest.

5.5Centers of graphs and trees

5.5.1Distances

Given a graph, we can define a “distance” between any two of its vertices, simply by counting edges on the shortest path from one to the other:

Definition 5.63.

LetGGbe a multigraph.

For any two verticesuuandvvofGG, we define thedistancebetweenuuandvvto be the smallest length of a path fromuutovv. If no such path exists, then this distance is defined to be∞\infty.

The distance betweenuuandvvis denoted byd⁡(u,v)d\left(u,v\right)or bydG(u,v)d_{G}\left(u,v\right)when the graphGGis not clear from the context.

Example 5.64.

IfGGis the multigraph from Example5.52, then

dG(1,9)=4,dG(4,13)=2,dG(4,4)=0.d_{G}\left(1,9\right)=4,\ \ \ \ \ \ \ \ \ \ d_{G}\left(4,13\right)=2,\ \ \ \ \ \ \ \ \ \ d_{G}\left(4,4\right)=0.

Proof 5.66.

Part(d)follows from Corollary3.23. The proofs of(a),(b)and(c)are then straightforward (the proof of(c)relies on part(d), because splicing two paths generally only yields a walk, not a path). Finally, in order to prove part(e), observe that any path of our multigraph has length≤|V|−1\leq\left|V\right|-1(since its vertices are distinct).

We note that the definition of a distance becomes simpler if our multigraph is a tree: Namely, ifTTis a tree, then the distanced⁡(u,v)d\left(u,v\right)between two verticesuuandvvis the length of theonlypath fromuutovvinTT. Thus, in a tree, we do not have to worry whether a given path is the shortest.

We also notice that ifGGis a multigraph, and ifuuandvvare two vertices ofGG, then the distancedG(u,v)d_{G}\left(u,v\right)inGGequals the distancedGsimp(u,v)d_{G^{\operatorname*{simp}}}\left(u,v\right)in the simple graphGsimpG^{\operatorname*{simp}}. (The reason for this is that any path ofGGcan be converted into a path ofGsimpG^{\operatorname*{simp}}having the same length, and vice versa. Of course, this is not a one-to-one correspondence, but it suffices for our purposes.) Thus, when studying distances on a multigraph, we can WLOG restrict ourselves to simple graphs.

The following few exercises give some curious properties of distances in various kinds of graphs.

Exercise 5.67.

Letaa,bbandccbe three vertices of a connected multigraphG=(V,E,φ)G=\left(V,E,\varphi\right). Prove thatd⁡(b,c)+d⁡(c,a)+d⁡(a,b)≤2|V|−2d\left(b,c\right)+d\left(c,a\right)+d\left(a,b\right)\leq 2\left|V\right|-2.

[Solution:This is Exercise 7 on midterm #1 from my Spring 2017 course, except that the simple graph has been replaced by a multigraph (but this makes no serious difference); seethe course pagefor solutions.]

Exercise 5.68.

Letaa,bbandccbe three vertices of a strongly connected multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right)such that|V|≥4\left|V\right|\geq 4. For any two verticesuuandvvofDD, we define the distanced⁡(u,v)d\left(u,v\right)to be the smallest length of a path fromuutovv. (This definition is the obvious analogue of Definition5.63for digraphs.)

(a)Prove thatd⁡(b,c)+d⁡(c,a)+d⁡(a,b)≤3|V|−4d\left(b,c\right)+d\left(c,a\right)+d\left(a,b\right)\leq 3\left|V\right|-4.
(b)For eachn≥5n\geq 5, construct an example in which|V|=n\left|V\right|=nandd⁡(b,c)+d⁡(c,a)+d⁡(a,b)=3|V|−4d\left(b,c\right)+d\left(c,a\right)+d\left(a,b\right)=3\left|V\right|-4. (No proof is required for the example.)

[Solution:This is Exercise 5 on homework set #3 from my Spring 2017 course, except that the simple digraph has been replaced by a multidigraph (but this makes no serious difference); seethe course pagefor solutions.]

Exercise 5.69.

LetGGbe a tree. Letxx,yy,zzandwwbe four vertices ofGG.

Show that the two largest ones among the three numbers

d⁡(x,y)+d⁡(z,w),d⁡(x,z)+d⁡(y,w)andd⁡(x,w)+d⁡(y,z)d\left(x,y\right)+d\left(z,w\right),\qquad d\left(x,z\right)+d\left(y,w\right)\qquad\text{and}\qquad d\left(x,w\right)+d\left(y,z\right)are equal.

[Solution:This is Exercise 6 on midterm #2 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 5.70.

LetGGbe a connected multigraph. Letxx,yy,zzandwwbe four vertices ofGG.

Assume that the two largest ones among the three numbers

Prove thatGGhas a cycle of length≤d⁡(x,z)+d⁡(y,w)+d⁡(x,w)+d⁡(y,z)\leq d\left(x,z\right)+d\left(y,w\right)+d\left(x,w\right)+d\left(y,z\right).

[Hint:This is a strengthening of Exercise5.69. Try deriving it by applying the latter exercise to a strategically chosen subgraph ofGG.]

[Solution:This is Exercise 1 on midterm #3 from my Spring 2017 course; seethe course pagefor solutions.]

5.5.2Eccentricity and centers

We can now define “eccentricities”:

Definition 5.71.

Letvvbe a vertex of a multigraphG=(V,E,φ)G=\left(V,E,\varphi\right). Theeccentricityofvv(with respect toGG) is defined to be the number

max⁡{d⁡(v,u)∣u∈V}∈ℕ∪{∞}.\max\left\{d\left(v,u\right)\ \mid\ u\in V\right\}\in\mathbb{N}\cup\left\{\infty\right\}.This eccentricity is denoted byeccv\operatorname*{ecc}voreccGv\operatorname*{ecc}\nolimits_{G}v.

Intuitively, the eccentricity of a vertexvvis “how far you can get away fromvv” onGG.

Definition 5.72.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Then, acenterofGGmeans a vertex ofGGwhose eccentricity is minimum (among all vertices).

(Some authors have a slightly different definition of a “center”: They define thecenterofGGto be thesetof all vertices ofGGwhose eccentricity is minimum. That is, what they call “center” is the set of what we call “centers”.)

Example 5.73.

LetGGbe the following multigraph:

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\lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{110.09691pt}{54.75275pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 152.34 75.76)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{\hbox{\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.15 0 L -12.47 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 0 12.47 L 0 67.22}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.47 0 L 67 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.76 78.74 L 11.52 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 78.74 66.76 L 78.74 11.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.72 78.74 L 144.57 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the eccentricities of its vertices are as follows (we are just labeling each vertex with its eccentricity):

432324.\hbox to189.09pt{\vbox to75.28pt{\pgfpicture\makeatletter\hbox{\hskip 66.0914pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -66.58 0 C -66.58 6.71 -72.03 12.16 -78.74 12.16 C -85.45 12.16 -90.9 6.71 -90.9 0 C -90.9 -6.71 -85.45 -12.16 -78.74 -12.16 C -72.03 -12.16 -66.58 -6.71 -66.58 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-59.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -82.2 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 78.74 C 12.16 85.45 6.71 90.9 0 90.9 C -6.71 90.9 -12.16 85.45 -12.16 78.74 C -12.16 72.03 -6.71 66.58 0 66.58 C 6.71 66.58 12.16 72.03 12.16 78.74 Z M 0 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 78.74 C 90.9 85.45 85.45 90.9 78.74 90.9 C 72.03 90.9 66.58 85.45 66.58 78.74 C 66.58 72.03 72.03 66.58 78.74 66.58 C 85.45 66.58 90.9 72.03 90.9 78.74 Z M 78.74 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope 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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 0 12.71 L 0 66.03}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.71 0 L 66.03 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 66.03 78.74 L 12.71 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 78.74 66.03 L 78.74 12.71}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 91.45 78.74 L 144.77 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Thus, the centers ofGGare the verticesrrandvv.

Example 5.74.

LetGGbe a complete graphKnK_{n}(withnnvertices). Then, each vertex ofGGhas the same eccentricity (which is11ifn≥2n\geq 2and00ifn=1n=1), and thus each vertex ofGGis a center ofGG.

Example 5.75.

LetGGbe a graph with more than one component. Then, each vertexvvofGGhas eccentricity∞\infty(because there exists at least one vertexuuthat lies in a different component ofGGthanvv, and thus this vertexuusatisfiesd⁡(v,u)=∞d\left(v,u\right)=\infty). Hence, each vertex ofGGis a center ofGG.

5.5.3The centers of a tree

As we see from Example5.75, eccentricity and centers are not very useful notions when the graph is disconnected. Even for a connected graph, Example5.73shows that the centers do not necessarily form a connected subgraph. However, in a tree, they behave a lot better:

Theorem 5.76.

LetTTbe a tree. Then:

(a)The treeTThas either11or22centers.
(b)IfTThas22centers, then these22centers are adjacent.
(c)Moreover, these centers can be found by the following algorithm: IfTThas more than22vertices, then we remove all leaves fromTT(simultaneously). What remains is again a tree. If that tree still has more than22vertices, we remove all leaves from it (simultaneously). The result is again a tree. If that tree still has more than22vertices, we remove all leaves from it (simultaneously), and continue doing so until we are left with a tree that has only11or22vertices. These vertices are the centers ofTT.

To prove Theorem5.76, we first study how a tree is affected when all its leaves are removed:

Lemma 5.77.

LetT=(V,E,φ)T=\left(V,E,\varphi\right)be a tree with more than22vertices.

LetLLbe the set of all leaves ofTT.

LetT∖LT\setminus Lbe the induced submultigraph ofTTon the setV∖LV\setminus L. (Thus,T∖LT\setminus Lis obtained fromTTby removing all the vertices inLLand all edges that contain a vertex inLL.)

Then:

(a)The multigraphT∖LT\setminus Lis a tree.
(b)For anyu∈V∖Lu\in V\setminus Landv∈V∖Lv\in V\setminus L, we have {paths ofTfromutov}={paths ofT∖Lfromutov}\left\{\text{paths of }T\text{ from }u\text{ to }v\right\}=\left\{\text{paths of }T\setminus L\text{ from }u\text{ to }v\right\}(that is, the paths ofTTfromuutovvare precisely the paths ofT∖LT\setminus Lfromuutovv).
(c)For anyu∈V∖Lu\in V\setminus Landv∈V∖Lv\in V\setminus L, we havedT(u,v)=dT∖L(u,v)d_{T}\left(u,v\right)=d_{T\setminus L}\left(u,v\right).
(d)Each vertexv∈V∖Lv\in V\setminus LsatisfieseccTv=eccT∖Lv+1\operatorname*{ecc}\nolimits_{T}v=\operatorname*{ecc}\nolimits_{T\setminus L}v+1.
(e)Each leafv∈Lv\in LsatisfieseccTv=eccTw+1\operatorname*{ecc}\nolimits_{T}v=\operatorname*{ecc}\nolimits_{T}w+1, wherewwis the unique neighbor ofvvinTT. (Aneighborofvvmeans a vertex that is adjacent tovv.)
(f)The centers ofTTare precisely the centers ofT∖LT\setminus L.

Example 5.78.

LetTTbe the following tree:

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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 269.91 67.37 L 243.04 13.63}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 281.27 67.37 L 308.15 13.63}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the setLLfrom Lemma5.77is{4,5,7,8,10,11}\left\{4,5,7,8,10,11\right\}, and the treeT∖LT\setminus Llooks as follows:

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Proof 5.79(Proof of Lemma5.77.).

First, we notice thatTTis a forest (sinceTTis a tree), and thus has no cycles. In particular,TTtherefore has no loops and no parallel edges. Also, for any two verticesuuandvvofTT, there is a unique path fromuutovvinTT.

Next, we introduce some terminology: If𝐩\mathbf{p}is a path of some multigraph, then anintermediate vertexof𝐩\mathbf{p}shall mean a vertex of𝐩\mathbf{p}that is neither the starting point nor the ending point of𝐩\mathbf{p}. In other words, if𝐩=(p0,e1,p1,e2,p2,…,ek,pk)\mathbf{p}=\left(p_{0},e_{1},p_{1},e_{2},p_{2},\ldots,e_{k},p_{k}\right)is a path of some multigraph, then the intermediate vertices of𝐩\mathbf{p}arep1,p2,…,pk−1p_{1},p_{2},\ldots,p_{k-1}. Clearly, any intermediate vertex of a path𝐩\mathbf{p}must have degree≥2\geq 2(since the path𝐩\mathbf{p}enters it along some edge, and leaves it along another). Hence, if𝐩\mathbf{p}is a path ofTT, then

any intermediate vertex of𝐩must belong toV∖L\text{any intermediate vertex of }\mathbf{p}\text{ must belong to }V\setminus L(14)(because it must have degree≥2\geq 2, thus cannot be a leaf ofTT; but this means that it cannot belong toLL; therefore, it must belong toV∖LV\setminus L).

(b)Letu∈V∖Lu\in V\setminus Landv∈V∖Lv\in V\setminus L. Let𝐩\mathbf{p}be a path ofTTfromuutovv. We shall show that𝐩\mathbf{p}is a path ofT∖LT\setminus Las well.

Indeed, let us first check that all vertices of𝐩\mathbf{p}belong toV∖LV\setminus L. This is clear for the verticesuuandvv(sinceu∈V∖Lu\in V\setminus Landv∈V∖Lv\in V\setminus L); but it also holds for every intermediate vertex of𝐩\mathbf{p}(by (14)). Thus, it does indeed hold for all vertices of𝐩\mathbf{p}.

We have thus shown that all vertices of𝐩\mathbf{p}belong toV∖LV\setminus L. Hence,𝐩\mathbf{p}is a path ofT∖LT\setminus L(sinceT∖LT\setminus Lis the induced submultigraph ofTTon the setV∖LV\setminus L).

Forget that we fixed𝐩\mathbf{p}. We have thus shown that every path𝐩\mathbf{p}ofTTfromuutovvis also a path ofT∖LT\setminus L. Hence,

{paths ofTfromutov}⊆{paths ofT∖Lfromutov}.\left\{\text{paths of }T\text{ from }u\text{ to }v\right\}\subseteq\left\{\text{paths of }T\setminus L\text{ from }u\text{ to }v\right\}.Conversely, we have

{paths ofT∖Lfromutov}⊆{paths ofTfromutov},\left\{\text{paths of }T\setminus L\text{ from }u\text{ to }v\right\}\subseteq\left\{\text{paths of }T\text{ from }u\text{ to }v\right\},since every path ofT∖LT\setminus Lis a path ofTT(becauseT∖LT\setminus Lis a submultigraph ofTT). Combining these two facts, we obtain

{paths ofTfromutov}={paths ofT∖Lfromutov}.\left\{\text{paths of }T\text{ from }u\text{ to }v\right\}=\left\{\text{paths of }T\setminus L\text{ from }u\text{ to }v\right\}.This proves Lemma5.77(b).

(c)This follows from Lemma5.77(b), since the distancedG(u,v)d_{G}\left(u,v\right)of two verticesuuandvvin a graphGGis defined to be the smallest length of a path fromuutovv.

(a)The graphTTis a tree, thus a forest. Hence, its submultigraphT∖LT\setminus Lis a forest as well (since any cycle ofT∖LT\setminus Lwould be a cycle ofTT). It thus remains to show thatT∖LT\setminus Lis connected.

First, it is easy to see thatT∖LT\setminus Lhas at least one vertex393939Proof.We assumed thatTThas more than22vertices. In other words, there exist three distinct verticesu,v,wu,v,wofTT. Consider theseu,v,wu,v,w. If all three distancesdT(u,v)d_{T}\left(u,v\right),dT(v,w)d_{T}\left(v,w\right)anddT(w,u)d_{T}\left(w,u\right)were equal to11, thenTTwould have a cycle (of the form(u,∗,v,∗,w,∗,u)\left(u,\ast,v,\ast,w,\ast,u\right), where each asterisk stands for some edge); but this would contradict the fact thatTThas no cycles. Thus, not all of these three distances are equal to11. Hence, at least one of them is≠1\neq 1. WLOG assume thatdT(u,v)≠1d_{T}\left(u,v\right)\neq 1(otherwise, we permuteu,v,wu,v,w). Hence, the path fromuutovvhas more than one edge (indeed, it must have at least one edge, sinceuuandvvare distinct). Therefore, this path has at least one intermediate vertex. This intermediate vertex then must belong toV∖LV\setminus L(by (14)). Hence, it is a vertex of the subgraphT∖LT\setminus L. This shows thatT∖LT\setminus Lhas at least one vertex.. It remains to show that any two vertices ofT∖LT\setminus Lare path-connected.

Letuuandvvbe two vertices ofT∖LT\setminus L. Then,u∈V∖Lu\in V\setminus Landv∈V∖Lv\in V\setminus L. Hence, Lemma5.77(b)yields

{paths ofTfromutov}={paths ofT∖Lfromutov}.\left\{\text{paths of }T\text{ from }u\text{ to }v\right\}=\left\{\text{paths of }T\setminus L\text{ from }u\text{ to }v\right\}.Thus,{paths ofT∖Lfromutov}={paths ofTfromutov}≠∅\left\{\text{paths of }T\setminus L\text{ from }u\text{ to }v\right\}=\left\{\text{paths of }T\text{ from }u\text{ to }v\right\}\neq\varnothing(since there exists a path ofTTfromuutovv(becauseTTis connected)). In other words, there exists a path ofT∖LT\setminus Lfromuutovv. In other words,uuandvvare path-connected inT∖LT\setminus L.

We have now shown that any two verticesuuandvvofT∖LT\setminus Lare path-connected inT∖LT\setminus L. This entails thatT∖LT\setminus Lis connected (sinceT∖LT\setminus Lhas at least one vertex). This proves Lemma5.77(a).

(d)Ifuuandvvare two vertices ofT∖LT\setminus L, then the two distancesdT(u,v)d_{T}\left(u,v\right)anddT∖L(u,v)d_{T\setminus L}\left(u,v\right)are equal (by Lemma5.77(c)); thus, we shall denote both distances byd⁡(u,v)d\left(u,v\right)(since there is no confusion to be afraid of).

Letv∈V∖Lv\in V\setminus L. We must show thateccTv=eccT∖Lv+1\operatorname*{ecc}\nolimits_{T}v=\operatorname*{ecc}\nolimits_{T\setminus L}v+1.

Letuube a vertex ofT∖LT\setminus Lsuch thatd⁡(v,u)d\left(v,u\right)is maximum. Thus,eccT∖Lv=d⁡(v,u)\operatorname*{ecc}\nolimits_{T\setminus L}v=d\left(v,u\right)(by the definition ofeccT∖Lv\operatorname*{ecc}\nolimits_{T\setminus L}v). However,uuis a vertex ofT∖LT\setminus L, and thus does not belong toLL. Hence,uuis not a leaf ofTT(sinceLLis the set of all leaves ofTT). Hence,uuhas degree≥2\geq 2inTT(since a vertex in a tree with more than11vertex cannot have degree00).

Now, consider the path𝐩\mathbf{p}fromvvtouuin the treeTT. This path𝐩\mathbf{p}has lengthd⁡(v,u)d\left(v,u\right). Sinceuuhas degree≥2\geq 2, there exist at least two edges ofTTthat containuu. Hence, in particular, there exists at least one edgeffthat containsuuand is distinct from the last edge of𝐩\mathbf{p}404040If the path𝐩\mathbf{p}has no edges, thenffcan be any edge that containsuu.. Consider this edgeff. Letwwbe the endpoint offfother thanuu. Appendingffandwwto the end of the path𝐩\mathbf{p}, we obtain a walk fromvvtoww. This walk is backtrack-free (sinceffis distinct from the last edge of𝐩\mathbf{p}) and thus must be a path (by Proposition5.2, sinceTThas no cycles). This path has lengthd⁡(v,u)+1d\left(v,u\right)+1(since it was obtained by appending an edge to the path𝐩\mathbf{p}, which has lengthd⁡(v,u)d\left(v,u\right)). Hence,d⁡(v,w)=d⁡(v,u)+1d\left(v,w\right)=d\left(v,u\right)+1(since there is only one path fromvvtoww, and we just showed that this path has lengthd⁡(v,u)+1d\left(v,u\right)+1). But the definition of eccentricity yields

eccTv≥d⁡(v,w)=d⁡(v,u)⏟=eccT∖Lv+1=eccT∖Lv+1.\operatorname*{ecc}\nolimits_{T}v\geq d\left(v,w\right)=\underbrace{d\left(v,u\right)}_{=\operatorname*{ecc}\nolimits_{T\setminus L}v}+\,1=\operatorname*{ecc}\nolimits_{T\setminus L}v+1.(15) On the other hand, letxxbe a vertex ofTTsuch thatd⁡(v,x)d\left(v,x\right)is maximum. Thus,eccTv=d⁡(v,x)\operatorname*{ecc}\nolimits_{T}v=d\left(v,x\right)(by the definition ofeccTv\operatorname*{ecc}\nolimits_{T}v). The path fromvvtoxxhas length≥1\geq 1(since otherwise, we would havex=vx=vand therefored⁡(v,x)=d⁡(v,v)=0d\left(v,x\right)=d\left(v,v\right)=0, which would easily contradict the maximality ofd⁡(v,x)d\left(v,x\right)). Thus, it has a second-to-last vertex. Letyybe this second-to-last vertex. Then, the path fromvvtoyyis simply the path fromvvtoxxwith its last edge removed. Consequently,d⁡(v,y)=d⁡(v,x)−1d\left(v,y\right)=d\left(v,x\right)-1. However, it is easy to see thaty∈V∖Ly\in V\setminus L414141Proof.Assume the contrary. Thus,y∉V∖Ly\notin V\setminus L. Hence,y≠vy\neq v(sincey∉V∖Ly\notin V\setminus Lbutv∈V∖Lv\in V\setminus L). However,yyis the second-to-last vertex of the path fromvvtoxx. Therefore,yyis either the starting pointvvof this path, or an intermediate vertex of this path. Sincey≠vy\neq v, we thus conclude thatyyis an intermediate vertex of this path. Hence, by (14), we see thatyymust belong toV∖LV\setminus L. But this contradictsy∉V∖Ly\notin V\setminus L. This contradiction shows that our assumption was false, qed.. In other words,yyis a vertex ofT∖LT\setminus L. Thus, the definition of eccentricity yields

eccT∖Lv≥d⁡(v,y)=d⁡(v,x)⏟=eccTv−1=eccTv−1,\operatorname*{ecc}\nolimits_{T\setminus L}v\geq d\left(v,y\right)=\underbrace{d\left(v,x\right)}_{=\operatorname*{ecc}\nolimits_{T}v}-\,1=\operatorname*{ecc}\nolimits_{T}v-1,so thateccTv≤eccT∖Lv+1\operatorname*{ecc}\nolimits_{T}v\leq\operatorname*{ecc}\nolimits_{T\setminus L}v+1. Combining this with (15), we obtaineccTv=eccT∖Lv+1\operatorname*{ecc}\nolimits_{T}v=\operatorname*{ecc}\nolimits_{T\setminus L}v+1. This proves Lemma5.77(d).

(e)Ifuuandvvare two vertices ofT∖LT\setminus L, then the two distancesdT(u,v)d_{T}\left(u,v\right)anddT∖L(u,v)d_{T\setminus L}\left(u,v\right)are equal (by Lemma5.77(c)); thus, we shall denote both distances byd⁡(u,v)d\left(u,v\right)(since there is no confusion to be afraid of).

Letv∈Lv\in Lbe a leaf. Letwwbe the unique neighbor ofvvinTT. We must prove thateccTv=eccTw+1\operatorname*{ecc}\nolimits_{T}v=\operatorname*{ecc}\nolimits_{T}w+1.

We first claim that

d(v,u)=d(w,u)+1for eachu∈V∖{v}.d\left(v,u\right)=d\left(w,u\right)+1\ \ \ \ \ \ \ \ \ \ \text{for each }u\in V\setminus\left\{v\right\}.(16) [Proof of (16):We havedeg⁡v=1\deg v=1(sincevvis a leaf). In other words, there is a unique edge ofTTthat containsvv. Leteebe this edge. The endpoints ofeearevvandww(sincewwis the unique neighbor ofvv). Thus,v≠wv\neq w(sinceTThas no loops) andd⁡(v,w)=1d\left(v,w\right)=1.

Now, letu∈V∖{v}u\in V\setminus\left\{v\right\}. Then, the path fromvvtouuinTTmust have length≥1\geq 1(sinceu≠vu\neq v), and therefore must begin with the edgeee(sinceeeis the only edge that containsvv). If we remove this edgeeefrom this path, we thus obtain a path fromwwtouu. As a consequence, the path fromvvtouuis longer by exactly11edge than the path fromwwtouu. In other words, we haved⁡(v,u)=d⁡(w,u)+1d\left(v,u\right)=d\left(w,u\right)+1. This proves (16).]

Now, the definition of eccentricity yields

eccTv=max⁡{d⁡(v,u)∣u∈V}.\operatorname*{ecc}\nolimits_{T}v=\max\left\{d\left(v,u\right)\ \mid\ u\in V\right\}.(17)This maximum is clearlynotattained foru=vu=v(sinced⁡(v,v)=0d\left(v,v\right)=0is smaller thand⁡(v,w)=1d\left(v,w\right)=1). Thus, this maximum does not change if we removevvfrom its indexing setVV. Hence, (17) rewrites as

eccTv\displaystyle\operatorname*{ecc}\nolimits_{T}v=max{d⁡(v,u)⏟=d⁡(w,u)+1(by (16))∣u∈V∖{v}}\displaystyle=\max\left\{\underbrace{d\left(v,u\right)}_{\begin{subarray}{c}=d\left(w,u\right)+1\\ \text{(by (\ref{pf.lem.eccent.tree-TL.e.d}))}\end{subarray}}\ \mid\ u\in V\setminus\left\{v\right\}\right\}=max⁡{d⁡(w,u)+1∣u∈V∖{v}}\displaystyle=\max\left\{d\left(w,u\right)+1\ \mid\ u\in V\setminus\left\{v\right\}\right\}=max⁡{d⁡(w,u)∣u∈V∖{v}}+1.\displaystyle=\max\left\{d\left(w,u\right)\ \mid\ u\in V\setminus\left\{v\right\}\right\}+1.(18) On the other hand, the definition of eccentricity yields

eccTw=max⁡{d⁡(w,u)∣u∈V}.\operatorname*{ecc}\nolimits_{T}w=\max\left\{d\left(w,u\right)\ \mid\ u\in V\right\}.(19)We shall now show that this maximum does not change if we removevvfrom its indexing setVV. In other words, we shall show that

max⁡{d⁡(w,u)∣u∈V}=max⁡{d⁡(w,u)∣u∈V∖{v}}.\max\left\{d\left(w,u\right)\ \mid\ u\in V\right\}=\max\left\{d\left(w,u\right)\ \mid\ u\in V\setminus\left\{v\right\}\right\}.(20) [Proof of (20):Assume that (20) is false. Then, the maximummax⁡{d⁡(w,u)∣u∈V}\max\left\{d\left(w,u\right)\ \mid\ u\in V\right\}is attainedonlyatu=vu=v. In other words, we have

d(w,v)>d(w,u)for allu∈V∖{v}.d\left(w,v\right)>d\left(w,u\right)\ \ \ \ \ \ \ \ \ \ \text{for all }u\in V\setminus\left\{v\right\}.(21)However, the treeTThas more than22vertices. Thus, it has a vertexuuthat is distinct from bothvvandww. Consider thisuu. Thus,u∈V∖{v}u\in V\setminus\left\{v\right\}, so that (21) yieldsd⁡(w,v)>d⁡(w,u)d\left(w,v\right)>d\left(w,u\right). In view ofd⁡(w,v)=d⁡(v,w)=1d\left(w,v\right)=d\left(v,w\right)=1, this rewrites as1>d⁡(w,u)1>d\left(w,u\right), so thatd⁡(w,u)<1d\left(w,u\right)<1. Therefore,w=uw=u. But this contradicts the fact thatwwis distinct fromuu. This contradiction shows that our assumption was false, and thus (20) is proved.]

Now, (18) becomes

eccTv\displaystyle\operatorname*{ecc}\nolimits_{T}v=max⁡{d⁡(w,u)∣u∈V∖{v}}⏟=max⁡{d⁡(w,u)∣u∈V}(by (20))+1\displaystyle=\underbrace{\max\left\{d\left(w,u\right)\ \mid\ u\in V\setminus\left\{v\right\}\right\}}_{\begin{subarray}{c}=\max\left\{d\left(w,u\right)\ \mid\ u\in V\right\}\\ \text{(by (\ref{pf.lem.eccent.tree-TL.e.eccTw2}))}\end{subarray}}+\,1=max⁡{d⁡(w,u)∣u∈V}⏟=eccTw(by (19))+1=eccTw+1.\displaystyle=\underbrace{\max\left\{d\left(w,u\right)\ \mid\ u\in V\right\}}_{\begin{subarray}{c}=\operatorname*{ecc}\nolimits_{T}w\\ \text{(by (\ref{pf.lem.eccent.tree-TL.e.eccTw}))}\end{subarray}}+\,1=\operatorname*{ecc}\nolimits_{T}w+1.This proves Lemma5.77(e).

(f)Lemma5.77(e)shows that any vertexv∈Lv\in Lhas a higher eccentricity than its unique neighbor. Thus, a vertexvvofTTthat minimizeseccTv\operatorname*{ecc}\nolimits_{T}vcannot belong toLL. In other words, a vertexvvofTTthat minimizeseccTv\operatorname*{ecc}\nolimits_{T}vmust belong toV∖LV\setminus L.

However, the centers ofTTare defined to be the vertices ofTTthat minimizeeccTv\operatorname*{ecc}\nolimits_{T}v. As we just proved, these vertices must belong toV∖LV\setminus L. Thus, the centers ofTTcan also be characterized as the verticesv∈V∖Lv\in V\setminus Lthat minimizeeccTv\operatorname*{ecc}\nolimits_{T}v. However, a vertexv∈V∖Lv\in V\setminus LminimizeseccTv\operatorname*{ecc}\nolimits_{T}vif and only if it minimizeseccT∖Lv\operatorname*{ecc}\nolimits_{T\setminus L}v(because Lemma5.77(d)yieldseccTv=eccT∖Lv+1\operatorname*{ecc}\nolimits_{T}v=\operatorname*{ecc}\nolimits_{T\setminus L}v+1for any such vertexvv). Thus, we conclude that the centers ofTTcan be characterized as the verticesv∈V∖Lv\in V\setminus Lthat minimizeeccT∖Lv\operatorname*{ecc}\nolimits_{T\setminus L}v. But this is precisely the definition of the centers ofT∖LT\setminus L. As a consequence, we see that the centers ofTTare precisely the centers ofT∖LT\setminus L. This proves Lemma5.77(f).

Proof 5.80(Proof of Theorem5.76.).

We shall prove parts(a)and(b)of Theorem5.76by strong induction on|V(T)|\left|\operatorname*{V}\left(T\right)\right|:

Induction step:Consider a treeTT. Assume that parts(a)and(b)of Theorem5.76are true for any tree with fewer than|V(T)|\left|\operatorname*{V}\left(T\right)\right|many vertices. We must now prove these parts for our treeTT.

If|V(T)|≤2\left|\operatorname*{V}\left(T\right)\right|\leq 2, then both parts are obvious. Hence, WLOG assume that|V(T)|>2\left|\operatorname*{V}\left(T\right)\right|>2. Thus, the treeTThas more than22vertices. LetLLbe the set of all leaves ofTT. Note that|L|≥2\left|L\right|\geq 2(since Theorem5.28(a)shows that any tree with at least22vertices has at least22leaves). Define the multigraphT∖LT\setminus Las in Lemma5.77. Then, Lemma5.77(f)shows that the centers ofTTare precisely the centers ofT∖LT\setminus L.

However, Lemma5.77(a)yields thatT∖LT\setminus Lis again a tree. This tree has fewer vertices thanTT(since|L|≥2>0\left|L\right|\geq 2>0). Hence, by the induction hypothesis, both parts(a)and(b)of Theorem5.76are true for the treeT∖LT\setminus Linstead ofTT. In other words, the treeT∖LT\setminus Lhas either11or22centers, and if it has22centers, then these22centers are adjacent. Since the centers ofTTare precisely the centers ofT∖LT\setminus L, we can rewrite this as follows: The treeTThas either11or22centers, and if it has22centers, then these22centers are adjacent. In other words, parts(a)and(b)of Theorem5.76hold for our treeTT. This completes the induction step. Thus, parts(a)and(b)of Theorem5.76are proved.

(c)This follows from Lemma5.77(f). Indeed, ifTThas at most22vertices, then all vertices ofTTare centers ofTT(this is trivial to check). If not, then each “leaf-removal” step of our algorithm leaves the set of centers ofTTunchanged (by Lemma5.77(f)), and thus the centers of the original treeTTare precisely the centers of the tree that remains at the end of the algorithm. But the latter tree has at most22vertices, and thus its centers are precisely its vertices. So the centers ofTTare precisely the vertices that remain at the end of the algorithm. Theorem5.76(c)is proven.

The following exercise shows another approach to the centers of a tree:

Exercise 5.81.

LetTTbe a tree. Let𝐩=(p0,∗,p1,∗,p2,…,∗,pm)\mathbf{p}=\left(p_{0},\ast,p_{1},\ast,p_{2},\ldots,\ast,p_{m}\right)be a longest path ofTT. (We write asterisks for the edges since we don’t need to name them.)

Prove the following:

(a)Ifmmis even, then the only center ofTTispm/2p_{m/2}.
(b)Ifmmis odd, then the two centers ofTTarep(m−1)/2p_{\left(m-1\right)/2}andp(m+1)/2p_{\left(m+1\right)/2}.

The notion of acentroidof a tree is a relative of the notion of a center. We briefly discuss it in the following exercise:

Exercise 5.83.

LetTTbe a tree. For any vertexvvofTT, we letcvc_{v}denote the size of the largest component of the graphT∖vT\setminus v. (Recall thatT∖vT\setminus vis the graph obtained fromTTby removing the vertexvvand all edges that containvv. Note that a component (according to our definition) is a set of vertices; thus, its size is the number of vertices in it.)

The verticesvvofTTthat minimize the numbercvc_{v}are called thecentroidsofTT.

(a)Prove thatTThas no more than two centroids, and furthermore, ifTThas two centroids, then these two centroids are adjacent.
(b)Find a treeTTsuch that the centroid(s) ofTTare distinct from the center(s) ofTT.

[Example:Here is an example of a treeTT, where each vertexvvis labelled with the corresponding numbercvc_{v}:

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\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 236.22 170.19 L 236.22 223.51}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 299.72 236.22 L 248.93 236.22}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Thus, the vertex labelled55is the only centroid of this treeTT.]

Note the analogy between Exercise5.83(a)and Theorem5.76(a)and(b).

5.6Arborescences

5.6.1Definitions

Enough about undirected graphs.

What would be a directed analogue of a tree? I.e., what kind of digraphs play the same role among digraphs that trees do among undirected graphs?

Trees are graphs that are connected and have no cycles. This suggests two directed versions:

•We can study digraphs that are strongly connected and have no cycles. Unfortunately, there is not much to study: Any such digraph has only11vertex and no arcs. (Make sure you understand why!)
•We can drop the connectedness requirement. Digraphs that have no cycles are calledacyclic, and more typically they are calleddags(short for “directed acyclic graphs”). However, these dags aren’t quite like trees. For example, a tree always has fewer edges than vertices, but a dag can have more arcs than vertices.424242For example, here is a dag with44vertices and55arcs:.\hbox to74.55pt{\vbox to74.55pt{\pgfpicture\makeatletter\hbox{\hskip 5.11386pt\lower-5.11386pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} 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Here is a more convincing analogue of trees for digraphs:434343We recall that we defined a multigraphDundD^{\operatorname*{und}}for every multidigraphDD(in Definition4.14). Roughly speaking, this multigraphDundD^{\operatorname*{und}}is obtained by “forgetting the directions” of the arcs ofDD. Parallel arcs are not merged into one. For example,11221122ifD=D=, thenDund=D^{\operatorname{und}}=.

Definition 5.84.

LetDDbe a multidigraph. Letrrbe a vertex ofDD.

(a)We say thatrris afrom-root(or, short,root) ofDDif for each vertexvvofDD, the digraphDDhas a path fromrrtovv.
(b)We say thatDDis anarborescence rooted fromrrifrris a from-root ofDDand the undirected multigraphDundD^{\operatorname*{und}}has no cycles. (Recall thatDundD^{\operatorname*{und}}is the multigraph obtained fromDDby turning each arc into an undirected edge. Parallel arcs are not merged into one!)

Of course, there are analogous notions of a “to-root” and an “arborescence rooted towardsrr”, but these are just the same notions that we just defined with all arrows reversed. So we need not study them separately; we can just take any property of “rooted from” and reverse all arcs to make it into a property of “rooted to”.

Example 5.85.

The multidigraph

has three from-roots (namely,00,11and22). It is not an arborescence rooted from any of them, because turning each arc into an undirected edge yields a graph with a cycle.

If we reverse the arc from00to11, then we obtain a multidigraph

which has only one from-root (namely,11) and is still not an arborescence (for the same reason as before).

Example 5.86.

Consider the following multidigraph:

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\lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This is an arborescence rooted from66. Indeed, it has paths from66to all vertices, and turning each arc into an undirected edge yields a tree.

If we reverse the arc from11to22, we obtain a multidigraph

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5.6.2Arborescences vs. trees: statement

The above examples suggest that an arborescence rooted fromrris basically the same as a tree, whose all edges have been “oriented away fromrr”. More precisely:

Theorem 5.87.

LetDDbe a multidigraph, and letrrbe a vertex ofDD. Then, the following two statements are equivalent:

•Statement C1:The multidigraphDDis an arborescence rooted fromrr.
•Statement C2:The undirected multigraphDundD^{\operatorname*{und}}is a tree, and each arc ofDDis “oriented away fromrr” (this means the following: the source of this arc lies on the unique path betweenrrand the target of this arc onDundD^{\operatorname*{und}}).

This is an easy theorem to believe, but an annoyingly hard one to formally prove in full detail! We shall prove this theorem later.

5.6.3The arborescence equivalence theorem

First, let us show another bunch of equivalent criteria for arborescences, imitating the tree equivalence theorem (Theorem5.21):

Theorem 5.88(The arborescence equivalence theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph with a from-rootrr. Then, the following six statements are equivalent:

•Statement A1:The multidigraphDDis an arborescence rooted fromrr.
•Statement A2:We have|A|=|V|−1\left|A\right|=\left|V\right|-1.
•Statement A3:The multigraphDundD^{\operatorname*{und}}is a tree.
•Statement A4:For each vertexv∈Vv\in V, the multidigraphDDhas a unique walk fromrrtovv.
•Statement A5:If we remove any arc fromDD, then the vertexrrwill no longer be a from-root of the resulting multidigraph.
•Statement A6:We havedeg−⁡r=0\deg^{-}r=0, and eachv∈V∖{r}v\in V\setminus\left\{r\right\}satisfiesdeg−⁡v=1\deg^{-}v=1.

Proof 5.89.

We will prove the implications A1⟹\LongrightarrowA4⟹\LongrightarrowA5⟹\LongrightarrowA6⟹\LongrightarrowA2⟹\LongrightarrowA3⟹\LongrightarrowA1. Since these implications form a cycle that includes all six statements, this will entail that all six statements are equivalent.

Before we prove these implications, we introduce a notation: Ifaais any arc ofDD, thenD∖aD\setminus ashall denote the multidigraph obtained fromDDby removing this arcaa. (Formally, this means thatD∖a:=(V,A∖{a},ψ∣A∖{a})D\setminus a:=\left(V,\ A\setminus\left\{a\right\},\ \psi\mid_{A\setminus\left\{a\right\}}\right).)

We now come to the proofs of the promised implications.

Proof of the implication A1⟹\LongrightarrowA4:Assume that Statement A1 holds. Thus,DDis an arborescence rooted fromrr. In other words,rris a from-root ofDDand the undirected multigraphDundD^{\operatorname*{und}}has no cycles.

We must show that for each vertexv∈Vv\in V, the multidigraphDDhas a unique walk fromrrtovv. The existence of such a walk is clear (becauserris a from-root ofDD). It is the uniqueness that we need to prove.

Assume the contrary. Thus, there exists a vertexv∈Vv\in Vsuch that two distinct walks𝐮\mathbf{u}and𝐯\mathbf{v}fromrrtovvexist. However, the multigraphDDhas no loops (since any loop ofDDwould be a loop ofDundD^{\operatorname*{und}}, and thus create a cycle ofDundD^{\operatorname*{und}}, but we know thatDundD^{\operatorname*{und}}has no cycles). Hence, any walk ofDDis automatically a backtrack-free walk ofDundD^{\operatorname*{und}}(indeed, it is backtrack-free because the only way two consecutive arcs of a walk in adigraphcan be equal is if they are loops). Therefore, the two walks𝐮\mathbf{u}and𝐯\mathbf{v}ofDDare two backtrack-free walks ofDundD^{\operatorname*{und}}. Thus, there are two distinct backtrack-free walks fromrrtovvinDundD^{\operatorname*{und}}(namely,𝐮\mathbf{u}and𝐯\mathbf{v}). Theorem5.4thus lets us conclude thatDundD^{\operatorname*{und}}has a cycle. But this contradicts the fact thatDundD^{\operatorname*{und}}has no cycles.

This contradiction shows that our assumption was wrong. Hence, we have proved that for each vertexv∈Vv\in V, the multidigraphDDhas a unique walk fromrrtovv. In other words, Statement A4 holds.

Proof of the implication A4⟹\LongrightarrowA5:Assume that Statement A4 holds.

Let nowaabe any arc ofDD. We shall show thatrris not a from-root of the multidigraphD∖aD\setminus a.

Indeed, letssbe the source andttthe target of the arcaa. We shall show that the digraphD∖aD\setminus ahas no path fromrrtott.

Indeed, assume the contrary. Thus,D∖aD\setminus ahas some path𝐩\mathbf{p}fromrrtott. This path does not use the arcaa(since it is a path ofD∖aD\setminus a).

On the other hand, we have assumed that Statement A4 holds. Applying this statement tov=sv=s, we conclude that the multidigraphDDhas a unique walk fromrrtoss. Let(v0,a1,v1,a2,v2,…,ak,vk)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)be this walk. By appending the arcaaand the vertexttto its end, we extend it to a longer walk

(v0,a1,v1,a2,v2,…,ak,vk,a,t),\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k},a,t\right),which is a walk fromrrtott. We denote this walk by𝐪\mathbf{q}.

We have now found two walks fromrrtottin the digraphDD: namely, the path𝐩\mathbf{p}and the walk𝐪\mathbf{q}. These two walks are distinct (since𝐪\mathbf{q}uses the arcaa, but𝐩\mathbf{p}does not). However, Statement A4 (applied tov=tv=t) yields that the multidigraphDDhas auniquewalk fromrrtott. This contradicts the fact that we just have found two distinct such walks.

This contradiction shows that our assumption was false. Hence, the digraphD∖aD\setminus ahas no path fromrrtott. Thus,rris not a from-root ofD∖aD\setminus a.

Forget that we fixedaa. We have now proved that ifaais any arc ofDD, thenrris not a from-root ofD∖aD\setminus a. In other words, if we remove any arc fromDD, then the vertexrrwill no longer be a from-root of the resulting multidigraph. Thus, Statement A5 holds.

Proof of the implication A5⟹\LongrightarrowA6:Assume that Statement A5 holds. We must prove that Statement A6 holds. In other words, we must prove thatdeg−⁡r=0\deg^{-}r=0, and that eachv∈V∖{r}v\in V\setminus\left\{r\right\}satisfiesdeg−⁡v=1\deg^{-}v=1.

Let us first prove thatdeg−⁡r=0\deg^{-}r=0. Indeed, assume the contrary. Thus,deg−⁡r≠0\deg^{-}r\neq 0, so that there exists an arcaawith targetrr. We shall show thatrris a from-root ofD∖aD\setminus a.

The arcaahas targetrr. Thus, a path that starts atrrcannot use this arcaa(because this arc would lead it back torr, but a path is not allowed to revisit any vertex), and therefore must be a path ofD∖aD\setminus a. Thus we have shown that any path ofDDthat starts atrris also a path ofD∖aD\setminus a. However, for each vertexvvofDD, the digraphDDhas a path fromrrtovv(sincerris a from-root ofDD). This path is also a path ofD∖aD\setminus a(since any path ofDDthat starts atrris also a path ofD∖aD\setminus a). Thus, for each vertexvvofD∖aD\setminus a, the digraphD∖aD\setminus ahas a path fromrrtovv. In other words,rris a from-root ofD∖aD\setminus a. However, we have assumed that Statement A5 holds. Thus, in particular, if we remove the arcaafromDD, then the vertexrrwill no longer be a from-root of the resulting multidigraph. In other words,rris not a from-root ofD∖aD\setminus a. But this contradicts the fact thatrris a from-root ofD∖aD\setminus a.

This contradiction shows that our assumption was false. Hence,deg−⁡r=0\deg^{-}r=0is proved.

Now, letv∈V∖{r}v\in V\setminus\left\{r\right\}be arbitrary. We must show thatdeg−⁡v=1\deg^{-}v=1.

Indeed, assume the contrary. Thus,deg−⁡v≠1\deg^{-}v\neq 1. Using the fact thatrris a from-root ofDD, it is thus easy to see thatdeg−⁡v≥2\deg^{-}v\geq 2444444Proof.Sincerris a from-root ofDD, we know that the digraphDDhas a path fromrrtovv. Sincev≠rv\neq r(becausev∈V∖{r}v\in V\setminus\left\{r\right\}), this path must have at least one arc. The last arc of this path is clearly an arc with targetvv. Thus, there exists at least one arc with targetvv. In other words,deg−⁡v≥1\deg^{-}v\geq 1. Combining this withdeg−⁡v≠1\deg^{-}v\neq 1, we obtaindeg−⁡v>1\deg^{-}v>1. In other words,deg−⁡v≥2\deg^{-}v\geq 2.. Hence, there exist two distinct arcsaaandbbwith targetvv. Consider these arcsaaandbb.

We are in one of the following three cases:

Case 1:The digraphD∖aD\setminus ahas a path fromrrtovv.

Case 2:The digraphD∖bD\setminus bhas a path fromrrtovv.

Case 3:Neither the digraphD∖aD\setminus anor the digraphD∖bD\setminus bhas a path fromrrtovv.

Let us first consider Case 1. In this case, the digraphD∖aD\setminus ahas a path fromrrtovv. Let𝐩\mathbf{p}be such a path.

We have assumed that Statement A5 holds. Thus, in particular, if we remove the arcaafromDD, then the vertexrrwill no longer be a from-root of the resulting multidigraph. In other words,rris not a from-root ofD∖aD\setminus a. In other words, there exists a vertexw∈Vw\in Vsuch that the digraphD∖aD\setminus ahas no path fromrrtoww(by the definition of a “from-root”). Consider this vertexww.

The digraphDDhas a path𝐪\mathbf{q}fromrrtoww(sincerris a from-root ofDD). Consider this path𝐪\mathbf{q}. If the path𝐪\mathbf{q}did not use the arcaa, then it would be a path ofD∖aD\setminus aas well, but this would contradict the fact thatD∖aD\setminus ahas no path fromrrtoww. Thus, the path𝐪\mathbf{q}must use the arcaa.

Consider the part of𝐪\mathbf{q}that comes after the arcaa. This part must be a path fromvvtoww(since the arcaahas targetvv, whereas the path𝐪\mathbf{q}has ending pointww). Let us denote this path by𝐪′\mathbf{q}^{\prime}. Thus, the path𝐪′\mathbf{q}^{\prime}does not use the arcaa(since it was defined as the part of𝐪\mathbf{q}that comes afteraa). Hence,𝐪′\mathbf{q}^{\prime}is a path ofD∖aD\setminus a.

Now, we know that the digraphD∖aD\setminus ahas a path𝐩\mathbf{p}fromrrtovvas well as a path𝐪′\mathbf{q}^{\prime}fromvvtoww. Splicing these paths together, we obtain a walk𝐩∗𝐪′\mathbf{p}\ast\mathbf{q}^{\prime}fromrrtoww. So we know thatD∖aD\setminus ahas a walk fromrrtoww. According to Corollary3.23, we thus conclude thatD∖aD\setminus ahas a path fromrrtoww. This contradicts the fact thatD∖aD\setminus ahas no path fromrrtoww.

We have thus obtained a contradiction in Case 1.

The same argument (but with the roles ofaaandbbinterchanged) results in a contradiction in Case 2.

Let us finally consider Case 3. In this case, neither the digraphD∖aD\setminus anor the digraphD∖bD\setminus bhas a path fromrrtovv. However, the digraphDDhas a path𝐩\mathbf{p}fromrrtovv(sincerris a from-root ofDD). Consider this path𝐩\mathbf{p}. If this path𝐩\mathbf{p}did not use the arcaa, then it would be a path ofD∖aD\setminus a, but this would contradict our assumption that the digraphD∖aD\setminus ahas no path fromrrtovv. Thus, this path𝐩\mathbf{p}must use the arcaa. For a similar reason, it must also use the arcbb. However, the two arcsaaandbbhave the same target (viz.,vv) and thus cannot both appear in the same path (since a path cannot visit a vertex more than once). This contradicts the fact that the path𝐩\mathbf{p}uses both arcsaaandbb. Hence, we have found a contradiction in Case 3.

We have now found contradictions in all three Cases 1, 2 and 3. This contradiction shows that our assumption was false. Hence,deg−⁡v=1\deg^{-}v=1is proved.

We have now proved that eachv∈V∖{r}v\in V\setminus\left\{r\right\}satisfiesdeg−⁡v=1\deg^{-}v=1. Since we have also shown thatdeg−⁡r=0\deg^{-}r=0, we thus have proved Statement A6.

Proof of the implication A6⟹\LongrightarrowA2:Assume that Statement A6 holds. We must prove that Statement A2 holds. However, Proposition4.11yields

|A|\displaystyle\left|A\right|=∑v∈Vdeg−⁡v=deg−⁡r⏟=0(by Statement A6)+∑v∈V∖{r}deg−⁡v⏟=1(by Statement A6)\displaystyle=\sum\limits_{v\in V}\deg^{-}v=\underbrace{\deg^{-}r}_{\begin{subarray}{c}=0\\ \text{(by Statement A6)}\end{subarray}}+\sum\limits_{v\in V\setminus\left\{r\right\}}\underbrace{\deg^{-}v}_{\begin{subarray}{c}=1\\ \text{(by Statement A6)}\end{subarray}}=0+∑v∈V∖{r}1=∑v∈V∖{r}1=|V∖{r}|=|V|−1.\displaystyle=0+\sum\limits_{v\in V\setminus\left\{r\right\}}1=\sum\limits_{v\in V\setminus\left\{r\right\}}1=\left|V\setminus\left\{r\right\}\right|=\left|V\right|-1.Hence, Statement A2 holds.

Proof of the implication A2⟹\LongrightarrowA3:Assume that Statement A2 holds. We must prove that Statement A3 holds.

For eachv∈Vv\in V, the digraphDDhas a path fromrrtovv(sincerris a from-root ofDD). Thus, for eachv∈Vv\in V, the graphDundD^{\operatorname*{und}}has a path fromrrtovv(since any path ofDDis a path ofDundD^{\operatorname*{und}}). Therefore, any two verticesuuandvvofDundD^{\operatorname*{und}}are path-connected inDundD^{\operatorname*{und}}(because we can get fromuutovvviarr, according to the previous sentence). Therefore, the graphDundD^{\operatorname*{und}}is connected (since it has at least one vertex454545This is becauser∈Vr\in V.). Moreover, its number of edges is|A|=|V|−1\left|A\right|=\left|V\right|-1(by Statement A2). Therefore, the multigraphDundD^{\operatorname*{und}}satisfies the Statement T4 of the tree equivalence theorem (Theorem5.21). Consequently, it satisfies Statement T1 of that theorem as well. In other words, it is a tree. This proves Statement A3.

Proof of the implication A3⟹\LongrightarrowA1:Assume that Statement A3 holds. We must prove that Statement A1 holds.

The multigraphDundD^{\operatorname*{und}}is a tree (by Statement A3), and thus is a forest; hence, it has no cycles. Since we also know thatrris a from-root ofDD, we thus conclude thatDDis an arborescence rooted fromrr(by the definition of an arborescence). In other words, Statement A1 is satisfied.

We have now proved all six implications in the chain A1⟹\LongrightarrowA4⟹\LongrightarrowA5⟹\LongrightarrowA6⟹\LongrightarrowA2⟹\LongrightarrowA3⟹\LongrightarrowA1. Thus, all six statements A1, A2,…\ldots, A6 are equivalent. This proves Theorem5.88.

Exercise 5.90.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph that has no cycles464646Recall that cycles in a digraph have to be directed cycles – i.e., each arc is traversed from its source to its target.. Letr∈Vr\in Vbe some vertex ofDD. Prove the following:

(a)Ifdeg−⁡u>0\deg^{-}u>0holds for allu∈V∖{r}u\in V\setminus\left\{r\right\}, thenrris a from-root ofDD.
(b)Ifdeg−⁡u=1\deg^{-}u=1holds for allu∈V∖{r}u\in V\setminus\left\{r\right\}, thenDDis an arborescence rooted fromrr.

5.7Arborescences vs. trees

Our next goal is to prove Theorem5.87, which connects arborescences with trees.

To prove it formally, we introduce a few notations regarding trees. First, we recall the notion of a distance (Definition5.63). We claim the following simple property of distances in trees:

Proposition 5.91.

LetT=(V,E,φ)T=\left(V,E,\varphi\right)be a tree. Letr∈Vr\in Vbe a vertex ofTT. Leteebe an edge ofTT, and letuuandvvbe its two endpoints. Then, the distancesd⁡(r,u)d\left(r,u\right)andd⁡(r,v)d\left(r,v\right)differ by exactly11(that is, we have eitherd⁡(r,u)=d⁡(r,v)+1d\left(r,u\right)=d\left(r,v\right)+1ord⁡(r,v)=d⁡(r,u)+1d\left(r,v\right)=d\left(r,u\right)+1).

Proof 5.92.

We recall that sinceTTis a tree, the distanced⁡(p,q)d\left(p,q\right)between two verticesppandqqofTTis simply the length of the path frompptoqq. (This path is unique, sinceTTis a tree.)

Let𝐩\mathbf{p}be the path fromrrtouu. Then, we are in one of the following two cases:

Case 1:The edgeeeis an edge of𝐩\mathbf{p}.

Case 2:The edgeeeis not an edge of𝐩\mathbf{p}.

Consider Case 1. In this case,eemust be thelastedge of𝐩\mathbf{p}(since otherwise,𝐩\mathbf{p}would visituumore than once, but𝐩\mathbf{p}cannot do this, since𝐩\mathbf{p}is a path). Thus, if we remove this last edgeee(and the vertexuu) from𝐩\mathbf{p}, then we obtain a path fromrrtovv. This path is exactly one edge shorter than𝐩\mathbf{p}. Thus,d⁡(r,v)=d⁡(r,u)−1d\left(r,v\right)=d\left(r,u\right)-1, so thatd⁡(r,u)=d⁡(r,v)+1d\left(r,u\right)=d\left(r,v\right)+1. So we are done in Case 1.

Now, consider Case 2. In this case, the edgeeeis not an edge of𝐩\mathbf{p}. Thus, we can appendeeandvvto the end of the path𝐩\mathbf{p}, and the result will be a backtrack-free walk𝐩′\mathbf{p}^{\prime}. However, a backtrack-free walk in a tree is always a path (since otherwise, it would contain a cycle474747by Proposition5.2, but a tree has no cycles). Thus,𝐩′\mathbf{p}^{\prime}is a path fromrrtovv, and it is exactly one edge longer than𝐩\mathbf{p}(by its construction). Therefore,d⁡(r,v)=d⁡(r,u)+1d\left(r,v\right)=d\left(r,u\right)+1. So we are done in Case 2.

Now, we are done in both cases, so that Proposition5.91is proven.

Definition 5.93.

LetT=(V,E,φ)T=\left(V,E,\varphi\right)be a tree. Letr∈Vr\in Vbe a vertex ofTT. Leteebe an edge ofTT. By Proposition5.91, the distances from the two endpoints ofeeto the vertexrrdiffer by exactly11. So one of them is smaller than the other.

(a)We define therr-parentofeeto be the endpoint ofeewhose distance torris the smallest. We denote this endpoint bye−re^{-r}.
(b)We define therr-childofeeto be the endpoint ofeewhose distance torris the largest. We denote this endpoint bye+re^{+r}.

Thus, by Proposition5.91, we have

d⁡(r,e+r)=d⁡(r,e−r)+1.d\left(r,e^{+r}\right)=d\left(r,e^{-r}\right)+1.

Example 5.94.

Here is a treeTT, a vertexrr, an edgeeeand itsrr-parente−re^{-r}and itsrr-childe+re^{+r}:

e+re^{+r}e−re^{-r}rr\vphantom{e^{-r}}ee

Definition 5.95.

LetT=(V,E,φ)T=\left(V,E,\varphi\right)be a tree. Letr∈Vr\in Vbe a vertex ofTT. Then, we define a multidigraphTr→T^{r\rightarrow}by

Tr→:=(V,E,ψ),T^{r\rightarrow}:=\left(V,E,\psi\right),whereψ:E→V×V\psi:E\rightarrow V\times Vis the map that sends each edgee∈Ee\in Eto the pair(e−r,e+r)\left(e^{-r},e^{+r}\right). Colloquially speaking, this means thatTr→T^{r\rightarrow}is the multidigraph obtained fromTTby turning each edgeeeinto an arc from itsrr-parente−re^{-r}to itsrr-childe+re^{+r}. This is what we mean when we speak of “orienting each edge ofTTaway fromrr” in Theorem5.87.

Example 5.96.

IfTTis the tree from Example5.94, thenTr→T^{r\rightarrow}is the following multidigraph:

Now, Theorem5.87can be rewritten as follows:

Theorem 5.97.

LetDDbe a multidigraph, and letrrbe a vertex ofDD. Then, the following two statements are equivalent:

•Statement C1:The multidigraphDDis an arborescence rooted fromrr.
•Statement C2:The undirected multigraphDundD^{\operatorname*{und}}is a tree, and we haveD=(Dund)r→D=\left(D^{\operatorname*{und}}\right)^{r\rightarrow}. (This is a honest equality, not just some isomorphism.)

The proof of this theorem is best organized by splitting into two lemmas:

Lemma 5.98.

LetT=(V,E,φ)T=\left(V,E,\varphi\right)be a tree. Letr∈Vr\in Vbe a vertex ofTT. Then, the multidigraphTr→T^{r\rightarrow}is an arborescence rooted fromrr.

Proof 5.99.

The idea is to show that if𝐩\mathbf{p}is a path fromrrto some vertexvvin the treeTT, then𝐩\mathbf{p}is also a path in the digraphTr→T^{r\rightarrow}, because all the edges of𝐩\mathbf{p}have been “oriented correctly” (i.e., their orientation matches how they are used in𝐩\mathbf{p}).

Here are the details: Clearly,(Tr→)und=T\left(T^{r\rightarrow}\right)^{\operatorname*{und}}=T. Hence, the graph(Tr→)und\left(T^{r\rightarrow}\right)^{\operatorname*{und}}is a tree and hence has no cycles. Thus, it suffices to prove thatrris a from-root ofTr→T^{r\rightarrow}. In other words, we must prove that

Tr→has a path fromrtovT^{r\rightarrow}\text{ has a path from }r\text{ to }v(22)for eachv∈Vv\in V.

We shall prove (22) by induction ond⁡(r,v)d\left(r,v\right)(whereddmeans the distance on the treeTT):

Base case:Ifv∈Vv\in Vsatisfiesd⁡(r,v)=0d\left(r,v\right)=0, thenv=rv=r, and thusTr→T^{r\rightarrow}has a path fromrrtovv(namely, the trivial path(r)\left(r\right)). Thus, (22) is proved ford⁡(r,v)=0d\left(r,v\right)=0.

Induction step:Letk∈ℕk\in\mathbb{N}. Assume (as the induction hypothesis) that (22) holds for eachv∈Vv\in Vsatisfyingd⁡(r,v)=kd\left(r,v\right)=k. We must now prove the same for eachv∈Vv\in Vsatisfyingd⁡(r,v)=k+1d\left(r,v\right)=k+1.

So letv∈Vv\in Vsatisfyd⁡(r,v)=k+1d\left(r,v\right)=k+1. Then, the path ofTTfromrrtovvhas lengthk+1k+1. Let𝐩\mathbf{p}be this path, leteebe its last edge, and letuube its second-to-last vertex (so that its last edgeeehas endpointsuuandvv). Then, by removing the last edgeeefrom the path𝐩\mathbf{p}, we obtain a path fromrrtouuthat is one edge shorter than𝐩\mathbf{p}. Hence,d⁡(r,u)=d⁡(r,v)−1<d⁡(r,v)d\left(r,u\right)=d\left(r,v\right)-1<d\left(r,v\right). Consequently, the edgeeehasrr-parentuuandrr-childvv(by Definition5.93). In other words,e−r=ue^{-r}=uande+r=ve^{+r}=v. Therefore, in the digraphTr→T^{r\rightarrow}, the edgeeeis an arc fromuutovv(by Definition5.95). Moreover, we haved⁡(r,u)=d⁡(r,v)−1=kd\left(r,u\right)=d\left(r,v\right)-1=k(sinced⁡(r,v)=k+1d\left(r,v\right)=k+1); therefore, the induction hypothesis tells us that (22) holds foruuinstead ofvv. In other words,Tr→T^{r\rightarrow}has a path fromrrtouu. Attaching the arceeand the vertexvvto this path, we obtain a walk ofTr→T^{r\rightarrow}fromrrtovv(sinceeeis an arc fromuutovvinTr→T^{r\rightarrow}). Thus, the digraphTr→T^{r\rightarrow}has a walk fromrrtovv, therefore also a path fromrrtovv. Hence, (22) holds for ourvv. This completes the induction step.

Thus, (22) is proved by induction. As we explained above, this yields Lemma5.98.

Lemma 5.100.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be an arborescence rooted fromrr(for somer∈Vr\in V). Leta∈Aa\in Abe an arc ofDD. Letssbe the source ofaa, and letttbe the target ofaa. Then:

(a)We haved⁡(r,s)<d⁡(r,t)d\left(r,s\right)<d\left(r,t\right), whereddmeans distance on the treeDundD^{\operatorname*{und}}.
(b)In the multidigraph(Dund)r→\left(D^{\operatorname*{und}}\right)^{r\rightarrow}, the arcaahas sourcessand targettt.

Proof 5.101.

(a)The vertexrris a from-root ofDD(sinceDDis an arborescence rooted fromrr). Thus,DDhas a path fromrrtott. Let𝐩\mathbf{p}be this path. Note thatdeg−⁡t≥1\deg^{-}t\geq 1, sincettis the target of at least one arc (namely, ofaa).

The digraphDDis an arborescence rooted fromrr, and thus satisfies Statement A6 in the arborescence equivalence theorem (Theorem5.88). In other words, we have

deg−r=0anddeg−v=1for eachv∈V∖{r}.\deg^{-}r=0\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \deg^{-}v=1\text{ for each }v\in V\setminus\left\{r\right\}.In particular, this entailsdeg−⁡v≤1\deg^{-}v\leq 1for eachv∈Vv\in V. Applying this tov=tv=t, we obtaindeg−⁡t≤1\deg^{-}t\leq 1. Hence, the arcaais theonlyarc whose target istt.

We havet≠rt\neq r(sincedeg−⁡r=0\deg^{-}r=0butdeg−⁡t≥1>0\deg^{-}t\geq 1>0). Thus, the path𝐩\mathbf{p}fromrrtotthas at least one arc. Its last arc is therefore an arc whose target istt. Hence, this last arc isaa(sinceaais theonlyarc whose target istt).

If we remove this last arc from the path𝐩\mathbf{p}, then we obtain a path𝐩′\mathbf{p}^{\prime}fromrrtoss(sincessis the source ofaa).

However, each path ofDDis a path ofDundD^{\operatorname*{und}}. Thus, in particular,𝐩\mathbf{p}is a path ofDundD^{\operatorname*{und}}fromrrtott, while𝐩′\mathbf{p}^{\prime}is a path ofDundD^{\operatorname*{und}}fromrrtoss. Since𝐩′\mathbf{p}^{\prime}is exactly one edge shorter than𝐩\mathbf{p}, we thus obtaind⁡(r,s)=d⁡(r,t)−1<d⁡(r,t)d\left(r,s\right)=d\left(r,t\right)-1<d\left(r,t\right). This proves Lemma5.100(a).

(b)The arcaaof the digraphDDhas sourcessand targettt. Hence, the edgeaaof the treeDundD^{\operatorname*{und}}has endpointsssandtt. Sinced⁡(r,s)<d⁡(r,t)d\left(r,s\right)<d\left(r,t\right)(by part(a)), this entails that itsrr-parent isssand itsrr-child istt(by Definition5.93). Thus, in the digraph(Dund)r→\left(D^{\operatorname*{und}}\right)^{r\rightarrow}, this edgeaabecomes an arc with sourcessand targettt(by Definition5.95). This proves Lemma5.100(b).

Proof 5.102(Proof of Theorem5.97.).

If(V,A,ψ)\left(V,A,\psi\right)is a multidigraph, then we shall refer to the mapψ:A→V×V\psi:A\rightarrow V\times V(which determines the source and the target of each arc) as the “psi-map” of this multidigraph.

Write the multidigraphDDasD=(V,A,ψ)D=\left(V,A,\psi\right). We shall now prove the implications C1⟹\LongrightarrowC2 and C2⟹\LongrightarrowC1 separately:

Proof of the implication C1⟹\LongrightarrowC2:Assume that Statement C1 holds. That is,DDis an arborescence rooted fromrr. We must prove Statement C2. In other words, we must prove that the undirected multigraphDundD^{\operatorname*{und}}is a tree, and thatD=(Dund)r→D=\left(D^{\operatorname*{und}}\right)^{r\rightarrow}.

It is clear (by the definition of an arborescence) thatDundD^{\operatorname*{und}}is a tree. It thus remains to prove thatD=(Dund)r→D=\left(D^{\operatorname*{und}}\right)^{r\rightarrow}.

The multidigraphsDDand(Dund)r→\left(D^{\operatorname*{und}}\right)^{r\rightarrow}have the same set of vertices (namely,VV) and the same set of arcs (namely,AA); we therefore just need to show that their psi-maps are the same. In other words, we need to show thatψ′=ψ\psi^{\prime}=\psi, whereψ′\psi^{\prime}is the psi-map of(Dund)r→\left(D^{\operatorname*{und}}\right)^{r\rightarrow}.

Leta∈Aa\in Abe arbitrary. Letψ⁡(a)=(s,t)\psi\left(a\right)=\left(s,t\right). Thus, the arcaaofDDhas sourcessand targettt. Lemma5.100(b)therefore shows that in the multidigraph(Dund)r→\left(D^{\operatorname*{und}}\right)^{r\rightarrow}, the arcaahas sourcessand targetttas well. In other words,ψ′(a)=(s,t)\psi^{\prime}\left(a\right)=\left(s,t\right)(sinceψ′\psi^{\prime}is the psi-map of this multidigraph). Hence,ψ′(a)=(s,t)=ψ⁡(a)\psi^{\prime}\left(a\right)=\left(s,t\right)=\psi\left(a\right).

Forget that we fixedaa. We thus have shown thatψ′(a)=ψ(a)\psi^{\prime}\left(a\right)=\psi\left(a\right)for eacha∈Aa\in A. In other words,ψ′=ψ\psi^{\prime}=\psi. As explained above, this completes the proof of Statement C2.

Proof of the implication C2⟹\LongrightarrowC1:Assume that Statement C2 holds. Thus, the undirected multigraphDundD^{\operatorname*{und}}is a tree, and we haveD=(Dund)r→D=\left(D^{\operatorname*{und}}\right)^{r\rightarrow}. Hence, Lemma5.98(applied toT=DundT=D^{\operatorname*{und}}) yields that the multidigraph(Dund)r→\left(D^{\operatorname*{und}}\right)^{r\rightarrow}is an arborescence rooted fromrr. In other words,DDis an arborescence rooted fromrr(sinceD=(Dund)r→D=\left(D^{\operatorname*{und}}\right)^{r\rightarrow}). This shows that Statement C1 holds.

Having now proved both implications C1⟹\LongrightarrowC2 and C2⟹\LongrightarrowC1, we conclude that Statements C1 and C2 are equivalent. Thus, Theorem5.97is proved.

Oof.

Let’s get one more consequence out of this. First, let us show that an arborescence can have only one root:

Proposition 5.103.

LetDDbe an arborescence rooted fromrr. Then,rris theonlyroot ofDD.

Proof 5.104(Proof of Proposition5.103.).

Assume the contrary. Thus,DDhas another rootssdistinct fromrr. Hence,DDhas a path fromrrtoss(sincerris a root) as well as a path fromsstorr(sincessis a root). Combining these paths gives a circuit of length>0>0. However, a circuit of length>0>0in a digraph must always contain a cycle (since Proposition4.30shows that it either is a path or contains a cycle; but it clearly cannot be a path). Hence,DDhas a cycle. Therefore,DundD^{\operatorname*{und}}also has a cycle (since any cycle ofDDis a cycle ofDundD^{\operatorname*{und}}). However,DundD^{\operatorname*{und}}has no cycles (sinceDDis an arborescence rooted fromrr). The preceding two sentences contradict each other. This shows that the assumption was wrong, and Proposition5.103is proven.

Definition 5.105.

A multidigraphDDis said to be anarborescenceif there exists a vertexrrofDDsuch thatDDis an arborescence rooted fromrr. In this case, thisrris uniquely determined as the only root ofDD(by Proposition5.103).

Theorem 5.106.

There are two mutually inverse maps

{pairs(T,r)of a treeTand a vertexrofT}\displaystyle\left\{\text{pairs }\left(T,r\right)\text{ of a tree }T\text{ and a vertex }r\text{ of }T\right\}→{arborescences},\displaystyle\rightarrow\left\{\text{arborescences}\right\},(T,r)\displaystyle\left(T,r\right)↦Tr→\displaystyle\mapsto T^{r\rightarrow}and

Proof 5.107.

The map

{arborescences}\displaystyle\left\{\text{arborescences}\right\}→{pairs(T,r)of a treeTand a vertexrofT},\displaystyle\rightarrow\left\{\text{pairs }\left(T,r\right)\text{ of a tree }T\text{ and a vertex }r\text{ of }T\right\},D\displaystyle D↦(Dund,D),\displaystyle\mapsto\left(D^{\operatorname*{und}},\sqrt{D}\right),is well-defined because ifDDis an arborescence, thenDundD^{\operatorname*{und}}is a tree. In order to show that these two maps are mutually inverse, we must check the following two statements:

1.Each arborescenceDDsatisfies(Dund)r→=D\left(D^{\operatorname*{und}}\right)^{r\rightarrow}=D, whererris the root ofDD;
2.Each pair(T,r)\left(T,r\right)of a treeTTand a vertexrrofTTsatisfies(Tr→)und=T\left(T^{r\rightarrow}\right)^{\operatorname*{und}}=Tand(Tr→)und=r\sqrt{\left(T^{r\rightarrow}\right)^{\operatorname*{und}}}=r.

However, Statement 1 follows from Theorem5.97(specifically, from the implication C1⟹\LongrightarrowC2 in Theorem5.97). Statement 2 follows from Lemma5.98(more precisely, the(Tr→)und=T\left(T^{r\rightarrow}\right)^{\operatorname*{und}}=Tpart of Statement 2 is obvious, whereas the(Tr→)und=r\sqrt{\left(T^{r\rightarrow}\right)^{\operatorname*{und}}}=rpart follows from Lemma5.98). Thus, Theorem5.106is proved.

Theorem5.106formalizes the idea that an arborescence is “just a tree with a chosen vertex”. For this reason, arborescences are sometimes called “oriented trees”, but this name is also shared with a more general notion, which is why I avoid it.

Exercise 5.108.

(In other words, show that we can assign to each vertex an edge that contains this vertex in such a way that no edge is assigned twice.)

5.8Spanning arborescences

In analogy to spanning subgraphs of a multigraph, we can define spanning subdigraphs of a multidigraph:

Definition 5.109.

Aspanning subdigraphof a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right)means a multidigraph of the form(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right), whereBBis a subset ofAA.

In other words, it means a submultidigraph ofDDwith the same vertex set asDD.

In other words, it means a multidigraph obtained fromDDby removing some arcs, but leaving all vertices untouched.

Definition 5.110.

LetDDbe a multidigraph. Letrrbe a vertex ofDD. Aspanning arborescence ofDDrooted fromrrmeans a spanning subdigraph ofDDthat is an arborescence rooted fromrr.

Example 5.111.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be the following multidigraph:

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{}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{197.63762pt}{-8.23856pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 273.47 -11.4)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ . Is there a spanning arborescence ofDDrooted from11? Yes, for instance,

(V,{a,c,e},ψ∣{a,c,e})=1234ace.\left(V,\ \left\{a,c,e\right\},\ \psi\mid_{\left\{a,c,e\right\}}\right)=\hbox to246pt{\vbox to75.28pt{\pgfpicture\makeatletter\hbox{\hskip 66.0914pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} 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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.98482}{-0.17365}{0.17365}{0.98482}{47.26822pt}{1.69923pt}\lxSVG@begingroup@{transform=matrix(0.98482 -0.17365 0.17365 0.98482 65.41 2.35)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} 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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.9397}{-0.34203}{0.34203}{0.9397}{161.52078pt}{3.347pt}\lxSVG@begingroup@{transform=matrix(0.9397 -0.34203 0.34203 0.9397 223.5 4.63)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{111.48285pt}{16.73288pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 154.26 23.15)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .By abuse of notation, we shall refer to this spanning arborescence simply as{a,c,e}\left\{a,c,e\right\}(since a spanning subdigraph ofDDis uniquely determined by its arc set). Another spanning arborescence ofDDrooted from11is{a,b,e}\left\{a,b,e\right\}. Yet another is{a,b,f}\left\{a,b,f\right\}. A non-example is{a,d,f}\left\{a,d,f\right\}(indeed, this is an arborescence rooted from33, not from11).

Is there a spanning arborescence ofDDrooted from22? Yes, for example{b,d,f}\left\{b,d,f\right\}.

Is there a spanning arborescence ofDDrooted from44? No, since44is not a from-root ofDD.

This illustrates a first obstruction to the existence of spanning arborescences: Namely, a digraphDDcan have a spanning arborescence rooted fromrronly ifrris a from-root. This necessary criterion is also sufficient:

Theorem 5.112.

LetDDbe a multidigraph. Letrrbe a from-root ofDD. Then,DDhas a spanning arborescence rooted fromrr.

Proof 5.113.

This is an analogue of the “every connected multigraph has a spanning tree” theorem (Theorem5.46) that we proved in 4 ways. At least the first proof easily adapts to the directed case:

Remove arcs fromDDone by one, but in such a way that the “rootness ofrr” (that is, the property thatrris a root of our multidigraph) is preserved. So we can only remove an arc ifrrremains a root afterwards.

Clearly, this removing process will eventually come to an end, sinceDDhas only finitely many arcs. LetD′D^{\prime}be the multidigraph obtained at the end of this process. Then,rris still a root ofD′D^{\prime}, but we cannot remove any more arcs fromD′D^{\prime}without breaking the rootness ofrr. That is, if we remove any arc fromD′D^{\prime}, then the vertexrrwill no longer be a from-root of the resulting multidigraph. This means thatD′D^{\prime}satisfies Statement A5 from the arborescence equivalence theorem (Theorem5.88). Thus,D′D^{\prime}satisfies Statement A1 as well (since all six statements A1, A2,…\ldots, A6 are equivalent). In other words,D′D^{\prime}is an arborescence rooted fromrr. SinceD′D^{\prime}is a spanning subdigraph ofDD, we thus conclude thatDDhas a spanning arborescence rooted fromrr(namely,D′D^{\prime}). This proves Theorem5.112.

Question 5.114.

Can the other three proofs of Theorem5.46be adapted to Theorem5.112, too?

Example 5.115.

Letnnbe a positive integer. Thenn-cycle digraphC→n\overrightarrow{C}_{n}is defined to be the simple digraph with vertices1,2,…,n1,2,\ldots,nand arcs12,23,34,…,(n−1)n,n112,\ 23,\ 34,\ \ldots,\ \left(n-1\right)n,\ n1. (Here is how it looks forn=5n=5:

)

Note that this digraphC→n\overrightarrow{C}_{n}is a directed analogue of the cycle graphCnC_{n}. As we recall from Example5.41, the cycle graphCnC_{n}hasnnspanning trees.

In contrast, the digraphC→n\overrightarrow{C}_{n}has only one spanning arborescence rooted from11. This spanning arborescence is the subdigraph ofC→n\overrightarrow{C}_{n}obtained by removing the arcn1n1.

Proof 5.116.

If we remove the arcn1n1fromC→n\overrightarrow{C}_{n}, then we obtain the simple digraphEEwith vertices1,2,…,n1,2,\ldots,nand arcs12,23,…,(n−1)n12,\ 23,\ \ldots,\ \left(n-1\right)n. This digraphEEis easily seen to be an arborescence rooted from11(indeed,11is a from-root ofEE, and the underlying undirected graphEund=PnE^{\operatorname*{und}}=P_{n}has no cycles). Thus,EEis a spanning arborescence ofC→n\overrightarrow{C}_{n}rooted from11.

We shall now prove that it is the only such arborescence. Indeed, letFFbe any spanning arborescence ofC→n\overrightarrow{C}_{n}rooted from11. Then,11is a from-root ofFF. Hence, for each vertexv∈{2,3,…,n}v\in\left\{2,3,\ldots,n\right\}, the digraphFFmust have a path from11tovv, and thus must contain an arc with targetvv(namely, the last arc of this path). This arc must be(v−1,v)\left(v-1,v\right)(since this is the only arc ofC→n\overrightarrow{C}_{n}with targetvv). Thus, for each vertexv∈{2,3,…,n}v\in\left\{2,3,\ldots,n\right\}, the digraphFFmust contain the arc(v−1,v)\left(v-1,v\right). In other words, the digraphFFmust contain alln−1n-1arcs12,23,…,(n−1)n12,\ 23,\ \ldots,\ \left(n-1\right)n. IfFFwere to also contain the remaining arcn1n1ofC→n\overrightarrow{C}_{n}, then the underlying undirected graphFund=CnF^{\operatorname*{und}}=C_{n}would contain a cycle, which would contradictFFbeing an arborescence. Hence,FFcannot contain the arcn1n1. Thus,FFcontains then−1n-1arcs12,23,…,(n−1)n12,\ 23,\ \ldots,\ \left(n-1\right)nand no others. In other words,F=EF=E. This shows that any spanning arborescence ofC→n\overrightarrow{C}_{n}rooted from11must beEE. In other words,EEis the only spanning arborescence ofC→n\overrightarrow{C}_{n}rooted from11. This completes the proof of Example5.115.

5.9The BEST theorem: statement

We now come to something much more surprising.

Recall that a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right)isbalancedif and only if each vertexvvsatisfiesdeg−⁡v=deg+⁡v\deg^{-}v=\deg^{+}v. This is necessary for the existence of an Eulerian circuit. IfDDis weakly connected, this is also sufficient (by Theorem4.69(a)).

Surprisingly, there is a formula for the number of these Eulerian circuits:

Theorem 5.117(The BEST theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a balanced multidigraph such that each vertex has indegree>0>0. Fix an arcaaofDD, and letrrbe its target. Letτ⁡(D,r)\tau\left(D,r\right)be the number of spanning arborescences ofDDrooted fromrr. Letε⁡(D,a)\varepsilon\left(D,a\right)be the number of Eulerian circuits ofDDwhose last arc isaa. Then,

ε⁡(D,a)=τ⁡(D,r)⋅∏u∈V(deg−⁡u−1)!.\varepsilon\left(D,a\right)=\tau\left(D,r\right)\cdot\prod\limits_{u\in V}\left(\deg^{-}u-1\right)!.

The “BEST” in the name of this theorem is an abbreviation for de Bruijn, van Aardenne--Ehrenfest, Smith and Tutte, who discovered it in the middle of the 20th century484848More precisely, van Aardenne–Ehrenfest and de Bruijn discovered it in 1951 (see[VanEhr51, §6]) generalizing an earlier result of Smith and Tutte..494949We note that the number of Eulerian circuits ofDDwhose last arc isaais precisely the number of all Eulerian circuits ofDDcounted up to rotation. Indeed, each Eulerian circuit ofDDcontains the arcaaexactly once, and thus can be rotated in a unique way to end withaa.

To prove this theorem, we shall restate it in terms of “arborescences to” (as opposed to “arborescences from”). Mathematically speaking, this restatement isn’t really necessary (the argument is the same in both cases up to reversing the directions of all arcs), but it helps make the proof more intuitive, since it lets us build our Eulerian circuits by moving forwards rather than backwards.

5.10Arborescences rooted torr

Here is the formal definition of “arborescences to”:

Definition 5.118.

LetDDbe a multidigraph. Letrrbe a vertex ofDD.

(a)We say thatrris ato-rootofDDif for each vertexvvofDD, the digraphDDhas a path fromvvtorr.
(b)We say thatDDis anarborescence rooted torrifrris a to-root ofDDand the undirected multigraphDundD^{\operatorname*{und}}has no cycles.

Clearly, Definition5.84and Definition5.118differ only in the direction of the arcs. In other words, if we reverse each arc of our digraph (turning its source into its target and vice versa), then a from-root becomes a to-root, and an arborescence rooted fromrrbecomes an arborescence rooted torr, and vice versa. Thus, every property that we have proved for arborescences rooted fromrrcan be translated into the language of arborescences rooted torrby reversing all arcs.

If you want to see this stated more rigorously, here is a formal definition of “reversing each arc”:

Definition 5.119.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Then,DrevD^{\operatorname*{rev}}shall denote the multidigraph(V,A,τ∘ψ)\left(V,A,\tau\circ\psi\right), whereτ:V×V→V×V\tau:V\times V\rightarrow V\times Vis the map that sends each pair(s,t)\left(s,t\right)to(t,s)\left(t,s\right). Thus, if an arcaaofDDhas sourcessand targettt, then it is also an arc ofDrevD^{\operatorname*{rev}}, but in this digraphDrevD^{\operatorname*{rev}}it has sourcettand targetss.

The multidigraphDrevD^{\operatorname*{rev}}is called thereversalof the multidigraphDD; we say that it is obtained fromDDby “reversing each arc”.

This notion of “reversing each arc” allows us to reverse walks in digraphs: If𝐰\mathbf{w}is a walk from a vertexsstottin some multidigraphDD, then its reversalrev⁡𝐰\operatorname{rev}\mathbf{w}(obtained by reading𝐰\mathbf{w}backwards) is a walk fromtttossin the multidigraphDrevD^{\operatorname*{rev}}. The same holds if we replace the word “walk” by “path”. Thus, we easily obtain the following:

Proposition 5.120.

LetDDbe a multidigraph. Letrrbe a vertex ofDD. Then:

(a)The vertexrris a to-root ofDDif and only ifrris a from-root ofDrevD^{\operatorname*{rev}}.
(b)The digraphDDis an arborescence rooted torrif and only ifDrevD^{\operatorname*{rev}}is an arborescence rooted fromrr.

Proof 5.121.

Completely straightforward unpacking of the definitions.

Note that when we reverse each arc in a digraphDD, the outdegrees of its vertices become their indegrees and vice versa. Hence, a balanced digraphDDremains balanced when this happens. In particular, the BEST theorem (Theorem5.117) thus gets translated as follows:

Theorem 5.122(The BEST’ theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a balanced multidigraph such that each vertex has outdegree>0>0. Fix an arcaaofDD, and letrrbe its source. Letτ⁡(D,r)\tau\left(D,r\right)be the number of spanning arborescences ofDDrooted torr. Letε⁡(D,a)\varepsilon\left(D,a\right)be the number of Eulerian circuits ofDDwhose first arc isaa. Then,

ε⁡(D,a)=τ⁡(D,r)⋅∏u∈V(deg+⁡u−1)!.\varepsilon\left(D,a\right)=\tau\left(D,r\right)\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!.

We will soon prove Theorem5.122, and then derive Theorem5.117from it by reversing the arcs.

First, however, let us state the analogue of the Arborescence Equivalence Theorem (Theorem5.88) for “arborescences rooted torr” (as opposed to “arborescences rooted fromrr”):

Theorem 5.123(The dual arborescence equivalence theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph with a to-rootrr. Then, the following six statements are equivalent:

•Statement A’1:The multidigraphDDis an arborescence rooted torr.
•Statement A’2:We have|A|=|V|−1\left|A\right|=\left|V\right|-1.
•Statement A’3:The multigraphDundD^{\operatorname*{und}}is a tree.
•Statement A’4:For each vertexv∈Vv\in V, the multidigraphDDhas a unique walk fromvvtorr.
•Statement A’5:If we remove any arc fromDD, then the vertexrrwill no longer be a to-root of the resulting multidigraph.
•Statement A’6:We havedeg+⁡r=0\deg^{+}r=0, and eachv∈V∖{r}v\in V\setminus\left\{r\right\}satisfiesdeg+⁡v=1\deg^{+}v=1.

Proof 5.124.

Upon reversing all arcs ofDD, this turns into the original Arborescence Equivalence Theorem (Theorem5.88).

5.11The BEST theorem: proof

We now come to the proof of the BEST theorem (Theorem5.117). As we said, we proceed by proving Theorem5.122first. We first outline the idea of the proof; then we will give the details.

Proof 5.125(Proof idea for Theorem5.122.).

Anaa-Eulerian circuitshall mean an Eulerian circuit ofDDwhose first arc isaa.

Let𝐞\mathbf{e}be anaa-Eulerian circuit. Its first arc isaa; therefore, its first and last vertex isrr.

Being an Eulerian circuit,𝐞\mathbf{e}must contain each arc ofDDand therefore contain each vertex ofDD(since each vertex has outdegree>0>0). For each vertexu≠ru\neq r, we lete⁡(u)e\left(u\right)be thelast exitof𝐞\mathbf{e}fromuu, that is, the last arc of𝐞\mathbf{e}that has sourceuu. LetExit𝐞\operatorname*{Exit}\mathbf{e}be the set of these last exitse⁡(u)e\left(u\right)for all verticesu≠ru\neq r. Then, we claim:

Claim 1:This setExit𝐞\operatorname*{Exit}\mathbf{e}(or, more precisely, the spanning subdigraph(V,Exit𝐞,ψ∣Exit𝐞)\left(V,\ \operatorname*{Exit}\mathbf{e},\ \psi\mid_{\operatorname*{Exit}\mathbf{e}}\right)) is a spanning arborescence ofDDrooted torr.

Let’s assume for the moment that Claim 1 is proven. Thus, given anyaa-Eulerian circuit𝐞\mathbf{e}, we have constructed a spanning arborescence ofDDrooted torr.

How manyaa-Eulerian circuits𝐞\mathbf{e}lead to a given arborescence in this way? The answer is rather nice:

Claim 2:For each spanning arborescence(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)ofDDrooted torr, there are exactly∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!manyaa-Eulerian circuits𝐞\mathbf{e}such thatExit𝐞=B\operatorname*{Exit}\mathbf{e}=B.

Let us again assume that this is proven. Combining Claim 1 with Claim 2, we obtain a∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!-to-11correspondence between theaa-Eulerian circuits and the spanning arborescences ofDDrooted torr. Thus, the number of the former is∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!times the number of the latter. But this is precisely the claim of Theorem5.122. Hence, in order to prove Theorem5.122, it remains to prove Claim 1 and Claim 2.

Here is the complete proof:

Proof 5.126(Proof of Theorem5.122.).

Some notations first:

Anoutgoing arcfrom a vertexuuwill mean an arc whose source isuu. Anincoming arcinto a vertexuuwill mean an arc whose target isuu.

Anaa-Eulerian circuitshall mean an Eulerian circuit ofDDwhose first arc isaa.

Asparbshall mean a spanning arborescence ofDDrooted torr.

A spanning subdigraph ofDDalways has the form(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)for some subsetBBofAA. Thus, it is uniquely determined by its arc setBB.

Hence, from now on, we shall identify a spanning subdigraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)ofDDwith its arc setBB. Conversely, any subsetBBofAAwill be identified with the corresponding spanning subdigraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)ofDD. Thus, for instance, when we say that a subsetBBofAA“is a sparb”, we shall actually mean that the corresponding spanning subdigraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)is a sparb.

For eachaa-Eulerian circuit𝐞\mathbf{e}, we define a subsetExit𝐞\operatorname*{Exit}\mathbf{e}ofAAas follows:

Let𝐞\mathbf{e}be anaa-Eulerian circuit. Its first arc isaa; thus, its first and last vertex isrr. Being an Eulerian circuit,𝐞\mathbf{e}must contain each arc ofDDand therefore also contain each vertex ofDD(since each vertex ofDDhas outdegree>0>0). For each vertexu∈V∖{r}u\in V\setminus\left\{r\right\}, we lete⁡(u)e\left(u\right)be thelast exitof𝐞\mathbf{e}fromuu; this means the last arc of𝐞\mathbf{e}that has sourceuu. We letExit𝐞\operatorname*{Exit}\mathbf{e}be the set of these last exitse⁡(u)e\left(u\right)for allu∈V∖{r}u\in V\setminus\left\{r\right\}. Thus, we have defined a subsetExit𝐞\operatorname*{Exit}\mathbf{e}ofAAfor eachaa-Eulerian circuit𝐞\mathbf{e}.

Example 5.127.

Here is an example of this construction: LetDDbe the multidigraph

1122334455aabbccddeeffgghhiijjkkllwithr=1r=1, and let𝐞\mathbf{e}be theaa-Eulerian circuit

(1,a,2,b,3,c,4,d,5,e,1,f,3,g,3,h,5,i,5,j,2,k,4,l,1)\left(1,a,2,b,3,c,4,d,5,e,1,f,3,g,3,h,5,i,5,j,2,k,4,l,1\right)(we have deliberately named the arcs in such a way that they appear on an Eulerian circuit in alphabetic order). Then,

e⁡(2)=k,e⁡(3)=h,e⁡(4)=l,e⁡(5)=j,e\left(2\right)=k,\ \ \ \ \ \ \ \ \ \ e\left(3\right)=h,\ \ \ \ \ \ \ \ \ \ e\left(4\right)=l,\ \ \ \ \ \ \ \ \ \ e\left(5\right)=j,so thatExit𝐞={k,h,l,j}\operatorname*{Exit}\mathbf{e}=\left\{k,h,l,j\right\}. Here isExit𝐞\operatorname*{Exit}\mathbf{e}as a spanning subdigraph:

Now, we claim the following:

Claim 1:Let𝐞\mathbf{e}be anaa-Eulerian circuit. Then, the setExit𝐞\operatorname*{Exit}\mathbf{e}is a sparb.

Claim 2:For each sparbBB(regarded as a subset ofAA), there are exactly∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!manyaa-Eulerian circuits𝐞\mathbf{e}such thatExit𝐞=B\operatorname*{Exit}\mathbf{e}=B.

[Proof of Claim 1:The setExit𝐞\operatorname*{Exit}\mathbf{e}contains exactly one outgoing arc (namely,e⁡(u)e\left(u\right)) from each vertexu∈V∖{r}u\in V\setminus\left\{r\right\}, and no outgoing arc fromrr. Thus,|Exit𝐞|=|V|−1\left|\operatorname*{Exit}\mathbf{e}\right|=\left|V\right|-1.

Let us number the arcs of𝐞\mathbf{e}asa1,a2,…,ama_{1},a_{2},\ldots,a_{m}, in the order in which they appear in𝐞\mathbf{e}. (Thus,a1=aa_{1}=a, since the first arc of𝐞\mathbf{e}isaa.)

Recall that the arcs inExit𝐞\operatorname*{Exit}\mathbf{e}are the arcse⁡(u)e\left(u\right)for allu∈V∖{r}u\in V\setminus\left\{r\right\}(defined as above – i.e., the arce⁡(u)e\left(u\right)is the last exit of𝐞\mathbf{e}fromuu). We shall refer to these arcs as thelast-exit arcs.

For eachu∈V∖{r}u\in V\setminus\left\{r\right\}, we letj⁡(u)j\left(u\right)be the unique numberi∈{1,2,…,m}i\in\left\{1,2,\ldots,m\right\}such thate⁡(u)=aie\left(u\right)=a_{i}. (Thisiiindeed exists and is unique, since each arc ofDDappears exactly once on𝐞\mathbf{e}.) Thus,j⁡(u)j\left(u\right)tells us how late in the Eulerian circuit𝐞\mathbf{e}the arce⁡(u)e\left(u\right)appears. Sincee⁡(u)e\left(u\right)is the last exit of𝐞\mathbf{e}fromuu, the Eulerian circuit𝐞\mathbf{e}never visits the vertexuuagain after this.

Thus, if a last-exit arce⁡(u)e\left(u\right)has targetv≠rv\neq r, then

j⁡(u)<j⁡(v)j\left(u\right)<j\left(v\right)(23)(because the arce⁡(u)e\left(u\right)leads the circuit𝐞\mathbf{e}into the vertexvv, which the circuit then has to exit at least once; therefore, the corresponding last-exit arce⁡(v)e\left(v\right)has to appear later in𝐞\mathbf{e}than the arce⁡(u)e\left(u\right)).

We shall now show thatrris a to-root ofExit𝐞\operatorname*{Exit}\mathbf{e}(that is, of the spanning subdigraph(V,Exit𝐞,ψ∣Exit𝐞)\left(V,\operatorname*{Exit}\mathbf{e},\psi\mid_{\operatorname*{Exit}\mathbf{e}}\right)). To this purpose, we must show that for each vertexv∈Vv\in V, there is a path fromvvtorrin the digraph(V,Exit𝐞,ψ∣Exit𝐞)\left(V,\operatorname*{Exit}\mathbf{e},\psi\mid_{\operatorname*{Exit}\mathbf{e}}\right).

Indeed, letv∈Vv\in Vbe any vertex. We must find a path fromvvtorrin the digraph(V,Exit𝐞,ψ∣Exit𝐞)\left(V,\operatorname*{Exit}\mathbf{e},\psi\mid_{\operatorname*{Exit}\mathbf{e}}\right). It will suffice to find a walk fromvvtorrin this digraph (by Corollary4.29). In other words, we must find a way to walk fromvvtorrinDDusing last-exit arcs only.

So we start walking atvv. Ifv=rv=r, then we are already done. Otherwise, we havev∈V∖{r}v\in V\setminus\left\{r\right\}, so that the arce⁡(v)e\left(v\right)and the numberj⁡(v)j\left(v\right)are well-defined. We thus take the arce⁡(v)e\left(v\right). This brings us to a vertexv′v^{\prime}(namely, the target ofe⁡(v)e\left(v\right)) that satisfiesj⁡(v)<j⁡(v′)j\left(v\right)<j\left(v^{\prime}\right)(by (23)). If this vertexv′v^{\prime}isrr, then we are done. If not, thene⁡(v′)e\left(v^{\prime}\right)andj⁡(v′)j\left(v^{\prime}\right)are well-defined, so we continue our walk by taking the arce⁡(v′)e\left(v^{\prime}\right). This brings us to a further vertexv′′v^{\prime\prime}(namely, the target ofe⁡(v′)e\left(v^{\prime}\right)) that satisfiesj⁡(v′)<j⁡(v′′)j\left(v^{\prime}\right)<j\left(v^{\prime\prime}\right)(by (23)). If this vertexv′′v^{\prime\prime}isrr, then we are done. Otherwise, we proceed as before. We thus construct a walk

(v,e⁡(v),v′,e⁡(v′),v′′,e⁡(v′′),…)\left(v,e\left(v\right),v^{\prime},e\left(v^{\prime}\right),v^{\prime\prime},e\left(v^{\prime\prime}\right),\ldots\right)that either goes on indefinitely or stops at the vertexrr.

However, this walking process cannot go on forever (since the chain of inequalitiesj⁡(v)<j⁡(v′)<j⁡(v′′)<⋯j\left(v\right)<j\left(v^{\prime}\right)<j\left(v^{\prime\prime}\right)<\cdotswould force the numbersj⁡(v),j⁡(v′),j⁡(v′′),…j\left(v\right),j\left(v^{\prime}\right),j\left(v^{\prime\prime}\right),\ldotsto be all distinct, but there are onlymmdistinct numbers in{1,2,…,m}\left\{1,2,\ldots,m\right\}). Thus, it must stop at the vertexrr. So we have found a walk fromvvtorrusing last-exit arcs only. Thus,Exit𝐞\operatorname*{Exit}\mathbf{e}has a walk fromvvtorr. Hence,Exit𝐞\operatorname*{Exit}\mathbf{e}has a path fromvvtorr.

Forget that we fixedvv. We thus have shown that for each vertexv∈Vv\in V, there is a path fromvvtorrin the digraph(V,Exit𝐞,ψ∣Exit𝐞)\left(V,\operatorname*{Exit}\mathbf{e},\psi\mid_{\operatorname*{Exit}\mathbf{e}}\right). In other words,rris a to-root ofExit𝐞\operatorname*{Exit}\mathbf{e}. Hence, we conclude (using the implication A’2⟹\LongrightarrowA’1 in Theorem5.123) thatExit𝐞\operatorname*{Exit}\mathbf{e}is an arborescence rooted torr(since|Exit𝐞|=|V|−1\left|\operatorname*{Exit}\mathbf{e}\right|=\left|V\right|-1). Therefore,Exit𝐞\operatorname*{Exit}\mathbf{e}is a sparb. This proves Claim 1.]

[Proof of Claim 2:LetBBbe a sparb. (As before,BBis a set of arcs, and we identify it with the spanning subdigraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right).)

We must prove that there are exactly∏u∈V(deg+⁡u−1)!\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!manyaa-Eulerian circuits𝐞\mathbf{e}such thatExit𝐞=B\operatorname*{Exit}\mathbf{e}=B.

We shall refer to the arcs inBBas theBB-arcs. Recall thatBBis an arborescence rooted torr(sinceBBis a sparb). Hence, by the implication A’1⟹\LongrightarrowA’6 in Theorem5.123, we see that the outdegrees of its vertices satisfy

degB+r=0,anddegB+v=1for allv∈V∖{r}\deg_{B}^{+}r=0,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \deg_{B}^{+}v=1\text{ for all }v\in V\setminus\left\{r\right\}(wheredegB+v\deg_{B}^{+}vmeans the outdegree of a vertex in the digraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)). In other words, there is noBB-arc with sourcerr; however, for each vertexu∈V∖{r}u\in V\setminus\left\{r\right\}, there is exactly oneBB-arc with sourceuu.

Now, we are trying to count theaa-Eulerian circuits𝐞\mathbf{e}such thatExit𝐞=B\operatorname*{Exit}\mathbf{e}=B.

Let us try to construct such anaa-Eulerian circuit𝐞\mathbf{e}as follows:

A turtle wants to walk through the digraphDDusing each arc ofDDat most once. It starts its walk by heading out from the vertexrralong the arcaa. From that point on, it proceeds in the usual way you would walk on a digraph: Each time it reaches a vertex, it chooses an arbitrary arc leading out of this vertex, observing the following two rules:

1.It never uses an arc that it has already used before.
2.It never uses aBB-arc unless it has to (i.e., unless thisBB-arc is the only outgoing arc from its current position that is still unused).

Clearly, the turtle will eventually get stuck at some vertex (with no more arcs left to continue walking along), sinceDDhas only finitely many arcs.

Let𝐰\mathbf{w}be the total walk that the turtle has traced by the time it got stuck. Thus,𝐰\mathbf{w}is a trail (i.e., a walk that uses no arc more than once) that starts with the vertexrrand the arcaa.

We will soon see that𝐰\mathbf{w}is anaa-Eulerian circuit satisfyingExit𝐰=B\operatorname*{Exit}\mathbf{w}=B. First, however, let us see an example:

Example 5.128.

LetDDbe the multidigraph

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purpose). LetBBbe the set{d,e,h,k}\left\{d,e,h,k\right\}, regarded as a spanning subdigraph ofDD. (The arcs ofBBare drawn bold and in red in the above picture.)

The turtle starts atr=1r=1and walks along the arcaa. This leads it to the vertex22. It now must choose between the arcsbbandkk, but since it must not use theBB-arckkunless it has to, it is actually forced to take the arcbbnext. This brings it to the vertex33. It now has to choose between the arcscc,ggandhh, but again the archhis disallowed because it is not yet time to use aBB-arc. Let us say that it takes the arcgg. This brings it back to the vertex33. Next, the turtle must walk alongcc(sinceggis already used, while theBB-arc still must wait until it is the only option). This brings it to the vertex44. Its next step is to take the arcllto the vertex11. From there, it follows the arcffto the vertex33. Now, it can finally take theBB-archh, since all the other outgoing arcs from33have already been used. This brings it to the vertex55. Now it has a choice between the arcsee,iiandjj, but the arceeis disallowed because it is aBB-arc. Let us say it decides to use the arcjj. This brings it to the vertex22. From there, it takes theBB-arckkto the vertex44(since it has no other options). From there, it continues along theBB-arcddto the vertex55. Now, it has to traverse the loopii, and then leave55along theBB-arceeto come back to11. At this point, the turtle is stuck, since it has nowhere left to go. The walk𝐰\mathbf{w}we obtained is thus

𝐰=(1,a,2,b,3,g,3,c,4,l,1,f,3,h,5,j,2,k,4,d,5,i,5,e,1).\mathbf{w}=\left(1,a,2,b,3,g,3,c,4,l,1,f,3,h,5,j,2,k,4,d,5,i,5,e,1\right).(Of course, other choices would have led to other walks.)

Returning to the general case, let us analyze the walk𝐰\mathbf{w}traversed by the turtle.

•First, we claim that𝐰\mathbf{w}is a closed walk (i.e., ends atrr). [Proof:Assume the contrary. Letuube the ending point of𝐰\mathbf{w}. Thus,uuis the vertex at which the turtle gets stuck. Moreover,u≠ru\neq r(since we just assumed that𝐰\mathbf{w}is not a closed walk). Hence, the walk𝐰\mathbf{w}enters the vertexuumore often than it leaves it (since it ends but does not start atuu). In other words, the turtle has entered the vertexuumore often than it has left it. However, sinceDDis balanced, we havedeg−⁡u=deg+⁡u\deg^{-}u=\deg^{+}u. The turtle has entered the vertexuuat mostdeg−⁡u\deg^{-}utimes (because it cannot use an arc twice, but there are onlydeg−⁡u\deg^{-}umany arcs with targetuu). Thus, it has left the vertexuulessthandeg−⁡u\deg^{-}utimes (because it has entered the vertexuumore often than it has left it). Sincedeg−⁡u=deg+⁡u\deg^{-}u=\deg^{+}u, this means that the turtle has left the vertexuuless thandeg+⁡u\deg^{+}utimes. Thus, by the time the turtle has gotten stuck atuu, there is at least one outgoing arc fromuuthat has not been used by the turtle. Therefore, the turtle is not actually stuck atuu. This is a contradiction. Thus, our assumption was wrong, so we have proved that𝐰\mathbf{w}is a closed walk.]

In other words,𝐰\mathbf{w}is a circuit. We shall next show that𝐰\mathbf{w}is an Eulerian circuit.

To do so, we introduce one more piece of notation: A vertexuuofDDwill be calledexhaustedif the turtle has used each outgoing arc fromuu(that is, if each outgoing arc fromuuis used in the circuit𝐰\mathbf{w}).

Since𝐰\mathbf{w}is a circuit, the ending point of𝐰\mathbf{w}is its starting point, i.e., the vertexrr. Thus, the turtle must have gotten stuck atrr. Hence, the vertexrris exhausted.

•We shall now show thatallvertices ofDDare exhausted. [Proof:Assume the contrary. Thus, there exists a vertexuuofDDthat is not exhausted. Consider thisuu. ButBBis a sparb, thus an arborescence rooted torr. Hence,rris a to-root ofBB. Therefore, there exists a path𝐩=(p0,b1,p1,b2,p2,…,bk,pk)\mathbf{p}=\left(p_{0},b_{1},p_{1},b_{2},p_{2},\ldots,b_{k},p_{k}\right)fromuutorrinBB. Consider this path. Thus, we havep0=up_{0}=uandpk=rp_{k}=r, and all the arcsb1,b2,…,bkb_{1},b_{2},\ldots,b_{k}belong toBB. There exists at least onei∈{0,1,…,k}i\in\left\{0,1,\ldots,k\right\}such that the vertexpip_{i}is exhausted (for instance,i=ki=kqualifies, sincepk=rp_{k}=ris exhausted). Consider thesmallestsuchii. Then,pi≠p0p_{i}\neq p_{0}(sincepip_{i}is exhausted, butp0=up_{0}=uis not). Hence,i≠0i\neq 0, so thati≥1i\geq 1. Therefore,pi−1p_{i-1}exists. Moreover, the vertexpi−1p_{i-1}is not exhausted (sinceiiwas defined to be thesmallestelement of{0,1,…,k}\left\{0,1,\ldots,k\right\}such thatpip_{i}is exhausted). The arcbib_{i}has sourcepi−1p_{i-1}and targetpip_{i}. Thus, it is an outgoing arc frompi−1p_{i-1}and incoming arc intopip_{i}. Furthermore, it belongs toBB(since all the arcsb1,b2,…,bkb_{1},b_{2},\ldots,b_{k}belong toBB). The digraphDDis balanced; thus,deg+⁡(pi)=deg−⁡(pi)\deg^{+}\left(p_{i}\right)=\deg^{-}\left(p_{i}\right). The vertexpip_{i}is exhausted. In other words, the turtle has used each outgoing arc frompip_{i}(by the definition of “exhausted”). Since the turtle never reuses an arc, this entails that the turtle has used exactlydeg+⁡(pi)\deg^{+}\left(p_{i}\right)many outgoing arcs frompip_{i}(sincedeg+⁡(pi)\deg^{+}\left(p_{i}\right)is the total number of outgoing arcs frompip_{i}inDD). In other words, it has used exactlydeg−⁡(pi)\deg^{-}\left(p_{i}\right)many outgoing arcs frompip_{i}(sincedeg+⁡(pi)=deg−⁡(pi)\deg^{+}\left(p_{i}\right)=\deg^{-}\left(p_{i}\right)). However, the turtle’s trajectory is a closed walk (in fact, it is the walk𝐰\mathbf{w}, which is closed). Thus, it must enter the vertexpip_{i}as often as it leaves this vertex. In other words, the number of incoming arcs intopip_{i}used by the turtle must equal the number of outgoing arcs frompip_{i}used by the turtle. Since we just found (in the preceding paragraph) that the latter number isdeg−⁡(pi)\deg^{-}\left(p_{i}\right), we thus conclude that the former number isdeg−⁡(pi)\deg^{-}\left(p_{i}\right)as well. In other words, the turtle must have used exactlydeg−⁡(pi)\deg^{-}\left(p_{i}\right)many incoming arcs intopip_{i}. Sincedeg−⁡(pi)\deg^{-}\left(p_{i}\right)is the total number of incoming arcs intopip_{i}inDD, we thus conclude that the turtle must have used all incoming arcs intopip_{i}(since the turtle never reuses an arc). Hence, in particular, the turtle must have used the arcbib_{i}(sincebib_{i}is an incoming arc intopip_{i}). This arcbib_{i}is an outgoing arc frompi−1p_{i-1}. Butbib_{i}is aBB-arc, and thus our turtle uses this arc only as a last resort (i.e., after using all other outgoing arcs frompi−1p_{i-1}). Hence, we conclude that the turtle must have used all outgoing arcs frompi−1p_{i-1}(since it has usedbib_{i}). In other words,pi−1p_{i-1}is exhausted. But this contradicts the fact thatpi−1p_{i-1}is not exhausted! This shows that our assumption was wrong, and our proof is finished.505050For the sake of diversity, let me sketch asecond proofof the same claim (i.e., that all vertices inDDare exhausted):Assume the contrary. Thus, there exists a non-exhausted vertexuuofDD. Consider thisuu. Then,u≠ru\neq r(sincerris exhausted butuuis not). Sinceuuis not exhausted, there is at least one outgoing arc fromuuthat the turtle has not used. Hence, the turtle has not used theBB-arc outgoing fromuu(since the turtle never uses aBB-arc before it has to). Letffbe thisBB-arc, and letu′u^{\prime}be its target. Thus, the turtle has not used all incoming arcs ofu′u^{\prime}(because it has not used the arcff). As a consequence, it has not used all outgoing arcs fromu′u^{\prime}either (because the turtle has leftu′u^{\prime}as often as it has enteredu′u^{\prime}, but the balancedness ofDDentails thatdeg−⁡(u′)=deg+⁡(u′)\deg^{-}\left(u^{\prime}\right)=\deg^{+}\left(u^{\prime}\right)). In other words, the vertexu′u^{\prime}is non-exhausted. Thus, by starting at the non-exhausted vertexuuand taking theBB-arc outgoing fromuu, we have arrived at a further non-exhausted vertexu′u^{\prime}. Applying the same argument tou′u^{\prime}instead ofuu, we can take a furtherBB-arc and arrive at a further non-exhausted vertexu′′u^{\prime\prime}. Continuing like this, we obtain an infinite sequence(u,u′,u′′,…)\left(u,u^{\prime},u^{\prime\prime},\ldots\right)of non-exhausted vertices such that any vertex in this sequence is reached from the previous one by traveling along aBB-arc. Clearly, this sequence must have two equal vertices (sinceDDhas only finitely many vertices). For example, let’s say thatu′′=u′′′′′u^{\prime\prime}=u^{\prime\prime\prime\prime\prime}. Then, if we consider only the part of the sequence betweenu′′u^{\prime\prime}andu′′′′′u^{\prime\prime\prime\prime\prime}, then we obtain a closed walk(u′′,∗,u′′′,∗,u′′′′,∗,u′′′′′),\left(u^{\prime\prime},\ast,u^{\prime\prime\prime},\ast,u^{\prime\prime\prime\prime},\ast,u^{\prime\prime\prime\prime\prime}\right),where each asterisk stands for someBB-arc (not the same one, of course). This is a closed walk of the digraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right). Since this closed walk has length>0>0, it cannot be a path; therefore, it contains a cycle (by Proposition4.30). Thus, we have found a cycle of the digraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right). However, the digraph(V,B,ψ∣B)\left(V,B,\psi\mid_{B}\right)is an arborescence, and thus has no cycles (because ifDDis an arborescence, then any cycle ofDDwould be a cycle ofDundD^{\operatorname*{und}}; but the multigraphDundD^{\operatorname*{und}}has no cycles by the definition of an arborescence). The previous two sentences contradict each other. This shows that our assumption was wrong, and our proof is finished..]

Thus, we have shown that all vertices ofDDare exhausted. In other words, the turtle has used all arcs ofDD. In other words, the trail𝐰\mathbf{w}contains all arcs ofDD. Since𝐰\mathbf{w}is a trail and a closed walk, this entails that𝐰\mathbf{w}is an Eulerian circuit ofDD. Since𝐰\mathbf{w}starts withrrandaa, this shows further that𝐰\mathbf{w}is anaa-Eulerian circuit. Since the turtle only usedBB-arcs as a last resort (and it used eachBB-arc eventually, because𝐰\mathbf{w}is Eulerian), we haveExit𝐰=B\operatorname*{Exit}\mathbf{w}=B.

Thus, the turtle’s walk has produced anaa-Eulerian circuit𝐞\mathbf{e}satisfyingExit𝐞=B\operatorname*{Exit}\mathbf{e}=B(namely, the walk𝐰\mathbf{w}). However, this circuit depends on some decisions the turtle made during its walk. Namely, every time the turtle was at some vertexu∈Vu\in V, it had to decide which arc to take next; this arc had to be an unused arc with sourceuu, subject to the conditions that

1.ifu≠ru\neq r, then theBB-arc515151We say “theBB-arc”, because there is exactly oneBB-arc with sourceuu.has to be used last;
2.ifu=ru=r, then the arcaahas to be used first.

Let us count how many options the turtle has had in total. To make the argument clearer, we modify the procedure somewhat: Instead of deciding ad-hoc which arc to take, the turtle should now make all these decisions before embarking on its journey. To do so, it chooses, for each vertexu∈Vu\in V, a total order on the set of all arcs with sourceuu, such that

1.ifu≠ru\neq r, then theBB-arc comes last in this order, and
2.ifu=ru=r, then the arcaacomes first in this order.

Note that this total order can be chosen in(deg+⁡u−1)!\left(\deg^{+}u-1\right)!many ways (since there aredeg+⁡u\deg^{+}uarcs with sourceuu, and we can freely choose their order except that one of them has a fixed position). Thus, in total, there are∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!many options for how the turtle can choose all these orders. Once these orders have been chosen, the turtle then uses them to decide which arcs to walk along: Namely, the first time it visits the vertexuu, it leaves it along the first arc (according to its chosen order); the second time, it uses the second arc; the third time, the third arc; and so on.

So the turtle has∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!many options, and each of these options leads to a differentaa-Eulerian circuit𝐞\mathbf{e}(because the total orders chosen by the turtle are reflected in𝐞\mathbf{e}: they are precisely the orders in which the respective arcs appear in𝐞\mathbf{e}). Moreover, eachaa-Eulerian circuit𝐞\mathbf{e}satisfyingExit𝐞=B\operatorname*{Exit}\mathbf{e}=Bcomes from one of these options525252Proof.Let𝐞\mathbf{e}be anaa-Eulerian circuit satisfyingExit𝐞=B\operatorname*{Exit}\mathbf{e}=B. We must show that, by choosing the appropriate total orders ahead of its journey, the turtle will trace this exact circuit𝐞\mathbf{e}. First, let me explain what the “appropriate total orders” are: They are the orders dictated by𝐞\mathbf{e}. That is, for each vertexu∈Vu\in V, the turtle must choose the total order on the set{all arcs with sourceu}\left\{\text{all arcs with source }u\right\}that coincides with the order in which these arcs appear on𝐞\mathbf{e}. This choice is legitimate, because the arcaais the first arc of𝐞\mathbf{e}(so it will certainly come first in its order), and because eachBB-arc appears in𝐞\mathbf{e}after all other arcs from the same source have appeared (so it will come last in its total order). Now, let me explain why the choice of these total orders will force the turtle to trace the circuit𝐞\mathbf{e}. Indeed, assume the contrary. Let𝐟\mathbf{f}be the circuit traced by the turtle. Thus, by our assumption,𝐟\mathbf{f}is distinct from𝐞\mathbf{e}. Hence,𝐟\mathbf{f}must diverge from𝐞\mathbf{e}at some point. Consider the first point at which this happens. Thus, at this point,𝐟\mathbf{f}leaves some vertexu∈Vu\in Valong an arce1e_{1}, whereas𝐞\mathbf{e}instead leaves it along a different arce2e_{2}. Since both𝐟\mathbf{f}and𝐞\mathbf{e}are Eulerian circuits, we know that bothe1e_{1}ande2e_{2}must appear in both𝐟\mathbf{f}and𝐞\mathbf{e}, and moreover, the circuits𝐟\mathbf{f}and𝐞\mathbf{e}agree until this point of divergence. Hence,e1e_{1}appears beforee2e_{2}in𝐟\mathbf{f}but appears aftere2e_{2}in𝐞\mathbf{e}. Therefore, the order in which the arcs with sourceuuappear in𝐟\mathbf{f}differs from the order in which they appear in𝐞\mathbf{e}(since bothe1e_{1}ande2e_{2}are arcs with sourceuu). However, this is absurd, since the turtle’s walk𝐟\mathbf{f}was constructed in such a way that it takes the arcs with sourceuuin the same order as𝐞\mathbf{e}does (because the total orders were appropriately chosen). So we found a contradiction, and our proof is complete..

Therefore, the total number ofaa-Eulerian circuits𝐞\mathbf{e}satisfyingExit𝐞=B\operatorname*{Exit}\mathbf{e}=Bis the total number of options, which is∏u∈V(deg+⁡u−1)!\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!as we know. This proves Claim 2.]

With Claims 1 and 2 proved, we are almost done. The map

{a-Eulerian circuits ofD}\displaystyle\left\{a\text{-Eulerian circuits of }D\right\}→{sparbs},\displaystyle\rightarrow\left\{\text{sparbs}\right\},𝐞\displaystyle\mathbf{e}↦Exit𝐞\displaystyle\mapsto\operatorname*{Exit}\mathbf{e}is well-defined (by Claim 1). Furthermore, Claim 2 shows that this map is a∏u∈V(deg+⁡u−1)!\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!-to-11correspondence535353Anmm-to-11correspondence(wheremmis a nonnegative integer) means a mapf:X→Yf:X\rightarrow Ybetween two sets such that each element ofYYhas exactlymmpreimages underff.(i.e., each sparbBBhas exactly ∏u∈V(deg+⁡u−1)!\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!many preimages under this map). Thus, by the multijection principle545454Themultijection principleis a basic counting principle that says the following: LetXXandYYbe two finite sets, and letm∈ℕm\in\mathbb{N}. Letf:X→Yf:X\rightarrow Ybe anmm-to-11correspondence (i.e., a map such that each element ofYYhas exactlymmpreimages underff). Then,|X|=m⋅|Y|\left|X\right|=m\cdot\left|Y\right|. For example,nn(intact) sheep have4n4nlegs in total, since the map that sends each leg to its sheep is a44-to-11correspondence., we conclude that555555The symbol “#\#” means “number”.

(#ofa-Eulerian circuits ofD)=(∏u∈V(deg+⁡u−1)!)⋅(#of sparbs).\left(\#\text{ of }a\text{-Eulerian circuits of }D\right)=\left(\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!\right)\cdot\left(\#\text{ of sparbs}\right).Sinceε(D,a)=(#ofa-Eulerian circuits ofD)\varepsilon\left(D,a\right)=\left(\#\text{ of }a\text{-Eulerian circuits of }D\right)andτ(D,r)=(#of sparbs)\tau\left(D,r\right)=\left(\#\text{ of sparbs}\right), we can rewrite this as follows:

ε⁡(D,a)=(∏u∈V(deg+⁡u−1)!)⋅τ⁡(D,r)=τ⁡(D,r)⋅∏u∈V(deg+⁡u−1)!.\varepsilon\left(D,a\right)=\left(\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!\right)\cdot\tau\left(D,r\right)=\tau\left(D,r\right)\cdot\prod\limits\limits_{u\in V}\left(\deg^{+}u-1\right)!.This proves Theorem5.122.

Proof 5.129(Proof of Theorem5.117.).

As we already mentioned, Theorem5.117follows from Theorem5.122by reversing each arc (i.e., by applying Theorem5.122to the digraphDrevD^{\operatorname*{rev}}instead ofDD).

The above proof of Theorem5.122originates from[VanEhr51, §6]; I have learned it from[Stanle18, Theorem 10.2]and[HLMPPW13, Corollary 4.10](where it appears in a mildly modified form).

5.12A corollary about spanning arborescences

Before we actually use the BEST (or BEST’) theorem to count the Eulerian circuits on any digraph, let us mention a neat corollary for the number of spanning arborescences:

Corollary 5.130.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a balanced multidigraph. For each vertexr∈Vr\in V, letτ⁡(D,r)\tau\left(D,r\right)be the number of spanning arborescences ofDDrooted torr. Then,τ⁡(D,r)\tau\left(D,r\right)does not depend onrr.

Proof 5.131(Proof of Corollary5.130.).

WLOG assume that|V|>1\left|V\right|>1(else, the claim is obvious). If there is a vertexv∈Vv\in Vwithdeg+⁡v=0\deg^{+}v=0, then this vertexvvsatisfiesdeg−⁡v=0\deg^{-}v=0as well (since the balancedness ofDDentailsdeg−⁡v=deg+⁡v=0\deg^{-}v=\deg^{+}v=0), and thereforeDDhas no spanning arborescences at all (since any spanning arborescence would have an arc with source or targetvv). Thus, we WLOG assume thatdeg+⁡v>0\deg^{+}v>0for allv∈Vv\in V. In other words, each vertex has outdegree>0>0.

Letrrandssbe two vertices ofDD. We must prove thatτ⁡(D,r)=τ⁡(D,s)\tau\left(D,r\right)=\tau\left(D,s\right).

Pick an arcaawith sourcerr. (This exists, sincedeg+⁡r>0\deg^{+}r>0.) Pick an arcbbwith sourcess. (This exists, sincedeg+⁡s>0\deg^{+}s>0.)

Applying the BEST’ theorem (Theorem5.122), we get

ε⁡(D,a)\displaystyle\varepsilon\left(D,a\right)=τ(D,r)⋅∏u∈V(deg+u−1)!and similarly\displaystyle=\tau\left(D,r\right)\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!\ \ \ \ \ \ \ \ \ \ \text{and similarly}ε⁡(D,b)\displaystyle\varepsilon\left(D,b\right)=τ⁡(D,s)⋅∏u∈V(deg+⁡u−1)!.\displaystyle=\tau\left(D,s\right)\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!.However,ε⁡(D,a)=ε⁡(D,b)\varepsilon\left(D,a\right)=\varepsilon\left(D,b\right), since counting Eulerian circuits that start withaais equivalent to counting Eulerian circuits that start withbb(because an Eulerian circuit can be rotated uniquely to start with any given arc). Thus, we obtain

τ⁡(D,r)⋅∏u∈V(deg+⁡u−1)!=ε⁡(D,a)=ε⁡(D,b)=τ⁡(D,s)⋅∏u∈V(deg+⁡u−1)!.\tau\left(D,r\right)\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!=\varepsilon\left(D,a\right)=\varepsilon\left(D,b\right)=\tau\left(D,s\right)\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!.Cancelling the (nonzero!) number∏u∈V(deg+⁡u−1)!\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!from this equality, we obtainτ⁡(D,r)=τ⁡(D,s)\tau\left(D,r\right)=\tau\left(D,s\right). This proves Corollary5.130.

5.13Spanning arborescences vs. spanning trees

The BEST theorem (Theorem5.122or Theorem5.117) connects the#\#of Eulerian circuits in a digraph with the#\#of spanning arborescences of the same digraph. Now let us try to find a way to compute the latter.

For example, let us try to do this for digraphs of the formGbidirG^{\operatorname*{bidir}}whereGGis a multigraph. I claim that the spanning arborescences ofGbidirG^{\operatorname*{bidir}}rooted to a given vertexrrare just the spanning trees ofGGin disguise:

Proposition 5.132.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Fix a vertexr∈Vr\in V. Recall that the arcs ofGbidirG^{\operatorname*{bidir}}are the pairs(e,i)∈E×{1,2}\left(e,i\right)\in E\times\left\{1,2\right\}. Identify each spanning tree ofGGwith its edge set, and each spanning arborescence ofGbidirG^{\operatorname*{bidir}}with its arc set.

IfBBis a spanning arborescence ofGbidirG^{\operatorname*{bidir}}rooted torr, then we set

B¯:={e∣(e,i)∈B}.\overline{B}:=\left\{e\ \mid\ \left(e,i\right)\in B\right\}.(Recall that we are identifying spanning arborescences with their arc sets, so that “(e,i)∈B\left(e,i\right)\in B” means “(e,i)\left(e,i\right)is an arc ofBB”.)

Then:

(a)IfBBis a spanning arborescence ofGbidirG^{\operatorname*{bidir}}rooted torr, thenB¯\overline{B}is a spanning tree ofGG.
(b)The map {spanning arborescences ofGbidirrooted tor}\displaystyle\left\{\text{spanning arborescences of }G^{\operatorname*{bidir}}\text{ rooted to }r\right\}→{spanning trees ofG},\displaystyle\rightarrow\left\{\text{spanning trees of }G\right\},B\displaystyle B↦B¯\displaystyle\mapsto\overline{B}is a bijection.

Example 5.133.

Here is a multigraphGG(on the left) with the corresponding multidigraphGbidirG^{\operatorname*{bidir}}(on the right):

11223344551122334455GGGbidirG^{\operatorname*{bidir}}Here is a spanning arborescenceBBofGbidirG^{\operatorname*{bidir}}rooted to11, and the corresponding spanning treeB¯\overline{B}ofGG:

11223344551122334455BBB¯\overline{B}(here, the arcs ofGbidirG^{\operatorname*{bidir}}thatdon’tbelong toBB, as well as the edges ofGGthatdon’tbelong toB¯\overline{B}, have been drawn as dotted arrows). It is fairly easy to see howBBcan be reconstructed fromB¯\overline{B}: You just need to replace each edge ofB¯\overline{B}by the appropriately directed arc (namely, the one that is “directed towards11”).

Proof 5.134(Proof of Proposition5.132.).

This is an exercise in yak-shaving (and we have, in fact, shaved a very similar yak in Section5.7; the only difference is that we are no longer dealing with trees in isolation, but rather with spanning trees ofGG).

(a)LetBBbe a spanning arborescence ofGbidirG^{\operatorname*{bidir}}rooted torr. Then,BundB^{\operatorname*{und}}is a tree (by the implication A’1⟹\LongrightarrowA’3 in Theorem5.123). However, it is easy to see thatBund≅B¯B^{\operatorname*{und}}\cong\overline{B}as multigraphs (indeed, each vertexvvofBundB^{\operatorname*{und}}corresponds to the same vertexvvofB¯\overline{B}, whereas any edge(e,i)\left(e,i\right)ofBundB^{\operatorname*{und}}corresponds to the edgeeeofB¯\overline{B})565656Here we need to use the fact that for each edgeeeofB¯\overline{B},exactly oneof the two pairs(e,1)\left(e,1\right)and(e,2)\left(e,2\right)is an edge ofBundB^{\operatorname*{und}}. But this is easy to check: At least one of the two pairs(e,1)\left(e,1\right)and(e,2)\left(e,2\right)must be an arc ofBB(sinceeeis an edge ofB¯\overline{B}). In other words, at least one of the two pairs(e,1)\left(e,1\right)and(e,2)\left(e,2\right)must be an edge ofBundB^{\operatorname*{und}}. But both of these pairs cannot be edges ofBundB^{\operatorname*{und}}at the same time (since this would create a cycle, butBundB^{\operatorname*{und}}is a tree and thus has no cycles). Hence, exactly one of these pairs is an edge ofBundB^{\operatorname*{und}}, qed.. Thus,B¯\overline{B}is a tree (sinceBundB^{\operatorname*{und}}is a tree)575757Alternatively, you can prove this as follows: The vertexrris a to-root ofBB(sinceBBis an arborescence rooted torr). Thus, for eachv∈Vv\in V, there is a path fromvvtorrinBB. By “projecting” this path ontoB¯\overline{B}(that is, replacing each arc(e,i)\left(e,i\right)of this path by the corresponding edgeeeofB¯\overline{B}), we obtain a path fromvvtorrinB¯\overline{B}. This shows that the multigraphB¯\overline{B}is connected. Furthermore, the definition ofB¯\overline{B}shows that|B¯|≤|B|=|V|−1\left|\overline{B}\right|\leq\left|B\right|=\left|V\right|-1(by Statement A’2 in Theorem5.123, sinceBBis an arborescence rooted torr). Hence,|B¯|<|V|\left|\overline{B}\right|<\left|V\right|. Thus, we can apply the implication T5⟹\LongrightarrowT1 of the Tree Equivalence Theorem (Theorem5.21) to conclude thatB¯\overline{B}is a tree., therefore a spanning tree ofGG(sinceB¯\overline{B}is clearly a spanning subgraph ofGG). This proves Proposition5.132(a).

(b)We must prove that this map is surjective and injective.

Surjectivity:LetTTbe a spanning tree ofGG. Then, the multidigraphTr→T^{r\rightarrow}(defined in Definition5.95) is an arborescence rooted fromrr(by Lemma5.98). Reversing each arc in this arborescenceTr→T^{r\rightarrow}, we obtain a new multidigraphTr←T^{r\leftarrow}, which is thus an arborescence rooted torr. Unfortunately,Tr←T^{r\leftarrow}is not a subdigraph ofGbidirG^{\operatorname*{bidir}}, for a rather stupid reason: The arcs ofTr←T^{r\leftarrow}are elements ofEE, whereas the arcs ofGbidirG^{\operatorname*{bidir}}are pairs of the form(e,i)\left(e,i\right)withe∈Ee\in Eandi∈{1,2}i\in\left\{1,2\right\}.

Fortunately, this is easily fixed: For each arceeofTr←T^{r\leftarrow}, we lete′e^{\prime}be the arc(e,i)\left(e,i\right)ofGbidirG^{\operatorname*{bidir}}that has the same source asee(and thus the same target asee). This is uniquely determined, since the arcs(e,1)\left(e,1\right)and(e,2)\left(e,2\right)ofGbidirG^{\operatorname*{bidir}}have different sources585858Proof.The edgeeeofTTis not a loop (becauseTTis a tree, but a tree cannot have any loops). Hence, its two endpoints are distinct. Thus, the arcs(e,1)\left(e,1\right)and(e,2)\left(e,2\right)ofGbidirG^{\operatorname*{bidir}}have different sources (since their sources are the two endpoints ofee).. If we replace each arceeofTr←T^{r\leftarrow}by the corresponding arce′e^{\prime}ofGbidirG^{\operatorname*{bidir}}, then we obtain a spanning subdigraphSSofGbidirG^{\operatorname*{bidir}}that is an arborescence rooted torr(sinceTr←T^{r\leftarrow}is an arborescence rooted torr, and we have only replaced its arcs by equivalent ones with the same sources and the same targets). In other words, we obtain a spanning arborescenceSSofGbidirG^{\operatorname*{bidir}}rooted torr. It is easy to see thatS¯=T\overline{S}=T. Hence, the map

{spanning arborescences ofGbidirrooted tor}\displaystyle\left\{\text{spanning arborescences of }G^{\operatorname*{bidir}}\text{ rooted to }r\right\}→{spanning trees ofG},\displaystyle\rightarrow\left\{\text{spanning trees of }G\right\},B\displaystyle B↦B¯\displaystyle\mapsto\overline{B}sendsSStoTT. This shows thatTTis a value of this map. Since we have proved this for every spanning treeTTofGG, we have thus shown that this map is surjective.

Injectivity:The main idea is that, in order to recover a spanning arborescenceBBback from the corresponding spanning treeB¯\overline{B}, we just need to “orient the edges of the tree towardsrr”. Here are the (annoyingly long) details:

LetBBandCCbe two sparbs595959Henceforth, “sparb” is short for “spanning arborescence ofGbidirG^{\operatorname*{bidir}}rooted torr”.such thatB¯=C¯\overline{B}=\overline{C}. We must show thatB=CB=C.

Assume the contrary. Thus,B≠CB\neq C. LetTTbe the treeB¯=C¯\overline{B}=\overline{C}. Thus, each edgeeeofTTcorresponds to either an arc(e,1)\left(e,1\right)or an arc(e,2)\left(e,2\right)inBB(sinceT=B¯T=\overline{B}), and likewise forCC. Conversely, each arc(e,i)\left(e,i\right)ofBBor ofCCcorresponds to an edgeeeofTT. Hence, fromB≠CB\neq C, we see that there must exist an edgeeeofTTsuch that

•eitherwe have(e,1)∈B\left(e,1\right)\in Band(e,2)∈C\left(e,2\right)\in C,
•orwe have(e,1)∈C\left(e,1\right)\in Cand(e,2)∈B\left(e,2\right)\in B.

Consider this edgeee. We WLOG assume that(e,1)∈B\left(e,1\right)\in Band(e,2)∈C\left(e,2\right)\in C(else, we can just swapBBwithCC). Let the arc(e,1)\left(e,1\right)ofGbidirG^{\operatorname*{bidir}}have sourcessand targettt, so that(e,2)\left(e,2\right)has sourcettand targetss. The edgeeethus has endpointsssandtt.

SinceBBis an arborescence rooted torr, the vertexrris a to-root ofBB. Hence, there exists a path𝐩\mathbf{p}fromsstorrinBB. This path𝐩\mathbf{p}must begin with the arc(e,1)\left(e,1\right)606060Proof.Sincerris a to-root ofBB, we know that there exists a path fromtttorrinBB. Let𝐭\mathbf{t}be this path. Extending this path𝐭\mathbf{t}by the vertexssand the arc(e,1)\left(e,1\right)(which we both insert at the start of𝐭\mathbf{t}), we obtain a walk𝐭′\mathbf{t}^{\prime}fromsstorrinBB. (So, if𝐭=(t,…,r)\mathbf{t}=\left(t,\ \ldots,\ r\right), then𝐭′=(s,(e,1),t,…,r)\mathbf{t}^{\prime}=\left(s,\ \left(e,1\right),\ t,\ \ldots,\ r\right).) However,BBis an arborescence rooted torr. Thus, Statement A’4 in the Dual Arborescence Equivalence Theorem (Theorem5.123) shows that for each vertexv∈Vv\in V, the digraphBBhas a unique walk fromvvtorr. Hence, in particular,BBhas a unique walk fromsstorr. Thus,𝐩=𝐭′\mathbf{p}=\mathbf{t}^{\prime}(since both𝐩\mathbf{p}and𝐭′\mathbf{t}^{\prime}are walks fromsstorrinBB). Since𝐭′\mathbf{t}^{\prime}begins with the arc(e,1)\left(e,1\right), we thus conclude that𝐩\mathbf{p}begins with the arc(e,1)\left(e,1\right).. Projecting this path𝐩\mathbf{p}down ontoTT, we obtain a path𝐩¯\overline{\mathbf{p}}fromsstorrinTT. (By the word “projecting”, we mean replacing each arc(e,i)\left(e,i\right)by the corresponding edgeee. Clearly, doing this to a path inBByields a path inTT, becauseT=B¯T=\overline{B}.) Since the path𝐩\mathbf{p}begins with the arc(e,1)\left(e,1\right), the “projected” path𝐩¯\overline{\mathbf{p}}begins with the edgeee. Thus, in the treeTT, the path fromsstorrbegins with the edgeee(because this path must be the path𝐩¯\overline{\mathbf{p}}). As a consequence,ttmust be the second vertex of this path (since the edgeeehas endpointsssandtt), so that removing the first edge from this path yields the path fromtttorr. Thus,d⁡(t,r)=d⁡(s,r)−1d\left(t,r\right)=d\left(s,r\right)-1, wheredddenotes distance on the treeTT. Hence,d⁡(t,r)<d⁡(s,r)d\left(t,r\right)<d\left(s,r\right).

A similar argument (but with the roles ofBBandCCswapped, as well as the roles ofssandttswapped, and the roles of(e,1)\left(e,1\right)and(e,2)\left(e,2\right)swapped) shows thatd⁡(s,r)<d⁡(t,r)d\left(s,r\right)<d\left(t,r\right). But this contradictsd⁡(t,r)<d⁡(s,r)d\left(t,r\right)<d\left(s,r\right).

This contradiction shows that our assumption was false. Thus, we have proved thatB=CB=C.

Forget that we fixedBBandCC. We thus have shown that ifBBandCCare two sparbs such thatB¯=C¯\overline{B}=\overline{C}, thenB=CB=C. In other words, our map

We have now shown that this map is both surjective and injective. Hence, it is a bijection. This proves Proposition5.132(b).

5.14The matrix-tree theorem

5.14.1Introduction

So counting spanning trees in a multigraph is a particular case of counting spanning arborescences (rooted to a given vertex) in a multidigraph. But how do we do either? Let us begin with some simple examples:

Example 5.135.

There is only one spanning tree of the complete graphK1K_{1}:

1.\begin{tabular}[c]{|c|}\hline\cr $\\hbox to18\.37pt\{\\vbox to18\.37pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 9\.18588pt\\lower\-9\.18588pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 0 C 12\.16 6\.71 6\.71 12\.16 0 12\.16 C \-6\.71 12\.16 \-12\.16 6\.71 \-12\.16 0 C \-12\.16 \-6\.71 \-6\.71 \-12\.16 0 \-12\.16 C 6\.71 \-12\.16 12\.16 \-6\.71 12\.16 0 Z M 0 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ \\ \hline\cr\end{tabular}\ \ . There is only one spanning tree of the complete graphK2K_{2}:

12.\begin{tabular}[c]{|c|}\hline\cr $\\hbox to69\.59pt\{\\vbox to18\.37pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 9\.18588pt\\lower\-9\.18588pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 0 C 12\.16 6\.71 6\.71 12\.16 0 12\.16 C \-6\.71 12\.16 \-12\.16 6\.71 \-12\.16 0 C \-12\.16 \-6\.71 \-6\.71 \-12\.16 0 \-12\.16 C 6\.71 \-12\.16 12\.16 \-6\.71 12\.16 0 Z M 0 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 83\.02 0 C 83\.02 6\.71 77\.58 12\.16 70\.87 12\.16 C 64\.15 12\.16 58\.71 6\.71 58\.71 0 C 58\.71 \-6\.71 64\.15 \-12\.16 70\.87 \-12\.16 C 77\.58 \-12\.16 83\.02 \-6\.71 83\.02 0 Z M 70\.87 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{48\.71504pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 67\.41 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.71 0 L 58\.16 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ \\ \hline\cr\end{tabular}\ \ \ . There are33spanning trees of the complete graphK3K_{3}:

123123123.\begin{tabular}[c]{|c|c|c|}\hline\cr $\\hbox to69\.59pt\{\\vbox to77\.51pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 26\.25749pt\\lower\-38\.75484pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 59\.4 0 C 59\.4 6\.71 53\.96 12\.16 47\.24 12\.16 C 40\.53 12\.16 35\.09 6\.71 35\.09 0 C 35\.09 \-6\.71 40\.53 \-12\.16 47\.24 \-12\.16 C 53\.96 \-12\.16 59\.4 \-6\.71 59\.4 0 Z M 47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{31\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 43\.78 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-11\.46 40\.91 C \-11\.46 47\.63 \-16\.91 53\.07 \-23\.62 53\.07 C \-30\.34 53\.07 \-35\.78 47\.63 \-35\.78 40\.91 C \-35\.78 34\.2 \-30\.34 28\.76 \-23\.62 28\.76 C \-16\.91 28\.76 \-11\.46 34\.2 \-11\.46 40\.91 Z M \-23\.62 40\.91\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-19\.57161pt\}\{26\.34674pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-27\.08 36\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-11\.46 \-40\.91 C \-11\.46 \-34\.2 \-16\.91 \-28\.76 \-23\.62 \-28\.76 C \-30\.34 \-28\.76 \-35\.78 \-34\.2 \-35\.78 \-40\.91 C \-35\.78 \-47\.63 \-30\.34 \-53\.07 \-23\.62 \-53\.07 C \-16\.91 \-53\.07 \-11\.46 \-47\.63 \-11\.46 \-40\.91 Z M \-23\.62 \-40\.91\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} 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53\.07 \-23\.62 53\.07 C \-30\.34 53\.07 \-35\.78 47\.63 \-35\.78 40\.91 C \-35\.78 34\.2 \-30\.34 28\.76 \-23\.62 28\.76 C \-16\.91 28\.76 \-11\.46 34\.2 \-11\.46 40\.91 Z M \-23\.62 40\.91\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-19\.57161pt\}\{26\.34674pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-27\.08 36\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-11\.46 \-40\.91 C \-11\.46 \-34\.2 \-16\.91 \-28\.76 \-23\.62 \-28\.76 C \-30\.34 \-28\.76 \-35\.78 \-34\.2 \-35\.78 \-40\.91 C \-35\.78 \-47\.63 \-30\.34 \-53\.07 \-23\.62 \-53\.07 C \-16\.91 \-53\.07 \-11\.46 \-47\.63 \-11\.46 \-40\.91 Z M \-23\.62 \-40\.91\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-19\.57161pt\}\{\-32\.79117pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-27\.08 \-45\.37$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-12\.61 34\.56 L 36\.24 6\.36\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-23\.62 28\.2 L \-23\.62 \-28\.2\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ & $\\hbox to69\.59pt\{\\vbox to77\.51pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 26\.25749pt\\lower\-38\.75484pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 59\.4 0 C 59\.4 6\.71 53\.96 12\.16 47\.24 12\.16 C 40\.53 12\.16 35\.09 6\.71 35\.09 0 C 35\.09 \-6\.71 40\.53 \-12\.16 47\.24 \-12\.16 C 53\.96 \-12\.16 59\.4 \-6\.71 59\.4 0 Z M 47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{31\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 43\.78 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-11\.46 40\.91 C \-11\.46 47\.63 \-16\.91 53\.07 \-23\.62 53\.07 C \-30\.34 53\.07 \-35\.78 47\.63 \-35\.78 40\.91 C \-35\.78 34\.2 \-30\.34 28\.76 \-23\.62 28\.76 C \-16\.91 28\.76 \-11\.46 34\.2 \-11\.46 40\.91 Z M \-23\.62 40\.91\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-19\.57161pt\}\{26\.34674pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-27\.08 36\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-11\.46 \-40\.91 C \-11\.46 \-34\.2 \-16\.91 \-28\.76 \-23\.62 \-28\.76 C \-30\.34 \-28\.76 \-35\.78 \-34\.2 \-35\.78 \-40\.91 C \-35\.78 \-47\.63 \-30\.34 \-53\.07 \-23\.62 \-53\.07 C \-16\.91 \-53\.07 \-11\.46 \-47\.63 \-11\.46 \-40\.91 Z M \-23\.62 \-40\.91\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-19\.57161pt\}\{\-32\.79117pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-27\.08 \-45\.37$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-12\.61 \-34\.56 L 36\.24 \-6\.36\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-23\.62 \-28\.2 L \-23\.62 28\.2\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ \\ \hline\cr\end{tabular}\ \ .(They are all isomorphic, but still distinct.)

There are1616spanning trees of the complete graphK4K_{4}:

123412341234123412341234123412341234123412341234.\begin{tabular}[c]{|c|c|c|c|}\hline\cr $\\hbox to86\.66pt\{\\vbox to86\.66pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 43\.3291pt\\lower\-43\.3291pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 59\.4 0 C 59\.4 6\.71 53\.96 12\.16 47\.24 12\.16 C 40\.53 12\.16 35\.09 6\.71 35\.09 0 C 35\.09 \-6\.71 40\.53 \-12\.16 47\.24 \-12\.16 C 53\.96 \-12\.16 59\.4 \-6\.71 59\.4 0 Z M 47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{31\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 43\.78 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 47\.24 C 12\.16 53\.96 6\.71 59\.4 0 59\.4 C \-6\.71 59\.4 \-12\.16 53\.96 \-12\.16 47\.24 C \-12\.16 40\.53 \-6\.71 35\.09 0 35\.09 C 6\.71 35\.09 12\.16 40\.53 12\.16 47\.24 Z M 0 47\.24\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{30\.921pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 42\.79$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-35\.09 0 C \-35\.09 6\.71 \-40\.53 12\.16 \-47\.24 12\.16 C \-53\.96 12\.16 \-59\.4 6\.71 \-59\.4 0 C \-59\.4 \-6\.71 \-53\.96 \-12\.16 \-47\.24 \-12\.16 C \-40\.53 \-12\.16 \-35\.09 \-6\.71 \-35\.09 0 Z M \-47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-36\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-50\.7 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 \-47\.24 C 12\.16 \-40\.53 6\.71 \-35\.09 0 \-35\.09 C \-6\.71 \-35\.09 \-12\.16 \-40\.53 \-12\.16 \-47\.24 C \-12\.16 \-53\.96 \-6\.71 \-59\.4 0 \-59\.4 C 6\.71 \-59\.4 12\.16 \-53\.96 12\.16 \-47\.24 Z M 0 \-47\.24\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{\-37\.36543pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 \-51\.7$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 38\.26 8\.99 L 8\.99 38\.26\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 34\.53 0 L \-34\.53 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 38\.26 \-8\.99 L 8\.99 \-38\.26\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope 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12\.16 C \-53\.96 12\.16 \-59\.4 6\.71 \-59\.4 0 C \-59\.4 \-6\.71 \-53\.96 \-12\.16 \-47\.24 \-12\.16 C \-40\.53 \-12\.16 \-35\.09 \-6\.71 \-35\.09 0 Z M \-47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-36\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-50\.7 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 \-47\.24 C 12\.16 \-40\.53 6\.71 \-35\.09 0 \-35\.09 C \-6\.71 \-35\.09 \-12\.16 \-40\.53 \-12\.16 \-47\.24 C \-12\.16 \-53\.96 \-6\.71 \-59\.4 0 \-59\.4 C 6\.71 \-59\.4 12\.16 \-53\.96 12\.16 \-47\.24 Z M 0 \-47\.24\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{\-37\.36543pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 \-51\.7$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\}\{\}\{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 8\.99 \-38\.26 L 38\.26 \-8\.99\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 34\.53 0 L \-34\.53 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-38\.26 8\.99 L \-8\.99 38\.26\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ & $\\hbox to86\.66pt\{\\vbox to86\.66pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 43\.3291pt\\lower\-43\.3291pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 59\.4 0 C 59\.4 6\.71 53\.96 12\.16 47\.24 12\.16 C 40\.53 12\.16 35\.09 6\.71 35\.09 0 C 35\.09 \-6\.71 40\.53 \-12\.16 47\.24 \-12\.16 C 53\.96 \-12\.16 59\.4 \-6\.71 59\.4 0 Z M 47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{31\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 43\.78 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 47\.24 C 12\.16 53\.96 6\.71 59\.4 0 59\.4 C \-6\.71 59\.4 \-12\.16 53\.96 \-12\.16 47\.24 C \-12\.16 40\.53 \-6\.71 35\.09 0 35\.09 C 6\.71 35\.09 12\.16 40\.53 12\.16 47\.24 Z M 0 47\.24\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{30\.921pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 42\.79$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-35\.09 0 C \-35\.09 6\.71 \-40\.53 12\.16 \-47\.24 12\.16 C \-53\.96 12\.16 \-59\.4 6\.71 \-59\.4 0 C \-59\.4 \-6\.71 \-53\.96 \-12\.16 \-47\.24 \-12\.16 C \-40\.53 \-12\.16 \-35\.09 \-6\.71 \-35\.09 0 Z M \-47\.24 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-36\.64322pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-50\.7 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 \-47\.24 C 12\.16 \-40\.53 6\.71 \-35\.09 0 \-35\.09 C \-6\.71 \-35\.09 \-12\.16 \-40\.53 \-12\.16 \-47\.24 C \-12\.16 \-53\.96 \-6\.71 \-59\.4 0 \-59\.4 C 6\.71 \-59\.4 12\.16 \-53\.96 12\.16 \-47\.24 Z M 0 \-47\.24\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{\-37\.36543pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 \-51\.7$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\}\{\}\{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 8\.99 \-38\.26 L 38\.26 \-8\.99\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 38\.26 8\.99 L 8\.99 38\.26\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\}\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-8\.99 38\.26 L \-38\.26 8\.99\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ \\ \hline\cr\end{tabular}\ \ .1234123412341234(There are only two non-isomorphic ones among them.)

This example suggests that the # of spanning trees of a complete graphKnK_{n}isnn−2n^{n-2}.

This is indeed true, and we will prove this later. First, however, let us discuss the more general problem of counting spanning arborescences of an arbitrary digraphDD.

5.14.2Notations

We will use the following convenient shorthand notation:

Definition 5.136.

We will use theIverson bracket notation: If𝒜\mathcal{A}is any logical statement, then we set

[𝒜]:={1,if𝒜is true;0,if𝒜is false.\left[\mathcal{A}\right]:=\begin{cases}1,&\text{if }\mathcal{A}\text{ is true;}\\ 0,&\text{if }\mathcal{A}\text{ is false.}\end{cases}

For example,[K2is a tree]=1\left[K_{2}\text{ is a tree}\right]=1whereas[K3is a tree]=0\left[K_{3}\text{ is a tree}\right]=0.

Definition 5.137.

LetMMbe a matrix. Letiiandjjbe two integers. Then,

Mi,jwill mean the entry ofMin rowiand columnj;\displaystyle M_{i,j}\text{ will mean the entry of }M\text{ in row }i\text{ and column }j\text{;}M∼i,∼jwill mean the matrixMwith rowiremoved and columnjremoved.\displaystyle M_{\sim i,\sim j}\text{ will mean the matrix }M\text{ with row }i\text{ removed and column }j\text{ removed.} For example,

(abcdefghi)2,3=fand(abcdefghi)∼2,∼3=(abgh).\left(\begin{array}[c]{ccc}a&b&c\\ d&e&f\\ g&h&i\end{array}\right)_{2,3}=f\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{ccc}a&b&c\\ d&e&f\\ g&h&i\end{array}\right)_{\sim 2,\sim 3}=\left(\begin{array}[c]{cc}a&b\\ g&h\end{array}\right).

5.14.3The Laplacian of a multidigraph

We shall now assign a matrix to (more or less) any multidigraph:616161Recall that the symbol “#\#” means “number”.

Definition 5.138.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}.

For anyi,j∈Vi,j\in V, we letai,ja_{i,j}be the#\#of arcs ofDDthat have sourceiiand targetjj.

TheLaplacianofDDis defined to be then×nn\times n-matrixL∈ℤn×nL\in\mathbb{Z}^{n\times n}whose entries are given by

Li,j=(deg+⁡i)⋅[i=j]⏟This is alsoknown asδi,j−ai,jfor alli,j∈V.L_{i,j}=\left(\deg^{+}i\right)\cdot\underbrace{\left[i=j\right]}_{\begin{subarray}{c}\text{This is also}\\ \text{known as }\delta_{i,j}\end{subarray}}-\,a_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in V.In other words, it is the matrix

L=(deg+⁡1−a1,1−a1,2⋯−a1,n−a2,1deg+⁡2−a2,2⋯−a2,n⋮⋮⋱⋮−an,1−an,2⋯deg+⁡n−an,n).L=\left(\begin{array}[c]{cccc}\deg^{+}1-a_{1,1}&-a_{1,2}&\cdots&-a_{1,n}\\ -a_{2,1}&\deg^{+}2-a_{2,2}&\cdots&-a_{2,n}\\ \vdots&\vdots&\ddots&\vdots\\ -a_{n,1}&-a_{n,2}&\cdots&\deg^{+}n-a_{n,n}\end{array}\right).

Example 5.139.

LetDDbe the digraph

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\lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{} {\lx@inpgf@ignorespaces{}{}{{}}{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{}{}\lx@inpgf@ignorespaces}{{}} {\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{\lx@inpgf@ignorespaces{}{}{{}}{}}{{{}}{{}}}{{}}{}{{}}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{{{{{{}}{}{}{}{}{{}}}}}{}{}{}{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 71.33 3.29 C 99.85 10.93 99.85 -10.93 72.94 -3.72}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{-0.96593}{0.25882}{-0.25882}{-0.96593}{52.13196pt}{-2.53278pt}\lxSVG@begingroup@{transform=matrix(-0.96593 0.25882 -0.25882 -0.96593 72.13 -3.5)} \lxSVG@begingroup@{_scopebegin=1} 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(2−1−1−0−01−0−1−0−01−1)=(1−1001−1000).\left(\begin{array}[c]{ccc}2-1&-1&-0\\ -0&1-0&-1\\ -0&-0&1-1\end{array}\right)=\left(\begin{array}[c]{ccc}1&-1&0\\ 0&1&-1\\ 0&0&0\end{array}\right).

One thing we notice from this example is that loops do not matter at all to the LaplacianLL. Indeed, a loop with sourceiiand targetiicounts once indeg+⁡i\deg^{+}iand once inai,ia_{i,i}, but these contributions cancel out.

Here is a simple property of Laplacians:

Proposition 5.140.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for some positive integernn.

Then, the LaplacianLLofDDis singular; i.e., we havedetL=0\det L=0.

Proof 5.141.

The sum of all columns ofLLis the zero vector, because for eachi∈Vi\in Vwe have

∑j=1nLi,j\displaystyle\sum\limits_{j=1}^{n}L_{i,j}=∑j=1n((deg+i)⋅[i=j]−ai,j)(by the definition ofL)\displaystyle=\sum\limits_{j=1}^{n}\left(\left(\deg^{+}i\right)\cdot\left[i=j\right]-a_{i,j}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of }L\right)=∑j=1n(deg+i)⋅[i=j]⏟=deg+⁡i(since only the addendforj=ican be nonzero)−∑j=1nai,j⏟=deg+⁡i(since this is countingall arcs with sourcei)\displaystyle=\underbrace{\sum\limits_{j=1}^{n}\left(\deg^{+}i\right)\cdot\left[i=j\right]}_{\begin{subarray}{c}=\deg^{+}i\\ \text{(since only the addend}\\ \text{for }j=i\text{ can be nonzero)}\end{subarray}}-\underbrace{\sum\limits_{j=1}^{n}a_{i,j}}_{\begin{subarray}{c}=\deg^{+}i\\ \text{(since this is counting}\\ \text{all arcs with source }i\text{)}\end{subarray}}=deg+⁡i−deg+⁡i=0.\displaystyle=\deg^{+}i-\deg^{+}i=0.In other words, we haveLe=0Le=0for the vectore:=(1,1,…,1)Te:=\left(1,1,\ldots,1\right)^{T}. Thus, this vectoreelies in the kernel (aka nullspace) ofLL, and soLLis singular.

(Note that we used the positivity ofnnhere! Ifn=0n=0, theneeis the zero vector, because a vector with00entries is automatically the zero vector.)

5.14.4The Matrix-Tree Theorem: statement

Proposition5.140shows that the determinant of the Laplacian of a digraph is not very interesting. It is common, however, that when a matrix has determinant00, its largest nonzero minors (= determinants of submatrices) often carry some interesting information; they are “the closest the matrix has” to a nonzero determinant. In the case of the Laplacian, they turn out to count spanning arborescences:

Theorem 5.142(Matrix-Tree Theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for some positive integernn.

LetLLbe the Laplacian ofDD. Letrrbe a vertex ofDD. Then,

(#of spanning arborescences ofDrooted tor)=det(L∼r,∼r).\left(\#\text{ of spanning arborescences of }D\text{ rooted to }r\right)=\det\left(L_{\sim r,\sim r}\right).

Before we prove this, some remarks:

•The determinantdet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)is the(r,r)\left(r,r\right)-th entry ofthe adjugate matrixofLL.
•TheV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}assumption is a typical “WLOG assumption”: If you have an arbitrary digraphDD, you can always rename its vertices as1,2,…,n1,2,\ldots,n, and then this assumption will be satisfied. Thus, Theorem5.142helps you count the spanning arborescences of any digraph. That said, you can also drop theV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}assumption from Theorem5.142if you are okay with matrices whose rows and columns are indexed not by numbers but by elements of an arbitrary finite set626262Such matrices are perfectly fine, just somewhat unusual and hard to write down (which row do you put on top?). Seehttps://mathoverflow.net/questions/317105for details..

5.14.5Application: Counting the spanning trees ofKnK_{n}

Now, let us use the Matrix-Tree Theorem to count the spanning trees ofKnK_{n}. This should provide some intuition for the theorem before we come to its proof.

We fix a positive integernn. LetLLbe the Laplacian of the multidigraphKnbidirK_{n}^{\operatorname*{bidir}}(whereKnK_{n}, as we recall, is the complete graph on the set{1,2,…,n}\left\{1,2,\ldots,n\right\}). Then, each vertex ofKnbidirK_{n}^{\operatorname*{bidir}}has outdegreen−1n-1, and thus we have

L=(n−1−1⋯−1−1n−1⋯−1⋮⋮⋱⋮−1−1⋯n−1)L=\left(\begin{array}[c]{cccc}n-1&-1&\cdots&-1\\ -1&n-1&\cdots&-1\\ \vdots&\vdots&\ddots&\vdots\\ -1&-1&\cdots&n-1\end{array}\right)(24)(this is then×nn\times n-matrix whose diagonal entries aren−1n-1and whose off-diagonal entries are−1-1). By Proposition5.132(b)(applied toG=KnG=K_{n}andr=1r=1), there is a bijection between{spanning arborescences ofKnbidirrooted to1}\left\{\text{spanning arborescences of }K_{n}^{\operatorname*{bidir}}\text{ rooted to }1\right\}and{spanning trees ofKn}\left\{\text{spanning trees of }K_{n}\right\}. Hence, by the bijection principle, we have

(#of spanning trees ofKn)\displaystyle\left(\#\text{ of spanning trees of }K_{n}\right)=(#of spanning arborescences ofKnbidirrooted to1)\displaystyle=\left(\#\text{ of spanning arborescences of }K_{n}^{\operatorname*{bidir}}\text{ rooted to }1\right)=det(L∼1,∼1)(by Theorem5.142, applied toD=Knbidirandr=1)\displaystyle=\det\left(L_{\sim 1,\sim 1}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by Theorem \ref{thm.MTT.MTT}, applied to }D=K_{n}^{\operatorname*{bidir}}\text{ and }r=1\right)=det(n−1−1⋯−1−1n−1⋯−1⋮⋮⋱⋮−1−1⋯n−1)⏟an(n−1)×(n−1)-matrix(by (24)).\displaystyle=\det\underbrace{\left(\begin{array}[c]{cccc}n-1&-1&\cdots&-1\\ -1&n-1&\cdots&-1\\ \vdots&\vdots&\ddots&\vdots\\ -1&-1&\cdots&n-1\end{array}\right)}_{\text{an }\left(n-1\right)\times\left(n-1\right)\text{-matrix}}\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{eq.MTT.Kn.L=})}\right).How do we compute this determinant? Here are three ways:636363All three ways requiren≥2n\geq 2, but this is okay since then=1n=1case is trivial.

•The most elementary approach is using row transformations: det(n−1−1⋯−1−1n−1⋯−1⋮⋮⋱⋮−1−1⋯n−1)\displaystyle\det\left(\begin{array}[c]{cccc}n-1&-1&\cdots&-1\\ -1&n-1&\cdots&-1\\ \vdots&\vdots&\ddots&\vdots\\ -1&-1&\cdots&n-1\end{array}\right)=det(n−1−1−1−1⋯−1−nn00⋯0−n0n0⋯0−n00n⋯0⋮⋮⋮⋮⋱⋮−n000⋯n)(here, we havesubtracted the1st rowfrom each other row)\displaystyle=\det\left(\begin{array}[c]{cccccc}n-1&-1&-1&-1&\cdots&-1\\ -n&n&0&0&\cdots&0\\ -n&0&n&0&\cdots&0\\ -n&0&0&n&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ -n&0&0&0&\cdots&n\end{array}\right)\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have}\\ \text{subtracted the }1\text{st row}\\ \text{from each other row}\end{array}\right)=nn−2det(n−1−1−1−1⋯−1−1100⋯0−1010⋯0−1001⋯0⋮⋮⋮⋮⋱⋮−1000⋯1)(here, we havefactored outannfrom eachrow except forthe first row)\displaystyle=n^{n-2}\det\left(\begin{array}[c]{cccccc}n-1&-1&-1&-1&\cdots&-1\\ -1&1&0&0&\cdots&0\\ -1&0&1&0&\cdots&0\\ -1&0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&0&\cdots&1\end{array}\right)\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have}\\ \text{factored out}\\ \text{an }n\text{ from each}\\ \text{row except for}\\ \text{the first row}\end{array}\right)=nn−2det(1000⋯0−1100⋯0−1010⋯0−1001⋯0⋮⋮⋮⋮⋱⋮−1000⋯1)⏟=1(since the matrix is triangularwith diagonal entries1,1,…,1)(here, we have added the2nd,3rd, etc. rows to the1st row)\displaystyle=n^{n-2}\underbrace{\det\left(\begin{array}[c]{cccccc}1&0&0&0&\cdots&0\\ -1&1&0&0&\cdots&0\\ -1&0&1&0&\cdots&0\\ -1&0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&0&\cdots&1\end{array}\right)}_{\begin{subarray}{c}=1\\ \text{(since the matrix is triangular}\\ \text{with diagonal entries }1,1,\ldots,1\text{)}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have added the }2\text{nd,}\\ 3\text{rd, etc. rows to the }1\text{st row}\end{array}\right)=nn−2.\displaystyle=n^{n-2}.
•The so-calledmatrix determinant lemma646464also known as (and probably much better titled) the “rank-one perturbation formula for the determinant”says that for anym×mm\times m-matrixA∈ℝm×mA\in\mathbb{R}^{m\times m}, any column vectoru∈ℝm×1u\in\mathbb{R}^{m\times 1}and any row vectorv∈ℝ1×mv\in\mathbb{R}^{1\times m}, we have det(A+uv)=detA+v⁡(adjA)u.\det\left(A+uv\right)=\det A+v\left(\operatorname*{adj}A\right)u.This helps us compute our determinant, since (n−1−1⋯−1−1n−1⋯−1⋮⋮⋱⋮−1−1⋯n−1)\displaystyle\left(\begin{array}[c]{cccc}n-1&-1&\cdots&-1\\ -1&n-1&\cdots&-1\\ \vdots&\vdots&\ddots&\vdots\\ -1&-1&\cdots&n-1\end{array}\right)=(n0⋯00n⋯0⋮⋮⋱⋮00⋯n)⏟=A+(−1−1⋮−1)⏟=u(11⋯1)⏟=v.\displaystyle=\underbrace{\left(\begin{array}[c]{cccc}n&0&\cdots&0\\ 0&n&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&n\end{array}\right)}_{=A}+\underbrace{\left(\begin{array}[c]{c}-1\\ -1\\ \vdots\\ -1\end{array}\right)}_{=u}\underbrace{\left(\begin{array}[c]{cccc}1&1&\cdots&1\end{array}\right)}_{=v}.
•Here is an approach that is heavier on linear algebra (specifically, eigenvectors and eigenvalues656565See[Treil17, Chapter 4]for a refresher.): Let(e1,e2,…,en−1)\left(e_{1},e_{2},\ldots,e_{n-1}\right)be the standard basis of theℝ\mathbb{R}-vector spaceℝn−1\mathbb{R}^{n-1}(so thateie_{i}is the column vector with itsii-th coordinate equal to11and all its other coordinates equal to00). Then, we can find the followingn−1n-1eigenvectors of our(n−1)×(n−1)\left(n-1\right)\times\left(n-1\right)-matrix(n−1−1⋯−1−1n−1⋯−1⋮⋮⋱⋮−1−1⋯n−1)\left(\begin{array}[c]{cccc}n-1&-1&\cdots&-1\\ -1&n-1&\cdots&-1\\ \vdots&\vdots&\ddots&\vdots\\ -1&-1&\cdots&n-1\end{array}\right): - –then−2n-2eigenvectorse1−eie_{1}-e_{i}for alli∈{2,3,…,n−1}i\in\left\{2,3,\ldots,n-1\right\}, each of them with eigenvaluenn(check this!); - –the eigenvectore1+e2+⋯+en−1e_{1}+e_{2}+\cdots+e_{n-1}with eigenvalue11(check this!). Since thesen−1n-1eigenvectors are linearly independent (check this!), they form a basis ofℝn−1\mathbb{R}^{n-1}. Hence, our matrix is similar to the diagonal matrix with diagonal entriesn,n,…,n⏟n−2times,1\underbrace{n,n,\ldots,n}_{n-2\text{ times}},1(by[Treil17, Chapter 4, Theorem 2.1]), and therefore has determinantnn⋯n⏟n−2times1=nn−2\underbrace{nn\cdots n}_{n-2\text{ times}}1=n^{n-2}.

There are other ways as well. Either way, the result we obtain isnn−2n^{n-2}. Thus, we have proved (relying on the Matrix-Tree Theorem, which we haven’t yet proved) the following result:

Theorem 5.143(Cayley’s formula).

Letnnbe a positive integer. Then, the#\#of spanning trees of the complete graphKnK_{n}isnn−2n^{n-2}.

In other words:

Corollary 5.144.

Letnnbe a positive integer. Then, the#\#of simple graphs with vertex set{1,2,…,n}\left\{1,2,\ldots,n\right\}that are trees isnn−2n^{n-2}.

Proof 5.145.

This is just Theorem5.143, since the simple graphs with vertex set{1,2,…,n}\left\{1,2,\ldots,n\right\}that are trees are precisely the spanning trees ofKnK_{n}.

Theorem5.143is widely known asCayley’s formula, despite its earlier discovery by Borchardt in 1860 ([Borcha60], giving a proof quite similar to ours). There are many other ways to prove it. I can particularly recommend the two combinatorial proofs given in[Galvin21, §2.4 and §2.5], as well as Joyal’s proof sketched in[Leinst19]. Most textbooks on enumerative combinatorics give one proof or another;[Stanle18, Appendix to Chapter 9]gives three. Cayley’s formula also appears in Aigner’s and Ziegler’s best-of compilation of mathematical proofs[AigZie18, Chapter 33]with four different proofs. Note that some of the sources use a matrix-tree theorem forundirectedgraphs; this is a particular case of our matrix-tree theorem.666666One moreremark:In Corollary5.144, we have counted the trees withnnvertices (i.e., simple graphs with vertex set{1,2,…,n}\left\{1,2,\ldots,n\right\}that are trees). It sounds equally natural to count the “unlabelled trees withnnvertices”, i.e., the equivalence classes of such trees up to isomorphism. Unfortunately, this is one of those “messy numbers” with no good expression: the best formula known is recursive. There is also an asymptotic formula (“Otter’s formula”,[Otter48]): the number of equivalence classes ofnn-vertex trees (up to isomorphism) is≈βαnn5/2withα≈2.955andβ≈0.5349.\approx\beta\dfrac{\alpha^{n}}{n^{5/2}}\ \ \ \ \ \ \ \ \ \ \text{with }\alpha\approx 2.955\text{ and }\beta\approx 0.5349.

However, in order to complete our proof, we still need to prove the Matrix-Tree Theorem.

5.14.6Preparations for the proof

In order to prepare for the proof of the Matrix-Tree Theorem, we state a simple lemma (yet another criterion for a digraph to be an arborescence):

Lemma 5.146.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Letrrbe a vertex ofDD. Assume thatDDhas no cycles. Assume moreover thatDDhas no arcs with sourcerr. Assume furthermore that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11. Then, the digraphDDis an arborescence rooted torr.

This lemma is precisely Exercise4.35(b), at least after reversing all arcs. But let us give a self-contained proof here:

Proof 5.147(Proof of Lemma5.146.).

Letuube any vertex ofDD. Clearly, the digraphDDhas at least one path that starts atuu(for example, the trivial path(u)\left(u\right)). Among all such paths, let𝐩=(v0,a1,v1,a2,v2,…,ak,vk)\mathbf{p}=\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)be alongestone.676767Such a longest path exists, sinceDDhas at least one path that starts atuu, and since a path ofDDcannot have length larger than|V|−1\left|V\right|-1.Thus,v0=uv_{0}=u.

We shall show thatvk=rv_{k}=r. Indeed, assume the contrary. Thus,vk≠rv_{k}\neq r, so thatvk∈V∖{r}v_{k}\in V\setminus\left\{r\right\}. Hence, the vertexvkv_{k}has outdegree11(since we assumed that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11). Thus, there exists an arcbbofDDthat has sourcevkv_{k}. Consider this arcbb, and letwwbe its target. Thus, appending the arcbband the vertexwwto the end of the path𝐩\mathbf{p}, we obtain a walk

𝐰=(v0,a1,v1,a2,v2,…,ak,vk,b,w)\mathbf{w}=\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k},b,w\right)ofDDthat starts atuu(sincev0=uv_{0}=u). Proposition4.30shows that this walk𝐰\mathbf{w}either is a path or contains a cycle. Hence,𝐰\mathbf{w}is a path (sinceDDhas no cycles). Thus,𝐰\mathbf{w}is a path ofDDthat starts atuu. Since𝐰\mathbf{w}is longer than𝐩\mathbf{p}(namely, longer by11), this shows that𝐩\mathbf{p}is not a longest path ofDDthat starts atuu. But this contradicts the very definition of𝐩\mathbf{p}.

This contradiction shows that our assumption was false. Hence,vk=rv_{k}=r. Thus,𝐩\mathbf{p}is a path fromuutorr(sincev0=uv_{0}=uandvk=rv_{k}=r). Therefore, the digraphDDhas a path fromuutorr(namely,𝐩\mathbf{p}).

Forget that we fixeduu. We thus have shown that for each vertexuuofDD, the digraphDDhas a path fromuutorr. In other words,rris a to-root ofDD. Furthermore, we havedeg+⁡r=0\deg^{+}r=0(sinceDDhas no arcs with sourcerr), and eachv∈V∖{r}v\in V\setminus\left\{r\right\}satisfiesdeg+⁡v=1\deg^{+}v=1(since we have assumed that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11). In other words, the digraphDDsatisfies Statement A’6 from the dual arborescence equivalence theorem (Theorem5.123). Therefore, it satisfies Statement A’1 from that theorem as well (since all six statements A’1, A’2,…\ldots, A’6 are equivalent). In other words,DDis an arborescence rooted torr. This proves Lemma5.146.

5.14.7The Matrix-Tree Theorem: proof

We shall now prove the Matrix-Tree Theorem (Theorem5.142), guided by the following battle plan:

1.First, we will prove it in the case when each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11. In this case, after removing all arcs with sourcerrfromDD(these arcs do not matter, since neither the submatrixL∼r,∼rL_{\sim r,\sim r}nor the spanning arborescences rooted torrdepend on them), we have essentially two options (subcases): eitherDDis itself an arborescence orDDhas a cycle.
2.Then, we will prove the matrix-tree theorem in the slightly more general case when eachv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree≤1\leq 1. This is easy, since a vertexv∈V∖{r}v\in V\setminus\left\{r\right\}having outdegree00trivializes the theorem.
3.Finally, we will prove the theorem in the general case. This is done by strong induction on the number of arcs ofDD. Every time you have a vertexv∈V∖{r}v\in V\setminus\left\{r\right\}with outdegree>1>1, you can pick such a vertex and color the outgoing arcs from it red and blue in such a way that each color is used at least once. Then, you can consider the subdigraph ofDDobtained by removing all blue arcs (call itDredD^{\operatorname*{red}}) and the subdigraph ofDDobtained by removing all red arcs (call itDblueD^{\operatorname*{blue}}). You can then apply the induction hypothesis toDredD^{\operatorname*{red}}and toDblueD^{\operatorname*{blue}}(since each of these two subdigraphs has fewer arcs thanDD), and add the results together. The good news is that both the # of spanning arborescences rooted torrand the determinantdet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)“behave additively” (we will soon see what this means).

So let us begin with Step 1. We first study a very special case:

Lemma 5.148.

(a)The digraphDDhas a unique spanning arborescence rooted torr.
(b)Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}. LetLLbe the Laplacian ofDD. Then,det(L∼r,∼r)=1\det\left(L_{\sim r,\sim r}\right)=1.

Proof 5.149.

(a)Lemma5.146shows that the digraphDDitself is an arborescence rooted torr.

As a consequence,DDitself is a spanning arborescence ofDDrooted torr.

(b)We WLOG assume thatr=nr=n(otherwise, we can rename the verticesr,r+1,r+2,…,nr,r+1,r+2,\ldots,nasn,r,r+1,…,n−1n,r,r+1,\ldots,n-1, so that the matrixL∼r,∼rL_{\sim r,\sim r}becomesL∼n,∼nL_{\sim n,\sim n}). Thus,V∖{r}=V∖{n}={1,2,…,n−1}V\setminus\left\{r\right\}=V\setminus\left\{n\right\}=\left\{1,2,\ldots,n-1\right\}(sinceV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}). Hence, each vertexi∈{1,2,…,n−1}i\in\left\{1,2,\ldots,n-1\right\}has outdegree11(since each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11, and we can apply this tov=iv=ibecausei∈{1,2,…,n−1}=V∖{r}i\in\left\{1,2,\ldots,n-1\right\}=V\setminus\left\{r\right\}). In other words,

deg+⁡i=1for eachi∈{1,2,…,n−1}.\deg^{+}i=1\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{1,2,\ldots,n-1\right\}.(25)Moreover,DDhas no cycles and thus no loops. Thus, for eachi∈{1,2,…,n−1}i\in\left\{1,2,\ldots,n-1\right\}, the numberai,ia_{i,i}defined in Definition5.138is

(sinceai,ia_{i,i}counts the loops with sourceii).

LetD′D^{\prime}be the digraphDDwith a loop added at each vertex – i.e., the multidigraph obtained fromDDby addingnnextra arcsℓ1,ℓ2,…,ℓn\ell_{1},\ell_{2},\ldots,\ell_{n}and letting each arcℓi\ell_{i}have sourceiiand targetii.

LetSn−1S_{n-1}denote the group of permutations of the set

{1,2,…,n−1}={1,2,…,n}⏟=V∖{n⏟=r}=V∖{r}.\left\{1,2,\ldots,n-1\right\}=\underbrace{\left\{1,2,\ldots,n\right\}}_{=V}\setminus\left\{\underbrace{n}_{=r}\right\}=V\setminus\left\{r\right\}. Now, fromr=nr=n, we have

det(L∼r,∼r)=det(L∼n,∼n)=∑σ∈Sn−1signσ⋅∏i=1n−1Li,σ⁡(i)\det\left(L_{\sim r,\sim r}\right)=\det\left(L_{\sim n,\sim n}\right)=\sum\limits_{\sigma\in S_{n-1}}\operatorname*{sign}\sigma\cdot\prod\limits_{i=1}^{n-1}L_{i,\sigma\left(i\right)}(27)(by the Leibniz formula for the determinant). We shall now study the addends in the sum on the right hand side of this equality. Specifically, we will show that the only addend whose product∏i=1n−1Li,σ⁡(i)\prod\limits_{i=1}^{n-1}L_{i,\sigma\left(i\right)}is nonzero is the addend forσ=id\sigma=\operatorname*{id}.

Indeed, letσ∈Sn−1\sigma\in S_{n-1}be a permutation such that the product∏i=1n−1Li,σ⁡(i)\prod\limits_{i=1}^{n-1}L_{i,\sigma\left(i\right)}is nonzero. We shall prove thatσ=id\sigma=\operatorname*{id}.

Consider an arbitraryv∈{1,2,…,n−1}v\in\left\{1,2,\ldots,n-1\right\}. Then,Lv,σ⁡(v)≠0L_{v,\sigma\left(v\right)}\neq 0(becauseLv,σ⁡(v)L_{v,\sigma\left(v\right)}is a factor in the product∏i=1n−1Li,σ⁡(i)\prod\limits_{i=1}^{n-1}L_{i,\sigma\left(i\right)}, which is nonzero). However, the definition ofLLyieldsLv,σ⁡(v)=(deg+v)⋅[v=σ(v)]−av,σ⁡(v)L_{v,\sigma\left(v\right)}=\left(\deg^{+}v\right)\cdot\left[v=\sigma\left(v\right)\right]-a_{v,\sigma\left(v\right)}. Thus,

(deg+v)⋅[v=σ(v)]−av,σ⁡(v)=Lv,σ⁡(v)≠0.\left(\deg^{+}v\right)\cdot\left[v=\sigma\left(v\right)\right]-a_{v,\sigma\left(v\right)}=L_{v,\sigma\left(v\right)}\neq 0.Hence, at least one of the numbers[v=σ(v)]\left[v=\sigma\left(v\right)\right]andav,σ⁡(v)a_{v,\sigma\left(v\right)}is nonzero. In other words, we havev=σ⁡(v)v=\sigma\left(v\right)(this is what it means for[v=σ(v)]\left[v=\sigma\left(v\right)\right]to be nonzero) or the digraphDDhas an arc with sourcevvand targetσ⁡(v)\sigma\left(v\right)(because this is what it means forav,σ⁡(v)a_{v,\sigma\left(v\right)}to be nonzero). In either case, the digraphD′D^{\prime}has an arc with sourcevvand targetσ⁡(v)\sigma\left(v\right)(because ifv=σ⁡(v)v=\sigma\left(v\right), then one of the loops we added toDDdoes the trick). We can apply the same argument toσ⁡(v)\sigma\left(v\right)instead ofvv, and obtain an arc with sourceσ⁡(v)\sigma\left(v\right)and targetσ⁡(σ⁡(v))\sigma\left(\sigma\left(v\right)\right). Similarly, we obtain an arc with sourceσ⁡(σ⁡(v))\sigma\left(\sigma\left(v\right)\right)and targetσ⁡(σ⁡(σ⁡(v)))\sigma\left(\sigma\left(\sigma\left(v\right)\right)\right). We can continue this reasoning indefinitely. By continuing it fornnsteps, we obtain a walk

(v,∗,σ⁡(v),∗,σ2(v),∗,σ3(v),…,∗,σn(v))\left(v,\ast,\sigma\left(v\right),\ast,\sigma^{2}\left(v\right),\ast,\sigma^{3}\left(v\right),\ldots,\ast,\sigma^{n}\left(v\right)\right)in the digraphD′D^{\prime}, where each asterisk means an arc (we don’t care about what these arcs are, so we are not giving them names). This walk cannot be a path (since it hasn+1n+1vertices, butD′D^{\prime}has onlynnvertices); thus, it must contain a cycle (by Proposition4.30). All arcs of this cycle must be loops (because otherwise, we could remove the loops from this cycle and obtain a cycle ofDD, but we know thatDDhas no cycles). In particular, its first arc is a loop. Thus, our above walk(v,∗,σ⁡(v),∗,σ2(v),∗,σ3(v),…,∗,σn(v))\left(v,\ast,\sigma\left(v\right),\ast,\sigma^{2}\left(v\right),\ast,\sigma^{3}\left(v\right),\ldots,\ast,\sigma^{n}\left(v\right)\right)contains a loop (since the arcs of the cycle come from this walk). In other words, we haveσi(v)=σi+1(v)\sigma^{i}\left(v\right)=\sigma^{i+1}\left(v\right)for somei∈{0,1,…,n−1}i\in\left\{0,1,\ldots,n-1\right\}. Sinceσ\sigmais injective, we can applyσ−i\sigma^{-i}to both sides of this equality, and conclude thatv=σ⁡(v)v=\sigma\left(v\right). In other words,σ⁡(v)=v\sigma\left(v\right)=v.

Forget that we fixedvv. We thus have shown thatσ⁡(v)=v\sigma\left(v\right)=vfor eachv∈{1,2,…,n−1}v\in\left\{1,2,\ldots,n-1\right\}. In other words,σ=id\sigma=\operatorname*{id}.

Forget that we fixedσ\sigma. We thus have proved thatσ=id\sigma=\operatorname*{id}for each permutationσ∈Sn−1\sigma\in S_{n-1}for which the product∏i=1n−1Li,σ⁡(i)\prod\limits_{i=1}^{n-1}L_{i,\sigma\left(i\right)}is nonzero. In other words, the only permutationσ∈Sn−1\sigma\in S_{n-1}for which the product∏i=1n−1Li,σ⁡(i)\prod\limits_{i=1}^{n-1}L_{i,\sigma\left(i\right)}is nonzero is the permutationid\operatorname*{id}.

Thus, the only nonzero addend on the right hand side of (27) is the addend corresponding toσ=id\sigma=\operatorname*{id}. Hence, (27) can be simplified as follows:

det(L∼n,∼n)=sign(id)⏟=1⋅∏i=1n−1Li,id(i)=∏i=1n−1Li,id(i).\det\left(L_{\sim n,\sim n}\right)=\underbrace{\operatorname*{sign}\left(\operatorname*{id}\right)}_{=1}\cdot\prod\limits_{i=1}^{n-1}L_{i,\operatorname*{id}\left(i\right)}=\prod\limits_{i=1}^{n-1}L_{i,\operatorname*{id}\left(i\right)}.Since eachi∈{1,2,…,n−1}i\in\left\{1,2,\ldots,n-1\right\}satisfies

Li,id(i)\displaystyle L_{i,\operatorname*{id}\left(i\right)}=Li,i=(deg+⁡i)⏟=1(by (25))⋅[i=i]⏟=1−ai,i⏟=0(by (26))(by the definition ofL)\displaystyle=L_{i,i}=\underbrace{\left(\deg^{+}i\right)}_{\begin{subarray}{c}=1\\ \text{(by (\ref{pf.lem.MTT.1.b.1}))}\end{subarray}}\cdot\underbrace{\left[i=i\right]}_{=1}-\underbrace{a_{i,i}}_{\begin{subarray}{c}=0\\ \text{(by (\ref{pf.lem.MTT.1.b.2}))}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of }L\right)=1⋅1−0=1,\displaystyle=1\cdot 1-0=1,this can be simplified todet(L∼n,∼n)=∏i=1n−11=1\det\left(L_{\sim n,\sim n}\right)=\prod\limits_{i=1}^{n-1}1=1. This proves Lemma5.148(b).

Next, we drop the “no cycles” condition:

Lemma 5.150.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Letrrbe a vertex ofDD. Assume that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11. Then, the MTT holds for theseDDandrr. (Here and in the following, “MTT” is short for “Matrix-Tree Theorem”, i.e., for Theorem5.142.)

Proof 5.151.

First of all, we note that an arc with sourcerrcannot appear in any spanning arborescence ofDDrooted torr(since any such arborescence satisfiesdeg+⁡r=0\deg^{+}r=0, according to Statement A’6 in the Dual Arborescence Equivalence Theorem (Theorem5.123)). Furthermore, the arcs with sourcerrdo not affect the matrixL∼r,∼rL_{\sim r,\sim r}, since they only appear in therr-th row of the matrixLL(but thisrr-th row is removed inL∼r,∼rL_{\sim r,\sim r}).

Hence, any arc with sourcerrcan be removed fromDDwithout disturbing anything we currently care about. Thus, we WLOG assume thatDDhas no arcs with sourcerr(else, we can just remove them fromDD).

We WLOG assume thatr=nr=n(otherwise, we can rename the verticesr,r+1,r+2,…,nr,r+1,r+2,\ldots,nasn,r,r+1,…,n−1n,r,r+1,\ldots,n-1, so that the matrixL∼r,∼rL_{\sim r,\sim r}becomesL∼n,∼nL_{\sim n,\sim n}).

We are in one of the following two cases:

Case 1:The digraphDDhas a cycle.

Case 2:The digraphDDhas no cycles.

Consider Case 1. In this case,DDhas a cycle𝐯=(v1,∗,v2,∗,…,∗,vm)\mathbf{v}=\left(v_{1},\ast,v_{2},\ast,\ldots,\ast,v_{m}\right)(where we again are putting asterisks in place of the arcs). This cycle cannot containrr(sinceDDhas no arcs with sourcerr). Thus, all its verticesv1,v2,…,vmv_{1},v_{2},\ldots,v_{m}belong toV∖{r}V\setminus\left\{r\right\}. Hence, for eachi∈{1,2,…,m−1}i\in\left\{1,2,\ldots,m-1\right\}, the vertexviv_{i}has outdegree11(since we assumed that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11). Consequently, for eachi∈{1,2,…,m−1}i\in\left\{1,2,\ldots,m-1\right\}, the only arc ofDDthat has sourceviv_{i}is the arc that followsviv_{i}on the cycle𝐯\mathbf{v}. Therefore, in the matrixLL, theviv_{i}-th row has a11in theviv_{i}-th position (becausedeg+⁡(vi)=1\deg^{+}\left(v_{i}\right)=1), a−1-1in thevi+1v_{i+1}-th position (since the arc that followsviv_{i}on the cycle𝐯\mathbf{v}has sourceviv_{i}and targetvi+1v_{i+1}), and00s in all other positions. Sincer=nr=n, the same must then be true for the matrixL∼r,∼rL_{\sim r,\sim r}: That is, theviv_{i}-th row of the matrixL∼r,∼rL_{\sim r,\sim r}has a11in theviv_{i}-th position, a−1-1in thevi+1v_{i+1}-th position, and00s in all other positions. Thus, the sum of thev1v_{1}-th,v2v_{2}-th,…\ldots,vm−1v_{m-1}-th rows ofL∼r,∼rL_{\sim r,\sim r}is the zero vector (since the11s and the−1-1s just cancel out)696969Namely, the−1-1in thevi+1v_{i+1}-th position of theviv_{i}-th row gets cancelled by the11in thevi+1v_{i+1}-th position of thevi+1v_{i+1}-th row. (We are using the fact thatvm=v1v_{m}=v_{1}here.).707070Let me illustrate this on a representative example: Assume that the numbersv1,v2,…,vm−1,vmv_{1},v_{2},\ldots,v_{m-1},v_{m}are1,2,…,m−1,11,2,\ldots,m-1,1(respectively). Then, the firstm−1m-1rows ofLLlook as follows:1−11−11−1⋱⋱1−1−11\begin{array}[c]{cccccccccc}1&-1&&&&&&&&\\ &1&-1&&&&&&&\\ &&1&-1&&&&&&\\ &&&\ddots&\ddots&&&&&\\ &&&&1&-1&&&&\\ -1&&&&&1&&&&\end{array}(where all the missing entries are zeroes). Thus, the sum of thesem−1m-1rows is the zero vector. The same is therefore true of the matrixL∼r,∼rL_{\sim r,\sim r}(since the firstm−1m-1rows of the latter matrix are just the firstm−1m-1rows ofLL, with theirrr-th entries removed). The general case is essentially the same as this example; the only difference is that the relevant rows are in other positions.

So we have found a nonempty set of rows ofL∼r,∼rL_{\sim r,\sim r}whose sum is the zero vector. This yields that the matrixL∼r,∼rL_{\sim r,\sim r}is singular (by basic properties of determinants717171Specifically, we are using the following fact: “LetMMbe a square matrix. If there is a certain nonempty set of rows ofMMwhose sum is the zero vector, then the matrixMMis singular.”. To prove this fact, we letSSbe this nonempty set. Choose one row from this set, and call it thechosen row. Now, add all the other rows fromSSto this one chosen row. This operation does not change the determinant ofMM(since the determinant of a matrix is unchanged when we add one row to another), but the resulting matrix has a zero row (namely, the chosen row) and thus has determinant00. Hence, the original matrixMMmust have had determinant00as well. In other words,MMwas singular, qed.), so its determinant isdet(L∼r,∼r)=0\det\left(L_{\sim r,\sim r}\right)=0. On the other hand, the digraphDDhas no spanning arborescence (because, in order to get a spanning arborescence ofDD, we would have to remove at least one arc of our cycle𝐯\mathbf{v}(since an arborescence cannot have a cycle); but then, the source of this arc would have outdegree00, and thus we could no longer find a path from this source torr, so we would not obtain a spanning arborescence). In other words,

(# of spanning arborescences ofDrooted tor)=0.\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=0.Comparing this withdet(L∼r,∼r)=0\det\left(L_{\sim r,\sim r}\right)=0, we conclude that the MTT holds in this case (since it claims that0=00=0). Thus, Case 1 is done.

Next, we consider Case 2. In this case,DDhas no cycles. Then,det(L∼r,∼r)=1\det\left(L_{\sim r,\sim r}\right)=1(by Lemma5.148(b)) and

(# of spanning arborescences ofDrooted tor)=1(by Lemma5.148(a)).\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=1\ \ \ \ \ \ \ \ \ \ \left(\text{by Lemma \ref{lem.MTT.1} {(a)}}\right).Thus, the MTT boils down to1=11=1, which is again true.

So Lemma5.150is proved.

Next, we venture into a mildly greater generality, corresponding to Step 2 in our above battle plan:

Lemma 5.152.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Letrrbe a vertex ofDD. Assume that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree≤1\leq 1. Then, the MTT (= Matrix-Tree Theorem) holds for theseDDandrr.

We give two proofs for this lemma: a short one using Lemma5.150, and a longer one that avoids Lemma5.150.

Proof 5.153(First proof of Lemma5.152.).

If each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree11, then this is true by Lemma5.150.

Thus, we WLOG assume that this is not the case. Hence, some vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree≠1\neq 1. Consider thisvv. The outdegree ofvvis≠1\neq 1, but also≤1\leq 1(by the hypothesis of the lemma). Hence, this outdegree must be00. That is, there is no arc with sourcevv.

WLOG assume thatr=nr=n(otherwise, rename the verticesr,r+1,r+2,…,nr,r+1,r+2,\ldots,nasn,r,r+1,…,n−1n,r,r+1,\ldots,n-1, so that the matrixL∼r,∼rL_{\sim r,\sim r}becomesL∼n,∼nL_{\sim n,\sim n}).

We havev≠rv\neq r. Hence, the digraphDDhas no path fromvvtorr(since any such path would include an arc with sourcevv, but there is no arc with sourcevv).

Therefore,DDhas no spanning arborescence rooted torr(because any such spanning arborescence would have to have a path fromvvtorr). In other words,

(# of spanning arborescences ofDrooted tor)=0.\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=0.Also,det(L∼r,∼r)=0\det\left(L_{\sim r,\sim r}\right)=0(since thevv-th row of the matrixL∼r,∼rL_{\sim r,\sim r}is00(because there is no arc with sourcevv)). So the MTT boils down to0=00=0again, and thus Lemma5.152is proved.

Proof 5.154(Second proof of Lemma5.152.).

Just as in our above proof of Lemma5.150, we can WLOG assume

•thatDDhas no arcs with sourcerr, and
•thatr=nr=n.

Let us make these two assumptions. We distinguish between two cases:

Case 1:The vertexrris a to-root ofDD.

Case 2:The vertexrris not a to-root ofDD.

Consider Case 1. In this case,rris a to-root ofDD. Each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}satisfiesdeg+⁡v≤1\deg^{+}v\leq 1(since we have assumed that each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree≤1\leq 1) and thereforedeg+⁡v=1\deg^{+}v=1(because otherwise, we would havedeg+⁡v=0\deg^{+}v=0, and thus there would be no path fromvvtorr, since any such path would have to start with an arc with sourcevv; but this would contradict the fact thatrris a to-root ofDD). Furthermore,deg+⁡r=0\deg^{+}r=0(sinceDDhas no arcs with sourcerr). Hence, the statement A’6 of Theorem5.123holds. Thus, the implication A’6⟹\LongrightarrowA’3 of Theorem5.123shows that the multigraphDundD^{\operatorname*{und}}is a tree. Hence,DundD^{\operatorname*{und}}has no cycles. Thus, the directed graphDDhas no cycles either. Therefore, Lemma5.148(a)shows thatDDhas a unique spanning arborescence rooted torr. In other words,

(# of spanning arborescences ofDrooted tor)=1.\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=1.Comparing this withdet(L∼r,∼r)=1\det\left(L_{\sim r,\sim r}\right)=1(which follows from Lemma5.148(b)), we obtain

(#of spanning arborescences ofDrooted tor)=det(L∼r,∼r).\left(\#\text{ of spanning arborescences of }D\text{ rooted to }r\right)=\det\left(L_{\sim r,\sim r}\right).Thus, the MTT is proved in Case 1.

Let us now consider Case 2. In this case, the vertexrris not a to-root ofDD. Hence, there exists a vertexv∈Vv\in Vsuch thatDDhas no path fromvvtorr. In other words, the set

S:={v∈V∣there exists no path fromvtor}S:=\left\{v\in V\ \mid\ \text{there exists no path from }v\text{ to }r\right\}is nonempty. Consider this setSS. Clearly,r∉Sr\notin S(since there is obviously a path fromrrtorr). In other words,n∉Sn\notin S(sincer=nr=n). Thus,S⊆{1,2,…,n}∖{n}={1,2,…,n−1}S\subseteq\left\{1,2,\ldots,n\right\}\setminus\left\{n\right\}=\left\{1,2,\ldots,n-1\right\}.

Recall thatr=nr=n. Hence, if we permute the vertices in{1,2,…,n−1}\left\{1,2,\ldots,n-1\right\}(that is, we rename them asσ⁡(1),σ⁡(2),…,σ⁡(n−1)\sigma\left(1\right),\sigma\left(2\right),\ldots,\sigma\left(n-1\right)for some permutationσ∈Sn−1\sigma\in S_{n-1}), then the#\#of spanning arborescences ofDDrooted torrremains unchanged, and the determinantdet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)remains unchanged as well (since the matrixL∼r,∼rL_{\sim r,\sim r}undergoes a permutation of its rows and the same permutation of its columns, which amounts to conjugating this matrix by a permutation matrix). Thus, the claim of the MTT does not change if we permute the vertices in{1,2,…,n−1}\left\{1,2,\ldots,n-1\right\}.

Hence, we can WLOG assume thatS={1,2,…,k}S=\left\{1,2,\ldots,k\right\}for some nonnegative integerk≤n−1k\leq n-1(indeed, we can achieve this by permuting the vertices in{1,2,…,n−1}\left\{1,2,\ldots,n-1\right\}, sinceS⊆{1,2,…,n−1}S\subseteq\left\{1,2,\ldots,n-1\right\}). Assume this. Then, the set{1,2,…,k}=S\left\{1,2,\ldots,k\right\}=Sis nonempty; thus,k>0k>0.

The following property ofSSwill be crucial:

Claim 0:If the source of an arcaaofDDbelongs toSS, then the target ofaaalso belongs toSS.

Proof 5.155(Proof of Claim 0.).

Letaabe an arc ofDDwhose source belongs toSS. We must show that the target ofaaalso belongs toSS.

Letssandttbe the source and the target ofaa. Then,s∈Ss\in S(since the source ofaabelongs toSS). In other words, there exists no path fromsstorr(by the definition ofSS).

If there was a path𝐩\mathbf{p}fromtttorr, then there would also be a walk fromsstorr(namely, the walk(s,a,t,…,r⏟the path𝐩)\left(s,a,\underbrace{t,\ldots,r}_{\text{the path }\mathbf{p}}\right)), and therefore there would also be a path fromsstorr(by Corollary4.29); but this would contradict the previous sentence. Hence, there exists no path fromtttorr. In other words,t∈St\in S(by the definition ofSS). In other words, the target ofaabelongs toSS(since this target istt). This proves Claim 0.

Now consider the induced subdigraphD⁡[S]D\left[S\right]ofDDon the setS={1,2,…,k}S=\left\{1,2,\ldots,k\right\}. LetLD⁡[S]∈ℤk×kL_{D\left[S\right]}\in\mathbb{Z}^{k\times k}be the Laplacian of this subdigraphD⁡[S]D\left[S\right]. We claim that the LaplacianLLofDDcan be written in block-matrix notation as follows:

L=(LD⁡[S]0k×(n−k)PQ)L=\left(\begin{array}[c]{cc}L_{D\left[S\right]}&0_{k\times\left(n-k\right)}\\ P&Q\end{array}\right)(28)for some(n−k)×k\left(n-k\right)\times k-matrixPPand some(n−k)×(n−k)\left(n-k\right)\times\left(n-k\right)-matrixQQ(where0k×(n−k)0_{k\times\left(n-k\right)}denotes the zero matrix withkkrows andn−kn-kcolumns). To prove this, we need to show the following two claims:

Claim 1:The top-leftk×kk\times kblock ofLLisLD⁡[S]L_{D\left[S\right]}.

Claim 2:The top-rightk×(n−k)k\times\left(n-k\right)block ofLLis the zero matrix0k×(n−k)0_{k\times\left(n-k\right)}.

Proof 5.156(Proof of Claim 1.).

We need to show thatLi,j=(LD⁡[S])i,jL_{i,j}=\left(L_{D\left[S\right]}\right)_{i,j}for alli,j∈{1,2,…,k}i,j\in\left\{1,2,\ldots,k\right\}.

So leti,j∈{1,2,…,k}i,j\in\left\{1,2,\ldots,k\right\}be arbitrary. Then,iiandjjare vertices of both digraphsDDandD⁡[S]D\left[S\right].

By the definition of the Laplacian, we have

Li,j=(deg+i)⋅[i=j]−ai,jL_{i,j}=\left(\deg^{+}i\right)\cdot\left[i=j\right]-a_{i,j}(29)and

(LD⁡[S])i,j=(deg+i)⋅[i=j]−ai,j,\left(L_{D\left[S\right]}\right)_{i,j}=\left(\deg^{+}i\right)\cdot\left[i=j\right]-a_{i,j},(30)whereai,ja_{i,j}denotes the # of arcs with sourceiiand targetjj. However, in theory, the outdegreedeg+⁡i\deg^{+}iin (29) must be distinguished from the outdegreedeg+⁡i\deg^{+}iin (30): Indeed, the former is defined with respect to the digraphDD, while the latter is defined with respect to the subdigraphD⁡[S]D\left[S\right]. Thus, the former outdegree counts all the arcs ofDDwith sourceii, whereas the latter outdegree counts only those arcs whose target belongs toSS. However, in our case, this distinction is unnecessary: Indeed, any arc ofDDwhose source isiimust also have its target belong toSS(by Claim 0, since its source isi∈{1,2,…,k}=Si\in\left\{1,2,\ldots,k\right\}=S), and thus must be an arc of the induced subdigraphD⁡[S]D\left[S\right]as well (since both its source and its target belong toSS). Therefore, the outdegreedeg+⁡i\deg^{+}iin (29) and the outdegreedeg+⁡i\deg^{+}iin (30) are equal. (Of course, theai,ja_{i,j}’s in both equations are also equal, sinceD⁡[S]D\left[S\right]contains all arcs fromiitojj.) Thus, the right hand sides of the equalities (29) and (30) are equal. Therefore, their left hand sides are also equal. In other words,Li,j=(LD⁡[S])i,jL_{i,j}=\left(L_{D\left[S\right]}\right)_{i,j}. This proves Claim 1.

Proof 5.157(Proof of Claim 2.).

We must prove thatLi,j=0L_{i,j}=0for alli∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}andj∈{k+1,k+2,…,n}j\in\left\{k+1,k+2,\ldots,n\right\}.

So leti∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}andj∈{k+1,k+2,…,n}j\in\left\{k+1,k+2,\ldots,n\right\}be arbitrary. Then,i∈Si\in S(sincei∈{1,2,…,k}=Si\in\left\{1,2,\ldots,k\right\}=S) andj∉Sj\notin S(sincej∈{k+1,k+2,…,n}={1,2,…,n}∖{1,2,…,k}⏟=S={1,2,…,n}∖Sj\in\left\{k+1,k+2,\ldots,n\right\}=\left\{1,2,\ldots,n\right\}\setminus\underbrace{\left\{1,2,\ldots,k\right\}}_{=S}=\left\{1,2,\ldots,n\right\}\setminus S). Thus,i≠ji\neq j, so that[i=j]=0\left[i=j\right]=0.

If some arcaaofDDhas sourceii, then the target ofaamust belong toSS(by Claim 0, since the source ofaaisi∈Si\in S), and therefore the target ofaacannot bejj(sincej∉Sj\notin S). In other words,DDhas no arc with sourceiiand targetjj.

Letai,ja_{i,j}be the # of arcs ofDDhaving sourceiiand targetjj. Then,ai,j=0a_{i,j}=0(sinceDDhas no arc with sourceiiand targetjj). But the definition of the LaplacianLLshows that

Li,j=(deg+⁡i)⋅[i=j]⏟=0−ai,j⏟=0=0.L_{i,j}=\left(\deg^{+}i\right)\cdot\underbrace{\left[i=j\right]}_{=0}-\underbrace{a_{i,j}}_{=0}=0.This proves Claim 2.

Combining Claim 1 with Claim 2, we see that then×nn\times n-matrixLLcan be written in the form (28) for some(n−k)×k\left(n-k\right)\times k-matrixPPand some(n−k)×(n−k)\left(n-k\right)\times\left(n-k\right)-matrixQQ. Thus, its submatrixL∼n,∼nL_{\sim n,\sim n}(which is obtained fromLLby removing the last row and the last column) can be written as

L∼n,∼n=(LD⁡[S]0k×(n−k)PQ)∼n,∼n=(LD⁡[S]0k×(n−k−1)P′Q′),L_{\sim n,\sim n}=\left(\begin{array}[c]{cc}L_{D\left[S\right]}&0_{k\times\left(n-k\right)}\\ P&Q\end{array}\right)_{\sim n,\sim n}=\left(\begin{array}[c]{cc}L_{D\left[S\right]}&0_{k\times\left(n-k-1\right)}\\ P^{\prime}&Q^{\prime}\end{array}\right),whereP′P^{\prime}andQ′Q^{\prime}are certain submatrices ofPPandQQ(namely,P′P^{\prime}isPPwithout its last row, whereasQ′Q^{\prime}isQQwithout its last row and its last column).727272We have usedk≤n−1k\leq n-1here (which is the reason why we aren’t cutting into theLD⁡[S]L_{D\left[S\right]}part ofLLwhen we are removing itsnn-th row and column).This is clearly a block-lower-triangular matrix with diagonal blocksLD⁡[S]L_{D\left[S\right]}andQ′Q^{\prime}. But the determinant of a block-lower-triangular matrix is known to equal the product of the determinants of its diagonal blocks737373This is a standard result about determinants; see, e.g.,[Grinbe20, Exercise 6.30].. Hence, we obtain

det(L∼n,∼n)=det(LD⁡[S])⋅det(Q′).\det\left(L_{\sim n,\sim n}\right)=\det\left(L_{D\left[S\right]}\right)\cdot\det\left(Q^{\prime}\right).(31) ButD⁡[S]D\left[S\right]is a digraph with vertex setS={1,2,…,k}S=\left\{1,2,\ldots,k\right\}, and the numberkkis a positive integer (sincek>0k>0). Hence, Proposition5.140(applied toD⁡[S]D\left[S\right]andkkinstead ofDDandnn) yields that the LaplacianLD⁡[S]L_{D\left[S\right]}ofD⁡[S]D\left[S\right]is singular; i.e., we havedet(LD⁡[S])=0\det\left(L_{D\left[S\right]}\right)=0. Thus, (31) becomes

det(L∼n,∼n)=det(LD⁡[S])⏟=0⋅det(Q′)=0.\det\left(L_{\sim n,\sim n}\right)=\underbrace{\det\left(L_{D\left[S\right]}\right)}_{=0}\cdot\det\left(Q^{\prime}\right)=0.In other words,det(L∼r,∼r)=0\det\left(L_{\sim r,\sim r}\right)=0(sincer=nr=n).

On the other hand,rris not a to-root ofDD. Therefore,rris not a to-root of any spanning subdigraph ofDD(since a to-root of a spanning subdigraph ofDDmust always be a to-root ofDD747474This is clear from the definition of a to-root: Ifssis a to-root of a spanning subdigraph ofDD, then for each vertexvvofDD, there must be a path fromvvtossin this subdigraph, and therefore also a path fromvvtossinDD.). Hence, the digraphDDhas no spanning arborescence rooted torr(becauserrwould be a to-root of any such arborescence). In other words,

(#of spanning arborescences ofDrooted tor)=0.\left(\#\text{ of spanning arborescences of }D\text{ rooted to }r\right)=0.Comparing this withdet(L∼r,∼r)=0\det\left(L_{\sim r,\sim r}\right)=0, we obtain

We have now proved the MTT in both Cases 1 and 2. So the MTT holds for ourDD. This proves Lemma5.152again.

We are now ready to prove the MTT in the general case (Step 3 of our battle plan):

Proof 5.158(Proof of Theorem5.142.).

First, we introduce a notation:

LetMMandNNbe twon×nn\times n-matrices that agree in all but one row. That is, there exists somej∈{1,2,…,n}j\in\left\{1,2,\ldots,n\right\}such that for eachi≠ji\neq j, we have (thei-th row ofM)=(thei-th row ofN).\left(\text{the }i\text{-th row of }M\right)=\left(\text{the }i\text{-th row of }N\right).Then, we writeM≡𝑗NM\overset{j}{\equiv}N, and we letM+𝑗NM\overset{j}{+}Nbe then×nn\times n-matrix that is obtained fromMMby adding thejj-th row ofNNto thejj-th row ofMM(while leaving all remaining rows unchanged).

For example, ifM=(abcdefghi)M=\left(\begin{array}[c]{ccc}a&b&c\\ d&e&f\\ g&h&i\end{array}\right)andN=(abcd′e′f′ghi)N=\left(\begin{array}[c]{ccc}a&b&c\\ d^{\prime}&e^{\prime}&f^{\prime}\\ g&h&i\end{array}\right), thenM≡2NM\overset{2}{\equiv}Nand

M+2N=(abcd+d′e+e′f+f′ghi).M\overset{2}{+}N=\left(\begin{array}[c]{ccc}a&b&c\\ d+d^{\prime}&e+e^{\prime}&f+f^{\prime}\\ g&h&i\end{array}\right). A well-known property of determinants (themultilinearity of the determinant) says that ifMMandNNare twon×nn\times n-matrices andj∈{1,2,…,n}j\in\left\{1,2,\ldots,n\right\}is a number such thatM≡𝑗NM\overset{j}{\equiv}N, then

det(M+𝑗N)=detM+detN.\det\left(M\overset{j}{+}N\right)=\det M+\det N. Now, let us prove the MTT. We proceed by strong induction on the # of arcs ofDD.

Induction step:Letm∈ℕm\in\mathbb{N}. Assume (as the induction hypothesis) that the MTT holds for all digraphsDDthat have<m<marcs. We must now prove it for our digraphDDwithmmarcs.

WLOG assume thatr=nr=n(otherwise, rename the verticesr,r+1,r+2,…,nr,r+1,r+2,\ldots,nasn,r,r+1,…,n−1n,r,r+1,\ldots,n-1).

If each vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree≤1\leq 1, then the MTT holds by Lemma5.152. Thus, we WLOG assume that some vertexv∈V∖{r}v\in V\setminus\left\{r\right\}has outdegree>1>1. Pick such a vertexvv. We color each arc with sourcevveither red or blue, making sure that at least one arc is red and at least one arc is blue. (We can do this, sincevvhas outdegree>1>1.) All arcs that do not have sourcevvremain uncolored.

Now, letDredD^{\operatorname*{red}}be the subdigraph obtained fromDDby removing all blue arcs. Then,DredD^{\operatorname*{red}}has fewer arcs thanDD. In other words,DredD^{\operatorname*{red}}has<m<marcs. Hence, the induction hypothesis yields that the MTT holds forDredD^{\operatorname*{red}}. That is, we have

(# of spanning arborescences ofDredrooted tor)=det(L∼r,∼rred),\left(\text{\# of spanning arborescences of }D^{\operatorname*{red}}\text{ rooted to }r\right)=\det\left(L_{\sim r,\sim r}^{\operatorname*{red}}\right),whereLredL^{\operatorname*{red}}means the Laplacian ofDredD^{\operatorname*{red}}.

Likewise, letDblueD^{\operatorname*{blue}}be the subdigraph obtained fromDDby removing all red arcs. Then,DblueD^{\operatorname*{blue}}has fewer arcs thanDD. Hence, the induction hypothesis yields that the MTT holds forDblueD^{\operatorname*{blue}}. That is,

(# of spanning arborescences ofDbluerooted tor)=det(L∼r,∼rblue),\left(\text{\# of spanning arborescences of }D^{\operatorname*{blue}}\text{ rooted to }r\right)=\det\left(L_{\sim r,\sim r}^{\operatorname*{blue}}\right),whereLblueL^{\operatorname*{blue}}means the Laplacian ofDblueD^{\operatorname*{blue}}.

Example 5.159.

LetDDbe the multidigraph

withr=1r=1. Its Laplacian is

L=(1−100001−100−103−1−10001−1−10001).L=\left(\begin{array}[c]{ccccc}1&-1&0&0&0\\ 0&1&-1&0&0\\ -1&0&3&-1&-1\\ 0&0&0&1&-1\\ -1&0&0&0&1\end{array}\right).Let us pickv=3v=3(this is a vertex with outdegree>1>1), and let us color the arcsaaandccred and the arcsbbandddblue (various other options are possible). Then,DredD^{\operatorname*{red}}andDblueD^{\operatorname*{blue}}look as follows (along with their LaplaciansLredL^{\operatorname*{red}}andLblueL^{\operatorname*{blue}}):

1122334455aacc1122334455bbddDredD^{\operatorname*{red}}DblueD^{\operatorname*{blue}}Lred=(1−100001−100001−100001−1−10001)L^{\operatorname*{red}}=\left(\begin{array}[c]{ccccc}1&-1&0&0&0\\ 0&1&-1&0&0\\ 0&0&1&-1&0\\ 0&0&0&1&-1\\ -1&0&0&0&1\end{array}\right)Lblue=(1−100001−100−1020−10001−1−10001)L^{\operatorname*{blue}}=\left(\begin{array}[c]{ccccc}1&-1&0&0&0\\ 0&1&-1&0&0\\ -1&0&2&0&-1\\ 0&0&0&1&-1\\ -1&0&0&0&1\end{array}\right)

Now, the digraphsDD,DblueD^{\operatorname*{blue}}andDredD^{\operatorname*{red}}differ only in the arcs with sourcevv, and as far as the latter arcs are concerned, the arcs ofDDare divided betweenDblueD^{\operatorname*{blue}}andDredD^{\operatorname*{red}}. Hence, by the definition of the Laplacian, we have

Lred≡𝑣LblueandLred+𝑣Lblue=L.L^{\operatorname*{red}}\overset{v}{\equiv}L^{\operatorname*{blue}}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ L^{\operatorname*{red}}\overset{v}{+}L^{\operatorname*{blue}}=L.Thus,

L∼r,∼rred≡𝑣L∼r,∼rblueandL∼r,∼rred+𝑣L∼r,∼rblue=L∼r,∼rL_{\sim r,\sim r}^{\operatorname*{red}}\overset{v}{\equiv}L_{\sim r,\sim r}^{\operatorname*{blue}}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ L_{\sim r,\sim r}^{\operatorname*{red}}\overset{v}{+}L_{\sim r,\sim r}^{\operatorname*{blue}}=L_{\sim r,\sim r}(here, we have used the fact thatr=nr=nandv≠rv\neq r, so that when we remove therr-th row and therr-th column of the matrixLL, thevv-th row remains thevv-th row). Hence,

det(L∼r,∼r⏟=L∼r,∼rred+𝑣L∼r,∼rblue)=det(L∼r,∼rred+𝑣L∼r,∼rblue)=det(L∼r,∼rred)+det(L∼r,∼rblue)\det\left(\underbrace{L_{\sim r,\sim r}}_{=L_{\sim r,\sim r}^{\operatorname*{red}}\overset{v}{+}L_{\sim r,\sim r}^{\operatorname*{blue}}}\right)=\det\left(L_{\sim r,\sim r}^{\operatorname*{red}}\overset{v}{+}L_{\sim r,\sim r}^{\operatorname*{blue}}\right)=\det\left(L_{\sim r,\sim r}^{\operatorname*{red}}\right)+\det\left(L_{\sim r,\sim r}^{\operatorname*{blue}}\right)(by the multilinearity of the determinant).

However, a similar equality holds for the # of spanning arborescences: namely, we have

(# of spanning arborescences ofDrooted tor)\displaystyle\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=(# of spanning arborescences ofDredrooted tor)\displaystyle=\left(\text{\# of spanning arborescences of }D^{\operatorname*{red}}\text{ rooted to }r\right)+(# of spanning arborescences ofDbluerooted tor).\displaystyle\ \ \ \ \ \ \ \ \ \ +\left(\text{\# of spanning arborescences of }D^{\operatorname*{blue}}\text{ rooted to }r\right).Here is why: Recall that an arborescence rooted torrmust satisfydeg+⁡v=1\deg^{+}v=1(by Statement A’6 in the Dual Arborescence Equivalence Theorem (Theorem5.123), sincev∈V∖{r}v\in V\setminus\left\{r\right\}). In other words, an arborescence rooted torrmust contain exactly one arc with sourcevv. In particular, a spanning arborescence ofDDrooted torrmust contain either a red arc or a blue arc, but not both at the same time. In the former case, it is a spanning arborescence ofDredD^{\operatorname*{red}}; in the latter, it is a spanning arborescence ofDblueD^{\operatorname*{blue}}. Conversely, any spanning arborescence ofDredD^{\operatorname*{red}}or ofDblueD^{\operatorname*{blue}}rooted torris automatically a spanning arborescence ofDDrooted torr. Thus,

(# of spanning arborescences ofDrooted tor)\displaystyle\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=(# of spanning arborescences ofDredrooted tor)⏟=det(L∼r,∼rred)(as we saw above)\displaystyle=\underbrace{\left(\text{\# of spanning arborescences of }D^{\operatorname*{red}}\text{ rooted to }r\right)}_{\begin{subarray}{c}=\det\left(L_{\sim r,\sim r}^{\operatorname*{red}}\right)\\ \text{(as we saw above)}\end{subarray}}+(# of spanning arborescences ofDbluerooted tor)⏟=det(L∼r,∼rblue)(as we saw above)\displaystyle\ \ \ \ \ \ \ \ \ \ +\underbrace{\left(\text{\# of spanning arborescences of }D^{\operatorname*{blue}}\text{ rooted to }r\right)}_{\begin{subarray}{c}=\det\left(L_{\sim r,\sim r}^{\operatorname*{blue}}\right)\\ \text{(as we saw above)}\end{subarray}}=det(L∼r,∼rred)+det(L∼r,∼rblue)=det(L∼r,∼r)\displaystyle=\det\left(L_{\sim r,\sim r}^{\operatorname*{red}}\right)+\det\left(L_{\sim r,\sim r}^{\operatorname*{blue}}\right)=\det\left(L_{\sim r,\sim r}\right)(since we proved thatdet(L∼r,∼r)=det(L∼r,∼rred)+det(L∼r,∼rblue)\det\left(L_{\sim r,\sim r}\right)=\det\left(L_{\sim r,\sim r}^{\operatorname*{red}}\right)+\det\left(L_{\sim r,\sim r}^{\operatorname*{blue}}\right)). That is, the MTT holds for our digraphDDand its vertexrr. This completes the induction step, and thus the MTT (Theorem5.142) is proved.

Our above proof of Theorem5.142has followed[Stanle18, Theorem 10.4]. Other proofs can be found across the literature, e.g., in[VanEhr51, Theorem 7], in[Margol10, Theorem 2.8], in[DeLeen19, Theorem 1]and in[Holzer22, Theorem 2.5.3]. (Some of these sources prove more general versions of the theorem. Confusingly, each source uses different notations and works in a slightly different setup, although most of them quickly reveal themselves to be equivalent upon some introspection.)

5.14.8Further exercises on the Laplacian

Exercise 5.160.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. LetLLbe the Laplacian of the digraphGbidirG^{\operatorname*{bidir}}. Prove thatLLis positive semidefinite.

[Hint:WriteLLasNTNN^{T}N, whereNNorNTN^{T}is some matrix you have seen before.

Note that the statement is not true if we replaceGbidirG^{\operatorname*{bidir}}by an arbitrary digraphDD.]

The following two exercises stand at the beginning of the theory ofchip-firingand related dynamical systems on a digraph (see[CorPer18],[Klivan19]and[JoyMel17]for much more). While the Laplacian is not mentioned in them directly, it is implicitly involved in the definition of a “donation” (how?).

Exercise 5.161.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a strongly connected multidigraph.

Awealth distributiononDDshall mean a family(kv)v∈V\left(k_{v}\right)_{v\in V}of integers (one for each vertexv∈Vv\in V). Ifk=(kv)v∈Vk=\left(k_{v}\right)_{v\in V}is a wealth distribution, then we refer to each valuekvk_{v}as thewealthof the vertexvv, and we define thetotal wealthofkkto be the sum∑v∈Vkv\sum\limits_{v\in V}k_{v}. We say that a vertexvvisin debtin a given wealth distributionk=(kv)v∈Vk=\left(k_{v}\right)_{v\in V}if its wealthkvk_{v}is negative.

For any verticesvvandww, we letav,wa_{v,w}denote the number of arcs that have sourcevvandww.

Adonationis an operation that transforms a wealth distribution as follows: We choose a vertexvv, and we decrease its wealth by its outdegreedeg+⁡v\deg^{+}v, and then increase the wealth of each vertexw∈Vw\in V(includingvvitself) byav,wa_{v,w}. (You can think ofvvas donating a unit of wealth for each arc that has sourcevv. This unit flows to the target to this arc. Note that a donation does not change the total wealth.)

[Example:For instance, consider the digraph

with wealth distribution(k1,k2,k3,k4,k5,k6)=(−1,−1,1,2,0,1)\left(k_{1},k_{2},k_{3},k_{4},k_{5},k_{6}\right)=\left(-1,-1,1,2,0,1\right). The vertices11and22are in debt here, but it is possible to get all vertices out of debt by having the vertices4,5,6,14,5,6,1donate in some order (the order clearly does not matter for the result757575Depending on the order, some vertices will go into debt in the process, but this is okay as long as they ultimately end up debt-free.).

Note that vertices are allowed to donate multiple times (although in the above example, this was unnecessary).]

[Hint:A donation will be calledsafeif its donorvv(that is, the vertex chosen to lose wealth) satisfieskv≥deg+⁡vk_{v}\geq\deg^{+}v, wherekkis the wealth distribution just before this donation. Start by showing that if the total wealth is larger than|A|−|V|\left|A\right|-\left|V\right|, then at least one vertexvvhas wealth≥deg+⁡v\geq\deg^{+}v(and thus can make a safe donation). Next, show that for any given wealth distributionkk, there are only finitely many wealth distributions that can be obtained fromkkby a sequence of safe donations. Finally, for any vertexvv, find a rational quantity that increases every time that a donor distinct fromvvmakes a donation. Conclude that in a sufficiently long sequence of safe donations, every vertex must appear as a donor. But a donor of a safe donation must be out of debt just before its safe donation, and will never go back into debt.]

Exercise 5.162.

We continue with the setting and terminology of Exercise5.161.

Aclawbackis an operation that transforms a wealth distribution as follows: We choose a vertexvv, and we increase its wealth by its outdegreedeg+⁡v\deg^{+}v, and then decrease the wealth of each vertexw∈Vw\in V(includingvvitself) byav,wa_{v,w}. (Thus, a clawback is the inverse of a donation.)

[Remark:Note that we are still assumingDDto be strongly connected. Otherwise, the truth of the claim is not guaranteed. For instance, for the digraph

with wealth distribution(k1,k2,k3,k4)=(0,0,−1,2)\left(k_{1},k_{2},k_{3},k_{4}\right)=\left(0,0,-1,2\right), no sequence of donations and clawbacks will result in every vertex being out of debt (since the wealth differencek4−k3k_{4}-k_{3}is preserved under any donation or clawback, but this difference is too large to come from a debt-free distribution with total weight11). ]

[Hint:Show that any donation is equivalent to an appropriately chosen composition of clawbacks. Something we know about the Laplacian may come useful here.]

5.14.9Application: Counting Eulerian circuits ofKnbidirK_{n}^{\operatorname*{bidir}}

Here is one more consequence of the MTT:

Proposition 5.163.

Letnnbe a positive integer. Pick any arcaaof the multidigraphKnbidirK_{n}^{\operatorname*{bidir}}. Then, the # of Eulerian circuits ofKnbidirK_{n}^{\operatorname*{bidir}}whose first arc isaaisnn−2⋅(n−2)!nn^{n-2}\cdot\left(n-2\right)!^{n}.

Proof 5.164.

Letrrbe the source of the arcaa. The digraphKnbidirK_{n}^{\operatorname*{bidir}}is balanced, and each of its vertices has outdegreen−1n-1. By the BEST’ theorem (Theorem5.122), we have

(# of Eulerian circuits ofKnbidirwhose first arc isa)\displaystyle\left(\text{\# of Eulerian circuits of }K_{n}^{\operatorname*{bidir}}\text{ whose first arc is }a\right)=(# of spanning arborescences ofKnbidirrooted tor)⏟=nn−2(as we saw in Subsection5.14.5in the case whenr=1,and can similarly prove for arbitraryr)⋅∏u=1n(deg+⁡u⏟=n−1−1)!\displaystyle=\underbrace{\left(\text{\# of spanning arborescences of }K_{n}^{\operatorname*{bidir}}\text{ rooted to }r\right)}_{\begin{subarray}{c}=n^{n-2}\\ \text{(as we saw in Subsection \ref{subsec.MTT.Kn} in the case when }r=1\text{,}\\ \text{and can similarly prove for arbitrary }r\text{)}\end{subarray}}\cdot\prod\limits_{u=1}^{n}\left(\underbrace{\deg^{+}u}_{=n-1}-\,1\right)!=nn−2⋅∏u=1n(n−2)!=nn−2⋅(n−2)!n,\displaystyle=n^{n-2}\cdot\prod\limits_{u=1}^{n}\left(n-2\right)!=n^{n-2}\cdot\left(n-2\right)!^{n},qed.

In comparison, there is no good formula known for the # of Eulerian circuits of the undirected graphKnK_{n}. Fornneven, this # is00of course (sinceKnK_{n}has vertices of odd degree in this case). Fornnodd, the # grows very fast, but little else is known about it (seehttps://oeis.org/A135388for some known values, and see Exercise5.166for a divisibility property).

Exercise 5.165.

Letnnbe a positive integer. LetN={1,2,…,n}N=\left\{1,2,\ldots,n\right\}. A mapf:N→Nf:N\rightarrow Nis said to benn-potentif eachi∈Ni\in Nsatisfiesfn−1(i)=nf^{n-1}\left(i\right)=n. (As usual,fkf^{k}denotes thekk-fold compositionf∘f∘⋯∘ff\circ f\circ\cdots\circ f.)

Prove that the # ofnn-potent mapsf:N→Nf:N\rightarrow Nisnn−2n^{n-2}.

[Hint:What do thesenn-potent maps have to do with trees?]

Exercise 5.166.

Letn=2m+1>2n=2m+1>2be an odd integer. Leteebe an edge of the (undirected) complete graphKnK_{n}. Prove that the # of Eulerian circuits ofKnK_{n}that start witheeis a multiple of(m−1)!n\left(m-1\right)!^{n}.

[Hint:Argue that each Eulerian circuit ofKnK_{n}is an Eulerian circuit of a unique balanced tournament. Here, a “balanced tournament” means a balanced digraph obtained fromKnK_{n}by orienting each edge.]

5.15The undirected Matrix-Tree Theorem

5.15.1The theorem

The Matrix-Tree Theorem becomes simpler if we apply it to a digraph of the formGbidirG^{\operatorname*{bidir}}:

Theorem 5.167(undirected Matrix-Tree Theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for some positive integernn.

LetLLbe the Laplacian of the digraphGbidirG^{\operatorname*{bidir}}. Explicitly, this is then×nn\times n-matrixL∈ℤn×nL\in\mathbb{Z}^{n\times n}whose entries are given by

Li,j=(degi)⋅[i=j]−ai,j,L_{i,j}=\left(\deg i\right)\cdot\left[i=j\right]-a_{i,j},whereai,ja_{i,j}is the#\#of edges ofGGthat have endpointsiiandjj(with loops counting twice). Then:

(a)For any vertexrrofGG, we have (# of spanning trees ofG)=det(L∼r,∼r).\left(\text{\# of spanning trees of }G\right)=\det\left(L_{\sim r,\sim r}\right).
(b)Letttbe an indeterminate. Expand the determinantdet(tIn+L)\det\left(tI_{n}+L\right)(here,InI_{n}denotes then×nn\times nidentity matrix) as a polynomial intt: det(tIn+L)=cntn+cn−1tn−1+⋯+c1t1+c0t0,\det\left(tI_{n}+L\right)=c_{n}t^{n}+c_{n-1}t^{n-1}+\cdots+c_{1}t^{1}+c_{0}t^{0},wherec0,c1,…,cnc_{0},c_{1},\ldots,c_{n}are numbers. (Note that this is the characteristic polynomial ofLLup to substituting−t-tforttand multiplying by a power of−1-1. Some of its coefficients arecn=1c_{n}=1andcn−1=TrLc_{n-1}=\operatorname*{Tr}Landc0=detLc_{0}=\det L.) Then, (# of spanning trees ofG)=1nc1.\left(\text{\# of spanning trees of }G\right)=\dfrac{1}{n}c_{1}.
(c)Letλ1,λ2,…,λn\lambda_{1},\lambda_{2},\ldots,\lambda_{n}be the eigenvalues ofLL, listed in such a way thatλn=0\lambda_{n}=0(we know that00is an eigenvalue ofLL, sinceLLis singular). Then, (# of spanning trees ofG)=1n⋅λ1λ2⋯λn−1.\left(\text{\# of spanning trees of }G\right)=\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}.

Proof 5.168.

(a)Letrrbe a vertex ofGG. Then, Proposition5.132(b)shows that there is a bijection

{spanning arborescences ofGbidirrooted tor}→{spanning trees ofG}.\left\{\text{spanning arborescences of }G^{\operatorname*{bidir}}\text{ rooted to }r\right\}\rightarrow\left\{\text{spanning trees of }G\right\}.Hence, by the bijection principle, we have

(# of spanning trees ofG)\displaystyle\left(\text{\# of spanning trees of }G\right)=(# of spanning arborescences ofGbidirrooted tor)\displaystyle=\left(\text{\# of spanning arborescences of }G^{\operatorname*{bidir}}\text{ rooted to }r\right)=det(L∼r,∼r)(by the Matrix-Tree Theorem (Theorem5.142)).\displaystyle=\det\left(L_{\sim r,\sim r}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by the Matrix-Tree Theorem (Theorem \ref{thm.MTT.MTT})}\right).This proves Theorem5.167(a).

(b)We claim that

c1=∑r=1ndet(L∼r,∼r).c_{1}=\sum\limits_{r=1}^{n}\det\left(L_{\sim r,\sim r}\right).(32)Note that this is a purely linear-algebraic result, and has nothing to do with the fact thatLLis the Laplacian of a digraph; it holds just as well ifLLis replaced by any square matrix.

Once (32) is proved, Theorem5.167(b)will easily follow, because (32) entails

1nc1\displaystyle\dfrac{1}{n}c_{1}=1n∑r=1ndet(L∼r,∼r)⏟=(# of spanning trees ofG)(by Theorem5.167(a))=1n∑r=1n(# of spanning trees ofG)⏟=n⋅(# of spanning trees ofG)\displaystyle=\dfrac{1}{n}\sum\limits_{r=1}^{n}\underbrace{\det\left(L_{\sim r,\sim r}\right)}_{\begin{subarray}{c}=\left(\text{\# of spanning trees of }G\right)\\ \text{(by Theorem \ref{thm.MTT.undir} {(a)})}\end{subarray}}=\dfrac{1}{n}\underbrace{\sum\limits_{r=1}^{n}\left(\text{\# of spanning trees of }G\right)}_{=n\cdot\left(\text{\# of spanning trees of }G\right)}=1n⋅n⋅(# of spanning trees ofG)=(# of spanning trees ofG).\displaystyle=\dfrac{1}{n}\cdot n\cdot\left(\text{\# of spanning trees of }G\right)=\left(\text{\# of spanning trees of }G\right).Thus, it remains to prove (32).

A rigorous proof of (32) can be found in[21s, Proposition 6.4.29]or inhttps://math.stackexchange.com/a/3989575/(both of these references actually describe all coefficientsc0,c1,…,cnc_{0},c_{1},\ldots,c_{n}of the polynomialdet(tIn+L)\det\left(tI_{n}+L\right), not just thet1t^{1}-coefficientc1c_{1}). We shall merely outline the proof of (32) on a convenient example. We want to computec1c_{1}. In other words, we want to compute the coefficient oft1t^{1}in the polynomialdet(tIn+L)\det\left(tI_{n}+L\right)(sincec1c_{1}is defined to be this very coefficient). Let us say thatn=4n=4, so thatLLhas the form

L=(abcda′b′c′d′a′′b′′c′′d′′a′′′b′′′c′′′d′′′).L=\left(\begin{array}[c]{cccc}a&b&c&d\\ a^{\prime}&b^{\prime}&c^{\prime}&d^{\prime}\\ a^{\prime\prime}&b^{\prime\prime}&c^{\prime\prime}&d^{\prime\prime}\\ a^{\prime\prime\prime}&b^{\prime\prime\prime}&c^{\prime\prime\prime}&d^{\prime\prime\prime}\end{array}\right).Thus,

det(tIn+L)=det(t+abcda′t+b′c′d′a′′b′′t+c′′d′′a′′′b′′′c′′′t+d′′′).\det\left(tI_{n}+L\right)=\det\left(\begin{array}[c]{cccc}t+a&b&c&d\\ a^{\prime}&t+b^{\prime}&c^{\prime}&d^{\prime}\\ a^{\prime\prime}&b^{\prime\prime}&t+c^{\prime\prime}&d^{\prime\prime}\\ a^{\prime\prime\prime}&b^{\prime\prime\prime}&c^{\prime\prime\prime}&t+d^{\prime\prime\prime}\end{array}\right).Imagine expanding the right hand side (using the Leibniz formula) and expanding the resulting products further. For instance, the product

(t+a)(t+b′)d′′c′′′\left(t+a\right)\left(t+b^{\prime}\right)d^{\prime\prime}c^{\prime\prime\prime}becomesttd′′c′′′+tb′d′′c′′′+atd′′c′′′+ab′d′′c′′′ttd^{\prime\prime}c^{\prime\prime\prime}+tb^{\prime}d^{\prime\prime}c^{\prime\prime\prime}+atd^{\prime\prime}c^{\prime\prime\prime}+ab^{\prime}d^{\prime\prime}c^{\prime\prime\prime}. In the huge sum that results, we are interested in those addends that contain exactly onett, because it is precisely these addends that contribute to the coefficient oft1t^{1}in the polynomialdet(tIn+L)\det\left(tI_{n}+L\right). Where do these addends come from? To pick up exactly onettfrom a product like(t+a)(t+b′)d′′c′′′\left(t+a\right)\left(t+b^{\prime}\right)d^{\prime\prime}c^{\prime\prime\prime}, we need to have at least one diagonal entry in our product (for example, we cannot pick up anyttfrom the productcd′b′′a′′′cd^{\prime}b^{\prime\prime}a^{\prime\prime\prime}), and we need to pick out thettfrom this diagonal entry (rather than, e.g., theaaorb′b^{\prime}orc′′c^{\prime\prime}ord′′′d^{\prime\prime\prime}). If we pick therr-th diagonal entry, then the rest of the product is part of the expansion ofdet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)(since we must not pick any furthertts and thus can pretend that they are not there in the first place). Thus, the totalt1t^{1}-coefficient indet(tIn+L)\det\left(tI_{n}+L\right)will be∑r=1ndet(L∼r,∼r)\sum\limits_{r=1}^{n}\det\left(L_{\sim r,\sim r}\right). This proves (32), and thus the proof of Theorem5.167(b)is complete.

(c)Consider the polynomialdet(tIn+L)\det\left(tI_{n}+L\right)introduced in part(b), and in particular itst1t^{1}-coefficientc1c_{1}.

It is known that the characteristic polynomialdet(tIn−L)\det\left(tI_{n}-L\right)ofLLis a monic polynomial of degreenn, and that its roots are the eigenvaluesλ1,λ2,…,λn\lambda_{1},\lambda_{2},\ldots,\lambda_{n}ofLL. Hence, it can be factored as follows:

det(tIn−L)=(t−λ1)(t−λ2)⋯(t−λn).\det\left(tI_{n}-L\right)=\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right)\cdots\left(t-\lambda_{n}\right).Substituting−t-tfortton both sides of this equality, we obtain

det(−tIn−L)=(−t−λ1)(−t−λ2)⋯(−t−λn).\det\left(-tI_{n}-L\right)=\left(-t-\lambda_{1}\right)\left(-t-\lambda_{2}\right)\cdots\left(-t-\lambda_{n}\right).Multiplying both sides of this equality by(−1)n\left(-1\right)^{n}, we find

det(tIn+L)\displaystyle\det\left(tI_{n}+L\right)=(t+λ1)(t+λ2)⋯(t+λn)\displaystyle=\left(t+\lambda_{1}\right)\left(t+\lambda_{2}\right)\cdots\left(t+\lambda_{n}\right)=(t+λ1)(t+λ2)⋯(t+λn−1)t(sinceλn=0).\displaystyle=\left(t+\lambda_{1}\right)\left(t+\lambda_{2}\right)\cdots\left(t+\lambda_{n-1}\right)t\ \ \ \ \ \ \ \ \ \ \left(\text{since }\lambda_{n}=0\right).Hence, thet1t^{1}-coefficient of the polynomialdet(tIn+L)\det\left(tI_{n}+L\right)isλ1λ2⋯λn−1\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}(since this is clearly thet1t^{1}-coefficient on the right hand side). Since we definedc1c_{1}to be thet1t^{1}-coefficient of the polynomialdet(tIn+L)\det\left(tI_{n}+L\right), we thus conclude thatc1=λ1λ2⋯λn−1c_{1}=\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}. However, Theorem5.167(b)yields

(# of spanning trees ofG)=1nc1⏟=λ1λ2⋯λn−1=1n⋅λ1λ2⋯λn−1.\left(\text{\# of spanning trees of }G\right)=\dfrac{1}{n}\underbrace{c_{1}}_{=\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}}=\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}.This proves Theorem5.167(c).

5.15.2Application: counting spanning trees ofKn,mK_{n,m}

Laplacians of digraphs often have computable eigenvalues, so Theorem5.167(c)is actually pretty useful. A striking example of a # of spanning trees (specifically, of thenn-hypercube graphQnQ_{n}, which we already met in Subsection2.14.4) that can be counted using eigenvalues will appear in Exercise5.176.

Here, however, let us give a simpler example, in which Theorem5.167(a)suffices:

Exercise 5.169.

Letnnandmmbe two positive integers. LetKn,mK_{n,m}be the simple graph withn+mn+mvertices

1,2,…,nand−1,−2,…,−m,1,2,\ldots,n\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ -1,-2,\ldots,-m,where two verticesiiandjjare adjacent if and only if they have opposite signs (i.e., each positive vertex is adjacent to each negative vertex, but no two vertices of the same sign are adjacent).

[For example, here is howK5,2K_{5,2}looks like:

How many spanning trees doesKn,mK_{n,m}have?

Proof 5.170(Solution.).

If we rename the negative vertices−1,−2,…,−m-1,-2,\ldots,-masn+1,n+2,…,n+mn+1,n+2,\ldots,n+m, then the LaplacianLLof the digraphKn,mbidirK_{n,m}^{\operatorname*{bidir}}can be written in block-matrix notation as follows:

L=(ABCD),L=\left(\begin{array}[c]{cc}A&B\\ C&D\end{array}\right),where

•AAis a diagonaln×nn\times n-matrix whose all diagonal entries are equal tomm(since there are no edges between positive vertices, and since each positive vertex has degreemm);
•BBis ann×mn\times m-matrix whose all entries equal−1-1;
•CCis anm×nm\times n-matrix whose all entries equal−1-1;
•DDis a diagonalm×mm\times m-matrix whose all diagonal entries are equal tonn.

For instance, ifn=3n=3andm=2m=2, then

L=(200−1−1020−1−1002−1−1−1−1−130−1−1−103).L=\left(\begin{array}[c]{ccccc}2&0&0&-1&-1\\ 0&2&0&-1&-1\\ 0&0&2&-1&-1\\ -1&-1&-1&3&0\\ -1&-1&-1&0&3\end{array}\right). Theorem5.167(a)yields

(# of spanning trees ofKn,m)=det(L∼r,∼r)for any vertexrofKn,m;\left(\text{\# of spanning trees of }K_{n,m}\right)=\det\left(L_{\sim r,\sim r}\right)\ \ \ \ \ \ \ \ \ \ \text{for any vertex }r\text{ of }K_{n,m};thus, we need to computedet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)for some vertexrr. We letr=1r=1. Then, the submatrixL∼r,∼r=L∼1,∼1L_{\sim r,\sim r}=L_{\sim 1,\sim 1}ofLLagain can be written in block-matrix notation as follows:

L∼r,∼r=(A~B~C~D),L_{\sim r,\sim r}=\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ \widetilde{C}&D\end{array}\right),(33)where

•A~\widetilde{A}is a diagonal(n−1)×(n−1)\left(n-1\right)\times\left(n-1\right)-matrix, whose all diagonal entries are equal tomm;
•B~\widetilde{B}is an(n−1)×m\left(n-1\right)\times m-matrix whose all entries equal−1-1;
•C~\widetilde{C}is anm×(n−1)m\times\left(n-1\right)-matrix whose all entries equal−1-1;
•DDis a diagonalm×mm\times m-matrix whose all diagonal entries are equal tonn.

Fortunately, determinants of block matrices are often not hard to compute, at least when some of the blocks are invertible. For example, the Schur complement providesa neat formula. Our life here is even easier, sinceA~\widetilde{A}andDDare multiples of identity matrices: namely,A~=mIn−1\widetilde{A}=mI_{n-1}andD=nImD=nI_{m}. We perform a “blockwise row transformation” on the block matrixL∼r,∼r=(A~B~C~D)L_{\sim r,\sim r}=\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ \widetilde{C}&D\end{array}\right), specifically subtracting theC~A~−1\widetilde{C}\widetilde{A}^{-1}-multiple of the first “block row”(A~B~)\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\end{array}\right)from the second “block row”(C~D)\left(\begin{array}[c]{cc}\widetilde{C}&D\end{array}\right)(yes, this is legitimate – it’s the same as left-multiplying by the block matrix(In−10−C~A~−1Im)\left(\begin{array}[c]{cc}I_{n-1}&0\\ -\widetilde{C}\widetilde{A}^{-1}&I_{m}\end{array}\right), which has determinant11because it is lower-triangular). As a result, we obtain

det(A~B~C~D)\displaystyle\det\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ \widetilde{C}&D\end{array}\right)=det(A~B~C~−C~A~−1A~D−C~A~−1B~)\displaystyle=\det\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ \widetilde{C}-\widetilde{C}\widetilde{A}^{-1}\widetilde{A}&D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\end{array}\right)=det(A~B~0D−C~A~−1B~).\displaystyle=\det\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ 0&D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\end{array}\right).The matrix on the right is “block-upper triangular”, so its determinant factors as follows:767676We are using the fact that if a matrix is block-triangular (with all diagonal blocks being square matrices), then its determinant is the product of the determinants of its diagonal blocks. See, e.g.,https://math.stackexchange.com/a/1221066/or[Grinbe20, Exercise 6.29]for a proof of this fact.

det(A~B~0D−C~A~−1B~)=detA~⋅det(D−C~A~−1B~).\det\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ 0&D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\end{array}\right)=\det\widetilde{A}\cdot\det\left(D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\right).Of course,detA~=mn−1\det\widetilde{A}=m^{n-1}, sinceA~\widetilde{A}is a diagonal matrix withm,m,…,mm,m,\ldots,mon the diagonal. Computingdet(D−C~A~−1B~)\det\left(D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\right)is a bit more complicated, but still doable: The matrixA~−1\widetilde{A}^{-1}is a diagonal matrix withm−1,m−1,…,m−1m^{-1},m^{-1},\ldots,m^{-1}on the diagonal; thus, its role in the productC~A~−1B~\widetilde{C}\widetilde{A}^{-1}\widetilde{B}is merely to multiply everything bym−1m^{-1}. Hence,C~A~−1B~=m−1C~B~\widetilde{C}\widetilde{A}^{-1}\widetilde{B}=m^{-1}\widetilde{C}\widetilde{B}. Since all entries ofC~\widetilde{C}andB~\widetilde{B}are−1-1’s, we see that all entries ofC~B~\widetilde{C}\widetilde{B}are(n−1)\left(n-1\right)’s. Putting all of this together, we see thatD−C~A~−1B~D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}is them×mm\times m-matrix whose all diagonal entries are equal ton−m−1(n−1)n-m^{-1}\left(n-1\right)and whose all off-diagonal entries are equal to−m−1(n−1)-m^{-1}\left(n-1\right). We have already computed the determinant of a matrix much like this back in our proof of Cayley’s Formula (Subsection5.14.5); let us deal with the general case:

Proposition 5.171.

Letn∈ℕn\in\mathbb{N}. Letxxandaabe two numbers. Then,

det(xaa⋯aaaxa⋯aaaax⋯aa⋮⋮⋮⋱⋮⋮aaa⋯xaaaa⋯ax)⏟then×n-matrixwhose diagonal entries arexand whose off-diagonal entries area=(x+(n−1)a)(x−a)n−1.\det\underbrace{\left(\begin{array}[c]{cccccc}x&a&a&\cdots&a&a\\ a&x&a&\cdots&a&a\\ a&a&x&\cdots&a&a\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ a&a&a&\cdots&x&a\\ a&a&a&\cdots&a&x\end{array}\right)}_{\begin{subarray}{c}\text{the }n\times n\text{-matrix}\\ \text{whose diagonal entries are }x\\ \text{and whose off-diagonal entries are }a\end{subarray}}=\left(x+\left(n-1\right)a\right)\left(x-a\right)^{n-1}.

Proposition5.171can be proved using similar reasoning as the determinant in Subsection5.14.5; we will say more about it later. For now, let us apply it tomm,n−m−1(n−1)n-m^{-1}\left(n-1\right)and−m−1(n−1)-m^{-1}\left(n-1\right)instead ofnn,xxandaa, to obtain

det(D−C~A~−1B~)\displaystyle\det\left(D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\right)=((n−m−1(n−1))+(m−1)(−m−1(n−1)))⏟=1\displaystyle=\underbrace{\left(\left(n-m^{-1}\left(n-1\right)\right)+\left(m-1\right)\left(-m^{-1}\left(n-1\right)\right)\right)}_{=1}⋅((n−m−1(n−1))−(−m−1(n−1))⏟=n)m−1\displaystyle\ \ \ \ \ \ \ \ \ \ \cdot\left(\underbrace{\left(n-m^{-1}\left(n-1\right)\right)-\left(-m^{-1}\left(n-1\right)\right)}_{=n}\right)^{m-1}=nm−1.\displaystyle=n^{m-1}. Now, it is time to combine everything we know. Theorem5.167(a)yields

(# of spanning trees ofKn,m)\displaystyle\left(\text{\# of spanning trees of }K_{n,m}\right)=det(L∼r,∼r)\displaystyle=\det\left(L_{\sim r,\sim r}\right)=det(A~B~C~D)(by (33))\displaystyle=\det\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ \widetilde{C}&D\end{array}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{sol.MTT.Knm.block2})}\right)=det(A~B~0D−C~A~−1B~)\displaystyle=\det\left(\begin{array}[c]{cc}\widetilde{A}&\widetilde{B}\\ 0&D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\end{array}\right)=detA~⏟=mn−1⋅det(D−C~A~−1B~)⏟=nm−1\displaystyle=\underbrace{\det\widetilde{A}}_{=m^{n-1}}\cdot\underbrace{\det\left(D-\widetilde{C}\widetilde{A}^{-1}\widetilde{B}\right)}_{=n^{m-1}}=mn−1⋅nm−1.\displaystyle=m^{n-1}\cdot n^{m-1}.

Thus, we have obtained the following:

Theorem 5.172.

Letnnandmmbe two positive integers. LetKn,mK_{n,m}be the simple graph withn+mn+mvertices

1,2,…,nand−1,−2,…,−m,1,2,\ldots,n\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ -1,-2,\ldots,-m,where two verticesiiandjjare adjacent if and only if they have opposite signs. Then,

(# of spanning trees ofKn,m)=mn−1⋅nm−1.\left(\text{\# of spanning trees of }K_{n,m}\right)=m^{n-1}\cdot n^{m-1}.

See[AbuSbe88]for a combinatorial proof of this theorem.

Exercise 5.173.

Letnnbe a positive integer. LetKn,2K_{n,2}be the simple graph with vertex set{1,2,…,n}∪{−1,−2}\left\{1,2,\ldots,n\right\}\cup\left\{-1,-2\right\}such that two vertices ofKn,2K_{n,2}are adjacent if and only if they have opposite signs (i.e., each positive vertex is adjacent to each negative vertex, but no two vertices of the same sign are adjacent). We regardKn,2K_{n,2}as a multigraph in the usual way.

(a)Without using the matrix-tree theorem, prove that the number of spanning trees ofKn,2K_{n,2}isn⋅2n−1n\cdot 2^{n-1}.
(b)LetKn,2′K_{n,2}^{\prime}be the graph obtained by adding a new edge{−1,−2}\left\{-1,-2\right\}toKn,2K_{n,2}. How many spanning trees doesKn,2′K_{n,2}^{\prime}have?

[Example:Here is the graphKn,2K_{n,2}forn=5n=5:

And here is the corresponding graphKn,2′K^{\prime}_{n,2}:

]

Exercise 5.174.

Letnnbe a positive integer. LetAAbe the(n−1)×(n−1)\left(n-1\right)\times\left(n-1\right)-matrix

(2−10⋯0−12−1⋯00−12⋯0⋮⋮⋮⋱⋮000⋯2),\left(\begin{array}[c]{ccccc}2&-1&0&\cdots&0\\ -1&2&-1&\cdots&0\\ 0&-1&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&2\end{array}\right),whose(i,j)\left(i,j\right)-th entry is

Ai,j:={2,ifi=j;−1,if|i−j|=1;0,otherwisefor alli,j∈{1,2,…,n−1}.A_{i,j}:=\begin{cases}2,&\text{if }i=j;\\ -1,&\text{if }\left|i-j\right|=1;\\ 0,&\text{otherwise}\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in\left\{1,2,\ldots,n-1\right\}.Prove thatdetA=n\det A=n.

[Hint:Recall Example5.41.]

Exercise 5.175.

LetGGbe a multigraph with an even number of vertices. Assume that each vertex ofGGhas an even degree. Prove thatGGhas an even number of spanning trees.

[Hint:First show a lemma: Letn∈ℕn\in\mathbb{N}be odd. LetBBbe ann×nn\times n-matrix with integer entries. Assume thatBBis symmetric (i.e., satisfiesBT=BB^{T}=B), and that all diagonal entries ofBBare even. Then,detB\det Bis even.]

Exercise 5.176.

Letnnbe a positive integer. LetQnQ_{n}be thenn-hypercube graph (as defined in Definition2.139). Recall that its vertex set is the setV:={0,1}nV:=\left\{0,1\right\}^{n}of length-nnbitstrings, and that two vertices are adjacent if and only if they differ in exactly one bit. Our goal is to compute the # of spanning trees ofQnQ_{n}.

LetDDbe the digraphQnbidirQ_{n}^{\operatorname*{bidir}}. LetLLbe the Laplacian ofDD. We regardLLas aV×VV\times V-matrix (i.e., as a2n×2n2^{n}\times 2^{n}-matrix whose rows and columns are indexed by bitstrings inVV).

We shall use the notationaia_{i}for theii-th entry of a bitstringaa. Thus, each bitstringa∈Va\in Vhas the forma=(a1,a2,…,an)a=\left(a_{1},a_{2},\ldots,a_{n}\right). (We shall avoid the shorthand notationa1a2⋯ana_{1}a_{2}\cdots a_{n}here, as it could be mistaken for an actual product.)

For any two bitstringsa,b∈Va,b\in V, we define the number⟨a,b⟩\left\langle a,b\right\rangleto be the integera1b1+a2b2+⋯+anbna_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}.

(a)Prove that every bitstringa∈Va\in Vsatisfies ∑b∈V(−1)⟨a,b⟩={2n,ifa=𝟎;0,otherwise.\sum\limits_{b\in V}\left(-1\right)^{\left\langle a,b\right\rangle}=\begin{cases}2^{n},&\text{ if $a=\\mathbf\{0\}$ ;}\\ 0,&\text{ otherwise.}\end{cases}Here,𝟎\mathbf{0}denotes the bitstring(0,0,…,0)∈V\left(0,0,\ldots,0\right)\in V.

Now, define a furtherV×VV\times V-matrixGGby requiring that its(a,b)\left(a,b\right)-th entry is

Ga,b=(−1)⟨a,b⟩for anya,b∈V.G_{a,b}=\left(-1\right)^{\left\langle a,b\right\rangle}\qquad\text{for any $a,b\\in V$ }.Furthermore, define a diagonalV×VV\times V-matrixDDby requiring that its(a,a)\left(a,a\right)-th entry is

Da,a\displaystyle D_{a,a}=2⋅(#ofi∈{1,2,…,n}such thatai=1)\displaystyle=2\cdot\left(\#\text{ of $i\\in\\left\\\{1,2,\\ldots,n\\right\\\}$ such that $a\_\{i\}=1$ }\right)=2⋅(the number of1s ina)for anya∈V\displaystyle=2\cdot\left(\text{the number of $1$ s in $a$ }\right)\qquad\text{for any $a\\in V$ }(and its off-diagonal entries are00).

Prove the following:

(b)We haveG2=2n⋅IG^{2}=2^{n}\cdot I, whereIIis the identityV×VV\times V-matrix.
(c)We haveGLG−1=DGLG^{-1}=D.
(d)The eigenvalues ofLLare2k2kfor allk∈{0,1,…,n}k\in\left\{0,1,\ldots,n\right\}, and each eigenvalue2k2kappears with multiplicity(nk)\dbinom{n}{k}.
(e)The # of spanning trees ofQnQ_{n}is 12n∏k=1n(2k)(nk).\dfrac{1}{2^{n}}\prod\limits_{k=1}^{n}\left(2k\right)^{\binom{n}{k}}.

[Example:As an example, here is the casen=3n=3. In this case, the graphQnQ_{n}looks as follows:

000000001001010010011011100100101101110110111111Q3=Q_{3}=The matricesLL,GGandDDare

L=(3−1−10−1000−130−10−100−103−100−100−1−13000−1−10003−1−100−100−130−100−10−103−1000−10−1−13),L=\begin{pmatrix}3&-1&-1&0&-1&0&0&0\\ -1&3&0&-1&0&-1&0&0\\ -1&0&3&-1&0&0&-1&0\\ 0&-1&-1&3&0&0&0&-1\\ -1&0&0&0&3&-1&-1&0\\ 0&-1&0&0&-1&3&0&-1\\ 0&0&-1&0&-1&0&3&-1\\ 0&0&0&-1&0&-1&-1&3\end{pmatrix},G=(111111111−11−11−11−111−1−111−1−11−1−111−1−111111−1−1−1−11−11−1−11−1111−1−1−1−1111−1−11−111−1),G=\begin{pmatrix}1&1&1&1&1&1&1&1\\ 1&-1&1&-1&1&-1&1&-1\\ 1&1&-1&-1&1&1&-1&-1\\ 1&-1&-1&1&1&-1&-1&1\\ 1&1&1&1&-1&-1&-1&-1\\ 1&-1&1&-1&-1&1&-1&1\\ 1&1&-1&-1&-1&-1&1&1\\ 1&-1&-1&1&-1&1&1&-1\end{pmatrix},D=(0000000002000000002000000002000000004000000004000000004000000006),D=\begin{pmatrix}0&0&0&0&0&0&0&0\\ 0&2&0&0&0&0&0&0\\ 0&0&2&0&0&0&0&0\\ 0&0&0&2&0&0&0&0\\ 0&0&0&0&4&0&0&0\\ 0&0&0&0&0&4&0&0\\ 0&0&0&0&0&0&4&0\\ 0&0&0&0&0&0&0&6\end{pmatrix},where the rows and the columns are ordered by listing the eight bitstringsa∈Va\in Vin the order000,001,010,011,100,101,110,111000,001,010,011,100,101,110,111. ]

As we promised, let us make a few more remarks about Proposition5.171. While this proposition can be proved by fairly straightforward row transformations (first subtracting the first row from all the other rows, then factoring anx−ax-afrom all the latter rows, then subtractingaatimes each of the latter rows to the first row to obtain a triangular matrix), it can also be viewed as a particular case of either of the following two determinantal identities:

Proposition 5.177.

Letn∈ℕn\in\mathbb{N}. Leta1,a2,…,ana_{1},a_{2},\ldots,a_{n}bennnumbers, and letxxbe a further number. Then,

det(xa1a2⋯an−1ana1xa2⋯an−1ana1a2x⋯an−1an⋮⋮⋮⋱⋮⋮a1a2a3⋯xana1a2a3⋯anx)⏟an(n+1)×(n+1)-matrix=(x+∑i=1nai)∏i=1n(x−ai).\det\underbrace{\left(\begin{array}[c]{cccccc}x&a_{1}&a_{2}&\cdots&a_{n-1}&a_{n}\\ a_{1}&x&a_{2}&\cdots&a_{n-1}&a_{n}\\ a_{1}&a_{2}&x&\cdots&a_{n-1}&a_{n}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ a_{1}&a_{2}&a_{3}&\cdots&x&a_{n}\\ a_{1}&a_{2}&a_{3}&\cdots&a_{n}&x\end{array}\right)}_{\text{an }\left(n+1\right)\times\left(n+1\right)\text{-matrix}}=\left(x+\sum\limits_{i=1}^{n}a_{i}\right)\prod\limits_{i=1}^{n}\left(x-a_{i}\right).

Proposition 5.178.

Letn∈ℕn\in\mathbb{N}. Letx1,x2,…,xnx_{1},x_{2},\ldots,x_{n}bennnumbers, and letaabe a further number. Then,

det(x1aa⋯aax2a⋯aaax3⋯a⋮⋮⋮⋱⋮aaa⋯xn)=∏i=1n(xi−a)+a∑i=1nyi,\det\left(\begin{array}[c]{ccccc}x_{1}&a&a&\cdots&a\\ a&x_{2}&a&\cdots&a\\ a&a&x_{3}&\cdots&a\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a&a&a&\cdots&x_{n}\end{array}\right)=\prod\limits_{i=1}^{n}\left(x_{i}-a\right)+a\sum\limits_{i=1}^{n}y_{i},where we setyi:=∏k∈{1,2,…,n};k≠i(xk−a)y_{i}:=\prod\limits_{\begin{subarray}{c}k\in\left\{1,2,\ldots,n\right\};\\ k\neq i\end{subarray}}\left(x_{k}-a\right)for eachi∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}.

Both of these propositions make good exercises in determinant evaluation. (Proposition5.177is[Grinbe20, Exercise 6.21], while Proposition5.178ishttps://math.stackexchange.com/a/2112473/.)

See[KleSta19]and[Rubey00]for more applications of the Matrix-Tree Theorem, and[Holzer22]for many more related results.

5.16de Bruijn sequences

5.16.1Definition

Let me move on to a more intricate application of what we have learned about arborescences.

A little puzzle first: What is special about the periodic sequence

||:0000 1111 0110 0101:||?||\vphantom{a}:\ 0000\ 1111\ 0110\ 0101\ :{||}\ \ \ \ \ \ \ \ \ \text{?}(This is an infinite sequence of00’s and11’s; the spaces between some of them are only for readability. The||:||\vphantom{a}:and:||:||symbols are “repeat signs” – they mean that everything that stands between them should be repeated over and over. So the sequence above is0000 1111 0110 0101 0000 1111…0000\ 1111\ 0110\ 0101\ 0000\ 1111\ \ldots.)

One nice property of this sequence is that if you slide a ”length-44window” (i.e., a window that shows four consecutive entries) along it, you get all1616possible bitstrings of length44depending on the position of the window, and these bitstrings do not repeat until you move1616steps to the right. Just see:

000011110110010100001111…\displaystyle\framebox{ $0000$ }11110110010100001111\ldots000011110110010100001111…\displaystyle 0\framebox{ $0001$ }1110110010100001111\ldots000011110110010100001111…\displaystyle 00\framebox{ $0011$ }110110010100001111\ldots000011110110010100001111…\displaystyle 000\framebox{ $0111$ }10110010100001111\ldots000011110110010100001111…\displaystyle 0000\framebox{ $1111$ }0110010100001111\ldots000011110110010100001111…\displaystyle 00001\framebox{ $1110$ }110010100001111\ldots000011110110010100001111…\displaystyle 000011\framebox{ $1101$ }10010100001111\ldots000011110110010100001111…\displaystyle 0000111\framebox{ $1011$ }0010100001111\ldots000011110110010100001111…\displaystyle 00001111\framebox{ $0110$ }010100001111\ldots000011110110010100001111…\displaystyle 000011110\framebox{ $1100$ }10100001111\ldots000011110110010100001111…\displaystyle 0000111101\framebox{ $1001$ }0100001111\ldots000011110110010100001111…\displaystyle 00001111011\framebox{ $0010$ }100001111\ldots000011110110010100001111…\displaystyle 000011110110\framebox{ $0101$ }00001111\ldots000011110110010100001111…\displaystyle 0000111101100\framebox{ $1010$ }0001111\ldots000011110110010100001111…\displaystyle 00001111011001\framebox{ $0100$ }001111\ldots000011110110010100001111…\displaystyle 000011110110010\framebox{ $1000$ }01111\ldotsNote that, as you slide the window along the sequence, at each step, the first bit is removed and a new bit is inserted at the end. Thus, by sliding a length-44window along the above sequence, you run through all1616possible length-44bitstrings in such a way that each bitstring is obtained from the previous one by removing the first bit and inserting a new bit at the end. This is nice and somewhat similar to Gray codes (in which you run through all bitstrings of a given length in such a way that only a single bit is changed at each step).

Can we find such nice sequences for any window length, not just44?

Here is an answer for window length33, for instance:

||:000 111 01:||.||\vphantom{a}:\ 000\ 111\ 01\ :{||}\ \ .What about higher window length?

Moreover, we can ask the same question with other alphabets. For instance, instead of bits, here is a similar sequence for the alphabet{0,1,2}\left\{0,1,2\right\}(that is, we use the numbers0,1,20,1,2instead of00and11) and window length22:

||:00 11 22 02 1:||.||\vphantom{a}:00\ 11\ 22\ 02\ 1\ :{||}\ . What about the general case? Let us give it a name:

Definition 5.179.

Letnnandkkbe two positive integers, and letKKbe akk-element set.

Ade Bruijn sequenceof ordernnonKKmeans aknk^{n}-tuple(c0,c1,…,ckn−1)\left(c_{0},c_{1},\ldots,c_{k^{n}-1}\right)of elements ofKKsuch that

(A)for eachnn-tuple(a1,a2,…,an)∈Kn\left(a_{1},a_{2},\ldots,a_{n}\right)\in K^{n}of elements ofKK, there is auniquer∈{0,1,…,kn−1}r\in\left\{0,1,\ldots,k^{n}-1\right\}such that (a1,a2,…,an)=(cr,cr+1,…,cr+n−1).\left(a_{1},a_{2},\ldots,a_{n}\right)=\left(c_{r},c_{r+1},\ldots,c_{r+n-1}\right).

Here, the indices under the letter “cc” are understood to be periodic moduloknk^{n}; that is, we setcq+kn=cqc_{q+k^{n}}=c_{q}for eachq∈ℤq\in\mathbb{Z}(so thatckn=c0c_{k^{n}}=c_{0}andckn+1=c1c_{k^{n}+1}=c_{1}and so on).

For example, forn=2n=2andk=3k=3andK={0,1,2}K=\left\{0,1,2\right\}, the99-tuple

(0,0,1,1,2,2,0,2,1)\left(0,0,1,1,2,2,0,2,1\right)is a de Bruijn sequence of ordernnonKK, because if we label the entries of this99-tuple asc0,c1,…,c8c_{0},c_{1},\ldots,c_{8}(and extend the indices periodically, so thatc9=c0c_{9}=c_{0}), then we have

(0,0)\displaystyle\left(0,0\right)=(c0,c1);(0,1)=(c1,c2);(0,2)=(c6,c7);\displaystyle=\left(c_{0},c_{1}\right);\qquad\left(0,1\right)=\left(c_{1},c_{2}\right);\qquad\left(0,2\right)=\left(c_{6},c_{7}\right);(1,0)\displaystyle\left(1,0\right)=(c8,c9);(1,1)=(c2,c3);(1,2)=(c3,c4);\displaystyle=\left(c_{8},c_{9}\right);\qquad\left(1,1\right)=\left(c_{2},c_{3}\right);\qquad\left(1,2\right)=\left(c_{3},c_{4}\right);(2,0)\displaystyle\left(2,0\right)=(c5,c6);(2,1)=(c7,c8);(2,2)=(c4,c5).\displaystyle=\left(c_{5},c_{6}\right);\qquad\left(2,1\right)=\left(c_{7},c_{8}\right);\qquad\left(2,2\right)=\left(c_{4},c_{5}\right).This de Bruijn sequence(0,0,1,1,2,2,0,2,1)\left(0,0,1,1,2,2,0,2,1\right)corresponds to the periodic sequence||:00 11 22 02 1:||||\vphantom{a}:00\ 11\ 22\ 02\ 1\ :{||}that we found above.

5.16.2Existence of de Bruijn sequences

It turns out that de Bruijn sequences always exist:

Theorem 5.180(de Bruijn, Sainte-Marie).

Letnnandkkbe positive integers. LetKKbe akk-element set. Then, a de Bruijn sequence of ordernnonKKexists.

Proof 5.181.

It looks reasonable to approach this using a digraph. For example, we can define a digraph whose vertices are thenn-tuples inKnK^{n}, and that has an arc from onenn-tupleiito anothernn-tuplejjifjjcan be obtained fromiiby dropping the first entry and adding a new entry at the end. Then, a de Bruijn sequence (of ordernnonKK) is the same as a Hamiltonian cycle of this digraph.

Unfortunately, we don’t have any useful criteria that would show that such a cycle exists. So this idea seems to be a dead end.

However, let us do something counterintuitive: We try to reinterpret de Bruijn sequences in terms of Eulerian circuits (rather than Hamiltonian cycles), since we have a good criterion for the existence of Eulerian circuits (unlike for that of Hamiltonian cycles)!

We need a different digraph for that. Namely, we letDDbe the multidigraph(Kn−1,Kn,ψ)\left(K^{n-1},K^{n},\psi\right), where the mapψ:Kn→Kn−1×Kn−1\psi:K^{n}\rightarrow K^{n-1}\times K^{n-1}is given by the formula

ψ(a1,a2,…,an)=((a1,a2,…,an−1),(a2,a3,…,an)).\psi\left(a_{1},a_{2},\ldots,a_{n}\right)=\left(\left(a_{1},a_{2},\ldots,a_{n-1}\right),\ \ \left(a_{2},a_{3},\ldots,a_{n}\right)\right).Thus, the vertices ofDDare the(n−1)\left(n-1\right)-tuples (not thenn-tuples!) of elements ofKK, whereas the arcs are thenn-tuples of elements ofKK, and each such arc(a1,a2,…,an)\left(a_{1},a_{2},\ldots,a_{n}\right)has source(a1,a2,…,an−1)\left(a_{1},a_{2},\ldots,a_{n-1}\right)and target(a2,a3,…,an)\left(a_{2},a_{3},\ldots,a_{n}\right). Hence, there is an arc from each(n−1)\left(n-1\right)-tuplei∈Kn−1i\in K^{n-1}to each(n−1)\left(n-1\right)-tuplej∈Kn−1j\in K^{n-1}that is obtained by dropping the first entry ofiiand adding a new entry at the end. (Be careful: Ifn=1n=1, thenDDhas only one vertex butnnarcs. If this confuses you, just do then=1n=1case by hand. For anyn>1n>1, there are no parallel arcs inDD.)

Example 5.182.

For example, ifn=3n=3andk=2k=2andK={0,1}K=\left\{0,1\right\}, thenDDlooks as follows (we again write our tuples without commas and without parentheses):

0000010110101111001001000000101101100100010010011011110110111111

Let us make a few observations aboutDD:

•The multidigraphDDis strongly connected. [Proof:We need to show that for any two verticesiiandjjofDD, there is a walk fromiitojj. But this is easy: Just insert the entries ofjjintoiione by one, pushing out the entries ofii. In other words, using the notationkpk_{p}for thepp-th entry of any tuplekk, we have the walk i\displaystyle i=(i1,i2,…,in−1)\displaystyle=\left(i_{1},i_{2},\ldots,i_{n-1}\right)→(i2,i3,…,in−1,j1)\displaystyle\rightarrow\left(i_{2},i_{3},\ldots,i_{n-1},j_{1}\right)→(i3,i4,…,in−1,j1,j2)\displaystyle\rightarrow\left(i_{3},i_{4},\ldots,i_{n-1},j_{1},j_{2}\right)→⋯\displaystyle\rightarrow\cdots→(in−1,j1,j2,…,jn−2)\displaystyle\rightarrow\left(i_{n-1},j_{1},j_{2},\ldots,j_{n-2}\right)→(j1,j2,…,jn−1)=j.\displaystyle\rightarrow\left(j_{1},j_{2},\ldots,j_{n-1}\right)=j.Note that this walk has lengthn−1n-1, and is the unique walk fromiitojjthat has lengthn−1n-1. Thus, the # of walks fromiitojjthat have lengthn−1n-1is11. This will come useful further below.]
•Thus, the multidigraphDDis weakly connected (since any strongly connected digraph is weakly connected).
•The multidigraphDDis balanced, and in fact each vertex ofDDhas outdegreekkand indegreekk. [Proof:Letiibe a vertex ofDD. The arcs with sourceiiare thenn-tuples whose firstn−1n-1entries form the(n−1)\left(n-1\right)-tupleiiwhile the last,nn-th entry is an arbitrary element ofKK. Thus, there are|K|\left|K\right|many such arcs. In other words,iihas outdegreekk. A similar argument shows thatiihas indegreekk. This entails thatdeg−⁡i=deg+⁡i\deg^{-}i=\deg^{+}i. Since this holds for every vertexii, we conclude thatDDis balanced.]
•The digraphDDhas an Eulerian circuit. [Proof:This follows from the directed Euler–Hierholzer theorem (Theorem4.69), sinceDDis weakly connected and balanced. Alternatively, we can derive this from the BEST theorem (Theorem5.117) as follows: Pick an arbitrary arcaaofDD, and letrrbe its source. Then,rris a from-root ofDD(sinceDDis strongly connected), and thusDDhas a spanning arborescence rooted fromrr(by Theorem5.112). In other words, using the notations of the BEST theorem (Theorem5.117), we haveτ⁡(D,r)≠0\tau\left(D,r\right)\neq 0. Moreover, each vertex ofDDhas indegreek>0k>0. Thus, the BEST theorem yields ε⁡(D,a)=τ⁡(D,r)⏟≠0⋅∏u∈V(deg−⁡u−1)!⏟≠0≠0.\varepsilon\left(D,a\right)=\underbrace{\tau\left(D,r\right)}_{\neq 0}\cdot\underbrace{\prod\limits_{u\in V}\left(\deg^{-}u-1\right)!}_{\neq 0}\neq 0.But this shows thatDDhas an Eulerian circuit whose last arc isaa.]

So we know thatDDhas an Eulerian circuit𝐜\mathbf{c}. This Eulerian circuit leads to a de Bruijn sequence as follows:

Letp0,p1,…,pkn−1p_{0},p_{1},\ldots,p_{k^{n}-1}be the arcs of𝐜\mathbf{c}(from first to last). Extend the subscripts periodically moduloknk^{n}(that is, setpq+kn=pqp_{q+k^{n}}=p_{q}for allq∈ℕq\in\mathbb{N}). Thus, we obtain an infinite walk777777We have never formally defined infinite walks, but it should be fairly clear what they are.with arcsp0,p1,p2,…p_{0},p_{1},p_{2},\ldots(since𝐜\mathbf{c}is a circuit). In other words, for eachi∈ℕi\in\mathbb{N}, the target of the arcpip_{i}is the source of the arcpi+1p_{i+1}.

In other words, for eachi∈ℕi\in\mathbb{N}, the lastn−1n-1entries ofpip_{i}are the firstn−1n-1entries ofpi+1p_{i+1}(since the target ofpip_{i}is the tuple consisting of the lastn−1n-1entries ofpip_{i}, whereas the source ofpi+1p_{i+1}is the tuple consisting of the firstn−1n-1entries ofpi+1p_{i+1}). Therefore, for eachi∈ℕi\in\mathbb{N}and eachj∈{2,3,…,n}j\in\left\{2,3,\ldots,n\right\}, we have

(thej-th entry ofpi)\displaystyle\left(\text{the }j\text{-th entry of }p_{i}\right)=(the(j−1)-st entry ofpi+1).\displaystyle=\left(\text{the }\left(j-1\right)\text{-st entry of }p_{i+1}\right).(34) Now, for eachi∈ℕi\in\mathbb{N}, we letxix_{i}denote the first entry of thenn-tuplepip_{i}. Then,xq+kn=xqx_{q+k^{n}}=x_{q}for allq∈ℕq\in\mathbb{N}(sincepq+kn=pqp_{q+k^{n}}=p_{q}for allq∈ℕq\in\mathbb{N}). In other words, the sequence(x0,x1,x2,…)\left(x_{0},x_{1},x_{2},\ldots\right)repeats itself everyknk^{n}terms. Note that theknk^{n}-tuple(x0,x1,…,xkn−1)\left(x_{0},x_{1},\ldots,x_{k^{n}-1}\right)consists of the first entries of the arcsp0,p1,…,pkn−1p_{0},p_{1},\ldots,p_{k^{n}-1}of𝐜\mathbf{c}(by the definition ofxix_{i}).

For eachi∈ℕi\in\mathbb{N}and eachs∈{1,2,…,n}s\in\left\{1,2,\ldots,n\right\}, we have

(thes-th entry ofpi)\displaystyle\left(\text{the }s\text{-th entry of }p_{i}\right)=(the(s−1)-st entry ofpi+1)(by (34))\displaystyle=\left(\text{the }\left(s-1\right)\text{-st entry of }p_{i+1}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.debr.exist.j-1})}\right)=(the(s−2)-nd entry ofpi+2)(by (34))\displaystyle=\left(\text{the }\left(s-2\right)\text{-nd entry of }p_{i+2}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.debr.exist.j-1})}\right)=(the(s−3)-rd entry ofpi+3)(by (34))\displaystyle=\left(\text{the }\left(s-3\right)\text{-rd entry of }p_{i+3}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.debr.exist.j-1})}\right)=⋯\displaystyle=\cdots=(the1-st entry ofpi+s−1)\displaystyle=\left(\text{the }1\text{-st entry of }p_{i+s-1}\right)=xi+s−1(sincexi+s−1was defined as the first entry ofpi+s−1).\displaystyle=x_{i+s-1}\ \ \ \ \ \ \ \ \ \ \left(\text{since }x_{i+s-1}\text{ was defined as the first entry of }p_{i+s-1}\right).In other words, for eachi∈ℕi\in\mathbb{N}, the entries ofpip_{i}(from first to last) are xi,xi+1,…,xi+n−1x_{i},x_{i+1},\ldots,x_{i+n-1}. In other words, for eachi∈ℕi\in\mathbb{N}, we have

pi=(xi,xi+1,…,xi+n−1).p_{i}=\left(x_{i},x_{i+1},\ldots,x_{i+n-1}\right).(35) Now, recall that𝐜\mathbf{c}is an Eulerian circuit. Thus, each arc ofDDappears exactly once among its arcsp0,p1,…,pkn−1p_{0},p_{1},\ldots,p_{k^{n}-1}. In other words, eachnn-tuple inKnK^{n}appears exactly once amongp0,p1,…,pkn−1p_{0},p_{1},\ldots,p_{k^{n}-1}(since the arcs ofDDare thenn-tuples inKnK^{n}). In other words, asiiranges from00tokn−1k^{n}-1, thenn-tuplepip_{i}takes each possible value inKnK^{n}exactly once.

In view of (35), we can rewrite this as follows: Asiiranges from00tokn−1k^{n}-1, thenn-tuple(xi,xi+1,…,xi+n−1)\left(x_{i},x_{i+1},\ldots,x_{i+n-1}\right)takes each possible value inKnK^{n}exactly once (since thisnn-tuple is preciselypip_{i}, as we have shown in the previous paragraph). In other words, for each(a1,a2,…,an)∈Kn\left(a_{1},a_{2},\ldots,a_{n}\right)\in K^{n}, there is auniquer∈{0,1,…,kn−1}r\in\left\{0,1,\ldots,k^{n}-1\right\}such that(a1,a2,…,an)=(xr,xr+1,…,xr+n−1)\left(a_{1},a_{2},\ldots,a_{n}\right)=\left(x_{r},x_{r+1},\ldots,x_{r+n-1}\right).

Hence, theknk^{n}-tuple(x0,x1,…,xkn−1)\left(x_{0},x_{1},\ldots,x_{k^{n}-1}\right)is a de Bruijn sequence of ordernnonKK. This shows that a de Bruijn sequence exists. Theorem5.180is thus proven.

Example 5.183.

Forn=3n=3andk=2k=2andK={0,1}K=\left\{0,1\right\}, one possible Eulerian circuit𝐜\mathbf{c}ofDDis

(00,001,01,010,10,101,01,011,11,111,11,110,10,100,00)\left(00,\ \mathbf{001},\ 01,\ \mathbf{010},\ 10,\ \mathbf{101},\ 01,\ \mathbf{011},\ 11,\ \mathbf{111},\ 11,\ \mathbf{110},\ 10,\ \mathbf{100},\ 00\right)(where we have written the arcs in bold for readability). The first entries of the arcs of this circuit form the sequence00101110010111, which is indeed a de Bruijn sequence of order33on{0,1}\left\{0,1\right\}. Any33consecutive entries of this sequence (extended periodically to the infinite sequence||:0010111:||||\vphantom{a}:0010111:||) form the respective arc of𝐜\mathbf{c}.

Theorem5.180is merely the starting point of a theory. Several specific de Bruijn sequences are known, many of them having peculiar properties. See[Freder82]for a survey of various such sequences787878Some of these sequences (the “prefer-one” and “prefer-opposite” generators) are just disguised implementations of the algorithm for finding an Eulerian circuit implicit in our proof of the BEST theorem.(note that they are called “full length nonlinear shift register sequences” in this survey).797979My favorite is the one obtained by concatenating all Lyndon words whose length dividesnnin lexicographically increasing order (assuming that the setKKis totally ordered). See[Moreno04]for the details of that construction.For the (somewhat confusing) history of de Bruijn sequences (and of Theorem5.180and of the Theorem5.186to be stated later), see[deBrui75].

There are also several variations on de Bruijn sequences. For some of them, see[ChDiGr92]. (Note that some of the open questions in that paper are still unsolved.) A variation that recently became quite popular is the notion of a “universal cycle for permutations” – a string that contains all “permutations” (more precisely,nn-tuples of distinct elements ofKK) as factors. See[EngVat18]for some recent progress on minimizing the length of such a string, including a contribution by a notorious hacker known as 4chan. (This is no longer really about Eulerian circuits, since some amount of duplication cannot be avoided in these strings.)

5.16.3Counting de Bruijn sequences

Let us move in a different direction. Having proved the existence of de Bruijn sequences in Theorem5.180, let us try to count them!

Question.Letnnandkkbe two positive integers. LetKKbe akk-element set. How many de Bruijn sequences of ordernnonKKare there?

To solve this, it makes sense to apply the BEST theorem to the digraphDDwe have constructed above. Alas,DDis not of the formGbidirG^{\operatorname*{bidir}}for some undirected graphGG, so we cannot apply the undirected MTT (Matrix-Tree Theorem). However,DDis a balanced multidigraph, and for such digraphs, a version of the undirected MTT still holds:

Theorem 5.184(balanced Matrix-Tree Theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a balanced multidigraph. Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for some positive integernn.

LetLLbe the Laplacian ofDD. Then:

(a)For any vertexrrofDD, we have (# of spanning arborescences ofDrooted tor)=det(L∼r,∼r).\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=\det\left(L_{\sim r,\sim r}\right).Moreover, this number does not depend onrr.
(b)Letttbe an indeterminate. Expand the determinantdet(tIn+L)\det\left(tI_{n}+L\right)(here,InI_{n}denotes then×nn\times nidentity matrix) as a polynomial intt: det(tIn+L)=cntn+cn−1tn−1+⋯+c1t1+c0t0,\det\left(tI_{n}+L\right)=c_{n}t^{n}+c_{n-1}t^{n-1}+\cdots+c_{1}t^{1}+c_{0}t^{0},wherec0,c1,…,cnc_{0},c_{1},\ldots,c_{n}are numbers. (Note that this is the characteristic polynomial ofLLup to substituting−t-tforttand multiplying by a power of−1-1. Some of its coefficients arecn=1c_{n}=1andcn−1=TrLc_{n-1}=\operatorname*{Tr}Landc0=detLc_{0}=\det L.) Then, for any vertexrrofDD, we have (# of spanning arborescences ofDrooted tor)=1nc1.\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=\dfrac{1}{n}c_{1}.
(c)Letλ1,λ2,…,λn\lambda_{1},\lambda_{2},\ldots,\lambda_{n}be the eigenvalues ofLL, listed in such a way thatλn=0\lambda_{n}=0. Then, for any vertexrrofDD, we have (# of spanning arborescences ofDrooted tor)=1n⋅λ1λ2⋯λn−1.\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}.
(d)Letλ1,λ2,…,λn\lambda_{1},\lambda_{2},\ldots,\lambda_{n}be the eigenvalues ofLL, listed in such a way thatλn=0\lambda_{n}=0. If all vertices ofDDhave outdegree>0>0, then (# of Eulerian circuits ofD)=|A|⋅1n⋅λ1λ2⋯λn−1⋅∏u∈V(deg+u−1)!.\left(\text{\# of Eulerian circuits of }D\right)=\left|A\right|\cdot\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!.(If you identify an Eulerian circuit with its cyclic rotations, then you should drop the|A|\left|A\right|factor on the right hand side.)

Proof 5.185.

(a)The equality comes from the MTT (Theorem5.142). It remains to prove that the # of spanning arborescences ofDDrooted torrdoes not depend onrr. But this is Corollary5.130.

(b)follows from(a)as in the undirected graph case (proof of Theorem5.167(b)).808080In more detail: Just as we proved in our above proof of Theorem5.167(for the undirected case), we havec1=∑r=1ndet(L∼r,∼r)c_{1}=\sum\limits_{r=1}^{n}\det\left(L_{\sim r,\sim r}\right). However, part(a)shows that the numberdet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)does not depend onrr. Thus, the sum∑r=1ndet(L∼r,∼r)\sum\limits_{r=1}^{n}\det\left(L_{\sim r,\sim r}\right)consists ofnnequal addends, each of which can be written asdet(L∼r,∼r)\det\left(L_{\sim r,\sim r}\right)for any vertexrrofDD. Therefore, this sum can be rewritten asn⋅det(L∼r,∼r)n\cdot\det\left(L_{\sim r,\sim r}\right)for any vertexrrofDD. Hence, the equalityc1=∑r=1ndet(L∼r,∼r)c_{1}=\sum\limits_{r=1}^{n}\det\left(L_{\sim r,\sim r}\right)can be rewritten asc1=n⋅det(L∼r,∼r)c_{1}=n\cdot\det\left(L_{\sim r,\sim r}\right)for any vertexrrofDD. Therefore,det(L∼r,∼r)=1nc1\det\left(L_{\sim r,\sim r}\right)=\dfrac{1}{n}c_{1}for any vertexrrofDD. But part(a)yields(# of spanning arborescences ofDrooted tor)=det(L∼r,∼r)=1nc1.\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)=\det\left(L_{\sim r,\sim r}\right)=\dfrac{1}{n}c_{1}.

(c)follows from(b)as in the undirected graph case (proof of Theorem5.167(c)).

(d)Assume that all vertices ofDDhave outdegree>0>0. Then,

(# of Eulerian circuits ofD)\displaystyle\left(\text{\# of Eulerian circuits of }D\right)=∑a∈A(# of Eulerian circuits ofDwhose first arc isa).\displaystyle=\sum\limits_{a\in A}\left(\text{\# of Eulerian circuits of }D\text{ whose first arc is }a\right).However, ifa∈Aa\in Ais any arc, and ifrris the source ofaa, then

(# of Eulerian circuits ofDwhose first arc isa)\displaystyle\left(\text{\# of Eulerian circuits of }D\text{ whose first arc is }a\right)=(# of spanning arborescences ofDrooted tor)⋅∏u∈V(deg+u−1)!\displaystyle=\left(\text{\# of spanning arborescences of }D\text{ rooted to }r\right)\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!(by the BEST’ theorem (Theorem5.122))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{by the BEST’ theorem (Theorem \ref{thm.BEST.to})}\right)=1n⋅λ1λ2⋯λn−1⋅∏u∈V(deg+u−1)!(by part(c)).\displaystyle=\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!\ \ \ \ \ \ \ \ \ \ \left(\text{by part {(c)}}\right).Hence,

(# of Eulerian circuits ofD)\displaystyle\left(\text{\# of Eulerian circuits of }D\right)=∑a∈A(# of Eulerian circuits ofDwhose first arc isa)⏟=1n⋅λ1λ2⋯λn−1⋅∏u∈V(deg+u−1)!\displaystyle=\sum\limits_{a\in A}\underbrace{\left(\text{\# of Eulerian circuits of }D\text{ whose first arc is }a\right)}_{=\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!}=∑a∈A1n⋅λ1λ2⋯λn−1⋅∏u∈V(deg+u−1)!\displaystyle=\sum\limits_{a\in A}\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!=|A|⋅1n⋅λ1λ2⋯λn−1⋅∏u∈V(deg+u−1)!.\displaystyle=\left|A\right|\cdot\dfrac{1}{n}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{n-1}\cdot\prod\limits_{u\in V}\left(\deg^{+}u-1\right)!.This proves part(d).

Now, let’s try to solve our question – i.e., let’s count the de Bruijn sequences of ordernnonKK.

Recall the digraphDDfrom our above proof of Theorem5.180. We constructed a de Bruijn sequence of ordernnonKKby finding an Eulerian circuit ofDD. This actually works both ways: The map

{Eulerian circuits ofD}\displaystyle\left\{\text{Eulerian circuits of }D\right\}→{de Bruijn sequences of ordernonK},\displaystyle\rightarrow\left\{\text{de Bruijn sequences of order }n\text{ on }K\right\},𝐜\displaystyle\mathbf{c}↦(the sequence of first entries of the arcs of𝐜)\displaystyle\mapsto\left(\text{the sequence of first entries of the arcs of }\mathbf{c}\right)is a bijection (make sure you understand why!). Hence, by the bijection principle, we have

(# of de Bruijn sequences of ordernonK)\displaystyle\left(\text{\# of de Bruijn sequences of order }n\text{ on }K\right)=(# of Eulerian circuits ofD).\displaystyle=\left(\text{\# of Eulerian circuits of }D\right).(36)By Theorem5.184(d), however, we have

(# of Eulerian circuits ofD)\displaystyle\left(\text{\# of Eulerian circuits of }D\right)=|Kn|⋅1kn−1⋅λ1λ2⋯λkn−1−1⋅∏u∈Kn−1(deg+u−1)!,\displaystyle=\left|K^{n}\right|\cdot\dfrac{1}{k^{n-1}}\cdot\lambda_{1}\lambda_{2}\cdots\lambda_{k^{n-1}-1}\cdot\prod\limits_{u\in K^{n-1}}\left(\deg^{+}u-1\right)!,(37)whereλ1,λ2,…,λkn−1\lambda_{1},\lambda_{2},\ldots,\lambda_{k^{n-1}}are the eigenvalues of the LaplacianLLofDD, indexed in such a way thatλkn−1=0\lambda_{k^{n-1}}=0. (Note that the digraphD=(Kn−1,Kn,ψ)D=\left(K^{n-1},K^{n},\psi\right)haskn−1k^{n-1}vertices, notnnvertices, so the “nn” in Theorem5.184iskn−1k^{n-1}here.)

As we know, each vertex ofDDhas outdegreekk. That is, we havedeg+⁡u=k\deg^{+}u=kfor eachu∈Kn−1u\in K^{n-1}. Thus,

∏u∈Kn−1(deg+⁡u−1)!=∏u∈Kn−1(k−1)!=((k−1)!)kn−1.\prod\limits_{u\in K^{n-1}}\left(\deg^{+}u-1\right)!=\prod\limits_{u\in K^{n-1}}\left(k-1\right)!=\left(\left(k-1\right)!\right)^{k^{n-1}}.Also,

|Kn|⋅1kn−1=kn⋅1kn−1=k.\left|K^{n}\right|\cdot\dfrac{1}{k^{n-1}}=k^{n}\cdot\dfrac{1}{k^{n-1}}=k.It remains to findλ1λ2⋯λkn−1−1\lambda_{1}\lambda_{2}\cdots\lambda_{k^{n-1}-1}. What are the eigenvalues ofLL?

The LaplacianLLof our digraphDDis akn−1×kn−1k^{n-1}\times k^{n-1}-matrix whose rows and columns are indexed by(n−1)\left(n-1\right)-tuples inKn−1K^{n-1}. Strictly speaking, we should relabel the vertices ofDDas1,2,…,kn−11,2,\ldots,k^{n-1}here, in order to have a “proper matrix” with a well-defined order on its rows and columns. But let’s not do this; instead, I trust you can do the relabeling yourself, or just use the more general notion of matrices that allows for the rows and the columns to be indexed by arbitrary things (seehttps://mathoverflow.net/questions/317105for details).

LetCCbe the adjacency matrix of the digraphDD; this is thekn−1×kn−1k^{n-1}\times k^{n-1}-matrix (again with rows and columns indexed by(n−1)\left(n-1\right)-tuples inKn−1K^{n-1}) whose(i,j)\left(i,j\right)-th entry is the # of arcs with sourceiiand targetjj. In particular, the trace ofCCis thus the # of loops ofDD. It is easy to see that the loops ofDDare precisely the arcs of the form(x,x,…,x)∈Kn\left(x,x,\ldots,x\right)\in K^{n}forx∈Kx\in K; thus,DDhas exactlykkloops. Hence, the trace ofCCiskk.

Recall the definition of the Laplacian matrixLL. We can restate it as follows:

whereΔ\Deltais the diagonal matrix whose diagonal entries are the outdegrees of the vertices ofDD. Since each vertex ofDDhas outdegreekk, the latter diagonal matrixΔ\Deltais simplyk⋅Ik\cdot I, whereIIis the identity matrix (of the appropriate size). Hence, (38) can be rewritten as

Thus, ifγ1,γ2,…,γkn−1\gamma_{1},\gamma_{2},\ldots,\gamma_{k^{n-1}}are the eigenvalues ofCC, thenk−γ1,k−γ2,…,k−γkn−1k-\gamma_{1},k-\gamma_{2},\ldots,k-\gamma_{k^{n-1}}are the eigenvalues ofLL. Computing the former will thus help us find the latter.

Furthermore, letJJbe thekn−1×kn−1k^{n-1}\times k^{n-1}-matrix (again with rows and columns indexed by(n−1)\left(n-1\right)-tuples inKn−1K^{n-1}) whose all entries are11. It is easy to see that the eigenvalues ofJJare

0,0,…,0⏟kn−1−1many zeroes,kn−1.\underbrace{0,0,\ldots,0}_{k^{n-1}-1\text{ many zeroes}},k^{n-1}.(The easiest way to see this is by noticing thatJJhas rank11and tracekn−1k^{n-1}.818181Here are the details: The matrixJJhas rank11(since all its rows are the same); thus, all but one of its eigenvalues are00. It remains to show that the remaining eigenvalue iskn−1k^{n-1}. However, it is known that the sum of the eigenvalues of a square matrix equals its trace. Thus, if all but one of the eigenvalues of a square matrix are00, then the remaining eigenvalue equals its trace. Applying this to our matrixJJ, we see that its remaining eigenvalue equals its trace, which iskn−1k^{n-1}.)

Now, here is something really underhanded: We observe that

[Proof:We need to show that all entries of the matrixCn−1C^{n-1}are11. So letiiandjjbe two vertices ofDD. We must then show that the(i,j)\left(i,j\right)-th entry ofCn−1C^{n-1}is11.

Recall the combinatorial interpretation of the powers of an adjacency matrix (Theorem4.38): For anyℓ∈ℕ\ell\in\mathbb{N}, the(i,j)\left(i,j\right)-th entry ofCℓC^{\ell}is the # of walks fromiitojj(inDD) that have lengthℓ\ell. Thus, in particular, the(i,j)\left(i,j\right)-th entry ofCn−1C^{n-1}is the # of walks fromiitojj(inDD) that have lengthn−1n-1. But this number is actually11, as we have already shown in our above proof of Theorem5.180. This completes the proof ofCn−1=JC^{n-1}=J.]

How does this help us compute the eigenvalues ofCC? Well, letγ1,γ2,…,γkn−1\gamma_{1},\gamma_{2},\ldots,\gamma_{k^{n-1}}be the eigenvalues ofCC. Then, for anyℓ∈ℕ\ell\in\mathbb{N}, the eigenvalues ofCℓC^{\ell}areγ1ℓ,γ2ℓ,…,γkn−1ℓ\gamma_{1}^{\ell},\gamma_{2}^{\ell},\ldots,\gamma_{k^{n-1}}^{\ell}(this is a fact that holds for any square matrix, and is probably easiest to prove using the Jordan canonical form or triangularization828282In fact this is a particular case of the Spectral Mapping Theorem ([Treil17, Chapter 9, Theorem 2.1]).). Hence, in particular,γ1n−1,γ2n−1,…,γkn−1n−1\gamma_{1}^{n-1},\gamma_{2}^{n-1},\ldots,\gamma_{k^{n-1}}^{n-1}are the eigenvalues ofCn−1=JC^{n-1}=J; but we know that the latter eigenvalues are0,0,…,0⏟kn−1−1many zeroes,kn−1\underbrace{0,0,\ldots,0}_{k^{n-1}-1\text{ many zeroes}},k^{n-1}. Hence, all but one of thekn−1k^{n-1}numbersγ1n−1,γ2n−1,…,γkn−1n−1\gamma_{1}^{n-1},\gamma_{2}^{n-1},\ldots,\gamma_{k^{n-1}}^{n-1}equal00. Thus, all but one of thekn−1k^{n-1}numbersγ1,γ2,…,γkn−1\gamma_{1},\gamma_{2},\ldots,\gamma_{k^{n-1}}equal00(we don’t know what the remaining number is, since(n−1)\left(n-1\right)-st roots are not uniquely determined inℂ\mathbb{C}). In other words, all but one of the eigenvalues ofCCequal00. The remaining eigenvalue must thus be the trace ofCC(because the sum of the eigenvalues of a square matrix is known to be the trace of that matrix), and therefore equalkk(since we know that the trace ofCCiskk).

So we have shown that the eigenvalues ofCCare0,0,…,0⏟kn−1−1many zeroes,k\underbrace{0,0,\ldots,0}_{k^{n-1}-1\text{ many zeroes}},k. Thus, the eigenvalues ofLLare

k−0,k−0,…,k−0⏟kn−1−1many(k−0)’s,k−k\underbrace{k-0,\ \ k-0,\ \ \ldots,\ \ k-0}_{k^{n-1}-1\text{ many }\left(k-0\right)\text{’s}},\ \ k-k(because ifγ1,γ2,…,γkn−1\gamma_{1},\gamma_{2},\ldots,\gamma_{k^{n-1}}are the eigenvalues ofCC, thenk−γ1,k−γ2,…,k−γkn−1k-\gamma_{1},k-\gamma_{2},\ldots,k-\gamma_{k^{n-1}}are the eigenvalues ofLL). In other words, the eigenvalues ofLLare

k,k,…,k⏟kn−1−1manyk’s,0.\underbrace{k,\ \ k,\ \ \ldots,\ \ k}_{k^{n-1}-1\text{ many }k\text{’s}},\ \ 0.Hence, the eigenvaluesλ1,λ2,…,λkn−1−1\lambda_{1},\lambda_{2},\ldots,\lambda_{k^{n-1}-1}in (37) all equalkk. Thus, (37) simplifies to

(# of Eulerian circuits ofD)\displaystyle\left(\text{\# of Eulerian circuits of }D\right)=|Kn|⋅1kn−1⏟=kn⋅1kn−1=k⋅kk⋯k⏟kn−1−1factors⋅∏u∈Kn−1(deg+⁡u−1)!⏟=((k−1)!)kn−1\displaystyle=\underbrace{\left|K^{n}\right|\cdot\dfrac{1}{k^{n-1}}}_{\begin{subarray}{c}=k^{n}\cdot\dfrac{1}{k^{n-1}}\\ =k\end{subarray}}\cdot\underbrace{kk\cdots k}_{k^{n-1}-1\text{ factors}}\cdot\underbrace{\prod\limits_{u\in K^{n-1}}\left(\deg^{+}u-1\right)!}_{=\left(\left(k-1\right)!\right)^{k^{n-1}}}=k⋅kk⋯k⏟kn−1−1factors⏟=kkn−1⋅((k−1)!)kn−1=kkn−1⋅((k−1)!)kn−1\displaystyle=\underbrace{k\cdot\underbrace{kk\cdots k}_{k^{n-1}-1\text{ factors}}}_{=k^{k^{n-1}}}\cdot\left(\left(k-1\right)!\right)^{k^{n-1}}=k^{k^{n-1}}\cdot\left(\left(k-1\right)!\right)^{k^{n-1}}=(k⋅(k−1)!⏟=k!)kn−1=k!kn−1.\displaystyle=\left(\underbrace{k\cdot\left(k-1\right)!}_{=k!}\right)^{k^{n-1}}=k!^{k^{n-1}}.In view of this, we can rewrite (36) as

(# of de Bruijn sequences of ordernonK)=k!kn−1.\left(\text{\# of de Bruijn sequences of order }n\text{ on }K\right)=k!^{k^{n-1}}.Thus, we have proved the following:

Theorem 5.186.

Letnnandkkbe positive integers. LetKKbe akk-element set. Then,

(# of de Bruijn sequences of ordernonK)=k!kn−1.\left(\text{\# of de Bruijn sequences of order }n\text{ on }K\right)=k!^{k^{n-1}}.

What a nice (and huge) answer!

Our above proof of Theorem5.186is essentially taken from[Stanle18, Chapter 10].

We note that a combinatorial proof of Theorem5.186(avoiding any use of linear algebra) has been recently given in[BidKis02].

5.17More on Laplacians

Much more can be said about the Laplacian of a digraph. The study of matrices associated to a graph or digraph is known asspectral graph theory; I’d say the Laplacian is probably the most prominent of these matrices (even though the adjacency matrix is somewhat easier to define). The original form of the matrix-tree theorem (actually a subtler variant of Theorem5.167(a)) was found by Gustav Kirchhoff in his study of electricity[Kirchh47](see[Holzer22, §2.1.1]for a modern exposition); the effective resistance between two nodes of an electrical network is a ratio of spanning-tree counts and thus can be computed using the Laplacian (see, e.g.,[Vos16, §2 and §3]). To be more precise, this relies on a “weighted count” of spanning trees, which is more general than the counting we have done so far; we will learn about it in the next section.

Another application of Laplacians is to drawing graphs: see “spectral layout” or “spectral graph drawing” (e.g.,[Gallie13]).

5.18On the left nullspace of the Laplacian

We shall now answer another natural question about Laplacians of digraphs. Recall that the LaplacianLLof a digraphDDalways satisfiesLe=0Le=0, wheree=(11⋮1)e=\left(\begin{array}[c]{c}1\\ 1\\ \vdots\\ 1\end{array}\right). Thus, the vectoreebelongs to the right nullspace (= right kernel) ofLL. It is not hard to see that ifDDhas a to-root and we are working over a characteristic-00field, theneespans this nullspace, i.e., there are no vectors in that nullspace other than scalar multiples ofee. (This is actually an “if and only if”.) What about the left nullspace ofLL? Can we explicitly find a nonzero vectorffwithfL=0fL=0? The answer is positive:

Theorem 5.187(harmonic vector theorem for Laplacians).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}.

For eachr∈Vr\in V, letτ⁡(D,r)\tau\left(D,r\right)be the # of spanning arborescences ofDDrooted torr.

Letffbe the row vector(τ⁡(D,1),τ⁡(D,2),…,τ⁡(D,n))\left(\tau\left(D,1\right),\ \tau\left(D,2\right),\ \ldots,\ \tau\left(D,n\right)\right). Then,fL=0fL=0.

Before we prove this, some remarks are in order. Theorem5.187(or, more precisely, its weighted version, which we will see in the next section) can be used to explicitly compute the steady state of a Markov chain (see[KrGrWi10]and also Corollary5.207below); a similar interpretation, but in economical terms (emergence of money in a barter economy), appears in[Sahi14, §1].

We shall give a proof of Theorem5.187based upon two lemmas. The first lemma is a general linear-algebraic result:

Lemma 5.189.

LetBBbe ann×nn\times n-matrix over an arbitrary commutative ring𝕂\mathbb{K}. (For example,𝕂\mathbb{K}can beℝ\mathbb{R}, in which caseBBis a real matrix.) Assume that the sum of all columns ofBBis the zero vector. Then, for anyr,s,t∈{1,2,…,n}r,s,t\in\left\{1,2,\ldots,n\right\}, we have

det(B∼r,∼t)=(−1)s−tdet(B∼r,∼s).\det\left(B_{\sim r,\sim t}\right)=\left(-1\right)^{s-t}\det\left(B_{\sim r,\sim s}\right).

Proof 5.190(Proof of Lemma5.189.).

There are various ways to prove this, but here is probably the most elegant one:

We WLOG assume thats≠ts\neq t, since otherwise the claim is obvious. Let us now change therr-th row of the matrixBBas follows:

•We replace thess-th entry of therr-th row by11.
•We replace thett-th entry of therr-th row by−1-1.
•We replace all other entries of therr-th row by00.

LetCCbe the resultingn×nn\times n-matrix.838383For example, ifn=4n=4andB=(abcda′b′c′d′a′′b′′c′′d′′a′′′b′′′c′′′d′′′)B=\left(\begin{array}[c]{cccc}a&b&c&d\\ a^{\prime}&b^{\prime}&c^{\prime}&d^{\prime}\\ a^{\prime\prime}&b^{\prime\prime}&c^{\prime\prime}&d^{\prime\prime}\\ a^{\prime\prime\prime}&b^{\prime\prime\prime}&c^{\prime\prime\prime}&d^{\prime\prime\prime}\end{array}\right)ands=1s=1andt=3t=3andr=2r=2, thenC=(abcd10−10a′′b′′c′′d′′a′′′b′′′c′′′d′′′)C=\left(\begin{array}[c]{cccc}a&b&c&d\\ 1&0&-1&0\\ a^{\prime\prime}&b^{\prime\prime}&c^{\prime\prime}&d^{\prime\prime}\\ a^{\prime\prime\prime}&b^{\prime\prime\prime}&c^{\prime\prime\prime}&d^{\prime\prime\prime}\end{array}\right).Thus,CCagrees withBBin all rows other than therr-th one. Hence, in particular,

C∼r,∼k=B∼r,∼kfor eachk∈{1,2,…,n}.C_{\sim r,\sim k}=B_{\sim r,\sim k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{1,2,\ldots,n\right\}.(39)Note also that the only nonzero entries in therr-th row ofCCare848484We are using the notationCr,kC_{r,k}for the entry ofCCin therr-th row and thekk-th column.Cr,s=1C_{r,s}=1andCr,t=−1C_{r,t}=-1. Hence, the entries in therr-th row ofCCadd up to00.

Recall that the sum of all columns ofBBis the zero vector. In other words, in each row ofBB, the entries add up to00. The matrixCCtherefore also has this property (because the only row ofCCthat differs from the corresponding row ofBBis therr-th row; however, we have shown above that in therr-th row, the entries ofCCalso add up to00). In other words, the sum of all columns ofCCis the zero vector. This easily entails thatdetC=0\det C=0858585Proof.It is well-known that the determinant of a matrix does not change if we add a column to another. Hence, the determinant ofCCwill not change if we add each column ofCCother than the first one to the first column ofCC. However, the result of this operation will be a matrix whose first column is00(since the sum of all columns ofCCis the zero vector), and therefore this matrix will have determinant00. Since the operation did not change the determinant, we thus conclude that the determinant ofCCwas00. In other words,detC=0\det C=0..

On the other hand, Laplace expansion along therr-th row yields

detC\displaystyle\det C=∑k=1n(−1)r+kCr,kdet(C∼r,∼k)\displaystyle=\sum\limits_{k=1}^{n}\left(-1\right)^{r+k}C_{r,k}\det\left(C_{\sim r,\sim k}\right)=(−1)r+s1det(C∼r,∼s)+(−1)r+t(−1)det(C∼r,∼t)\displaystyle=\left(-1\right)^{r+s}1\det\left(C_{\sim r,\sim s}\right)+\left(-1\right)^{r+t}\left(-1\right)\det\left(C_{\sim r,\sim t}\right)(since the only nonzero entriesCr,kC_{r,k}in therr-th row ofCCareCr,s=1C_{r,s}=1andCr,t=−1C_{r,t}=-1). Comparing this withdetC=0\det C=0, we obtain

0\displaystyle 0=(−1)r+s1det(C∼r,∼s)+(−1)r+t(−1)det(C∼r,∼t)\displaystyle=\left(-1\right)^{r+s}1\det\left(C_{\sim r,\sim s}\right)+\left(-1\right)^{r+t}\left(-1\right)\det\left(C_{\sim r,\sim t}\right)=(−1)r+sdet(C∼r,∼s)⏟=B∼r,∼s(by (39))−(−1)r+tdet(C∼r,∼t)⏟=B∼r,∼t(by (39))\displaystyle=\left(-1\right)^{r+s}\det\underbrace{\left(C_{\sim r,\sim s}\right)}_{\begin{subarray}{c}=B_{\sim r,\sim s}\\ \text{(by (\ref{pf.lem.sol.7.1.matrix.2}))}\end{subarray}}-\left(-1\right)^{r+t}\det\underbrace{\left(C_{\sim r,\sim t}\right)}_{\begin{subarray}{c}=B_{\sim r,\sim t}\\ \text{(by (\ref{pf.lem.sol.7.1.matrix.2}))}\end{subarray}}=(−1)r+sdet(B∼r,∼s)−(−1)r+tdet(B∼r,∼t).\displaystyle=\left(-1\right)^{r+s}\det\left(B_{\sim r,\sim s}\right)-\left(-1\right)^{r+t}\det\left(B_{\sim r,\sim t}\right).In other words,(−1)r+tdet(B∼r,∼t)=(−1)r+sdet(B∼r,∼s)\left(-1\right)^{r+t}\det\left(B_{\sim r,\sim t}\right)=\left(-1\right)^{r+s}\det\left(B_{\sim r,\sim s}\right). Dividing both sides of this by(−1)r+t\left(-1\right)^{r+t}, we obtaindet(B∼r,∼t)=(−1)s−tdet(B∼r,∼s)\det\left(B_{\sim r,\sim t}\right)=\left(-1\right)^{s-t}\det\left(B_{\sim r,\sim s}\right). This proves Lemma5.189.

Our next lemma is the following generalization of Theorem5.142:

Theorem 5.191(Matrix-Tree Theorem, off-diagonal version).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for some positive integernn.

LetLLbe the Laplacian ofDD. Letrrandssbe two vertices ofDD. Then,

(#of spanning arborescences ofDrooted tor)=(−1)r+sdet(L∼r,∼s).\left(\#\text{ of spanning arborescences of }D\text{ rooted to }r\right)=\left(-1\right)^{r+s}\det\left(L_{\sim r,\sim s}\right).

Note that Theorem5.142is the particular case of Theorem5.191fors=rs=r. Fortunately, using Lemma5.189, we can easily derive the general case from the particular:

Proof 5.192(Proof of Theorem5.191.).

We have seen (in the proof of Proposition5.140) that the sum of all columns of the LaplacianLLis the zero vector. Hence, Lemma5.189(applied to𝕂=ℚ\mathbb{K}=\mathbb{Q}andB=LB=Landt=rt=r) yields

det(L∼r,∼r)=(−1)s−r⏟=(−1)r+sdet(L∼r,∼s)=(−1)r+sdet(L∼r,∼s).\det\left(L_{\sim r,\sim r}\right)=\underbrace{\left(-1\right)^{s-r}}_{=\left(-1\right)^{r+s}}\det\left(L_{\sim r,\sim s}\right)=\left(-1\right)^{r+s}\det\left(L_{\sim r,\sim s}\right).However, the Matrix-Tree Theorem (Theorem5.142) yields

(#of spanning arborescences ofDrooted tor)\displaystyle\left(\#\text{ of spanning arborescences of }D\text{ rooted to }r\right)=det(L∼r,∼r)\displaystyle=\det\left(L_{\sim r,\sim r}\right)=(−1)r+sdet(L∼r,∼s).\displaystyle=\left(-1\right)^{r+s}\det\left(L_{\sim r,\sim s}\right).This proves Theorem5.191.

We are now ready to prove Theorem5.187:

Proof 5.193(Proof of Theorem5.187.).

For eachr,s∈{1,2,…,n}r,s\in\left\{1,2,\ldots,n\right\}, we have

τ⁡(D,r)\displaystyle\tau\left(D,r\right)=(#of spanning arborescences ofDrooted tor)\displaystyle=\left(\#\text{ of spanning arborescences of }D\text{ rooted to }r\right)(by the definition ofτ(D,r))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of }\tau\left(D,r\right)\right)=(−1)r+sdet(L∼r,∼s)\displaystyle=\left(-1\right)^{r+s}\det\left(L_{\sim r,\sim s}\right)(40)(by Theorem5.191).

However, we havef=(τ⁡(D,1),τ⁡(D,2),…,τ⁡(D,n))f=\left(\tau\left(D,1\right),\tau\left(D,2\right),\ldots,\tau\left(D,n\right)\right). Thus, for eachs∈{1,2,…,n}s\in\left\{1,2,\ldots,n\right\}, thess-th entry of the column vectorfLfLis868686We are using the notationLr,sL_{r,s}for the entry of the matrixLLin therr-th row and thess-th column.

∑r=1nτ⁡(D,r)⏟=(−1)r+sdet(L∼r,∼s)(by (40))Lr,s\displaystyle\sum\limits_{r=1}^{n}\underbrace{\tau\left(D,r\right)}_{\begin{subarray}{c}=\left(-1\right)^{r+s}\det\left(L_{\sim r,\sim s}\right)\\ \text{(by (\ref{sol.7.1.b.1}))}\end{subarray}}L_{r,s}=∑r=1n(−1)r+sdet(L∼r,∼s)Lr,s\displaystyle=\sum\limits_{r=1}^{n}\left(-1\right)^{r+s}\det\left(L_{\sim r,\sim s}\right)L_{r,s}=∑r=1n(−1)r+sLr,sdet(L∼r,∼s)=detL\displaystyle=\sum\limits_{r=1}^{n}\left(-1\right)^{r+s}L_{r,s}\det\left(L_{\sim r,\sim s}\right)=\det L(since Laplace expansion along thes-th columnyieldsdetL=∑r=1n(−1)r+sLr,sdet(L∼r,∼s))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since Laplace expansion along the }s\text{-th column}\\ \text{yields }\det L=\sum\limits_{r=1}^{n}\left(-1\right)^{r+s}L_{r,s}\det\left(L_{\sim r,\sim s}\right)\end{array}\right)=0\displaystyle=0(by Proposition5.140). This shows that all entries offLfLare00. In other words,fL=0fL=0. Theorem5.187is thus proved.

Other proofs of Theorem5.187exist. In particular, a combinatorial proof is sketched in[Sahi14, Theorem 1]. (More precisely,[Sahi14, Theorem 1]in this paper is the claim of Theorem5.187upon reversing all the arcs and replacing all matrices by their transposes.)878787I tried to explain this proof in more detail inthe solutions to Spring 2018 Math 4707 midterm #3– see the proof of Theorem 0.7 in those solutions; you be the judge if I succeeded.

5.19A weighted Matrix-Tree Theorem

5.19.1Definitions

We have so far beencountingarborescences. A natural generalization of counting isweighted counting– i.e., you assign a certain number (a “weight”) to each arborescence (or whatever object you are interested in), and then yousumthe weights of all arborescences (instead of merely counting them). This generalizes counting, because if all weights are11, then you get the # of arborescences.

If you pick the weights to be completely random, then the sum won’t usually be particularly interesting. However, some choices of weights lead to good behavior. Let us see what we get if we assign a weight to eacharcof our digraph, and then define the weight of an arborescence to be theproductof the weights of the arcs that appear in this arborescence.

Definition 5.194.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph.

Let𝕂\mathbb{K}be a commutative ring. Assume that an elementwa∈𝕂w_{a}\in\mathbb{K}is assigned to each arca∈Aa\in A. We call thiswaw_{a}theweightof the arcaa. (You can assume that𝕂=ℝ\mathbb{K}=\mathbb{R}, so that the weights are just numbers.)

(a)For any two verticesi,j∈Vi,j\in V, we letai,jwa_{i,j}^{w}be the sum of the weights of all arcs ofDDthat have sourceiiand targetjj.
(b)For any vertexi∈Vi\in V, we define theweighted outdegreedeg+w⁡i\deg^{+w}iofiito be the sum ∑a∈A;the source ofaisiwa.\sum\limits_{\begin{subarray}{c}a\in A;\\ \text{the source of }a\text{ is }i\end{subarray}}w_{a}.
(c)IfBBis a subdigraph ofDD, then theweightw⁡(B)w\left(B\right)ofBBis defined to be the product∏ais an arc ofBwa\prod\limits\limits_{a\text{ is an arc of }B}w_{a}. This is the product of the weights of all arcs ofBB.
(d)Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}. Theweighted LaplacianofDD(with respect to the weightswaw_{a}) is defined to be then×nn\times n-matrixLw∈𝕂n×nL^{w}\in\mathbb{K}^{n\times n}(note that the “ww” here is a superscript, not an exponent) whose entries are given by Li,jw=(deg+wi)⋅[i=j]−ai,jwfor alli,j∈V.L_{i,j}^{w}=\left(\deg^{+w}i\right)\cdot\left[i=j\right]-a_{i,j}^{w}\ \ \ \ \ \ \ \ \ \ \text{for all }i,j\in V.

These definitions generalize analogous definitions in the “unweighted case”. Indeed, if we take all the arc weightswaw_{a}to be11, then the weighted outdegreedeg+w⁡i\deg^{+w}iof a vertexiibecomes its usual outdegreedeg⁡i\deg i, and the weighted LaplacianLwL^{w}becomes the usual LaplacianLL. The weightw⁡(B)w\left(B\right)of a subdigraphBBsimply becomes11in this case.

5.19.2The weighted Matrix-Tree Theorem

We now can generalize the original MTT (= Matrix-Tree Theorem)888888To remind: The original MTT is Theorem5.142.as follows:

Theorem 5.195(weighted Matrix-Tree Theorem).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph.

Let𝕂\mathbb{K}be a commutative ring. Assume that an elementwa∈𝕂w_{a}\in\mathbb{K}is assigned to each arca∈Aa\in A. We call thiswaw_{a}theweightof the arcaa.

Assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}. LetLwL^{w}be the weighted Laplacian ofDD.

Letrrbe a vertex ofDD. Then,

Example 5.196.

LetDDbe the following multidigraph:

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\lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.22221pt}{-29.78717pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.07 -41.22)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,\ \ \ \ \ \ \ \ \ \ \text{and let }r=3.Then,DDhas two spanning arborescences rooted torr. One of the two has arcsα\alphaandβ\beta(and thus has weightwαwβw_{\alpha}w_{\beta}); the other has arcsγ\gammaandβ\beta(and thus has weightwγwβw_{\gamma}w_{\beta}). Hence,

∑Bis a spanningarborescenceofDrooted torw(B)=wαwβ+wγwβ,\sum\limits_{\begin{subarray}{c}B\text{ is a spanning}\\ \text{arborescence}\\ \text{of }D\text{ rooted to }r\end{subarray}}w\left(B\right)=w_{\alpha}w_{\beta}+w_{\gamma}w_{\beta},(41)The weighted LaplacianLwL^{w}is

Lw=(wα+wγ−wα−wγ0wβ−wβ−wδ0wδ)L^{w}=\left(\begin{array}[c]{ccc}w_{\alpha}+w_{\gamma}&-w_{\alpha}&-w_{\gamma}\\ 0&w_{\beta}&-w_{\beta}\\ -w_{\delta}&0&w_{\delta}\end{array}\right)(since, for example,deg+w⁡1=wα+wγ\deg^{+w}1=w_{\alpha}+w_{\gamma}anda1,1w=0a_{1,1}^{w}=0anda1,2w=wαa_{1,2}^{w}=w_{\alpha}). Thus,

L∼3,∼3w\displaystyle L_{\sim 3,\sim 3}^{w}=(wα+wγ−wα0wβ)and therefore\displaystyle=\left(\begin{array}[c]{cc}w_{\alpha}+w_{\gamma}&-w_{\alpha}\\ 0&w_{\beta}\end{array}\right)\ \ \ \ \ \ \ \ \ \ \text{and therefore}det(L∼3,∼3w)\displaystyle\det\left(L_{\sim 3,\sim 3}^{w}\right)=(wα+wγ)wβ=wαwβ+wγwβ.\displaystyle=\left(w_{\alpha}+w_{\gamma}\right)w_{\beta}=w_{\alpha}w_{\beta}+w_{\gamma}w_{\beta}.The right hand side of this agrees with that of (41). This confirms the weighted MTT for ourDDandrr.

As we already said, the weighted MTT generalizes the original MTT, because if we take allwaw_{a}’s to be11, we just recover the original MTT.

However, we can also go backwards: we can derive the weighted MTT from the original MTT. Let us do this.

5.19.3The polynomial identity trick

First, we recall a standard result in algebra, known as theprinciple of permanence of polynomial identitiesor as thepolynomial identity trick(it also goes under several other names). Here is one incarnation of this principle:

Theorem 5.197(principle of permanence of polynomial identities).

LetP⁡(x1,x2,…,xm)P\left(x_{1},x_{2},\ldots,x_{m}\right)andQ⁡(x1,x2,…,xm)Q\left(x_{1},x_{2},\ldots,x_{m}\right)be two polynomials with integer coefficients in several indeterminatesx1,x2,…,xmx_{1},x_{2},\ldots,x_{m}. Assume that the equality

P⁡(k1,k2,…,km)=Q⁡(k1,k2,…,km)P\left(k_{1},k_{2},\ldots,k_{m}\right)=Q\left(k_{1},k_{2},\ldots,k_{m}\right)(42)holds for everymm-tuple(k1,k2,…,km)∈ℕm\left(k_{1},k_{2},\ldots,k_{m}\right)\in\mathbb{N}^{m}of nonnegative integers. Then,P⁡(x1,x2,…,xm)P\left(x_{1},x_{2},\ldots,x_{m}\right)andQ⁡(x1,x2,…,xm)Q\left(x_{1},x_{2},\ldots,x_{m}\right)are identical as polynomials (so that, in particular, the equality (42) holds not only for every(k1,k2,…,km)∈ℕm\left(k_{1},k_{2},\ldots,k_{m}\right)\in\mathbb{N}^{m}, but also for every(k1,k2,…,km)∈ℂm\left(k_{1},k_{2},\ldots,k_{m}\right)\in\mathbb{C}^{m}, and more generally, for every(k1,k2,…,km)∈𝕂m\left(k_{1},k_{2},\ldots,k_{m}\right)\in\mathbb{K}^{m}where𝕂\mathbb{K}is an arbitrary commutative ring).

Theorem5.197is often summarized as “in order to prove that two polynomials are equal, it suffices to show that they are equal on all nonnegative integer points” (where a “nonnegative integer point” means a point – i.e., a tuple of inputs – whose all entries are nonnegative integers). Even shorter, one says that “a polynomial identity (i.e., an equality between two polynomials) needs only to be checked on nonnegative integers”. For example, if you can prove the equality

(x+y)4+(x−y)4=2x4+12x2y2+2y4\left(x+y\right)^{4}+\left(x-y\right)^{4}=2x^{4}+12x^{2}y^{2}+2y^{4}for all nonnegative integersxxandyy, then you automatically conclude that this equality holds as a polynomial identity, and thus is true for any elementsxxandyyof a commutative ring.

A typical application of Theorem5.197is to argue that a polynomial identity you have proved for all nonnegative integers must automatically hold for all inputs (because of Theorem5.197). Some examples of such reasoning can be found in[19fco, §2.6.3 and §2.6.4]and in[20f, §7.5.3]. A variant of Theorem5.197is[Conrad21, Theorem 2.6]; actually, the proof of[Conrad21, Theorem 2.6]can be trivially adapted to prove Theorem5.197(just replace “nonempty open set inℂk\mathbb{C}^{k}” by “ℕk\mathbb{N}^{k}”). In truth, there is nothing special about nonnegative integers and the setℕ\mathbb{N}; you could replaceℕ\mathbb{N}by any infinite set of numbers (or even any sufficiently large set of numbers, where “sufficiently large” means “more thanmax⁡{deg⁡P,deg⁡Q}\max\left\{\deg P,\deg Q\right\}many”). See[Alon02, Lemma 2.1]or[Lason10, Theorem 2]for a fairly general version of Theorem5.197that includes such cases898989To be precise,[Alon02, Lemma 2.1]and[Lason10, Theorem 2]are not concerned with two polynomials being identical, but rather with one polynomial being identically zero. But this is an equivalent question: Two polynomialsPPandQQare identical if and only if their differenceP−QP-Qis identically zero..

5.19.4Proof of the weighted MTT

We can now deduce the weighted MTT from the original MTT (Theorem5.142):

Proof 5.198(Proof of Theorem5.195.).

The claim of Theorem5.195(for fixedDDandrr) is an equality between two polynomials in the arc weightswaw_{a}. (For instance, in Example5.196, this equality iswαwβ+wγwβ=det(wα+wγ−wα0wβ)w_{\alpha}w_{\beta}+w_{\gamma}w_{\beta}=\det\left(\begin{array}[c]{cc}w_{\alpha}+w_{\gamma}&-w_{\alpha}\\ 0&w_{\beta}\end{array}\right).)

Therefore, thanks to Theorem5.197, it suffices to prove this equality in the case when all arc weightswaw_{a}are nonnegative integers. So let us WLOG assume that all arc weightswaw_{a}are nonnegative integers.

Let us now replace each arcaaofDDbywaw_{a}many copies of the arcaa(having the same source asaaand the same target asaa). The result is a new digraphD′D^{\prime}. Here is an example:

Example 5.199.

LetDDbe the digraph

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\lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.58865pt}{-8.23856pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.58 -11.4)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,and let the arc weights bewα=2w_{\alpha}=2andwβ=3w_{\beta}=3andwγ=2w_{\gamma}=2. Then,D′D^{\prime}looks as follows:

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\hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,whereα1,α2\alpha_{1},\alpha_{2}are the two arcs obtained fromα\alpha, and so on.

Now, recall that the digraphD′D^{\prime}has the same vertices asDD, but each arcaaofDDhas turned intowaw_{a}arcs ofD′D^{\prime}. Thus, the weighted outdegreedeg+w⁡i\deg^{+w}iof a vertexiiofDDequals the (usual, i.e., non-weighted) outdegreedeg+⁡i\deg^{+}iof the same vertexiiofD′D^{\prime}. Hence, the weighted LaplacianLwL^{w}ofDDis the (usual, i.e., non-weighted) Laplacian ofD′D^{\prime}.

Recall again that the digraphD′D^{\prime}has the same vertices asDD, but each arcaaofDDhas turned intowaw_{a}arcs ofD′D^{\prime}. Thus, each subdigraphBBofDDgives rise tow⁡(B)w\left(B\right)many subdigraphs ofD′D^{\prime}(because we can replace each arcaaofBBby any of thewaw_{a}many copies of this arc inD′D^{\prime}). Moreover, this correspondence takes spanning arborescences to spanning arborescences909090More precisely: LetBBbe a subdigraph ofDD, and letB′B^{\prime}be any of thew⁡(B)w\left(B\right)many subdigraphs ofD′D^{\prime}that are obtained fromBBthrough this correspondence. Then,BBis a spanning arborescence ofDDrooted torrif and only ifB′B^{\prime}is a spanning arborescence ofD′D^{\prime}rooted torr., and we can obtain any spanning arborescence ofD′D^{\prime}in this way from exactly oneBB. Hence,

∑Bis a spanningarborescenceofDrooted torw(B)=(# of spanning arborescences ofD′rooted tor).\sum\limits_{\begin{subarray}{c}B\text{ is a spanning}\\ \text{arborescence}\\ \text{of }D\text{ rooted to }r\end{subarray}}w\left(B\right)=\left(\text{\# of spanning arborescences of }D^{\prime}\text{ rooted to }r\right).Thus, applying the original MTT (Theorem5.142) toD′D^{\prime}yields the weighted MTT forDD(since the weighted LaplacianLwL^{w}ofDDis the (usual, i.e., non-weighted) Laplacian ofD′D^{\prime}). This completes the proof of Theorem5.195.

[Remark:Alternatively, it is not hard to adapt our above proof of the original MTT to the weighted case.]

5.19.5Application: Counting trees by their degrees

The weighted MTT has some applications that wouldn’t be obvious from the original MTT. Here is one:

Exercise 5.200.

Letn≥2n\geq 2be an integer, and letd1,d2,…,dnd_{1},d_{2},\ldots,d_{n}bennpositive integers. Annn-treeshall mean a simple graph with vertex set{1,2,…,n}\left\{1,2,\ldots,n\right\}that is a tree. We know from Corollary5.144that there arenn−2n^{n-2}manynn-trees. How many of thesenn-trees have the property that

deg⁡i=difor each vertexi?\deg i=d_{i}\ \ \ \ \ \ \ \ \ \ \text{for each vertex }i\text{ ?}

Proof 5.201(Solution.).

Thenn-trees are just the spanning trees of the complete graphKnK_{n}.

To incorporate thedeg⁡i=di\deg i=d_{i}condition into our count, we use a generating function. So let usnotfix the numbersd1,d2,…,dnd_{1},d_{2},\ldots,d_{n}, but rather consider the polynomial

P(x1,x2,…,xn):=∑Tis ann-treex1deg⁡1x2deg⁡2⋯xndeg⁡nP\left(x_{1},x_{2},\ldots,x_{n}\right):=\sum\limits_{T\text{ is an }n\text{-tree}}x_{1}^{\deg 1}x_{2}^{\deg 2}\cdots x_{n}^{\deg n}(43)innnindeterminatesx1,x2,…,xnx_{1},x_{2},\ldots,x_{n}(wheredeg⁡i\deg imeans the degree ofiiinTT). Then, thex1d1x2d2⋯xndnx_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}-coefficient of this polynomialP⁡(x1,x2,…,xn)P\left(x_{1},x_{2},\ldots,x_{n}\right)is the # ofnn-treesTTsatisfying the property that

deg⁡i=difor each vertexi\deg i=d_{i}\ \ \ \ \ \ \ \ \ \ \text{for each vertex }i(because each suchnn-treeTTcontributes a monomialx1d1x2d2⋯xndnx_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}to the sum on the right hand side of (43), whereas any othernn-treeTTcontributes a different monomial to this sum).

Let us assign to each edgeijijofKnK_{n}the weightwij:=xixjw_{ij}:=x_{i}x_{j}. Then, the definition ofP⁡(x1,x2,…,xn)P\left(x_{1},x_{2},\ldots,x_{n}\right)rewrites as follows:

P⁡(x1,x2,…,xn)=∑Tis ann-treew⁡(T),P\left(x_{1},x_{2},\ldots,x_{n}\right)=\sum\limits_{T\text{ is an }n\text{-tree}}w\left(T\right),wherew⁡(T)w\left(T\right)denotes the product of the weights of all edges ofTT. (Indeed, for any subgraphTTofKnK_{n}, the weightw⁡(T)w\left(T\right)equalsx1deg⁡1x2deg⁡2⋯xndeg⁡nx_{1}^{\deg 1}x_{2}^{\deg 2}\cdots x_{n}^{\deg n}, wheredeg⁡i\deg imeans the degree ofiiinTT.)

We have assigned weights to the edges of the graphKnK_{n}; let us now assign the same weights to the arcs of the digraphKnbidirK_{n}^{\operatorname*{bidir}}. That is, the two arcs(ij,1)\left(ij,1\right)and(ij,2)\left(ij,2\right)corresponding to an edgeijijofKnK_{n}shall both have the weight

w(ij,1)=w(ij,2)=wij=xixj.w_{\left(ij,1\right)}=w_{\left(ij,2\right)}=w_{ij}=x_{i}x_{j}.(44) As we are already used to, we can replace spanning trees ofKnK_{n}by spanning arborescences ofKnbidirK_{n}^{\operatorname*{bidir}}rooted to11, since the former are in bijection with the latter. Thus, we have

(# of spanning trees ofKn)\displaystyle\left(\text{\# of spanning trees of }K_{n}\right)=(# of spanning arborescences ofKnbidirrooted to1).\displaystyle=\left(\text{\# of spanning arborescences of }K_{n}^{\operatorname*{bidir}}\text{ rooted to }1\right).Moreover, since this bijection preserves weights (because of (44)), we also have

∑Tis a spanningtree ofKnw(T)=∑Bis a spanningarborescence ofKnbidirrooted to1w(B).\sum\limits_{\begin{subarray}{c}T\text{ is a spanning}\\ \text{tree of }K_{n}\end{subarray}}w\left(T\right)=\sum\limits_{\begin{subarray}{c}B\text{ is a spanning}\\ \text{arborescence of }K_{n}^{\operatorname*{bidir}}\\ \text{rooted to }1\end{subarray}}w\left(B\right).In other words,

∑Tis ann-treew(T)=∑Bis a spanningarborescence ofKnbidirrooted to1w(B)\sum\limits_{T\text{ is an }n\text{-tree}}w\left(T\right)=\sum\limits_{\begin{subarray}{c}B\text{ is a spanning}\\ \text{arborescence of }K_{n}^{\operatorname*{bidir}}\\ \text{rooted to }1\end{subarray}}w\left(B\right)(since the spanning trees ofKnK_{n}are precisely thenn-trees).

To compute the right hand side, we shall use the weighted Matrix-Tree Theorem. The weighted Laplacian ofKnbidirK_{n}^{\operatorname*{bidir}}(with the weights we have just defined) is then×nn\times n-matrixLwL^{w}with entries given by

Li,jw\displaystyle L_{i,j}^{w}=(deg+wi)⋅[i=j]−ai,jw\displaystyle=\left(\deg^{+w}i\right)\cdot\left[i=j\right]-a_{i,j}^{w}={deg+w⁡i−ai,jw,ifi=j;−ai,jw,ifi≠j\displaystyle=\begin{cases}\deg^{+w}i-a_{i,j}^{w},&\text{if }i=j;\\ -a_{i,j}^{w},&\text{if }i\neq j\end{cases}={deg+w⁡i,ifi=j;−ai,jw,ifi≠j(sinceai,jw=0wheni=j(becauseKnbidirhas no loops))\displaystyle=\begin{cases}\deg^{+w}i,&\text{if }i=j;\\ -a_{i,j}^{w},&\text{if }i\neq j\end{cases}\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since }a_{i,j}^{w}=0\text{ when }i=j\\ \text{(because }K_{n}^{\operatorname*{bidir}}\text{ has no loops)}\end{array}\right)={xi(x1+x2+⋯+xn)−xixj,ifi=j;−xixj,ifi≠j\displaystyle=\begin{cases}x_{i}\left(x_{1}+x_{2}+\cdots+x_{n}\right)-x_{i}x_{j},&\text{if }i=j;\\ -x_{i}x_{j},&\text{if }i\neq j\end{cases}(sincedeg+wi=xix1+xix2+⋯+xixi−1+xixi+1+⋯+xixn=xi(x1+x2+⋯+xi−1+xi+1+⋯+xn)=xi(x1+x2+⋯+xn)−xixi=xi(x1+x2+⋯+xn)−xixjwheneveri=j,and sinceai,jw=xixjwheneveri≠j)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since }\deg^{+w}i=x_{i}x_{1}+x_{i}x_{2}+\cdots+x_{i}x_{i-1}+x_{i}x_{i+1}+\cdots+x_{i}x_{n}\\ =x_{i}\left(x_{1}+x_{2}+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n}\right)\\ =x_{i}\left(x_{1}+x_{2}+\cdots+x_{n}\right)-x_{i}x_{i}\\ =x_{i}\left(x_{1}+x_{2}+\cdots+x_{n}\right)-x_{i}x_{j}\text{ whenever }i=j\text{,}\\ \text{and since }a_{i,j}^{w}=x_{i}x_{j}\text{ whenever }i\neq j\end{array}\right)=[i=j]xi(x1+x2+⋯+xn)−xixj\displaystyle=\left[i=j\right]x_{i}\left(x_{1}+x_{2}+\cdots+x_{n}\right)-x_{i}x_{j}=xi([i=j](x1+x2+⋯+xn)−xj).\displaystyle=x_{i}\left(\left[i=j\right]\left(x_{1}+x_{2}+\cdots+x_{n}\right)-x_{j}\right).We can find its minordet(L∼1,∼1w)\det\left(L_{\sim 1,\sim 1}^{w}\right)without too much trouble (e.g., using row transformations similar to the ones we have done back in the proof of Cayley’s formula919191The first step, of course, is to factor anxix_{i}out of theii-th row for eachii.); the result is

det(L∼1,∼1w)=x1x2⋯xn(x1+x2+⋯+xn)n−2.\det\left(L_{\sim 1,\sim 1}^{w}\right)=x_{1}x_{2}\cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}. Summarizing what we have done so far,

P⁡(x1,x2,…,xn)\displaystyle P\left(x_{1},x_{2},\ldots,x_{n}\right)=∑Tis ann-treew(T)=∑Bis a spanningarborescence ofKnbidirrooted to1w(B)\displaystyle=\sum\limits_{T\text{ is an }n\text{-tree}}w\left(T\right)=\sum\limits_{\begin{subarray}{c}B\text{ is a spanning}\\ \text{arborescence of }K_{n}^{\operatorname*{bidir}}\\ \text{rooted to }1\end{subarray}}w\left(B\right)=det(L∼1,∼1w)(by the weighted Matrix-Tree Theorem)\displaystyle=\det\left(L_{\sim 1,\sim 1}^{w}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by the weighted Matrix-Tree Theorem}\right)=x1x2⋯xn(x1+x2+⋯+xn)n−2.\displaystyle=x_{1}x_{2}\cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}.(45)As we recall, we are looking for thex1d1x2d2⋯xndnx_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}-coefficient in this polynomial. From (45), we see that

(thex1d1x2d2⋯xndn-coefficient ofP(x1,x2,…,xn))\displaystyle\left(\text{the }x_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}\text{-coefficient of }P\left(x_{1},x_{2},\ldots,x_{n}\right)\right)=(thex1d1x2d2⋯xndn-coefficient ofx1x2⋯xn(x1+x2+⋯+xn)n−2)\displaystyle=\left(\text{the }x_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}\text{-coefficient of }x_{1}x_{2}\cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}\right)=(thex1d1−1x2d2−1⋯xndn−1-coefficient of(x1+x2+⋯+xn)n−2)\displaystyle=\left(\text{the }x_{1}^{d_{1}-1}x_{2}^{d_{2}-1}\cdots x_{n}^{d_{n}-1}\text{-coefficient of }\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}\right)(because when we multiply a polynomial byx1x2⋯xnx_{1}x_{2}\cdots x_{n}, all the exponents in it get incremented by11, so its coefficients just shift by a11in each exponent).

Now, how can we describe the coefficients of(x1+x2+⋯+xn)n−2\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}, or, more generally, of(x1+x2+⋯+xn)m\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{m}for somem∈ℕm\in\mathbb{N}? These are the so-calledmultinomial coefficients(named in analogy to the binomial coefficients, which are their particular case forn=2n=2). Their definition is as follows: Ifp1,p2,…,pn,qp_{1},p_{2},\ldots,p_{n},qare nonnegative integers withq=p1+p2+⋯+pnq=p_{1}+p_{2}+\cdots+p_{n}, then themultinomial coefficient(qp1,p2,…,pn)\dbinom{q}{p_{1},p_{2},\ldots,p_{n}}is defined to beq!p1!p2!⋯pn!\dfrac{q!}{p_{1}!p_{2}!\cdots p_{n}!}. Ifq≠p1+p2+⋯+pnq\neq p_{1}+p_{2}+\cdots+p_{n}, then it is defined to be00instead. In either case, this coefficient is easily seen to be an integer.929292See[24wd, Section 6.7]for an introduction to multinomial coefficients.Themultinomial formula (aka multinomial theorem)says that for eachk∈ℕk\in\mathbb{N}, we have

(x1+x2+⋯+xn)k\displaystyle\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}=∑i1,i2,…,in∈ℕ;i1+i2+⋯+in=k(ki1,i2,…,in)x1i1x2i2⋯xnin\displaystyle=\sum\limits_{\begin{subarray}{c}i_{1},i_{2},\ldots,i_{n}\in\mathbb{N};\\ i_{1}+i_{2}+\cdots+i_{n}=k\end{subarray}}\dbinom{k}{i_{1},i_{2},\ldots,i_{n}}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}=∑i1,i2,…,in∈ℕ(ki1,i2,…,in)x1i1x2i2⋯xnin\displaystyle=\sum\limits_{i_{1},i_{2},\ldots,i_{n}\in\mathbb{N}}\dbinom{k}{i_{1},i_{2},\ldots,i_{n}}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}(it does not matter whether we restrict the sum by the conditioni1+i2+⋯+in=ki_{1}+i_{2}+\cdots+i_{n}=kor not, since the coefficient(ki1,i2,…,in)\dbinom{k}{i_{1},i_{2},\ldots,i_{n}}is defined to be00when this condition is violated anyway). Hence,

(thex1i1x2i2⋯xnin-coefficient of(x1+x2+⋯+xn)k)=(ki1,i2,…,in)\left(\text{the }x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}\text{-coefficient of }\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}\right)=\dbinom{k}{i_{1},i_{2},\ldots,i_{n}}for anyk∈ℕk\in\mathbb{N}and anyi1,i2,…,in∈ℕi_{1},i_{2},\ldots,i_{n}\in\mathbb{N}. In particular,

(thex1d1−1x2d2−1⋯xndn−1-coefficient of(x1+x2+⋯+xn)n−2)\displaystyle\left(\text{the }x_{1}^{d_{1}-1}x_{2}^{d_{2}-1}\cdots x_{n}^{d_{n}-1}\text{-coefficient of }\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}\right)=(n−2d1−1,d2−1,…,dn−1).\displaystyle=\dbinom{n-2}{d_{1}-1,\ d_{2}-1,\ \ldots,\ d_{n}-1}. Summarizing, we find

(thex1d1x2d2⋯xndn-coefficient ofP(x1,x2,…,xn))\displaystyle\left(\text{the }x_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}\text{-coefficient of }P\left(x_{1},x_{2},\ldots,x_{n}\right)\right)=(thex1d1−1x2d2−1⋯xndn−1-coefficient of(x1+x2+⋯+xn)n−2)\displaystyle=\left(\text{the }x_{1}^{d_{1}-1}x_{2}^{d_{2}-1}\cdots x_{n}^{d_{n}-1}\text{-coefficient of }\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n-2}\right)=(n−2d1−1,d2−1,…,dn−1).\displaystyle=\dbinom{n-2}{d_{1}-1,\ d_{2}-1,\ \ldots,\ d_{n}-1}.However, thex1d1x2d2⋯xndnx_{1}^{d_{1}}x_{2}^{d_{2}}\cdots x_{n}^{d_{n}}-coefficient ofP⁡(x1,x2,…,xn)P\left(x_{1},x_{2},\ldots,x_{n}\right)is the # ofnn-treesTTsatisfying the property that

deg⁡i=difor each vertexi\deg i=d_{i}\ \ \ \ \ \ \ \ \ \ \text{for each vertex }i(as we have seen above). Thus, we have proved the following:

Theorem 5.202(refined Cayley’s formula).

Letn≥2n\geq 2be an integer, and letd1,d2,…,dnd_{1},d_{2},\ldots,d_{n}bennpositive integers. Then, the # ofnn-trees with the property that

deg⁡i=difor eachi∈{1,2,…,n}\deg i=d_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{1,2,\ldots,n\right\}is the multinomial coefficient

(n−2d1−1,d2−1,…,dn−1).\dbinom{n-2}{d_{1}-1,\ d_{2}-1,\ \ldots,\ d_{n}-1}.

This theorem can also be proved via a combinatorial bijection (see, e.g.,[Galvin21, §6]), or through an elementary induction argument using Theorem5.30and Theorem5.32(see, e.g.,[Galvin21, §5]).

5.19.6The weighted harmonic vector theorem

The harmonic vector theorem for Laplacians (Theorem5.187) also has a weighted version:

Theorem 5.203(harmonic vector theorem for weighted Laplacians).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}. Let𝕂\mathbb{K}be a commutative ring. Assume that an elementwa∈𝕂w_{a}\in\mathbb{K}is assigned to each arca∈Aa\in A. For eachr∈Vr\in V, letτw(D,r)\tau^{w}\left(D,r\right)be the sum of the weights of all the spanning arborescences ofDDrooted torr. Letfwf^{w}be the row vector(τw(D,1),τw(D,2),…,τw(D,n))\left(\tau^{w}\left(D,1\right),\ \tau^{w}\left(D,2\right),\ \ldots,\ \tau^{w}\left(D,n\right)\right). LetLwL^{w}be the weighted Laplacian ofDD. Then,fwLw=0f^{w}L^{w}=0.

Proof 5.204.

Similar to the unweighted case (Theorem5.187).

When the digraphDDis strongly connected, and the weightswaw_{a}in Theorem5.203are positive reals, we can make a slightly stronger claim:

Corollary 5.205.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a strongly connected multidigraph, whereV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}for somen∈ℕn\in\mathbb{N}. Assume that a positive real numberwa∈ℝw_{a}\in\mathbb{R}is assigned to each arca∈Aa\in A. For eachr∈Vr\in V, letτw(D,r)\tau^{w}\left(D,r\right)be the sum of the weights of all the spanning arborescences ofDDrooted torr. Letfwf^{w}be the row vector(τw(D,1),τw(D,2),…,τw(D,n))\left(\tau^{w}\left(D,1\right),\ \tau^{w}\left(D,2\right),\ \ldots,\ \tau^{w}\left(D,n\right)\right). LetLwL^{w}be the weighted Laplacian ofDD. Then:

(a)We havefwLw=0f^{w}L^{w}=0.
(b)All entries of the vectorfwf^{w}are positive reals.
(c)If all numberswaw_{a}are rational, then all entries of the vectorfwf^{w}are rational.
(d)The matrixLwL^{w}has rankn−1n-1.
(e)Letsws_{w}be the sum of all entries offwf^{w}. Then, there is a unique row vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxLw=0xL^{w}=0and such that the sum of all entries ofxxis11. This vectorxxis1swfw\dfrac{1}{s_{w}}f^{w}.

Proof 5.206.

(a)This follows from Theorem5.203(applied to𝕂=ℝ\mathbb{K}=\mathbb{R}).

(b)We must show thatτw(D,r)\tau^{w}\left(D,r\right)is positive for eachr∈Vr\in V. So let us fixr∈Vr\in V. Recall thatτw(D,r)\tau^{w}\left(D,r\right)is the sum of the weights of all the spanning arborescences ofDDrooted torr. This sum is not an empty sum (since there exists a spanning arborescence ofDDrooted fromrr939393This follows from the analogue of Theorem5.112for to-roots instead of from-roots (sinceDDis strongly connected, and thusrris a to-root ofDD).), and all its addends are positive (since the weight of a spanning arborescence ofDDis a product of some of the positive realswaw_{a}, and thus itself positive). Hence, this sum is positive. In other words,τw(D,r)\tau^{w}\left(D,r\right)is positive. This proves part(b).

(c)This follows from the construction offwf^{w}.

(d)Pick any vertexr∈Vr\in V. Theorem5.195(applied to𝕂=ℝ\mathbb{K}=\mathbb{R}) shows that

∑Bis a spanningarborescenceofDrooted torw(B)=det(L∼r,∼rw).\sum\limits_{\begin{subarray}{c}B\text{ is a spanning}\\ \text{arborescence}\\ \text{of }D\text{ rooted to }r\end{subarray}}w\left(B\right)=\det\left(L_{\sim r,\sim r}^{w}\right).The left hand side of this equality equals the numberτw(D,r)\tau^{w}\left(D,r\right), which is positive (as we showed in the proof of part(b)above). Thus, so is the right hand side. In other words,det(L∼r,∼rw)>0\det\left(L_{\sim r,\sim r}^{w}\right)>0. Hence, in particular,det(L∼r,∼rw)≠0\det\left(L_{\sim r,\sim r}^{w}\right)\neq 0. This shows that then×nn\times n-matrixLwL^{w}has an invertible(n−1)×(n−1)\left(n-1\right)\times\left(n-1\right)-submatrix (namely,L∼r,∼rwL_{\sim r,\sim r}^{w}). Hence, its rank is at leastn−1n-1. In other words,rank(Lw)≥n−1\operatorname*{rank}\left(L^{w}\right)\geq n-1.

But the vectorfwf^{w}is nonzero (by part(b)) and lies in the left nullspace949494Theleft nullspaceof a matrixAAmeans the vector space of all row vectorsvvsatisfyingvA=0vA=0.ofLwL^{w}(since part(a)saysfwLw=0f^{w}L^{w}=0). Hence, the matrixLwL^{w}is singular, and thus its rank isrank(Lw)<n\operatorname*{rank}\left(L^{w}\right)<n. Combined withrank(Lw)≥n−1\operatorname*{rank}\left(L^{w}\right)\geq n-1, this yieldsrank(Lw)=n−1\operatorname*{rank}\left(L^{w}\right)=n-1. Thus, part(d)is proved.

(e)First, we observe thatsws_{w}is positive (by part(b)), thus nonzero. The vector1swfw\dfrac{1}{s_{w}}f^{w}really is a row vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxLw=0xL^{w}=0(since(1swfw)Lw=1swfwLw⏟=0(by part(a))=0\left(\dfrac{1}{s_{w}}f^{w}\right)L^{w}=\dfrac{1}{s_{w}}\underbrace{f^{w}L^{w}}_{\begin{subarray}{c}=0\\ \text{(by part {(a)})}\end{subarray}}=0) and such that the sum of all entries ofxxis11(since the sum of all entries offwf^{w}issws_{w}). It remains to show that it is the only such vector.

Indeed, assume the contrary. Thus, there are two distinct such vectorsxx. Let them beyyandzz. These two vectorsyyandzzbelong to the left nullspace ofLwL^{w}(since they are vectorsxxsatisfyingxLw=0xL^{w}=0) but cannot be linearly dependent (since each of them has its sum of entries equal to11). Thus, they are linearly independent. This shows that the left nullspace ofLwL^{w}has dimension≥2\geq 2(since it contains the two linearly independent vectorsyyandzz). In other words, the matrixLwL^{w}has rank≤n−2\leq n-2. But this contradicts part(d)of the corollary. This contradiction completes our proof of part(e).

A consequence of Corollary5.205is a graph-theoretical proof of a famous fact from matrix analysis (often worded in the language of Markov chains, cf.[GriSne07, Theorem 11.10]):

Corollary 5.207.

Letnnbe a positive integer. LetP=(pi,j)1≤i,j≤n∈ℝn×nP=\left(p_{i,j}\right)_{1\leq i,j\leq n}\in\mathbb{R}^{n\times n}be a matrix whose entriespi,jp_{i,j}are nonnegative reals. Assume that

∑j=1npi,j=1for eachi∈{1,2,…,n}.\sum\limits_{j=1}^{n}p_{i,j}=1\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{1,2,\ldots,n\right\}.(MatricesPPsatisfying these conditions are calledstochastic.) LetDDbe the simple digraph(V,A)\left(V,A\right)with vertex setV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}and arc setA={(i,j)∈V×V∣pi,j>0}A=\left\{\left(i,j\right)\in V\times V\ \mid\ p_{i,j}>0\right\}. Assume that this digraphDDis strongly connected. (MatricesPPsatisfying this condition are calledirreducible.) Then:

(a)There is a unique row vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxP=xxP=xand such that the sum of all entries ofxxis11. (In the lingo of probabilists, this vectorxxis called thestationary distributionor thesteady stateof the Markov chain defined byPP.)
(b)The entries of this vectorxxare positive reals.
(c)If all thepi,jp_{i,j}are rational, then the entries of this vectorxxare rational.

Proof 5.208.

We identify our simple digraphD=(V,A)D=\left(V,A\right)with the corresponding multidigraphDmult=(V,A,ψ)D^{\operatorname*{mult}}=\left(V,A,\psi\right)(see Definition4.17). This is a strongly connected multidigraph.

To each arc(i,j)\left(i,j\right)ofDD, we assign the positive real numberw(i,j):=pi,jw_{\left(i,j\right)}:=p_{i,j}(this is positive, because(i,j)∈A\left(i,j\right)\in Aguarantees thatpi,j>0p_{i,j}>0). We call this number the “weight” of the arc(i,j)\left(i,j\right). Then, Definition5.194(a)(applied to𝕂=ℝ\mathbb{K}=\mathbb{R}) defines a numberai,jwa_{i,j}^{w}for any two verticesi,j∈V={1,2,…,n}i,j\in V=\left\{1,2,\ldots,n\right\}. We claim that any two verticesi,j∈V={1,2,…,n}i,j\in V=\left\{1,2,\ldots,n\right\}satisfy

ai,jw=pi,j.a_{i,j}^{w}=p_{i,j}.(46) [Proof of (46):Leti,j∈V={1,2,…,n}i,j\in V=\left\{1,2,\ldots,n\right\}be two vertices. Then,pi,j≥0p_{i,j}\geq 0(since all entries ofPPare nonnegative). Hence, we are in one of the following two cases:

Case 1:We havepi,j>0p_{i,j}>0.

Case 2:We havepi,j=0p_{i,j}=0.

Let us consider Case 1. In this case, we havepi,j>0p_{i,j}>0. Thus,(i,j)∈A\left(i,j\right)\in A(by the definition ofAA). Hence, the digraphDDhas exactly one arc fromiitojj, namely the arc(i,j)\left(i,j\right). The definition ofai,jwa_{i,j}^{w}thus shows thatai,jw=w(i,j)=pi,ja_{i,j}^{w}=w_{\left(i,j\right)}=p_{i,j}(by the definition ofw(i,j)w_{\left(i,j\right)}). This proves (46) in Case 1.

Let us now consider Case 2. In this case, we havepi,j=0p_{i,j}=0. Thus,(i,j)∉A\left(i,j\right)\notin A(by the definition ofAA). Hence, the digraphDDhas no arc fromiitojj. The definition ofai,jwa_{i,j}^{w}thus shows thatai,jw=(empty sum)=0=pi,ja_{i,j}^{w}=\left(\text{empty sum}\right)=0=p_{i,j}. This proves (46) in Case 2.

Thus, the proof of (46) is complete (since we have covered both cases).]

Furthermore, Definition5.194(b)(applied to𝕂=ℝ\mathbb{K}=\mathbb{R}) yields that the weighted outdegreedeg+w⁡i\deg^{+w}iof any vertexi∈V={1,2,…,n}i\in V=\left\{1,2,\ldots,n\right\}is

deg+w⁡i\displaystyle\deg^{+w}i=∑a∈A;the source ofaisiwa=∑j∈V⏟=∑j=1n∑a∈Ais anarc fromitojwa⏟=ai,jw(by the definition ofai,jw)=∑j=1nai,jw⏟=pi,j(by (46))\displaystyle=\sum\limits_{\begin{subarray}{c}a\in A;\\ \text{the source of }a\text{ is }i\end{subarray}}w_{a}=\underbrace{\sum\limits_{j\in V}}_{=\sum\limits_{j=1}^{n}}\ \ \underbrace{\sum\limits_{\begin{subarray}{c}a\in A\text{ is an}\\ \text{arc from }i\text{ to }j\end{subarray}}w_{a}}_{\begin{subarray}{c}=a_{i,j}^{w}\\ \text{(by the definition of }a_{i,j}^{w}\text{)}\end{subarray}}=\sum\limits_{j=1}^{n}\underbrace{a_{i,j}^{w}}_{\begin{subarray}{c}=p_{i,j}\\ \text{(by (\ref{pf.cor.MTT.markov.awij=}))}\end{subarray}}=∑j=1npi,j=1\displaystyle=\sum\limits_{j=1}^{n}p_{i,j}=1(47)(by one of the assumptions of the corollary).

LetLwL^{w}be the weighted Laplacian ofDD. Then, for eachi,j∈V={1,2,…,n}i,j\in V=\left\{1,2,\ldots,n\right\}, we have

Li,jw\displaystyle L_{i,j}^{w}=(deg+w⁡i)⏟=1(by (47))⋅[i=j]−ai,jw⏟=pi,j(by (46))(by Definition5.194(d))\displaystyle=\underbrace{\left(\deg^{+w}i\right)}_{\begin{subarray}{c}=1\\ \text{(by (\ref{pf.cor.MTT.markov.deg=1}))}\end{subarray}}\cdot\left[i=j\right]-\underbrace{a_{i,j}^{w}}_{\begin{subarray}{c}=p_{i,j}\\ \text{(by (\ref{pf.cor.MTT.markov.awij=}))}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\text{by Definition \ref{def.wMTT.wt-defs} {(d)}}\right)=[i=j]−pi,j.\displaystyle=\left[i=j\right]-p_{i,j}.(48)Note that the numbers[i=j]\left[i=j\right]andpi,jp_{i,j}on the right hand side here are the(i,j)\left(i,j\right)-th entries of the matricesInI_{n}andPP, respectively. Thus, (48) shows that

Lw=In−P.L^{w}=I_{n}-P.(49) For eachr∈Vr\in V, letτw(D,r)\tau^{w}\left(D,r\right)be the sum of the weights of all the spanning arborescences ofDDrooted torr. Letfwf^{w}be the row vector (τw(D,1),τw(D,2),…,τw(D,n))\left(\tau^{w}\left(D,1\right),\ \tau^{w}\left(D,2\right),\ \ldots,\ \tau^{w}\left(D,n\right)\right). Letsws_{w}be the sum of all entries offwf^{w}.

Corollary5.205(e)shows that there is a unique row vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxLw=0xL^{w}=0and such that the sum of all entries ofxxis11. Since the equationxLw=0xL^{w}=0is equivalent toxP=xxP=x(because (49) shows thatxLw=x⁡(In−P)=x−xPxL^{w}=x\left(I_{n}-P\right)=x-xP, and thus we have the chain of logical equivalences(xLw=0)⟺(x−xP=0)⟺(xP=x)\left(xL^{w}=0\right)\ \Longleftrightarrow\ \left(x-xP=0\right)\ \Longleftrightarrow\ \left(xP=x\right)), we can rewrite this as follows: There is a unique row vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxP=xxP=xand such that the sum of all entries ofxxis11. This proves Corollary5.207(a).

(b)Consider the unique row vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxP=xxP=xand such that the sum of all entries ofxxis11. This is precisely the unique vectorx∈ℝ1×nx\in\mathbb{R}^{1\times n}such thatxLw=0xL^{w}=0and such that the sum of all entries ofxxis11(because, as we saw above, the equationxLw=0xL^{w}=0is equivalent toxP=xxP=x). By Corollary5.205(e), this unique vectorxxis therefore1swfw\dfrac{1}{s_{w}}f^{w}. However, the entries offwf^{w}are positive reals (by Corollary5.205(b)), and thus their sumsws_{w}is a positive real as well. Therefore, the entries of1swfw\dfrac{1}{s_{w}}f^{w}are positive reals. In other words, the entries ofxxare positive reals (sincexxis1swfw\dfrac{1}{s_{w}}f^{w}). This proves Corollary5.207(b).

(c)Assume that all thepi,jp_{i,j}are rational. Hence, all entries of the vectorfwf^{w}are rational (by Corollary5.205(c)), whence their sumsws_{w}is rational as well. Therefore, the entries of1swfw\dfrac{1}{s_{w}}f^{w}are rational, too. In other words, the entries ofxxare rational (sincexxis1swfw\dfrac{1}{s_{w}}f^{w}). This proves Corollary5.207(c).

Here ends our study of spanning trees and their enumeration. An interested reader can learn more from[Rubey00],[Holzer22],[Moon70]and[GrSaSu14].

6Colorings

Now to something different: Let’s color the vertices of a graph!

6.1Definition

This is a serious course, so our colors are positive integers. Coloring the vertices thus means assigning a color (= a positive integer) to each vertex. Here are the details:

Definition 6.1.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Letk∈ℕk\in\mathbb{N}.

(a)Akk-coloringofGGmeans a mapf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}. Given such akk-coloringff, we refer to the numbers1,2,…,k1,2,\ldots,kas thecolors, and we refer to each valuef⁡(v)f\left(v\right)as thecolorof the vertexvvin thekk-coloringff.
(b)Akk-coloringffofGGis said to beproperif no two adjacent vertices ofGGhave the same color. (In other words, akk-coloringffofGGis proper if there exists no edge ofGGwhose endpointsuuandvvsatisfyf⁡(u)=f⁡(v)f\left(u\right)=f\left(v\right).)

Example 6.2.

Here are two77-colorings of a graph:

112244225511113344223311(where the numbers on the nodes are not the vertices, but rather the colors of the vertices). The77-coloring on the left (yes, it is a77-coloring, even though it does not actually use the colors33,66and77) is not proper, because the two adjacent vertices on the top left have the same color. The77-coloring on the right, however, is proper.

Example 6.3.

Here is a bunch of graphs:

12345A=12345B=12345C=12345D=.\begin{tabular}[c]{|c|c|}\hline\cr&\\ \hline\cr $\\hbox to175\.37pt\{\\vbox to75\.28pt\{\\pgfpicture\\makeatletter\\hbox\{\\hskip 109\.2812pt\\lower\-37\.63864pt\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@begingroup@\{fill=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.4pt\} \\lx@inpgf@ignorespaces\\nullfont\\hbox to0\.0pt\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\\lx@inpgf@ignorespaces\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M \-66\.58 0 C \-66\.58 6\.71 \-72\.03 12\.16 \-78\.74 12\.16 C \-85\.45 12\.16 \-90\.9 6\.71 \-90\.9 0 C \-90\.9 \-6\.71 \-85\.45 \-12\.16 \-78\.74 \-12\.16 C \-72\.03 \-12\.16 \-66\.58 \-6\.71 \-66\.58 0 Z M \-78\.74 0\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-59\.40552pt\}\{\-3\.22221pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-82\.2 \-4\.46$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 39\.37 C 12\.16 46\.08 6\.71 51\.53 0 51\.53 C \-6\.71 51\.53 \-12\.16 46\.08 \-12\.16 39\.37 C \-12\.16 32\.66 \-6\.71 27\.21 0 27\.21 C 6\.71 27\.21 12\.16 32\.66 12\.16 39\.37 Z M 0 39\.37\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-2\.5pt\}\{25\.23055pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-3\.46 34\.91$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 90\.9 39\.37 C 90\.9 46\.08 85\.45 51\.53 78\.74 51\.53 C 72\.03 51\.53 66\.58 46\.08 66\.58 39\.37 C 66\.58 32\.66 72\.03 27\.21 78\.74 27\.21 C 85\.45 27\.21 90\.9 32\.66 90\.9 39\.37 Z M 78\.74 39\.37\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} 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\-46\.08 90\.9 \-39\.37 Z M 78\.74 \-39\.37\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \}\{\{\{\{\\lx@inpgf@ignorespaces\}\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{54\.40552pt\}\{\-31\.67497pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 75\.28 \-43\.83$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \{\{\}\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\{\}\{\{\{\}\}\}\{\{\}\}\{\}\{\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\}\{\}\{\}\{\}\{\}\{\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=0\.8pt\} \\lx@inpgf@ignorespaces\\lxSVG@begingroup@\{stroke=\#009900\} \{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.16 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\\lxSVG@transformcm\{1\.0\}\{0\.0\}\{0\.0\}\{1\.0\}\{\-105\.9482pt\}\{\-3\.41666pt\}\\lxSVG@begingroup@\{transform=matrix$1\.0 0\.0 0\.0 1\.0 \-146\.6 \-4\.73$\} \\pgfsys@hbox\{62\}\\lxSVG@closescope \}\}\} \\lxSVG@closescope \}\}\} \\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 0 26\.66 L 0 \-26\.66\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 12\.71 39\.37 L 66\.03 39\.37\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 66\.03 \-39\.37 L 12\.71 \-39\.37\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \{\{\}\}\{\}\\lx@inpgf@ignorespaces\\hbox\{\\hbox\{\{\\lxSVG@begingroup@\{\_scopebegin=1\} \{\{\}\}\{\}\{\{\}\} \{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\}\{\{\{\{\{\}\}\{\}\{\}\{\}\{\}\{\{\}\}\}\}\}\{\{\}\}\{\}\{\}\{\}\\lxSVG@begingroup@\{\_scopebegin=1\} \\lxSVG@begingroup@\{stroke=\#000000\} \\lxSVG@setlinewidth\{\\the\\pgflinewidth\}\\lxSVG@begingroup@\{stroke\-width=1\.2pt\} \\lx@inpgf@ignorespaces\{\}\\lxSVG@stroke\\lxSVG@drawpath@unclipped\{M 78\.74 26\.66 L 78\.74 \-26\.66\}\{fill:none\} \\lx@inpgf@ignorespaces \\lxSVG@closescope \\lxSVG@closescope \}\}\} \\lxSVG@closescope \\lxSVG@closescope \{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\{\\lx@inpgf@ignorespaces\}\\hss\}\\lxSVG@discardpath\\lxSVG@closescope \\hss\}\}\\lxSVG@closescope\\endpgfpicture\}\}$ &\\ \hline\cr\end{tabular}\ \ \ \ \ \ .Which of them have proper33-colorings?

•The graphAAhas a proper33-coloring. For example, the mapffthat sends the vertices1,2,3,4,51,2,3,4,5to the colors1,2,1,2,31,2,1,2,3(respectively) is a proper33-coloring.
•The graphBBhas no proper33-coloring. Indeed, the four vertices2,3,4,52,3,4,5are mutually adjacent, so they would have to have44distinct colors in a properkk-coloring; but this is not possible unlessk≥4k\geq 4.
•The graphCChas a proper33-coloring and even a proper22-coloring (e.g., assigning color11to each odd vertex and color22to each even vertex).
•The graphDDhas no proper33-coloring and, in fact, no properkk-coloring for anyk∈ℕk\in\mathbb{N}. The reason is that the vertex33is adjacent to itself, but obviously has the same color as itself no matter what thekk-coloring is. More generally, a graph with a loop cannot have a properkk-coloring for anyk∈ℕk\in\mathbb{N}.

Example 6.4.

Letn∈ℕn\in\mathbb{N}. Thenn-hypercubeQnQ_{n}(introduced in Definition2.139) has a proper22-coloring: Namely, the map

f:{0,1}n\displaystyle f:\left\{0,1\right\}^{n}→{1,2},\displaystyle\rightarrow\left\{1,2\right\},(a1,a2,…,an)\displaystyle\left(a_{1},a_{2},\ldots,a_{n}\right)↦{1,ifa1+a2+⋯+anis odd;2,ifa1+a2+⋯+anis even\displaystyle\mapsto\begin{cases}1,&\text{if }a_{1}+a_{2}+\cdots+a_{n}\text{ is odd};\\ 2,&\text{if }a_{1}+a_{2}+\cdots+a_{n}\text{ is even}\end{cases}is a proper22-coloring ofQnQ_{n}. (Check this! It boils down to the fact that if two bitstrings(a1,a2,…,an)\left(a_{1},a_{2},\ldots,a_{n}\right)and(b1,b2,…,bn)\left(b_{1},b_{2},\ldots,b_{n}\right)differ in exactly one entry, then the corresponding sumsa1+a2+⋯+ana_{1}+a_{2}+\cdots+a_{n}andb1+b2+⋯+bnb_{1}+b_{2}+\cdots+b_{n}differ by exactly11.)

Example 6.5.

Letnnandmmbe two positive integers. The Cartesian productPn×PmP_{n}\times P_{m}of thenn-th path graphPnP_{n}and themm-th path graphPmP_{m}is known as the(n,m)\left(n,m\right)-grid graph, as it looks as follows:

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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 116.48 362.21 L 155.17 362.21}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 207.03 362.21 L 245.72 362.21}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{ {}{}{}} {{{{{}}{ {}{}}{}{}{{}{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 297.59 362.21 L 352.13 362.21}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} { {}{}{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{ {}{}}{}{}{{}{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 372.28 362.21 L 426.28 362.21}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This(n,m)\left(n,m\right)-grid graphPn×PmP_{n}\times P_{m}has a proper22-coloring: namely, the map that sends each vertex(i,j)\left(i,j\right)to{1,ifi+jis even;2,ifi+jis odd.\begin{cases}1,&\text{if }i+j\text{ is even};\\ 2,&\text{if }i+j\text{ is odd}.\end{cases}

This22-coloring is called the “chessboard coloring” for a fairly obvious reason (view each vertex as a square of a chessboard).

More generally, ifGGandHHare two simple graphs each having a proper22-coloring, then their Cartesian productG×HG\times Hhas a proper22-coloring as well. (See Exercise6.29for the proof.)

Example 6.6.

Here is the Petersen graph (as defined in Subsection2.6.3):

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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 75.49 -167.05 L 182.19 -20.17}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .I claim that it has a proper33-coloring. Can you find it?

As we see, some graphs have proper33-colorings, while others don’t. Clearly, having44mutually adjacent vertices makes a proper33-coloring impossible (indeed, by the pigeonhole principle, two of them must have the same color), but this is far from an “if and only if”. The question of determining whether a given graph has a proper33-coloring is NP-complete (see, e.g.,[Goldre10, Proposition 4.11]).

6.22-colorings

6.2.1The undirected case

In contrast, the existence of proper22-colorings is a much simpler question. The following is a nice criterion:

Theorem 6.7(2-coloring equivalence theorem).

LetGGbe a multigraph. Then, the following three statements are equivalent:

•Statement B1:The graphGGhas a proper22-coloring.
•Statement B2:The graphGGhas no cycles of odd length.
•Statement B3:The graphGGhas no circuits of odd length.

To prove this theorem, we will need a fact that is somewhat similar to Proposition3.29:

Proposition 6.8.

LetGGbe a multigraph. Letuuandvvbe two vertices ofGG. Let𝐰\mathbf{w}be an odd-length walk fromuutovv. Then,𝐰\mathbf{w}contains either an odd-lengthpathfromuutovvor an odd-lengthcycle(or both).

Here, we are using the following rather intuitive terminology:

•A walk is said to beodd-lengthif its length is odd.
•A walk𝐰\mathbf{w}is said tocontaina walk𝐯\mathbf{v}if each edge of𝐯\mathbf{v}is an edge of𝐰\mathbf{w}. (This does not necessarily mean that𝐯\mathbf{v}appears in𝐰\mathbf{w}as a contiguous block.)
•We remind the reader once again that a “circuit” just means a closed walk to us; we impose no further requirements.

Example 6.9.

Consider the following simple graph (which we treat as a multigraph):

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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ . (a)The odd-length walk(1,∗,2,∗,3,∗,4,∗,5,∗,2,∗,6,∗,7)\left(1,\ast,2,\ast,3,\ast,4,\ast,5,\ast,2,\ast,6,\ast,7\right)(we are using asterisks for the edges, since they can be trivially recovered from the vertices) contains the odd-length path(1,∗,2,∗,6,∗,7)\left(1,\ast,2,\ast,6,\ast,7\right)from11to77.

(b)The odd-length walk(3,∗,2,∗,1,∗,6,∗,2,∗,3)\left(3,\ast,2,\ast,1,\ast,6,\ast,2,\ast,3\right)contains the odd-length cycle(2,∗,1,∗,6,∗,2)\left(2,\ast,1,\ast,6,\ast,2\right).

Proof 6.10(Proof of Proposition6.8.).

We apply strong induction on the length of𝐰\mathbf{w}.

Thus, we fix ak∈ℕk\in\mathbb{N}, and we assume (as the induction hypothesis) that Proposition6.8is already proved for all odd-length walks of length<k<k. Now, we must prove it for an odd-length walk𝐰\mathbf{w}of lengthkk.

Write this walk𝐰\mathbf{w}as𝐰=(w0,∗,w1,∗,w2,…,∗,wk)\mathbf{w}=\left(w_{0},\ast,w_{1},\ast,w_{2},\ldots,\ast,w_{k}\right). Hence,kkis the length of𝐰\mathbf{w}, and thus is odd.

We must prove that𝐰\mathbf{w}contains either an odd-length path fromuutovvor an odd-length cycle.

If𝐰\mathbf{w}itself is a path, then we are done. So WLOG assume that𝐰\mathbf{w}is not a path. Thus, two of the verticesw0,w1,…,wkw_{0},w_{1},\ldots,w_{k}of𝐰\mathbf{w}are equal. In other words, there exists a pair(i,j)\left(i,j\right)of integersiiandjjwith0≤i<j≤k0\leq i<j\leq kandwi=wjw_{i}=w_{j}. Among all such pairs, we pick one withminimumdifferencej−ij-i. Then, the verticeswi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct (sincej−ij-iis minimum).

Let𝐜\mathbf{c}be the part of𝐰\mathbf{w}betweenwiw_{i}andwjw_{j}; thus,959595Here is an illustration (which, however, is a bit simplistic: the walk𝐰\mathbf{w}can intersect itself arbitrarily many times, not just once as shown here):w0w1wi=wjwj+1wkwi+1wj−1⋯⋯⋯.\hbox to381.85pt{\vbox to148.69pt{\pgfpicture\makeatletter\hbox{\hskip 11.56552pt\lower-21.3563pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} 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𝐜=(wi,∗,wi+1,∗,…,∗,wj).\mathbf{c}=\left(w_{i},\ast,w_{i+1},\ast,\ldots,\ast,w_{j}\right).This𝐜\mathbf{c}is clearly a closed walk (sincewi=wjw_{i}=w_{j}). Ifj−ij-iis odd, then this closed walk𝐜\mathbf{c}is a cycle (indeed, its verticeswi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct, and therefore its edges are distinct as well969696For the very skeptical, here is aproofof this: Assume (for the sake of contradiction) that the walk𝐜\mathbf{c}has two equal edges. Let the first of them be an edge betweenwpw_{p}andwp+1w_{p+1}, and let the second be an edge betweenwqw_{q}andwq+1w_{q+1}, for some distinct elementsppandqqof{i,i+1,…,j−1}\left\{i,i+1,\ldots,j-1\right\}. Since equal edges have equal endpoints, we thus have{wp,wp+1}={wq,wq+1}\left\{w_{p},w_{p+1}\right\}=\left\{w_{q},w_{q+1}\right\}, so thatwp∈{wp,wp+1}={wq,wq+1}w_{p}\in\left\{w_{p},w_{p+1}\right\}=\left\{w_{q},w_{q+1}\right\}. In other words,wpw_{p}equals eitherwqw_{q}orwq+1w_{q+1}. Sincewp≠wqw_{p}\neq w_{q}(becausewi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct), this entails thatwp=wq+1w_{p}=w_{q+1}. Similarly,wq=wp+1w_{q}=w_{p+1}. However,ppandqqare distinct. Thus, at least one ofppandqqis distinct fromj−1j-1. We WLOG assume thatq≠j−1q\neq j-1(otherwise, we can swapppwithqq). Hence,q+1≠jq+1\neq j, so thatq+1∈{i,i+1,…,j−1}q+1\in\left\{i,i+1,\ldots,j-1\right\}. Thus, fromwp=wq+1w_{p}=w_{q+1}, we conclude thatp=q+1p=q+1(sincewi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct). Thus,p=q+1>qp=q+1>q, so thatp+1>p>qp+1>p>qand thereforep+1≠qp+1\neq q. However,wq=wp+1w_{q}=w_{p+1}. Ifp+1p+1was an element of{i,i+1,…,j−1}\left\{i,i+1,\ldots,j-1\right\}, then this would entailq=p+1q=p+1(sincewi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct), which would contradictp+1≠qp+1\neq q. Thus,p+1p+1cannot be an element of{i,i+1,…,j−1}\left\{i,i+1,\ldots,j-1\right\}. Hence,p+1=jp+1=j(sincep+1p+1clearly belongs to{i,i+1,…,j}\left\{i,i+1,\ldots,j\right\}). Thus,wp+1=wj=wiw_{p+1}=w_{j}=w_{i}, so thatwi=wp+1=wqw_{i}=w_{p+1}=w_{q}. This entailsi=qi=q(sincewi,wi+1,…,wj−1w_{i},w_{i+1},\ldots,w_{j-1}are distinct). Hence,i=q=p−1i=q=p-1(sincep=q+1p=q+1). Therefore,j⏟=p+1−i⏟=p−1=(p+1)−(p−1)=2\underbrace{j}_{=p+1}-\underbrace{i}_{=p-1}=\left(p+1\right)-\left(p-1\right)=2. This contradicts the fact thatj−ij-iis odd. This contradiction shows that our assumption (that the walk𝐜\mathbf{c}has two equal edges) was false. Hence, the edges of𝐜\mathbf{c}are distinct.), and thus we have found an odd-length cycle contained in𝐰\mathbf{w}(namely,𝐜\mathbf{c}is such a cycle, since its length isj−ij-i, which is odd). This means that we are done ifj−ij-iis odd.

Thus, we WLOG assume thatj−ij-iis even. Hence, cutting out the closed walk𝐜\mathbf{c}from the original walk𝐰\mathbf{w}, we obtain a walk

𝐰′:=(w0,∗,w1,∗,…,∗,wi=wj,∗,wj+1,∗,wj+2,…,wk)\mathbf{w}^{\prime}:=\left(w_{0},\ast,w_{1},\ast,\ldots,\ast,w_{i}=w_{j},\ast,w_{j+1},\ast,w_{j+2},\ldots,w_{k}\right)fromuutovv. This new walk𝐰′\mathbf{w}^{\prime}has lengthk−(j−i)k-\left(j-i\right), which is odd (sincekkis odd butj−ij-iis even) and smaller thankk(sincei<ji<j). Hence, we can apply the induction hypothesis to this walk𝐰′\mathbf{w}^{\prime}. As a consequence, we conclude that this walk𝐰′\mathbf{w}^{\prime}contains either an odd-length path fromuutovvor an odd-length cycle. Therefore, the walk𝐰\mathbf{w}also contains either an odd-length path fromuutovvor an odd-length cycle (since anything contained in𝐰′\mathbf{w}^{\prime}is automatically contained in𝐰\mathbf{w}). But this is precisely what we set out to prove. This completes the induction step, and so we have proved Proposition6.8.

Now, let us prove the 2-coloring equivalence theorem:

Proof 6.11(Proof of Theorem6.7.).

Write the multigraphGGasG=(V,E,φ)G=\left(V,E,\varphi\right). We shall prove the implications B1⟹\LongrightarrowB2⟹\LongrightarrowB3⟹\LongrightarrowB1.

Proof of the implication B1⟹\LongrightarrowB2:Assume that Statement B1 holds. We must prove that Statement B2 holds.

We have assumed that B1 holds. In other words, the graphGGhas a proper22-coloring. Letffbe this22-coloring. Thus,ffis a map fromVVto{1,2}\left\{1,2\right\}such that any two adjacent verticesxxandyyofGGsatisfyf⁡(x)≠f⁡(y)f\left(x\right)\neq f\left(y\right).

Assume (for contradiction) thatGGhas a cycle of odd length. Let

(v0,∗,v1,∗,v2,∗,…,∗,vk)\left(v_{0},\ast,v_{1},\ast,v_{2},\ast,\ldots,\ast,v_{k}\right)be this cycle. Thus,kkis odd, and we havevk=v0v_{k}=v_{0}, so thatf⁡(vk)=f⁡(v0)f\left(v_{k}\right)=f\left(v_{0}\right). Moreover, for eachi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}, the vertexviv_{i}is adjacent tovi−1v_{i-1}(since(v0,∗,v1,∗,v2,∗,…,∗,vk)\left(v_{0},\ast,v_{1},\ast,v_{2},\ast,\ldots,\ast,v_{k}\right)is a cycle) and therefore satisfies

f⁡(vi)≠f⁡(vi−1)f\left(v_{i}\right)\neq f\left(v_{i-1}\right)(50)(sinceffis a proper22-coloring).

We WLOG assume thatf⁡(v0)=1f\left(v_{0}\right)=1(otherwise, we “rename” the colors11and22so that the colorf⁡(v0)f\left(v_{0}\right)becomes11). Then, (50) (applied toi=1i=1) yieldsf⁡(v1)≠f⁡(v0)=1f\left(v_{1}\right)\neq f\left(v_{0}\right)=1, so thatf⁡(v1)=2f\left(v_{1}\right)=2(sincef⁡(v1)f\left(v_{1}\right)must be either11or22). Hence, (50) (applied toi=2i=2) yieldsf⁡(v2)≠f⁡(v1)=2f\left(v_{2}\right)\neq f\left(v_{1}\right)=2, so thatf⁡(v2)=1f\left(v_{2}\right)=1(sincef⁡(v2)f\left(v_{2}\right)must be either11or22). For similar reasons, we can successively obtainf⁡(v3)=2f\left(v_{3}\right)=2andf⁡(v4)=1f\left(v_{4}\right)=1andf⁡(v5)=2f\left(v_{5}\right)=2and so on. The general formula we obtain (strictly speaking, it needs to be proved by induction onii) says that

f⁡(vi)={1,ifiis even;2,ifiis oddfor eachi∈{0,1,…,k}.f\left(v_{i}\right)=\begin{cases}1,&\text{if }i\text{ is even;}\\ 2,&\text{if }i\text{ is odd}\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{0,1,\ldots,k\right\}.Applying this toi=ki=k, we conclude thatf⁡(vk)=2f\left(v_{k}\right)=2(sincekkis odd). However, this contradictsf⁡(vk)=f⁡(v0)=1≠2f\left(v_{k}\right)=f\left(v_{0}\right)=1\neq 2. This contradiction shows that our assumption was false. Hence,GGhas no cycle of odd length. In other words, Statement B2 holds. This proves the implication B1⟹\LongrightarrowB2.

Proof of the implication B2⟹\LongrightarrowB3:Assume that Statement B2 holds. We must prove that Statement B3 holds. In other words, we must show thatGGhas no odd-length circuits.

Assume the contrary. Thus,GGhas an odd-length circuit𝐰\mathbf{w}. Letuube the starting and ending point of𝐰\mathbf{w}. Thus, Proposition6.8(applied tov=uv=u) shows that this odd-length circuit𝐰\mathbf{w}contains either an odd-lengthpathfromuutouuor an odd-length cycle. SinceGGhas no odd-length cycle (because we assumed that Statement B2 holds), we thus concludes that𝐰\mathbf{w}contains an odd-lengthpathfromuutouu. However, an odd-length path fromuutouuis impossible (since the only path fromuutouuhas length00). Thus, we obtain a contradiction, which shows thatGGhas no odd-length circuits. This proves the implication B2⟹\LongrightarrowB3.

Proof of the implication B3⟹\LongrightarrowB1:Assume that Statement B3 holds. We must prove that Statement B1 holds.

We have assumed that Statement B3 holds. In other words,GGhas no odd-length circuits. We must find a proper22-coloring ofGG.

We WLOG assume thatGGis connected (otherwise, letC1,C2,…,CkC_{1},C_{2},\ldots,C_{k}be the components ofGG, and apply the implication B3⟹\LongrightarrowB1 to each of the smaller graphsG⁡[C1],G⁡[C2],…,G⁡[Ck]G\left[C_{1}\right],\ G\left[C_{2}\right],\ \ldots,\ G\left[C_{k}\right], and then combine the resulting proper22-colorings of these smaller graphs into a single proper22-coloring ofGG). Fix any vertexrrofGG. Define a mapf:V→{1,2}f:V\rightarrow\left\{1,2\right\}by setting

f⁡(v)={1,ifd(v,r)is even;2,ifd(v,r)is oddfor eachv∈Vf\left(v\right)=\begin{cases}1,&\text{if }d\left(v,r\right)\text{ is even};\\ 2,&\text{if }d\left(v,r\right)\text{ is odd}\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for each }v\in V(whered⁡(v,r)d\left(v,r\right)denotes the distance fromvvtorr, that is, the smallest length of a path fromvvtorr).

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\lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 274.47 -8.99 L 245.21 -38.26}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(Of course, the numbers on the nodes here are not the vertices, but rather the colors of these vertices.) Note that all values offfcan be easily found by the following recursive algorithm: Start by assigning the color11torr. Then, assign the color22to all neighbors ofrr. Then, assign the color11to all neighbors of these neighbors (unless they have already been colored). Then, assign the color22to all neighbors of these neighbors of these neighbors, and so on.Indeed, assume the contrary. Thus, some two adjacent verticesuuandvvhave the same colorf⁡(u)=f⁡(v)f\left(u\right)=f\left(v\right). Consider theseuuandvv. Sincef⁡(u)=f⁡(v)f\left(u\right)=f\left(v\right), we are in one of the following two cases:

Case 1:We havef⁡(u)=f⁡(v)=1f\left(u\right)=f\left(v\right)=1.

Case 2:We havef⁡(u)=f⁡(v)=2f\left(u\right)=f\left(v\right)=2.

Let us consider Case 2. In this case, we havef⁡(u)=f⁡(v)=2f\left(u\right)=f\left(v\right)=2. This means thatd⁡(u,r)d\left(u,r\right)andd⁡(v,r)d\left(v,r\right)are both odd (by the definition offf). Hence, there is an odd-length path𝐩\mathbf{p}fromuutorrand an odd-length path𝐪\mathbf{q}fromvvtorr. Consider these𝐩\mathbf{p}and𝐪\mathbf{q}. Also, there is an edgeeethat joinsuuandvv(sinceuuandvvare adjacent). Consider this edgeee. By combining the paths𝐩\mathbf{p}and𝐪\mathbf{q}and inserting the edgeeeinto the result, we obtain a circuit fromrrtorr(which starts by following the path𝐩\mathbf{p}backwards touu, then takes the edgeeetovv, then follows the path𝐪\mathbf{q}back torr). This circuit has odd length (since𝐩\mathbf{p}and𝐪\mathbf{q}have odd lengths, and since the edgeeeadds11to the length). Thus, we have found an odd-length circuit ofGG. However, we assumed thatGGhas no odd-length circuits. Contradiction!

Thus, we have found a contradiction in Case 2. Similarly, we can find a contradiction in Case 1. Thus, we always get a contradiction. This shows thatffis indeed a proper22-coloring. Thus, Statement B1 holds. This proves the implication B3⟹\LongrightarrowB1.989898Note that this proof provides a reasonably efficient algorithm for constructing a proper22-coloring ofGG, as long as you know how to compute distances in a graph (we have done this, e.g., in homework set #4 exercise 5) and how to compute the components of a graph (this is not hard).

For aesthetical reasons, let me give asecond proof of the implication B3⟹\LongrightarrowB1, which avoids the awkward “breakGGup into components” step:

Assume again that Statement B3 holds. We must prove that Statement B1 holds.

We assumed that Statement B3 holds. In other words,GGhas no odd-length cycles.

Two verticesuuandvvofGGwill be calledoddly connectedifGGhas an odd-length path fromuutovv. By Proposition6.8, this condition is equivalent to “GGhas an odd-length walk fromuutovv”, sinceGGhas no odd-length cycles. Moreover, a vertexuucannot be oddly connected to itself (since the only path fromuutouuis the trivial length-00path(u)\left(u\right), which is not odd-length).

A subsetAAofVVwill be calledodd-path-lessif no two vertices inAAare oddly connected. (Note that “two vertices” doesn’t mean “two distinct vertices”.)

Pick a maximum-size odd-path-less subsetAAofVV(such anAAexists, since∅\varnothingis clearly odd-path-less). Now, letf:V→{1,2}f:V\rightarrow\left\{1,2\right\}be the22-coloring ofGGthat assigns the color11to all vertices inAAand assigns the color22to all vertices not inAA.

We shall show that this22-coloringffis proper.

To prove this, we must show that no two adjacent vertices have color11and that no two adjacent vertices have color22. The first of these two claims is obvious999999Proof.An edge always makes a walk of length11, which is odd. Thus, two adjacent vertices are automatically oddly connected. Hence, two adjacent vertices cannot both be contained in the odd-path-less subsetAA. In other words, two adjacent vertices cannot both have color11.. It thus remains to prove the second claim – i.e., to prove that no two adjacent vertices have color22.

Assume the contrary. Thus, there exist two adjacent verticesuuandvvthat both have color22. Consider theseuuandvv. These verticesuuandvvhave color22; in other words, neither of them belongs toAA.

The vertexuuis not oddly connected to itself (as we already saw). Hence, the vertexuuis oddly connected to at least one vertexa∈Aa\in A(because otherwise, we could insertuuinto the odd-path-less setAAand obtain a larger odd-path-less subsetA∪{u}A\cup\left\{u\right\}ofVV; but this would contradict the fact thatAAis amaximum-sizeodd-path-less subset ofVV). For similar reasons, the vertexvvis oddly connected to at least one vertexb∈Ab\in A. Consider these verticesaaandbb. Sinceuuis oddly connected toaa, there exists an odd-length walk𝐩\mathbf{p}fromuutoaa. Reversing this walk𝐩\mathbf{p}yields an odd-length walk𝐩′\mathbf{p}^{\prime}fromaatouu. Sincevvis oddly connected tobb, there exists an odd-length walk𝐪\mathbf{q}fromvvtobb. Finally, there is an edgeeewith endpointsuuandvv(sinceuuandvvare adjacent). Combine the two walks𝐩′\mathbf{p}^{\prime}and𝐪\mathbf{q}and insert this edgeeebetween them; this yields a walk fromaatobb(viauuandvv) that has odd length (since𝐩′\mathbf{p}^{\prime}and𝐪\mathbf{q}have odd length each, and insertingeeadds11to the length). Thus,GGhas an odd-length walk fromaatobb. In other words, the verticesaaandbbare oddly connected. This contradicts the fact that the setAAis odd-path-less (sinceaaandbbbelong toAA).

This contradiction shows that our assumption was false. Thus, we have shown that no two adjacent vertices have color22. This completes our proof thatffis a proper22-coloring. Thus, Statement B1 holds. This proves the implication B3⟹\LongrightarrowB1 once again.

Having proved all three implications B1⟹\LongrightarrowB2 and B2⟹\LongrightarrowB3 and B3⟹\LongrightarrowB1, we now conclude that the three statements B1, B2 and B3 are equivalent. This proves Theorem6.7.

A further simple property of proper22-colorings is the following:100100100Recall thatconnG\operatorname*{conn}Gdenotes the number of components of a graphGG.

Proposition 6.13.

LetGGbe a multigraph that has a proper22-coloring. Then,GGhas exactly2connG2^{\operatorname*{conn}G}many proper22-colorings.

Proof 6.14(Proof sketch.).

For each componentCCofGG, let us fix an arbitrary vertexrC∈Cr_{C}\in C. When constructing a proper22-coloringffofGG, we can freely choose the colorsf⁡(rC)f\left(r_{C}\right)of these verticesrCr_{C}; the colors of all other vertices are then uniquely determined (see the first proof of the implication B3⟹\LongrightarrowB1 in our above proof of Theorem6.7for the details). Thus, we have2connG2^{\operatorname*{conn}G}many options (sinceGGhasconnG\operatorname*{conn}Gmany components). The proposition follows.

6.2.2The directed case

There is an interesting variant of Theorem6.7for directed graphs ([Langlo23, Lemma 1.1.5]):

Theorem 6.15.

LetDDbe a strongly connected multidigraph. Then, the following three statements are equivalent:

•Statement B’1:The underlying undirected graphDundD^{\operatorname*{und}}has a proper22-coloring.
•Statement B’2:The digraphDDhas no cycles of odd length.
•Statement B’3:The digraphDDhas no circuits of odd length.

Keep in mind that cycles and circuits in a digraph are directed cycles, traversing each arc from source to target; thus, a digraphDDwill usually have a lot fewer cycles than the underlying undirected graphDundD^{\operatorname*{und}}. So the implication B’2⟹\LongrightarrowB’1 in Theorem6.15is much stronger than the corresponding implication B2⟹\LongrightarrowB1 in Theorem6.7, seeing that statement B’2 looks weaker than B2 forG=DundG=D^{\operatorname*{und}}. The “strongly connected” requirement in Theorem6.15is the price to pay for this extra strength: No connectedness was needed in Theorem6.7, but we do need to requireDDto be strongly connected in Theorem6.15, since otherwise the implication B’2⟹\LongrightarrowB’1 would fail for the first digraph in Example4.90(a)(which has no cycles at all, but whose underlying undirected graphK3K_{3}certainly has no proper22-coloring).

The equivalence B’2⟺\LongleftrightarrowB’3 in Theorem6.15does indeed hold even whenDDis not strongly connected. This follows easily from the following directed analogue of Proposition6.8:

Proposition 6.16.

LetDDbe a multidigraph. Letuuandvvbe two vertices ofDD. Let𝐰\mathbf{w}be an odd-length walk fromuutovv. Then,𝐰\mathbf{w}contains either an odd-lengthpathfromuutovvor an odd-lengthcycle(or both).

Proof 6.17.

Literally the same as for Proposition6.8. (It is even a bit easier, because proving that a closed walk is a cycle in a multidigraph does not require showing that its arcs are distinct.)

Corollary 6.18.

LetDDbe a multidigraph. Then,DDhas no cycles of odd length if and only ifDDhas no circuits of odd length.

Proof 6.19.

⟹:\Longrightarrow:Assume thatDDhas no cycles of odd length. We must show thatDDhas no circuits of odd length.

Let𝐜\mathbf{c}be a circuit of odd length. Letuube its starting point. Then,uuis also its ending point (since𝐜\mathbf{c}is a circuit). Hence,𝐜\mathbf{c}is an odd-length walk fromuutouu. Thus, Proposition6.16(applied to𝐰=𝐜\mathbf{w}=\mathbf{c}andv=uv=u) shows that𝐜\mathbf{c}contains either an odd-lengthpathfromuutouuor an odd-lengthcycle(or both). The former option is impossible, since the only path fromuutouuis the length-00path(u)\left(u\right), which is certainly not odd-length. But the latter option is also impossible, since we have assumed thatDDhas no cycles of odd length. Thus, we obtain a contradiction in each case.

So we have found a contradiction whenever𝐜\mathbf{c}is a circuit of odd length. Hence,DDhas no circuits of odd length. This proves the “⟹\Longrightarrow” direction of Corollary6.18.

⟸:\Longleftarrow:Obvious (since any cycle is a circuit).

We can now prove Theorem6.15:

Proof 6.20(Proof of Theorem6.15.).

Corollary6.18shows that statements B’2 and B’3 are equivalent.

We shall now prove the implications B’1⟹\LongrightarrowB’3⟹\LongrightarrowB’1.

Proof of the implication B’1⟹\LongrightarrowB’3:Assume that Statement B’1 holds. We must prove that Statement B’3 holds.

We have assumed that B’1 holds. In other words, the graphDundD^{\operatorname*{und}}has a proper22-coloring. Hence, the implication B1⟹\LongrightarrowB3 of Theorem6.7(applied toG=DundG=D^{\operatorname*{und}}) shows that the graphDundD^{\operatorname*{und}}has no circuits of odd length. Hence, the digraphDDhas no circuits of odd length either (since each circuit ofDDis also a circuit ofDundD^{\operatorname*{und}}). In other words, Statement B’3 holds. This proves the implication B’1⟹\LongrightarrowB’3.

Proof of the implication B’3⟹\LongrightarrowB’1:Assume that Statement B’3 holds. We must prove that Statement B’1 holds.

We have assumed that B’3 holds. In other words, the digraphDDhas no circuits of odd length.

LetVVbe the vertex set ofDD. For any two verticesu,v∈Vu,v\in V, we letd⁡(u,v)d\left(u,v\right)be the smallest length of a path fromuutovvinDD(such a path exists, sinceDDis strongly connected). Now we claim the following:

Claim 1:Letu,v∈Vu,v\in Vbe two vertices. Let𝐰\mathbf{w}be any walk fromvvtouuinDD. Then, the length of𝐰\mathbf{w}is≡d⁡(u,v)mod⁡2\equiv d\left(u,v\right)\operatorname{mod}2.

Proof 6.21(Proof of Claim 1.).

Letkkbe the length of𝐰\mathbf{w}. We must then show thatk≡d⁡(u,v)mod⁡2k\equiv d\left(u,v\right)\operatorname{mod}2.

The digraphDDhas a path𝐩\mathbf{p}fromuutovvwhose length isd⁡(u,v)d\left(u,v\right)(by the definition ofd⁡(u,v)d\left(u,v\right)). Consider this𝐩\mathbf{p}. Splicing the path𝐩\mathbf{p}(fromuutovv) with the walk𝐰\mathbf{w}(fromvvtouu), we obtain a walk𝐩∗𝐰\mathbf{p}\ast\mathbf{w}fromuutouu. This walk𝐩∗𝐰\mathbf{p}\ast\mathbf{w}is a circuit (being a walk fromuutouu), and thus cannot have odd length (sinceDDhas no circuits of odd length). Hence,𝐩∗𝐰\mathbf{p}\ast\mathbf{w}has even length. But the length of𝐩∗𝐰\mathbf{p\ast w}isd⁡(u,v)+kd\left(u,v\right)+k(since𝐩\mathbf{p}has lengthd⁡(u,v)d\left(u,v\right)while𝐰\mathbf{w}has lengthkk). Combining the previous two sentences, we conclude thatd⁡(u,v)+kd\left(u,v\right)+kis even. In other words,d⁡(u,v)≡−k≡kmod⁡2d\left(u,v\right)\equiv-k\equiv k\operatorname{mod}2, so thatk≡d⁡(u,v)mod⁡2k\equiv d\left(u,v\right)\operatorname{mod}2. This proves Claim 1.

Now, pick any vertexrrofDD(this exists, sinceDDis strongly connected and thus has at least11vertex). Define a mapf:V→{1,2}f:V\rightarrow\left\{1,2\right\}by setting

f⁡(v)={1,ifd(v,r)is even;2,ifd(v,r)is oddfor eachv∈V.f\left(v\right)=\begin{cases}1,&\text{if }d\left(v,r\right)\text{ is even};\\ 2,&\text{if }d\left(v,r\right)\text{ is odd}\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for each }v\in V.I claim thatffis a proper22-coloring ofDundD^{\operatorname*{und}}. Indeed, assume the contrary. Thus, some two adjacent verticesuuandvvofDundD^{\operatorname*{und}}have the same colorf⁡(u)=f⁡(v)f\left(u\right)=f\left(v\right). Consider theseuuandvv. Sincef⁡(u)=f⁡(v)f\left(u\right)=f\left(v\right), we are in one of the following two cases:

Case 1:We havef⁡(u)=f⁡(v)=1f\left(u\right)=f\left(v\right)=1.

Case 2:We havef⁡(u)=f⁡(v)=2f\left(u\right)=f\left(v\right)=2.

The digraphDDis strongly connected, and thus has a path𝐩\mathbf{p}fromrrtouuas well as a path from𝐪\mathbf{q}fromrrtovv. Consider these paths𝐩\mathbf{p}and𝐪\mathbf{q}. Claim 1 (applied to𝐩\mathbf{p}andrrinstead of𝐰\mathbf{w}andvv) yields that the length of the path𝐩\mathbf{p}is≡d⁡(u,r)mod⁡2\equiv d\left(u,r\right)\operatorname{mod}2. In other words, the length of the path𝐩\mathbf{p}is odd (sinced⁡(u,r)d\left(u,r\right)is odd). Likewise, the length of the path𝐪\mathbf{q}is odd.

But the fact thatuuandvvare adjacent inDundD^{\operatorname*{und}}shows that the digraphDDhas either an arc fromuutovvor an arc fromvvtouu(or both). We WLOG assume thatDDhas an arc fromuutovv(otherwise, we can just swapuuwithvv). Letaabe this arc. Attaching the arcaato the path𝐩\mathbf{p}fromrrtouuat its end, we obtain a walk𝐰\mathbf{w}fromrrtovv, and moreover this walk has even length (since the path𝐩\mathbf{p}has odd length, and we have obtained𝐰\mathbf{w}by attaching one more arc to it).

But Claim 1 (applied tovvandrrinstead ofuuandvv) shows that the length of𝐰\mathbf{w}is≡d⁡(v,r)mod⁡2\equiv d\left(v,r\right)\operatorname{mod}2. In other words, the length of𝐰\mathbf{w}is odd (sinced⁡(v,r)d\left(v,r\right)is odd). But this contradicts the fact that𝐰\mathbf{w}has even length.

Having proved the implications B’1⟹\LongrightarrowB’3 and B’3⟹\LongrightarrowB’1 and the equivalence B’2⟺\LongleftrightarrowB’3, we conclude that all three statements B’1, B’2 and B’3 are equivalent. Thus, Theorem6.15holds.

6.3The Brooks theorems

As we said, the existence of a properkk-coloring for a given graphGGis a hard computational problem unlessk≤2k\leq 2. The same holds for theoretical criteria: Fork>2k>2, I am not aware of any good criteria that are simultaneously necessary and sufficient for the existence of a properkk-coloring. However, some sufficient criteria are known. Here is one:101101101Recall that a multigraph is calledlooplessif it has no loops.

Theorem 6.22(Little Brooks theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a loopless multigraph with at least one vertex. Let

α:=max⁡{deg⁡v∣v∈V}.\alpha:=\max\left\{\deg v\ \mid\ v\in V\right\}.Then,GGhas a proper(α+1)\left(\alpha+1\right)-coloring.

Proof 6.23(Proof sketch.).

Letv1,v2,…,vnv_{1},v_{2},\ldots,v_{n}be the vertices ofVV, listed in some order (with no repetitions). We construct a proper(α+1)\left(\alpha+1\right)-coloringf:V→{1,2,…,α+1}f:V\rightarrow\left\{1,2,\ldots,\alpha+1\right\}ofGGrecursively as follows:

•First, we choosef⁡(v1)f\left(v_{1}\right)arbitrarily.
•Then, we choosef⁡(v2)f\left(v_{2}\right)to be distinct from the colors of all already-colored neighbors ofv2v_{2}.
•Then, we choosef⁡(v3)f\left(v_{3}\right)to be distinct from the colors of all already-colored neighbors ofv3v_{3}.
•Then, we choosef⁡(v4)f\left(v_{4}\right)to be distinct from the colors of all already-colored neighbors ofv4v_{4}.
•And so on, until all valuesf⁡(v1),f⁡(v2),…,f⁡(vn)f\left(v_{1}\right),\ f\left(v_{2}\right),\ \ldots,\ f\left(v_{n}\right)have been chosen.

Why do we never run out of colors in this process? Well: When choosingf⁡(vi)f\left(v_{i}\right), we must choose a color distinct from the colors of all already-colored neighbors ofviv_{i}. Sinceviv_{i}has at mostα\alphaneighbors (becausedeg⁡(vi)≤α\deg\left(v_{i}\right)\leq\alpha), this means that we have at mostα\alphacolors to avoid. Since there areα+1\alpha+1colors in total, this leaves us at least11color that we can choose; therefore, we don’t run out of colors.

The resulting(α+1)\left(\alpha+1\right)-coloringf:V→{1,2,…,α+1}f:V\rightarrow\left\{1,2,\ldots,\alpha+1\right\}is called agreedy coloring. This(α+1)\left(\alpha+1\right)-coloringffis indeed proper, because if an edge has endpointsviv_{i}andvjv_{j}withi>ji>j, then the construction off⁡(vi)f\left(v_{i}\right)ensures thatf⁡(vi)f\left(v_{i}\right)is distinct fromf⁡(vj)f\left(v_{j}\right). (Note how we are using the fact thatGGis loopless here! IfGGhad a loop, then the endpoints of this loop could not be written asviv_{i}andvjv_{j}withi>ji>j.)

In general, theα+1\alpha+1in Theorem6.22cannot be improved. Here are two examples:

•Ifn≥2n\geq 2, then the cycle graphCnC_{n}102102102See Definition3.18for the proper definition ofCnC_{n}whenn=2n=2.has maximum degree α=max⁡{deg⁡v∣v∈V}=2\alpha=\max\left\{\deg v\ \mid\ v\in V\right\}=2. Thus, Theorem6.22shows thatCnC_{n}has a proper33-coloring. Whennnis even,CnC_{n}has a proper22-coloring as well, but this is not the case whennnis odd (by Theorem6.7).
•Ifn≥1n\geq 1, then the complete graphKnK_{n}has maximum degree α=max⁡{deg⁡v∣v∈V}=n−1\alpha=\max\left\{\deg v\ \mid\ v\in V\right\}=n-1. Thus, Theorem6.22shows thatKnK_{n}has a propernn-coloring. By the pigeonhole principle, it is clear thatKnK_{n}has no proper(n−1)\left(n-1\right)-coloring.

Interestingly, these two examples are in fact the only cases when a connected loopless multigraph with maximum degreeα\alphacan fail to have a properα\alpha-coloring. In all other cases, we can improve theα+1\alpha+1toα\alpha:

Theorem 6.24(Brooks theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a connected loopless multigraph. Let

α:=max⁡{deg⁡v∣v∈V}.\alpha:=\max\left\{\deg v\ \mid\ v\in V\right\}.Assume thatGGis neither a complete graph nor an odd-length cycle. Then,GGhas a properα\alpha-coloring.

Proof 6.25.

Despite the seemingly little difference, this is significantly harder to prove than Theorem6.22. Various proofs can be found in[CraRab15]and in most serious textbooks on graph theory.

6.4Exercises on proper colorings

Exercise 6.26.

LetGGbe a simple graph withnnvertices. Letkkbe a positive integer.

Prove the following:

(a)IfGGhas a properkk-coloring, thenGGhas no subgraph isomorphic toKk+1K_{k+1}.
(b)Ifk≥n−2k\geq n-2, then the converse to part(a)also holds: IfGGhas no subgraph isomorphic toKk+1K_{k+1}, thenGGhas a properkk-coloring.
(c)Does the converse to part(a)hold fork<n−2k<n-2as well? Specifically, does it hold forn=5n=5andk=2k=2?

Exercise 6.27.

LetGGbe a connected loopless multigraph. Prove thatGGhas a proper22-coloring if and only if every three verticesu,v,wu,v,wofGGsatisfy

2|d⁡(u,v)+d⁡(v,w)+d⁡(w,u).2\mid d\left(u,v\right)+d\left(v,w\right)+d\left(w,u\right).

Exercise 6.28.

Fix two positive integersnnandkkwithn≥2k>0n\geq 2k>0. LetS={1,2,…,n}S=\left\{1,2,\ldots,n\right\}. Consider thekk-Kneser graphKS,kK_{S,k}as defined in Subsection2.6.3. Prove thatKS,kK_{S,k}has a proper(n−2k+2)\left(n-2k+2\right)-coloring.

[Hint:What can you say about the minima (i.e., smallest elements) of two disjoint subsets ofSS? (Being distinct is a good first step.)]

[Remark:Lóvasz has proved in 1978 (using topology!) that this result is optimal – in the sense thatn−2k+2n-2k+2is the smallest integerqqsuch thatKS,kK_{S,k}has a properqq-coloring.]

Exercise 6.29.

Letk∈ℕk\in\mathbb{N}. LetGGandHHbe two simple graphs. Assume that each ofGGandHHhas a properkk-coloring. Prove that the Cartesian productG×HG\times H(defined in Definition2.144) has a properkk-coloring as well.

[Remark:It is easy to see that the converse holds as well (i.e., ifG×HG\times Hhas a properkk-coloring, then so doGGandHH), provided that the vertex setsV(G)\operatorname*{V}\left(G\right)andV(H)\operatorname*{V}\left(H\right)are both nonempty.]

Exercise 6.30.

Letn∈ℕn\in\mathbb{N}. LetGGbe thenn-th coprimality graphCopn\operatorname*{Cop}\nolimits_{n}defined in Example2.3. Letk∈ℕk\in\mathbb{N}. Letmmbe the number of prime numbers in the set{1,2,…,n}\left\{1,2,\ldots,n\right\}. Prove the following:

(a)The graphGGhas a properkk-coloring if and only ifk≥m+1k\geq m+1.
(b)The graphGGhas a subgraph isomorphic toKkK_{k}if and only ifk≤m+1k\leq m+1.

Exercise 6.31.

Letnnandkkbe two positive integers. LetKKbe a set of sizekk. LetDDbe the de Bruijn digraph – i.e., the multidigraph constructed in the proof of Theorem5.180. LetGGbe the result of removing all loops from the undirected graphDundD^{\operatorname{und}}. Prove thatGGhas a proper(k+1)\left(k+1\right)-coloring.

Exercise 6.32.

Letk∈ℕk\in\mathbb{N}. LetGGbe a simple graph with fewer thank⁡(k+1)2\dfrac{k\left(k+1\right)}{2}edges. Prove thatGGhas a properkk-coloring.

Colorings and edge-colorings can be interesting even when they are not proper. Here is a sample result:

Exercise 6.33.

Letnnbe a positive integer. LetEEbe the edge set of the complete graphKnK_{n}. A mapf:E→ℤf:E\rightarrow\mathbb{Z}is said to be anedge-coloringofKnK_{n}. The values of this mapffare called thecolorsof the edge-coloringff. (We imagine each valuef⁡(e)f\left(e\right)to be a color assigned to the edgeee.)

An edge-coloringf:E→ℤf:E\rightarrow\mathbb{Z}ofKnK_{n}is said to beGallaiif for any three distinct verticesi,j,ki,j,k, at least two of the three numbersf⁡(ij)f\left(ij\right),f⁡(jk)f\left(jk\right)andf⁡(ki)f\left(ki\right)are equal (where we use the shorthanduvuvfor the edge{u,v}\left\{u,v\right\}, as usual). (Visually speaking, this means that each triangle has at least two equally-colored sides.)

Prove that the maximum number of colors (i.e., the maximum size off⁡(E)f\left(E\right)) in a Gallai edge-coloringffofKnK_{n}isn−1n-1.

6.5The chromatic polynomial

Here is another surprise: The number of properkk-colorings of a given multigraphGGturns out to be a polynomial function inkk(with integer coefficients). More precisely:

Theorem 6.34(Whitney’s chromatic polynomial theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. LetχG\chi_{G}be the polynomial in the single indeterminatexxwith coefficients inℤ\mathbb{Z}defined as follows:

χG=∑F⊆E(−1)|F|xconn(V,F,φ∣F)=∑His a spanningsubgraph ofG(−1)|E(H)|xconnH.\chi_{G}=\sum\limits_{F\subseteq E}\left(-1\right)^{\left|F\right|}x^{\operatorname*{conn}\left(V,F,\varphi\mid_{F}\right)}=\sum\limits_{\begin{subarray}{c}H\text{ is a spanning}\\ \text{subgraph of }G\end{subarray}}\left(-1\right)^{\left|\operatorname*{E}\left(H\right)\right|}x^{\operatorname*{conn}H}.(The symbol “∑F⊆E\sum\limits_{F\subseteq E}” means “sum over all subsetsFFofEE”.)

Then, for anyk∈ℕk\in\mathbb{N}, we have

(# of properk-colorings ofG)=χG(k).\left(\text{\# of proper }k\text{-colorings of }G\right)=\chi_{G}\left(k\right).

The proper place for this theorem is probably a course on enumerative combinatorics, but let us give here a proof for the sake of completeness (optional material). The following proof is essentially due to Hassler Whitney in 1930 ([Whitne32, §6]), and I am mostly copypasting it from my own writeup[17s-mt2s, §0.5](with some changes stemming from the fact that we are here working with multigraphs rather than simple graphs).

We are going to use theIverson bracket notation:

Definition 6.35.

If𝒜\mathcal{A}is any logical statement, then[𝒜]\left[\mathcal{A}\right]shall denote the truth value of𝒜\mathcal{A}; this is the number{1,if𝒜is true;0,if𝒜is false.\begin{cases}1,&\text{if }\mathcal{A}\text{ is true};\\ 0,&\text{if }\mathcal{A}\text{ is false}.\end{cases}

For instance,[2+2=4]=1\left[2+2=4\right]=1and[2+2=5]=0\left[2+2=5\right]=0.

We next recall a well-known combinatorial identity (see[17s, Lemma 3.3.5]or[19fco, Proposition 2.9.10]):

Lemma 6.36.

LetPPbe a finite set. Then,

∑A⊆P(−1)|A|=[P=∅].\sum\limits_{\begin{subarray}{c}A\subseteq P\end{subarray}}\left(-1\right)^{\left|A\right|}=\left[P=\varnothing\right].(The symbol “∑A⊆P\sum\limits_{\begin{subarray}{c}A\subseteq P\end{subarray}}” means “sum over all subsetsAAofPP”.)

Next, we introduce a specific notation related to colorings:

Definition 6.37.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Letk∈ℕk\in\mathbb{N}. Letf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}be akk-coloring. We then define a subsetEfE_{f}ofEEby

Ef:={e∈E∣the two endpoints ofehave the same color inf}.E_{f}:=\left\{e\in E\ \mid\ \text{the two endpoints of }e\text{ have the same color in }f\right\}.(Recall that the “color inff” of a vertexvvmeans the valuef⁡(v)f\left(v\right). If an edgee∈Ee\in Eis a loop, theneealways belongs toEfE_{f}, since we think of the two endpoints ofeeas being equal.)

The elements ofEfE_{f}are called theff-monochromaticedges ofGG. (“Monochromatic” means “one-colored”, so no surprises here.)

Example 6.38.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be the following multigraph:

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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Letf:V→{1,2}f:V\rightarrow\left\{1,2\right\}be the22-coloring ofGGthat sends each odd vertex to11and each even vertex to22. (Here, an “odd vertex” means a vertex that is odd as an integer. Thus, the odd vertices are1,3,51,3,5. “Even vertices” are understood similarly.) Then,Ef={a,b}E_{f}=\left\{a,b\right\}.

Notice the following simple fact:

Proposition 6.39.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Letk∈ℕk\in\mathbb{N}. Letf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}be akk-coloring. Then, thekk-coloringffis proper if and only ifEf=∅E_{f}=\varnothing.

Proof 6.40(Proof of Proposition6.39.).

We have the following chain of equivalences:

(thek-coloringfis proper)\displaystyle\ \left(\text{the }k\text{-coloring }f\text{ is proper}\right)⟺(no two adjacent vertices have the same color)\displaystyle\Longleftrightarrow\ \left(\text{no two adjacent vertices have the same color}\right)(by the definition of “proper”)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of \textquotedblleft proper\textquotedblright}\right)⟺(there is no edgee∈Esuch that the two endpoints ofehave the same color)\displaystyle\Longleftrightarrow\ \left(\text{there is no edge }e\in E\text{ such that the two endpoints of }e\text{ have the same color}\right)(since adjacent vertices are vertices thatare the two endpoints of an edge)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since adjacent vertices are vertices that}\\ \text{are the two endpoints of an edge}\end{array}\right)⟺(there exists no element ofEf)\displaystyle\Longleftrightarrow\ \left(\text{there exists no element of }E_{f}\right)(since the elements ofEfare precisely the edgese∈Esuch that the two endpoints ofehave the same color(by the definition ofEf))\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since the elements of }E_{f}\text{ are precisely the edges }e\in E\\ \text{such that the two endpoints of }e\text{ have the same color}\\ \text{(by the definition of }E_{f}\text{)}\end{array}\right)⟺(Ef=∅).\displaystyle\Longleftrightarrow\ \left(E_{f}=\varnothing\right).This proves Proposition6.39.

Lemma 6.41.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. LetBBbe a subset ofEE. Letk∈ℕk\in\mathbb{N}. Then, the number of allkk-coloringsf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}satisfyingB⊆EfB\subseteq E_{f}iskconn⁡(V,B,φ∣B)k^{\operatorname{conn}\left(V,B,\varphi\mid_{B}\right)}.

Proof 6.42(Proof of Lemma6.41.).

IfCCis a nonempty subset ofVV, and iff:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}is anykk-coloring ofGG, then we shall say thatffisconstant onCCif the restrictionf∣Cf\mid_{C}is a constant map (i.e., if the colorsf⁡(c)f\left(c\right)for allc∈Cc\in Care equal). We shall show the following claim:

Claim 1:Letf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}be anykk-coloring ofGG. Then, we haveB⊆EfB\subseteq E_{f}if and only ifffis constant on each component of the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right).

[Proof of Claim 1:This is an “if and only if” statement; we shall prove its “⟹\Longrightarrow” and “⟸\Longleftarrow” directions separately:

⟹:\Longrightarrow:Assume thatB⊆EfB\subseteq E_{f}. We must prove thatffis constant on each component of the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right).

LetCCbe a component of(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right). We must prove thatffis constant onCC. In other words, we must prove thatf⁡(c)=f⁡(d)f\left(c\right)=f\left(d\right)for anyc,d∈Cc,d\in C.

So let us fixc,d∈Cc,d\in C. Then, the verticesccandddbelong to the same component of the graph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right)(namely, toCC). Hence, these verticesccandddare path-connected in this graph. In other words, the graph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right)has a path fromcctodd. Let

𝐩=(v0,e1,v1,e2,v2,…,es,vs)\mathbf{p}=\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{s},v_{s}\right)be this path. Hence,v0=cv_{0}=candvs=dv_{s}=dande1,e2,…,es∈Be_{1},e_{2},\ldots,e_{s}\in B.

Leti∈{1,2,…,s}i\in\left\{1,2,\ldots,s\right\}. Then, the endpoints of the edgeeie_{i}arevi−1v_{i-1}andviv_{i}(sinceeie_{i}is surrounded byvi−1v_{i-1}andviv_{i}on the path𝐩\mathbf{p}). However, frome1,e2,…,es∈Be_{1},e_{2},\ldots,e_{s}\in B, we obtainei∈B⊆Efe_{i}\in B\subseteq E_{f}. Hence, the two endpoints ofeie_{i}have the same color inff(by the definition ofEfE_{f}). In other words,f⁡(vi−1)=f⁡(vi)f\left(v_{i-1}\right)=f\left(v_{i}\right)(since the endpoints of the edgeeie_{i}arevi−1v_{i-1}andviv_{i}).

Forget that we fixedii. We thus have proved the equalityf⁡(vi−1)=f⁡(vi)f\left(v_{i-1}\right)=f\left(v_{i}\right)for eachi∈{1,2,…,s}i\in\left\{1,2,\ldots,s\right\}. Combining these equalities, we obtain

f⁡(v0)=f⁡(v1)=f⁡(v2)=⋯=f⁡(vs).f\left(v_{0}\right)=f\left(v_{1}\right)=f\left(v_{2}\right)=\cdots=f\left(v_{s}\right).Hence,f⁡(v0)=f⁡(vs)f\left(v_{0}\right)=f\left(v_{s}\right). In other words,f⁡(c)=f⁡(d)f\left(c\right)=f\left(d\right)(sincev0=cv_{0}=candvs=dv_{s}=d).

Forget that we fixedccanddd. We thus have shown thatf⁡(c)=f⁡(d)f\left(c\right)=f\left(d\right)for anyc,d∈Cc,d\in C. In other words,ffis constant onCC. SinceCCwas allowed to be an arbitrary component of(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right), we thus conclude thatffis constant on each component of the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right). This proves the “⟹\Longrightarrow” direction of Claim 1.

⟸:\Longleftarrow:Assume thatffis constant on each component of the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right). We must prove thatB⊆EfB\subseteq E_{f}.

Indeed, lete∈Be\in B. Letuuandvvbe the two endpoints ofee. Then,(u,e,v)\left(u,e,v\right)is a walk fromuutovvin the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right)(sincee∈Be\in B). Hence,uuis path-connected tovvin this multigraph. In other words,uuandvvbelong to the same component of the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right). Therefore,f⁡(u)=f⁡(v)f\left(u\right)=f\left(v\right)(sinceffis constant on each component of the multigraph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right)). This means that the two endpoints ofeehave the same color inff(sinceuuandvvare the endpoints ofee). Combining this with the fact thate∈Ee\in E(becausee∈B⊆Ee\in B\subseteq E), we conclude thate∈Efe\in E_{f}(by the definition ofEfE_{f}).

Forget that we fixedee. We thus have shown thate∈Efe\in E_{f}for eache∈Be\in B. In other words,B⊆EfB\subseteq E_{f}. This proves the “⟸\Longleftarrow” direction of Claim 1. The proof of Claim 1 is now complete.]

Now, Claim 1 shows that thekk-coloringsf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}satisfyingB⊆EfB\subseteq E_{f}are precisely thekk-coloringsf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}that are constant on each component of the graph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right). Hence, all suchkk-coloringsffcan be obtained by the following procedure:

•For eachcomponentCCof the graph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right), pick a colorcCc_{C}(that is, an elementcCc_{C}of{1,2,…,k}\left\{1,2,\ldots,k\right\}) and then assign this colorcCc_{C}to each vertex inCC(that is, setf⁡(v)=cCf\left(v\right)=c_{C}for eachv∈Cv\in C).

This procedure involves choices (because for each componentCCof(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right), we get to pick a color): Namely, for each of theconn⁡(V,B,φ∣B)\operatorname{conn}\left(V,B,\varphi\mid_{B}\right)many components of the graph(V,B,φ∣B)\left(V,B,\varphi\mid_{B}\right), we must choose a color from the set{1,2,…,k}\left\{1,2,\ldots,k\right\}. Thus, we have a total ofkconn⁡(V,B,φ∣B)k^{\operatorname{conn}\left(V,B,\varphi\mid_{B}\right)}many options (since we are choosing amongkkcolors for each of theconn⁡(V,B,φ∣B)\operatorname{conn}\left(V,B,\varphi\mid_{B}\right)components). Each of these options gives rise to a differentkk-coloringf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}. Therefore, the number of allkk-coloringsf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}satisfyingB⊆EfB\subseteq E_{f}iskconn⁡(V,B,φ∣B)k^{\operatorname{conn}\left(V,B,\varphi\mid_{B}\right)}(because all of thesekk-colorings can be obtained by this procedure). This proves Lemma6.41.

Corollary 6.43.

Let(V,E,φ)\left(V,E,\varphi\right)be a multigraph. LetFFbe a subset ofEE. Letk∈ℕk\in\mathbb{N}. Then,

kconn⁡(V,F,φ∣F)=∑f:V→{1,2,…,k};F⊆Ef1.k^{\operatorname{conn}\left(V,F,\varphi\mid_{F}\right)}=\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ F\subseteq E_{f}\end{subarray}}1.

Proof 6.44(Proof of Corollary6.43.).

We have

∑f:V→{1,2,…,k};F⊆Ef1\displaystyle\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ F\subseteq E_{f}\end{subarray}}1=(the number of allf:V→{1,2,…,k}satisfyingF⊆Ef)⋅1\displaystyle=\left(\text{the number of all }f:V\rightarrow\left\{1,2,\ldots,k\right\}\text{ satisfying }F\subseteq E_{f}\right)\cdot 1=(the number of allf:V→{1,2,…,k}satisfyingF⊆Ef)\displaystyle=\left(\text{the number of all }f:V\rightarrow\left\{1,2,\ldots,k\right\}\text{ satisfying }F\subseteq E_{f}\right)=kconn⁡(V,F,φ∣F)\displaystyle=k^{\operatorname{conn}\left(V,F,\varphi\mid_{F}\right)}(because Lemma6.41(applied toB=FB=F) shows that the number of allkk-coloringsf:V→{1,2,…,k}f:V\rightarrow\left\{1,2,\ldots,k\right\}satisfyingF⊆EfF\subseteq E_{f}iskconn⁡(V,F,φ∣F)k^{\operatorname{conn}\left(V,F,\varphi\mid_{F}\right)}). This proves Corollary6.43.

Proof 6.45(Proof of Theorem6.34.).

First of all, the equality

∑F⊆E(−1)|F|xconn(V,F,φ∣F)=∑His a spanningsubgraph ofG(−1)|E(H)|xconnH\sum\limits_{F\subseteq E}\left(-1\right)^{\left|F\right|}x^{\operatorname*{conn}\left(V,F,\varphi\mid_{F}\right)}=\sum\limits_{\begin{subarray}{c}H\text{ is a spanning}\\ \text{subgraph of }G\end{subarray}}\left(-1\right)^{\left|\operatorname*{E}\left(H\right)\right|}x^{\operatorname*{conn}H}is clear, because the spanning subgraphs ofGGare precisely the subgraphs of the form(V,F,φ∣F)\left(V,F,\varphi\mid_{F}\right)for someF⊆EF\subseteq E.

Now, letk∈ℕk\in\mathbb{N}. We must prove that(# of properk-colorings ofG)=χG(k)\left(\text{\# of proper }k\text{-colorings of }G\right)=\chi_{G}\left(k\right).

Let us substitutekkforxxin the equality

χG=∑F⊆E(−1)|F|xconn⁡(V,F,φ∣F).\chi_{G}=\sum\limits_{F\subseteq E}\left(-1\right)^{\left|F\right|}x^{\operatorname{conn}\left(V,F,\varphi\mid_{F}\right)}.We thus obtain

χG(k)\displaystyle\chi_{G}\left(k\right)=∑F⊆E(−1)|F|kconn⁡(V,F,φ∣F)⏟=∑f:V→{1,2,…,k};F⊆Ef1(by Corollary6.43)=∑F⊆E(−1)|F|∑f:V→{1,2,…,k};F⊆Ef1\displaystyle=\sum\limits_{F\subseteq E}\left(-1\right)^{\left|F\right|}\underbrace{k^{\operatorname{conn}\left(V,F,\varphi\mid_{F}\right)}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ F\subseteq E_{f}\end{subarray}}1\\ \text{(by Corollary \ref{cor.mt2.chrompoly.kconn2})}\end{subarray}}=\sum\limits_{F\subseteq E}\left(-1\right)^{\left|F\right|}\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ F\subseteq E_{f}\end{subarray}}1=∑F⊆E∑f:V→{1,2,…,k};F⊆Ef⏟=∑f:V→{1,2,…,k}∑F⊆E;F⊆Ef(−1)|F|1⏟=(−1)|F|=∑f:V→{1,2,…,k}∑F⊆E;F⊆Ef⏟=∑F⊆Ef(sinceEf⊆E)(−1)|F|\displaystyle=\underbrace{\sum\limits_{F\subseteq E}\ \ \sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ F\subseteq E_{f}\end{subarray}}}_{=\sum\limits_{f:V\rightarrow\left\{1,2,\ldots,k\right\}}\ \ \sum\limits_{\begin{subarray}{c}F\subseteq E;\\ F\subseteq E_{f}\end{subarray}}}\underbrace{\left(-1\right)^{\left|F\right|}1}_{=\left(-1\right)^{\left|F\right|}}=\sum\limits_{f:V\rightarrow\left\{1,2,\ldots,k\right\}}\ \ \underbrace{\sum\limits_{\begin{subarray}{c}F\subseteq E;\\ F\subseteq E_{f}\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{F\subseteq E_{f}}\\ \text{(since }E_{f}\subseteq E\text{)}\end{subarray}}\left(-1\right)^{\left|F\right|}=∑f:V→{1,2,…,k}∑F⊆Ef(−1)|F|=∑f:V→{1,2,…,k}∑A⊆Ef(−1)|A|⏟=[Ef=∅](by Lemma6.36,applied toP=Ef)\displaystyle=\sum\limits_{f:V\rightarrow\left\{1,2,\ldots,k\right\}}\ \ \sum\limits_{F\subseteq E_{f}}\left(-1\right)^{\left|F\right|}=\sum\limits_{f:V\rightarrow\left\{1,2,\ldots,k\right\}}\ \ \underbrace{\sum\limits_{A\subseteq E_{f}}\left(-1\right)^{\left|A\right|}}_{\begin{subarray}{c}=\left[E_{f}=\varnothing\right]\\ \text{(by Lemma \ref{lem.dominating.heinrich-lemma1},}\\ \text{applied to }P=E_{f}\text{)}\end{subarray}}(here, we have renamed the summation indexFin the inner sum asA)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have renamed the summation index }F\\ \text{in the inner sum as }A\end{array}\right)=∑f:V→{1,2,…,k}[Ef=∅]\displaystyle=\sum\limits_{f:V\rightarrow\left\{1,2,\ldots,k\right\}}\left[E_{f}=\varnothing\right]=∑f:V→{1,2,…,k};Ef=∅[Ef=∅]⏟=1(sinceEf=∅is true)+∑f:V→{1,2,…,k};notEf=∅[Ef=∅]⏟=0(sinceEf=∅is false)\displaystyle=\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ E_{f}=\varnothing\end{subarray}}\underbrace{\left[E_{f}=\varnothing\right]}_{\begin{subarray}{c}=1\\ \text{(since }E_{f}=\varnothing\text{ is true)}\end{subarray}}+\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ \text{not }E_{f}=\varnothing\end{subarray}}\underbrace{\left[E_{f}=\varnothing\right]}_{\begin{subarray}{c}=0\\ \text{(since }E_{f}=\varnothing\text{ is false)}\end{subarray}}(since eachf:V→{1,2,…,k}either satisfiesEf=∅or does not)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since each }f:V\rightarrow\left\{1,2,\ldots,k\right\}\text{ either satisfies }E_{f}=\varnothing\\ \text{or does not}\end{array}\right)=∑f:V→{1,2,…,k};Ef=∅1+∑f:V→{1,2,…,k};notEf=∅0⏟=0=∑f:V→{1,2,…,k};Ef=∅1\displaystyle=\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ E_{f}=\varnothing\end{subarray}}1+\underbrace{\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ \text{not }E_{f}=\varnothing\end{subarray}}0}_{\begin{subarray}{c}=0\end{subarray}}=\sum\limits_{\begin{subarray}{c}f:V\rightarrow\left\{1,2,\ldots,k\right\};\\ E_{f}=\varnothing\end{subarray}}1=(the number of allf:V→{1,2,…,k}such thatEf=∅)⋅1\displaystyle=\left(\text{the number of all }f:V\rightarrow\left\{1,2,\ldots,k\right\}\text{ such that }E_{f}=\varnothing\right)\cdot 1=(the number of allf:V→{1,2,…,k}such thatEf=∅)\displaystyle=\left(\text{the number of all }f:V\rightarrow\left\{1,2,\ldots,k\right\}\text{ such that }E_{f}=\varnothing\right)=(the number of allf:V→{1,2,…,k}such that thek-coloringfis proper)\displaystyle=\left(\text{the number of all }f:V\rightarrow\left\{1,2,\ldots,k\right\}\text{ such that the }k\text{-coloring }f\text{ is proper}\right)(since Proposition6.39shows that the condition “Ef=∅”is equivalent to “thek-coloringfis proper”)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since Proposition \ref{prop.mt2.chrompoly.EcapEf} shows that the condition \textquotedblleft}E_{f}=\varnothing\text{\textquotedblright}\\ \text{is equivalent to \textquotedblleft the }k\text{-coloring }f\text{ is proper\textquotedblright}\end{array}\right)=(the number of all properk-colorings).\displaystyle=\left(\text{the number of all proper }k\text{-colorings}\right).In other words, the number of properkk-colorings ofGGisχG(k)\chi_{G}\left(k\right). This completes the proof of Theorem6.34.

Definition 6.46.

The polynomialχG\chi_{G}in Theorem6.34is known as thechromatic polynomialofGG.

Here are the chromatic polynomials of some graphs:

Proposition 6.47.

Letn≥1n\geq 1be an integer.

(a)For the path graphPnP_{n}withnnvertices, we have χPn=x(x−1)n−1.\chi_{P_{n}}=x\left(x-1\right)^{n-1}.
(b)More generally, for any treeTTwithnnvertices, we have χT=x(x−1)n−1.\chi_{T}=x\left(x-1\right)^{n-1}.
(c)For the complete graphKnK_{n}withnnvertices, we have χKn=x(x−1)(x−2)⋯(x−n+1).\chi_{K_{n}}=x\left(x-1\right)\left(x-2\right)\cdots\left(x-n+1\right).
(d)For the empty graphEnE_{n}withnnvertices, we have χEn=xn.\chi_{E_{n}}=x^{n}.
(e)Assume thatn≥2n\geq 2. For the cycle graphCnC_{n}withnnvertices, we have χCn=(x−1)n+(−1)n(x−1).\chi_{C_{n}}=\left(x-1\right)^{n}+\left(-1\right)^{n}\left(x-1\right).

Proof 6.48(Proof sketch.).

(c)In order to prove that two polynomials with real coefficients are identical, it suffices to show that they agree on all nonnegative integers (this is an instance of the “principle of permanence of polynomial identities” that we have already stated as Theorem5.197). Thus, in order to prove thatχKn=x(x−1)(x−2)⋯(x−n+1)\chi_{K_{n}}=x\left(x-1\right)\left(x-2\right)\cdots\left(x-n+1\right), it suffices to show thatχKn(k)=k(k−1)(k−2)⋯(k−n+1)\chi_{K_{n}}\left(k\right)=k\left(k-1\right)\left(k-2\right)\cdots\left(k-n+1\right)for eachk∈ℕk\in\mathbb{N}.

So let us do this. Fixk∈ℕk\in\mathbb{N}. Theorem6.34(applied toG=KnG=K_{n}) yields

(# of properk-colorings ofKn)=χKn(k).\left(\text{\# of proper }k\text{-colorings of }K_{n}\right)=\chi_{K_{n}}\left(k\right).(51)Now, how many properkk-colorings doesKnK_{n}have? We can construct such a properkk-coloring as follows:

•First, choose the color of the vertex11. There arekkoptions for this.
•Then, choose the color of the vertex22. There arek−1k-1options for this, since it must differ from the color of11.
•Then, choose the color of the vertex33. There arek−2k-2options for this, since it must differ from the colors of11and of22(and the latter two colors are distinct, so we must subtract22, not11).
•And so on, until allnnvertices are colored.

The total number of options to perform this construction is therefore k(k−1)(k−2)⋯(k−n+1)k\left(k-1\right)\left(k-2\right)\cdots\left(k-n+1\right). Hence,

(# of properk-colorings ofKn)=k(k−1)(k−2)⋯(k−n+1).\left(\text{\# of proper }k\text{-colorings of }K_{n}\right)=k\left(k-1\right)\left(k-2\right)\cdots\left(k-n+1\right).Comparing this with (51), we obtainχKn(k)=k(k−1)(k−2)⋯(k−n+1)\chi_{K_{n}}\left(k\right)=k\left(k-1\right)\left(k-2\right)\cdots\left(k-n+1\right). As we already explained, this completes the proof of Proposition6.47(c).

(d)This is similar to part(c), but easier. We leave the proof to the reader. Alternatively, it follows easily from the definition ofχEn\chi_{E_{n}}, sinceEnE_{n}has only one spanning subgraph (namely,EnE_{n}itself).

(b)(This is an outline; see[17s-mt2s, §0.6]for details.)

We proceed by induction onnn. Ifn=1n=1, then this is easily checked by hand. Ifn>1n>1, then the treeTThas at least one leaf (by Theorem5.28(a)). Thus, we can fix a leafℓ\ellofTT. The graphT∖ℓT\setminus\ellthen is a tree (by Theorem5.30) and hasn−1n-1vertices, and therefore (by the induction hypothesis) its chromatic polynomial isχT∖ℓ=x(x−1)n−2\chi_{T\setminus\ell}=x\left(x-1\right)^{n-2}. However, for any givenk∈ℕk\in\mathbb{N}, we can construct a properkk-coloring ofTTby first choosing a properkk-coloring ofT∖ℓT\setminus\elland then choosing the color of the remaining leafℓ\ell(there arek−1k-1choices for it, since it has to differ from the color of the unique neighbor ofℓ\ell). Therefore, for eachk∈ℕk\in\mathbb{N}, we have

(# of properk-colorings ofT)=(# of properk-colorings ofT∖ℓ)⋅(k−1).\left(\text{\# of proper }k\text{-colorings of }T\right)=\left(\text{\# of proper }k\text{-colorings of }T\setminus\ell\right)\cdot\left(k-1\right).In view of Theorem6.34, this equality can be rewritten as

χT(k)=χT∖ℓ(k)⋅(k−1).\chi_{T}\left(k\right)=\chi_{T\setminus\ell}\left(k\right)\cdot\left(k-1\right).Since this holds for allk∈ℕk\in\mathbb{N}, we thus conclude that

χT=χT∖ℓ⏟=x(x−1)n−2⋅(x−1)=x(x−1)n−2⋅(x−1)=x(x−1)n−1.\chi_{T}=\underbrace{\chi_{T\setminus\ell}}_{=x\left(x-1\right)^{n-2}}\cdot\left(x-1\right)=x\left(x-1\right)^{n-2}\cdot\left(x-1\right)=x\left(x-1\right)^{n-1}.This completes the induction step.

Alternatively, Proposition6.47(b)can also be derived from the definition ofχT\chi_{T}, using the fact that every spanning subgraphHHofTThas no cycles and therefore satisfiesconnH=n−|E(H)|\operatorname*{conn}H=n-\left|\operatorname*{E}\left(H\right)\right|(by Corollary5.11).

(a)This is a particular case of part(b), sincePnP_{n}is a tree withnnvertices.

(e)There are different ways to prove this; see[LeeShi19]for four different proofs. The simplest one is probably by induction onnn: Letn≥2n\geq 2. Fixk∈ℕk\in\mathbb{N}. A properkk-coloring ofCnC_{n}is the same as a properkk-coloring ofPnP_{n}that assigns different colors to the vertices11andnn. Hence,

(# of properk-colorings ofCn)\displaystyle\left(\text{\# of proper }k\text{-colorings of }C_{n}\right)=(# of properk-colorings ofPnthat assign different colors to1andn)\displaystyle=\left(\text{\# of proper }k\text{-colorings of }P_{n}\text{ that assign different colors to }1\text{ and }n\right)=(# of properk-colorings ofPn)⏟=k(k−1)n−1(by part(a))\displaystyle=\underbrace{\left(\text{\# of proper }k\text{-colorings of }P_{n}\right)}_{\begin{subarray}{c}=k\left(k-1\right)^{n-1}\\ \text{(by part {(a)})}\end{subarray}}−(# of properk-colorings ofPnthat assign the same color to1andn)⏟=(# of properk-colorings ofCn−1)(why?)\displaystyle\ \ \ \ \ \ \ \ \ \ -\underbrace{\left(\text{\# of proper }k\text{-colorings of }P_{n}\text{ that assign the same color to }1\text{ and }n\right)}_{\begin{subarray}{c}=\left(\text{\# of proper }k\text{-colorings of }C_{n-1}\right)\\ \text{(why?)}\end{subarray}}=k(k−1)n−1−(# of properk-colorings ofCn−1).\displaystyle=k\left(k-1\right)^{n-1}-\left(\text{\# of proper }k\text{-colorings of }C_{n-1}\right).In view of Theorem6.34, this equality can be rewritten as

χCn(k)=k(k−1)n−1−χCn−1(k).\chi_{C_{n}}\left(k\right)=k\left(k-1\right)^{n-1}-\chi_{C_{n-1}}\left(k\right).Since this holds for allk∈ℕk\in\mathbb{N}, we thus obtain

χCn=x(x−1)n−1−χCn−1.\chi_{C_{n}}=x\left(x-1\right)^{n-1}-\chi_{C_{n-1}}.This is a recursion that is easily solved forχCn\chi_{C_{n}}, yielding the claim of part(e).

(Proposition6.47(e)also appeared as Exercise 2(a)on midterm #3 in my Spring 2017 course; seethe course websitefor solutions.)

Exercise 6.49.

Letg∈ℕg\in\mathbb{N}. LetGGbe the simple graph whose vertices are the2g+12g+1integers−g,−g+1,…,g−1,g-g,-g+1,\ldots,g-1,g, and whose edges are

{0,i}for alli∈{1,2,…,g};\displaystyle\left\{0,i\right\}\qquad\text{ for all }i\in\left\{1,2,\ldots,g\right\};{0,−i}for alli∈{1,2,…,g};\displaystyle\left\{0,-i\right\}\qquad\text{ for all }i\in\left\{1,2,\ldots,g\right\};{i,−i}for alli∈{1,2,…,g}\displaystyle\left\{i,-i\right\}\qquad\text{ for all }i\in\left\{1,2,\ldots,g\right\}(these are3g3gedges in total).

Compute the chromatic polynomialχG\chi_{G}ofGG.

[Here is howGGlooks like in the case wheng=4g=4:

0011−1-122−2-233−3-344−4-4]

[Solution:This is Exercise 2(b)on midterm #3 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 6.50.

LetGGbe a multigraph. LetχG\chi_{G}be its chromatic polynomial. For eachi∈ℕi\in\mathbb{N}, letaia_{i}be the coefficient ofxix^{i}inχG\chi_{G}. Prove thata1a_{1}is odd if and only if the graphGGis connected and has a proper22-coloring.

6.6Vizing’s theorem

So far we have been coloring the vertices of a graph. We can also color the edges:

Definition 6.51.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Letk∈ℕk\in\mathbb{N}.

Akk-edge-coloringofGGmeans a mapf:E→{1,2,…,k}f:E\rightarrow\left\{1,2,\ldots,k\right\}.

Such akk-edge-coloringffis calledproperif no two distinct edges that have a common endpoint have the same color.

The most prominent fact about edge-colorings is the following theorem:

Theorem 6.52(Vizing’s theorem).

LetGGbe a simple graph with at least one vertex. Let

α:=max⁡{deg⁡v∣v∈V}.\alpha:=\max\left\{\deg v\ \mid\ v\in V\right\}.Then,GGhas a proper(α+1)\left(\alpha+1\right)-edge-coloring.

Proof 6.53.

See, e.g.,[Schrij04]or various textbooks on graph theory.103103103Note that[Schrij04]uses some standard graph-theoretical notations: What we callα\alphais denoted byΔ⁡(G)\Delta\left(G\right)in[Schrij04], whereasχ′(G)\chi^{\prime}\left(G\right)denotes the minimumk∈ℕk\in\mathbb{N}for whichGGhas a properkk-edge-coloring.

Two remarks:

•Theα+1\alpha+1in Vizing’s theorem cannot be improved in general (e.g., takeGGto be an odd-length cycle graphCnC_{n}).
•Vizing’s theorem can be adapted to work for multigraphs instead of simple graphs. However, this requires replacing theα+1\alpha+1byα+m\alpha+m, wheremmis the maximum number of distinct mutually parallel edges inGG(since otherwise, the multigraph(K3bidir)und\left(K_{3}^{\operatorname*{bidir}}\right)^{\operatorname*{und}}would be a counterexample, as it hasα=4\alpha=4but has no proper55-edge-coloring). For a proof of this, see[BerFou91, Corollary 2].

6.7Further exercises

Some interesting things can be said about colorings of graphs, even about non-proper colorings:

Exercise 6.54.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

Prove that there exists a22-coloringffofGGwith the following property: For each vertexv∈Vv\in V, at most12deg⁡v\dfrac{1}{2}\deg vamong the neighbors ofvvhave the same color asvv.

[Remark:This problem is often restated as follows: You are given a (finite) set of politicians; some politicians are mutual enemies. (No politician is his own enemy. Ifuuis an enemy ofvv, thenvvis an enemy ofuu. An enemy of an enemy is not necessarily a friend. So this is just a simple graph.) Prove that it is possible to subdivide this set into two (disjoint) parties such that no politician has more than half of his enemies in his own party.]

[Hint:First, pick an arbitrary22-coloringffofGG. Then, gradually improve it until it satisfies the required property.]

[Solution:This is Exercise 1 on homework set #0 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise6.54can be generalized to multiple colors:

Exercise 6.55.

Letk∈ℕk\in\mathbb{N}. Letp1,p2,…,pkp_{1},p_{2},\ldots,p_{k}bekknonnegative real numbers such thatp1+p2+⋯+pk≥1p_{1}+p_{2}+\cdots+p_{k}\geq 1.

LetG=(V,E)G=\left(V,E\right)be a simple graph.

Prove that there exists akk-coloringffofGGwith the following property: For each vertexv∈Vv\in V, at mostpf⁡(v)deg⁡vp_{f\left(v\right)}\deg vneighbors ofvvhave the same color asvv.

[Solution:This is Exercise 5 on midterm #1 from my Spring 2017 course; seethe course pagefor solutions.]

6.8Some recent results

The theory of properkk-colorings can be surprisingly deep and difficult. The following theorem was conjectured by Ioan Tomescu in 1971, and only proved in 2019 by Fox, He and Manners using sophisticated probabilistic arguments[FoHeMa19, Theorem 1]:

Theorem 6.56(Tomescu, Fox, He, Manners).

LetGGbe a graph withnnvertices, and letkkbe a positive integer. Assume thatGGhas no proper(k−1)\left(k-1\right)-coloring. Then:

(a)The number of properkk-colorings ofGGis at mostk!⋅kn−kk!\cdot k^{n-k}.
(b)IfGGis connected andk≥4k\geq 4, then the number of properkk-colorings ofGGis at mostk!⋅(k−1)n−kk!\cdot\left(k-1\right)^{n-k}.

(Part(a)is[Tomesc85, problem 10.22]; seehttps://mathoverflow.net/a/492188/for a clearer version of the proof.) An even more general conjecture of Tomescu says that ifGGis a connected graph withnnvertices, and ifssandkkare two integers such thats≥k≥4s\geq k\geq 4and such thatGGhas no proper(k−1)\left(k-1\right)-coloring, then the number of properss-colorings ofGGis at mosts(s−1)(s−2)⋯(s−k+1)⋅(s−1)n−ks\left(s-1\right)\left(s-2\right)\cdots\left(s-k+1\right)\cdot\left(s-1\right)^{n-k}; this is still unsolved ([FoHeMa19, Conjecture 12]).

The coefficients of the chromatic polynomialχG\chi_{G}defined in Theorem6.34have also been the topic of much research, from which I shall only mention a few highlights:

Theorem 6.57(Whitney, Huh).

LetGGbe a graph withnnvertices. Write its chromatic polynomialχG\chi_{G}as∑i∈ℕaixi\sum\limits_{i\in\mathbb{N}}a_{i}x^{i}withai∈ℤa_{i}\in\mathbb{Z}(so thata0,a1,a2,…a_{0},a_{1},a_{2},\ldotsare its coefficients). Then:

(a)We haveai=0a_{i}=0for alli>ni>n.
(b)Ifn>0n>0, thena0=0a_{0}=0.
(c)For eachi∈ℕi\in\mathbb{N}, we have(−1)n−iai≥0\left(-1\right)^{n-i}a_{i}\geq 0. (Thus, the coefficientsa0,a1,a2,…a_{0},a_{1},a_{2},\ldotsalternate in sign; in other words, all coefficients of the polynomial(−1)nχG(−x)\left(-1\right)^{n}\chi_{G}\left(-x\right)are nonnegative.)
(d)For eachi>0i>0, we haveai2≥ai−1ai+1a_{i}^{2}\geq a_{i-1}a_{i+1}.

Parts(a)and(b)of Theorem6.57are very easy consequences of Theorem6.34, and are only mentioned for completeness’s sake. Part(c)was proved by Whitney[Whitne32, §7]combinatorially (i.e., by constructing a set with size(−1)n−iai\left(-1\right)^{n-i}a_{i}). Part(d)is a famous conjecture, stated by Hoggar in 1974, and only proved in 2012 by June Huh[Huh12]using deep algebraic geometry. A later proof by Bränden and Leake[BräLea21, Theorem 4.1]relies on multivariate analysis, which is still far from elementary. Huh’s work has brought the subject to the fore of research again, and many similar results have since been proven using his methods. More about the chromatic polynomial can be found in the recent text[Tittma25].

Another famous result on proper colorings isthefour-color theorem, which says that every planar graph (i.e., any graph that can be embedded in the real plane in such a way that no two edges intersect; cf. Example2.7) has a proper44-coloring. This was conjectured by Guthrie in 1852 and only proved by Appel and Haken in 1976 using a long and complex computer-aided case analysis; see[FriFri98]for the details and the history of this result.

7Independent sets

7.1Definition and the Caro–Wei theorem

Next, we define one of the most fundamental notions in graph theory:

Definition 7.1.

Anindependent setof a multigraphGGmeans a subsetSSofV(G)\operatorname*{V}\left(G\right)such that no two elements ofSSare adjacent.

In other words, an independent set ofGGmeans an induced subgraph ofGGthat has no edges104104104This is a somewhat sloppy statement. Of course, an independent set is not literally an induced subgraph, since the former is just a set, while the latter is a graph. What I mean is that a subsetSSofV(G)\operatorname*{V}\left(G\right)is independent if and only if the induced subgraphG⁡[S]G\left[S\right]has no edges.. Note that “no two elements ofSS” doesn’t mean “no two distinct elements ofSS”.

Thus, for example, what we called an “anti-triangle” (back in Definition2.9) is an independent set of size33.

Example 7.2.

LetGGbe the graphGGfrom Example2.49. Then, the independent sets ofGGare the sets∅\varnothing,{1}\left\{1\right\},{2}\left\{2\right\},{3}\left\{3\right\},{4}\left\{4\right\}and{2,4}\left\{2,4\right\}.

One classical computational problem in graph theory is to find a maximum-size independent set of a given graph. This problem is NP-hard105105105And, in fact, NP-complete; see[Goldre10, Proposition 4.10]., so don’t expect a quick algorithm or even a good formula for the maximum size of an independent set. However, there are some lower bounds for this maximum size. Here is one, known as theCaro–Wei theorem([AloSpe16, Chapter 6, Probabilistic Lens]):

Theorem 7.4(Caro–Wei theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a loopless multigraph. Then,GGhas an independent set of size

≥∑v∈V11+deg⁡v.\geq\sum\limits_{v\in V}\dfrac{1}{1+\deg v}.

Example 7.5.

LetGGbe the following loopless multigraph:

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L 26.66 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 45.73 -57.18 L 72.38 -11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the degrees of the vertices ofGGare3,2,3,2,2,23,2,3,2,2,2. Hence, Theorem7.4yields thatGGhas an independent set of size

≥11+3+11+2+11+3+11+2+11+2+11+2=116≈1.83.\geq\dfrac{1}{1+3}+\dfrac{1}{1+2}+\dfrac{1}{1+3}+\dfrac{1}{1+2}+\dfrac{1}{1+2}+\dfrac{1}{1+2}=\dfrac{11}{6}\approx 1.83.Since the size of an independent set is always an integer, we can round this up and conclude thatGGhas an independent set of size≥2\geq 2. In truth,GGactually has an independent set of size33(namely,{2,4,6}\left\{2,4,6\right\}), but there is no way to tell this from the degrees of its vertices alone. For example, the vertices of the graph

have the same degrees as those ofGG, butHHhas no independent set of size33.

We shall give two proofs of Theorem7.4, both of them illustrating useful techniques.106106106Note that the looplessness requirement in Theorem7.4is important: IfGGhas a loop at each vertex, then the only independent set ofGGis∅\varnothing.

Proof 7.6(First proof of Theorem7.4.).

Assume the contrary. Thus, each independent setSSofGGhas size

|S|<∑v∈V11+deg⁡v.\left|S\right|<\sum\limits_{v\in V}\dfrac{1}{1+\deg v}.(52) AVV-listingshall mean a list of all vertices inVV, with each vertex occurring exactly once in the list. Ifσ\sigmais aVV-listing, then we define a subsetJσJ_{\sigma}ofVVas follows:

Jσ:={v∈V∣voccursbeforeall neighbors ofvinσ}.J_{\sigma}:=\left\{v\in V\ \mid\ v\text{ occurs {before} all neighbors of }v\text{ in }\sigma\right\}. [Example:LetGGbe the following graph:

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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 64.11 8.99 L 101.25 46.13}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Letσ\sigmabe theVV-listing(1,2,7,5,3,4,6)\left(1,2,7,5,3,4,6\right). Then, the vertex11occurs before all its neighbors (22,44and55) inσ\sigma, and thus we have1∈Jσ1\in J_{\sigma}. Likewise, the vertex77occurs before all its neighbors (33and66) inσ\sigma, so that we have7∈Jσ7\in J_{\sigma}. But the vertex22does not occur before all its neighbors inσ\sigma(indeed, it occurs after its neighbor11), so that we have2∉Jσ2\notin J_{\sigma}. Likewise, the vertices5,3,4,65,3,4,6don’t belong toJσJ_{\sigma}. Altogether, we thus obtainJσ={1,7}J_{\sigma}=\left\{1,7\right\}.]

The setJσJ_{\sigma}is an independent set ofGG(because if two verticesuuandvvinJσJ_{\sigma}were adjacent, thenuuwould have to occur beforevvinσ\sigma, butvvwould have to occur beforeuuinσ\sigma; but these two statements clearly contradict each other). Thus, (52) (applied toS=JσS=J_{\sigma}) yields

|Jσ|<∑v∈V11+deg⁡v.\left|J_{\sigma}\right|<\sum\limits_{v\in V}\dfrac{1}{1+\deg v}. This inequality holds foreachVV-listingσ\sigma. Thus, summing this inequality over allVV-listingsσ\sigma, we obtain

∑σis aV-listing|Jσ|\displaystyle\sum\limits_{\sigma\text{ is a }V\text{-listing}}\left|J_{\sigma}\right|<∑σis aV-listing∑v∈V11+deg⁡v\displaystyle<\sum\limits_{\sigma\text{ is a }V\text{-listing}}\ \ \sum\limits_{v\in V}\dfrac{1}{1+\deg v}=(# of allV-listings)⋅∑v∈V11+deg⁡v.\displaystyle=\left(\text{\# of all }V\text{-listings}\right)\cdot\sum\limits_{v\in V}\dfrac{1}{1+\deg v}.(53) On the other hand, I claim the following:

Claim 1:For eachv∈Vv\in V, we have (# of allV-listingsσsatisfyingv∈Jσ)≥(# of allV-listings)1+deg⁡v.\left(\text{\# of all }V\text{-listings }\sigma\text{ satisfying }v\in J_{\sigma}\right)\geq\dfrac{\left(\text{\# of all }V\text{-listings}\right)}{1+\deg v}.

[Proof of Claim 1:Fix a vertexv∈Vv\in V. Definedeg′⁡v\deg^{\prime}vto be the # of all neighbors ofvv. Clearly,deg′⁡v≤deg⁡v\deg^{\prime}v\leq\deg v.

We shall call aVV-listingσ\sigmagoodif the vertexvvoccurs in it before all its neighbors. In other words, aVV-listingσ\sigmais good if and only if it satisfiesv∈Jσv\in J_{\sigma}(becausev∈Jσv\in J_{\sigma}means that the vertexvvoccurs inσ\sigmabefore all its neighbors107107107This follows straight from the definition ofJσJ_{\sigma}.). Thus, we must show that

(# of all goodV-listings)≥(# of allV-listings)1+deg⁡v.\left(\text{\# of all good }V\text{-listings}\right)\geq\dfrac{\left(\text{\# of all }V\text{-listings}\right)}{1+\deg v}. We define a map

Γ:{allV-listings}→{all goodV-listings}\Gamma:\left\{\text{all }V\text{-listings}\right\}\rightarrow\left\{\text{all good }V\text{-listings}\right\}as follows: Wheneverτ\tauis aVV-listing, we letΓ⁡(τ)\Gamma\left(\tau\right)be theVV-listing obtained fromτ\tauby swappingvvwith the first neighbor ofvvthat occurs inτ\tau(or, ifτ\tauis already good, then we just do nothing, i.e., we setΓ⁡(τ)=τ\Gamma\left(\tau\right)=\tau). This mapΓ\Gammais a(1+deg′⁡v)\left(1+\deg^{\prime}v\right)-to-11correspondence – i.e., for each goodVV-listingσ\sigma, there are exactly1+deg′⁡v1+\deg^{\prime}vmanyVV-listingsτ\tauthat satisfyΓ⁡(τ)=σ\Gamma\left(\tau\right)=\sigma(in fact, one of theseτ\tau’s isσ\sigmaitself, and the remainingdeg′⁡v\deg^{\prime}vmany of theseτ\tau’s are obtained fromσ\sigmaby switchingvvwith some neighbor ofvv). Hence, by the multijection principle108108108See a footnote in the proof of Theorem5.122for the statement of the multijection principle., we conclude that

|{allV-listings}|=(1+deg′v)⋅|{all goodV-listings}|.\left|\left\{\text{all }V\text{-listings}\right\}\right|=\left(1+\deg^{\prime}v\right)\cdot\left|\left\{\text{all good }V\text{-listings}\right\}\right|.In other words,

(# of allV-listings)=(1+deg′⁡v)⋅(# of all goodV-listings).\left(\text{\# of all }V\text{-listings}\right)=\left(1+\deg^{\prime}v\right)\cdot\left(\text{\# of all good }V\text{-listings}\right).Hence,

(# of all goodV-listings)=(# of allV-listings)1+deg′⁡v≥(# of allV-listings)1+deg⁡v\left(\text{\# of all good }V\text{-listings}\right)=\dfrac{\left(\text{\# of all }V\text{-listings}\right)}{1+\deg^{\prime}v}\geq\dfrac{\left(\text{\# of all }V\text{-listings}\right)}{1+\deg v}(sincedeg′⁡v≤deg⁡v\deg^{\prime}v\leq\deg v). This proves Claim 1 (since the goodVV-listings are precisely theVV-listingsσ\sigmasatisfyingv∈Jσv\in J_{\sigma}).]

Next, we recall a basic property of the Iverson bracket notation109109109See, e.g., Definition5.136for the definition of the Iverson bracket notation.: IfTTis a subset of a finite setSS, then

|T|=∑v∈S[v∈T].\left|T\right|=\sum\limits_{v\in S}\left[v\in T\right].(54)(Indeed, the sum∑v∈S[v∈T]\sum\limits_{v\in S}\left[v\in T\right]contains an addend equal to11for eachv∈Tv\in T, and an addend equal to00for eachv∈S∖Tv\in S\setminus T. Thus, this sum amounts to|T|⋅1+|S∖T|⋅0=|T|\left|T\right|\cdot 1+\left|S\setminus T\right|\cdot 0=\left|T\right|.)

Now, (53) yields

(# of allV-listings)⋅∑v∈V11+deg⁡v\displaystyle\left(\text{\# of all }V\text{-listings}\right)\cdot\sum\limits_{v\in V}\dfrac{1}{1+\deg v}>∑σis aV-listing|Jσ|⏟=∑v∈V[v∈Jσ](by (54))=∑σis aV-listing∑v∈V⏟=∑v∈V∑σis aV-listing[v∈Jσ]\displaystyle>\sum\limits_{\sigma\text{ is a }V\text{-listing}}\underbrace{\left|J_{\sigma}\right|}_{\begin{subarray}{c}=\sum\limits\limits_{v\in V}\left[v\in J_{\sigma}\right]\\ \text{(by (\ref{pf.thm.indep.lower-bd.4}))}\end{subarray}}=\underbrace{\sum\limits_{\sigma\text{ is a }V\text{-listing}}\ \ \sum\limits\limits_{v\in V}}_{=\sum\limits\limits_{v\in V}\ \ \sum\limits_{\sigma\text{ is a }V\text{-listing}}}\left[v\in J_{\sigma}\right]=∑v∈V∑σis aV-listing[v∈Jσ]⏟=(# of allV-listingsσsatisfyingv∈Jσ)(because the sum∑σis aV-listing[v∈Jσ]contains an addend equal to1for eachV-listingσsatisfyingv∈Jσ,and an addend equal to0for each otherV-listingσ)\displaystyle=\sum\limits\limits_{v\in V}\ \ \underbrace{\sum\limits_{\sigma\text{ is a }V\text{-listing}}\left[v\in J_{\sigma}\right]}_{\begin{subarray}{c}{}_{\begin{subarray}{c}=\left(\text{\# of all }V\text{-listings }\sigma\text{ satisfying }v\in J_{\sigma}\right)\end{subarray}}\\ \text{(because the sum }\sum\limits_{\sigma\text{ is a }V\text{-listing}}\left[v\in J_{\sigma}\right]\\ \text{contains an addend equal to }1\text{ for each }V\text{-listing }\sigma\text{ satisfying }v\in J_{\sigma}\text{,}\\ \text{and an addend equal to }0\text{ for each other }V\text{-listing }\sigma\text{)}\end{subarray}}=∑v∈V(# of allV-listingsσsatisfyingv∈Jσ)⏟≥(# of allV-listings)1+deg⁡v(by Claim 1)\displaystyle=\sum\limits\limits_{v\in V}\underbrace{\left(\text{\# of all }V\text{-listings }\sigma\text{ satisfying }v\in J_{\sigma}\right)}_{\begin{subarray}{c}\geq\dfrac{\left(\text{\# of all }V\text{-listings}\right)}{1+\deg v}\\ \text{(by Claim 1)}\end{subarray}}≥∑v∈V(# of allV-listings)1+deg⁡v=(# of allV-listings)⋅∑v∈V11+deg⁡v.\displaystyle\geq\sum\limits\limits_{v\in V}\dfrac{\left(\text{\# of all }V\text{-listings}\right)}{1+\deg v}=\left(\text{\# of all }V\text{-listings}\right)\cdot\sum\limits_{v\in V}\dfrac{1}{1+\deg v}.This is absurd (since no real numberxxcan satisfyx>xx>x). So we got a contradiction, and our proof of Theorem7.4is complete.

Let us now give a second proof of the theorem, which does provide a good algorithm:

Proof 7.8(Second proof of Theorem7.4.).

We proceed by strong induction on|V|\left|V\right|. Thus, we fixp∈ℕp\in\mathbb{N}, and we assume (as the induction hypothesis) that Theorem7.4is already proved for all loopless multigraphsGGwith<p<pvertices. We must now prove it for a loopless multigraphG=(V,E,φ)G=\left(V,E,\varphi\right)withppvertices.

If|V|=0\left|V\right|=0, then this is clear (since∅\varnothingis an independent set of appropriate size). Thus, we WLOG assume that|V|≠0\left|V\right|\neq 0. We furthermore assume WLOG thatGGis a simple graph (because otherwise, we can replaceGGbyGsimpG^{\operatorname*{simp}}; this can only decrease the degreesdeg⁡v\deg vof the verticesv∈Vv\in V, and thus our claim only becomes stronger).

Since|V|≠0\left|V\right|\neq 0, there exists a vertexu∈Vu\in VwithdegG⁡u\deg_{G}uminimum110110110Here, the notationdegH⁡u\deg_{H}umeans the degree of a vertexuuin a graphHH.. Pick such auu. Thus,

degG⁡v≥degG⁡ufor eachv∈V.\deg_{G}v\geq\deg_{G}u\ \ \ \ \ \ \ \ \ \ \text{for each }v\in V.(55) LetU:={u}∪{all neighbors ofu}U:=\left\{u\right\}\cup\left\{\text{all neighbors of }u\right\}. Thus,U⊆VU\subseteq Vand|U|=1+degG⁡u\left|U\right|=1+\deg_{G}u(this is a honest equality, sinceGGis a simple graph).

LetG′G^{\prime}be the induced subgraph ofGGon the setV∖UV\setminus U. This is the simple graph obtained fromGGby removing all vertices belonging toUU(that is, removing the vertexuualong with all its neighbors) and removing all edges that require these vertices. Then,G′G^{\prime}has fewer vertices thanGG. Hence,G′G^{\prime}has<p<pvertices (sinceGGhasppvertices). Hence, by the induction hypothesis, Theorem7.4is already proved forG′G^{\prime}. In other words,G′G^{\prime}has an independent set of size

≥∑v∈V∖U11+degG′⁡v.\geq\sum\limits_{v\in V\setminus U}\dfrac{1}{1+\deg_{G^{\prime}}v}.LetTTbe such an independent set. SetS:={u}∪TS:=\left\{u\right\}\cup T. Then,SSis an independent set ofGG(sinceT⊆V∖UT\subseteq V\setminus U, so thatTTcontains no neighbors ofuu). Moreover, I claim that|S|≥∑v∈V11+degG⁡v\left|S\right|\geq\sum\limits\limits_{v\in V}\dfrac{1}{1+\deg_{G}v}. Indeed, this follows from

∑v∈V11+degG⁡v\displaystyle\sum\limits\limits_{v\in V}\dfrac{1}{1+\deg_{G}v}=∑v∈U11+degG⁡v⏟≤11+degG⁡u(sincedegGv≥degG⁡u(by (55)))+∑v∈V∖U11+degG⁡v⏟≤11+degG′⁡v(sincedegGv≥degG′⁡v(becauseG′is a subgraph ofG))\displaystyle=\sum\limits\limits_{v\in U}\underbrace{\dfrac{1}{1+\deg_{G}v}}_{\begin{subarray}{c}\leq\dfrac{1}{1+\deg_{G}u}\\ \text{(since }\deg_{G}v\geq\deg_{G}u\\ \text{(by (\ref{pf.thm.indep.lower-bd.2nd.mindeg})))}\end{subarray}}+\sum\limits\limits_{v\in V\setminus U}\underbrace{\dfrac{1}{1+\deg_{G}v}}_{\begin{subarray}{c}\leq\dfrac{1}{1+\deg_{G^{\prime}}v}\\ \text{(since }\deg_{G}v\geq\deg_{G^{\prime}}v\\ \text{(because }G^{\prime}\text{ is a subgraph of }G\text{))}\end{subarray}}≤∑v∈U11+degG⁡u⏟=|U|⋅11+degG⁡u=1(since|U|=1+degG⁡u)+∑v∈V∖U11+degG′⁡v⏟≤|T|(sinceThas size≥∑v∈V∖U11+degG′⁡v)\displaystyle\leq\underbrace{\sum\limits\limits_{v\in U}\dfrac{1}{1+\deg_{G}u}}_{\begin{subarray}{c}=\left|U\right|\cdot\dfrac{1}{1+\deg_{G}u}\\ =1\\ \text{(since }\left|U\right|=1+\deg_{G}u\text{)}\end{subarray}}+\underbrace{\sum\limits\limits_{v\in V\setminus U}\dfrac{1}{1+\deg_{G^{\prime}}v}}_{\begin{subarray}{c}\leq\left|T\right|\\ \text{(since }T\text{ has size }\geq\sum\limits_{v\in V\setminus U}\dfrac{1}{1+\deg_{G^{\prime}}v}\text{)}\end{subarray}}≤1+|T|=|S|(sinceS={u}∪T).\displaystyle\leq 1+\left|T\right|=\left|S\right|\ \ \ \ \ \ \ \ \ \ \left(\text{since }S=\left\{u\right\}\cup T\right).So we have found an independent set ofGGhaving size≥∑v∈V11+degG⁡v\geq\sum\limits\limits_{v\in V}\dfrac{1}{1+\deg_{G}v}(namely,SS). This means that Theorem7.4holds for ourGG. This completes the induction step, and Theorem7.4is proved.

See also[Chen14]and[AloSpe16]for more about probabilistic proofs in combinatorics and in general. Here are two more applications of probabilistic proofs:

Exercise 7.10.

LetG=(V,E)G=\left(V,E\right)be a simple graph such that each vertex ofGGhas degree≥1\geq 1. Prove that there exists a subsetSSofVVhaving size≥∑v∈V21+deg⁡v\geq\sum\limits_{v\in V}\dfrac{2}{1+\deg v}and with the property that the induced subgraphG⁡[S]G\left[S\right]is a forest.

[Hint:As the example of112233shows, this claim is not true for loopless multigraphs (unlike the similar Theorem7.4).]

Exercise 7.11.

Letnnbe a positive integer. Prove that there exists a tournament withnnvertices and at leastn!2n−1\dfrac{n!}{2^{n-1}}Hamiltonian paths.

7.2A weaker (but simpler) lower bound

Let us now weaken Theorem7.4a bit:

Corollary 7.12.

LetGGbe a loopless multigraph withnnvertices andmmedges. Then,GGhas an independent set of size

≥n2n+2m.\geq\dfrac{n^{2}}{n+2m}.

In order to prove this, we will need the following inequality:

Lemma 7.13.

Leta1,a2,…,ana_{1},a_{2},\ldots,a_{n}bennpositive reals. Then,

1a1+1a2+⋯+1an≥n2a1+a2+⋯+an.\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\geq\dfrac{n^{2}}{a_{1}+a_{2}+\cdots+a_{n}}.

Proof 7.14(Proof of Lemma7.13.).

There are several ways to prove this:111111111For unexplained terminology used in the bullet points below, see any textbook on inequalities, such as[Steele04]. (That said, notation is not completely standardized; what I call “AM-HM inequality” is dubbed “HM-AM inequality” in[Steele04].)

•ApplyJensen’s inequalityto the convex functionℝ+→ℝ+,x↦1x\mathbb{R}^{+}\rightarrow\mathbb{R}^{+},\ x\mapsto\dfrac{1}{x}.
•Applythe Cauchy-Schwarz inequalityto get (a1+a2+⋯+an)(1a1+1a2+⋯+1an)\displaystyle\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\right)≥(a11a1+a21a2+⋯+an1an⏟=n)2=n2.\displaystyle\geq\left(\underbrace{\sqrt{a_{1}\dfrac{1}{a_{1}}}+\sqrt{a_{2}\dfrac{1}{a_{2}}}+\cdots+\sqrt{a_{n}\dfrac{1}{a_{n}}}}_{=n}\right)^{2}=n^{2}.
•
•
•There is a direct proof, too: First, recall the famous inequality uv+vu≥2,\dfrac{u}{v}+\dfrac{v}{u}\geq 2,(56)which holds for any two positive realsuuandvv. (This follows by observing thatuv+vu−2=(u−v)2uv≥0\dfrac{u}{v}+\dfrac{v}{u}-2=\dfrac{\left(u-v\right)^{2}}{uv}\geq 0.) Now, (a1+a2+⋯+an)(1a1+1a2+⋯+1an)\displaystyle\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\right)=(∑i=1nai)(∑j=1n1aj)=∑i=1n∑j=1nai1aj=∑i=1n∑j=1naiaj\displaystyle=\left(\sum\limits_{i=1}^{n}a_{i}\right)\left(\sum\limits_{j=1}^{n}\dfrac{1}{a_{j}}\right)=\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}a_{i}\dfrac{1}{a_{j}}=\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\dfrac{a_{i}}{a_{j}}=12(∑i=1n∑j=1naiaj+∑i=1n∑j=1naiaj)(sincex=12(x+x)for anyx∈ℝ)\displaystyle=\dfrac{1}{2}\left(\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\dfrac{a_{i}}{a_{j}}+\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\dfrac{a_{i}}{a_{j}}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{since }x=\dfrac{1}{2}\left(x+x\right)\text{ for any }x\in\mathbb{R}\right)=12(∑i=1n∑j=1naiaj+∑j=1n∑i=1najai)(here, we renamediandjasjandiin the second double sum)\displaystyle=\dfrac{1}{2}\left(\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\dfrac{a_{i}}{a_{j}}+\sum\limits_{j=1}^{n}\ \ \sum\limits_{i=1}^{n}\dfrac{a_{j}}{a_{i}}\right)\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we renamed }i\text{ and }j\text{ as }j\text{ and }i\\ \text{in the second double sum}\end{array}\right)=12(∑i=1n∑j=1naiaj+∑i=1n∑j=1najai)(here, we swapped the twosummation signs in thesecond double sum)\displaystyle=\dfrac{1}{2}\left(\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\dfrac{a_{i}}{a_{j}}+\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\dfrac{a_{j}}{a_{i}}\right)\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we swapped the two}\\ \text{summation signs in the}\\ \text{second double sum}\end{array}\right)=12∑i=1n∑j=1n(aiaj+ajai)⏟≥2(by (56))≥12∑i=1n∑j=1n2⏟=n2⋅2=12n2⋅2=n2,\displaystyle=\dfrac{1}{2}\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}\underbrace{\left(\dfrac{a_{i}}{a_{j}}+\dfrac{a_{j}}{a_{i}}\right)}_{\begin{subarray}{c}\geq 2\\ \text{(by (\ref{pf.lem.indep.cauchy-lame.uvvu}))}\end{subarray}}\geq\dfrac{1}{2}\underbrace{\sum\limits_{i=1}^{n}\ \ \sum\limits_{j=1}^{n}2}_{=n^{2}\cdot 2}=\dfrac{1}{2}n^{2}\cdot 2=n^{2},from which the claim of Lemma7.13follows.

Proof 7.15(Proof of Corollary7.12.).

Write the multigraphGGasG=(V,E,φ)G=\left(V,E,\varphi\right). Thus,|V|=n\left|V\right|=nand|E|=m\left|E\right|=m. We WLOG assume thatV={1,2,…,n}V=\left\{1,2,\ldots,n\right\}(since|V|=n\left|V\right|=n). Hence,

∑v=1ndeg⁡v\displaystyle\sum\limits_{v=1}^{n}\deg v=∑v∈Vdegv=2⋅|E|⏟=m(by Proposition2.18)\displaystyle=\sum\limits_{v\in V}\deg v=2\cdot\underbrace{\left|E\right|}_{=m}\ \ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.deg.euler}}\right)=2m.\displaystyle=2m.However, Theorem7.4yields thatGGhas an independent set of size

≥∑v∈V11+deg⁡v=∑v=1n11+deg⁡v(sinceV={1,2,…,n})\displaystyle\geq\sum\limits_{v\in V}\dfrac{1}{1+\deg v}=\sum\limits_{v=1}^{n}\dfrac{1}{1+\deg v}\ \ \ \ \ \ \ \ \ \ \left(\text{since }V=\left\{1,2,\ldots,n\right\}\right)≥n2∑v=1n(1+deg⁡v)(by Lemma7.13, applied to thenpositiverealsav=1+deg⁡vfor allv∈{1,2,…,n})\displaystyle\geq\dfrac{n^{2}}{\sum\limits_{v=1}^{n}\left(1+\deg v\right)}\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{by Lemma \ref{lem.indep.cauchy-lame}, applied to the }n\text{ positive}\\ \text{reals }a_{v}=1+\deg v\text{ for all }v\in\left\{1,2,\ldots,n\right\}\end{array}\right)=n2n+2m(since∑v=1n(1+degv)=n+∑v∈Vdeg⁡v⏟=2m=n+2m).\displaystyle=\dfrac{n^{2}}{n+2m}\ \ \ \ \ \ \ \ \ \ \left(\text{since }\sum\limits_{v=1}^{n}\left(1+\deg v\right)=n+\underbrace{\sum\limits_{v\in V}\deg v}_{=2m}=n+2m\right).This proves Corollary7.12.

7.3A proof of Turan’s theorem

Recall Turan’s theorem (Theorem2.28), whose proof we have not given so far. Now is the time. For the sake of convenience, let me repeat the statement of the theorem:

Theorem 7.16(Turan’s theorem).

Letrrbe a positive integer. LetGGbe a simple graph withnnvertices andeeedges. Assume that

e>r−1r⋅n22.e>\dfrac{r-1}{r}\cdot\dfrac{n^{2}}{2}.Then, there existr+1r+1distinct vertices ofGGthat are mutually adjacent (i.e., any two distinct vertices among theser+1r+1vertices are adjacent).

We can now easily derive it from Corollary7.12:

Proof 7.17(Proof of Theorem7.16.).

Write the simple graphGGasG=(V,E)G=\left(V,E\right). Thus,|V|=n\left|V\right|=nand|E|=e\left|E\right|=eandE⊆𝒫2(V)E\subseteq\mathcal{P}_{2}\left(V\right).

LetE¯:=𝒫2(V)∖E\overline{E}:=\mathcal{P}_{2}\left(V\right)\setminus E. Thus, the setE¯\overline{E}consists of all “non-edges” ofGG– that is, of all22-element subsets ofVVthat are not edges ofGG. Clearly,

|E¯|=|𝒫2(V)∖E|=|𝒫2(V)|⏟=(n2)−|E|⏟=e=(n2)−e.\left|\overline{E}\right|=\left|\mathcal{P}_{2}\left(V\right)\setminus E\right|=\underbrace{\left|\mathcal{P}_{2}\left(V\right)\right|}_{=\dbinom{n}{2}}-\underbrace{\left|E\right|}_{=e}=\dbinom{n}{2}-e. Now, letG¯\overline{G}be the simple graph(V,E¯)\left(V,\overline{E}\right). This simple graphG¯\overline{G}is called thecomplementary graphofGG; it hasnnvertices and|E¯|=(n2)−e\left|\overline{E}\right|=\dbinom{n}{2}-eedges.112112112For example, if1122334455G=G=, then1122334455G¯=\overline{G}=.Hence, Corollary7.12(applied toG¯\overline{G}and(n2)−e\dbinom{n}{2}-einstead ofGGandmm) yields thatG¯\overline{G}has an independent set of size

≥n2n+2⋅((n2)−e).\geq\dfrac{n^{2}}{n+2\cdot\left(\dbinom{n}{2}-e\right)}.LetSSbe this independent set. Its size is

|S|≥n2n+2⋅((n2)−e)=n2n+n⁡(n−1)−2e=n2n2−2e>r\left|S\right|\geq\dfrac{n^{2}}{n+2\cdot\left(\dbinom{n}{2}-e\right)}=\dfrac{n^{2}}{n+n\left(n-1\right)-2e}=\dfrac{n^{2}}{n^{2}-2e}>r(this inequality follows by high-school algebra frome>r−1r⋅n22e>\dfrac{r-1}{r}\cdot\dfrac{n^{2}}{2}). Hence,|S|≥r+1\left|S\right|\geq r+1(since|S|\left|S\right|andrrare integers). However,SSis an independent set ofG¯\overline{G}. Thus, any two distinct vertices inSSare non-adjacent inG¯\overline{G}and therefore adjacent inGG(by the definition ofG¯\overline{G}). Since|S|≥r+1\left|S\right|\geq r+1, we have thus foundr+1r+1(or more) distinct vertices ofGGthat are mutually adjacent inGG. This proves Theorem7.16.

Several other beautiful proofs of Theorem7.16can be found in[AigZie18, Chapter 41]and[Zhao23, §1.2].

7.4A brief introduction to digraph kernels

7.4.1Definitions and main results

We shall now curtly touch on the topic of independent sets in directed graphs, and particularly on the so-calledkernels, which combine independence withabsorptiveness(a digraph analogue of domination). The relevant definitions are very simple:

Definition 7.18.

Two verticesuuandvvof a multidigraphDDare said to beadjacentif they are adjacent in the undirected graphDundD^{\operatorname*{und}}. (In other words, they are adjacent if and only ifDDhas an arc with sourceuuand targetvvor an arc with sourcevvand targetuu.)

Definition 7.19.

Letuube a vertex of a multidigraphDD. Anoutneighborofuumeans a vertexvvofDDsuch thatDDhas an arc with sourceuuand targetvv.

Definition 7.20.

LetDDbe a multidigraph with vertex setVV. LetSSbe a subset ofVV. Then:

(a)The setSSis calledindependent(or anindependent setofDD) if no two elements ofSSare adjacent. In other words, it is called independent if it is an independent set of the undirected graphDundD^{\operatorname*{und}}.
(b)The setSSis calledabsorbing(or anabsorbing setofDD) if it has the following property: Each vertexv∈V∖Sv\in V\setminus Shas at least one outneighbor inSS. (Informally, this property is saying that each vertex ofDDthat does not itself belong toSSis just “one step away” fromSS.)
(c)The setSSis called akernel ofDDif it is both independent and absorbing.

The concept of a kernel originates in the game-theoretical work of von Neumann and Morgenstern in[NeuMor04, §65].

Example 7.21.

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\lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 157.48 66.03 L 157.48 14.37}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.0}{-1.0}{1.0}{0.0}{113.81104pt}{9.78587pt}\lxSVG@begingroup@{transform=matrix(0.0 -1.0 1.0 0.0 157.48 13.54)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 87.73 69.75 L 147.32 10.16}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.7071}{-0.7071}{0.7071}{0.7071}{106.89139pt}{6.91965pt}\lxSVG@begingroup@{transform=matrix(0.7071 -0.7071 0.7071 0.7071 147.91 9.57)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the set{1,3}\left\{1,3\right\}is independent but not absorbing (since the vertex44has no outneighbor in{1,3}\left\{1,3\right\}). On the other hand, the set{2,4}\left\{2,4\right\}is absorbing but not independent (since its two elements22and44are adjacent). However, the set{1,4}\left\{1,4\right\}is both independent and absorbing, thus a kernel ofDD. It is easy to see that this is the only kernel ofDD.
(b)Consider the55-cycle digraphC→5\overrightarrow{C}_{5}, as defined (and drawn) in Example5.115. Then, the set{1,3}\left\{1,3\right\}is independent but not absorbing (since the vertex44has no outneighbor in{1,3}\left\{1,3\right\}). On the other hand, the set{1,3,5}\left\{1,3,5\right\}is absorbing but not independent (since the vertices11and55are adjacent). It is easy to see that each independent set ofC→5\overrightarrow{C}_{5}has at most22elements, while each absorbing set ofC→5\overrightarrow{C}_{5}has at least33elements. Hence,C→5\overrightarrow{C}_{5}has no kernel. Likewise, thenn-th cycle digraphC→n\overrightarrow{C}_{n}for any oddnnhas no kernel.
(c)On the other hand, thenn-th cycle digraphC→n\overrightarrow{C}_{n}for any evennnhas two kernels:{1,3,5,…,n−1}\left\{1,3,5,\ldots,n-1\right\}and{2,4,6,…,n}\left\{2,4,6,\ldots,n\right\}.
(d)Now letDDbe the simple digraph 12345.\hbox to132.18pt{\vbox to75.28pt{\pgfpicture\makeatletter\hbox{\hskip 9.18588pt\lower-9.18588pt\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@begingroup@{fill=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.4pt} \lx@inpgf@ignorespaces\nullfont\hbox to0.0pt{\lxSVG@begingroup@{_scopebegin=1} {{\lx@inpgf@ignorespaces}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 78.74 C 12.16 85.45 6.71 90.9 0 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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 78.74 12.71 L 78.74 64.37}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.0}{1.0}{-1.0}{0.0}{56.90552pt}{47.11964pt}\lxSVG@begingroup@{transform=matrix(0.0 1.0 -1.0 0.0 78.74 65.2)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then,DDhas two kernels:{1,3,5}\left\{1,3,5\right\}and{2,4}\left\{2,4\right\}.

As this example shows, a digraphDDcan have multiple kernels or just one or none at all. Thus it is natural to ask for necessary and sufficient criteria for the existence of kernels. Example7.21(b)might suggest that an odd-length cycle would always prevent a kernel from existing (just like an odd cycle in an undirected graph makes a proper22-coloring impossible), but this is not the case: e.g., the digraph

has a kernel{0}\left\{0\right\}despite having a length-55cycle. However, the converse is true, as Moses Richardson proved in 1946[Richar46, Theorem]:

Theorem 7.22(Richardson).

LetDDbe a multidigraph. IfDDhas no odd-length cycles, thenDDhas a kernel.

Moreover, ifDDhas no cycles at all, then the kernel is unique, as shown by von Neumann and Morgenstern in 1944[NeuMor04, §65]:

Theorem 7.23(von Neumann, Morgenstern).

LetDDbe a multidigraph. IfDDhas no cycles, thenDDhas a unique kernel.

In fact, there is a “dual” to Theorem7.22for uniqueness instead of existence:

Theorem 7.24.

LetDDbe a multidigraph. IfDDhas no even-length cycles113113113An “even-length cycle” means a cycle that has even length., thenDDhas at most one kernel.

We shall prove these three theorems soon. The interested reader can consult Langlois’s thesis[Langlo23]for a comprehensive survey of recent and older work on kernels, including several generalizations of Theorem7.22.

7.4.2Proofs

Proof 7.25(Proof of Theorem7.22.).

We proceed by strong induction on|V(D)|\left|\operatorname*{V}\left(D\right)\right|(that is, the # of vertices ofDD).

Base case:Theorem7.22is obvious if|V(D)|=0\left|\operatorname*{V}\left(D\right)\right|=0, because in this case the empty set∅\varnothingis a kernel ofDD.

Induction step:LetNNbe a positive integer. Assume (as the induction hypothesis) that Theorem7.22holds whenever|V(D)|<N\left|\operatorname*{V}\left(D\right)\right|<N. We must now prove that Theorem7.22holds whenever|V(D)|=N\left|\operatorname*{V}\left(D\right)\right|=N.

So letDDbe a multidigraph with|V(D)|=N\left|\operatorname*{V}\left(D\right)\right|=Nthat has no odd-length cycles. We must show thatDDhas a kernel.

The digraphDDhas at least one vertex (since|V(D)|=N\left|\operatorname*{V}\left(D\right)\right|=Nis positive). Thus, Theorem4.64shows thatDDhas at least one sink component. Pick such a sink component, and call itCC. Then,CCis a strong component ofDD. Hence, the induced subdigraphD⁡[C]D\left[C\right]is strongly connected (by Proposition4.53). Moreover, this subdigraphD⁡[C]D\left[C\right]has no odd-length cycles (since it is a subdigraph ofDD, which has no odd-length cycles). In other words, the statement B’2 of Theorem6.15is satisfied forD⁡[C]D\left[C\right]instead ofDD. Hence, the implication B’2⟹\LongrightarrowB’1 in Theorem6.15(applied toD⁡[C]D\left[C\right]instead ofDD) shows that the underlying undirected graph(D⁡[C])und\left(D\left[C\right]\right)^{\operatorname*{und}}has a proper22-coloring. Letf:C→{1,2}f:C\rightarrow\left\{1,2\right\}be this proper22-coloring.

The mapf:C→{1,2}f:C\rightarrow\left\{1,2\right\}allows us to decompose the setCCinto two disjoint subsets

C1\displaystyle C_{1}:={v∈C∣f(v)=1}and\displaystyle:=\left\{v\in C\ \mid\ f\left(v\right)=1\right\}\ \ \ \ \ \ \ \ \ \ \text{and}C2\displaystyle C_{2}:={v∈C∣f⁡(v)=2},\displaystyle:=\left\{v\in C\ \mid\ f\left(v\right)=2\right\},which satisfyC1∪C2=CC_{1}\cup C_{2}=C. SinceC1∪C2=CC_{1}\cup C_{2}=Cis nonempty (becauseCCis a strong component ofDD), we conclude that at least one of these two subsetsC1C_{1}andC2C_{2}is nonempty. We WLOG assume thatC1C_{1}is nonempty (indeed, ifC2C_{2}is nonempty, then we can swap the two colors11and22in the proper22-coloringff, and thus makeC1C_{1}nonempty instead).

Now, letL:=V(D)∖CL:=\operatorname*{V}\left(D\right)\setminus Cbe the complement ofCCinV(D)\operatorname*{V}\left(D\right). Thus,C∩L=∅C\cap L=\varnothingandC∪L=V(D)C\cup L=\operatorname*{V}\left(D\right).

Define a subsetPPofLLby

P:={v∈L∣vhasnooutneighbor inC1}.P:=\left\{v\in L\ \mid\ v\text{ has {no} outneighbor in }C_{1}\right\}. Now consider the induced subdigraphD⁡[P]D\left[P\right]ofDD. Its number of vertices is

|V(D⁡[P])|\displaystyle\left|\operatorname*{V}\left(D\left[P\right]\right)\right|=|P|≤|L|(sinceP⊆L)\displaystyle=\left|P\right|\leq\left|L\right|\ \ \ \ \ \ \ \ \ \ \left(\text{since }P\subseteq L\right)=|V(D)|⏟=N−|C|⏟>0(sinceCis nonempty)(sinceL=V(D)∖C)\displaystyle=\underbrace{\left|\operatorname*{V}\left(D\right)\right|}_{=N}-\underbrace{\left|C\right|}_{\begin{subarray}{c}>0\\ \text{(since }C\text{ is nonempty)}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\text{since }L=\operatorname*{V}\left(D\right)\setminus C\right)<N.\displaystyle<N.Moreover, this digraphD⁡[P]D\left[P\right]has no odd-length cycles (since it is a subdigraph ofDD, which has no odd-length cycles). Hence, our induction hypothesis ensures that Theorem7.22can be applied toD⁡[P]D\left[P\right]instead ofDD. As a result, we conclude thatD⁡[P]D\left[P\right]has a kernel. LetKKbe this kernel ofD⁡[P]D\left[P\right]. Thus,K⊆PK\subseteq Pis an independent set ofD⁡[P]D\left[P\right]and simultaneously an absorbing set ofD⁡[P]D\left[P\right].

We now claim thatK∪C1K\cup C_{1}is a kernel ofDD. In order to prove this, we must verify the following two claims:

Claim 1:The setK∪C1K\cup C_{1}is independent (with respect toDD).

Claim 2:The setK∪C1K\cup C_{1}is absorbing (with respect toDD).

Proof 7.26(Proof of Claim 1.).

Assume the contrary. Thus,K∪C1K\cup C_{1}contains two adjacent verticesuuandvv. Consider theseuuandvv.

Sinceuuandvvare adjacent, we know thatDDhas an arc fromuutovvor an arc fromvvtouu. We WLOG assume thatDDhas an arc fromuutovv(because in the other case, we can just swapuuwithvv). Letaabe this arc. The existence of this arc shows thatvvis an outneighbor ofuu.

Each of the verticesuuandvvbelongs toK∪C1K\cup C_{1}, hence belongs toKKor toC1C_{1}. Thus, we are in one of the following four cases:

Case 1:We haveu∈Ku\in Kandv∈Kv\in K.

Case 2:We haveu∈Ku\in Kandv∈C1v\in C_{1}.

Case 3:We haveu∈C1u\in C_{1}andv∈Kv\in K.

Case 4:We haveu∈C1u\in C_{1}andv∈C1v\in C_{1}.

Let us first consider Case 1. In this case, we haveu∈Ku\in Kandv∈Kv\in K. Hence,u∈K⊆Pu\in K\subseteq Pandv∈K⊆Pv\in K\subseteq P. Thus,uuandvvare adjacent vertices ofDDboth belonging toPP. In other words,uuandvvare adjacent vertices ofD⁡[P]D\left[P\right]. These two vertices both belong toKK(sinceu∈Ku\in Kandv∈Kv\in K). But this impossible, sinceKKis an independent set ofD⁡[P]D\left[P\right]and thus contains no two adjacent vertices. Thus, we obtained a contradiction in Case 1.

Let us next consider Case 2. In this case, we haveu∈Ku\in Kandv∈C1v\in C_{1}. Butvvis an outneighbor ofuu. Hence,uuhas an outneighbor inC1C_{1}(namely,vv). However,u∈K⊆Pu\in K\subseteq Pshows thatuuhas no outneighbor inC1C_{1}(by the definition ofPP). These two sentences contradict each other. Thus, we have found a contradiction in Case 2.

Next, let us consider Case 3. In this case, we haveu∈C1u\in C_{1}andv∈Kv\in K. We haveu∈C1⊆Cu\in C_{1}\subseteq C. In other words, the source of the arcaalies inCC(since this source isuu). ButCCis a sink component ofDD. Hence, if the source of an arc ofDDlies inCC, then its target also lies inCC(by the definition of “sink component”). Applying this to the arcaa, we conclude that the target ofaalies inCC(since the source ofaalies inCC). But this is absurd, since the target ofaaisv∈K⊆P⊆L=V(D)∖Cv\in K\subseteq P\subseteq L=\operatorname*{V}\left(D\right)\setminus C. Thus, we have found a contradiction in Case 3.

Finally, let us consider Case 4. In this case, we haveu∈C1u\in C_{1}andv∈C1v\in C_{1}. By the definition ofC1C_{1}, this entails thatf⁡(u)=1f\left(u\right)=1andf⁡(v)=1f\left(v\right)=1. Butuuandvvare two vertices inCC(sinceu∈C1⊆Cu\in C_{1}\subseteq Candv∈C1⊆Cv\in C_{1}\subseteq C) that are adjacent inDD, and thus are two adjacent vertices in the induced subdigraphD⁡[C]D\left[C\right]. In other words,uuandvvare two adjacent vertices in the undirected graph(D⁡[C])und\left(D\left[C\right]\right)^{\operatorname*{und}}. Sincef:C→{1,2}f:C\rightarrow\left\{1,2\right\}is a proper22-coloring of this graph, we thus obtainf⁡(u)≠f⁡(v)f\left(u\right)\neq f\left(v\right). But this contradictsf⁡(u)=1=f⁡(v)f\left(u\right)=1=f\left(v\right). Hence, we have found a contradiction in Case 4.

We now have found contradictions in all four cases. Hence, our assumption was wrong, and Claim 1 is proved.

Proof 7.27(Proof of Claim 2.).

We must prove that each vertexv∈V(D)∖(K∪C1)v\in\operatorname*{V}\left(D\right)\setminus\left(K\cup C_{1}\right)has at least one outneighbor inK∪C1K\cup C_{1}. So letv∈V(D)∖(K∪C1)v\in\operatorname*{V}\left(D\right)\setminus\left(K\cup C_{1}\right)be a vertex. We must show thatvvhas an outneighbor inK∪C1K\cup C_{1}. This is the claim that we shall now prove.

We are in one of the following two cases:

Case 1:We havev∈Cv\in C.

Case 2:We havev∉Cv\notin C.

We first consider Case 1. In this case, we havev∈Cv\in C. Butv∈V(D)∖(K∪C1)v\in\operatorname*{V}\left(D\right)\setminus\left(K\cup C_{1}\right)and thusv∉K∪C1v\notin K\cup C_{1}, so thatv∉C1v\notin C_{1}. Combining this withv∈C=C1∪C2v\in C=C_{1}\cup C_{2}, we obtainv∈(C1∪C2)∖C1⊆C2v\in\left(C_{1}\cup C_{2}\right)\setminus C_{1}\subseteq C_{2}. Therefore,f⁡(v)=2f\left(v\right)=2(by the definition ofC2C_{2}).

The setC1C_{1}is nonempty, thus contains some vertexw∈C1w\in C_{1}. Consider thisww. Thus,w∈C1⊆Cw\in C_{1}\subseteq C. But the digraphD⁡[C]D\left[C\right]is strongly connected, and thus has a path fromvvtoww(since bothvvandwwbelong toCC). This path cannot have length00(sincev∉C1v\notin C_{1}andw∈C1w\in C_{1}entailv≠wv\neq w), and thus has a first arc. This first arc has sourcevv; letuube its target. Thus,uuis an outneighbor ofvvinD⁡[C]D\left[C\right]. Hence,uuandvvare adjacent vertices of the digraphD⁡[C]D\left[C\right], hence adjacent vertices of the undirected graph(D⁡[C])und\left(D\left[C\right]\right)^{\operatorname*{und}}. Sinceffis a proper22-coloring of(D⁡[C])und\left(D\left[C\right]\right)^{\operatorname*{und}}, we thus obtainf⁡(u)≠f⁡(v)=2f\left(u\right)\neq f\left(v\right)=2. Hence,f⁡(u)=1f\left(u\right)=1(sincef⁡(u)f\left(u\right)is either11or22). Therefore,u∈C1u\in C_{1}(by the definition ofC1C_{1}), so thatu∈C1⊆K∪C1u\in C_{1}\subseteq K\cup C_{1}. Thus,vvhas an outneighbor inK∪C1K\cup C_{1}(namely,uu). This proves our claim in Case 1.

Let us now consider Case 2. In this case, we havev∉Cv\notin C. Hence,v∈V(D)∖C=Lv\in\operatorname*{V}\left(D\right)\setminus C=L. On the other hand,v∈V(D)∖(K∪C1)v\in\operatorname*{V}\left(D\right)\setminus\left(K\cup C_{1}\right)and thusv∉K∪C1v\notin K\cup C_{1}, so thatv∉Kv\notin K.

Ifv∉Pv\notin P, thenvvhas an outneighbor inC1C_{1}(by the definition ofPP) and thus has an outneighbor inK∪C1K\cup C_{1}(sinceC1⊆K∪C1C_{1}\subseteq K\cup C_{1}). Thus, we are done ifv∉Pv\notin P. Hence, for the rest of this proof, we WLOG assume thatv∈Pv\in P. Thus,v∈P∖Kv\in P\setminus K(sincev∉Kv\notin K).

Recall thatKKis an absorbing set ofD⁡[P]D\left[P\right]. Hence, fromv∈P∖Kv\in P\setminus K, we conclude thatvvhas at least one outneighbor inKK. Thus,vvhas at least one outneighbor inK∪C1K\cup C_{1}(sinceK⊆K∪C1K\subseteq K\cup C_{1}). Thus, we have proved our claim in Case 2.

Thus, in both cases, we have proved our claim thatvvhas an outneighbor inK∪C1K\cup C_{1}. This completes the proof of Claim 2.

Now, the subsetK∪C1K\cup C_{1}ofV(D)\operatorname*{V}\left(D\right)is independent (by Claim 1) and absorbing (by Claim 2); hence, it is a kernel ofDD(by the definition of a kernel). Thus,DDhas a kernel. This proves Theorem7.22for ourDD, and thus completes the induction step.

Proof 7.28(Proof of Theorem7.24.).

Assume thatDDhas no even-length cycles. We must show thatDDhas at most one kernel.

Assume the contrary. Thus,DDhas two distinct kernelsKKandLL.

A vertex will be calleddisputedif it belongs toK∖LK\setminus Lor toL∖KL\setminus K. LetΔ=(K∖L)∪(L∖K)\Delta=\left(K\setminus L\right)\cup\left(L\setminus K\right)be the set of all disputed vertices ofDD. This setΔ\Deltais nonempty (sinceKKandLLare distinct). Now we shall show the following:

Claim 1:Each vertexv∈K∖Lv\in K\setminus Lhas at least one outneighbor inL∖KL\setminus K.

Proof 7.29(Proof of Claim 1.).

Letv∈K∖Lv\in K\setminus Lbe a vertex. We must show thatvvhas at least one outneighbor inL∖KL\setminus K.

Indeed,v∈K∖L⊆V(D)∖Lv\in K\setminus L\subseteq\operatorname*{V}\left(D\right)\setminus L. ButLLis a kernel ofDD, and thus an absorbing set. Hence, fromv∈V(D)∖Lv\in\operatorname*{V}\left(D\right)\setminus L, we conclude thatvvhas an outneighbor inLL. Letw∈Lw\in Lbe this outneighbor. Then,vvandwware two adjacent vertices. ButKKis a kernel ofDD, and thus an independent set. Hence,KKcontains no two adjacent vertices. If we hadw∈Kw\in K, thenvvandwwwould be two adjacent vertices inKK(sincev∈K∖L⊆Kv\in K\setminus L\subseteq Kandw∈Kw\in K), which would contradict the previous sentence. Thus, we cannot havew∈Kw\in K. Combining this withw∈Lw\in L, we obtainw∈L∖Kw\in L\setminus K. Hence,vvhas an outneighbor inL∖KL\setminus K(namely,ww). This proves Claim 1.

Claim 2:Each vertexv∈L∖Kv\in L\setminus Khas at least one outneighbor inK∖LK\setminus L.

Proof 7.30(Proof of Claim 2.).

Analogous to Claim 1 (just swapKKwithLL).

Claim 3:Each vertexv∈Δv\in\Deltahas at least one outneighbor inΔ\Delta.

Proof 7.31(Proof of Claim 3.).

This follows from Claim 1 and Claim 2 (sinceΔ=(K∖L)∪(L∖K)\Delta=\left(K\setminus L\right)\cup\left(L\setminus K\right)).

Now, consider the induced subdigraphD⁡[Δ]D\left[\Delta\right]ofDDon the setΔ\Delta. This subdigraphD⁡[Δ]D\left[\Delta\right]has at least one vertex (sinceΔ\Deltais nonempty). Moreover, each vertexvvof this subdigraphD⁡[Δ]D\left[\Delta\right]has at least one outneighbor in this subdigraph (by Claim 3), i.e., has outdegreedegD⁡[Δ]+v>0\deg_{D\left[\Delta\right]}^{+}v>0(wheredegD⁡[Δ]+v\deg_{D\left[\Delta\right]}^{+}vdenotes the outdegree ofvvwith respect to the digraphD⁡[Δ]D\left[\Delta\right]). Hence, Exercise4.32(a)(applied toD⁡[Δ]D\left[\Delta\right]instead ofDD) shows that the digraphD⁡[Δ]D\left[\Delta\right]has a cycle.

On the other hand, each vertexv∈Δv\in\Deltabelongs either toK∖LK\setminus Lor toL∖KL\setminus K(sinceΔ=(K∖L)∪(L∖K)\Delta=\left(K\setminus L\right)\cup\left(L\setminus K\right)), and these two possibilities are mutually exclusive (since the setsK∖LK\setminus LandL∖KL\setminus Kare disjoint). Hence, we can define a mapf:Δ→{1,2}f:\Delta\rightarrow\left\{1,2\right\}by

f⁡(v)={1,ifv∈K∖L;2,ifv∈L∖Kfor eachv∈Δ.f\left(v\right)=\begin{cases}1,&\text{if }v\in K\setminus L;\\ 2,&\text{if }v\in L\setminus K\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for each }v\in\Delta.Consider this mapff. We now claim the following:

Claim 4:The mapffis a proper22-coloring of the graph(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}.

Proof 7.32(Proof of Claim 4.).

Letuuandvvbe two adjacent vertices of(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}. We must show thatf⁡(u)≠f⁡(v)f\left(u\right)\neq f\left(v\right).

The verticesuuandvvare adjacent in the graph(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}, hence also in the digraphD⁡[Δ]D\left[\Delta\right], thus also in the digraphDD. ButKKis a kernel ofDD, thus an independent set ofDD. Hence,KKcontains no two adjacent vertices. Therefore, the two adjacent verticesuuandvvcannot both belong toKK. Hence, we have eitheru∉Ku\notin Korv∉Kv\notin K(or both). We WLOG assume thatu∉Ku\notin K(otherwise, swapuuwithvv). Hence,u∉K∖Lu\notin K\setminus L. Butuuis a vertex of(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}; thus,u∈Δ=(K∖L)∪(L∖K)u\in\Delta=\left(K\setminus L\right)\cup\left(L\setminus K\right). Combining this withu∉K∖Lu\notin K\setminus L, we obtainu∈((K∖L)∪(L∖K))∖(K∖L)⊆L∖Ku\in\left(\left(K\setminus L\right)\cup\left(L\setminus K\right)\right)\setminus\left(K\setminus L\right)\subseteq L\setminus K. Therefore, the definition offfyieldsf⁡(u)=2f\left(u\right)=2.

We have shown thatu∉Ku\notin Korv∉Kv\notin K. Likewise, we can show thatu∉Lu\notin Lorv∉Lv\notin L(sinceLLis also a kernel). Sinceu∉Lu\notin Lis impossible (becauseu∈L∖K⊆Lu\in L\setminus K\subseteq L), we thus obtainv∉Lv\notin L. Hence,v∉L∖Kv\notin L\setminus K. Butvvis a vertex of(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}; thus,v∈Δ=(K∖L)∪(L∖K)v\in\Delta=\left(K\setminus L\right)\cup\left(L\setminus K\right). Combining this withv∉L∖Kv\notin L\setminus K, we obtainv∈((K∖L)∪(L∖K))∖(L∖K)⊆K∖Lv\in\left(\left(K\setminus L\right)\cup\left(L\setminus K\right)\right)\setminus\left(L\setminus K\right)\subseteq K\setminus L. Therefore, the definition offfyieldsf⁡(v)=1f\left(v\right)=1. Thus,f⁡(u)=2≠1=f⁡(v)f\left(u\right)=2\neq 1=f\left(v\right). Claim 4 is thus proved.

Claim 4 shows that the graph(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}has a proper22-coloring. Hence, the implication B1⟹\LongrightarrowB2 in Theorem6.7(applied toG=(D⁡[Δ])undG=\left(D\left[\Delta\right]\right)^{\operatorname*{und}}) shows that(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}has no cycles of odd length. Thus, the digraphD⁡[Δ]D\left[\Delta\right]has no cycles of odd length either (since any such cycle would also be a cycle of(D⁡[Δ])und\left(D\left[\Delta\right]\right)^{\operatorname*{und}}). But we know that the digraphD⁡[Δ]D\left[\Delta\right]has a cycle. Thus, this cycle must have even length (sinceD⁡[Δ]D\left[\Delta\right]has no cycles of odd length). So we have shown thatD⁡[Δ]D\left[\Delta\right]has an even-length cycle. Thus,DDhas an even-length cycle (sinceD⁡[Δ]D\left[\Delta\right]is a subgraph ofDD). This contradicts the assumption thatDDhas no even-length cycles. This contradiction shows that our assumption was false. Hence, Theorem7.24is proved.

Proof 7.33(Proof of Theorem7.23.).

The digraphDDhas no cycles, hence no odd-length cycles and no even-length cycles. Thus, Theorem7.22shows thatDDhas a kernel, whereas Theorem7.24shows thatDDhas at most one kernel. Combining these, we conclude thatDDhas a unique kernel. This proves Theorem7.23. (Note that there are much easier proofs of Theorem7.23around.)

8Matchings

8.1Introduction

Independent sets of a graph consist of vertices that “have no edges in common” (i.e., no two belong to the same edge).

In a sense,matchingsare the dual notion to this: they consist of edges that “have no vertices in common” (i.e., no two contain the same vertex). Here is the formal definition:

Definition 8.1.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a loopless multigraph.

(a)AmatchingofGGmeans a subsetMMofEEsuch that no two distinct edges inMMhave a common endpoint.
(b)IfMMis a matching ofGG, then anMM-edgeshall mean an edge that belongs toMM.
(c)IfMMis a matching ofGG, and ifv∈Vv\in Vis any vertex, then we say thatvvismatchedinMM(orsaturatedinMM) ifvvis an endpoint of anMM-edge. In this case, this latterMM-edge is necessarily unique (sinceMMis a matching), and is called theMM-edgeofvv. The other endpoint of thisMM-edge (i.e., its endpoint different fromvv) is called theMM-partnerofvv.
(d)A matchingMMofGGis said to beperfectif each vertex ofGGis matched inMM.
(e)LetAAbe a subset ofVV. A matchingMMofGGis said to beAA-completeif each vertex inAAis matched inMM.

Thus, a matchingMMof a multigraphG=(V,E,φ)G=\left(V,E,\varphi\right)is perfect if and only if it isVV-complete.

Example 8.2.

LetGGbe the following simple graph:

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•The set{12,36,47}\left\{12,\ 36,\ 47\right\}is a matching ofGG. If we call this setMM, then the vertices matched inMMare1,2,3,4,6,71,2,3,4,6,7, and their respectiveMM-partners are2,1,6,7,3,42,1,6,7,3,4. This matching is not perfect, but it is (for example){1,3,4}\left\{1,3,4\right\}-complete and{1,2,3,4,6,7}\left\{1,2,3,4,6,7\right\}-complete.
•The set{12,36,67}\left\{12,\ 36,\ 67\right\}is not a matching ofGG, since the two distinct edges3636and6767from this set have a common endpoint.
•The sets∅\varnothing,{36}\left\{36\right\},{15,29,36,47}\left\{15,\ 29,\ 36,\ 47\right\}are matchings ofGGas well.

We see that any matching “pairs up” some vertices using the existing edges of the graph. Clearly, theMM-partner of theMM-partner of a vertexvvisvvitself. Also, no two distinct vertices have the sameMM-partner (since otherwise, theirMM-edges would have a common endpoint).

Warning 8.1.1.

If a multigraphGGhas loops, then most authors additionally require that a matching must not contain any loops. This ensures that Remark8.3remains valid.

Here are some natural questions:

•Does a given graphGGhave a perfect matching?
•If not, can we find a maximum-size matching?
•What about anAA-complete matching for a givenA⊆VA\subseteq V?

Here are some examples:

Example 8.4.

(a)Ifnnis even, then {{(i,j),(i+1,j)}∣iis odd, whilejis arbitrary}\left\{\left\{\left(i,j\right),\ \left(i+1,j\right)\right\}\ \mid\ i\text{ is odd, while }j\text{ is arbitrary}\right\}is a perfect matching ofPn×PmP_{n}\times P_{m}. For example, here is this perfect matching forn=4n=4andm=3m=3(we have drawn all edges that donotbelong to this matching as dotted lines): (1,1)\left(1,1\right)(1,2)\left(1,2\right)(1,3)\left(1,3\right)(2,1)\left(2,1\right)(2,2)\left(2,2\right)(2,3)\left(2,3\right)(3,1)\left(3,1\right)(3,2)\left(3,2\right)(3,3)\left(3,3\right)(4,1)\left(4,1\right)(4,2)\left(4,2\right)(4,3)\left(4,3\right)
(b)Likewise, ifmmis even, then {{(i,j),(i,j+1)}∣jis odd, whileiis arbitrary}\left\{\left\{\left(i,j\right),\ \left(i,j+1\right)\right\}\ \mid\ j\text{ is odd, while }i\text{ is arbitrary}\right\}is a perfect matching ofPn×PmP_{n}\times P_{m}.
(c)Ifnnandmmare both odd, thenPn×PmP_{n}\times P_{m}has no perfect matching. Indeed, any loopless multigraphGGwith an odd number of vertices cannot have a perfect matching, since each edge of the matching covers exactly22vertices.

Example 8.5.

The “pentagon with two antlers”C5′′C_{5}^{\prime\prime}(this is my notation, hopefully sufficiently natural) is the following graph:

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\lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 22.18 68.25 L 32.57 100.24}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -58.06 42.18 L -85.27 61.95}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .It has no perfect matching. This is easiest to see as follows: The graphC5′′C_{5}^{\prime\prime}is loopless, so each edge contains exactly two vertices. Thus, any matchingMMofC5′′C_{5}^{\prime\prime}matches exactly2⋅|M|2\cdot\left|M\right|vertices. In particular, any matching ofC5′′C_{5}^{\prime\prime}matches an even number of vertices. Since the total number of verticesC5′′C_{5}^{\prime\prime}is odd, this entails thatC5′′C_{5}^{\prime\prime}has no perfect matching.

What is the maximum size of a matching ofC5′′C_{5}^{\prime\prime}? The matching{12,34}\left\{12,\ 34\right\}ofC5′′C_{5}^{\prime\prime}has size22and cannot be improved by adding any new edges. Thus, one is tempted to believe that the maximum size of a matching is22. However, this is not the case. Indeed, the matching{12,37,45}\left\{12,\ 37,\ 45\right\}has size33. This latter matching is actually maximum-size.

Example8.5shows that when searching for a maximum-size matching, it is not sufficient to just keep adding edges until no further edges can be added; this strategy may lead to a non-improvable but non-maximum matching. This suggests that finding a maximum-size matching may be one of those hard problems like finding a maximum-size independent set. But no – there is a polynomial-time algorithm! It’s known as theEdmonds blossom algorithm, and it has a running time ofO⁡(|E|⋅|V|2)O\left(\left|E\right|\cdot\left|V\right|^{2}\right); however, it is too complicated to be covered in this course. We shall here focus on a simple case of the problem that is already interesting enough and almost as useful as the general case.

Namely, we shall study matchings ofbipartite graphs.

8.2Bipartite graphs

8.2.1Definition and examples

Definition 8.6.

Abipartite graphmeans a triple(G,X,Y)\left(G,X,Y\right), where

•G=(V,E,φ)G=\left(V,E,\varphi\right)is a multigraph, and
•XXandYYare two disjoint subsets ofVVsuch thatX∪Y=VX\cup Y=Vand such that each edge ofGGhas one endpoint inXXand one endpoint inYY.

Example 8.7.

Consider the66-th cycle graphC6C_{6}:

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-27.21 -68.19 Z M -39.37 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-30.95276pt}{-52.50397pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -42.83 -72.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 51.53 -68.19 C 51.53 -61.48 46.08 -56.03 39.37 -56.03 C 32.66 -56.03 27.21 -61.48 27.21 -68.19 C 27.21 -74.91 32.66 -80.35 39.37 -80.35 C 46.08 -80.35 51.53 -74.91 51.53 -68.19 Z M 39.37 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{25.95276pt}{-52.50397pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 35.91 -72.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{}{{}}{}{{}}{}{{}}{}{{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{\hbox{\hbox{\hbox{\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 72.38 11.01 L 45.73 57.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 26.66 68.19 L -26.66 68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -45.73 57.18 L -72.38 11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -72.38 -11.01 L -45.73 -57.18}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -26.66 -68.19 L 26.66 -68.19}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 45.73 -57.18 L 72.38 -11.01}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then,(C6,{1,3,5},{2,4,6})\left(C_{6},\ \left\{1,3,5\right\},\ \left\{2,4,6\right\}\right)is a bipartite graph, since each edge ofGGhas one endpoint in{1,3,5}\left\{1,3,5\right\}and one endpoint in{2,4,6}\left\{2,4,6\right\}. Also,(C6,{2,4,6},{1,3,5})\left(C_{6},\ \left\{2,4,6\right\},\ \left\{1,3,5\right\}\right)is a bipartite graph.

Note that a bipartite graph(G,X,Y)\left(G,X,Y\right)is not just the graphGGbut rather the whole package consisting of the graphGGand the subsetsXXandYY. Two different bipartite graphs can have the same underlying graphGGbut different choices ofXXandYY. For example, the two bipartite graphs(C6,{1,3,5},{2,4,6})\left(C_{6},\ \left\{1,3,5\right\},\ \left\{2,4,6\right\}\right)and(C6,{2,4,6},{1,3,5})\left(C_{6},\ \left\{2,4,6\right\},\ \left\{1,3,5\right\}\right)are different.

We typically draw a bipartite graph(G,X,Y)\left(G,X,Y\right)by drawing the graphGGin such a way that the vertices inXXare aligned along one vertical line and the verticesYYare aligned along another, with the former line being left of the latter. Thus, for example, the bipartite graph(C6,{1,3,5},{2,4,6})\left(C_{6},\ \left\{1,3,5\right\},\ \left\{2,4,6\right\}\right)can be drawn as follows:

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-12.16 78.74 C -12.16 72.03 -6.71 66.58 0 66.58 C 6.71 66.58 12.16 72.03 12.16 78.74 Z M 0 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 78.74 C 90.9 85.45 85.45 90.9 78.74 90.9 C 72.03 90.9 66.58 85.45 66.58 78.74 C 66.58 72.03 72.03 66.58 78.74 66.58 C 85.45 66.58 90.9 72.03 90.9 78.74 Z M 78.74 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -3.46 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 90.9 0 C 90.9 6.71 85.45 12.16 78.74 12.16 C 72.03 12.16 66.58 6.71 66.58 0 C 66.58 -6.71 72.03 -12.16 78.74 -12.16 C 85.45 -12.16 90.9 -6.71 90.9 0 Z M 78.74 0}{fill:none} 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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 69.75 69.75 L 8.99 8.99}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.71 0 L 66.03 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 69.75 -8.99 L 8.99 -69.75}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.71 -78.74 L 66.03 -78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 73.06 -67.37 L 5.68 67.37}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Similarly, the bipartite graph(C6,{2,4,6},{1,3,5})\left(C_{6},\ \left\{2,4,6\right\},\ \left\{1,3,5\right\}\right)can be drawn as follows:

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66.58 78.74 C 66.58 72.03 72.03 66.58 78.74 66.58 C 85.45 66.58 90.9 72.03 90.9 78.74 Z M 78.74 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{53.6833pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 74.28)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 78.74 C 12.16 85.45 6.71 90.9 0 90.9 C -6.71 90.9 -12.16 85.45 -12.16 78.74 C -12.16 72.03 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-6.71 90.9 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{54.40552pt}{-3.22221pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 75.28 -4.46)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 12.16 0 C 12.16 6.71 6.71 12.16 0 12.16 C -6.71 12.16 -12.16 6.71 -12.16 0 C -12.16 -6.71 -6.71 -12.16 0 -12.16 C 6.71 -12.16 12.16 -6.71 12.16 0 Z M 0 0}{fill:none} 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\lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .

This example suggests the following terminology:

Definition 8.8.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. We shall refer to the vertices inXXas theleft verticesof this bipartite graph. We shall refer to the vertices inYYas theright verticesof this bipartite graph. Moreover, the edges ofGGwill be called theedgesof this bipartite graph.

Thus, each edge of a bipartite graph joins one left vertex with one right vertex.114114114Another example of a bipartite graph is(Kn,m,{1,2,…,n},{−1,−2,…,−m})\left(K_{n,m},\ \left\{1,2,\ldots,n\right\},\ \left\{-1,-2,\ldots,-m\right\}\right), whereKn,mK_{n,m}is as in Exercise5.169.

8.2.2Bipartite graphs as graphs with a proper22-coloring

Bipartite graphs are “the same as” multigraphs with a proper22-coloring. To wit:

Proposition 8.9.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph.

(a)If(G,X,Y)\left(G,X,Y\right)is a bipartite graph, then the map f:V\displaystyle f:V→{1,2},\displaystyle\rightarrow\left\{1,2\right\},v\displaystyle v↦{1,ifv∈X;2,ifv∈Y\displaystyle\mapsto\begin{cases}1,&\text{if }v\in X;\\ 2,&\text{if }v\in Y\end{cases}is a proper22-coloring ofGG.
(b)Conversely, iff:V→{1,2}f:V\rightarrow\left\{1,2\right\}is a proper22-coloring ofGG, then(G,V1,V2)\left(G,V_{1},V_{2}\right)is a bipartite graph, where we set Vi:={all vertices with colori}for eachi∈{1,2}.V_{i}:=\left\{\text{all vertices with color }i\right\}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{1,2\right\}.
(c)These constructions are mutually inverse. (That is, going from a bipartite graph to a proper22-coloring and back again results in the original bipartite graph, whereas going from a proper22-coloring to a bipartite graph and back again results in the original22-coloring.)

Proof 8.10.

An exercise in understanding the definitions.

Proposition 8.11.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. Then, the graphGGhas no circuits of odd length. In particular,GGhas no loops or triangles.

Proof 8.12.

By Proposition8.9(a), we know thatGGhas a proper22-coloring. Hence, the22-coloring equivalence theorem (Theorem6.7) shows thatGGhas no circuits of odd length. In particular,GGhas no loops or triangles (since these would yield circuits of length11or33, respectively).

8.2.3Neighbor sets

We need another piece of notation:

Definition 8.13.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be any multigraph. LetUUbe a subset ofVV. Then, we define

N(U):={v∈V∣vhas a neighbor inU}.N\left(U\right):=\left\{v\in V\ \mid\ v\text{ has a neighbor in }U\right\}.This is called theneighbor setofUU.

Example 8.14.

IfGGis the “pentagon with antlers”C5′′C_{5}^{\prime\prime}from Example8.5, then

N⁡({1,5,6})\displaystyle N\left(\left\{1,5,6\right\}\right)={1,2,4,5};\displaystyle=\left\{1,2,4,5\right\};N⁡({1})\displaystyle N\left(\left\{1\right\}\right)={2,5};\displaystyle=\left\{2,5\right\};N⁡(∅)\displaystyle N\left(\varnothing\right)=∅.\displaystyle=\varnothing.

For bipartite graphs, the neighbor set has a nice property:

Proposition 8.15.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. LetA⊆XA\subseteq X. Then,

N⁡(A)⊆Y.N\left(A\right)\subseteq Y.

Proof 8.16.

Letv∈N⁡(A)v\in N\left(A\right). Thus, the vertexvvhas a neighbor inAA(by definition ofN⁡(A)N\left(A\right)). Letwwbe this neighbor. Then,w∈A⊆Xw\in A\subseteq X, so thatw∉Yw\notin Y(since the bipartiteness of(G,X,Y)\left(G,X,Y\right)shows that the setsXXandYYare disjoint).

There exists some edge that has endpointsvvandww(sincewwis a neighbor ofvv). This edge must have an endpoint inYY(since the bipartiteness of(G,X,Y)\left(G,X,Y\right)shows that each edge ofGGhas one endpoint inYY). In other words, one ofvvandwwmust belong toYY(since the endpoints of this edge arevvandww). Sincew∉Yw\notin Y, we thus conclude thatv∈Yv\in Y.

Thus, we have shown thatv∈Yv\in Yfor eachv∈N⁡(A)v\in N\left(A\right). In other words,N⁡(A)⊆YN\left(A\right)\subseteq Y.

Exercise 8.17.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. Prove that

∑A⊆X(−1)|A|[N(A)=Y]=∑B⊆Y(−1)|B|[N(B)=X]\sum\limits_{A\subseteq X}\left(-1\right)^{\left|A\right|}\left[N\left(A\right)=Y\right]=\sum\limits_{B\subseteq Y}\left(-1\right)^{\left|B\right|}\left[N\left(B\right)=X\right](where we are using the Iverson bracket notation).

8.3Hall’s marriage theorem

How can we tell whether a bipartite graph has a perfect matching? anXX-complete matching?

8.3.1Generalities

First, to keep the suspense, let us prove some trivialities:

Proposition 8.18.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. LetMMbe a matching ofGG. Then:

(a)TheMM-partner of a vertexx∈Xx\in X(if it exists) belongs toYY. TheMM-partner of a vertexy∈Yy\in Y(if it exists) belongs toXX.
(b)We have|M|≤|X|\left|M\right|\leq\left|X\right|and|M|≤|Y|\left|M\right|\leq\left|Y\right|.
(c)IfMMisXX-complete, then|X|≤|Y|\left|X\right|\leq\left|Y\right|.
(d)IfMMis perfect, then|X|=|Y|\left|X\right|=\left|Y\right|.
(e)If|M|≥|X|\left|M\right|\geq\left|X\right|, thenMMisXX-complete.
(f)IfMMisXX-complete and we have|X|=|Y|\left|X\right|=\left|Y\right|, thenMMis perfect.

Proof 8.19.

Each edge ofGGhas an endpoint inXXand an endpoint inYY(since(G,X,Y)\left(G,X,Y\right)is a bipartite graph). Thus, in particular, eachMM-edge has an endpoint inXXand an endpoint inYY. Moreover, no twoMM-edges share a common endpoint (sinceMMis a matching).

(a)This follows from the fact that eachMM-edge has an endpoint inXXand an endpoint inYY.

(e)Assume that|M|≥|X|\left|M\right|\geq\left|X\right|.

(f)Assume thatMMisXX-complete and that we have|X|=|Y|\left|X\right|=\left|Y\right|.

Example 8.20.

Consider the bipartite graph

(drawn as explained in Example8.7). Does this graph have a perfect matching? No, because the two left vertices11and33would necessarily have the same partner in such a matching (since their only possible partner is22).

Similarly, the bipartite graph

has no perfect matching, since the three left vertices11,55and77have only two potential partners (viz.,22and66).

Proposition 8.21.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. LetAAbe a subset ofXX. Assume thatGGhas anXX-complete matching. Then,|N⁡(A)|≥|A|\left|N\left(A\right)\right|\geq\left|A\right|.

Proof 8.22.

LetVVbe the vertex set ofGG. We assumed thatGGhas anXX-complete matching. LetMMbe such a matching. Thus, eachx∈Xx\in Xhas anMM-partner. The map

𝐩:X\displaystyle\mathbf{p}:X→V,\displaystyle\rightarrow V,x\displaystyle x↦(theM-partner ofx)\displaystyle\mapsto\left(\text{the }M\text{-partner of }x\right)is injective (since two distinct vertices cannot have the sameMM-partner). Thus,|𝐩⁡(A)|=|A|\left|\mathbf{p}\left(A\right)\right|=\left|A\right|(because any injective map preserves the size of a subset). However,𝐩⁡(A)⊆N⁡(A)\mathbf{p}\left(A\right)\subseteq N\left(A\right), because theMM-partner of an element ofAAwill always belong toN⁡(A)N\left(A\right). Hence,|𝐩⁡(A)|≤|N⁡(A)|\left|\mathbf{p}\left(A\right)\right|\leq\left|N\left(A\right)\right|. Thus,|N⁡(A)|≥|𝐩⁡(A)|=|A|\left|N\left(A\right)\right|\geq\left|\mathbf{p}\left(A\right)\right|=\left|A\right|, qed.

8.3.2Hall’s marriage theorem

Proposition8.21gives a necessary condition for the existence of anXX-complete matching in a bipartite graph(G,X,Y)\left(G,X,Y\right). Interestingly, this condition is also sufficient:

Theorem 8.23(Hall’s marriage theorem, short: HMT).

Then,GGhas anXX-complete matching.

This is called “marriage theorem” because one can interpret a bipartite graph as a dating scene, withXXbeing the guys andYYthe ladies. A guyxxand a ladyyyare adjacent if and only if they are interested in one another. Thus, anXX-complete matching is a way of marrying off each guy to some lady he is mutually interested in (without allowing polygamy). This is a classical model for bipartite graphs and appears all across the combinatorics literature; to my knowledge, however, no real-life applications have been found along these lines. Nevertheless, Hall’s marriage theorem can be applied in many other situations, such as logistics (although its generalizations, which we will soon see, are even more useful in that). Philip Hall has originally invented the theorem in 1935 (in a somewhat obfuscated form), motivated (I believe) by a problem about finite groups. So did Wilhelm Maak, also in 1935, for use in analysis (defining a notion of integrals for almost-periodic functions).

There are many proofs of Hall’s marriage theorem, some pretty easy. Two short and self-contained proofs can be found in[LeLeMe18, §12.5.2]and in[Harju14, Theorem 3.9]. I will tease you by leaving the theorem unproved for several pages, while exploring some of its many consequences. Afterwards, I will give two proofs of Hall’s marriage theorem:

•one proof using the theory ofnetwork flows(Section9.5) -- an elegant theory created for use in logistics116116116and, more generally, operations researchin the 1950s that has proved to be quite useful in combinatorics. Among other consequences, this proof will also provide a polynomial-time algorithm for actually finding a maximum matching in a bipartite graph (Theorem8.23by itself does not help here).
•another proof using theGallai–Milgram theorem(Subsection10.2.3) – an elegant and surprising property of paths in digraphs.

8.4König and Hall–König

Hall’s marriage theorem is famous for its many forms and versions, most of which are “secretly” equivalent to it (i.e., can be derived from it and conversely can be used to derive it without too much trouble). We will start with one that is known asKönig’s theorem(discovered independently by Dénes Kőnig and Jenő Egerváry in 1931). This relies on the notion of avertex cover. Here is its definition:

Definition 8.24.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. Avertex coverofGGmeans a subsetCCofVVsuch that each edge ofGGcontains at least one vertex inCC.

Example 8.25.

Letn≥1n\geq 1. What are the vertex covers of the complete graphKnK_{n}?

A quick thought reveals that any subsetSSof{1,2,…,n}\left\{1,2,\ldots,n\right\}that has at leastn−1n-1elements is a vertex cover ofKnK_{n}. (In fact,KnK_{n}has no loops, so that each edge ofKnK_{n}contains two different vertices, and thus at least one of these two vertices belongs toSS.) On the other hand, a subsetSSwith fewer thann−1n-1vertices will never be a vertex cover ofKnK_{n}(since there will be at least two distinct vertices that don’t belong toSS, and the edge that joins these two vertices contains no vertex inSS).

Example 8.26.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be the graph from Example8.20. Then, the set{2,5}\left\{2,5\right\}is a vertex cover ofGG. Of course, any subset ofVVthat contains{2,5}\left\{2,5\right\}as a subset will thus also be a vertex cover ofGG.

Note that the notion of a vertex cover is (in some sense) “dual” to the notion of an edge cover, which we defined in Exercise2.92. For those getting confused, here is a convenient table (courtesy ofNadia Lafrenière, Math 38, Spring 2021):

a …is a set of …that contains …matchingedgesat most one edge per vertexedge coveredgesat least one edge per vertexindependent setverticesat most one vertex per edgevertex coververticesat least one vertex per edge The notion of vertex covers is also somewhat reminiscent of the notion of dominating sets; here is the precise relation:

Proposition 8.28.

LetGGbe a loopless multigraph.

Letmmbe the largest size of a matching ofGG.

Letccbe the smallest size of a vertex cover ofGG.

Then,m≤cm\leq c.

Proof 8.29.

By the definition ofmm, we know thatGGhas a matchingMMof sizemm.

By the definition ofcc, we know thatGGhas a vertex coverCCof sizecc.

In general, we can havem<cm<cin Proposition8.28. However, for abipartitegraph, equality reigns:

Theorem 8.30(König’s theorem).

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph.

Letmmbe the largest size of a matching ofGG.

Letccbe the smallest size of a vertex cover ofGG.

Then,m=cm=c.

Both Hall’s and König’s theorems easily follow from the following theorem:

Theorem 8.31(Hall–König matching theorem).

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. Then, there exist a matchingMMofGGand a subsetUUofXXsuch that

|M|≥|N⁡(U)|+|X|−|U|.\left|M\right|\geq\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|.

We will prove this theorem in Section9.5and again in Subsection10.2.3. For now, let us show that Hall’s marriage theorem (Theorem8.23), König’s theorem (Theorem8.30) and the Hall–König matching theorem (Theorem8.31) are mutually equivalent. More precisely, we will explain how to derive the first two from the third, and outline the reverse derivations.

Proof 8.32(Proof of Theorem8.23using Theorem8.31.).

Assume that Theorem8.31has already been proved.

Theorem8.31yields that there exist a matchingMMofGGand a subsetUUofXXsuch that

|M|≥|N⁡(U)|+|X|−|U|.\left|M\right|\geq\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|.Consider theseMMandUU. The Hall condition shows that each subsetAAofXXsatisfies|N⁡(A)|≥|A|\left|N\left(A\right)\right|\geq\left|A\right|. Applying this toA=UA=U, we obtain|N⁡(U)|≥|U|\left|N\left(U\right)\right|\geq\left|U\right|. Thus,

|M|≥|N⁡(U)|⏟≥|U|+|X|−|U|≥|X|.\left|M\right|\geq\underbrace{\left|N\left(U\right)\right|}_{\geq\left|U\right|}+\left|X\right|-\left|U\right|\geq\left|X\right|.Hence, the matchingMMisXX-complete (by Proposition8.18(e)). Thus, we have found anXX-complete matching. This proves Theorem8.23(assuming that Theorem8.31is true).

Proof 8.33(Proof of Theorem8.30using Theorem8.31.).

Assume that Theorem8.31has already been proved.

Write the multigraphGGasG=(V,E,φ)G=\left(V,E,\varphi\right). Theorem8.31yields that there exist a matchingMMofGGand a subsetUUofXXsuch that

LetC:=(X∖U)∪N⁡(U)C:=\left(X\setminus U\right)\cup N\left(U\right). This is a subset ofVV. Moreover, each edge ofGGhas at least one endpoint inCC(this is easy to see117117117Proof.Leteebe an edge ofGG. We must show thateehas at least one endpoint inCC. Clearly, the edgeeehas an endpoint inXX(since(G,X,Y)\left(G,X,Y\right)is a bipartite graph). Letxxbe this endpoint. Thisxxeither belongs toUUor doesn’t.•Ifxxbelongs toUU, then the other endpoint ofee(that is, the endpoint distinct fromxx) belongs toN⁡(U)N\left(U\right)(since its neighborxxbelongs toUU) and therefore toCC(sinceN⁡(U)⊆(X∖U)∪N⁡(U)=CN\left(U\right)\subseteq\left(X\setminus U\right)\cup N\left(U\right)=C).•Ifxxdoes not belong toUU, thenxxbelongs toX∖UX\setminus U(sincex∈Xx\in X) and therefore toCC(sinceX∖U⊆(X∖U)∪N⁡(U)=CX\setminus U\subseteq\left(X\setminus U\right)\cup N\left(U\right)=C).In either of these two cases, we have found an endpoint ofeethat belongs toCC. Thus,eehas at least one endpoint inCC, qed.). Hence,CCis a vertex cover ofGG. Therefore,|C|≥c\left|C\right|\geq c(sinceccis the smallest size of a vertex cover ofGG). The definition ofCCyields

|C|\displaystyle\left|C\right|=|(X∖U)∪N⁡(U)|\displaystyle=\left|\left(X\setminus U\right)\cup N\left(U\right)\right|≤|X∖U|⏟=|X|−|U|(sinceU⊆X)+|N(U)|(actually an equality, but we don’t care)\displaystyle\leq\underbrace{\left|X\setminus U\right|}_{\begin{subarray}{c}=\left|X\right|-\left|U\right|\\ \text{(since }U\subseteq X\text{)}\end{subarray}}+\left|N\left(U\right)\right|\ \ \ \ \ \ \ \ \ \ \left(\text{actually an equality, but we don’t care}\right)=|X|−|U|+|N(U)|=|N(U)|+|X|−|U|≤|M|(by (57))\displaystyle=\left|X\right|-\left|U\right|+\left|N\left(U\right)\right|=\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|\leq\left|M\right|\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.match.konig.1})}\right)≤m.\displaystyle\leq m.Hence,m≥|C|≥cm\geq\left|C\right|\geq c. Combining this withm≤cm\leq c(which follows from Proposition8.28), we obtainm=cm=c. Thus, Theorem8.30follows.

Conversely, it is not hard to derive the HKMT from either Hall or König:

Proof 8.34(Proof of Theorem8.31using Theorem8.23(sketched).).

Assume that Theorem8.23has already been proved.

Add a bunch of “dummy vertices” toYYand join each of these “dummy vertices” by a new edge to each vertex inXX. How many “dummy vertices” should we add? As many as it takes to ensure that every subsetAAofXXsatisfies the Hall condition – i.e., exactlymax⁡{|A|−|N⁡(A)|∣Ais a subset ofX}\max\left\{\left|A\right|-\left|N\left(A\right)\right|\ \mid\ A\text{ is a subset of }X\right\}many.

LetG′G^{\prime}be the resulting graph. Let alsoDDbe the set of all dummy vertices that were added toYY, and letY′=Y∪DY^{\prime}=Y\cup Dbe the set of all right vertices ofG′G^{\prime}. (The set of left vertices ofG′G^{\prime}is stillXX.) Then, the bipartite graph(G′,X,Y′)\left(G^{\prime},X,Y^{\prime}\right)satisfies the Hall condition, and therefore we can apply Theorem8.23to(G′,X,Y′)\left(G^{\prime},X,Y^{\prime}\right)instead of(G,X,Y)\left(G,X,Y\right), and conclude that the graphG′G^{\prime}has anXX-complete matching. LetM′M^{\prime}be this matching. By removing fromM′M^{\prime}all edges that contain dummy vertices, we obtain a matchingMMofGG. This matchingMMhas size

|M|\displaystyle\left|M\right|=|M′|−(the number of edges that were removed fromM′)⏟≤(the number of dummy vertices)(since each dummy vertex is contained in at most oneM′-edge)\displaystyle=\left|M^{\prime}\right|-\underbrace{\left(\text{the number of edges that were removed from }M^{\prime}\right)}_{\begin{subarray}{c}\leq\left(\text{the number of dummy vertices}\right)\\ \text{(since each dummy vertex is contained in at most one }M^{\prime}\text{-edge)}\end{subarray}}≥|M′|−(the number of dummy vertices)⏟=max{|A|−|N(A)|∣Ais a subset ofX}(by the construction of the dummy vertices)\displaystyle\geq\left|M^{\prime}\right|-\underbrace{\left(\text{the number of dummy vertices}\right)}_{\begin{subarray}{c}=\max\left\{\left|A\right|-\left|N\left(A\right)\right|\ \mid\ A\text{ is a subset of }X\right\}\\ \text{(by the construction of the dummy vertices)}\end{subarray}}=|M′|−max{|A|−|N(A)|∣Ais a subset ofX}.\displaystyle=\left|M^{\prime}\right|-\max\left\{\left|A\right|-\left|N\left(A\right)\right|\ \mid\ A\text{ is a subset of }X\right\}.(58) However, the maximum of a set is always an element of this set. Hence, there exists a subsetUUofXXsuch that

max⁡{|A|−|N⁡(A)|∣Ais a subset ofX}=|U|−|N⁡(U)|.\max\left\{\left|A\right|-\left|N\left(A\right)\right|\ \mid\ A\text{ is a subset of }X\right\}=\left|U\right|-\left|N\left(U\right)\right|.Consider thisUU. Then, (58) becomes

|M|\displaystyle\left|M\right|≥|M′|⏟≥|X|(sinceM′isX-complete,and thus eachx∈XhasanM′-edge (and theseedges are distinct))−max⁡{|A|−|N⁡(A)|∣Ais a subset ofX}⏟=|U|−|N⁡(U)|\displaystyle\geq\underbrace{\left|M^{\prime}\right|}_{\begin{subarray}{c}\geq\left|X\right|\\ \text{(since }M^{\prime}\text{ is }X\text{-complete,}\\ \text{and thus each }x\in X\text{ has}\\ \text{an }M^{\prime}\text{-edge (and these}\\ \text{edges are distinct))}\end{subarray}}-\underbrace{\max\left\{\left|A\right|-\left|N\left(A\right)\right|\ \mid\ A\text{ is a subset of }X\right\}}_{=\left|U\right|-\left|N\left(U\right)\right|}≥|X|−(|U|−|N⁡(U)|)=|N⁡(U)|+|X|−|U|.\displaystyle\geq\left|X\right|-\left(\left|U\right|-\left|N\left(U\right)\right|\right)=\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|.Hence, we have found a matchingMMofGGand a subsetUUofXXsuch that|M|≥|N⁡(U)|+|X|−|U|\left|M\right|\geq\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|. This proves Theorem8.31(assuming that Theorem8.23is true).

Proof 8.35(Proof of Theorem8.31from Theorem8.30(sketched).).

Assume that Theorem8.30has already been proved.

LetMMbe a maximum-size matching ofGG. LetCCbe a minimum-size vertex cover ofGG. Then, Theorem8.30says that|M|=|C|\left|M\right|=\left|C\right|.

|N⁡(U)|⏟≤|C∖X|+|X|−|U⏟=X∖C|≤|C∖X|+|X|−|X∖C|⏟=|C∩X|=|C∖X|+|C∩X|=|C|=|M|.\underbrace{\left|N\left(U\right)\right|}_{\leq\left|C\setminus X\right|}+\left|X\right|-\left|\underbrace{U}_{=X\setminus C}\right|\leq\left|C\setminus X\right|+\underbrace{\left|X\right|-\left|X\setminus C\right|}_{=\left|C\cap X\right|}=\left|C\setminus X\right|+\left|C\cap X\right|=\left|C\right|=\left|M\right|.Hence,|M|≥|N⁡(U)|+|X|−|U|\left|M\right|\geq\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|. This proves Theorem8.31(assuming that Theorem8.30is true).

Theorem8.31thus occupies a convenient “high ground” between the Hall and König theorems, allowing easy access to both of them. We shall prove Theorem8.31in Section9.5and again in Subsection10.2.3.

8.5Systems of representatives

There are two more equivalent form of the HMT that have the “advantage” that they do not rely on the notion of a graph. When non-combinatorialists use the HMT, they often use it in one of these forms. Here is the first form:

Theorem 8.36(existence of SDR).

LetA1,A2,…,AnA_{1},A_{2},\ldots,A_{n}be anynnsets. Assume that the union of anyppof these sets has size≥p\geq p, for allp∈{0,1,…,n}p\in\left\{0,1,\ldots,n\right\}. (In other words, assume that

|Ai1∪Ai2∪⋯∪Aip|≥pfor any1≤i1<i2<⋯<ip≤n.\left|A_{i_{1}}\cup A_{i_{2}}\cup\cdots\cup A_{i_{p}}\right|\geq p\ \ \ \ \ \ \ \ \ \ \text{for any }1\leq i_{1}<i_{2}<\cdots<i_{p}\leq n.)

Then, we can findnndistinctelements

a1∈A1,a2∈A2,…,an∈An.a_{1}\in A_{1},\ \ \ \ \ \ \ \ \ \ a_{2}\in A_{2},\ \ \ \ \ \ \ \ \ \ \ldots,\ \ \ \ \ \ \ \ \ \ a_{n}\in A_{n}.

Example 8.38.

Takea standard deck of cards, and deal them out into1313piles of44cards each – e.g., as follows:

{2♠,2♡,9♢,K♢},{A♠,A♡,3♠,3♢},{A♢,4♣,5♣,Q♣},\displaystyle\left\{2\spadesuit,\ 2\heartsuit,\ 9\diamondsuit,\ K\diamondsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{A\spadesuit,\ A\heartsuit,\ 3\spadesuit,\ 3\diamondsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{A\diamondsuit,\ 4\clubsuit,\ 5\clubsuit,\ Q\clubsuit\right\},{2♢,4♡,5♡,5♠},{A♣,7♣,7♠,7♡},{4♠,6♠,6♢,6♣},\displaystyle\left\{2\diamondsuit,\ 4\heartsuit,\ 5\heartsuit,\ 5\spadesuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{A\clubsuit,\ 7\clubsuit,\ 7\spadesuit,\ 7\heartsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{4\spadesuit,\ 6\spadesuit,\ 6\diamondsuit,\ 6\clubsuit\right\},{3♡,3♣,8♠,8♡},{2♣,K♣,K♡,10♡},{4♢,5♢,9♠,9♡},\displaystyle\left\{3\heartsuit,\ 3\clubsuit,\ 8\spadesuit,\ 8\heartsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{2\clubsuit,\ K\clubsuit,\ K\heartsuit,\ 10\heartsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{4\diamondsuit,\ 5\diamondsuit,\ 9\spadesuit,\ 9\heartsuit\right\},{Q♠,Q♡,Q♢,Q♣},{6♡,J♠,J♢,J♣},{7♢,8♢,8♣,9♣},\displaystyle\left\{Q\spadesuit,\ Q\heartsuit,\ Q\diamondsuit,\ Q\clubsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{6\heartsuit,\ J\spadesuit,\ J\diamondsuit,\ J\clubsuit\right\},\ \ \ \ \ \ \ \ \ \ \left\{7\diamondsuit,\ 8\diamondsuit,\ 8\clubsuit,\ 9\clubsuit\right\},{10♠,J♡,10♢,10♣}\displaystyle\left\{10\spadesuit,\ J\heartsuit,\ 10\diamondsuit,\ 10\clubsuit\right\}(you can distribute the cards among the piles randomly; this is just one example). Then, I claim that it is possible to select exactly11card from each pile so that the1313selected cards contain exactly11card of each rank (i.e., exactly one ace, exactly one22, exactly one33, and so on).

Indeed, this follows from Theorem8.36(applied toAi={ranks that occur in thei-th pile}A_{i}=\left\{\text{ranks that occur in the }i\text{-th pile}\right\}) because anypppiles contain cards of at leastppdifferent ranks.

Proof 8.39(Proof of Theorem8.36.).

First, we WLOG assume that allnnsetsA1,A2,…,AnA_{1},A_{2},\ldots,A_{n}are finite. (If not, then we can just replace each infinite one by annn-element subset thereof. The assumption|Ai1∪Ai2∪⋯∪Aip|≥p\left|A_{i_{1}}\cup A_{i_{2}}\cup\cdots\cup A_{i_{p}}\right|\geq pwill not be disturbed by this change – make sure you understand why!)

Furthermore, we WLOG assume that no integer belongs to any of thennsetsA1,A2,…,AnA_{1},A_{2},\ldots,A_{n}(otherwise, we just rename the elements of these sets so that they aren’t integers any more).

Now, letX={1,2,…,n}X=\left\{1,2,\ldots,n\right\}andY=A1∪A2∪⋯∪AnY=A_{1}\cup A_{2}\cup\cdots\cup A_{n}. Both setsXXandYYare finite, and are disjoint.

We define a simple graphGGas follows:

•The vertices ofGGare the elements ofX∪YX\cup Y.
•A vertexx∈Xx\in Xis adjacent to a vertexy∈Yy\in Yif and only ify∈Axy\in A_{x}. There are no further adjacencies.

Thus,(G,X,Y)\left(G,X,Y\right)is a bipartite graph. The assumption|Ai1∪Ai2∪⋯∪Aip|≥p\left|A_{i_{1}}\cup A_{i_{2}}\cup\cdots\cup A_{i_{p}}\right|\geq pensures that it satisfies the Hall condition. Hence, by the HMT (Theorem8.23), we conclude that this graphGGhas anXX-complete matching. This matching must have the form

{{1,a1},{2,a2},…,{n,an}}\left\{\left\{1,a_{1}\right\},\ \left\{2,a_{2}\right\},\ \ldots,\ \left\{n,a_{n}\right\}\right\}for somea1,a2,…,an∈Ya_{1},a_{2},\ldots,a_{n}\in Y(since(G,X,Y)\left(G,X,Y\right)is bipartite, so that the partners of the vertices1,2,…,n∈X1,2,\ldots,n\in Xmust belong toYY). These elementsa1,a2,…,an∈Ya_{1},a_{2},\ldots,a_{n}\in Yare distinct (since two edges in a matching cannot have a common endpoint), and eachi∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}satisfiesai∈Aia_{i}\in A_{i}(since the vertexaia_{i}is adjacent toiiinGG). Thus, thesea1,a2,…,ana_{1},a_{2},\ldots,a_{n}are precisely thenndistinct elements we are looking for. This proves Theorem8.36.

Conversely, it is not hard to derive the HMT from Theorem8.36. Thus, Theorem8.36is an equivalent version of the HMT. It is Theorem8.36that Hall originally discovered ([Hall35, Theorem 1]).

Here is the second set-theoretical restatement of the HMT:

Theorem 8.40(existence of SCR).

LetA1,A2,…,AnA_{1},A_{2},\ldots,A_{n}bennsets. LetB1,B2,…,BmB_{1},B_{2},\ldots,B_{m}bemmsets. Assume that for any numbers1≤i1<i2<⋯<ip≤n1\leq i_{1}<i_{2}<\cdots<i_{p}\leq n, there exist at leastppelementsj∈{1,2,…,m}j\in\left\{1,2,\ldots,m\right\}such that the unionAi1∪Ai2∪⋯∪AipA_{i_{1}}\cup A_{i_{2}}\cup\cdots\cup A_{i_{p}}has nonempty intersection withBjB_{j}. Then, there exists an injective mapσ:{1,2,…,n}→{1,2,…,m}\sigma:\left\{1,2,\ldots,n\right\}\rightarrow\left\{1,2,\ldots,m\right\}such that alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}satisfyAi∩Bσ⁡(i)≠∅A_{i}\cap B_{\sigma\left(i\right)}\neq\varnothing.

Proof 8.41.

We leave this to the reader. Again, construct an appropriate bipartite graph and apply the HMT.

(The “SCR” in the name of the theorem is short for “system of common representatives”.)

See[MirPer66]for much more about systems of representatives.

8.6Regular bipartite graphs

The HMT gives a necessary and sufficient criterion for the existence of anXX-complete matching in an arbitrary bipartite graph. In the more restrictive setting ofregularbipartite graphs – i.e., bipartite graphs where each vertex has the same degree –, there is a simpler sufficient condition: such a matching always exists! We shall soon prove this surprising fact (which is not hard using the HMT), but first let us get the definition in order:

Definition 8.42.

Letk∈ℕk\in\mathbb{N}. A multigraphGGis said to bekk-regularif all its vertices have degreekk.

Example 8.43.

A11-regular graph is a graph whose entire edge set is a perfect matching. In other words, a11-regular graph is a graph that is a disjoint union of copies of the22-nd path graphP2P_{2}. Here is an example of such a graph:

Example 8.44.

A22-regular graph is a graph that is a disjoint union of cycle graphs. Here is an example of such a graph:

(yes,C1C_{1}andC2C_{2}are allowed).

Example 8.45.

The33-regular graphs are known ascubic graphsortrivalent graphs. An example is the Petersen graph (defined in Subsection2.6.3). Here is another example (known astheFrucht graph):

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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .More examples of cubic graphs can be found onthe Wikipedia page. There is no hope of describing them all.

Recall the Kneser graphs defined in Subsection2.6.3. They are all regular:

Example 8.46.

Any Kneser graphKS,kK_{S,k}is(|S|−kk)\dbinom{\left|S\right|-k}{k}-regular.

Proof 8.47.

This is saying that ifAAis akk-element subset of a finite setSS, then there are precisely(|S|−kk)\dbinom{\left|S\right|-k}{k}manykk-element subsets ofSSthat are disjoint fromAA. But this is clear, since the latter subsets are just thekk-element subsets of the(|S|−k)\left(\left|S\right|-k\right)-element setS∖AS\setminus A.

Proposition 8.48.

Letk>0k>0. Let(G,X,Y)\left(G,X,Y\right)be akk-regular bipartite graph (i.e., a bipartite graph such thatGGiskk-regular). Then,|X|=|Y|\left|X\right|=\left|Y\right|.

Proof 8.49.

Write the multigraphGGasG=(V,E,φ)G=\left(V,E,\varphi\right). Each edgee∈Ee\in Econtains exactly one vertexx∈Xx\in X(since(G,X,Y)\left(G,X,Y\right)is a bipartite graph). Hence,

|E|\displaystyle\left|E\right|=∑x∈X(# of edges that contain the vertexx)⏟=deg⁡x=∑x∈Xdeg⁡x⏟=k(sinceGisk-regular)\displaystyle=\sum\limits_{x\in X}\underbrace{\left(\text{\# of edges that contain the vertex }x\right)}_{=\deg x}=\sum\limits_{x\in X}\underbrace{\deg x}_{\begin{subarray}{c}=k\\ \text{(since }G\text{ is }k\text{-regular)}\end{subarray}}=∑x∈Xk=k⋅|X|.\displaystyle=\sum\limits_{x\in X}k=k\cdot\left|X\right|.Similarly,|E|=k⋅|Y|\left|E\right|=k\cdot\left|Y\right|. Comparing these two equalities, we obtaink⋅|X|=k⋅|Y|k\cdot\left|X\right|=k\cdot\left|Y\right|. Sincek>0k>0, we can divide this bykk, and conclude|X|=|Y|\left|X\right|=\left|Y\right|.

Theorem 8.50(Frobenius matching theorem).

Letk>0k>0. Let(G,X,Y)\left(G,X,Y\right)be akk-regular bipartite graph (i.e., a bipartite graph such thatGGiskk-regular). Then,GGhas a perfect matching.

Proof 8.51.

First, we claim that each subsetAAofXXsatisfies|N⁡(A)|≥|A|\left|N\left(A\right)\right|\geq\left|A\right|.

Indeed, letAAbe a subset ofXX. Consider the edges ofGGthat have at least one endpoint inAA. We shall call such edges “AA-edges”. How manyAA-edges are there?

On the one hand, eachAA-edge contains exactly one vertex inAA(why?118118118Here we are using the fact thatA⊆XA\subseteq X, so that no two vertices inAAcan be adjacent.). Thus,

(# ofA-edges)\displaystyle\left(\text{\# of }A\text{-edges}\right)=∑x∈A(# ofA-edges containing the vertexx)⏟=deg⁡x(since each edge that contains the vertexxis anA-edge)\displaystyle=\sum\limits_{x\in A}\underbrace{\left(\text{\# of }A\text{-edges containing the vertex }x\right)}_{\begin{subarray}{c}=\deg x\\ \text{(since each edge that contains the vertex }x\\ \text{is an }A\text{-edge)}\end{subarray}}=∑x∈Adeg⁡x⏟=k(sinceGisk-regular)=∑x∈Ak=k⋅|A|.\displaystyle=\sum\limits_{x\in A}\underbrace{\deg x}_{\begin{subarray}{c}=k\\ \text{(since }G\text{ is }k\text{-regular)}\end{subarray}}=\sum\limits_{x\in A}k=k\cdot\left|A\right|. On the other hand, eachAA-edge contains exactly one vertex inN⁡(A)N\left(A\right)(why?119119119Here we are using the fact thatN⁡(A)⊆YN\left(A\right)\subseteq Y(which follows fromA⊆XA\subseteq Xusing Proposition8.15), so that no two vertices inN⁡(A)N\left(A\right)can be adjacent.). Thus,

(# ofA-edges)\displaystyle\left(\text{\# of }A\text{-edges}\right)=∑y∈N⁡(A)(# ofA-edges containing the vertexy)⏟≤deg⁡y\displaystyle=\sum\limits_{y\in N\left(A\right)}\underbrace{\left(\text{\# of }A\text{-edges containing the vertex }y\right)}_{\begin{subarray}{c}\leq\deg y\end{subarray}}≤∑y∈N⁡(A)deg⁡y⏟=k(sinceGisk-regular)=∑y∈N⁡(A)k=k⋅|N⁡(A)|.\displaystyle\leq\sum\limits_{y\in N\left(A\right)}\underbrace{\deg y}_{\begin{subarray}{c}=k\\ \text{(since }G\text{ is }k\text{-regular)}\end{subarray}}=\sum\limits_{y\in N\left(A\right)}k=k\cdot\left|N\left(A\right)\right|.Hence,

k⋅|N(A)|≥(# ofA-edges)=k⋅|A|.k\cdot\left|N\left(A\right)\right|\geq\left(\text{\# of }A\text{-edges}\right)=k\cdot\left|A\right|.Sincek>0k>0, we can divide this inequality bykk, and thus find|N⁡(A)|≥|A|\left|N\left(A\right)\right|\geq\left|A\right|.

For a rather surprising alternative proof of Theorem8.50, using Eulerian circuits instead of Hall’s marriage theorem, see example 2 in the MathOverflow questionhttps://mathoverflow.net/q/271608/.

8.7Latin squares

One of many applications of Theorem8.50is to the study of Latin squares. Here is the definition of this concept:

Definition 8.52.

Letn∈ℕn\in\mathbb{N}. ALatin squareof ordernnis ann×nn\times n-matrixMMthat satisfies the following conditions:

1.The entries ofMMare the numbers1,2,…,n1,2,\ldots,n, each appearing exactlynntimes.
2.In each row ofMM, the entries are distinct.
3.In each column ofMM, the entries are distinct.

Example 8.53.

Here is a Latin square of order55:

(1234523451345124512351234).\left(\begin{array}[c]{ccccc}1&2&3&4&5\\ 2&3&4&5&1\\ 3&4&5&1&2\\ 4&5&1&2&3\\ 5&1&2&3&4\end{array}\right).Similarly, for eachn∈ℕn\in\mathbb{N}, the matrix(ci+j−1)1≤i≤n,1≤j≤n\left(c_{i+j-1}\right)_{1\leq i\leq n,\ 1\leq j\leq n}, where

ck={k,ifk≤n;k−n,else,c_{k}=\begin{cases}k,&\text{if }k\leq n;\\ k-n,&\text{else,}\end{cases}is a Latin square of ordernn.

A popular example of Latin squares of order99are Sudokus (but they have to satisfy an additional requirement, concerning certain3×33\times 3subsquares). Seethe Wikipedia pageand the book[LayMul98]for much more about Latin squares.

The Latin squares in Example8.53are rather boring. What would be a good algorithm to construct general Latin squares?

Here is an attempt at a recursive algorithm: We just start by filling in the first row, then the second row, then the third row, and so on, making sure at each step that the distinctness conditions (Conditions 2 and 3 in Definition8.52) are satisfied.

Example 8.54.

Let us construct a Latin square of order55by this algorithm. We begin (e.g.) with the first row

(31425).\left(\begin{array}[c]{ccccc}3&1&4&2&5\end{array}\right).Then, we append a second row(24153)\left(\begin{array}[c]{ccccc}2&4&1&5&3\end{array}\right)to it, chosen in such a way that its five entries are distinct and also each entry is distinct from the entry above (again, there are many possibilities; we have just picked one). Thus, we have our first two rows:

(3142524153).\left(\begin{array}[c]{ccccc}3&1&4&2&5\\ 2&4&1&5&3\end{array}\right).We continue along the same lines, ending up with the Latin square

(3142524153152345234143512)\left(\begin{array}[c]{ccccc}3&1&4&2&5\\ 2&4&1&5&3\\ 1&5&2&3&4\\ 5&2&3&4&1\\ 4&3&5&1&2\end{array}\right)(or another, depending on the choices we have made).

Does this algorithm always work?

To be fully honest, it’s not a fully specified algorithm, since I haven’t explained how to fill a row (it’s not straightforward). But let’s assume that we know how to do this, if it is at all possible. The natural question is: Will we always be able to produce a complete Latin square using this algorithm, or will we get stuck somewhere (having constructedkkrows for somek<nk<n, but being unable to produce a(k+1)\left(k+1\right)-st row)?

It turns out that we won’t get stuck this way. In other words, the following holds:

Proposition 8.55.

Letn∈ℕn\in\mathbb{N}andk∈{0,1,…,n−1}k\in\left\{0,1,\ldots,n-1\right\}. Then, anyk×nk\times nLatin rectangle(i.e., anyk×nk\times n-matrix that contains the entries1,2,…,n1,2,\ldots,n, each appearing exactlykktimes, and satisfies the Conditions 2 and 3 from Definition8.52) can be extended to a(k+1)×n\left(k+1\right)\times nLatin rectangle by adding an appropriately chosen extra row at the bottom.

Proof 8.56.

LetMMbe ak×nk\times nLatin rectangle120120120For example, ifn=5n=5andk=3k=3, thenMMcan be(314252415315234)\left(\begin{array}[c]{ccccc}3&1&4&2&5\\ 2&4&1&5&3\\ 1&5&2&3&4\end{array}\right).. We want to find a new row that we can append toMMat the bottom, such that the result will be a(k+1)×n\left(k+1\right)\times nLatin rectangle.

This new row should contain the numbers1,2,…,n1,2,\ldots,nin some order. Moreover, for eachi∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}, itsii-th entry should be distinct from all entries of theii-th column ofMM. How do we find such a new row?

LetX={1,2,…,n}X=\left\{1,2,\ldots,n\right\}andY={−1,−2,…,−n}Y=\left\{-1,-2,\ldots,-n\right\}.

LetGGbe the simple graph with vertex setX∪YX\cup Y, where we let a vertexi∈Xi\in Xbe adjacent to a vertex−j∈Y-j\in Yif and only if the numberjjdoes not appear in theii-th column ofMM. There should be no further adjacencies.

Thus,(G,X,Y)\left(G,X,Y\right)is a bipartite graph. Moreover, the graphGGis(n−k)\left(n-k\right)-regular (this is not hard to see121121121Proof.Each vertexi∈Xi\in Xhas degreen−kn-k(after all, there arekknumbers in{1,2,…,n}\left\{1,2,\ldots,n\right\}that appear in theii-th column ofMM, thusn−kn-knumbers in{1,2,…,n}\left\{1,2,\ldots,n\right\}thatdo notappear in this column). It remains to show that each vertex−j∈Y-j\in Yhas degreen−kn-kas well. To see this, consider some vertex−j∈Y-j\in Y. Then, the numberjjappears exactly once in each row ofMM(since Condition 2 forces each row to contain the numbers1,2,…,n1,2,\ldots,nin some order). Hence, the numberjjappears a total ofkktimes inMM. Thesekkappearances ofjjmust be inkkdistinct columns (since having two of them in the same column would conflict with Condition 3). Thus, there arekkcolumns ofMMthat containjj, and thereforen−kn-kcolumns that don’t. In other words, the vertex−j∈Y-j\in Yhas degreen−kn-k.). Thus, by the Frobenius matching theorem (Theorem8.50), the graphGGhas a perfect matching. Let

{{1,−a1},{2,−a2},…,{n,−an}}\left\{\left\{1,\ -a_{1}\right\},\ \left\{2,\ -a_{2}\right\},\ \ldots,\ \left\{n,\ -a_{n}\right\}\right\}be this perfect matching. Then, the numbersa1,a2,…,ana_{1},a_{2},\ldots,a_{n}are distinct (since two edges in a matching cannot have a common endpoint), and the numberaia_{i}does not appear in theii-th column ofMM(since{i,−ai}\left\{i,\ -a_{i}\right\}is an edge ofGG). Thus, we can append the row

(a1a2⋯an)\left(\begin{array}[c]{cccc}a_{1}&a_{2}&\cdots&a_{n}\end{array}\right)toMMat the bottom and obtain a(k+1)×n\left(k+1\right)\times nLatin rectangle. This proves Proposition8.55.

Proposition8.55is a result of Marshall Hall (no relation to Philip Hall) from 1945 (see[Hall45]), and the proof given above is exactly his.

8.8Magic matrices and the Birkhoff–von Neumann theorem

Let us now apply the HMT to linear algebra.

Recall thatℕ={0,1,2,…}\mathbb{N}=\left\{0,1,2,\ldots\right\}. We also setℝ+:={all nonnegative reals}\mathbb{R}_{+}:=\left\{\text{all nonnegative reals}\right\}.

Here are three very similar definitions:

Definition 8.57.

Anℕ\mathbb{N}-magic matrixmeans ann×nn\times n-matrixMMthat satisfies the following three conditions:

1.All entries ofMMare nonnegative integers.
2.The sum of the entries in each row ofMMis equal.
3.The sum of the entries in each column ofMMis equal.

Definition 8.58.

Anℝ+\mathbb{R}_{+}-magic matrixmeans ann×nn\times n-matrixMMthat satisfies the following three conditions:

1.All entries ofMMare nonnegative reals.
2.The sum of the entries in each row ofMMis equal.
3.The sum of the entries in each column ofMMis equal.

Definition 8.59.

Adoubly stochastic matrixmeans ann×nn\times n-matrixMMthat satisfies the following three conditions:

1.All entries ofMMare nonnegative reals.
2.The sum of the entries in each row ofMMis11.
3.The sum of the entries in each column ofMMis11.

Clearly, these three concepts are closely related (in particular, allℕ\mathbb{N}-magic matrices and all doubly stochastic matrices areℝ+\mathbb{R}_{+}-magic). The most important of them is the last; in particular, majorization theory (one of the main methods for proving inequalities) is deeply connected to the properties of doubly stochastic matrices (see[MaOlAr11, Chapter 2]). See[BapRag97, Chapter 2]for a chapter-length treatment of doubly stochastic matrices. We shall only prove some of their most basic properties. First, some examples:

Example 8.60.

For anyn>0n>0, then×nn\times n-matrix

(11⋯111⋯1⋮⋮⋱⋮11⋯1)\left(\begin{array}[c]{cccc}1&1&\cdots&1\\ 1&1&\cdots&1\\ \vdots&\vdots&\ddots&\vdots\\ 1&1&\cdots&1\end{array}\right)isℕ\mathbb{N}-magic and alsoℝ+\mathbb{R}_{+}-magic. This matrix is not doubly stochastic (unlessn=1n=1), since the sum of the entries in a row or column isnn, not11. However, if we divide this matrix bynn, it becomes doubly stochastic.

Example 8.61.

Here is anℕ\mathbb{N}-magic3×33\times 3-matrix:

(705264363).\left(\begin{array}[c]{ccc}7&0&5\\ 2&6&4\\ 3&6&3\end{array}\right).Dividing this matrix by1212gives a doubly stochastic matrix.

Example 8.62.

Apermutation matrixis ann×nn\times n-matrix whose entries are00’s and11’s, and which has exactly one11in each row and exactly one11in each column. For example,(0010100001000001)\left(\begin{array}[c]{cccc}0&0&1&0\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\end{array}\right)is a permutation matrix of size44.

For anyn∈ℕn\in\mathbb{N}, there aren!n!many permutation matrices (of sizenn), since they are in bijection with the permutations of{1,2,…,n}\left\{1,2,\ldots,n\right\}. Namely, ifσ\sigmais a permutation of{1,2,…,n}\left\{1,2,\ldots,n\right\}, then the corresponding permutation matrixP⁡(σ)P\left(\sigma\right)has its(i,σ⁡(i))\left(i,\sigma\left(i\right)\right)-th entries equal to11for alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}, while its remainingn2−nn^{2}-nentries are00. For example, ifσ\sigmais the permutation of{1,2,3}\left\{1,2,3\right\}sending1,2,31,2,3to2,3,12,3,1, then the corresponding permutation matrixP⁡(σ)P\left(\sigma\right)is(010001100)\left(\begin{array}[c]{ccc}0&1&0\\ 0&0&1\\ 1&0&0\end{array}\right).

Any permutation matrix isℕ\mathbb{N}-magic,ℝ+\mathbb{R}_{+}-magic and doubly stochastic.

It turns out that these permutation matrices are (in a sense) the “building blocks” of all magic (and doubly stochastic) matrices! Namely, the following holds:

Theorem 8.63(Birkhoff–von Neumann theorem).

Letn∈ℕn\in\mathbb{N}. Then:

(a)Anyℕ\mathbb{N}-magicn×nn\times n-matrix can be expressed as a finite sum of permutation matrices.
(b)Anyℝ+\mathbb{R}_{+}-magicn×nn\times n-matrix can be expressed as anℝ+\mathbb{R}_{+}-linear combination of permutation matrices (i.e., it can be expressed in the formλ1P1+λ2P2+⋯+λkPk\lambda_{1}P_{1}+\lambda_{2}P_{2}+\cdots+\lambda_{k}P_{k}, whereλ1,λ2,…,λk∈ℝ+\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in\mathbb{R}_{+}are numbers and whereP1,P2,…,PkP_{1},P_{2},\ldots,P_{k}are permutation matrices).
(c)Letn>0n>0. Any doubly stochasticn×nn\times n-matrix can be expressed as aconvex combinationof permutation matrices (i.e., it can be expressed in the formλ1P1+λ2P2+⋯+λkPk\lambda_{1}P_{1}+\lambda_{2}P_{2}+\cdots+\lambda_{k}P_{k}, whereλ1,λ2,…,λk∈ℝ+\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in\mathbb{R}_{+}are numbers satisfyingλ1+λ2+⋯+λk=1\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}=1and whereP1,P2,…,PkP_{1},P_{2},\ldots,P_{k}are permutation matrices).

Soon we will sketch a proof of this theorem using the HMT. First, two simple results that will be used in the proof.

Proposition 8.64.

LetAAbe anℕ\mathbb{N}-magic orℝ+\mathbb{R}_{+}-magicn×nn\times n-matrix. Then, the sum of all entries in a row ofAAequals the sum of all entries in a column ofAA.

Proof 8.65.

Both sums equal1n\dfrac{1}{n}times the sum of all entries ofAA(sinceAAhasnnrows andnncolumns).

Lemma 8.66.

LetMMbe anℕ\mathbb{N}-magic orℝ+\mathbb{R}_{+}-magic matrix that is not the zero matrix. Then, there exists a permutationσ\sigmaof{1,2,…,n}\left\{1,2,\ldots,n\right\}such that all entriesM1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}are nonzero.

Example 8.67.

Ifn=3n=3andM=(271019820)M=\left(\begin{array}[c]{ccc}2&7&1\\ 0&1&9\\ 8&2&0\end{array}\right), then the permutationσ\sigmathat sends1,2,31,2,3to3,2,13,2,1has this property.

Proof 8.68(Proof of Lemma8.66.).

Letssdenote the sum of the entries in any given row ofMM(it doesn’t matter which row we take, sinceMMis magic). Then,ssis also the sum of the entries in any given column ofMM(by Proposition8.64). Also, the sum of all entries ofMMisnsns. Hence,ns>0ns>0(sinceMMhas nonnegative entries and is not the zero matrix). Thus,s>0s>0.

LetX={1,2,…,n}X=\left\{1,2,\ldots,n\right\}andY={−1,−2,…,−n}Y=\left\{-1,-2,\ldots,-n\right\}.

LetGGbe the simple graph with vertex setX∪YX\cup Yand with edges defined as follows: A vertexi∈Xi\in Xshall be adjacent to a vertex−j∈Y-j\in Yif and only ifMi,j>0M_{i,j}>0(here,Mi,jM_{i,j}denotes the(i,j)\left(i,j\right)-th entry ofMM). There shall be no further adjacencies.

Thus,(G,X,Y)\left(G,X,Y\right)is a bipartite graph.

We shall now prove that it satisfies the Hall condition. That is, we shall prove that every subsetAAof{1,2,…,n}\left\{1,2,\ldots,n\right\}satisfies|N⁡(A)|≥|A|\left|N\left(A\right)\right|\geq\left|A\right|.

Assume the contrary. Thus, there exists a subsetAAof{1,2,…,n}\left\{1,2,\ldots,n\right\}that satisfies|N⁡(A)|<|A|\left|N\left(A\right)\right|<\left|A\right|. Consider thisAA. WLOG assume thatA={1,2,…,k}A=\left\{1,2,\ldots,k\right\}for somek∈{0,1,…,n}k\in\left\{0,1,\ldots,n\right\}(otherwise, we permute the rows ofMM). Thus, all positive entries in the firstkkrows ofAAare concentrated in fewer thankkcolumns (since the columns in which they lie are thejj-th columns forj∈N⁡(A)j\in N\left(A\right), but we have|N⁡(A)|<|A|=k\left|N\left(A\right)\right|<\left|A\right|=k). Therefore, the sum of these entries is smaller thanksks(since the sum of all entries in any given column isss). On the other hand, however, the sum of these entries equalsksks, because they are all the positive entries in the firstkkrows ofAA(and the sum of all positive entries in a given row equals the sum ofallentries in this row, which isss). The two preceding sentences clearly contradict each other. This contradiction shows that our assumption was false.

Hence, the Hall condition is satisfied. Thus, the HMT yields thatGGhas a perfect matching. Let

{{1,−a1},{2,−a2},…,{n,−an}}\left\{\left\{1,-a_{1}\right\},\ \left\{2,-a_{2}\right\},\ \ldots,\ \left\{n,-a_{n}\right\}\right\}be this perfect matching. Then,a1,a2,…,ana_{1},a_{2},\ldots,a_{n}are distinct, so we can find a permutationσ\sigmaof{1,2,…,n}\left\{1,2,\ldots,n\right\}such thatai=σ⁡(i)a_{i}=\sigma\left(i\right)for alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}. This permutationσ\sigmathen satisfiesMi,σ⁡(i)>0M_{i,\sigma\left(i\right)}>0for alli∈{1,2,…,n}i\in\left\{1,2,\ldots,n\right\}, which is what we wanted. Thus, Lemma8.66is proved.

Proof 8.69(Proof of Theorem8.63(sketched).).

(a)LetMMbe anℕ\mathbb{N}-magicn×nn\times n-matrix. How can we expressMMas a sum of permutation matrices?

We can try the following method: Try to subtract a permutation matrix fromMMin such a way that the result will still be anℕ\mathbb{N}-magic matrix. Then do this again, and again and again… until we reach the zero matrix. Once we have arrived at the zero matrix, the sum of all the permutation matrices that we have subtracted along the way must beMM.

Let us experience this method on an example: Letn=3n=3and122122122We are here omitting zero entries from matrices. Thus,(2711982)\left(\begin{array}[c]{ccc}2&7&1\\ &1&9\\ 8&2&\end{array}\right)means the matrix(271019820)\left(\begin{array}[c]{ccc}2&7&1\\ 0&1&9\\ 8&2&0\end{array}\right).M=(2711982)M=\left(\begin{array}[c]{ccc}2&7&1\\ &1&9\\ 8&2&\end{array}\right). If we subtract a permutation matrix fromMM, then the resulting matrix will still satisfy Conditions 2 and 3 of Definition8.57(since the sum of the entries in any row has been decreased by11, and the sum of the entries in any column has also been decreased by11); however, Condition 1 is not guaranteed, since the subtraction may turn an entry ofMMnegative (which is not allowed). For example, this would happen if we tried to subtract the permutation matrix(111)\left(\begin{array}[c]{ccc}1&&\\ &1&\\ &&1\end{array}\right)fromMM. Fortunately, Lemma8.66tells us that there is a permutationσ\sigmaof{1,2,…,n}\left\{1,2,\ldots,n\right\}such that all entriesM1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}are nonzero. If we choose such aσ\sigma, and subtract the corresponding permutation matrixP⁡(σ)P\left(\sigma\right)fromMM, then we obtain anℕ\mathbb{N}-magic matrix, because subtracting11from the nonzero entriesM1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}cannot render any of these entries negative. In our example, we can pickσ\sigmato be the permutation that sends1,2,31,2,3to3,2,13,2,1. The corresponding permutation matrixP⁡(σ)P\left(\sigma\right)is(111)\left(\begin{array}[c]{ccc}&&1\\ &1&\\ 1&&\end{array}\right). Subtracting this matrix fromMM, we find

(2711982)−(111)=(27972).\left(\begin{array}[c]{ccc}2&7&1\\ &1&9\\ 8&2&\end{array}\right)-\left(\begin{array}[c]{ccc}&&1\\ &1&\\ 1&&\end{array}\right)=\left(\begin{array}[c]{ccc}2&7&\\ &&9\\ 7&2&\end{array}\right).This is again anℕ\mathbb{N}-magic matrix. Thus, let us do the same to it that we did toMM: We again subtract a permutation matrix.

This time, we can actually do better: We can subtract the permutation matrix(111)\left(\begin{array}[c]{ccc}&1&\\ &&1\\ 1&&\end{array}\right)from(27972)\left(\begin{array}[c]{ccc}2&7&\\ &&9\\ 7&2&\end{array}\right)not just once, but77times, without rendering any entry negative, because the relevant entries7,9,77,9,7are all≥7\geq 7. The result is

(27972)−7⋅(111)=(222).\left(\begin{array}[c]{ccc}2&7&\\ &&9\\ 7&2&\end{array}\right)-7\cdot\left(\begin{array}[c]{ccc}&1&\\ &&1\\ 1&&\end{array}\right)=\left(\begin{array}[c]{ccc}2&&\\ &&2\\ &2&\end{array}\right). Now, we follow the same recipe and again subtract a permutation matrix. This time, we can do it22times, and obtain

(222)−2⋅(111)=()=03×3\left(\begin{array}[c]{ccc}2&&\\ &&2\\ &2&\end{array}\right)-2\cdot\left(\begin{array}[c]{ccc}1&&\\ &&1\\ &1&\end{array}\right)=\left(\begin{array}[c]{ccc}&&\\ &&\\ &&\end{array}\right)=0_{3\times 3}(the zero matrix, in case you’re wondering).

Thus, we have arrived at the zero matrix by successively subtracting permutation matrices fromMM. Hence,MMis the sum of all the permutation matrices that have been subtracted: namely,

M=(111)+7⋅(111)+2⋅(111),M=\left(\begin{array}[c]{ccc}&&1\\ &1&\\ 1&&\end{array}\right)+7\cdot\left(\begin{array}[c]{ccc}&1&\\ &&1\\ 1&&\end{array}\right)+2\cdot\left(\begin{array}[c]{ccc}1&&\\ &&1\\ &1&\end{array}\right),which is a sum of1+7+21+7+2permutation matrices.

This method works in general, because:

•IfMMis anℕ\mathbb{N}-magic matrix that is not the zero matrix, then Lemma8.66tells us that there is a permutationσ\sigmaof{1,2,…,n}\left\{1,2,\ldots,n\right\}such that all entriesM1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}are nonzero. We can then choose such aσ\sigmaand subtract the corresponding permutation matrixP⁡(σ)P\left(\sigma\right)fromMM.
•Better yet, we can subtractm⋅P⁡(σ)m\cdot P\left(\sigma\right)fromMM, where m=min⁡{M1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)}.m=\min\left\{M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}\right\}.This results in anℕ\mathbb{N}-magic matrix (since the sum of the entries decreases bymmin each row and bymmin each column, and since we are only subtractingmmfrom a bunch of entries that are≥m\geq m) that has at least one fewer nonzero entry thanMM(since at least one of the nonzero entriesM1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}becomes00whenmmis subtracted from it).
•This way, in each step of our process, the number of nonzero entries of our matrix decreases by at least11(but the matrix remains anℕ\mathbb{N}-magic matrix throughout the process). Hence, we eventually (after at mostn2n^{2}steps) will end up with the zero matrix.

This proves Theorem8.63(a).

(b)This is analogous to the proof of part(a)(but this time, wehaveto subtractm⋅P⁡(σ)m\cdot P\left(\sigma\right)rather thanP⁡(σ)P\left(\sigma\right)in our procedure, since the nonzero entriesM1,σ⁡(1),M2,σ⁡(2),…,Mn,σ⁡(n)M_{1,\sigma\left(1\right)},\ M_{2,\sigma\left(2\right)},\ \ldots,\ M_{n,\sigma\left(n\right)}are not necessarily≥1\geq 1).

(c)LetMMbe a doubly stochasticn×nn\times n-matrix. Then,MMis alsoℝ+\mathbb{R}_{+}-magic. Hence, part(b)shows thatMMcan be expressed in the formλ1P1+λ2P2+⋯+λkPk\lambda_{1}P_{1}+\lambda_{2}P_{2}+\cdots+\lambda_{k}P_{k}, whereλ1,λ2,…,λk∈ℝ+\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in\mathbb{R}_{+}are numbers and whereP1,P2,…,PkP_{1},P_{2},\ldots,P_{k}are permutation matrices. Consider theseλ1,λ2,…,λk\lambda_{1},\lambda_{2},\ldots,\lambda_{k}and theseP1,P2,…,PkP_{1},P_{2},\ldots,P_{k}.

Now, consider the sum of all entries in the first row ofMM. It is easy to see that this sum isλ1+λ2+⋯+λk\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}(becauseM=λ1P1+λ2P2+⋯+λkPkM=\lambda_{1}P_{1}+\lambda_{2}P_{2}+\cdots+\lambda_{k}P_{k}, but each permutation matrixPiP_{i}contributes a11to the sum of all entries in the first row). But we know that this sum is11, sinceMMis doubly stochastic. Comparing these, we conclude thatλ1+λ2+⋯+λk=1\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}=1. Thus, we have expressedMMin the formλ1P1+λ2P2+⋯+λkPk\lambda_{1}P_{1}+\lambda_{2}P_{2}+\cdots+\lambda_{k}P_{k}, whereλ1,λ2,…,λk∈ℝ+\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in\mathbb{R}_{+}are numbers satisfyingλ1+λ2+⋯+λk=1\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}=1and whereP1,P2,…,PkP_{1},P_{2},\ldots,P_{k}are permutation matrices. This proves Theorem8.63(c).

8.9Further uses of Hall’s marriage theorem

The following few exercises illustrate other applications of Hall’s marriage theorem:

Exercise 8.70.

Exercise 8.71.

LetAAandBBbe two finite sets such that|B|≥|A|\left|B\right|\geq\left|A\right|. Letdi,jd_{i,j}be a real number for each(i,j)∈A×B\left(i,j\right)\in A\times B. Let

m1=minσ:A→Binjectivemaxi∈Adi,σ⁡(i)m_{1}=\min\limits_{\sigma:A\rightarrow B\text{ injective}}\ \max\limits_{i\in A}d_{i,\sigma\left(i\right)}and

m2=maxI⊆A;J⊆B;|I|+|J|=|B|+1⁡min(i,j)∈I×Jdi,j.m_{2}=\max\limits_{\begin{subarray}{c}I\subseteq A;\ J\subseteq B;\\ \left|I\right|+\left|J\right|=\left|B\right|+1\end{subarray}}\ \min\limits_{\left(i,j\right)\in I\times J}d_{i,j}.(The notation “minsome kind of objectssome kind of value\min_{\text{some kind of objects}}\text{some kind of value}” means the minimum of the given value over all objects of the given kind. An analogous notation is used for a maximum.) Prove thatm1=m2m_{1}=m_{2}.

Exercise 8.72.

(In other words, show that we can assign to each vertex an edge that does not contain this vertex in such a way that no edge is assigned twice.)

[Remark:This is, in some sense, an “evil twin” to Exercise5.108. However, it requires a simple graph, not a multigraph, since a multigraph with a single vertex and a single loop would constitute a counterexample. Incidentally, Exercise5.108can also be solved using Hall’s marriage theorem.]

[Solution:This is Exercise 1 on midterm #2 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 8.73.

LetSSbe a finite set, and letk∈ℕk\in\mathbb{N}. LetA1,A2,…,AkA_{1},A_{2},\ldots,A_{k}bekksubsets ofSSsuch that each element ofSSlies in exactly one of thesekksubsets. Prove that the following statements are equivalent:

•Statement 1:There exists a bijectionσ:S→S\sigma:S\to Ssuch that eachi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}satisfiesσ⁡(Ai)∩Ai=∅\sigma\left(A_{i}\right)\cap A_{i}=\varnothing.
•Statement 2:Eachi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}satisfies|Ai|≤|S|/2\left|A_{i}\right|\leq\left|S\right|/2.

[Solution:This is Exercise 5 on homework set #4 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 8.74.

LetSSbe a finite set. Letk∈ℕk\in\mathbb{N}be such that|S|≥2k+1\left|S\right|\geq 2k+1. Prove that there exists an injective mapf:𝒫k(S)→𝒫k+1(S)f:\mathcal{P}_{k}\left(S\right)\rightarrow\mathcal{P}_{k+1}\left(S\right)such that eachX∈𝒫k(S)X\in\mathcal{P}_{k}\left({S}\right)satisfiesf⁡(X)⊇Xf\left(X\right)\supseteq X.

(In other words, prove that we can add to eachkk-element subsetXXofSSan additional element fromS∖XS\setminus Xsuch that the resulting(k+1)\left(k+1\right)-element subsets are distinct.)

[Example:ForS={1,2,3,4,5}S=\left\{1,2,3,4,5\right\}andk=2k=2, we can (for instance) have the mapffsend

{1,2}\displaystyle\left\{1,2\right\}↦{1,2,4},{1,3}↦{1,3,4},{1,4}↦{1,4,5},\displaystyle\mapsto\left\{1,2,4\right\},\ \ \ \ \ \ \ \ \ \ \left\{1,3\right\}\mapsto\left\{1,3,4\right\},\ \ \ \ \ \ \ \ \ \ \left\{1,4\right\}\mapsto\left\{1,4,5\right\},{1,5}\displaystyle\left\{1,5\right\}↦{1,3,5},{2,3}↦{1,2,3},{2,4}↦{2,4,5},\displaystyle\mapsto\left\{1,3,5\right\},\ \ \ \ \ \ \ \ \ \ \left\{2,3\right\}\mapsto\left\{1,2,3\right\},\ \ \ \ \ \ \ \ \ \ \left\{2,4\right\}\mapsto\left\{2,4,5\right\},{2,5}\displaystyle\left\{2,5\right\}↦{1,2,5},{3,4}↦{2,3,4},{3,5}↦{2,3,5},\displaystyle\mapsto\left\{1,2,5\right\},\ \ \ \ \ \ \ \ \ \ \left\{3,4\right\}\mapsto\left\{2,3,4\right\},\ \ \ \ \ \ \ \ \ \ \left\{3,5\right\}\mapsto\left\{2,3,5\right\},{4,5}\displaystyle\left\{4,5\right\}↦{3,4,5}.\displaystyle\mapsto\left\{3,4,5\right\}.Do you see any pattern behind these values? (I don’t).]

[Hint:First, reduce the problem to the case when|S|=2k+1\left|S\right|=2k+1. Then, in that case, restate it as a claim about matchings in a certain bipartite graph.]

[Solution:This is Exercise 4 on homework set #4 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 8.75.

A subsetAAofXXwill be calledneighbor-criticalif|N⁡(A)|=|A|\left|N\left(A\right)\right|=\left|A\right|.

LetAAandBBbe two neighbor-critical subsets ofXX. Prove that the subsetsA∪BA\cup BandA∩BA\cap Bare also neighbor-critical.

[Solution:This is Exercise 6 on homework set #4 from my Spring 2017 course; seethe course pagefor solutions.]

For more about applications, restatements and variants of Hall’s marriage theorem, see the recent survey[Camero25].

8.10Further exercises on matchings

Exercise 8.76.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph. LetMMbe a matching ofGG.

Anaugmenting pathforMMshall mean a path(v0,e1,v1,e2,v2,…,ek,vk)\left(v_{0},e_{1},v_{1},e_{2},v_{2},\ldots,e_{k},v_{k}\right)ofGGsuch thatkkis odd (note thatk=1k=1is allowed) and such that

•the even-indexed edgese2,e4,…,ek−1e_{2},e_{4},\ldots,e_{k-1}belong toMM(note that this condition is vacuously true ifk=1k=1);
•the odd-indexed edgese1,e3,…,eke_{1},e_{3},\ldots,e_{k}belong toE∖ME\setminus M;
•neither the starting pointv0v_{0}nor the ending pointvkv_{k}is matched inMM.

Prove thatMMhas maximum size among all matchings ofGGif and only if there exists no augmenting path forMM.

[Hint:IfMMandM′M^{\prime}are two matchings ofGG, what can you say about the symmetric difference(M∪M′)∖(M∩M′)\left(M\cup M^{\prime}\right)\setminus\left(M\cap M^{\prime}\right)?]

Exercise 8.77.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. LetAAbe a subset ofXX, and letBBbe a subset ofYY. Assume thatGGhas anAA-complete matching, and thatGGhas aBB-complete matching. Prove thatGGhas anA∪BA\cup B-complete matching.

Exercise 8.78.

Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph withX≠∅X\neq\varnothing. Assume thatGGhas anXX-complete matching.

An edgeeeofGGwill be calleduselessifGGhas noXX-complete matching that containsee.

Prove that there exists a vertexx∈Xx\in Xsuch that no edge that containsxxis useless.

Exercise 8.79.

•Statement 1:The verticesxxandf⁡(x)f\left(x\right)ofGGare adjacent.
•Statement 2:There exists nox′∈Xx^{\prime}\in Xsuch that the verticesxxandf⁡(x′)f\left(x^{\prime}\right)ofGGare adjacent.

[Remark:In vaguely matrimonial terminology, this is saying that in a group ofmmmen andwwwomen satisfyingw≥2m−1w\geq 2m-1, we can always marry each man (monogamously) to a woman in such a way that either he likes his partner or all women he likes are unmarried.]

[Solution:This is Exercise 3 on midterm #3 from my Spring 2017 course; seethe course pagefor solutions.]

Exercise 8.80.

Let(G,X,Y)\left(G,X,Y\right)and(H,U,V)\left(H,U,V\right)be bipartite graphs.

Assume thatGGis a simple graph and has anXX-complete matching.

Assume thatHHis a simple graph and has aUU-complete matching.

Consider the Cartesian productG×HG\times HofGGandHHdefined in Definition2.144. (Note that we requiredGGandHHto be simple graphs only to avoid having to defineG×HG\times Hfor multigraphs.)

(a)Show that(G×H,(X×V)∪(Y×U),(X×U)∪(Y×V))\left(G\times H,\ \left(X\times V\right)\cup\left(Y\times U\right),\ \left(X\times U\right)\cup\left(Y\times V\right)\right)is a bipartite graph.
(b)Prove that the graphG×HG\times Hhas an(X×V)∪(Y×U)\left(X\times V\right)\cup\left(Y\times U\right)-complete matching.

[Solution:This is Exercise 3 on homework set #4 from my Spring 2017 course; seethe course pagefor solutions.]

9Networks and flows

In this chapter, I will give an introduction tonetwork flowsand their optimization. This is a topic of great interest to logisticians, as even the simplest results have obvious applications to scheduling trains and trucks. It also has lots of purely mathematical consequences; in particular, we will use network flows to finally prove the Hall–König matching theorem (and thus the HMT, König’s theorem, and their many consequences).

I will follow my notes[17s-lec16], which are a good place to look up the details of some proofs that I will only sketch. That said, I will be using multidigraphs instead of simple digraphs, so some adaptations will be necessary (since[17s-lec16]only works with simple digraphs). These adaptations are generally easy.

I will only cover the very basics of network flow optimization, leading to a proof of the max-flow-min-cut theorem (for integer-valued flows) and to a proof of the Hall–König matching theorem. For the deeper reaches of the theory, see[ForFul74](a classical textbook written by the inventors of the subject),[Schrij17, Chapter 4]and[Schrij03, Part I].

9.1Definitions

9.1.1Networks

Recall that we use the notationℕ={0,1,2,…}\mathbb{N}=\left\{0,1,2,\ldots\right\}.

Definition 9.1.

Anetworkconsists of

•a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right);
•two distinct verticess∈Vs\in Vandt∈Vt\in V, called thesourceand thesink, respectively;
•a functionc:A→ℕc:A\rightarrow\mathbb{N}, called thecapacity function.

Example 9.2.

Here is an example of a network:

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9.1.2The notationsS¯\overline{S},[P,Q]\left[P,Q\right]andd⁡(P,Q)d\left(P,Q\right)

Definition 9.5.

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}. Then:

(a)For any arca∈Aa\in A, we call the numberc⁡(a)∈ℕc\left(a\right)\in\mathbb{N}thecapacityof the arcaa.
(b)For any subsetSSofVV, we letS¯\overline{S}denote the subsetV∖SV\setminus SofVV.
(c)IfPPandQQare two subsets ofVV, then[P,Q]\left[P,Q\right]shall mean the set of all arcs ofDDwhose source belongs toPPand whose target belongs toQQ. That is, [P,Q]:={a∈A∣ψ⁡(a)∈P×Q}.\left[P,Q\right]:=\left\{a\in A\ \mid\ \psi\left(a\right)\in P\times Q\right\}.
(d)IfPPandQQare two subsets ofVV, and ifd:A→ℕd:A\rightarrow\mathbb{N}is any function, then the numberd⁡(P,Q)∈ℕd\left(P,Q\right)\in\mathbb{N}is defined by d⁡(P,Q):=∑a∈[P,Q]d⁡(a).d\left(P,Q\right):=\sum\limits_{a\in\left[P,Q\right]}d\left(a\right).(In particular, we can apply this tod=cd=c, and then getc⁡(P,Q)=∑a∈[P,Q]c⁡(a)c\left(P,Q\right)=\sum\limits\limits_{a\in\left[P,Q\right]}c\left(a\right).)

Example 9.6.

Let us again consider the network from Example9.2. For the subset{s,u}\left\{s,u\right\}ofVV, we have{s,u}¯={p,v,q,t}\overline{\left\{s,u\right\}}=\left\{p,v,q,t\right\}and

[{s,u},{s,u}¯]={sp,uv,uq}\left[\left\{s,u\right\},\overline{\left\{s,u\right\}}\right]=\left\{sp,\ uv,\ uq\right\}(recall that ourDDis a simple digraph, so an arc is just a pair of two vertices) and

c⁡({s,u},{s,u}¯)\displaystyle c\left(\left\{s,u\right\},\overline{\left\{s,u\right\}}\right)=∑a∈[{s,u},{s,u}¯]c⁡(a)=c⁡(sp)⏟=3+c⁡(uv)⏟=1+c⁡(uq)⏟=1\displaystyle=\sum\limits_{a\in\left[\left\{s,u\right\},\overline{\left\{s,u\right\}}\right]}c\left(a\right)=\underbrace{c\left(sp\right)}_{=3}+\underbrace{c\left(uv\right)}_{=1}+\underbrace{c\left(uq\right)}_{=1}=3+1+1=5.\displaystyle=3+1+1=5.We can make this visually clearer if we draw a “border” between the sets{s,u}\left\{s,u\right\}and{s,u}¯\overline{\left\{s,u\right\}}:

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arcs that cross this border from{s,u}\left\{s,u\right\}to{s,u}¯\overline{\left\{s,u\right\}}. (Of course, this visualization works only for sums of the formd⁡(P,P¯)d\left(P,\overline{P}\right), not for the more general case ofd⁡(P,Q)d\left(P,Q\right)wherePPandQQcan have elements in common. But thed⁡(P,P¯)d\left(P,\overline{P}\right)are the most useful sums.)

Exercise 9.7.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a balanced multidigraph. For any subsetSSofVV, we setS¯:=V∖S\overline{S}:=V\setminus S. For any two subsetsSSandTTofVV, we set

[S,T]\displaystyle\left[S,T\right]:={a∈A∣the source ofabelongs toS,\displaystyle:=\left\{a\in A\ \mid\ \text{the source of }a\text{ belongs to }S\text{,}\right.and the target ofabelongs toT}.\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{and the target of }a\text{ belongs to }T\right\}.Prove that|[S,S¯]|=|[S¯,S]|\left|\left[S,\overline{S}\right]\right|=\left|\left[\overline{S},S\right]\right|for any subsetSSofVV.

9.1.3Flows

Let us now define flows on a network:

Definition 9.8.

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}.

Aflow(on the networkNN) means a functionf:A→ℕf:A\rightarrow\mathbb{N}with the following properties:

•We have0≤f⁡(a)≤c⁡(a)0\leq f\left(a\right)\leq c\left(a\right)for each arca∈Aa\in A. This condition is called thecapacity constraints(we are using the plural form, since there is one constraint for each arca∈Aa\in A).
•For any vertexv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}, we have f−(v)=f+(v),f^{-}\left(v\right)=f^{+}\left(v\right),where we set f−(v):=∑a∈Ais an arcwith targetvf(a)andf+(v):=∑a∈Ais an arcwith sourcevf(a).f^{-}\left(v\right):=\sum\limits_{\begin{subarray}{c}a\in A\text{ is an arc}\\ \text{with target }v\end{subarray}}f\left(a\right)\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ f^{+}\left(v\right):=\sum\limits_{\begin{subarray}{c}a\in A\text{ is an arc}\\ \text{with source }v\end{subarray}}f\left(a\right).This is called theconservation constraints.

Iff:A→ℕf:A\rightarrow\mathbb{N}is a flow anda∈Aa\in Ais an arc, then the nonnegative integerf⁡(a)f\left(a\right)will be called thearc flowofffonaa.

Example 9.9.

To draw a flowffon a networkNN, we draw the networkNN, with one little tweak: Instead of writing the capacityc⁡(a)c\left(a\right)atop each arca∈Aa\in A, we write “f⁡(a)f\left(a\right)ofc⁡(a)c\left(a\right)” atop each arca∈Aa\in A. For example, here is a flowffon the networkNNfrom Example9.2:

ssuuppvvqqtt2of22\text{ of }21of31\text{ of }31of31\text{ of }31of11\text{ of }11of11\text{ of }10of10\text{ of }12of22\text{ of }21of21\text{ of }2(so, for example,f⁡(su)=2f\left(su\right)=2,f⁡(pq)=1f\left(pq\right)=1andf⁡(qv)=0f\left(qv\right)=0).

For another example, here is a different flowggon the same networkNN:

ssuuppvvqqtt1of21\text{ of }20of30\text{ of }30of30\text{ of }30of10\text{ of }11of11\text{ of }11of11\text{ of }10of20\text{ of }21of21\text{ of }2

There are several intuitive ways to think of a networkNNand of a flow on it:

•We can visualizeNNas a collection of one-way roads: Each arca∈Aa\in Ais a one-way road, and its capacityc⁡(a)c\left(a\right)is how much traffic it can (maximally) handle per hour. A flowffonNNcan then be understood as traffic flowing through these roads, wheref⁡(a)f\left(a\right)is the amount of traffic that travels through the arcaain an hour. The conservation constraints say that the traffic out of a given vertexvvequals the traffic intovvunlessvvis one ofssandtt. (We imagine that traffic can arbitrarily materialize or dematerialize atssandtt.)
•We can visualizeNNas a collection of pipes: Each arca∈Aa\in Ais a pipe, and its capacityc⁡(a)c\left(a\right)is how much water it can maximally transport in a second. A flowffonNNcan then be viewed as water flowing through the pipes, wheref⁡(a)f\left(a\right)is the amount of water traveling through a pipeaain a second. The capacity constraints say that no pipe is over its capacity or carries a negative amount of water. The conservation constraints say that at every vertexvvother thanssandtt, the amount of water coming in (that is,f−(v)f^{-}\left(v\right)) equals the amount of water moving out (that is,f+(v)f^{+}\left(v\right)); that is, there are no leaks and no water being injected into the system other than atssandtt. This is whyssis called the “source” andttis the “sink”. A slightly counterintuitive aspect of this visualization is that each pipe has a direction, and water can only flow in that one direction (from source to target). That said, you can always model an undirected pipe by having two pipes of opposite directions.
•We can regardNNas a money transfer scheme: Each vertexv∈Vv\in Vis a bank account, and the goal is to transfer some money fromsstott. All other verticesvvact as middlemen. Each arca∈Aa\in Acorresponds to a possibility of transfer from its source to its target; the maximum amount that can be transferred on this arc isc⁡(a)c\left(a\right). A flow describes a way in which money is transferred such that each middleman vertexv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}ends up receiving exactly as much money as it gives away.

Needless to say, these visualizations have been chosen for their intuitive grasp; the real-life applications of network flows are somewhat different.

9.1.4Inflow, outflow and value of a flow

Next, we define certain numbers related to any flow on a network:

Definition 9.11.

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}. Letf:A→ℕf:A\rightarrow\mathbb{N}be an arbitrary map (e.g., a flow onNN). Then:

(a)For each vertexv∈Vv\in V, we set f−(v):=∑a∈Ais an arcwith targetvf(a)andf+(v):=∑a∈Ais an arcwith sourcevf(a).f^{-}\left(v\right):=\sum\limits_{\begin{subarray}{c}a\in A\text{ is an arc}\\ \text{with target }v\end{subarray}}f\left(a\right)\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ f^{+}\left(v\right):=\sum\limits_{\begin{subarray}{c}a\in A\text{ is an arc}\\ \text{with source }v\end{subarray}}f\left(a\right).We callf−(v)f^{-}\left(v\right)theinflowofffintovv, and we callf+(v)f^{+}\left(v\right)theoutflowoffffromvv. Note that ifffis a flow, then each vertexv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}satisfiesf−(v)=f+(v)f^{-}\left(v\right)=f^{+}\left(v\right); this is the conservation constraints.
(b)We define thevalueof the mapffto be the numberf+(s)−f−(s)f^{+}\left(s\right)-f^{-}\left(s\right). This value is denoted by|f|\left|f\right|.

Example 9.12.

The flowffin Example9.9satisfies

f+(s)\displaystyle f^{+}\left(s\right)=3,f−(s)=0,\displaystyle=3,\ \ \ \ \ \ \ \ \ \ f^{-}\left(s\right)=0,f+(u)\displaystyle f^{+}\left(u\right)=f−(u)=2,\displaystyle=f^{-}\left(u\right)=2,f+(p)\displaystyle f^{+}\left(p\right)=f−(p)=1,\displaystyle=f^{-}\left(p\right)=1,f+(v)\displaystyle f^{+}\left(v\right)=f−(v)=1,\displaystyle=f^{-}\left(v\right)=1,f+(q)\displaystyle f^{+}\left(q\right)=f−(q)=2,,\displaystyle=f^{-}\left(q\right)=2,,f+(t)\displaystyle f^{+}\left(t\right)=0,f−(t)=3\displaystyle=0,\ \ \ \ \ \ \ \ \ \ f^{-}\left(t\right)=3and has value|f|=3\left|f\right|=3. The flowggin Example9.9has value|g|=1\left|g\right|=1. More generally, the flowf𝐩f_{\mathbf{p}}in Remark9.10always has value|f𝐩|=1\left|f_{\mathbf{p}}\right|=1.

Example 9.13.

For any networkNN, we can define thezero flowonNN. This is the flow0A:A→ℕ0_{A}:A\rightarrow\mathbb{N}that sends each arca∈Aa\in Ato00. This flow has value|0A|=0\left|0_{A}\right|=0.

9.2The maximum flow problem and bipartite graphs

Now we can state an important optimization problem, known as themaximum flow problem: Given a networkNN, how can we find a flow of maximum possible value?

Example 9.14.

Finding a maximum matching in a bipartite graph is a particular case of the maximum flow problem.

Indeed, let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. Then, we can transform this graph into a networkNNas follows:

•Add two new verticesssandtt.
•Turn each edgeeeofGGinto an arce→\overrightarrow{e}whose source is theXX-endpoint ofee(that is, the endpoint ofeethat belongs toXX) and whose target is theYY-endpoint ofee(that is, the endpoint ofeethat belongs toYY).
•Add an arc fromssto each vertex inXX.
•Add an arc from each vertex inYYtott.
•Assign to each arc the capacity11.

Here is an example of a bipartite graph(G,X,Y)\left(G,X,Y\right)(as usual, drawn with theXX-vertices on the left and with theYY-vertices on the right) and the corresponding networkNN:

112233445566112233445566ssttGGNN(we are not showing the capacities of the arcs, since they are all equal to11).

The flows of this networkNNare in bijection with the matchings ofGG. Namely, ifffis a flow onNN, then the set

{e∈E(G)∣f⁡(e→)=1}\left\{e\in\operatorname*{E}\left(G\right)\ \mid\ f\left(\overrightarrow{e}\right)=1\right\}is a matching ofGG. Conversely, ifMMis a matching ofGG, then we obtain a flowffonNNby assigning the arc flow11to all arcs of the forme→\overrightarrow{e}wheree∈Me\in M, as well as assigning the arc flow11to every new arc that joinsssorttto a vertex matched inMM. All other arcs are assigned the arc flow00. For instance, in our above example, the matching{15,36}\left\{15,\ 36\right\}corresponds to the following flow:

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\lxSVG@transformcm{0.7071}{0.7071}{-0.7071}{0.7071}{107.02095pt}{-6.79008pt}\lxSVG@begingroup@{transform=matrix(0.7071 0.7071 -0.7071 0.7071 148.08 -9.4)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -5.53 7.32 C -4.52 2.93 -2.27 0.85 0 0 C -2.27 -0.85 -4.52 -2.93 -5.53 -7.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,where we are using the convention that an arcaawithf⁡(a)=0f\left(a\right)=0is drawn dashed whereas an arcaawithf⁡(a)=1f\left(a\right)=1is drawn boldfaced (thankfully, the only possibilities forf⁡(a)f\left(a\right)are00and11, because all capacities are11).

(See[17s-lec16, Proposition 1.36 till Proposition 1.40]for details and proofs; that said, the proofs are straightforward and you will probably “see” them just by starting at an example.)

9.3Basic properties of flows

Before we approach the maximum flow problem, let us prove some simple observations about flows:

Proposition 9.15.

|f|\displaystyle\left|f\right|=f+(s)−f−(s)\displaystyle=f^{+}\left(s\right)-f^{-}\left(s\right)=f−(t)−f+(t).\displaystyle=f^{-}\left(t\right)-f^{+}\left(t\right).

Proof 9.16.

Eachv∈Vv\in Vsatisfiesf+(v)=∑a∈Ais an arcwith sourcevf(a)f^{+}\left(v\right)=\sum\limits_{\begin{subarray}{c}a\in A\text{ is an arc}\\ \text{with source }v\end{subarray}}f\left(a\right)(by the definition off+(v)f^{+}\left(v\right)). Summing this equality over allv∈Vv\in V, we obtain

∑v∈Vf+(v)=∑v∈V∑a∈Ais an arcwith sourcev⏟=∑a∈Af(a)=∑a∈Af(a)\sum\limits\limits_{v\in V}f^{+}\left(v\right)=\underbrace{\sum\limits\limits_{v\in V}\ \ \sum\limits_{\begin{subarray}{c}a\in A\text{ is an arc}\\ \text{with source }v\end{subarray}}}_{=\sum\limits_{a\in A}}f\left(a\right)=\sum\limits_{a\in A}f\left(a\right)(note that this is a generalization of the familiar fact that∑v∈Vdeg+⁡v=|A|\sum\limits\limits_{v\in V}\deg^{+}v=\left|A\right|). Similarly,∑v∈Vf−(v)=∑a∈Af⁡(a)\sum\limits\limits_{v\in V}f^{-}\left(v\right)=\sum\limits\limits_{a\in A}f\left(a\right). Hence,

∑v∈V(f−(v)−f+(v))=∑v∈Vf−(v)⏟=∑a∈Af⁡(a)−∑v∈Vf+(v)⏟=∑a∈Af⁡(a)=0.\sum\limits\limits_{v\in V}\left(f^{-}\left(v\right)-f^{+}\left(v\right)\right)=\underbrace{\sum\limits\limits_{v\in V}f^{-}\left(v\right)}_{=\sum\limits\limits_{a\in A}f\left(a\right)}-\underbrace{\sum\limits_{v\in V}f^{+}\left(v\right)}_{=\sum\limits\limits_{a\in A}f\left(a\right)}=0.(59) However, by the conservation constraints, we havef−(v)=f+(v)f^{-}\left(v\right)=f^{+}\left(v\right)for eachv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}. In other words,f−(v)−f+(v)=0f^{-}\left(v\right)-f^{+}\left(v\right)=0for eachv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}. Thus, in the sum∑v∈V(f−(v)−f+(v))\sum\limits\limits_{v\in V}\left(f^{-}\left(v\right)-f^{+}\left(v\right)\right), all addends are00except for the addends forv=sv=sand forv=tv=t. Hence, the sum boils down to these two addends:

∑v∈V(f−(v)−f+(v))=(f−(s)−f+(s))+(f−(t)−f+(t)).\sum\limits\limits_{v\in V}\left(f^{-}\left(v\right)-f^{+}\left(v\right)\right)=\left(f^{-}\left(s\right)-f^{+}\left(s\right)\right)+\left(f^{-}\left(t\right)-f^{+}\left(t\right)\right).Comparing this with (59), we obtain

(f−(s)−f+(s))+(f−(t)−f+(t))=0,\left(f^{-}\left(s\right)-f^{+}\left(s\right)\right)+\left(f^{-}\left(t\right)-f^{+}\left(t\right)\right)=0,so that

f−(t)−f+(t)=−(f−(s)−f+(s))=f+(s)−f−(s)=|f|f^{-}\left(t\right)-f^{+}\left(t\right)=-\left(f^{-}\left(s\right)-f^{+}\left(s\right)\right)=f^{+}\left(s\right)-f^{-}\left(s\right)=\left|f\right|(by the definition of|f|\left|f\right|). This proves Proposition9.15.

Proposition 9.17.

(a)We have f⁡(S,S¯)−f⁡(S¯,S)=∑v∈S(f+(v)−f−(v)).f\left(S,\overline{S}\right)-f\left(\overline{S},S\right)=\sum\limits_{v\in S}\left(f^{+}\left(v\right)-f^{-}\left(v\right)\right).(Recall that we are using Definition9.5here, so thatf⁡(P,Q)f\left(P,Q\right)means∑a∈[P,Q]f⁡(a)\sum\limits_{a\in\left[P,Q\right]}f\left(a\right).)
(b)Assume thats∈Ss\in Sandt∉St\notin S. Then, |f|=f⁡(S,S¯)−f⁡(S¯,S).\left|f\right|=f\left(S,\overline{S}\right)-f\left(\overline{S},S\right).
(c)Assume thats∈Ss\in Sandt∉St\notin S. Then, |f|≤c⁡(S,S¯).\left|f\right|\leq c\left(S,\overline{S}\right).
(d)Assume thats∈Ss\in Sandt∉St\notin S. Then,|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right)if and only if (f⁡(a)=0for alla∈[S¯,S])\left(f\left(a\right)=0\text{ for all }a\in\left[\overline{S},S\right]\right)and (f⁡(a)=c⁡(a)for alla∈[S,S¯]).\left(f\left(a\right)=c\left(a\right)\text{ for all }a\in\left[S,\overline{S}\right]\right).

Proof 9.18.

Let me first make these claims intuitive in terms of the “money transfer scheme” model for our network. ConsiderSSas a country. Then,f⁡(S,S¯)f\left(S,\overline{S}\right)is the “export” from this countrySS(that is, the total wealth exported fromSS), whereasf⁡(S¯,S)f\left(\overline{S},S\right)is the “import” into this countrySS(that is, the total wealth imported intoSS). Thus, part(a)of the proposition is saying that the “net export” ofSS(that is, the export fromSSminus the import intoSS) can be computed by summing the “outflow minus inflow” values of all accounts inSS. This should match the intuition for exports and imports (in particularly, any transfers that happen withinSSshould cancel out when we sum the “outflow minus inflow” values of all accounts inSS). Part(b)says that if the countrySScontains the sourcessbut not the sinktt(that is, the goal of the network is to transfer money out of the country), then the total value transferred is actually the net export ofSS. Part(c)claims that this total value is no larger than the total “export capacity”c⁡(S,S¯)c\left(S,\overline{S}\right)(that is, the total capacity of the “export arcs”a∈[S,S¯]a\in\left[S,\overline{S}\right]). Part(d)says that if equality holds in this inequality (i.e., if the total value equals the total export capacity), then each “import arc”a∈[S¯,S]a\in\left[\overline{S},S\right]is unused (i.e., nothing is imported intoSS), whereas each “export arc”a∈[S,S¯]a\in\left[S,\overline{S}\right]is used to its full capacity.

I hope this demystifies all claims of the proposition. But for the sake of completeness, here are rigorous proofs (though rather terse ones, since I assume you have seen enough manipulations of sum to fill in the details):

(a)This follows from

∑v∈S(f+(v)−f−(v))\displaystyle\sum\limits_{v\in S}\left(f^{+}\left(v\right)-f^{-}\left(v\right)\right)=∑v∈Sf+(v)⏟=f⁡(S,V)(why?)−∑v∈Sf−(v)⏟=f⁡(V,S)(why?)\displaystyle=\underbrace{\sum\limits_{v\in S}f^{+}\left(v\right)}_{\begin{subarray}{c}=f\left(S,V\right)\\ \text{(why?)}\end{subarray}}-\underbrace{\sum\limits_{v\in S}f^{-}\left(v\right)}_{\begin{subarray}{c}=f\left(V,S\right)\\ \text{(why?)}\end{subarray}}=f⁡(S,V)⏟=f⁡(S,S)+f⁡(S,S¯)(sinceVis the union of thetwo disjoint setsSandS¯)−f⁡(V,S)⏟=f⁡(S,S)+f⁡(S¯,S)(sinceVis the union of thetwo disjoint setsSandS¯)\displaystyle=\underbrace{f\left(S,V\right)}_{\begin{subarray}{c}=f\left(S,S\right)+f\left(S,\overline{S}\right)\\ \text{(since }V\text{ is the union of the}\\ \text{two disjoint sets }S\text{ and }\overline{S}\text{)}\end{subarray}}-\underbrace{f\left(V,S\right)}_{\begin{subarray}{c}=f\left(S,S\right)+f\left(\overline{S},S\right)\\ \text{(since }V\text{ is the union of the}\\ \text{two disjoint sets }S\text{ and }\overline{S}\text{)}\end{subarray}}=f⁡(S,S)+f⁡(S,S¯)−(f⁡(S,S)+f⁡(S¯,S))=f⁡(S,S¯)−f⁡(S¯,S).\displaystyle=f\left(S,S\right)+f\left(S,\overline{S}\right)-\left(f\left(S,S\right)+f\left(\overline{S},S\right)\right)=f\left(S,\overline{S}\right)-f\left(\overline{S},S\right). (b)We haveS∖{s}⊆V∖{s,t}S\setminus\left\{s\right\}\subseteq V\setminus\left\{s,t\right\}(sincet∉St\notin S). From part(a), we obtain

f⁡(S,S¯)−f⁡(S¯,S)\displaystyle f\left(S,\overline{S}\right)-f\left(\overline{S},S\right)=∑v∈S(f+(v)−f−(v))\displaystyle=\sum\limits_{v\in S}\left(f^{+}\left(v\right)-f^{-}\left(v\right)\right)=(f+(s)−f−(s))⏟=|f|(by the definition of|f|)+∑v∈S∖{s}(f+(v)−f−(v))⏟=0(by the conservation constraints,sincev∈S∖{s}⊆V∖{s,t})(sinces∈S)\displaystyle=\underbrace{\left(f^{+}\left(s\right)-f^{-}\left(s\right)\right)}_{\begin{subarray}{c}=\left|f\right|\\ \text{(by the definition of }\left|f\right|\text{)}\end{subarray}}+\sum\limits_{v\in S\setminus\left\{s\right\}}\underbrace{\left(f^{+}\left(v\right)-f^{-}\left(v\right)\right)}_{\begin{subarray}{c}=0\\ \text{(by the conservation constraints,}\\ \text{since }v\in S\setminus\left\{s\right\}\subseteq V\setminus\left\{s,t\right\}\text{)}\end{subarray}}\ \ \ \ \ \ \ \ \ \ \left(\text{since }s\in S\right)=|f|.\displaystyle=\left|f\right|.This proves part(b).

(c)The capacity constraints yield thatf⁡(a)≤c⁡(a)f\left(a\right)\leq c\left(a\right)for each arca∈Aa\in A. Summing up these inequalities over alla∈[S,S¯]a\in\left[S,\overline{S}\right], we obtainf⁡(S,S¯)≤c⁡(S,S¯)f\left(S,\overline{S}\right)\leq c\left(S,\overline{S}\right). The capacity constraints furthermore yield thatf⁡(a)≥0f\left(a\right)\geq 0for each arca∈Aa\in A. Summing up these inequalities over alla∈[S¯,S]a\in\left[\overline{S},S\right], we obtainf⁡(S¯,S)≥0f\left(\overline{S},S\right)\geq 0. Hence, part(b)yields

|f|=f⁡(S,S¯)⏟≤c⁡(S,S¯)−f⁡(S¯,S)⏟≥0≤c⁡(S,S¯).\left|f\right|=\underbrace{f\left(S,\overline{S}\right)}_{\leq c\left(S,\overline{S}\right)}-\underbrace{f\left(\overline{S},S\right)}_{\geq 0}\leq c\left(S,\overline{S}\right).This proves part(c).

(d)We must characterize the equality case in part(c). However, recall the proof of part(c): We obtained the inequality|f|≤c⁡(S,S¯)\left|f\right|\leq c\left(S,\overline{S}\right)by summing up the inequalitiesf⁡(a)≤c⁡(a)f\left(a\right)\leq c\left(a\right)over all arcsa∈[S,S¯]a\in\left[S,\overline{S}\right]and subtracting the sum of the inequalitiesf⁡(a)≥0f\left(a\right)\geq 0over all arcsa∈[S¯,S]a\in\left[\overline{S},S\right]. Hence, in order for the inequality|f|≤c⁡(S,S¯)\left|f\right|\leq c\left(S,\overline{S}\right)to become an equality, it is necessary and sufficient that all the inequalities involved – i.e., the inequalitiesf⁡(a)≤c⁡(a)f\left(a\right)\leq c\left(a\right)for all arcsa∈[S,S¯]a\in\left[S,\overline{S}\right]as well as the inequalitiesf⁡(a)≥0f\left(a\right)\geq 0for all arcsa∈[S¯,S]a\in\left[\overline{S},S\right]– become equalities. In other words, it is necessary and sufficient that we have

(f⁡(a)=0for alla∈[S¯,S])\left(f\left(a\right)=0\text{ for all }a\in\left[\overline{S},S\right]\right)and

(f⁡(a)=c⁡(a)for alla∈[S,S¯]).\left(f\left(a\right)=c\left(a\right)\text{ for all }a\in\left[S,\overline{S}\right]\right).This proves Proposition9.17(d).

9.4The max-flow-min-cut theorem

9.4.1Cuts and their capacities

One more definition, before we get to the hero of this story:

Definition 9.19.

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}. Then:

(a)AcutofNNshall mean a subset ofAAthat has the form[S,S¯]\left[S,\overline{S}\right], whereSSis a subset ofVVsatisfyings∈Ss\in Sandt∉St\notin S.
(b)Thecapacityof a cut[S,S¯]\left[S,\overline{S}\right]is defined to be the numberc⁡(S,S¯)=∑a∈[S,S¯]c⁡(a)c\left(S,\overline{S}\right)=\sum\limits\limits_{a\in\left[S,\overline{S}\right]}c\left(a\right).

Example 9.20.

Let us again consider the network from Example9.2. Then,[{s,u},{s,u}¯]={sp,uv,uq}\left[\left\{s,u\right\},\overline{\left\{s,u\right\}}\right]=\left\{sp,\ uv,\ uq\right\}is a cut of this network, and its capacity isc⁡({s,u},{s,u}¯)=5c\left(\left\{s,u\right\},\overline{\left\{s,u\right\}}\right)=5.

9.4.2The max-flow-min-cut theorem: statement

Now, Proposition9.17(c)says that the value of any flowffcan never be larger than the capacity of any cut[S,S¯]\left[S,\overline{S}\right]. Thus, in particular, the maximum value of a flow is≤\leqto the minimum capacity of a cut.

Furthermore, Proposition9.17(d)says that if this inequality is an equality – i.e., if the value of some flowffequals the capacity of some cut[S,S¯]\left[S,\overline{S}\right]–, then the flowffmust use each arc that crosses the cut in the right direction (fromSStoS¯\overline{S}) to its full capacity and must not use any of the arcs that cross the cut in the wrong direction (fromS¯\overline{S}toSS).

It turns out that this inequality actuallyisan equality for any maximum flow and any minimum cut:

Theorem 9.21(max-flow-min-cut theorem).

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}. Then,

max{|f|∣fis a flow}=min{c(S,S¯)∣S⊆V;s∈S;t∉S}.\max\left\{\left|f\right|\ \mid\ f\text{ is a flow}\right\}=\min\left\{c\left(S,\overline{S}\right)\ \mid\ S\subseteq V;\ s\in S;\ t\notin S\right\}.In other words, the maximum value of a flow equals the minimum capacity of a cut.

We shall soon sketch a proof of this theorem that doubles as a fairly efficient (polynomial-time) algorithm for finding both a maximum flow (i.e., a flow of maximum value) and a minimum cut (i.e., a cut of minimum capacity). The algorithm is known as theFord-Fulkerson algorithm, and is sufficiently fast to be useful in practice.

9.4.3How to augment a flow

The idea of this algorithm is to start by havingffbe the zero flow (i.e., the flow from Example9.13), and then gradually increase its value|f|\left|f\right|by making changes to some of its arc flowsf⁡(a)f\left(a\right).

Of course, we cannot unilaterally change the arc flowf⁡(a)f\left(a\right)on a single arc, since this will (usually) mess up the conservation constraints. Thus, if we changef⁡(a)f\left(a\right), then we will also have to changef⁡(b)f\left(b\right)for some other arcsb∈Ab\in Ato make the result a flow again. One way to do this is to increase all arc flowsf⁡(a)f\left(a\right)along some path fromsstott. Here is an example of such an increase:

Example 9.22.

Consider the flowfffrom Example9.9. We can increase the arc flowsf⁡(sp),f⁡(pq),f⁡(qv),f⁡(vt)f\left(sp\right),\ f\left(pq\right),\ f\left(qv\right),\ f\left(vt\right)offfon all the arcs of the path(s,p,q,v,t)\left(s,p,q,v,t\right)(since neither of these arcs is used to its full capacity). As a result, we obtain the following flowhh:

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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.89442}{-0.4472}{0.4472}{0.89442}{162.48363pt}{4.11539pt}\lxSVG@begingroup@{transform=matrix(0.89442 -0.4472 0.4472 0.89442 224.83 5.69)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{129.4578pt}{23.38075pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 179.13 32.35)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,whose value|h|\left|h\right|is44. It is easy to see that this is actually the maximum value of a flow on our network (since|h|=4\left|h\right|=4equals the capacityc⁡({t}¯,{t})c\left(\overline{\left\{t\right\}},\left\{t\right\}\right)of the cut[{t}¯,{t}]\left[\overline{\left\{t\right\}},\left\{t\right\}\right], but Proposition9.17(c)tells us that the value of any flow is≤\leqto the capacity of any cut).

However, simple increases like the one we just did are not always enough to find a maximum flow. They can leave us stuck at a “local maximum” – i.e., at a flow which does not have any more paths fromsstottthat can be used for any further increases (i.e., any path fromsstottcontains an arc that is already used to its full capacity), yet is not a maximum flow. Here is an example:

Example 9.23.

Consider the following network and flow:

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-1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{150.79999pt}{21.03108pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 208.66 29.1)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This flow is not maximum, but each path fromsstotthas at least one arc that is used to its full capacity. Thus, we cannot improve this flow by increasing all its arc flows on any given path fromsstott.

The trick to get past this hurdle is to use a “zig-zag path” – i.e., not a literal path, but rather a sequence(v0,a1,v1,a2,v2,…,ak,vk)\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)of vertices and arcs that can use arcs both in the forward and backward directions (i.e., anyi∈{1,2,…,k}i\in\left\{1,2,\ldots,k\right\}has to satisfy eitherψ⁡(ai)=(vi−1,vi)\psi\left(a_{i}\right)=\left(v_{i-1},v_{i}\right)orψ⁡(ai)=(vi,vi−1)\psi\left(a_{i}\right)=\left(v_{i},v_{i-1}\right)). Instead of increasing the flow on all arcs of this “path”, we do something slightly subtler: On the forward arcs, we increase the flow; on the backward arcs, we decrease it (all by the same amount). This, too, preserves the conservation constraints (think about why; we will soon see a rigorous proof), so it is a valid way of increasing the value of a flow. Here is an example:

Example 9.24.

Consider the flow in Example9.23. The underlying digraph has a “zig-zag path”(s,p,q,u,v,t)\left(s,p,q,u,v,t\right), which uses the arcuquqin the backward direction. We can decrease the arc flows offfon all forward arcsspsp,pqpq,uvuvandvtvtof this “zig-zag path”, and decrease it on the backward arcuquq. As a result, we obtain the flow

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-1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{150.79999pt}{21.03108pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 208.66 29.1)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This new flow has value22, and can easily be seen to be a maximum flow.

Good news: Allowing ourselves to use “zig-zag paths” like this (rather than literal paths only), we never get stuck at a non-maximum flow; we can always increase the value further and further until we eventually arrive at a maximum flow.

In order to prove this, we introduce some convenient notations. We prefer not to talk about “zig-zag paths”, but rather reinterpret these “zig-zag paths” as (literal) paths of an appropriately chosen digraph (notDD). This has the advantage of allowing us to use known properties of paths without having to first generalize them to “zig-zag paths”.

9.4.4The residual digraph

The appropriately chosen digraph is the so-calledresidual digraphof a flow; it is defined as follows:

Definition 9.25.

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}.

(a)For each arca∈Aa\in A, we introduce a new arca−1a^{-1}, which should act like a reversal of the arcaa(that is, its source should be the target ofaa, and its target should be the source ofaa). We don’t add these new arcsa−1a^{-1}to our digraphDD, but we keep them ready for use in a different digraph (which we will define below). Here is what this means in rigorous terms: For each arca∈Aa\in A, we introduce a new object, which we calla−1a^{-1}. We letA−1A^{-1}be the set of these new objectsa−1a^{-1}fora∈Aa\in A. We extend the mapψ:A→V×V\psi:A\rightarrow V\times Vto a mapψ^:A∪A−1→V×V\widehat{\psi}:A\cup A^{-1}\rightarrow V\times Vas follows: For eacha∈Aa\in A, we let ψ^(a)=(u,v)andψ^(a−1)=(v,u),\widehat{\psi}\left(a\right)=\left(u,v\right)\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \widehat{\psi}\left(a^{-1}\right)=\left(v,u\right),whereuuandvvare defined by(u,v)=ψ⁡(a)\left(u,v\right)=\psi\left(a\right). For each arca∈Aa\in A, we shall refer to the new arca−1a^{-1}as thereversalofaa, and conversely, we shall refer to the original arcaaas thereversalofa−1a^{-1}. We set(a−1)−1:=a\left(a^{-1}\right)^{-1}:=afor eacha∈Aa\in A. We shall refer to the arcsa∈Aa\in Aasforward arcs, and to their reversalsa−1a^{-1}asbackward arcs.
(b)Letf:A→ℕf:A\rightarrow\mathbb{N}be any flow onNN. We define theresidual digraphDfD_{f}of this flowffto be the multidigraph(V,Af,ψf)\left(V,A_{f},\psi_{f}\right), where Af={a∈A∣f(a)<c(a)}∪{a−1∣a∈A;f(a)>0}A_{f}=\left\{a\in A\ \mid\ f\left(a\right)<c\left(a\right)\right\}\cup\left\{a^{-1}\ \mid\ a\in A;\ f\left(a\right)>0\right\}andψf:=ψ^∣Af\psi_{f}:=\widehat{\psi}\mid_{A_{f}}. (This is usually not a subdigraph ofDD.) Thus, the residual digraphDfD_{f}has the same vertices asVV, but its arcs are those arcs ofDDthat are not used to their full capacity byffas well as the reversals of all arcs ofDDthat are used byff.

Example 9.26.

Letffbe the flowfffrom Example9.9. Then, the residual digraphDfD_{f}is

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Example 9.27.

Letffbe the non-maximum flow from Example9.23. Then, the residual digraphDfD_{f}is

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\hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This digraphDfD_{f}has a path fromsstott, which corresponds precisely to the “zig-zag path”(s,p,q,u,v,t)\left(s,p,q,u,v,t\right)we found in Example9.24.

You can think of the residual digraphDfD_{f}as follows: Each arc ofDfD_{f}corresponds to an opportunity to change an arc flowf⁡(a)f\left(a\right); namely, a forward arcaaofDfD_{f}means thatf⁡(a)f\left(a\right)can be increased, whereas a backward arca−1a^{-1}ofDfD_{f}means thatf⁡(a)f\left(a\right)can be decreased. Hence, the paths of the residual digraphDfD_{f}are the “zig-zag paths” ofDDthat allow the flowffto be increased (on forward arcs) or decreased (on backward arcs) as in Example9.24. Thus, usingDfD_{f}, we can avoid talking about “zig-zag paths”.

9.4.5The augmenting path lemma

The following crucial lemma tells us that such “zig-zag path increases” are valid (i.e., turn flows into flows), and are sufficient to find a maximum flow (i.e., if no more “zig-zag path increases” are possible, then our flow is already maximal):

Lemma 9.28(augmenting path lemma).

(a)If the digraphDfD_{f}has a path fromsstott, then the networkNNhas a flowf′f^{\prime}with a larger value thanff.
(b)If the digraphDfD_{f}has no path fromsstott, then the flowffhas maximum value (among all flows onNN), and there exists a subsetSSofVVsatisfyings∈Ss\in Sandt∉St\notin Sand|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right).

Proof 9.29.

(a)Assume that the digraphDfD_{f}has a path fromsstott. Pick such a path, and call it𝐩\mathbf{p}. Each arc of𝐩\mathbf{p}is an arc ofDfD_{f}.

For each forward arca∈Aa\in Athat appears in𝐩\mathbf{p}, we havef⁡(a)<c⁡(a)f\left(a\right)<c\left(a\right)(sinceaais an arc ofDfD_{f}), and thus we can increase the arc flowf⁡(a)f\left(a\right)by some positiveε∈ℕ\varepsilon\in\mathbb{N}(namely, by anyε≤c⁡(a)−f⁡(a)\varepsilon\leq c\left(a\right)-f\left(a\right)) without violating the capacity constraints.123123123Of course, such a unilateral increase will likely violate the conservation constraints.

For each backward arca−1∈A−1a^{-1}\in A^{-1}that appears in𝐩\mathbf{p}, we havef⁡(a)>0f\left(a\right)>0(sincea−1a^{-1}is an arc ofDfD_{f}), and thus we can decrease the arc flowf⁡(a)f\left(a\right)by some positiveε∈ℕ\varepsilon\in\mathbb{N}(namely, by anyε≤f⁡(a)\varepsilon\leq f\left(a\right)) without violating the capacity constraints.

Let now

ε\displaystyle\varepsilon:=min⁡({c⁡(a)−f⁡(a)∣a∈Ais a forward arc that appears in𝐩}CLOSE\displaystyle:=\min\Big(\left\{c\left(a\right)-f\left(a\right)\ \mid\ a\in A\text{ is a forward arc that appears in }\mathbf{p}\right\}OPEN∪{f⁡(a)∣a−1∈A−1is a backward arc that appears in𝐩}).\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cup\left\{f\left(a\right)\ \mid\ a^{-1}\in A^{-1}\text{ is a backward arc that appears in }\mathbf{p}\right\}\Big).Thisε\varepsilonis a positive integer (since it is a minimum of a set of positive integers124124124because•for each forward arca∈Aa\in Athat appears in𝐩\mathbf{p}, we havef⁡(a)<c⁡(a)f\left(a\right)<c\left(a\right)and thusc⁡(a)−f⁡(a)>0c\left(a\right)-f\left(a\right)>0;•for each backward arca−1∈A−1a^{-1}\in A^{-1}that appears in𝐩\mathbf{p}, we havef⁡(a)>0f\left(a\right)>0.). Letf′:A→ℕf^{\prime}:A\rightarrow\mathbb{N}be the map obtained fromffas follows:

•For each forward arca∈Aa\in Athat appears in𝐩\mathbf{p}, we increase the arc flowf⁡(a)f\left(a\right)byε\varepsilon(that is, we setf′(a):=f⁡(a)+εf^{\prime}\left(a\right):=f\left(a\right)+\varepsilon).
•For each backward arca−1∈A−1a^{-1}\in A^{-1}that appears in𝐩\mathbf{p}, we decrease the arc flowf⁡(a)f\left(a\right)byε\varepsilon(that is, we setf′(a):=f⁡(a)−εf^{\prime}\left(a\right):=f\left(a\right)-\varepsilon).
•For all other arcsaaofDD, we keep the arc flowf⁡(a)f\left(a\right)unchanged (i.e., we setf′(a):=f(a)f^{\prime}\left(a\right):=f\left(a\right)).

This new mapf′f^{\prime}still satisfies the capacity constraints125125125since the definition ofε\varepsilonshows that•for each forward arcaathat appears in𝐩\mathbf{p}, we haveε≤c⁡(a)−f⁡(a)\varepsilon\leq c\left(a\right)-f\left(a\right)and thusf⁡(a)+ε≤c⁡(a)f\left(a\right)+\varepsilon\leq c\left(a\right);•for each backward arca−1∈A−1a^{-1}\in A^{-1}that appears in𝐩\mathbf{p}, we haveε≤f⁡(a)\varepsilon\leq f\left(a\right)and thusf⁡(a)−ε≥0f\left(a\right)-\varepsilon\geq 0.. We claim that it also satisfies the conservation constraints. To check this, we have to verify that(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right)for each vertexv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}. So let us do this.

Letv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}be a vertex. We know thatf−(v)=f+(v)f^{-}\left(v\right)=f^{+}\left(v\right)(sinceffis a flow). We must prove that(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right).

The path𝐩\mathbf{p}is a path fromsstott. Thus, it neither starts nor ends atvv(sincev∈V∖{s,t}v\in V\setminus\left\{s,t\right\}). Hence, ifvvis a vertex of𝐩\mathbf{p}, then the path𝐩\mathbf{p}entersvvby some arc and exitsvvby another. Hence, we are in one of the following five cases:

Case 1:The vertexvvis not a vertex of the path𝐩\mathbf{p}.

Case 2:The path𝐩\mathbf{p}entersvvby a forward arc and exitsvvby a forward arc.

Case 3:The path𝐩\mathbf{p}entersvvby a forward arc and exitsvvby a backward arc.

Case 4:The path𝐩\mathbf{p}entersvvby a backward arc and exitsvvby a forward arc.

Case 5:The path𝐩\mathbf{p}entersvvby a backward arc and exitsvvby a backward arc.

Now, we can prove(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right)in each of these five cases by hand. Here is how this can be done in the first three cases:

First, we consider Case 1. In this case,vvis not a vertex of the path𝐩\mathbf{p}. Hence, each arca∈Aa\in Awith targetvvsatisfiesf′(a)=f(a)f^{\prime}\left(a\right)=f\left(a\right)(because neitheraanora−1a^{-1}appears in𝐩\mathbf{p}). Therefore,(f′)−(v)=f−(v)\left(f^{\prime}\right)^{-}\left(v\right)=f^{-}\left(v\right). Similarly,(f′)+(v)=f+(v)\left(f^{\prime}\right)^{+}\left(v\right)=f^{+}\left(v\right). Hence,(f′)−(v)=f−(v)=f+(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=f^{-}\left(v\right)=f^{+}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right). Thus, we have proved(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right)in Case 1.

Let us now consider Case 2. In this case, the path𝐩\mathbf{p}entersvvby a forward arc and exitsvvby a forward arc. Letbbbe the former arc, andccthe latter. Then, bothbbandccare arcs ofDD, and the vertexvvis the target ofbband the source ofcc. The definition off′f^{\prime}yields thatf′(b)=f⁡(b)+εf^{\prime}\left(b\right)=f\left(b\right)+\varepsilon, whereas each other arca∈Aa\in Awith targetvvsatisfiesf′(a)=f(a)f^{\prime}\left(a\right)=f\left(a\right). Hence,(f′)−(v)=f−(v)+ε\left(f^{\prime}\right)^{-}\left(v\right)=f^{-}\left(v\right)+\varepsilon. Similarly, using the arccc, we see that(f′)+(v)=f+(v)+ε\left(f^{\prime}\right)^{+}\left(v\right)=f^{+}\left(v\right)+\varepsilon. Hence,(f′)−(v)=f−(v)⏟=f+(v)+ε=f+(v)+ε=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\underbrace{f^{-}\left(v\right)}_{=f^{+}\left(v\right)}+\varepsilon=f^{+}\left(v\right)+\varepsilon=\left(f^{\prime}\right)^{+}\left(v\right). Thus, we have proved(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right)in Case 2.

Let us next consider Case 3. In this case, the path𝐩\mathbf{p}entersvvby a forward arc and exitsvvby a backward arc. Letbbbe the former arc, andc−1c^{-1}the latter. Then, bothbbandccare arcs ofDD, and the vertexvvis the target of bothbbandcc. The definition off′f^{\prime}yields thatf′(b)=f⁡(b)+εf^{\prime}\left(b\right)=f\left(b\right)+\varepsilon(since𝐩\mathbf{p}uses the forward arcbb) andf′(c)=f⁡(c)−εf^{\prime}\left(c\right)=f\left(c\right)-\varepsilon(since𝐩\mathbf{p}uses the backward arcc−1c^{-1}), whereas each other arca∈Aa\in Awith targetvvsatisfiesf′(a)=f(a)f^{\prime}\left(a\right)=f\left(a\right). Hence,(f′)−(v)=f−(v)+ε−ε=f−(v)\left(f^{\prime}\right)^{-}\left(v\right)=f^{-}\left(v\right)+\varepsilon-\varepsilon=f^{-}\left(v\right). Moreover,(f′)+(v)=f+(v)\left(f^{\prime}\right)^{+}\left(v\right)=f^{+}\left(v\right)(since none of the arcs ofDDwith sourcevvappears in𝐩\mathbf{p}, nor does its reversal). Hence,(f′)−(v)=f−(v)=f+(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=f^{-}\left(v\right)=f^{+}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right). Thus, we have proved(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right)in Case 3.

The other two cases are similar (Case 4 is analogous to Case 3, while Case 5 is analogous to Case 2). Thus, altogether, we have proved(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right)in all five cases.

Forget that we fixedvv. We thus have shown that each vertexv∈V∖{s,t}v\in V\setminus\left\{s,t\right\}satisfies(f′)−(v)=(f′)+(v)\left(f^{\prime}\right)^{-}\left(v\right)=\left(f^{\prime}\right)^{+}\left(v\right). In other words, the mapf′f^{\prime}satisfies the conservation constraints. Sincef′f^{\prime}also satisfies the capacity constraints, we thus conclude thatf′f^{\prime}is a flow.

What is the value|f′|\left|f^{\prime}\right|of this flow? The path𝐩\mathbf{p}starts atss, so it exitsssby some arcγ\gamma(it must have at least one arc, sinces≠ts\neq t) and never comes back tossagain. If this arcγ\gammais a forward arcbb, thenf′(b)=f⁡(b)+εf^{\prime}\left(b\right)=f\left(b\right)+\varepsilonand therefore(f′)+(s)=f+(s)+ε\left(f^{\prime}\right)^{+}\left(s\right)=f^{+}\left(s\right)+\varepsilonand(f′)−(s)=f−(s)\left(f^{\prime}\right)^{-}\left(s\right)=f^{-}\left(s\right). If this arcγ\gammais a backward arcc−1c^{-1}, thenf′(c)=f⁡(c)−εf^{\prime}\left(c\right)=f\left(c\right)-\varepsilonand therefore(f′)−(s)=f−(s)−ε\left(f^{\prime}\right)^{-}\left(s\right)=f^{-}\left(s\right)-\varepsilonand(f′)+(s)=f+(s)\left(f^{\prime}\right)^{+}\left(s\right)=f^{+}\left(s\right). Thus,

(f′)+(s)−(f′)−(s)\displaystyle\left(f^{\prime}\right)^{+}\left(s\right)-\left(f^{\prime}\right)^{-}\left(s\right)={(f+(s)+ε)−f−(s),ifγis a forward arc;f+(s)−(f−(s)−ε),ifγis a backward arc\displaystyle=\begin{cases}\left(f^{+}\left(s\right)+\varepsilon\right)-f^{-}\left(s\right),&\text{if }\gamma\text{ is a forward arc};\\ f^{+}\left(s\right)-\left(f^{-}\left(s\right)-\varepsilon\right),&\text{if }\gamma\text{ is a backward arc}\end{cases}={f+(s)−f−(s)+ε,ifγis a forward arc;f+(s)−f−(s)+ε,ifγis a backward arc\displaystyle=\begin{cases}f^{+}\left(s\right)-f^{-}\left(s\right)+\varepsilon,&\text{if }\gamma\text{ is a forward arc};\\ f^{+}\left(s\right)-f^{-}\left(s\right)+\varepsilon,&\text{if }\gamma\text{ is a backward arc}\end{cases}=f+(s)−f−(s)⏟=|f|(by the definition of|f|)+ε=|f|+ε.\displaystyle=\underbrace{f^{+}\left(s\right)-f^{-}\left(s\right)}_{\begin{subarray}{c}=\left|f\right|\\ \text{(by the definition of }\left|f\right|\text{)}\end{subarray}}+\,\varepsilon=\left|f\right|+\varepsilon.However, the definition of the value|f′|\left|f^{\prime}\right|yields

|f′|=(f′)+(s)−(f′)−(s)=|f|+ε>|f|(sinceε>0).\left|f^{\prime}\right|=\left(f^{\prime}\right)^{+}\left(s\right)-\left(f^{\prime}\right)^{-}\left(s\right)=\left|f\right|+\varepsilon>\left|f\right|\ \ \ \ \ \ \ \ \ \ \left(\text{since }\varepsilon>0\right).In other words, the flowf′f^{\prime}has a larger value thanff. Thus, we have found a flowf′f^{\prime}with a larger value thanff. This proves Lemma9.28(a).

(b)Assume that the digraphDfD_{f}has no path fromsstott. Define a subsetSSofVVby

S={v∈V∣the digraphDfhas a path fromstov}.S=\left\{v\in V\ \mid\ \text{the digraph }D_{f}\text{ has a path from }s\text{ to }v\right\}.Then,s∈Ss\in S(because the trivial path(s)\left(s\right)is a path fromsstoss) andt∉St\notin S(since we assumed thatDfD_{f}has no path fromsstott). We shall next show that|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right).

Indeed, we shall obtain this from Proposition9.17(d). To do so, we will first show that

(f⁡(a)=0for alla∈[S¯,S])\left(f\left(a\right)=0\text{ for all }a\in\left[\overline{S},S\right]\right)(60)and

(f⁡(a)=c⁡(a)for alla∈[S,S¯]).\left(f\left(a\right)=c\left(a\right)\text{ for all }a\in\left[S,\overline{S}\right]\right).(61) [Proof of (60):Leta∈[S¯,S]a\in\left[\overline{S},S\right]. Assume thatf⁡(a)≠0f\left(a\right)\neq 0. Thus,f⁡(a)>0f\left(a\right)>0(since the capacity constraints yieldf⁡(a)≥0f\left(a\right)\geq 0). Hence, the backward arca−1a^{-1}is an arc of the residual digraphDfD_{f}. Letuube the source ofaa, and letvvbe the target ofaa. Sincea∈[S¯,S]a\in\left[\overline{S},S\right], we thus haveu∈S¯u\in\overline{S}andv∈Sv\in S. Fromv∈Sv\in S, we see that the digraphDfD_{f}has a path fromsstovv. Let𝐪\mathbf{q}be this path. Appending the backward arca−1a^{-1}(which is an arc fromvvtouu) and the vertexuuto this path𝐪\mathbf{q}(at the end), we obtain a walk fromsstouuinDfD_{f}. Hence,DfD_{f}has a walk fromsstouu, thus also a path fromsstouu(by Corollary4.29). This entailsu∈Su\in S(by the definition ofSS). However, this contradictsu∈S¯=V∖Su\in\overline{S}=V\setminus S. This contradiction shows that our assumption (thatf⁡(a)≠0f\left(a\right)\neq 0) was wrong. Therefore,f⁡(a)=0f\left(a\right)=0. This proves (60).]

[Proof of (61):Leta∈[S,S¯]a\in\left[S,\overline{S}\right]. Assume thatf⁡(a)≠c⁡(a)f\left(a\right)\neq c\left(a\right). Thus,f⁡(a)<c⁡(a)f\left(a\right)<c\left(a\right)(since the capacity constraints yieldf⁡(a)≤c⁡(a)f\left(a\right)\leq c\left(a\right)). Hence, the forward arcaais an arc of the residual digraphDfD_{f}. Letuube the source ofaa, and letvvbe the target ofaa. Sincea∈[S,S¯]a\in\left[S,\overline{S}\right], we thus haveu∈Su\in Sandv∈S¯v\in\overline{S}. Fromu∈Su\in S, we see that the digraphDfD_{f}has a path fromsstouu. Let𝐪\mathbf{q}be this path. Appending the forward arcaa(which is an arc fromuutovv) and the vertexvvto this path𝐪\mathbf{q}(at the end), we obtain a walk fromsstovvinDfD_{f}. Hence,DfD_{f}has a walk fromsstovv, thus also a path fromsstovv(by Corollary4.29). This entailsv∈Sv\in S(by the definition ofSS). However, this contradictsv∈S¯=V∖Sv\in\overline{S}=V\setminus S. This contradiction shows that our assumption (thatf⁡(a)≠c⁡(a)f\left(a\right)\neq c\left(a\right)) was wrong. Therefore,f⁡(a)=c⁡(a)f\left(a\right)=c\left(a\right). This proves (61).]

Now, Proposition9.17(d)yields that|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right)holds (since (60) and (61) hold).

We have now found a subsetSSofVVsatisfyings∈Ss\in Sandt∉St\notin Sand|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right). In order to prove Lemma9.28(b), it suffices to show that the flowffhas maximum value (among all flows onNN). However, this is now easy: Any flowggonNNhas value|g|≤c⁡(S,S¯)\left|g\right|\leq c\left(S,\overline{S}\right)(by Proposition9.17(c), applied togginstead offf). In other words, any flowggonNNhas value|g|≤|f|\left|g\right|\leq\left|f\right|(since|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right)). Thus, the flowffhas maximum value. This completes the proof of Lemma9.28(b).

9.4.6Proof of max-flow-min-cut

We are now ready to prove the max-flow-min-cut theorem (Theorem9.21):

Proof 9.30(Proof of Theorem9.21.).

We letf:A→ℕf:A\rightarrow\mathbb{N}be the zero flow onNN(see Example9.13for its definition). Now, we shall incrementally increase the value|f|\left|f\right|of this flow by the following algorithm (known as theFord-Fulkerson algorithm):

1.Construct the residual digraphDfD_{f}.
2.If the digraphDfD_{f}has a path fromsstott, then Lemma9.28(a)shows that the networkNNhas a flowf′f^{\prime}with a larger value thanff(and furthermore, the proof of Lemma9.28(a)shows how to find such anf′f^{\prime}efficiently126126126Of course, this requires an algorithm for finding a path fromsstottinDfD_{f}. But there are many efficient algorithms for this (see, e.g., homework set #4 exercise 5).). Fix such anf′f^{\prime}, and replaceffbyf′f^{\prime}. Then, go back to step 1.
3.If the digraphDfD_{f}has no path fromsstott, then we end the algorithm.

The replacement offfbyf′f^{\prime}in Step 2 of this algorithm will be called anaugmentation. Thus, the algorithm proceeds by repeatedly performing augmentations until this is no longer possible.

I claim that the algorithm will eventually end – i.e., it cannot keep performing augmentations forever. Indeed, each augmentation increases the value|f|\left|f\right|of the flowff, and therefore it increases this value|f|\left|f\right|by at least11(because increasing an integer always means increasing it by at least11). However, the value|f|\left|f\right|is bounded from above by the capacityc⁡(S,S¯)c\left(S,\overline{S}\right)of an arbitrary cut[S,S¯]\left[S,\overline{S}\right](by Proposition9.17(c)), and thus cannot get increased by11more thanc⁡(S,S¯)c\left(S,\overline{S}\right)many times (since its initial value is00). Therefore, we cannot perform more thanc⁡(S,S¯)c\left(S,\overline{S}\right)many augmentations in sequence.

Thus, the algorithm eventually ends. Let us consider the flowffthat is obtained once the algorithm has ended. This flowffhas the property that the digraphDfD_{f}has no path fromsstott. Thus, Lemma9.28(b)shows that the flowffhas maximum value (among all flows onNN), and there exists a subsetSSofVVsatisfyings∈Ss\in Sandt∉St\notin Sand|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right). Consider thisSS.

Since the flowffhas maximum value, we have

|f|=max{|g|∣gis a flow}.\left|f\right|=\max\left\{\left|g\right|\ \mid\ g\text{ is a flow}\right\}. On the other hand, for each subsetTTofVVsatisfyings∈Ts\in Tandt∉Tt\notin T, we have

c⁡(S,S¯)=|f|≤c⁡(T,T¯)c\left(S,\overline{S}\right)=\left|f\right|\leq c\left(T,\overline{T}\right)(by Proposition9.17(c), applied toTTinstead ofSS). Hence,

c(S,S¯)=min{c(T,T¯)∣T⊆V;s∈T;t∉T}.c\left(S,\overline{S}\right)=\min\left\{c\left(T,\overline{T}\right)\ \mid\ T\subseteq V;\ s\in T;\ t\notin T\right\}.Comparing this with

c⁡(S,S¯)=|f|=max⁡{|g|∣gis a flow},c\left(S,\overline{S}\right)=\left|f\right|=\max\left\{\left|g\right|\ \mid\ g\text{ is a flow}\right\},we obtain

max{|g|∣gis a flow}=min{c(T,T¯)∣T⊆V;s∈T;t∉T}.\max\left\{\left|g\right|\ \mid\ g\text{ is a flow}\right\}=\min\left\{c\left(T,\overline{T}\right)\ \mid\ T\subseteq V;\ s\in T;\ t\notin T\right\}.In other words, the maximum value of a flow equals the minimum capacity of a cut. This proves Theorem9.21. (Of course, we cannot use the lettersffandSSfor the bound variables inmax⁡{|g|∣gis a flow}\max\left\{\left|g\right|\ \mid\ g\text{ is a flow}\right\}and min{c(T,T¯)∣T⊆V;s∈T;t∉T}\min\left\{c\left(T,\overline{T}\right)\ \mid\ T\subseteq V;\ s\in T;\ t\notin T\right\}, sinceffandSSalready stand for a specific flow and a specific set.)

9.5Application: Deriving Hall–König

Now, let us apply the max-flow-min-cut theorem to prove the Hall–König matching theorem (8.31):

Proof 9.32(Proof of Theorem8.31(sketched).).

(This is an outline; see[17s-lec16, proof of Lemma 1.42]for details.127127127Note that[17s-lec16, Lemma 1.42]is stated only for a simple graphGG, not for a multigraphGG. However, this really makes no difference here: If(G,X,Y)\left(G,X,Y\right)is a bipartite graph withGGbeing a multigraph, then(Gsimp,X,Y)\left(G^{\operatorname*{simp}},X,Y\right)is a bipartite graph as well, and clearly any matching ofGsimpG^{\operatorname*{simp}}yields a matching ofGGhaving the same size (and the setN⁡(U)N\left(U\right)does not change fromGGtoGsimpG^{\operatorname*{simp}}either). Thus, in proving Theorem8.31, we can WLOG assume thatGGis a simple graph.) As explained in Example9.14, we can turn the bipartite graph(G,X,Y)\left(G,X,Y\right)into a network so that the matchings ofGGbecome the flowsffof this network. The max-flow-min-cut theorem (Theorem9.21) yields that

max{|f|∣fis a flow}=min{c(S,S¯)∣S⊆V;s∈S;t∉S},\max\left\{\left|f\right|\ \mid\ f\text{ is a flow}\right\}=\min\left\{c\left(S,\overline{S}\right)\ \mid\ S\subseteq V;\ s\in S;\ t\notin S\right\},whereVVis the vertex set of the digraph that underlies our network. Thus, there exist a flowffand a cut[S,S¯]\left[S,\overline{S}\right]of this network such that|f|=c⁡(S,S¯)\left|f\right|=c\left(S,\overline{S}\right). Consider theseffandSS. Thus,SSis a subset ofVVsuch thats∈Ss\in Sandt∉St\notin S.

LetMMbe the matching ofGGcorresponding to the flowff(that is, we letMMbe the set of all edgeseeofGGsuch thatf⁡(e→)=1f\left(\overrightarrow{e}\right)=1). Thus,|M|=|f|\left|M\right|=\left|f\right|.

LetU:=X∩SU:=X\cap S. Then,UUis a subset ofXX. Here is an illustration of the cut[S,S¯]\left[S,\overline{S}\right]on a simple example (the flowffis not shown):

112233445566ssttXXYYS\color[rgb]{1,0,0}{S}S¯\color[rgb]{1,0,0}{\overline{S}}(the orange oval is the setUU).

Now, we have

|M|\displaystyle\left|M\right|=|f|=c⁡(S,S¯)=c⁡({s},S¯)⏟=|X∖U|(why?)+c⁡(X∩S⏟=U,S¯)+c⁡(Y∩S,S¯)⏟=|Y∩S|(why?)\displaystyle=\left|f\right|=c\left(S,\ \overline{S}\right)=\underbrace{c\left(\left\{s\right\},\ \overline{S}\right)}_{\begin{subarray}{c}=\left|X\setminus U\right|\\ \text{(why?)}\end{subarray}}+c\left(\underbrace{X\cap S}_{=U},\ \overline{S}\right)+\underbrace{c\left(Y\cap S,\ \overline{S}\right)}_{\begin{subarray}{c}=\left|Y\cap S\right|\\ \text{(why?)}\end{subarray}}(sinceSis the union of the disjoint sets{s},X∩SandY∩S)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\text{since }S\text{ is the union of the disjoint sets }\left\{s\right\}\text{, }X\cap S\text{ and }Y\cap S\right)=|X∖U|⏟=|X|−|U|+c⁡(U,S¯)+|Y∩S|⏟≥|N⁡(U)|(since each vertexy∈N(U)either belongs toY∩Sand thus contributes to|Y∩S|, or belongs toS¯and thus contributes toc(U,S¯))\displaystyle=\underbrace{\left|X\setminus U\right|}_{=\left|X\right|-\left|U\right|}+\underbrace{c\left(U,\ \overline{S}\right)+\left|Y\cap S\right|}_{\begin{subarray}{c}\geq\left|N\left(U\right)\right|\\ \text{(since each vertex }y\in N\left(U\right)\text{ either belongs to }Y\cap S\\ \text{and thus contributes to }\left|Y\cap S\right|\text{, or belongs to }\overline{S}\\ \text{and thus contributes to }c\left(U,\ \overline{S}\right)\text{)}\end{subarray}}≥|X|−|U|+|N⁡(U)|=|N⁡(U)|+|X|−|U|.\displaystyle\geq\left|X\right|-\left|U\right|+\left|N\left(U\right)\right|=\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|.This proves Theorem8.31.

Having proved the Hall–König matching theorem (Theorem8.31), we have thus completed the proofs of Hall’s marriage theorem (Theorem8.23) and of König’s theorem (Theorem8.30) as well, because we already know how to derive the latter two theorems from the former.

9.6Other applications

Further applications of the max-flow-min-cut theorem include:

•A curious fact about rounding matrix entries (stated in terms of a digraph in[Schrij17, Exercise 4.13]): LetAAbe anm×nm\times n-matrix with real entries. Assume that all row sums128128128Arow sumof a matrix means the sum of all entries in some row of this matrix. Thus, anm×nm\times n-matrix hasmmrow sums.ofAAand all column sums129129129Acolumn sumof a matrix means the sum of all entries in some column of this matrix. Thus, anm×nm\times n-matrix hasnncolumn sums.ofAAare integers. Then, we can round each non-integer entry ofAA(that is, replace it either by the next-smaller integer or the next-larger integer) in such a way that the resulting matrix has the same row sums asAAand the same column sums asAA.
•An Euler-Hierholzer-like criterion for the existence of an Eulerian circuit in a “mixed graph” (a general notion of a graph that can contain both undirected edges and directed arcs)[ForFul74, §II.7].
•A proof[Berge91, §6.3]of the Erdös–Gallai theorem, which states that for a given weakly decreasingnn-tuple(d1≥d2≥⋯≥dn)\left(d_{1}\geq d_{2}\geq\cdots\geq d_{n}\right)of nonnegative integers, there exists a simple graph withnnvertices whosennvertices have degreesd1,d2,…,dnd_{1},d_{2},\ldots,d_{n}if and only if the sumd1+d2+⋯+dnd_{1}+d_{2}+\cdots+d_{n}is even and eachk∈{1,2,…,n}k\in\left\{1,2,\ldots,n\right\}satisfies ∑i=1kdi≤k⁡(k−1)+∑i=k+1nmin⁡{di,k}.\sum\limits_{i=1}^{k}d_{i}\leq k\left(k-1\right)+\sum\limits_{i=k+1}^{n}\min\left\{d_{i},k\right\}.(The “only if” part of this theorem was Exercise2.34= Exercise 6 on homework set #2.)

The following exercise can be solved both with and without using the max-flow-min-cut theorem; it should make good practice to solve it in both ways.

Exercise 9.33.

Consider a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in Vand a sinkt∈Vt\in V, and a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}such thats≠ts\neq t. (You can replaceℕ\mathbb{N}byℚ+\mathbb{Q}_{+}orℝ+\mathbb{R}_{+}here.)

Anss-tt-cutting subsetshall mean a subsetSSofVVsatisfyings∈Ss\in Sandt∉St\notin S.

Letmmdenote the minimum possible value ofc⁡(S,S¯)c\left(S,\overline{S}\right)whereSSranges over thess-tt-cutting subsets. (Recall that this is the maximum value of a flow, according to Theorem9.21.)

Anss-tt-cutting subsetSSis said to becut-minimalif it satisfiesc⁡(S,S¯)=mc\left(S,\overline{S}\right)=m.

LetXXandYYbe two cut-minimalss-tt-cutting subsets. Prove thatX∩YX\cap YandX∪YX\cup Yalso are cut-minimalss-tt-cutting subsets.

[Solution:This is Exercise 7 on homework set #5 from my Spring 2017 course (except that the simple digraph has been replaced by a multidigraph); seethe course pagefor solutions.]

10More about paths

In this chapter, we will learn a few more things about paths in graphs and digraphs.

10.1Menger’s theorems

We begin with a series of fundamental results known asMenger’s theorems(named after Karl Menger, who discovered one of them in 1927 as an auxiliary result in a topological study of curves130130130See[Schrij03, §9.6e]for more about its history.).

Imagine you have44different ways to get from Philadelphia to NYC, all using different roads (i.e., no piece of road is used by more than one of your44ways). Then, if33arbitrary roads get blocked, then you still have a way to get to NYC.

This is obvious (indeed, each blocked road destroys at most one of your44paths, so you still have at least one path left undisturbed after33roads have been blocked). A more interesting question is the converse: If the road network is sufficiently robust that blocking33arbitrary roads will not disconnect you from NYC, does this mean that you can find44different ways to NYC all using different roads?

Menger’s theorems answer this question (and various questions of this kind) in the positive, in several different setups. Each of these theorems can be roughly described as “the maximum number of pairwise independent paths from some place to another place equals the minimum size of a bottleneck that separates the former from the latter”. Here, the “places” can be vertices or sets of vertices; the word “independent” can mean “having no arcs in common” or “having no intermediate vertices in common” or “having no vertices at all in common”; and the word “bottleneck” can mean a set of arcs or of vertices whose removal would disconnect the former place from the latter. Here is a quick overview of all Menger’s theorems that we will prove:131131131All undefined terminology used here will be defined further below.

•for directed graphs: Theorem …the places are …the paths must be …the bottleneck consists of …10.6verticesarc-disjointarcs10.11verticesarc-disjointarcs of a cut10.24sets of verticesarc-disjointarcs of a cut10.43verticesinternally vertex-disjointvertices∈V∖{s,t}\in V\setminus\left\{s,t\right\}10.48sets of verticesinternally vertex-disjointvertices∈V∖(X∪Y)\in V\setminus\left(X\cup Y\right)10.52sets of verticesvertex-disjointvertices∈V\in V
•for undirected graphs: Theorem …the places are …the paths must be …the bottleneck consists of …10.29verticesedge-disjointedges10.33verticesedge-disjointedges of a cut10.56verticesinternally vertex-disjointvertices∈V∖{s,t}\in V\setminus\left\{s,t\right\}10.58sets of verticesinternally vertex-disjointvertices∈V∖(X∪Y)\in V\setminus\left(X\cup Y\right)10.59sets of verticesvertex-disjointvertices∈V\in V(I could state more, but I don’t want this to go on forever.)

10.1.1The arc-Menger theorem for directed graphs

We begin with the most natural setup: a directed graph (one-way roads) with roads being arcs. The following definitions will help keep the theorems short:

Definition 10.1.

Two walks𝐩\mathbf{p}and𝐪\mathbf{q}in a digraph are said to bearc-disjointif they have no arc in common.

Example 10.2.

The following picture shows two arc-disjoint paths𝐩\mathbf{p}and𝐪\mathbf{q}(they can be told apart by their labels: each arc of𝐩\mathbf{p}is labelled with a “𝐩\mathbf{p}”, and likewise for𝐪\mathbf{q}):

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\lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ . The following picture shows two paths𝐫\mathbf{r}and𝐬\mathbf{s}that arenotarc-disjoint (the common arc is marked with “𝐫,𝐬\mathbf{r},\mathbf{s}”):

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-6.52 -3.6 -3.6 -6.52 0 -6.52 C 3.6 -6.52 6.52 -3.6 6.52 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{0.0pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 0 0)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 45.89 -39.37 C 45.89 -35.77 42.97 -32.85 39.37 -32.85 C 35.77 -32.85 32.85 -35.77 32.85 -39.37 C 32.85 -42.97 35.77 -45.89 39.37 -45.89 C 42.97 -45.89 45.89 -42.97 45.89 -39.37 Z M 39.37 -39.37}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{28.45276pt}{-28.45276pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 39.37 -39.37)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 85.26 0 C 85.26 3.6 82.34 6.52 78.74 6.52 C 75.14 6.52 72.22 3.6 72.22 0 C 72.22 -3.6 75.14 -6.52 78.74 -6.52 C 82.34 -6.52 85.26 -3.6 85.26 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{56.90552pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 78.74 0)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 164 0 C 164 3.6 161.08 6.52 157.48 6.52 C 153.88 6.52 150.96 3.6 150.96 0 C 150.96 -3.6 153.88 -6.52 157.48 -6.52 C 161.08 -6.52 164 -3.6 164 0 Z M 157.48 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{113.81104pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 157.48 0)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@begingroup@{_scopebegin=1} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 7.08 0 L 70 0}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{51.19167pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 70.83 0)} \lxSVG@begingroup@{_scopebegin=1} 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{{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 85.82 0 L 148.74 0}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{108.09718pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 149.57 0)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{79.3333pt}{-8.37746pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 109.77 -11.59)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 1.83 -6.83 C 6.23 -23.26 16.11 -33.14 30.93 -37.11}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.96593}{-0.25882}{0.25882}{0.96593}{22.9336pt}{-26.97389pt}\lxSVG@begingroup@{transform=matrix(0.96593 -0.25882 0.25882 0.96593 31.73 -37.32)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{6.89233pt}{-27.66974pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 9.54 -38.29)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}{}{}{}{}{{}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 46.21 -37.54 C 62.63 -33.14 72.51 -23.26 76.48 -8.44}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.25882}{0.96593}{-0.96593}{0.25882}{55.42665pt}{-5.51917pt}\lxSVG@begingroup@{transform=matrix(0.25882 0.96593 -0.96593 0.25882 76.69 -7.64)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{45.47702pt}{-27.66974pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 62.93 -38.29)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .

Definition 10.3.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letssandttbe two vertices ofDD. A subsetBBofAAis said to be anss-tt-arc-separatorif each path fromsstottcontains at least one arc fromBB. Equivalently, a subsetBBofAAis said to be anss-tt-arc-separatorif the multidigraph(V,A∖B,ψ∣A∖B)\left(V,\ A\setminus B,\ \psi\mid_{A\setminus B}\right)has no path fromsstott(in other words, removing fromDDall arcs contained inBBdestroys all paths fromsstott).

Example 10.4.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be the following multidigraph:

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\lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{82.70969pt}{-39.15068pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 114.45 -54.17)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the set{α,γ}\left\{\alpha,\gamma\right\}is not anss-tt-arc-separator (since the path drawn in blue contains no arc from this set). However, the set{β,γ}\left\{\beta,\gamma\right\}is anss-tt-arc-separator, and so is the set{δ,ε}\left\{\delta,\varepsilon\right\}. Of course, any set that contains any of{β,γ}\left\{\beta,\gamma\right\}and{δ,ε}\left\{\delta,\varepsilon\right\}as a subset is therefore anss-tt-arc-separator as well.

Example 10.5.

LetDDbe a multidigraph. Letssandttbe two vertices ofDD. Then, the empty set∅\varnothingis anss-tt-arc-separator if and only ifDDhas no path fromsstott. This degenerate case should not be forgotten!

We can now state the first Menger’s theorem:

Theorem 10.6(arc-Menger theorem for directed graphs, version 1).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letssandttbe two distinct vertices ofDD. Then, the maximum number of pairwise arc-disjoint paths fromsstottequals the minimum size of anss-tt-arc-separator.

Example 10.7.

LetDDbe the multidigraph from Example10.4. Then, the minimum size of anss-tt-arc-separator is22(indeed,{β,γ}\left\{\beta,\gamma\right\}is anss-tt-arc-separator of size22, and it is easy to see that there are noss-tt-arc-separators of smaller size). Hence, Theorem10.6yields that the maximum number of pairwise arc-disjoint paths fromsstottis22as well. And indeed, we can easily find22arc-disjoint paths fromsstott, namely the red and the blue paths in the following figure:

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Before proving Theorem10.6, let me state another variant of this theorem, which is closer to the proof. First, some notations:

Definition 10.8.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letssandttbe two distinct vertices ofDD.

(a)For each subsetSSofVV, we setS¯:=V∖S\overline{S}:=V\setminus Sand [S,S¯]\displaystyle\left[S,\overline{S}\right]:={a∈A∣the source ofabelongs toS,\displaystyle:=\left\{a\in A\ \mid\ \text{the source of }a\text{ belongs to }S\text{,}\right.and the target ofabelongs toS¯}.\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{and the target of }a\text{ belongs to }\overline{S}\right\}.(These are the same definitions that we introduced for networks in Definition9.5.)
(b)Anss-tt-cutmeans a subset ofAAthat has the form[S,S¯]\left[S,\overline{S}\right], whereSSis a subset ofVVthat satisfiess∈Ss\in Sandt∉St\notin S. (This was just called a “cut” back in Definition9.19(a).)

Anss-tt-cut is called this way because its removal would cut the vertexssfrom the vertextt. More precisely:

Proof 10.10.

LetBBbe anss-tt-cut. We must prove thatBBis anss-tt-arc-separator. In other words, we must prove that each path fromsstottcontains at least one arc fromBB.

We know thatBBis anss-tt-cut. In other words,B=[S,S¯]B=\left[S,\overline{S}\right], whereSSis a subset ofVVthat satisfiess∈Ss\in Sandt∉St\notin S. Consider this subsetSS.

Each path fromsstottstarts at a vertex inSS(sinces∈Ss\in S) and ends at a vertex outside ofSS(sincet∉St\notin S). Thus, each such path has to escape the setSSat some point – i.e., it must contain an arc whose source is inSSand whose target is outside ofSS. But such an arc must necessarily belong to[S,S¯]\left[S,\overline{S}\right](by the definition of[S,S¯]\left[S,\overline{S}\right]). Thus, each path fromsstottmust contain an arc from[S,S¯]\left[S,\overline{S}\right]. In other words, each path fromsstottmust contain an arc fromBB(sinceB=[S,S¯]B=\left[S,\overline{S}\right]). In other words,BBis anss-tt-arc-separator (by the definition of anss-tt-arc-separator). This proves Remark10.9.

Theorem 10.11(arc-Menger theorem for directed graphs, version 2).

Example 10.12.

LetDDbe the following multidigraph:

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{{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 168.95 73.39 C 200.15 58.73 216.71 42.07 230.49 12.3}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.41997}{-0.90755}{0.90755}{0.41997}{166.8289pt}{8.34114pt}\lxSVG@begingroup@{transform=matrix(0.41997 -0.90755 0.90755 0.41997 230.84 11.54)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the maximum number of pairwise arc-disjoint paths fromsstottis33. Indeed, the following picture shows33such paths in red, blue and brown, respectively:

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}}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}}{{}}{{{{}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{{{}}{{\lx@inpgf@ignorespaces}}{{\lx@inpgf@ignorespaces}}{{}}{{}}}{{{{}}{}{}{}{}{{}}}} }{{}{}\lx@inpgf@ignorespaces}{{}} {}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#BF8040} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 170.53 -72.66 C 201.09 -58.03 217.17 -41.64 230.5 -12.3}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#BF8040} \lxSVG@transformcm{0.41367}{0.91043}{-0.91043}{0.41367}{166.83267pt}{-8.34288pt}\lxSVG@begingroup@{transform=matrix(0.41367 0.91043 -0.91043 0.41367 230.85 -11.54)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .More than33pairwise arc-disjoint paths fromsstottcannot exist inDD, since (e.g.) there are only33arcs outgoing fromss.

By Theorem10.11, this shows that the minimum size of anss-tt-cut inDDis33as well. There are manyss-tt-cuts of size33(for instance, the “obvious” cut[{s},{s}¯]\left[\left\{s\right\},\overline{\left\{s\right\}}\right]has this property, as does thess-tt-cut[{s,a,f},{s,a,f}¯]\left[\left\{s,a,f\right\},\overline{\left\{s,a,f\right\}}\right]).

Let us now reverse of the direction of the arc fromcctoeeinDD(thus destroying the brown path). The resulting multidigraphD′D^{\prime}looks as follows:

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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.41997}{-0.90755}{0.90755}{0.41997}{166.8289pt}{8.34114pt}\lxSVG@begingroup@{transform=matrix(0.41997 -0.90755 0.90755 0.41997 230.84 11.54)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .This digraphD′D^{\prime}has no more than22pairwise arc-disjoint paths fromsstott. This can be seen by observing that thess-tt-cut[{s,c},{s,c}¯]\left[\left\{s,c\right\},\overline{\left\{s,c\right\}}\right]has size22(it consists of the arc fromsstoaaand the arc fromsstobb), so that the minimum size of anss-tt-cut is at most22, and therefore (by Theorem10.11) the maximum number of pairwise arc-disjoint paths fromsstottis at most22as well. It is easy to see that the latter number is exactly22(since our red and blue paths still exist inD′D^{\prime}).

To prove the above two arc-Menger theorems, we need one more lemma about networks. We recall the notations from Section9.1and from Definition9.19, and introduce a couple more:

Definition 10.13.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Letf,g:A→ℕf,g:A\rightarrow\mathbb{N}be two maps. Then:

(a)We letf+gf+gdenote the map fromAAtoℕ\mathbb{N}that sends each arca∈Aa\in Atof⁡(a)+g⁡(a)f\left(a\right)+g\left(a\right). (This is the pointwise sum offfandgg.)
(b)We writeg≤fg\leq fif and only if each arca∈Aa\in Asatisfiesg⁡(a)≤f⁡(a)g\left(a\right)\leq f\left(a\right).
(c)Ifg≤fg\leq f, then we letf−gf-gdenote the map fromAAtoℕ\mathbb{N}that sends each arca∈Aa\in Atof⁡(a)−g⁡(a)f\left(a\right)-g\left(a\right). (This is really a map toℕ\mathbb{N}, sinceg≤fg\leq fentailsg⁡(a)≤f⁡(a)g\left(a\right)\leq f\left(a\right).)

These notations satisfy the properties that you’d expect: e.g., the pointwise sum of maps fromAAtoℕ\mathbb{N}is associative (meaning that(f+g)+h=f+(g+h)\left(f+g\right)+h=f+\left(g+h\right), so that you can writef+g+hf+g+hfor both sides); inequalities can be manipulated in the usual way (e.g., we havef−g≤hf-g\leq hif and only iff≤g+hf\leq g+h). Verifying this all is straightforward.

The following definition codifies the flows that we constructed in Remark9.10:

Definition 10.14.

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}. Let𝐩\mathbf{p}be a path fromsstottinDD. Then, we define a mapf𝐩:A→ℕf_{\mathbf{p}}:A\rightarrow\mathbb{N}by setting

f𝐩(a)={1,ifais an arc of𝐩;0,otherwisefor eacha∈A.f_{\mathbf{p}}\left(a\right)=\begin{cases}1,&\text{if }a\text{ is an arc of }\mathbf{p};\\ 0,&\text{otherwise}\end{cases}\ \ \ \ \ \ \ \ \ \ \text{for each }a\in A.We call this mapf𝐩f_{\mathbf{p}}thepath flowof𝐩\mathbf{p}. It is an actual flow of value11if all the arcs of𝐩\mathbf{p}have capacity≥1\geq 1.

Example 10.15.

Consider the following network:

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\lxSVG@transformcm{0.7071}{0.7071}{-0.7071}{0.7071}{107.3038pt}{-6.50723pt}\lxSVG@begingroup@{transform=matrix(0.7071 0.7071 -0.7071 0.7071 148.48 -9)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,where each arc has capacity11. Then, the path𝐩=(s,2,6,t)\mathbf{p}=\left(s,2,6,t\right)leads to the following path flowf𝐩f_{\mathbf{p}}:

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The path flow thus turns any path fromsstottin a network into a flow, provided that the arcs have enough capacity to carry this flow. If we havemmpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}fromsstott, then we can add their path flows together, and obtain a flowf𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}of valuemm, provided (again) that the arcs have enough capacity for it. (In general, we cannot uniquely reconstruct𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}back from this latter flow, as they might have gotten “mixed together”.)

Our next lemma can be viewed as a (partial) converse of this observation: Any flowffof valuemm“contains” a sumf𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}ofmmpath flowsf𝐩1,f𝐩2,…,f𝐩mf_{\mathbf{p}_{1}},f_{\mathbf{p}_{2}},\ldots,f_{\mathbf{p}_{m}}corresponding tomm(not necessarily distinct) paths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}fromsstott. Here, the word “contains” signals thatffis not necessarily equal tof𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}, but only satisfiesf𝐩1+f𝐩2+⋯+f𝐩m≤ff_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq fin general. So here is the lemma:

Lemma 10.16(flow path decomposition lemma).

LetNNbe a network consisting of a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right), a sources∈Vs\in V, a sinkt∈Vt\in Vand a capacity functionc:A→ℕc:A\rightarrow\mathbb{N}. Letffbe a flow onNNthat has valuemm. Then, there existmmpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}fromsstottinDDsuch that

f𝐩1+f𝐩2+⋯+f𝐩m≤f.f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq f.

Proof 10.17.

We induct onmm.

Thebase case(m=0m=0) is obvious (since the empty sumf𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}is the zero flow, and thus is≤f\leq fbecause of the capacity constraints).

Induction step:Letmmbe a positive integer. Assume (as the induction hypothesis) that the lemma holds form−1m-1. We must prove the lemma formm.

So we consider a flowffonNNthat has valuemm. We need to show that there existmmpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}fromsstottinDDsuch thatf𝐩1+f𝐩2+⋯+f𝐩m≤ff_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq f.

We shall first find some path𝐩\mathbf{p}fromsstottsuch thatf𝐩≤ff_{\mathbf{p}}\leq f.

We shall refer to the arcsa∈Aa\in Asatisfyingf⁡(a)>0f\left(a\right)>0as theactivearcs. LetA′:={a∈A∣f⁡(a)>0}A^{\prime}:=\left\{a\in A\ \mid\ f\left(a\right)>0\right\}be the set of these active arcs. Consider the spanning subdigraphD′:=(V,A′,ψ∣A′)D^{\prime}:=\left(V,A^{\prime},\psi\mid_{A^{\prime}}\right)ofDD.

LetSSbe the set of all verticesv∈Vv\in Vsuch thatD′D^{\prime}has a path fromsstovv. Then,s∈Ss\in S(since the trivial path(s)\left(s\right)is a path ofD′D^{\prime}).

We next claim that each arcb∈[S,S¯]b\in\left[S,\overline{S}\right]satisfiesf⁡(b)=0f\left(b\right)=0.

[Proof:Assume the contrary. Thus, some arcb∈[S,S¯]b\in\left[S,\overline{S}\right]satisfiesf⁡(b)≠0f\left(b\right)\neq 0. Consider thisbb. Fromf⁡(b)≠0f\left(b\right)\neq 0, we obtainf⁡(b)>0f\left(b\right)>0(sinceffis a flow), thusb∈A′b\in A^{\prime}(by the definition ofA′A^{\prime}). Hence,bbis an arc ofD′D^{\prime}(by the definition ofD′D^{\prime}).

Letuube the source of the arcbb, andvvits target. Sinceb∈[S,S¯]b\in\left[S,\overline{S}\right], we therefore haveu∈Su\in Sandv∈S¯v\in\overline{S}. Sinceu∈Su\in S, the digraphD′D^{\prime}has a path𝐩\mathbf{p}fromsstouu(by the definition ofSS). Consider this path𝐩\mathbf{p}. Appending the arcbband the vertexvvat the end of this path𝐩\mathbf{p}, we obtain a walk fromsstovvinD′D^{\prime}(sincebbis an arc ofD′D^{\prime}with sourceuuand targetvv). Hence, the digraphD′D^{\prime}has a walk fromsstovv, thus also a path fromsstovv(by Corollary4.29). This means thatv∈Sv\in S(by the definition ofSS). But this contradictsv∈S¯=V∖Sv\in\overline{S}=V\setminus S. This contradiction shows that our assumption was wrong, qed.]

We thus have proved that eachb∈[S,S¯]b\in\left[S,\overline{S}\right]satisfiesf⁡(b)=0f\left(b\right)=0. Therefore,f⁡(S,S¯)=0f\left(S,\overline{S}\right)=0(using the notations of Definition9.5(d)). However, recall thats∈Ss\in S. Thus, if we hadt∉St\notin S, then Proposition9.17(b)would yield

|f|=f⁡(S,S¯)⏟=0−f⁡(S¯,S)⏟≥0≤0−0=0,\left|f\right|=\underbrace{f\left(S,\overline{S}\right)}_{=0}-\underbrace{f\left(\overline{S},S\right)}_{\geq 0}\leq 0-0=0,which would contradict|f|=m>0\left|f\right|=m>0. Hence, we must havet∈St\in S. In other words, the digraphD′D^{\prime}has a path fromsstott(by the definition ofSS). Let𝐩\mathbf{p}be this path. Then,𝐩\mathbf{p}is also a path inDDand satisfiesf𝐩≤ff_{\mathbf{p}}\leq f132132132Proof.We need to prove that each arca∈Aa\in Asatisfiesf𝐩(a)≤f(a)f_{\mathbf{p}}\left(a\right)\leq f\left(a\right). So leta∈Aa\in Abe an arc. Ifaais not an arc of𝐩\mathbf{p}, then the definition off𝐩f_{\mathbf{p}}yieldsf𝐩(a)=0≤f⁡(a)f_{\mathbf{p}}\left(a\right)=0\leq f\left(a\right)(sinceffis a flow), so we are done in this case. Hence, assume WLOG thataais an arc of𝐩\mathbf{p}. Thus,aais an arc ofD′D^{\prime}(since𝐩\mathbf{p}is a path ofD′D^{\prime}). In other words,a∈A′a\in A^{\prime}. By the definition ofA′A^{\prime}, this means thatf⁡(a)>0f\left(a\right)>0. Sincef⁡(a)f\left(a\right)is an integer, we thus havef⁡(a)≥1=f𝐩(a)f\left(a\right)\geq 1=f_{\mathbf{p}}\left(a\right)(sinceaais an arc of𝐩\mathbf{p}). In other words,f𝐩(a)≤f(a)f_{\mathbf{p}}\left(a\right)\leq f\left(a\right). This is precisely what we wanted to prove.. Therefore,f−f𝐩f-f_{\mathbf{p}}is a map fromAAtoℕ\mathbb{N}. Moreover,f−f𝐩f-f_{\mathbf{p}}is again a flow133133133Here, we are using the fact (which is straightforward to prove) that ifggandhhare two flows withh≤gh\leq g, theng−hg-his again a flow., and has value|f−f𝐩|=m−1\left|f-f_{\mathbf{p}}\right|=m-1134134134Here, we are using the fact (which is straightforward to prove) that ifggandhhare two flows satisfyingh≤gh\leq g, then|g−h|=|g|−|h|\left|g-h\right|=\left|g\right|-\left|h\right|. Applying this fact tog=fg=fandh=f𝐩h=f_{\mathbf{p}}, we obtain|f−f𝐩|=|f|⏟=m−|f𝐩|⏟=1=m−1\left|f-f_{\mathbf{p}}\right|=\underbrace{\left|f\right|}_{=m}-\underbrace{\left|f_{\mathbf{p}}\right|}_{=1}=m-1.. Thus, by the induction hypothesis, we can apply Lemma10.16tom−1m-1andf−f𝐩f-f_{\mathbf{p}}instead ofmmandff. As a result, we conclude that there existm−1m-1paths𝐩1,𝐩2,…,𝐩m−1\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m-1}fromsstottinDDsuch thatf𝐩1+f𝐩2+⋯+f𝐩m−1≤f−f𝐩f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m-1}}\leq f-f_{\mathbf{p}}. Consider thesem−1m-1paths𝐩1,𝐩2,…,𝐩m−1\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m-1}, and set𝐩m:=𝐩\mathbf{p}_{m}:=\mathbf{p}. Then,f𝐩1+f𝐩2+⋯+f𝐩m−1≤f−f𝐩=f−f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m-1}}\leq f-f_{\mathbf{p}}=f-f_{\mathbf{p}_{m}}(since𝐩=𝐩m\mathbf{p}=\mathbf{p}_{m}), so thatf𝐩1+f𝐩2+⋯+f𝐩m≤ff_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq f.

Thus, we have foundmmpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}fromsstottinDDsuch thatf𝐩1+f𝐩2+⋯+f𝐩m≤ff_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq f. But this is precisely what we wanted. Thus, the induction step is complete, and Lemma10.16is proved.

Proof 10.20(Proof of Theorem10.11.).

We makeDDinto a networkNN(with sourcessand sinktt) by assigning the capacity11to each arca∈Aa\in A. Clearly, a cut of this network is the same as what we call anss-tt-cut. Moreover, the capacityc⁡(S,S¯)c\left(S,\overline{S}\right)of a cut[S,S¯]\left[S,\overline{S}\right]is simply the size of this cut (since each arc has capacity11).

The max-flow-min-cut theorem (Theorem9.21) tells us that the maximum value of a flow equals the minimum capacity of a cut, i.e., the minimum size of anss-tt-cut (because, as we just explained, a cut is the same as anss-tt-cut, and its capacity is simply its size). It thus remains to show that the maximum value of a flow is the maximum number of pairwise arc-disjoint paths fromsstott. But this is easy by now:

•If you have a flowffof valuemm, then you can findmmpairwise arc-disjoint paths fromsstott(because Lemma10.16gives yoummpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}such thatf𝐩1+f𝐩2+⋯+f𝐩m≤ff_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq f, and the latter inequality tells you that thesemmpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}are arc-disjoint135135135Proof.Assume the contrary. Thus, thesemmpaths are not arc-disjoint. In other words, there exists an arcaathat is used by two paths𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}withi≠ji\neq j. Consider this arcaaand the corresponding indicesiiandjj. Sinceaais used by𝐩i\mathbf{p}_{i}, we havef𝐩i(a)=1f_{\mathbf{p}_{i}}\left(a\right)=1. Likewise,f𝐩j(a)=1f_{\mathbf{p}_{j}}\left(a\right)=1. However,f𝐩1+f𝐩2+⋯+f𝐩m≤ff_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\leq f, so that(f𝐩1+f𝐩2+⋯+f𝐩m)(a)\displaystyle\left(f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\right)\left(a\right)≤f(a)≤c(a)(by the capacity constraints)\displaystyle\leq f\left(a\right)\leq c\left(a\right)\ \ \ \ \ \ \ \ \ \ \left(\text{by the capacity constraints}\right)=1(since each arc has capacity1).\displaystyle=1\ \ \ \ \ \ \ \ \ \ \left(\text{since each arc has capacity }1\right).Thus,1\displaystyle 1≥(f𝐩1+f𝐩2+⋯+f𝐩m)(a)=f𝐩1(a)+f𝐩2(a)+⋯+f𝐩m(a)\displaystyle\geq\left(f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\right)\left(a\right)=f_{\mathbf{p}_{1}}\left(a\right)+f_{\mathbf{p}_{2}}\left(a\right)+\cdots+f_{\mathbf{p}_{m}}\left(a\right)≥f𝐩i(a)⏟=1+f𝐩j(a)⏟=1(sincef𝐩i(a)andf𝐩j(a)are two distinct addendsof the sumf𝐩1(a)+f𝐩2(a)+⋯+f𝐩m(a)(becausei≠j), and since all the remainingaddends are≥0(sincef𝐩(a)≥0for each path𝐩))\displaystyle\geq\underbrace{f_{\mathbf{p}_{i}}\left(a\right)}_{=1}+\underbrace{f_{\mathbf{p}_{j}}\left(a\right)}_{=1}\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since }f_{\mathbf{p}_{i}}\left(a\right)\text{ and }f_{\mathbf{p}_{j}}\left(a\right)\text{ are two distinct addends}\\ \text{of the sum }f_{\mathbf{p}_{1}}\left(a\right)+f_{\mathbf{p}_{2}}\left(a\right)+\cdots+f_{\mathbf{p}_{m}}\left(a\right)\\ \text{(because }i\neq j\text{), and since all the remaining}\\ \text{addends are }\geq 0\text{ (since }f_{\mathbf{p}}\left(a\right)\geq 0\text{ for each path }\mathbf{p}\text{)}\end{array}\right)=1+1>1,\displaystyle=1+1>1,which is absurd. This contradiction shows that our assumption was false, qed.). Thus, (the maximum number of pairwise arc-disjoint paths fromstot)\displaystyle\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\right)≥(the maximum value of a flow).\displaystyle\geq\left(\text{the maximum value of a flow}\right).(62)
•Conversely, if you havemmpairwise arc-disjoint paths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}fromsstott, then you obtain a flow of valuemm(namely,f𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}is such a flow136136136Proof.First, we observe that the mapf𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}satisfies the conservation constraints (because it is the sum of the functionsf𝐩1,f𝐩2,…,f𝐩mf_{\mathbf{p}_{1}},f_{\mathbf{p}_{2}},\ldots,f_{\mathbf{p}_{m}}, each of which satisfies the conservation constraints). Let us now check that it satisfies the capacity constraints. Indeed, leta∈Aa\in Abe an arc. Then,aabelongs to at most one of themmpaths𝐩1,𝐩2,…,𝐩m\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{m}(since thesemmpaths are arc-disjoint). In other words, at most one of themmnumbersf𝐩1(a),f𝐩2(a),…,f𝐩m(a)f_{\mathbf{p}_{1}}\left(a\right),\ f_{\mathbf{p}_{2}}\left(a\right),\ \ldots,\ f_{\mathbf{p}_{m}}\left(a\right)equals11; all the remaining numbers equal00. Hence, the sumf𝐩1(a)+f𝐩2(a)+⋯+f𝐩m(a)f_{\mathbf{p}_{1}}\left(a\right)+f_{\mathbf{p}_{2}}\left(a\right)+\cdots+f_{\mathbf{p}_{m}}\left(a\right)of thesemmnumbers equals either11or00; in either case, we thus havef𝐩1(a)+f𝐩2(a)+⋯+f𝐩m(a)∈{0,1}f_{\mathbf{p}_{1}}\left(a\right)+f_{\mathbf{p}_{2}}\left(a\right)+\cdots+f_{\mathbf{p}_{m}}\left(a\right)\in\left\{0,1\right\}. Now,(f𝐩1+f𝐩2+⋯+f𝐩m)(a)=f𝐩1(a)+f𝐩2(a)+⋯+f𝐩m(a)∈{0,1},\left(f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\right)\left(a\right)=f_{\mathbf{p}_{1}}\left(a\right)+f_{\mathbf{p}_{2}}\left(a\right)+\cdots+f_{\mathbf{p}_{m}}\left(a\right)\in\left\{0,1\right\},so that0≤(f𝐩1+f𝐩2+⋯+f𝐩m)(a)≤1=c⁡(a)0\leq\left(f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\right)\left(a\right)\leq 1=c\left(a\right)(since each arc has capacity11). Since we have proved this for each arca∈Aa\in A, we thus have shown that the mapf𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}satisfies the capacity constraints. Hence, this map is a flow (since it also satisfies the conservation constraints). It remains to show that the value of this flow ismm. But this is easy: For any flowsg1,g2,…,gkg_{1},g_{2},\ldots,g_{k}, we have|g1+g2+⋯+gk|=|g1|+|g2|+⋯+|gk|\left|g_{1}+g_{2}+\cdots+g_{k}\right|=\left|g_{1}\right|+\left|g_{2}\right|+\cdots+\left|g_{k}\right|(this is straightforward to see from the definition of value). Thus,|f𝐩1+f𝐩2+⋯+f𝐩m|=|f𝐩1|+|f𝐩2|+⋯+|f𝐩m|=∑k=1m|f𝐩k|⏟=1=∑k=1m1=m.\left|f_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}\right|=\left|f_{\mathbf{p}_{1}}\right|+\left|f_{\mathbf{p}_{2}}\right|+\cdots+\left|f_{\mathbf{p}_{m}}\right|=\sum\limits_{k=1}^{m}\underbrace{\left|f_{\mathbf{p}_{k}}\right|}_{=1}=\sum\limits_{k=1}^{m}1=m.In other words, the value of the flowf𝐩1+f𝐩2+⋯+f𝐩mf_{\mathbf{p}_{1}}+f_{\mathbf{p}_{2}}+\cdots+f_{\mathbf{p}_{m}}ismm.). Thus, (the maximum value of a flow)\displaystyle\left(\text{the maximum value of a flow}\right)≥(the maximum number of pairwise arc-disjoint paths fromstot).\displaystyle\geq\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\right).

Combining this last inequality with (62), we obtain

(the maximum number of pairwise arc-disjoint paths fromstot)\displaystyle\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\right)=(the maximum value of a flow)\displaystyle=\left(\text{the maximum value of a flow}\right)=(the minimum size of ans-t-cut)(as we have proved before).\displaystyle=\left(\text{the minimum size of an }s\text{-}t\text{-cut}\right)\ \ \ \ \ \ \ \ \ \ \left(\text{as we have proved before}\right).Thus, Theorem10.11is proved.

Theorem10.11can also be proved without using network flows (see, e.g.,[Schrij17, Corollary 4.1b]for such a proof).

Proof 10.21(Proof of Theorem10.6.).

Letxxdenote the maximum number of pairwise arc-disjoint paths fromsstott.

Letncn_{c}denote the minimum size of anss-tt-cut.

Letnsn_{s}denote the minimum size of anss-tt-arc-separator.137137137If you are wondering why we chose the baroque notations “xx”, “ncn_{c}” and “nsn_{s}” for these three numbers: The letter “xx” appears in “maximum”, whereas the letter “nn” appears in “minimum”. The subscripts “cc” and “ss” should be reasonably clear.

Theorem10.11says thatx=ncx=n_{c}. Our goal is to prove thatx=nsx=n_{s}.

Remark10.9shows that anyss-tt-cut is anss-tt-arc-separator. Thus,ns≤ncn_{s}\leq n_{c}.

The inequalityx≤nsx\leq n_{s}follows easily from the pigeonhole principle138138138Proof.We know that there existxxpairwise arc-disjoint paths fromsstott(by the definition ofxx). Let𝐩1,𝐩2,…,𝐩x\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{x}be thesexxpaths. We know that there exists anss-tt-arc-separator of sizensn_{s}(by the definition ofnsn_{s}). LetBBbe thisss-tt-arc-separator. Thus, each path fromsstottcontains at least one arc fromBB(by the definition of anss-tt-arc-separator). Hence, in particular, each of thexxpaths𝐩1,𝐩2,…,𝐩x\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{x}contains at least one arc fromBB. These altogetherxxarcs must be distinct (since thexxpaths𝐩1,𝐩2,…,𝐩x\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{x}are arc-disjoint); thus, we have found at leastxxarcs that belong toBB. This shows that|B|≥x\left|B\right|\geq x. However,BBhas sizensn_{s}; in other words, we have|B|=ns\left|B\right|=n_{s}. Thus,ns=|B|≥xn_{s}=\left|B\right|\geq x, so thatx≤nsx\leq n_{s}.. Combining this withns≤nc=xn_{s}\leq n_{c}=x(sincex=ncx=n_{c}), we obtainx=nsx=n_{s}. Thus, Theorem10.6is proved.

Exercise 10.22.

LetDDbe a balanced multidigraph. Letssandttbe two vertices ofDD. Letk∈ℕk\in\mathbb{N}. Assume thatDDhaskkpairwise arc-disjoint paths fromsstott. Show thatDDhaskkpairwise arc-disjoint paths fromtttoss.

Exercise 10.23.

LetDDbe a multidigraph. Letk∈ℕk\in\mathbb{N}. Letuu,vvandwwbe three vertices ofDD. Assume that there existkkarc-disjoint paths fromuutovv. Assume furthermore that there existkkarc-disjoint paths fromvvtoww.

Prove that there existkkarc-disjoint paths fromuutoww.

[Note:Ifu=wu=w, then the trivial path(u)\left(u\right)counts as being arc-disjoint from itself (so in this case, there exist arbitrarily many arc-disjoint paths fromuutoww).]

[Solution:This is Exercise 3 on midterm #2 from my Spring 2017 course (except that it is stated for multidigraphs instead of simple digraphs); seethe course pagefor solutions.]

We can also extend the arc-Menger theorem to paths between different pairs of vertices:

Theorem 10.24(arc-Menger theorem for directed graphs, multi-terminal version).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letXXandYYbe two disjoint subsets ofVV.

Apath fromXXtoYYshall mean a path whose starting point belongs toXXand whose ending point belongs toYY.

AnXX-YY-cutshall mean a subset ofAAthat has the form[S,S¯]\left[S,\overline{S}\right], whereSSis a subset ofVVthat satisfiesX⊆SX\subseteq SandY⊆S¯Y\subseteq\overline{S}.

Then, the maximum number of pairwise arc-disjoint paths fromXXtoYYequals the minimum size of anXX-YY-cut.

Example 10.25.

Here is an example of a digraphD=(V,A,ψ)D=\left(V,A,\psi\right), with two disjoint subsetsXXandYYofVVdrawn as ovals:

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\lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#0000FF} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 5 -5 L 72.56 -72.56}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#0000FF} \lxSVG@transformcm{0.7071}{-0.7071}{0.7071}{0.7071}{52.8652pt}{-52.8652pt}\lxSVG@begingroup@{transform=matrix(0.7071 -0.7071 0.7071 0.7071 73.15 -73.15)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .According to Theorem10.24, the minimum size of anXX-YY-cut must thus also be22. And indeed, here is such anXX-YY-cut:

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Proof 10.26(Proof of Theorem10.24.).

We transform our digraphD=(V,A,ψ)D=\left(V,A,\psi\right)into a new multidigraphD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right)as follows:

•We replace all the vertices inXXby a single (new) vertexss, and replace all the vertices inYYby a single (new) vertextt. (Thus, formally speaking, we setV′=(V∖(X∪Y))∪{s,t}V^{\prime}=\left(V\setminus\left(X\cup Y\right)\right)\cup\left\{s,t\right\}, wheressandttare two objects not inVV.) For any vertexp∈Vp\in V, we define a vertexp′∈V′p^{\prime}\in V^{\prime}by p′={s,ifp∈X;t,ifp∈Y;p,otherwise.p^{\prime}=\begin{cases}s,&\text{if }p\in X;\\ t,&\text{if }p\in Y;\\ p,&\text{otherwise.}\end{cases}We refer to this vertexp′p^{\prime}as theprojectionofpp.
•We keep all the arcs ofDDaround, but we replace all their endpoints (i.e., sources and targets) by their projections (thus, any endpoint inXXgets replaced byss, and any endpoint inYYgets replaced bytt, while an endpoint that belongs neither toXXnor toYYstays unchanged). For example, an arc with source inXXbecomes an arc with source inss. (Formally speaking, this means the following: We setA′=AA^{\prime}=Aand we define the mapψ′:A′→V′×V′\psi^{\prime}:A^{\prime}\rightarrow V^{\prime}\times V^{\prime}as follows: For anya∈A′=Aa\in A^{\prime}=A, we setψ′(a)=(u′,v′)\psi^{\prime}\left(a\right)=\left(u^{\prime},v^{\prime}\right), where(u,v)=ψ⁡(a)\left(u,v\right)=\psi\left(a\right).)

For instance, ifDDis the digraph from Example10.25, thenD′D^{\prime}looks as follows:

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Now, Theorem10.11(applied toD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right)instead ofD=(V,A,ψ)D=\left(V,A,\psi\right)) shows that the maximum number of pairwise arc-disjoint paths fromsstottinD′D^{\prime}equals the minimum size of anss-tt-cut inD′D^{\prime}.

Let us now connect this with the claim that we want to prove. It is easy to see that the minimum size of anss-tt-cut inD′D^{\prime}equals the minimum size of anXX-YY-cut inDD(indeed, thess-tt-cuts inD′D^{\prime}are precisely theXX-YY-cuts inDD139139139In more detail:•Anyss-tt-cut inD′D^{\prime}has the form[S,S¯]\left[S,\overline{S}\right]for some subsetSSofV′V^{\prime}satisfyings∈Ss\in Sandt∉St\notin S; it is therefore equal to the set[S′,S′¯]\left[S^{\prime},\overline{S^{\prime}}\right], whereS′S^{\prime}is the subset ofVVgiven byS′:=(S∖{s})∪XS^{\prime}:=\left(S\setminus\left\{s\right\}\right)\cup X. Therefore, it is anXX-YY-cut inDD.•Conversely, anyXX-YY-cut inDDhas the form[S,S¯]\left[S,\overline{S}\right]for some subsetSSofVVsatisfyingX⊆SX\subseteq SandY⊆S¯Y\subseteq\overline{S}; it is therefore equal to the set[S′,S′¯]\left[S^{\prime},\overline{S^{\prime}}\right], whereS′S^{\prime}is the subset ofV′V^{\prime}given byS′:=(S∖X)∪{s}S^{\prime}:=\left(S\setminus X\right)\cup\left\{s\right\}. Therefore, it is anss-tt-cut inD′D^{\prime}.). If we can also show that the maximum number of pairwise arc-disjoint paths fromsstottinD′D^{\prime}equals the maximum number of pairwise arc-disjoint paths fromXXtoYYinDD, then the result of the preceding paragraph will thus become the claim of Theorem10.24, so we will be done.

So how can we show that the maximum number of pairwise arc-disjoint paths fromsstottinD′D^{\prime}equals the maximum number of pairwise arc-disjoint paths fromXXtoYYinDD? It would be easy if there was a well-behaved bijection between the former paths and the latter paths that preserves the arcs of any path, but this is not quite the case. Each path fromXXtoYYinDDbecomes awalkfromsstottinD′D^{\prime}if we replace each of its vertices by its projection. However, the latter walk is not necessarily a path, since different vertices can have the same projection.

Fortunately, this is easy to fix. If we havekkpairwise arc-disjoint paths fromXXtoYYinDD, then we can turn them intokkpairwise arc-disjointwalksfromsstottinD′D^{\prime}, and then we also obtainkkpairwise arc-disjointpathsfromsstottinD′D^{\prime}(since any walk fromsstottcontains a path fromsstott). Thus,

(the maximum number of pairwise arc-disjoint paths fromstotinD′)\displaystyle\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\text{ in }D^{\prime}\right)≥(the maximum number of pairwise arc-disjoint paths fromXtoYinD).\displaystyle\geq\left(\text{the maximum number of pairwise arc-disjoint paths from }X\text{ to }Y\text{ in }D\right).Conversely, if we havekkpairwise arc-disjoint paths fromsstottinD′D^{\prime}, then we can “lift” thesekkpaths back to the digraphDD(preserving the arcs, and replacing the verticesssandttby appropriate vertices inXXandYYto make them belong to the right arcs), and thus obtainkkpairwise arc-disjoint paths fromXXtoYYinDD. Thus,

(the maximum number of pairwise arc-disjoint paths fromXtoYinD)\displaystyle\left(\text{the maximum number of pairwise arc-disjoint paths from }X\text{ to }Y\text{ in }D\right)≥(the maximum number of pairwise arc-disjoint paths fromstotinD′).\displaystyle\geq\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\text{ in }D^{\prime}\right).Combining these two inequalities, we obtain

10.1.2The edge-Menger theorem for undirected graphs

We shall now state analogues of Theorem10.6and Theorem10.11for undirected graphs. First, the unsurprising definitions:

Definition 10.27.

Two walks𝐩\mathbf{p}and𝐪\mathbf{q}in a graph are said to beedge-disjointif they have no edge in common.

Definition 10.28.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letssandttbe two vertices ofGG. A subsetBBofEEis said to be anss-tt-edge-separatorif each path fromsstottcontains at least one edge fromBB. Equivalently, a subsetBBofEEis said to be anss-tt-edge-separatorif the multigraph(V,E∖B,φ∣E∖B)\left(V,\ E\setminus B,\ \varphi\mid_{E\setminus B}\right)has no path fromsstott(in other words, removing fromGGall edges contained inBBdestroys all paths fromsstott).

Now comes the analogue of Theorem10.6:

Theorem 10.29(edge-Menger theorem for undirected graphs, version 1).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letssandttbe two distinct vertices ofGG. Then, the maximum number of pairwise edge-disjoint paths fromsstottequals the minimum size of anss-tt-edge-separator.

To state the analogue of Theorem10.11, we need to first adopt Definition10.8to undirected graphs:

Definition 10.30.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letssandttbe two distinct vertices ofGG.

(a)For each subsetSSofVV, we setS¯:=V∖S\overline{S}:=V\setminus Sand [S,S¯]und\displaystyle\left[S,\overline{S}\right]_{\operatorname*{und}}:={e∈E∣one endpoint ofebelongs toS,\displaystyle:=\left\{e\in E\ \mid\ \text{one endpoint of }e\text{ belongs to }S\text{,}\right.while the other belongs toS¯}.\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{while the other belongs to }\overline{S}\right\}.
(b)An (undirected)ss-tt-cutmeans a subset ofEEthat has the form[S,S¯]und\left[S,\overline{S}\right]_{\operatorname*{und}}, whereSSis a subset ofVVthat satisfiess∈Ss\in Sandt∉St\notin S.

The following remark is an analogue of Remark10.9:

Proof 10.32.

Analogous to the proof of Remark10.9.

And here is the analogue of Theorem10.11:

Theorem 10.33(edge-Menger theorem for undirected graphs, version 2).

Proof 10.34(Proof of Theorem10.33.).

We shall not prove this from scratch, but rather derive this from the directed version (Theorem10.11).

Namely, we apply Theorem10.11to140140140Recall thatGbidirG^{\operatorname*{bidir}}is the multidigraph obtained fromGGby replacing each edge by two arcs in opposite directions. (If the edge has endpointsuuandvv, then one of the two arcs has sourceuuand targetvv, while the other has sourcevvand targetuu.) See Definition4.15for a formal definition.D=GbidirD=G^{\operatorname*{bidir}}. We thus see that the maximum number of pairwise arc-disjoint paths fromsstott(inGbidirG^{\operatorname*{bidir}}) equals the minimum size of anss-tt-cut (inGbidirG^{\operatorname*{bidir}}). This is similar to the claim that we want to prove, but not quite the same statement, becauseGbidirG^{\operatorname*{bidir}}is notGG. To obtain the claim that we want to prove, we must prove the following two claims:

Claim 1:The maximum number of pairwise arc-disjoint paths fromsstott(inGbidirG^{\operatorname*{bidir}}) equals the maximum number of pairwise edge-disjoint paths fromsstott(inGG).

Claim 2:The minimum size of a directedss-tt-cut141141141A “directedss-tt-cut” here simply means anss-tt-cut in a digraph.(inGbidirG^{\operatorname*{bidir}}) equals the minimum size of an (undirected)ss-tt-cut (inGG).

Claim 2 is very easy to verify, since the directedss-tt-cuts inGbidirG^{\operatorname*{bidir}}are essentially the same as the undirectedss-tt-cuts inGG142142142In more detail: IfSSis a subset ofVVthat satisfiess∈Ss\in Sandt∉St\notin S, then the directedss-tt-cut[S,S¯]\left[S,\overline{S}\right]inGbidirG^{\operatorname*{bidir}}and the undirectedss-tt-cut[S,S¯]und\left[S,\overline{S}\right]_{\operatorname*{und}}inGGhave the same size (because each edge in[S,S¯]und\left[S,\overline{S}\right]_{\operatorname*{und}}corresponds to exactly one arc in[S,S¯]\left[S,\overline{S}\right]). Thus, the sizes of the directedss-tt-cuts inGbidirG^{\operatorname*{bidir}}are exactly the sizes of the undirectedss-tt-cuts inGG. In particular, the minimum size of a former cut equals the minimum size of a latter cut. This proves Claim 2..

It remains to verify Claim 1. The simplest approach is to argue that each path fromsstottinGbidirG^{\operatorname*{bidir}}becomes a path fromsstottinGG(just replace each arc of the path by the corresponding undirected edge). Unfortunately, this alone does not suffice, since two arc-disjoint paths inGbidirG^{\operatorname*{bidir}}won’t necessarily become edge-disjoint paths inGG. Here is an example of how this can go wrong (imagine that the two arcs betweenuuandvvcome from the same edge ofGG, and the two paths are marked red and blue):

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{}{}{}{{{}}{{\lx@inpgf@ignorespaces}}{{}}} {{{}}{{\lx@inpgf@ignorespaces}}{{}}} {\lx@inpgf@ignorespaces}{{}}{}{{\lx@inpgf@ignorespaces}}{\lx@inpgf@ignorespaces}{{\lx@inpgf@ignorespaces}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#0000FF} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 5.99 -10.38 C 39 -67.49 142.19 -67.45 174.32 -11.75}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#0000FF} \lxSVG@transformcm{0.49963}{0.86623}{-0.86623}{0.49963}{126.2823pt}{-7.96985pt}\lxSVG@begingroup@{transform=matrix(0.49963 0.86623 -0.86623 0.49963 174.74 -11.03)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(63)If we replace each arc by the corresponding edge here, then the two paths will no longer be edge-disjoint (since the edge betweenuuandvvwill be used by both paths).

However, this kind of situation can be averted. To do this, we letkkbe the maximum number of pairwise arc-disjoint paths fromsstottinGbidirG^{\operatorname*{bidir}}. We now choosekkpairwise arc-disjoint paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}fromsstottinGbidirG^{\operatorname*{bidir}}in such a way that theirtotal length(i.e., the sum of the lengths of𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}) isas small as possible. Then, it is easy to see that these paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}become edge-disjoint paths inGGwhen we replace each arc by the corresponding edge.

[Proof:Assume the contrary. Thus, two of these paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}end up sharing an edge when we replace each arc by the corresponding edge. Let𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}be these two paths (wherei≠ji\neq j, of course). Leteebe the edge that they end up sharing, and letuuandvvbe the two endpoints ofee, in the order in which they appear on𝐩i\mathbf{p}_{i}. Hence, the path𝐩i\mathbf{p}_{i}uses the edgeee(or, more precisely, one of the two arcs ofGbidirG^{\operatorname*{bidir}}corresponding toee) to get fromuutovv.

Since the paths𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}are arc-disjoint, they cannot both use the edgeeein the same direction (because this would mean that𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}share the same arc ofGbidirG^{\operatorname*{bidir}}). Hence, the path𝐩j\mathbf{p}_{j}uses the edgeeeto get fromvvtouu(since the path𝐩i\mathbf{p}_{i}uses the edgeeeto get fromuutovv). Hence, the paths𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}have the following forms:

𝐩i\displaystyle\mathbf{p}_{i}=(…,u,e1,v,…);\displaystyle=\left(\ldots,u,e_{1},v,\ldots\right);𝐩j\displaystyle\mathbf{p}_{j}=(…,v,e2,u,…),\displaystyle=\left(\ldots,v,e_{2},u,\ldots\right),wheree1e_{1}ande2e_{2}are the two arcs ofGbidirG^{\operatorname*{bidir}}that correspond to the edgeee. Now, let us replace the two paths𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}by two new walks143143143Here is an illustration:𝐩i(in red) and𝐩j(in blue):suvte1e2𝐩i′(in red) and𝐩j′(in blue):suvt.\begin{tabular}[c]{|cc|}\hline\cr $\\mathbf\{p\}\_\{i\}$ (in red) and $\\mathbf\{p\}\_\{j\}$ (in blue):&\\ \hline\cr $\\mathbf\{p\}\_\{i\}^\{\\prime\}$ (in red) and $\\mathbf\{p\}\_\{j\}^\{\\prime\}$ (in blue):&\\ \hline\cr\end{tabular}\ \ \ .(The wavy arrows stand not for single arcs, but for sequences of multiple arcs.)

𝐩i′\displaystyle\mathbf{p}_{i}^{\prime}=(…⏟the part of𝐩ibeforeu,u,…⏟the part of𝐩jafteru)and\displaystyle=\left(\underbrace{\ldots}_{\begin{subarray}{c}\text{the part of }\mathbf{p}_{i}\\ \text{before }u\end{subarray}},u,\underbrace{\ldots}_{\begin{subarray}{c}\text{the part of }\mathbf{p}_{j}\\ \text{after }u\end{subarray}}\right)\ \ \ \ \ \ \ \ \ \ \text{and}𝐩j′\displaystyle\mathbf{p}_{j}^{\prime}=(…⏟the part of𝐩jbeforev,v,…⏟the part of𝐩iafterv).\displaystyle=\left(\underbrace{\ldots}_{\begin{subarray}{c}\text{the part of }\mathbf{p}_{j}\\ \text{before }v\end{subarray}},v,\underbrace{\ldots}_{\begin{subarray}{c}\text{the part of }\mathbf{p}_{i}\\ \text{after }v\end{subarray}}\right).These walks𝐩i′\mathbf{p}_{i}^{\prime}and𝐩j′\mathbf{p}_{j}^{\prime}are two walks fromsstott, and they don’t use any arcs that were not already used by𝐩i\mathbf{p}_{i}or𝐩j\mathbf{p}_{j}. Thus, they are arc-disjoint from all of the paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}except for𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}. Moreover, they are arc-disjoint from each other (since𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}were arc-disjoint, and since the arcs of any path are distinct). Furthermore, their total length is smaller by22than the total length of𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}(since they use all the arcs of𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}except fore1e_{1}ande2e_{2}). They are not necessarily paths, but we can turn them into paths fromsstottby successively removing cycles (as in the proof of Corollary4.29). If we do this, we end up with two paths𝐩i′′\mathbf{p}_{i}^{\prime\prime}and𝐩j′′\mathbf{p}_{j}^{\prime\prime}fromsstottthat are arc-disjoint from each other and from all of the paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}except for𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}, and whose total length is smaller by at least22than the total length of𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}.

Thus, if we replace𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}by these two paths𝐩i′′\mathbf{p}_{i}^{\prime\prime}and𝐩j′′\mathbf{p}_{j}^{\prime\prime}(while leaving the remainingk−2k-2of ourkkpaths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}unchanged), then we obtainkkmutually arc-disjoint paths fromsstottwhose total length is smaller than the total length of our originalkkpaths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}. However, this is absurd, because we chose our originalkkpairwise arc-disjoint paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}fromsstottin such a way that their total length is as small as possible. This contradiction shows that our assumption was wrong. Thus, we have proved that the paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}become edge-disjoint paths inGGwhen we replace each arc by the corresponding edge.]

Hence, we have foundkkpairwise edge-disjoint paths fromsstottinGG(namely, thekkpaths that are obtained from the paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}when we replace each arc by the corresponding edge). This shows that

(the maximum number of pairwise edge-disjoint paths fromstotinG)\displaystyle\left(\text{the maximum number of pairwise edge-disjoint paths from }s\text{ to }t\text{ in }G\right)≥k\displaystyle\geq k=(the maximum number of pairwise arc-disjoint paths fromstotinGbidir)\displaystyle=\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\text{ in }G^{\operatorname*{bidir}}\right)(by the definition ofkk). Conversely, we can easily see that

(the maximum number of pairwise arc-disjoint paths fromstotinGbidir)\displaystyle\left(\text{the maximum number of pairwise arc-disjoint paths from }s\text{ to }t\text{ in }G^{\operatorname*{bidir}}\right)≥(the maximum number of pairwise edge-disjoint paths fromstotinG)\displaystyle\geq\left(\text{the maximum number of pairwise edge-disjoint paths from }s\text{ to }t\text{ in }G\right)(since there is an obvious way to transform paths inGGinto paths inGbidirG^{\operatorname*{bidir}}(just replace each edge by one of the two corresponding arcs ofGbidirG^{\operatorname*{bidir}}), and applying this transformation to edge-disjoint paths ofGGyields arc-disjoint paths ofGbidirG^{\operatorname*{bidir}}). Combining these two inequalities, we obtain

Proof 10.35(Proof of Theorem10.29.).

This can be derived from Theorem10.33and Remark10.31in the same way as we derived Theorem10.6from Theorem10.11and Remark10.9.

Exercise 10.36.

LetGGbe a multigraph such that each vertex ofGGhas even degree. Letssandttbe two distinct vertices ofGG. Prove that the maximum number of pairwise edge-disjoint paths fromsstottis even.

10.1.3The vertex-Menger theorem for directed graphs

The Menger theorems we have seen so far have been concerned with paths not having arcs in common. What if we want to avoid common vertices too?

Definition 10.37.

Let𝐩\mathbf{p}be a path of some graph or digraph. Then, anintermediate vertexof𝐩\mathbf{p}shall mean a vertex of𝐩\mathbf{p}that is neither the starting point nor the ending point of𝐩\mathbf{p}.

Definition 10.38.

Two paths𝐩\mathbf{p}and𝐪\mathbf{q}in a graph or digraph are said to beinternally vertex-disjointif they have no common intermediate vertices.

Example 10.39.

The two paths𝐩\mathbf{p}and𝐪\mathbf{q}in Example10.2are arc-disjoint, but not internally vertex-disjoint.

Here are two internally vertex-disjoint paths𝐩\mathbf{p}and𝐪\mathbf{q}:

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\hss}}\lxSVG@closescope\endpgfpicture}}\ \ . One trivial case of internally vertex-disjoint paths is a path of length≤1\leq 1: Namely, a path of length≤1\leq 1is internally vertex-disjoint from any path, including itself (since it has no intermediate vertices).

Definition 10.40.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letssandttbe two vertices ofDD. A subsetWWofV∖{s,t}V\setminus\left\{s,t\right\}is said to be aninternalss-tt-vertex-separatorif each path fromsstottcontains at least one vertex fromWW. Equivalently, a subsetWWofV∖{s,t}V\setminus\left\{s,t\right\}is said to be aninternalss-tt-vertex-separatorif the induced subdigraph ofDDon the setV∖WV\setminus Whas no path fromsstott(in other words, removing fromDDall vertices contained inWWdestroys all paths fromsstott).

Example 10.41.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be the following multidigraph:

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\lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the sets{a,b}\left\{a,b\right\}and{a,c}\left\{a,c\right\}are internalss-tt-vertex-separators (indeed, removing the verticesaaandbbcuts offssfrom the rest of the digraph, whereas removing the verticesaaandccdoes the same tott), but the sets{a}\left\{a\right\}and{b,c}\left\{b,c\right\}are not (since the path fromsstottviaccandbbavoidsaa, whereas the path fromsstottviaaaavoidsbbandcc).

Example 10.42.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Letssandttbe two distinct vertices ofDD. Then:

(a)The empty set∅\varnothingis an internalss-tt-vertex-separator if and only ifDDhas no path fromsstott.
(b)IfDDhas no arc with sourcessand targettt, then the setV∖{s,t}V\setminus\left\{s,t\right\}is an internalss-tt-vertex-separator (since any path fromsstottcontains at least one intermediate vertex, and such a vertex must belong toV∖{s,t}V\setminus\left\{s,t\right\}).
(c)IfDDhas an arc with sourcessand targettt, then there exists no internalss-tt-vertex-separator (since the “direct” length-11path fromsstottcontains no vertices besidesssandtt).

Now, we state the analogue of Theorem10.6and Theorem10.11for internally vertex-disjoint paths:

Theorem 10.43(vertex-Menger theorem for directed graphs).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letssandttbe two distinct vertices ofDD. Assume thatDDhas no arc with sourcessand targettt. Then, the maximum number of pairwise internally vertex-disjoint paths fromsstottequals the minimum size of an internalss-tt-vertex-separator.

Example 10.44.

LetDDbe the following multidigraph:

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\lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the maximum number of pairwise internally vertex-disjoint paths fromsstottis22. Indeed, the following picture shows22such paths in red and blue, respectively:

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-3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Why can there be no33such paths? This is not obvious from a quick look, but can be easily derived from Theorem10.43. Indeed, Theorem10.43yields that the maximum number of pairwise internally vertex-disjoint paths fromsstottequals the minimum size of an internalss-tt-vertex-separator. Thus, if the former number was larger than22, then so would be the latter number. But this cannot be the case, since the22-element set{a,f}\left\{a,f\right\}is easily checked to be an internalss-tt-vertex-separator. Hence, we see that both of these numbers are22.

Example 10.45.

Consider again the digraphDDfrom Example10.12. In that example, we found33pairwise arc-disjoint paths fromsstott. These33paths are not internally vertex-disjoint (in fact, the brown path has non-starting and non-ending vertices in common with both the red and the blue path). However, there do exist33pairwise internally vertex-disjoint paths fromsstott. Can you find them?

Proof 10.46(Proof of Theorem10.43.).

We will again derive this from the arc-Menger theorem (Theorem10.11), applied to an appropriate multidigraphD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right).

What is this multidigraphD′D^{\prime}? The idea is to modify the digraphDDin such a way that paths having a common vertex become paths having a common arc. The most natural way to achieve this is to “stretch out” each vertexvvofDDinto a little arc. In order to do this in a systematic manner, we replace each vertexvvofDDby two distinct verticesviv^{i}andvov^{o}(the notations stand for “vv-in” and “vv-out”, and we can think ofviv^{i}as the “entrance” tovvwhilevov^{o}is the “exit” fromvv) and an arcviov^{io}that goes fromviv^{i}tovov^{o}. Any existing arcaaofDDbecomes a new arcaoia^{oi}ofDD, whose source and target are specified as follows: Ifaahas sourceuuand targetvv, thenaoia^{oi}will have sourceuou^{o}and targetviv^{i}.

Here is an example: If

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sos^{o}xox^{o}yoy^{o}zoz^{o}wow^{o}tot^{o}sis^{i}xix^{i}yiy^{i}ziz^{i}wiw^{i}tit^{i}D′=D^{\prime}=aoi\color[rgb]{0,0,1}{a^{oi}}boi\color[rgb]{0,0,1}{b^{oi}}doi\color[rgb]{0,0,1}{d^{oi}}eoi\color[rgb]{0,0,1}{e^{oi}}foi\color[rgb]{0,0,1}{f^{oi}}coi\color[rgb]{0,0,1}{c^{oi}}goi\color[rgb]{0,0,1}{g^{oi}}hoi\color[rgb]{0,0,1}{h^{oi}}ioi\color[rgb]{0,0,1}{i^{oi}}sio\color[rgb]{1,0,0}{s^{io}}xio\color[rgb]{1,0,0}{x^{io}}yio\color[rgb]{1,0,0}{y^{io}}zio\color[rgb]{1,0,0}{z^{io}}wio\color[rgb]{1,0,0}{w^{io}}tio\color[rgb]{1,0,0}{t^{io}}(where all arcs of the formaoia^{oi}fora∈Aa\in Ahave been colored blue, whereas all arcs of the formviov^{io}forv∈Vv\in Vhave been colored red). ThisD′D^{\prime}satisfies the property that we want it to satisfy: For instance, the two paths

(s,a,x,d,y,g,t)and\displaystyle\left(s,a,x,d,y,g,t\right)\ \ \ \ \ \ \ \ \ \ \text{and}(s,b,z,i,y,e,w,h,t)\displaystyle\left(s,b,z,i,y,e,w,h,t\right)ofDDhave the vertexyyin common, so the corresponding two paths

(so,aoi,xi,xio,xo,doi,yi,yio,yo,goi,ti)and\displaystyle\left(s^{o},a^{oi},x^{i},x^{io},x^{o},d^{oi},y^{i},y^{io},y^{o},g^{oi},t^{i}\right)\ \ \ \ \ \ \ \ \ \ \text{and}(so,boi,zi,zio,zo,ioi,yi,yio,yo,eoi,wi,wio,wo,hoi,ti)\displaystyle\left(s^{o},b^{oi},z^{i},z^{io},z^{o},i^{oi},y^{i},y^{io},y^{o},e^{oi},w^{i},w^{io},w^{o},h^{oi},t^{i}\right)ofD′D^{\prime}have the arcyioy^{io}in common. If you think ofDDas a railway network with the vertices being train stations and the arcs being train rides, thenD′D^{\prime}is a more detailed version of this network that records a change of platforms as an arc as well.

Here is a formal definition of the multidigraphD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right)in full generality:

•We replace each vertexvvofDDby two new verticesviv^{i}andvov^{o}. We callviv^{i}an “in-vertex” andvov^{o}an “out-vertex”. The vertex set ofD′D^{\prime}will be the set V′:={vi∣v∈V}⏟in-vertices∪{vo∣v∈V}⏟out-vertices.V^{\prime}:=\underbrace{\left\{v^{i}\ \mid\ v\in V\right\}}_{\text{in-vertices}}\cup\underbrace{\left\{v^{o}\ \mid\ v\in V\right\}}_{\text{out-vertices}}.
•Each arca∈Aa\in Ais replaced by a new arcaoia^{oi}, which is defined as follows: If the arca∈Aa\in Ahas sourceuuand targetvv, then we replace it by a new arcaoia^{oi}, which has sourceuou^{o}and targetviv^{i}. This arcaoia^{oi}will be called an “arc-arc” ofD′D^{\prime}(since it originates from an arc ofDD).
•For any vertexv∈Vv\in VofDD, we introduce a new arcviov^{io}, which has sourceviv^{i}and targetvov^{o}. This arcviov^{io}will be called a “vertex-arc” ofD′D^{\prime}(since it originates from a vertex ofDD).
•The arc set ofD′D^{\prime}will be the set A′:={aoi∣a∈A}⏟the arc-arcs∪{vio∣v∈V}⏟the vertex-arcs.A^{\prime}:=\underbrace{\left\{a^{oi}\ \mid\ a\in A\right\}}_{\text{the arc-arcs}}\cup\underbrace{\left\{v^{io}\ \mid\ v\in V\right\}}_{\text{the vertex-arcs}}.The mapψ′:A′→V′×V′\psi^{\prime}:A^{\prime}\rightarrow V^{\prime}\times V^{\prime}is defined as we already explained: - –For any arc-arcaoi∈Aa^{oi}\in A, we letψ′(aoi):=(uo,vi)\psi^{\prime}\left(a^{oi}\right):=\left(u^{o},v^{i}\right), where(u,v)=ψ⁡(a)\left(u,v\right)=\psi\left(a\right). - –For any vertex-arcvio∈Vv^{io}\in V, we letψ′(vio):=(vi,vo)\psi^{\prime}\left(v^{io}\right):=\left(v^{i},v^{o}\right).

Note thatD′D^{\prime}is something like a “bipartite digraph”: Each of its arcs goes either from an out-vertex to an in-vertex or vice versa. Namely, each arc-arc goes from an out-vertex to an in-vertex, whereas each vertex-arc goes from an in-vertex to an out-vertex. Thus, on any walk ofD′D^{\prime}, the arc-arcs and the vertex-arcs have to alternate.

If𝐩=(v0,a1,v1,a2,v2,…,ak,vk)\mathbf{p}=\left(v_{0},a_{1},v_{1},a_{2},v_{2},\ldots,a_{k},v_{k}\right)is any nontrivial144144144We say that a path isnontrivialif it has length>0>0.path ofDD, then we can define a corresponding path𝐩oi\mathbf{p}^{oi}ofD′D^{\prime}by

𝐩oi:=(v0o,a1oi,v1i,v1io,v1o,a2oi,v2i,v2io,v2o,…,akoi,vki).\mathbf{p}^{oi}:=\left(v_{0}^{o},a_{1}^{oi},v_{1}^{i},v_{1}^{io},v_{1}^{o},a_{2}^{oi},v_{2}^{i},v_{2}^{io},v_{2}^{o},\ldots,a_{k}^{oi},v_{k}^{i}\right).This path𝐩oi\mathbf{p}^{oi}is obtained from𝐩\mathbf{p}by

•replacing the starting pointv0v_{0}byv0ov_{0}^{o};
•replacing the ending pointvkv_{k}byvkiv_{k}^{i};
•replacing each other vertexvjv_{j}by the sequencevji,vjio,vjov_{j}^{i},v_{j}^{io},v_{j}^{o};
•replacing each arcaja_{j}byajoia_{j}^{oi}.

Informally speaking, this simply means that we stretch out each intermediate vertex of𝐩\mathbf{p}to the corresponding arc.

If𝐩\mathbf{p}is a path fromsstottinDD, then𝐩oi\mathbf{p}^{oi}is a path fromsos^{o}totit^{i}inD′D^{\prime}. Conversely, any path fromsos^{o}totit^{i}inD′D^{\prime}must have the form𝐩oi\mathbf{p}^{oi}, where𝐩\mathbf{p}is some path fromsstottinDD(because on any walk ofD′D^{\prime}, the arc-arcs and the vertex-arcs have to alternate). Therefore, the map

{paths fromstotinD}\displaystyle\left\{\text{paths from }s\text{ to }t\text{ in }D\right\}→{paths fromsototiinD′},\displaystyle\rightarrow\left\{\text{paths from }s^{o}\text{ to }t^{i}\text{ in }D^{\prime}\right\},𝐩\displaystyle\mathbf{p}↦𝐩oi\displaystyle\mapsto\mathbf{p}^{oi}(64)is a bijection. Moreover, two paths𝐩\mathbf{p}and𝐪\mathbf{q}ofDDare internally vertex-disjoint if and only if the paths𝐩oi\mathbf{p}^{oi}and𝐪oi\mathbf{q}^{oi}are arc-disjoint (since each vertex of a path𝐩\mathbf{p}except for its starting and ending points is represented by an arc in𝐩oi\mathbf{p}^{oi}).

Now, letkkbe the maximum number of pairwise arc-disjoint paths fromsos^{o}totit^{i}inD′D^{\prime}. Thus,D′D^{\prime}haskkpairwise arc-disjoint paths fromsos^{o}totit^{i}. Applying the inverse of the bijection (64) to thesekkpaths, we obtainkkpairwise internally vertex-disjoint paths fromsstottinDD(because two paths𝐩\mathbf{p}and𝐪\mathbf{q}ofDDare internally vertex-disjoint if and only if the paths𝐩oi\mathbf{p}^{oi}and𝐪oi\mathbf{q}^{oi}are arc-disjoint). Hence,

(the maximum number of pairwise internally vertex-disjoint\displaystyle\left(\text{the maximum number of pairwise internally vertex-disjoint }\right.paths fromstotinD)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{paths from }s\text{ to }t\text{ in }D\right)≥k.\displaystyle\geq k.(65) Our next goal is to find an internalss-tt-vertex-separatorW⊆V∖{s,t}W\subseteq V\setminus\left\{s,t\right\}of size|W|≤k\left|W\right|\leq k.

First, we simplify our setting a bit.

A path fromsstottcannot contain a loop; nor can it contain an arc with sourcettand targetss(since the vertices of a path must be distinct). Hence, we can remove such arcs (i.e., loops as well as arcs with sourcettand targetss) fromDDwithout affecting the meaning of the claim we are proving. Thus, we WLOG assume that the digraphDDhas no such arcs. Since we also know (by assumption) thatDDhas no arc with sourcessand targettt, we thus conclude thatDDhas no arc with source∈{s,t}\in\left\{s,t\right\}and target∈{s,t}\in\left\{s,t\right\}(because each such arc would either have sourcessand targettt, or have sourcettand targetss, or be a loop). In other words, each arc ofDDhas at least one endpoint145145145Anendpointof an arc means a vertex that is either the source or the target of this arc.distinct from bothssandtt.

However,kkis the maximum number of pairwise arc-disjoint paths fromsos^{o}totit^{i}inD′D^{\prime}. Therefore, by Theorem10.11(applied toD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right),sos^{o}andtit^{i}instead ofD=(V,A,ψ)D=\left(V,A,\psi\right),ssandtt), this numberkkequals the minimum size of ansos^{o}-tit^{i}-cut inD′D^{\prime}. Hence, there exists ansos^{o}-tit^{i}-cut[S,S¯]\left[S,\overline{S}\right]inD′D^{\prime}such that|[S,S¯]|=k\left|\left[S,\overline{S}\right]\right|=k. Consider thissos^{o}-tit^{i}-cut[S,S¯]\left[S,\overline{S}\right]. Since[S,S¯]\left[S,\overline{S}\right]is ansos^{o}-tit^{i}-cut, we haveS⊆V′S\subseteq V^{\prime}andso∈Ss^{o}\in Sandti∉St^{i}\notin S.

LetB:=[S,S¯]B:=\left[S,\overline{S}\right]. Then,|B|=|[S,S¯]|=k\left|B\right|=\left|\left[S,\overline{S}\right]\right|=k. Moreover, it is easy to see thatsio∉Bs^{io}\notin B146146146Proof.If we hadsio∈[S,S¯]s^{io}\in\left[S,\overline{S}\right], then we would havesi∈Ss^{i}\in Sandso∈S¯s^{o}\in\overline{S}; however,so∈S¯s^{o}\in\overline{S}would contradictso∈Ss^{o}\in S. Thus, we cannot havesio∈[S,S¯]s^{io}\in\left[S,\overline{S}\right]. In other words, we cannot havesio∈Bs^{io}\in B(sinceB=[S,S¯]B=\left[S,\overline{S}\right]). Hence,sio∉Bs^{io}\notin B.andtio∉Bt^{io}\notin B147147147Proof.If we hadtio∈[S,S¯]t^{io}\in\left[S,\overline{S}\right], then we would haveti∈St^{i}\in Sandto∈S¯t^{o}\in\overline{S}; however,ti∈St^{i}\in Swould contradictti∈S¯t^{i}\in\overline{S}. Thus, we cannot havetio∈[S,S¯]t^{io}\in\left[S,\overline{S}\right]. In other words, we cannot havetio∈Bt^{io}\in B(sinceB=[S,S¯]B=\left[S,\overline{S}\right]). Hence,tio∉Bt^{io}\notin B..

To each vertexw∈V′w\in V^{\prime}ofD′D^{\prime}, we assign a vertexβ⁡(w)∈V\beta\left(w\right)\in VofDDas follows: Ifw=viw=v^{i}orw=vow=v^{o}for somev∈Vv\in V, then we setβ⁡(w):=v\beta\left(w\right):=v. In other words,β⁡(w)\beta\left(w\right)is the vertexvvsuch thatw∈{vi,vo}w\in\left\{v^{i},v^{o}\right\}. We shall callβ⁡(w)\beta\left(w\right)thebaseof the vertexww.

For each arcb∈Bb\in B, there exists at least one endpointwwofbbsuch thatβ⁡(w)∈V∖{s,t}\beta\left(w\right)\in V\setminus\left\{s,t\right\}148148148Proof:Letb∈Bb\in Bbe an arc. We must prove that there exists at least one endpointwwofbbsuch thatβ⁡(w)∈V∖{s,t}\beta\left(w\right)\in V\setminus\left\{s,t\right\}. The arcbbis either a vertex-arc or an arc-arc. Thus, we are in one of the following two cases:Case 1:The arcbbis a vertex-arc.Case 2:The arcbbis an arc-arc. Let us first consider Case 1. In this case, the arcbbis a vertex-arc. In other words,b=viob=v^{io}for somev∈Vv\in V. Consider thisvv. Then,vio=b∈Bv^{io}=b\in B. Hence,v≠sv\neq s(sincev=sv=swould entailvio=sio∉Bv^{io}=s^{io}\notin B, which would contradictvio∈Bv^{io}\in B) andv≠tv\neq t(sincev=tv=twould entailvio=tio∉Bv^{io}=t^{io}\notin B, which would contradictvio∈Bv^{io}\in B). Therefore,v∈V∖{s,t}v\in V\setminus\left\{s,t\right\}. Also, clearly,viv^{i}is an endpoint ofbband satisfiesβ⁡(vi)=v∈V∖{s,t}\beta\left(v^{i}\right)=v\in V\setminus\left\{s,t\right\}. Hence, there exists at least one endpointwwofbbsuch thatβ⁡(w)∈V∖{s,t}\beta\left(w\right)\in V\setminus\left\{s,t\right\}(namely,w=viw=v^{i}). Thus, our proof is complete in Case 1. Let us now consider Case 2. In this case, the arcbbis an arc-arc. In other words,b=aoib=a^{oi}for somea∈Aa\in A. Consider thisaa. Now,aais an arc ofDD(sincea∈Aa\in A), and thus has at least one endpoint distinct from bothssandtt(since we have shown above that each arc ofDDhas at least one endpoint distinct from bothssandtt). Letxxbe this endpoint. Then,x∈V∖{s,t}x\in V\setminus\left\{s,t\right\}(sincexxis distinct from bothssandtt). Butxxis an endpoint ofaa. In other words,xxis either the source or the target ofaa. Hence, the arcaoia^{oi}ofD′D^{\prime}either has sourcexox^{o}or has targetxix^{i}(by the definition ofaoia^{oi}). In other words, the arcbbofD′D^{\prime}either has sourcexox^{o}or has targetxix^{i}(sinceb=aoib=a^{oi}). Sinceβ⁡(xo)=x∈V∖{s,t}\beta\left(x^{o}\right)=x\in V\setminus\left\{s,t\right\}andβ⁡(xi)=x∈V∖{s,t}\beta\left(x^{i}\right)=x\in V\setminus\left\{s,t\right\}, we thus conclude that the arcbbofD′D^{\prime}has at least one endpointwwsuch thatβ⁡(w)∈V∖{s,t}\beta\left(w\right)\in V\setminus\left\{s,t\right\}(namely,w=xow=x^{o}ifbbhas sourcexox^{o}, andw=xiw=x^{i}ifbbhas targetxix^{i}). This completes our proof in Case 2. Thus, we are done in both Cases 1 and 2, so that our proof is complete.. We choose such an endpointwwarbitrarily, and we denote its baseβ⁡(w)\beta\left(w\right)byβ⁡(b)\beta\left(b\right). We shall callβ⁡(b)\beta\left(b\right)thebasepointof the arcbb. Thus, by definition, we have

β⁡(b)∈V∖{s,t}for eachb∈B.\beta\left(b\right)\in V\setminus\left\{s,t\right\}\ \ \ \ \ \ \ \ \ \ \text{for each }b\in B.(66) We letβ⁡(B)\beta\left(B\right)denote the set{β⁡(b)∣b∈B}\left\{\beta\left(b\right)\ \mid\ b\in B\right\}. Clearly,|β⁡(B)|≤|B|=k\left|\beta\left(B\right)\right|\leq\left|B\right|=k.

Now, we claim that

every path fromstot(inD) contains a vertex inβ(B).\text{every path from }s\text{ to }t\text{ (in }D\text{) contains a vertex in }\beta\left(B\right).(67) [Proof of (67):Let𝐩\mathbf{p}be a path fromsstott(inDD). We must prove that𝐩\mathbf{p}contains a vertex inβ⁡(B)\beta\left(B\right).

Recall that we have assigned a path𝐩oi\mathbf{p}^{oi}ofD′D^{\prime}to the path𝐩\mathbf{p}ofDD. The definition of𝐩oi\mathbf{p}^{oi}shows that the base of any vertex of𝐩oi\mathbf{p}^{oi}is a vertex of𝐩\mathbf{p}(indeed, ifv0,v1,…,vkv_{0},v_{1},\ldots,v_{k}are the vertices of𝐩\mathbf{p}, then the vertices of𝐩oi\mathbf{p}^{oi}arev0o,v1i,v1o,v2i,v2o,…,vk−1i,vk−1o,vkiv_{0}^{o},v_{1}^{i},v_{1}^{o},v_{2}^{i},v_{2}^{o},\ldots,v_{k-1}^{i},v_{k-1}^{o},v_{k}^{i}, and their respective bases arev0,v1,v1,v2,v2,…,vk−1,vk−1,vkv_{0},v_{1},v_{1},v_{2},v_{2},\ldots,v_{k-1},v_{k-1},v_{k}).

The path𝐩oi\mathbf{p}^{oi}is a path fromsos^{o}totit^{i}(since𝐩\mathbf{p}is a path fromsstott). Hence, it starts at a vertex inSS(sinceso∈Ss^{o}\in S) and ends at a vertex that is not inSS(sinceti∉St^{i}\notin S). Thus, this path𝐩oi\mathbf{p}^{oi}must cross fromSSintoS¯\overline{S}somewhere. In other words, there exists an arcbbof𝐩oi\mathbf{p}^{oi}such that the source ofbbbelongs toSSbut the target ofbbbelongs toS¯\overline{S}. Consider this arcbb. Thus,b∈[S,S¯]=Bb\in\left[S,\overline{S}\right]=B, so thatβ⁡(b)∈β⁡(B)\beta\left(b\right)\in\beta\left(B\right)(by the definition ofβ⁡(B)\beta\left(B\right)). Both endpoints ofbbare vertices of𝐩oi\mathbf{p}^{oi}(sincebbis an arc of𝐩oi\mathbf{p}^{oi}).

Now, consider the basepointβ⁡(b)\beta\left(b\right)of this arcbb. This basepointβ⁡(b)\beta\left(b\right)is the base of an endpoint ofbb(by the definition ofβ⁡(b)\beta\left(b\right)). Thus,β⁡(b)\beta\left(b\right)is the base of a vertex of𝐩oi\mathbf{p}^{oi}(since both endpoints ofbbare vertices of𝐩oi\mathbf{p}^{oi}). Hence,β⁡(b)\beta\left(b\right)is a vertex of𝐩\mathbf{p}(since the base of any vertex of𝐩oi\mathbf{p}^{oi}is a vertex of𝐩\mathbf{p}). In other words, the path𝐩\mathbf{p}contains the vertexβ⁡(b)\beta\left(b\right). Sinceβ⁡(b)∈β⁡(B)\beta\left(b\right)\in\beta\left(B\right), we thus conclude that𝐩\mathbf{p}contains a vertex inβ⁡(B)\beta\left(B\right). This proves (67).]

Now, the setβ⁡(B)\beta\left(B\right)is a subset ofV∖{s,t}V\setminus\left\{s,t\right\}(sinceβ⁡(b)∈V∖{s,t}\beta\left(b\right)\in V\setminus\left\{s,t\right\}for eachb∈Bb\in B) and has the property that every path fromsstottcontains a vertex inβ⁡(B)\beta\left(B\right)(by (67)). In other words,β⁡(B)\beta\left(B\right)is a subsetW⊆V∖{s,t}W\subseteq V\setminus\left\{s,t\right\}such that every path fromsstottcontains a vertex inWW. In other words,β⁡(B)\beta\left(B\right)is an internalss-tt-vertex-separator (by the definition of an “internalss-tt-vertex-separator”). Thus,

(the minimum size of an internals-t-vertex-separator)\displaystyle\left(\text{the minimum size of an internal }s\text{-}t\text{-vertex-separator}\right)≤|β⁡(B)|=k\displaystyle\leq\left|\beta\left(B\right)\right|=k≤(the maximum number of pairwise internally vertex-disjoint\displaystyle\leq\left(\text{the maximum number of pairwise internally vertex-disjoint }\right.paths fromstotinD)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{paths from }s\text{ to }t\text{ in }D\right)(by (65)).

On the other hand, we have

(the minimum size of an internals-t-vertex-separator)\displaystyle\left(\text{the minimum size of an internal }s\text{-}t\text{-vertex-separator}\right)≥(the maximum number of pairwise internally vertex-disjoint\displaystyle\geq\left(\text{the maximum number of pairwise internally vertex-disjoint}\right.paths fromstotinD)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{paths from }s\text{ to }t\text{ in }D\right)(by the pigeonhole principle149149149Proofin more detail:Letnnbe the minimum size of an internalss-tt-vertex-separator. Letxxbe the maximum number of pairwise internally vertex-disjoint paths fromsstottinDD. We must show thatn≥xn\geq x. Assume the contrary. Thus,n<xn<x. The definition ofnnshows that there exists an internalss-tt-vertex-separatorWWthat has sizenn. The setWWis an internalss-tt-vertex-separator. In other words,WWis a subset ofV∖{s,t}V\setminus\left\{s,t\right\}such that every path fromsstottcontains a vertex inWW. Moreover,WWhas sizenn; thus,|W|=n<x\left|W\right|=n<x. The definition ofxxshows that there existxxpairwise internally vertex-disjoint paths fromsstottinDD. Let𝐩1,𝐩2,…,𝐩x\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{x}be thesexxpaths. Each of thesexxpaths𝐩1,𝐩2,…,𝐩x\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{x}must contain at least one vertex inWW(since every path fromsstottcontains a vertex inWW). Since|W|<x\left|W\right|<x, we thus conclude by the pigeonhole principle that at least two of thexxpaths𝐩1,𝐩2,…,𝐩x\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{x}must contain the same vertex inWW. In other words, there exist two distinct elementsi,j∈{1,2,…,x}i,j\in\left\{1,2,\ldots,x\right\}such that𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}contain the same vertex inWW. Letwwbe the latter vertex. Thus,w∈W⊆V∖{s,t}w\in W\subseteq V\setminus\left\{s,t\right\}. Hence,wwis distinct from bothssandtt. Therefore,wwis an intermediate vertex of𝐩i\mathbf{p}_{i}(since the path𝐩i\mathbf{p}_{i}has starting pointssand ending pointtt). Likewise,wwis an intermediate vertex of𝐩j\mathbf{p}_{j}. However, the paths𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}are internally vertex-disjoint, and thus have no common intermediate vertex. This contradicts the fact thatwwis an intermediate vertex of both paths𝐩i\mathbf{p}_{i}and𝐩j\mathbf{p}_{j}. This contradiction shows that our assumption was false. Hence,n≥xn\geq xis proved, qed.). Combining this inequality with the preceding one, we obtain

(the minimum size of an internals-t-vertex-separator)\displaystyle\left(\text{the minimum size of an internal }s\text{-}t\text{-vertex-separator}\right)=(the maximum number of pairwise internally vertex-disjointCLOSE\displaystyle=\left(\text{the maximum number of pairwise internally vertex-disjoint}\right.paths fromstotinD).\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{paths from }s\text{ to }t\text{ in }D\right).This proves Theorem10.43.

There is also a variant of the vertex-Menger theorem similar to what Theorem10.24did for the arc-Menger theorem. Again, we need some notations first:

Definition 10.47.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letXXandYYbe two subsets ofVV.

(a)Apath fromXXtoYYshall mean a path whose starting point belongs toXXand whose ending point belongs toYY.
(b)A subsetWWofVVis said to be anXX-YY-vertex-separatorif each path fromXXtoYYcontains at least one vertex fromWW. Equivalently, a subsetWWofVVis said to be anXX-YY-vertex-separatorif the induced subdigraph ofDDon the setV∖WV\setminus Whas no path fromXXtoYY(in other words, removing fromDDall vertices contained inWWdestroys all paths fromXXtoYY).
(c)AnXX-YY-vertex-separatorWWis said to beinternalif it is a subset ofV∖(X∪Y)V\setminus\left(X\cup Y\right)(that is, if it is disjoint fromXXand fromYY).

Theorem 10.48(vertex-Menger theorem for directed graphs, multi-terminal version 1).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letXXandYYbe two disjoint subsets ofVV. Assume thatDDhas no arc with source inXXand target inYY.

Then, the maximum number of pairwise internally vertex-disjoint paths fromXXtoYYequals the minimum size of an internalXX-YY-vertex-separator.

Example 10.49.

LetDDbe the following multidigraph:

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\lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 11.75 78.74 L 70 78.74}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{51.19167pt}{56.90552pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 70.83 78.74)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the maximum number of pairwise internally vertex-disjoint paths fromXXtoYYis22; here are two such paths (drawn in red and blue):

X\color[rgb]{1,0,0}{X}Y\color[rgb]{1,0,0}{Y}aabbcc(there are other choices, of course). The minimum size of an internalXX-YY-vertex-separator is22as well; indeed,{a,b}\left\{a,b\right\}is such an internalXX-YY-vertex-separator. These two numbers are equal, just as Theorem10.48predicts.

Proof 10.50(Proof of Theorem10.48.).

We define a new multidigraphD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right)as in the proof of Theorem10.24. Then,D′D^{\prime}has no arc with sourcessand targettt(sinceDDhas no arc with source inXXand target inYY).

Hence, Theorem10.43(applied toD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right)instead ofD=(V,A,ψ)D=\left(V,A,\psi\right)) shows that the maximum number of pairwise internally vertex-disjoint paths fromsstottinD′D^{\prime}equals the minimum size of an internalss-tt-vertex-separator inD′D^{\prime}.

Let us now see what this result means for our original digraphDD. Indeed:

•The minimum size of an internalss-tt-vertex-separator inD′D^{\prime}equals the minimum size of an internalXX-YY-vertex-separator inDD(indeed, the internalss-tt-vertex-separators inD′D^{\prime}are precisely the internalXX-YY-vertex-separators inDD150150150Proof.We recall the definitions of internalss-tt-vertex-separators inD′D^{\prime}and of internalXX-YY-vertex-separators inDD:–An internalss-tt-vertex-separator inD′D^{\prime}is a subsetWWofV′∖{s,t}V^{\prime}\setminus\left\{s,t\right\}such that each path fromsstottcontains at least one vertex fromWW.–An internalXX-YY-vertex-separator inDDis a subsetWWofV∖(X∪Y)V\setminus\left(X\cup Y\right)such that each path fromXXtoYYcontains at least one vertex fromWW.These two definitions describe the same object, because of the following two reasons:–We haveV′∖{s,t}=V∖(X∪Y)V^{\prime}\setminus\left\{s,t\right\}=V\setminus\left(X\cup Y\right).–The paths fromsstottare in bijection with the paths fromXXtoYY(indeed, any path of the latter kind can be transformed into a path of the former kind by replacing the starting point byssand replacing the ending point bytt). This bijection preserves the intermediate vertices (i.e., the vertices other than the starting point and the ending point). Thus, a path𝐩\mathbf{p}fromsstottcontains at least one vertex fromWWif and only if the corresponding path fromXXtoYY(that is, the image of𝐩\mathbf{p}under our bijection) contains at least one vertex fromWW.Thus, the internalss-tt-vertex-separators inD′D^{\prime}are precisely the internalXX-YY-vertex-separators inDD.).
•The maximum number of pairwise internally vertex-disjoint paths fromsstottinD′D^{\prime}equals the maximum number of pairwise internally vertex-disjoint paths fromXXtoYYinDD151151151Proof.We make the following two observations:Observation 1:Letk∈ℕk\in\mathbb{N}. IfD′D^{\prime}haskkpairwise internally vertex-disjoint paths fromsstott, thenDDhaskkpairwise internally vertex-disjoint paths fromXXtoYY.[Proof of Observation 1:Assume thatD′D^{\prime}haskkpairwise internally vertex-disjoint paths fromsstott. We can “lift” thesekkpaths tokkpaths fromXXtoYYinDD(preserving the arcs, and replacing the verticesssandttby appropriate vertices inXXandYYto make them belong to the right arcs). The resultingkkpaths fromXXtoYYinDDare still pairwise internally vertex-disjoint (since our “lifting” operation has not changed the intermediate vertices of our paths). Thus,DDhaskkpairwise internally vertex-disjoint paths fromXXtoYY. This proves Observation 1.]Observation 2:Letk∈ℕk\in\mathbb{N}. IfDDhaskkpairwise internally vertex-disjoint paths fromXXtoYY, thenD′D^{\prime}haskkpairwise internally vertex-disjoint paths fromsstott.[Proof of Observation 2:Assume thatDDhaskkpairwise internally vertex-disjoint paths fromXXtoYY. We can replace thesekkpaths bykkpairwise internally vertex-disjointwalksfromsstottinD′D^{\prime}(by replacing their starting points withssand replacing their ending points withtt). Thus,D′D^{\prime}haskkpairwise internally vertex-disjointwalksfromsstott. Therefore,D′D^{\prime}haskkpairwise internally vertex-disjointpathsfromsstottas well (since each walk contains a path, and of course we don’t lose internal vertex-disjointness if we restrict our walk to a path contained in it). This proves Observation 2.] Observation 2 shows that the maximum number of pairwise internally vertex-disjoint paths fromsstottinD′D^{\prime}is≥\geqto the maximum number of pairwise internally vertex-disjoint paths fromXXtoYYinDD. But Observation 1 shows the reverse inequality (i.e., it shows that the former number is≤\leqto the latter number). Thus, the inequality is an equality, i.e., the two numbers are equal. Qed..

Hence, the result of the preceding paragraph is precisely the claim of Theorem10.48, and our proof is thus complete.

Another variant of this result can be stated for vertex-disjoint (as opposed to internally vertex-disjoint) paths. These are even easier to define:

Definition 10.51.

Two paths𝐩\mathbf{p}and𝐪\mathbf{q}in a graph or digraph are said to bevertex-disjointif they have no common vertices.

Theorem 10.52(vertex-Menger theorem for directed graphs, multi-terminal version 2).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph, and letXXandYYbe two subsets ofVV.

Then, the maximum number of pairwise vertex-disjoint paths fromXXtoYYequals the minimum size of anXX-YY-vertex-separator.

Example 10.53.

LetDDbe the following multidigraph:

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\lxSVG@transformcm{0.7071}{0.7071}{-0.7071}{0.7071}{106.88583pt}{49.98032pt}\lxSVG@begingroup@{transform=matrix(0.7071 0.7071 -0.7071 0.7071 147.9 69.16)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 49.34 -72.09 L 150.21 -4.85}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{0.83205}{0.5547}{-0.5547}{0.83205}{109.05684pt}{-3.16948pt}\lxSVG@begingroup@{transform=matrix(0.83205 0.5547 -0.5547 0.83205 150.9 -4.39)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} {{}}{}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}{}{}{}{{}}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 51.35 -78.74 L 148.74 -78.74}{fill:none} {{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{108.09718pt}{-56.90552pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 149.57 -78.74)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then, the maximum number of pairwise vertex-disjoint paths fromXXtoYYis22. Here are two such paths (drawn in red and blue):

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157.48 118.11 C 135.74 118.11 118.11 65.23 118.11 0 C 118.11 -65.23 135.74 -118.11 157.48 -118.11 C 179.22 -118.11 196.85 -65.23 196.85 0 Z M 157.48 0}{} \lx@inpgf@ignorespaces \lxSVG@closescope {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-61.44023pt}{-97.3104pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -85.01 -134.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{109.79715pt}{-97.3104pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 151.93 -134.65)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@begingroup@{_scopebegin=1} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -72.22 78.74 C -72.22 82.34 -75.14 85.26 -78.74 85.26 C -82.34 85.26 -85.26 82.34 -85.26 78.74 C -85.26 75.14 -82.34 72.22 -78.74 72.22 C -75.14 72.22 -72.22 75.14 -72.22 78.74 Z M -78.74 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-56.90552pt}{56.90552pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -78.74 78.74)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -72.22 0 C -72.22 3.6 -75.14 6.52 -78.74 6.52 C -82.34 6.52 -85.26 3.6 -85.26 0 C -85.26 -3.6 -82.34 -6.52 -78.74 -6.52 C -75.14 -6.52 -72.22 -3.6 -72.22 0 Z M -78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-56.90552pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -78.74 0)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M -72.22 -78.74 C -72.22 -75.14 -75.14 -72.22 -78.74 -72.22 C -82.34 -72.22 -85.26 -75.14 -85.26 -78.74 C -85.26 -82.34 -82.34 -85.26 -78.74 -85.26 C -75.14 -85.26 -72.22 -82.34 -72.22 -78.74 Z M -78.74 -78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{-56.90552pt}{-56.90552pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 -78.74 -78.74)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 6.52 0 C 6.52 3.6 3.6 6.52 0 6.52 C -3.6 6.52 -6.52 3.6 -6.52 0 C -6.52 -3.6 -3.6 -6.52 0 -6.52 C 3.6 -6.52 6.52 -3.6 6.52 0 Z M 0 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{0.0pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 0 0)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 6.52 78.74 C 6.52 82.34 3.6 85.26 0 85.26 C -3.6 85.26 -6.52 82.34 -6.52 78.74 C -6.52 75.14 -3.6 72.22 0 72.22 C 3.6 72.22 6.52 75.14 6.52 78.74 Z M 0 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{0.0pt}{56.90552pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 0 78.74)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 85.26 0 C 85.26 3.6 82.34 6.52 78.74 6.52 C 75.14 6.52 72.22 3.6 72.22 0 C 72.22 -3.6 75.14 -6.52 78.74 -6.52 C 82.34 -6.52 85.26 -3.6 85.26 0 Z M 78.74 0}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{56.90552pt}{0.0pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 78.74 0)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 85.26 78.74 C 85.26 82.34 82.34 85.26 78.74 85.26 C 75.14 85.26 72.22 82.34 72.22 78.74 C 72.22 75.14 75.14 72.22 78.74 72.22 C 82.34 72.22 85.26 75.14 85.26 78.74 Z M 78.74 78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{56.90552pt}{56.90552pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 78.74 78.74)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} {{}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {{}{{{}}}{{}}{}{}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{}{}{}{}{}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=0.8pt} \lx@inpgf@ignorespaces\lxSVG@begingroup@{stroke=#009900} {}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 50.8 -78.74 C 50.8 -72.43 45.68 -67.31 39.37 -67.31 C 33.06 -67.31 27.94 -72.43 27.94 -78.74 C 27.94 -85.05 33.06 -90.17 39.37 -90.17 C 45.68 -90.17 50.8 -85.05 50.8 -78.74 Z M 39.37 -78.74}{fill:none} \lx@inpgf@ignorespaces \lxSVG@closescope }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} 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{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .If we were only looking forinternallyvertex-disjoint paths, then we could add a third path to these two (namely, the path that starts at the topmost vertex ofXXand ends at the topmost vertex ofYY). However, this path and our red paths are onlyinternallyvertex-disjoint, not vertex-disjoint. A little bit of thought shows thatDDhas no more than22vertex-disjoint paths fromXXtoYY.

The minimum size of anXX-YY-vertex-separator is22as well; indeed,{u,y}\left\{u,y\right\}is such anXX-YY-vertex-separator. This number equals the maximum number of pairwise vertex-disjoint paths fromXXtoYY, just as Theorem10.52predicts.

Proof 10.54(Proof of Theorem10.52.).

We will reduce this to Theorem10.43, again by tweaking our digraph appropriately. This time, the tweak is pretty simple: We add two new verticesssandtttoDD, and we furthermore add an arc fromssto eachx∈Xx\in Xand an arc from eachy∈Yy\in Ytott(thus, we add a total of|X|+|Y|\left|X\right|+\left|Y\right|new arcs). We denote the resulting digraph byD′D^{\prime}. In more detail, the definition ofD′D^{\prime}is as follows:

•We introduce two new verticesssandtt, and we setV′:=V∪{s,t}V^{\prime}:=V\cup\left\{s,t\right\}. This setV′V^{\prime}will be the vertex set ofD′D^{\prime}.
•For eachx∈Xx\in X, we introduce a new arcaxa_{x}, which shall have sourcessand targetxx.
•For eachy∈Yy\in Y, we introduce a new arcbyb_{y}, which shall have sourceyyand targettt.
•We letA′:=A∪{ax∣x∈X}∪{by∣y∈Y}A^{\prime}:=A\cup\left\{a_{x}\ \mid\ x\in X\right\}\cup\left\{b_{y}\ \mid\ y\in Y\right\}. This setA′A^{\prime}will be the arc set ofD′D^{\prime}.
•We extend our mapψ:A→V×V\psi:A\rightarrow V\times Vto a mapψ′:A′→V′×V′\psi^{\prime}:A^{\prime}\rightarrow V^{\prime}\times V^{\prime}by setting ψ′(ax)=(s,x)for eachx∈X\psi^{\prime}\left(a_{x}\right)=\left(s,x\right)\ \ \ \ \ \ \ \ \ \ \text{for each }x\in Xand ψ′(by)=(y,t)for eachy∈Y\psi^{\prime}\left(b_{y}\right)=\left(y,t\right)\ \ \ \ \ \ \ \ \ \ \text{for each }y\in Y(and, of course,ψ′(c)=ψ(c)\psi^{\prime}\left(c\right)=\psi\left(c\right)for eachc∈Ac\in A).
•We defineD′D^{\prime}to be the multidigraph(V′,A′,ψ′)\left(V^{\prime},A^{\prime},\psi^{\prime}\right).

For instance, ifDDis the multidigraph from Example10.53, thenD′D^{\prime}looks as follows:

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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#0000FF} \lxSVG@transformcm{0.7071}{0.7071}{-0.7071}{0.7071}{164.20932pt}{-6.50723pt}\lxSVG@begingroup@{transform=matrix(0.7071 0.7071 -0.7071 0.7071 227.22 -9)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .(The arcsaxa_{x}are drawn in red; the arcsbyb_{y}are drawn in blue.)

By its construction, the digraphD′D^{\prime}has no arc with sourcessand targettt. Hence, Theorem10.43(applied toD′=(V′,A′,ψ′)D^{\prime}=\left(V^{\prime},A^{\prime},\psi^{\prime}\right)instead ofD=(V,A,ψ)D=\left(V,A,\psi\right)) yields that the maximum number of pairwise internally vertex-disjoint paths fromsstottequals the minimum size of an internalss-tt-vertex-separator. However, it is easy to see the following two claims:

Claim 1:The maximum number of pairwise internally vertex-disjoint paths fromsstottequals the maximum number of pairwise vertex-disjoint paths fromXXtoYY(inDD).

Claim 2:The minimum size of an internalss-tt-vertex-separator equals the minimum size of anXX-YY-vertex-separator (inDD).

[Proof of Claim 1 (sketched):Given any path𝐩\mathbf{p}fromsstott, we can remove the starting point and the ending point of this path; the result will always be a path fromXXtoYY(inDD). Let us denote the latter path by𝐩¯\overline{\mathbf{p}}. Thus, we obtain a map

{paths fromstot}\displaystyle\left\{\text{paths from }s\text{ to }t\right\}→{paths fromXtoY(inD)},\displaystyle\rightarrow\left\{\text{paths from }X\text{ to }Y\text{ (in }D\text{)}\right\},𝐩\displaystyle\mathbf{p}↦𝐩¯.\displaystyle\mapsto\overline{\mathbf{p}}.This map is easily seen to be a bijection (indeed, if𝐪\mathbf{q}is a path fromXXtoYY(inDD), then we can easily extend it to a path𝐩\mathbf{p}fromsstottby inserting the appropriate arcaxa_{x}at its beginning and the appropriate arcbyb_{y}at its end; this latter path𝐩\mathbf{p}will then satisfy𝐩¯=𝐪\overline{\mathbf{p}}=\mathbf{q}). Moreover, two paths𝐩\mathbf{p}and𝐪\mathbf{q}fromsstottare internally vertex-disjoint if and only if the corresponding paths𝐩¯\overline{\mathbf{p}}and𝐪¯\overline{\mathbf{q}}are vertex-disjoint (because the intermediate vertices of𝐩\mathbf{p}are the vertices of𝐩¯\overline{\mathbf{p}}, whereas the intermediate vertices of𝐪\mathbf{q}are the vertices of𝐪¯\overline{\mathbf{q}}). This proves Claim 1.]

[Proof of Claim 2 (sketched):It is easy to see that the internalss-tt-vertex-separators are precisely theXX-YY-vertex-separators (inDD). (To show this, compare the definitions of these two objects using the bijection from the proof of Claim 1, and observe thatV′∖{s,t}=VV^{\prime}\setminus\left\{s,t\right\}=V.) From this, Claim 2 follows.]

Recall that the maximum number of pairwise internally vertex-disjoint paths fromsstottequals the minimum size of an internalss-tt-vertex-separator. In view of Claim 1 and Claim 2, we can rewrite this as follows: The maximum number of pairwise vertex-disjoint paths fromXXtoYYequals the minimum size of anXX-YY-vertex-separator. Thus, Theorem10.52is proved.

We note that Hall’s Marriage Theorem (Theorem8.23) can be easily derived from any of the directed Menger theorems (exercise!). I have heard that this can also be done in reverse. This places the Menger theorems in the cluster of theorems equivalent to Hall’s Marriage Theorem (such as König’s theorem).

10.1.4The vertex-Menger theorem for undirected graphs

Vertex-Menger theorems also exist for undirected graphs. Here are the undirected analogues of Theorem10.43, Theorem10.48and Theorem10.52, along with the definitions they rely on:

Definition 10.55.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letssandttbe two vertices ofGG. A subsetWWofV∖{s,t}V\setminus\left\{s,t\right\}is said to be aninternalss-tt-vertex-separatorif each path fromsstottcontains at least one vertex fromWW. Equivalently, a subsetWWofV∖{s,t}V\setminus\left\{s,t\right\}is said to be aninternalss-tt-vertex-separatorif the induced subgraph ofGGon the setV∖WV\setminus Whas no path fromsstott(in other words, removing fromGGall vertices contained inWWdestroys all paths fromsstott).

Theorem 10.56(vertex-Menger theorem for undirected graphs).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letssandttbe two distinct vertices ofGG. Assume thatGGhas no edge with endpointsssandtt. Then, the maximum number of pairwise internally vertex-disjoint paths fromsstottequals the minimum size of an internalss-tt-vertex-separator.

Definition 10.57.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letXXandYYbe two subsets ofVV.

(a)Apath fromXXtoYYshall mean a path whose starting point belongs toXXand whose ending point belongs toYY.
(b)A subsetWWofVVis said to be anXX-YY-vertex-separatorif each path fromXXtoYYcontains at least one vertex fromWW. Equivalently, a subsetWWofVVis said to be anXX-YY-vertex-separatorif the induced subgraph ofGGon the setV∖WV\setminus Whas no path fromXXtoYY(in other words, removing fromGGall vertices contained inWWdestroys all paths fromXXtoYY).
(c)AnXX-YY-vertex-separatorWWis said to beinternalif it is a subset ofV∖(X∪Y)V\setminus\left(X\cup Y\right)(that is, if it is disjoint fromXXand fromYY).

Theorem 10.58(vertex-Menger theorem for undirected graphs, multi-terminal version 1).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letXXandYYbe two disjoint subsets ofVV. Assume thatGGhas no edge with one endpoint inXXand the other endpoint inYY.

Then, the maximum number of pairwise internally vertex-disjoint paths fromXXtoYYequals the minimum size of an internalXX-YY-vertex-separator.

Theorem 10.59(vertex-Menger theorem for undirected graphs, multi-terminal version 2).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph, and letXXandYYbe two subsets ofVV.

Then, the maximum number of pairwise vertex-disjoint paths fromXXtoYYequals the minimum size of anXX-YY-vertex-separator.

Theorem10.56, Theorem10.58and Theorem10.59follow immediately by applying the analogous theorems for directed graphs (i.e., Theorem10.43, Theorem10.48and Theorem10.52) to the digraphGbidirG^{\operatorname*{bidir}}instead ofDD(since the paths ofGGare in bijection with the paths ofGbidirG^{\operatorname*{bidir}}).

10.2The Gallai–Milgram theorem

Next, we proceed to some more obscure properties of paths in digraphs and graphs.

10.2.1Definitions

In order to state the first of these properties, we need the following three definitions (the first two of which were already made in Section7.4):

Definition 10.60.

Definition 10.61.

Anindependent setof a multidigraphDDmeans a subsetSSofV(D)\operatorname*{V}\left(D\right)such that no two elements ofSSare adjacent. In other words, it means an independent set of the undirected graphDundD^{\operatorname*{und}}.

Definition 10.62.

Apath coverof a multidigraphDDmeans a set of paths ofDDsuch that each vertex ofDDis contained in exactly one of these paths.

Example 10.63.

LetDDbe the following digraph:

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\lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Then,{(1,∗,5,∗,4),(3,∗,2)}\left\{\left(1,\ast,5,\ast,4\right),\ \left(3,\ast,2\right)\right\}is a path cover ofDD(we are again writing asterisks for the arcs, since the arcs ofDDare uniquely determined by their sources and their targets). Another path cover ofDDis{(1,∗,3,∗,4),(2),(5)}\left\{\left(1,\ast,3,\ast,4\right),\ \left(2\right),\ \left(5\right)\right\}. Yet another path cover ofDDis{(1),(2),(3),(4),(5)}\left\{\left(1\right),\ \left(2\right),\ \left(3\right),\ \left(4\right),\ \left(5\right)\right\}. There are many more.

Note that the set{(1,∗,5,∗,4),(3,∗,2,∗,4)}\left\{\left(1,\ast,5,\ast,4\right),\ \left(3,\ast,2,\ast,4\right)\right\}is not a path cover ofDD, since the vertex44is contained in two (not one) of its paths.

Let us draw the three path covers we have mentioned (by simply drawing the arcs of the paths they contain, while omitting all other arcs ofDD):

112233445511223344551122334455{(1,∗,5,∗,4),(3,∗,2)}\left\{\left(1,\ast,5,\ast,4\right),\ \left(3,\ast,2\right)\right\}{(1,∗,3,∗,4),(2),(5)}\left\{\left(1,\ast,3,\ast,4\right),\ \left(2\right),\ \left(5\right)\right\}{(1),(2),(3),(4),(5)}\left\{\left(1\right),\ \left(2\right),\ \left(3\right),\ \left(4\right),\ \left(5\right)\right\}

(Note that we have already seen path covers of a “complete” simple digraph(V,V×V)\left(V,\ V\times V\right)in the proof of Theorem4.86; we called them “path covers ofVV”.)

10.2.2The Gallai–Milgram theorem

Now, theGallai–Milgram theoremstates the following:

Theorem 10.65(Gallai–Milgram theorem).

LetDDbe a loopless digraph. Then, there exist a path cover𝒫\mathcal{P}ofDDand an independent setSSofDDsuch thatSShas exactly one vertex from each path in𝒫\mathcal{P}(in other words, for each path𝐩∈𝒫\mathbf{p}\in\mathcal{P}, exactly one vertex of𝐩\mathbf{p}belongs toSS).

Example 10.66.

LetDDbe the digraph from Example10.63. Then, Theorem10.65tells us that there exist a path cover𝒫\mathcal{P}ofDDand an independent setSSofDDsuch thatSShas exactly one vertex from each path in𝒫\mathcal{P}. For example, we can take𝒫={(1,∗,5,∗,4),(3,∗,2)}\mathcal{P}=\left\{\left(1,\ast,5,\ast,4\right),\ \left(3,\ast,2\right)\right\}andS={5,3}S=\left\{5,3\right\}.

We will now prove Theorem10.65, following Diestel’s book[Dieste17, Theorem 2.5.1]:

Proof 10.67(Proof of Theorem10.65.).

Write the multidigraphDDasD=(V,A,ψ)D=\left(V,A,\psi\right). We introduce a notation:

•If𝒫\mathcal{P}is a path cover ofDD, then across-cutof𝒫\mathcal{P}means a subsetSSofVVthat contains exactly one vertex from each path in𝒫\mathcal{P}.

Thus, the claim we must prove is saying that there exist a path cover𝒫\mathcal{P}ofDDand an independent cross-cut of𝒫\mathcal{P}.

We will show something stronger:

Claim 1:Any minimum-size path cover ofDDhas an independent cross-cut.

Note that the size of a path cover means the number of paths in it. Thus, a minimum-size path cover means a path cover with the smallest possible number of paths.

We will show something even stronger than Claim 1. To state this stronger claim, we need more notations:

•If𝒫\mathcal{P}is a path cover, thenEnds𝒫\operatorname*{Ends}\mathcal{P}means the set of the ending points of all paths in𝒫\mathcal{P}. Note that|Ends𝒫|=|𝒫|\left|\operatorname*{Ends}\mathcal{P}\right|=\left|\mathcal{P}\right|.
•A path cover𝒫\mathcal{P}is said to beend-minimalif no proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}can be written asEnds𝒬\operatorname*{Ends}\mathcal{Q}for a path cover𝒬\mathcal{Q}.

Example 10.68.

For instance, ifDDis as in Example10.63, and if

𝒫\displaystyle\mathcal{P}={(1,∗,5,∗,4),(3,∗,2)},\displaystyle=\left\{\left(1,\ast,5,\ast,4\right),\ \left(3,\ast,2\right)\right\},𝒬\displaystyle\mathcal{Q}={(1,∗,3,∗,4),(2),(5)},\displaystyle=\left\{\left(1,\ast,3,\ast,4\right),\ \left(2\right),\ \left(5\right)\right\},ℛ\displaystyle\mathcal{R}={(1),(2),(3),(4),(5)}\displaystyle=\left\{\left(1\right),\ \left(2\right),\ \left(3\right),\ \left(4\right),\ \left(5\right)\right\}are the three path covers from Example10.63, then

Ends𝒫={4,2},Ends𝒬={4,2,5},Endsℛ={1,2,3,4,5},\operatorname*{Ends}\mathcal{P}=\left\{4,2\right\},\ \ \ \ \ \ \ \ \ \ \operatorname*{Ends}\mathcal{Q}=\left\{4,2,5\right\},\ \ \ \ \ \ \ \ \ \ \operatorname*{Ends}\mathcal{R}=\left\{1,2,3,4,5\right\},which shows immediately that neither𝒬\mathcal{Q}norℛ\mathcal{R}is end-minimal (sinceEnds𝒫\operatorname*{Ends}\mathcal{P}is a proper subset of each ofEnds𝒬\operatorname*{Ends}\mathcal{Q}andEndsℛ\operatorname*{Ends}\mathcal{R}). It is easy to see that𝒫\mathcal{P}is end-minimal (and also minimum-size).

Back to the general case. Clearly, any minimum-size path cover ofDDis also end-minimal152152152Proof.Let𝒫\mathcal{P}be a minimum-size path cover ofDD. If𝒫\mathcal{P}was not end-minimal, then there would be a path cover𝒬\mathcal{Q}with|Ends𝒬|<|Ends𝒫|\left|\operatorname*{Ends}\mathcal{Q}\right|<\left|\operatorname*{Ends}\mathcal{P}\right|and therefore|𝒬|=|Ends𝒬|<|Ends𝒫|=|𝒫|\left|\mathcal{Q}\right|=\left|\operatorname*{Ends}\mathcal{Q}\right|<\left|\operatorname*{Ends}\mathcal{P}\right|=\left|\mathcal{P}\right|; but this would contradict the fact that𝒫\mathcal{P}is minimum-size. Hence,𝒫\mathcal{P}is end-minimal.. Thus, the following claim is stronger than Claim 1:

Claim 2:Any end-minimal path cover ofDDhas an independent cross-cut.

It is Claim 2 that we will be proving.153153153On a sidenote: Is Claim 2 really stronger than Claim 1? Yes, because it can happen that some end-minimal path cover fails to be minimum-size. For example, the path cover{(1,∗,2,∗,3),(4)}\left\{\left(1,\ast,2,\ast,3\right),\ \left(4\right)\right\}in the digraph11223344has this property.

[Proof of Claim 2:We proceed by induction on|V|\left|V\right|.

Base case:Claim 2 is obvious when|V|=0\left|V\right|=0(since∅\varnothingis an independent cross-cut in this case).

Induction step:Consider a multidigraphD=(V,A,ψ)D=\left(V,A,\psi\right)with|V|=N\left|V\right|=N. Assume (as the induction hypothesis) that Claim 2 is already proved for any multidigraph withN−1N-1vertices.

Let𝒫\mathcal{P}be an end-minimal path cover ofDD. We must show that𝒫\mathcal{P}has an independent cross-cut.

Let𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}be the paths in𝒫\mathcal{P}(listed without repetitions), and letv1,v2,…,vkv_{1},v_{2},\ldots,v_{k}be their respective ending points. Thus,{v1,v2,…,vk}=Ends𝒫\left\{v_{1},v_{2},\ldots,v_{k}\right\}=\operatorname*{Ends}\mathcal{P}andk=|Ends𝒫|k=\left|\operatorname*{Ends}\mathcal{P}\right|.

Recall that we must find an independent cross-cut of𝒫\mathcal{P}. If the set{v1,v2,…,vk}\left\{v_{1},v_{2},\ldots,v_{k}\right\}is independent, then we are done (since this set{v1,v2,…,vk}\left\{v_{1},v_{2},\ldots,v_{k}\right\}is clearly a cross-cut of𝒫\mathcal{P}). Thus, we WLOG assume that this is not the case. Hence, there is an arc from some vertexviv_{i}to some vertexvjv_{j}. These two verticesviv_{i}andvjv_{j}are distinct (becauseDDis loopless). Since we can swap our paths𝐩1,𝐩2,…,𝐩k\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}(and thus their ending pointsv1,v2,…,vkv_{1},v_{2},\ldots,v_{k}) at will, we can thus WLOG assume thati=2i=2andj=1j=1. Assume this. Thus, there is an arc fromv2v_{2}tov1v_{1}. We shall refer to this arc as theblue arc, and we will draw it accordingly:154154154This picture illustrates just one representative case, withk=4k=4. The four columns (from left to right) are the four paths𝐩1,𝐩2,𝐩3,𝐩4\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{p}_{3},\mathbf{p}_{4}. Of course, the digraphDDcan have many more arcs than we have drawn on this picture, but we are not interested in them right now.

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We can extend the path𝐩2\mathbf{p}_{2}beyond its ending pointv2v_{2}by inserting the blue arc and the vertexv1v_{1}at its end. This results in a new path, which we denote by𝐩2+v1\mathbf{p}_{2}+v_{1}; this path has ending pointv1v_{1}.

Ifv1v_{1}is the only vertex on the path𝐩1\mathbf{p}_{1}(that is, if the path𝐩1\mathbf{p}_{1}has length00), then we can therefore replace the path𝐩2\mathbf{p}_{2}by𝐩2+v1\mathbf{p}_{2}+v_{1}and remove the length-00path𝐩1\mathbf{p}_{1}from our path cover𝒫\mathcal{P}, and we thus obtain a new path cover𝒬\mathcal{Q}such thatEnds𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}. But this is impossible, since we assumed that𝒫\mathcal{P}is end-minimal. Therefore,v1v_{1}is not the only vertex on𝐩1\mathbf{p}_{1}.

Thus, letvvbe the second-to-last vertex on𝐩1\mathbf{p}_{1}(that is, the vertex that is immediately followed byv1v_{1}). Then, the path𝐩1\mathbf{p}_{1}contains an arc fromvvtov1v_{1}. We shall refer to this arc as thered arc, and we will draw it accordingly:

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\lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -6.2 8.32 C -5.07 3.33 -2.54 0.97 0 0 C -2.54 -0.97 -5.07 -3.33 -6.2 -8.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ . LetD′D^{\prime}be the digraphD∖v1D\setminus v_{1}(that is, the digraph obtained fromDDby removing the vertexv1v_{1}and all arcs that havev1v_{1}as source or target). Let𝐩1′\mathbf{p}_{1}^{\prime}be the result of removing the vertexv1v_{1}and the red arc from the path𝐩1\mathbf{p}_{1}. Then,𝒫′:={𝐩1′,𝐩2,𝐩3,…,𝐩k}\mathcal{P}^{\prime}:=\left\{\mathbf{p}_{1}^{\prime},\mathbf{p}_{2},\mathbf{p}_{3},\ldots,\mathbf{p}_{k}\right\}is a path cover ofD′D^{\prime}. Note that the path𝐩1′\mathbf{p}_{1}^{\prime}has ending pointvv(since it is obtained from𝐩1\mathbf{p}_{1}by removing the last vertex and the last arc, but we know that the second-to-last vertex on𝐩1\mathbf{p}_{1}isvv), whereas the paths𝐩2,𝐩3,…,𝐩k\mathbf{p}_{2},\mathbf{p}_{3},\ldots,\mathbf{p}_{k}have ending pointsv2,v3,…,vkv_{2},v_{3},\ldots,v_{k}. Thus,Ends(𝒫′)={v,v2,v3,…,vk}\operatorname*{Ends}\left(\mathcal{P}^{\prime}\right)=\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}. Here is an illustration of the digraphD′=D∖v1D^{\prime}=D\setminus v_{1}and its path cover𝒫′\mathcal{P}^{\prime}:

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\lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ . Consider the path cover𝒫′\mathcal{P}^{\prime}ofD′D^{\prime}. If we can find an independent cross-cut of𝒫′\mathcal{P}^{\prime}, then we will be done, because any such cross-cut will also be an independent cross-cut of our original path cover{𝐩1,𝐩2,…,𝐩k}=𝒫\left\{\mathbf{p}_{1},\mathbf{p}_{2},\ldots,\mathbf{p}_{k}\right\}=\mathcal{P}. Since the digraphD∖v1D\setminus v_{1}hasN−1N-1vertices155155155because the digraphDDhas|V|=N\left|V\right|=Nvertices, we can find such an independent cross-cut by our induction hypothesis if we can prove that the path cover𝒫′\mathcal{P}^{\prime}is end-minimal (as a path cover ofD′D^{\prime}).

So let us prove this now. Indeed, assume the contrary. Thus,D′D^{\prime}has a path cover𝒬′\mathcal{Q}^{\prime}such thatEnds(𝒬′)\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)is a proper subset ofEnds(𝒫′)\operatorname*{Ends}\left(\mathcal{P}^{\prime}\right). Consider this𝒬′\mathcal{Q}^{\prime}. Note that156156156The symbol “⊊\subsetneq” (note that the stroke only crosses the straight line, not the curved one) means “proper subset of”.

Ends(𝒬′)⊊Ends(𝒫′)={v,v2,v3,…,vk}.\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subsetneq\operatorname*{Ends}\left(\mathcal{P}^{\prime}\right)=\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}.As a consequence,|Ends(𝒬′)|<|{v,v2,v3,…,vk}|=k\left|\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\right|<\left|\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}\right|=k.

Now, we are in one of the following three cases:

Case 1:We havev∈Ends(𝒬′)v\in\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right).

Case 2:We havev∉Ends(𝒬′)v\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)butv2∈Ends(𝒬′)v_{2}\in\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right).

Case 3:We havev∉Ends(𝒬′)v\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)andv2∉Ends(𝒬′)v_{2}\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right).

Let us consider these cases one by one:

•We first consider Case 1. In this case, we havev∈Ends(𝒬′)v\in\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right). In other words, some path𝐩∈𝒬′\mathbf{p}\in\mathcal{Q}^{\prime}ends atvv. Let us extend this path𝐩\mathbf{p}beyondvvby inserting the red arc and the vertexv1v_{1}at its end. Thus, we obtain a path ofDD, which we call𝐩+v1\mathbf{p}+v_{1}. Replacing𝐩\mathbf{p}by𝐩+v1\mathbf{p}+v_{1}in𝒬′\mathcal{Q}^{\prime}, we obtain a path cover𝒬\mathcal{Q}ofDDsuch thatEnds𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}157157157Proof.We obtained𝒬\mathcal{Q}from𝒬′\mathcal{Q}^{\prime}by replacing𝐩\mathbf{p}by𝐩+v1\mathbf{p}+v_{1}. As a consequence of this replacement, the ending pointvvof𝐩\mathbf{p}has been replaced by the ending pointv1v_{1}of𝐩+v1\mathbf{p}+v_{1}. Thus,Ends𝒬\displaystyle\operatorname*{Ends}\mathcal{Q}=(Ends(𝒬′)∖{v})⏟⊆{v2,v3,…,vk}(sinceEnds(𝒬′)⊆{v,v2,v3,…,vk})∪{v1}\displaystyle=\underbrace{\left(\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\setminus\left\{v\right\}\right)}_{\begin{subarray}{c}\subseteq\left\{v_{2},v_{3},\ldots,v_{k}\right\}\\ \text{(since }\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subseteq\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}\text{)}\end{subarray}}\cup\left\{v_{1}\right\}⊆{v2,v3,…,vk}∪{v1}={v1,v2,…,vk}=Ends𝒫.\displaystyle\subseteq\left\{v_{2},v_{3},\ldots,v_{k}\right\}\cup\left\{v_{1}\right\}=\left\{v_{1},v_{2},\ldots,v_{k}\right\}=\operatorname*{Ends}\mathcal{P}.For the same reason, we have|Ends𝒬|=|Ends(𝒬′)|<k=|Ends𝒫|\left|\operatorname*{Ends}\mathcal{Q}\right|=\left|\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\right|<k=\left|\operatorname*{Ends}\mathcal{P}\right|, so thatEnds𝒬≠Ends𝒫\operatorname*{Ends}\mathcal{Q}\neq\operatorname*{Ends}\mathcal{P}. Combining this withEnds𝒬⊆Ends𝒫\operatorname*{Ends}\mathcal{Q}\subseteq\operatorname*{Ends}\mathcal{P}, we conclude thatEnds𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}.. But this contradicts the fact that𝒫\mathcal{P}is end-minimal. Thus, we have obtained a contradiction in Case 1.
•Next, we consider Case 2. In this case, we havev∉Ends(𝒬′)v\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)butv2∈Ends(𝒬′)v_{2}\in\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right). CombiningEnds(𝒬′)⊆{v,v2,v3,…,vk}\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subseteq\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}withv∉Ends(𝒬′)v\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right), we obtain Ends(𝒬′)⊆{v,v2,v3,…,vk}∖{v}={v2,v3,…,vk}.\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subseteq\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}\setminus\left\{v\right\}=\left\{v_{2},v_{3},\ldots,v_{k}\right\}.Fromv2∈Ends(𝒬′)v_{2}\in\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right), we see that some path𝐩∈𝒬′\mathbf{p}\in\mathcal{Q}^{\prime}ends atv2v_{2}. Let us extend this path𝐩\mathbf{p}beyondv2v_{2}by inserting the blue arc and the vertexv1v_{1}at its end. Thus, we obtain a path ofDD, which we call𝐩+v1\mathbf{p}+v_{1}. Replacing𝐩\mathbf{p}by𝐩+v1\mathbf{p}+v_{1}in𝒬′\mathcal{Q}^{\prime}, we obtain a path cover𝒬\mathcal{Q}ofDDsuch thatEnds𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}158158158Proof.We obtained𝒬\mathcal{Q}from𝒬′\mathcal{Q}^{\prime}by replacing𝐩\mathbf{p}by𝐩+v1\mathbf{p}+v_{1}. As a consequence of this replacement, the ending pointv2v_{2}of𝐩\mathbf{p}has been replaced by the ending pointv1v_{1}of𝐩+v1\mathbf{p}+v_{1}. Thus,Ends𝒬\displaystyle\operatorname*{Ends}\mathcal{Q}=(Ends(𝒬′)∖{v2})⏟⊆{v3,v4,…,vk}(sinceEnds(𝒬′)⊆{v2,v3,…,vk})∪{v1}\displaystyle=\underbrace{\left(\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\setminus\left\{v_{2}\right\}\right)}_{\begin{subarray}{c}\subseteq\left\{v_{3},v_{4},\ldots,v_{k}\right\}\\ \text{(since }\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subseteq\left\{v_{2},v_{3},\ldots,v_{k}\right\}\text{)}\end{subarray}}\cup\left\{v_{1}\right\}⊆{v3,v4,…,vk}∪{v1}={v1,v3,v4,…,vk}\displaystyle\subseteq\left\{v_{3},v_{4},\ldots,v_{k}\right\}\cup\left\{v_{1}\right\}=\left\{v_{1},v_{3},v_{4},\ldots,v_{k}\right\}⊊{v1,v2,…,vk}=Ends𝒫.\displaystyle\subsetneq\left\{v_{1},v_{2},\ldots,v_{k}\right\}=\operatorname*{Ends}\mathcal{P}.In other words,Ends𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}.. But this contradicts the fact that𝒫\mathcal{P}is end-minimal. Thus, we have obtained a contradiction in Case 2.
•Finally, we consider Case 3. In this case, we havev∉Ends(𝒬′)v\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)andv2∉Ends(𝒬′)v_{2}\notin\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right). Combining this withEnds(𝒬′)⊆{v,v2,v3,…,vk}\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subseteq\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}, we obtain Ends(𝒬′)⊆{v,v2,v3,…,vk}∖{v,v2}={v3,v4,…,vk},\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\subseteq\left\{v,v_{2},v_{3},\ldots,v_{k}\right\}\setminus\left\{v,v_{2}\right\}=\left\{v_{3},v_{4},\ldots,v_{k}\right\},so that|Ends(𝒬′)|≤|{v3,v4,…,vk}|=k−2\left|\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)\right|\leq\left|\left\{v_{3},v_{4},\ldots,v_{k}\right\}\right|=k-2. Now, adding the trivial path(v1)\left(v_{1}\right)to𝒬′\mathcal{Q}^{\prime}yields a path cover𝒬\mathcal{Q}ofDDsuch thatEnds𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}159159159Proof.We obtained𝒬\mathcal{Q}from𝒬′\mathcal{Q}^{\prime}by adding the trivial path(v1)\left(v_{1}\right), whose ending point isv1v_{1}. Thus,Ends𝒬\displaystyle\operatorname*{Ends}\mathcal{Q}=Ends(𝒬′)⏟⊆{v3,v4,…,vk}∪{v1}⊆{v3,v4,…,vk}∪{v1}={v1,v3,v4,…,vk}\displaystyle=\underbrace{\operatorname*{Ends}\left(\mathcal{Q}^{\prime}\right)}_{\subseteq\left\{v_{3},v_{4},\ldots,v_{k}\right\}}\cup\left\{v_{1}\right\}\subseteq\left\{v_{3},v_{4},\ldots,v_{k}\right\}\cup\left\{v_{1}\right\}=\left\{v_{1},v_{3},v_{4},\ldots,v_{k}\right\}⊊{v1,v2,…,vk}=Ends𝒫.\displaystyle\subsetneq\left\{v_{1},v_{2},\ldots,v_{k}\right\}=\operatorname*{Ends}\mathcal{P}.In other words,Ends𝒬\operatorname*{Ends}\mathcal{Q}is a proper subset ofEnds𝒫\operatorname*{Ends}\mathcal{P}.. But this contradicts the fact that𝒫\mathcal{P}is end-minimal. Thus, we have found a contradiction in Case 3.

So we have obtained a contradiction in each case. Thus, our assumption was false. This shows that the path cover𝒫′\mathcal{P}^{\prime}is end-minimal. As we already said above, this allows us to apply the induction hypothesis toD′D^{\prime}instead ofDD, and conclude that the end-minimal path cover𝒫′\mathcal{P}^{\prime}ofD′D^{\prime}has an independent cross-cut. This independent cross-cut is clearly an independent cross-cut of𝒫\mathcal{P}as well, and thus we have shown that𝒫\mathcal{P}has an independent cross-cut. This proves Claim 2.]

As explained above, this completes the proof of Theorem10.65.

10.2.3Applications

Here are two simple applications of the Gallai–Milgram theorem:

•Remember the Easy Rédei theorem (Theorem4.95), which we proved long ago. It says that each tournament has a Hamiltonian path. We can now prove it again using the Gallai–Milgram theorem: New proof of the Easy Rédei theorem:Indeed, letDDbe a tournament. The Gallai–Milgram theorem shows thatDDhas a path cover with an independent cross-cut160160160See the above proof of Theorem10.65for the definition of a “cross-cut”.. Consider this path cover and this cross-cut. But sinceDDis a tournament, any independent set ofDDhas size≤1\leq 1. Thus, our independent cross-cut must have size≤1\leq 1. Hence, our path cover must consist of11path only (because the size of the path cover equals the size of its cross-cut). But this means that it is a Hamiltonian path (or, more precisely, it consists of a single path, which is necessarily a Hamiltonian path). Hence,DDhas a Hamiltonian path. So we have proved the Easy Rédei theorem (Theorem4.95) again.
•Less obviously, Hall’s Marriage Theorem (Theorem8.23) and the Hall–König matching theorem (Theorem8.31) can be proved again using Gallai–Milgram. Here is how: New proof of the Hall–König matching theorem:Let(G,X,Y)\left(G,X,Y\right)be a bipartite graph. LetDDbe the digraph obtained fromGGby directing each edge so that it goes fromXXtoYY(in other words, each edge with endpointsx∈Xx\in Xandy∈Yy\in Ybecomes an arc with sourcexxand targetyy). Thus, in the digraphDD, no vertex can simultaneously be the source of some arc and the target of some arc. Thus, any path ofDDhas length≤1\leq 1. 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\lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}$\\ $G$ & $D$ \\ \hline\cr\end{tabular}\ \ \ \ \ . As we said, any path ofDDhas length≤1\leq 1. Thus, any path ofDDcorresponds either to a vertex ofGGor to an edge ofGG(depending on whether its length is00or11). Hence, any path cover𝒫\mathcal{P}ofDDnecessarily consists of length-00paths (corresponding to vertices ofGG) and length-11paths (corresponding to edges ofGG); moreover, the edges of𝒫\mathcal{P}(that is, the edges corresponding to the length-11paths in𝒫\mathcal{P}) form a matching ofGG, and the vertices of𝒫\mathcal{P}(that is, the vertices corresponding to length-00paths in𝒫\mathcal{P}) are precisely the vertices that are not matched in this matching. Now, Theorem10.65shows that there exist a path cover𝒫\mathcal{P}ofDDand an independent cross-cutSSof𝒫\mathcal{P}. Consider these𝒫\mathcal{P}andSS. 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We have|S|=|X∩S|+|Y∩S|\left|S\right|=\left|X\cap S\right|+\left|Y\cap S\right|(since the setSSis the union of its two disjoint subsetsX∩SX\cap SandY∩SY\cap S). The setSSis an independent set of the digraphDD, thus also an independent set of the graphDund=GD^{\operatorname*{und}}=G. From this, we easily obtainN⁡(X∩S)⊆Y∖SN\left(X\cap S\right)\subseteq Y\setminus S(since(G,X,Y)\left(G,X,Y\right)is a bipartite graph)161161161Proof.Letv∈N⁡(X∩S)v\in N\left(X\cap S\right). Thus,vvis a vertex with a neighbor inX∩SX\cap S. Letxxbe this neighbor. Then,x∈X∩S⊆Xx\in X\cap S\subseteq X, so that the vertexvvhas a neighbor inXX(namely,xx). Since(G,X,Y)\left(G,X,Y\right)is a bipartite graph, this entails thatv∈Yv\in Y. Furthermore, we havex∈X∩S⊆Sx\in X\cap S\subseteq S. If we hadv∈Sv\in S, then the setSSwould contain two adjacent vertices (namely,vvandxx), which would contradict the fact thatSSis an independent set ofGG. Thus, we havev∉Sv\notin S. Combiningv∈Yv\in Ywithv∉Sv\notin S, we obtainv∈Y∖Sv\in Y\setminus S. Forget that we fixedvv. We thus have shown thatv∈Y∖Sv\in Y\setminus Sfor eachv∈N⁡(X∩S)v\in N\left(X\cap S\right). In other words,N⁡(X∩S)⊆Y∖SN\left(X\cap S\right)\subseteq Y\setminus S.. Therefore,|N⁡(X∩S)|≤|Y∖S|\left|N\left(X\cap S\right)\right|\leq\left|Y\setminus S\right|, so that|Y∖S|≥|N⁡(X∩S)|\left|Y\setminus S\right|\geq\left|N\left(X\cap S\right)\right|. Hence, |Y|\displaystyle\left|Y\right|=|Y∖S|⏟≥|N⁡(X∩S)|+|Y∩S|≥|N⁡(X∩S)|+|Y∩S|⏟=|S|−|X∩S|(since|S|=|X∩S|+|Y∩S|)\displaystyle=\underbrace{\left|Y\setminus S\right|}_{\geq\left|N\left(X\cap S\right)\right|}+\left|Y\cap S\right|\geq\left|N\left(X\cap S\right)\right|+\underbrace{\left|Y\cap S\right|}_{\begin{subarray}{c}=\left|S\right|-\left|X\cap S\right|\\ \text{(since }\left|S\right|=\left|X\cap S\right|+\left|Y\cap S\right|\text{)}\end{subarray}}=|N⁡(X∩S)|+|S|−|X∩S|.\displaystyle=\left|N\left(X\cap S\right)\right|+\left|S\right|-\left|X\cap S\right|.(68) Now, letMMbe the set of edges ofGGcorresponding to the length-11paths in our path cover𝒫\mathcal{P}. As we already mentioned, this setMMis a matching ofGG(since two paths in𝒫\mathcal{P}cannot have a vertex in common). The vertices that are not matched inMMare precisely the vertices that don’t belong to any of the length-11paths in𝒫\mathcal{P}; in other words, they are the vertices that belong to length-00paths in𝒫\mathcal{P}(since𝒫\mathcal{P}is a path cover, and any path has length≤1\leq 1). We letppbe the number of such vertices that lie inXX, and we letqqbe the number of such vertices that lie inYY. Thus, our path cover𝒫\mathcal{P}contains exactlyp+qp+qlength-00paths: namely,pplength-00paths consisting of a vertex inXXandqqlength-00paths consisting of a vertex inYY. Hence, the path cover𝒫\mathcal{P}consists of|M|+p+q\left|M\right|+p+qpaths in total (since it contains|M|\left|M\right|many length-11paths). The setSScontains exactly one vertex from each of these|M|+p+q\left|M\right|+p+qpaths (sinceSSis a cross-cut of𝒫\mathcal{P}); therefore, |S|=|M|+p+q.\left|S\right|=\left|M\right|+p+q.(69) Each vertexy∈Yy\in Ythat is matched inMMbelongs to exactly oneMM-edge (namely, to itsMM-edge), and conversely, eachMM-edge contains exactly one vertex inYY(which, of course, is matched inMM). Thus, the map {vertices inYthat are matched inM}\displaystyle\left\{\text{vertices in }Y\text{ that are matched in }M\right\}→M,\displaystyle\rightarrow M,y\displaystyle y↦(theM-edge ofy)\displaystyle\mapsto\left(\text{the }M\text{-edge of }y\right)is a bijection. Hence, the bijection principle yields (# of vertices inYthat are matched inM)=|M|.\left(\text{\# of vertices in }Y\text{ that are matched in }M\right)=\left|M\right|.(70) On the other hand, the setYYcontains exactlyqqvertices that are not matched inMM(by the definition ofqq). Therefore,YYcontains exactly|Y|−q\left|Y\right|-qvertices that are matched inMM. In other words, (# of vertices inYthat are matched inM)=|Y|−q.\left(\text{\# of vertices in }Y\text{ that are matched in }M\right)=\left|Y\right|-q. Comparing this with (70), we obtain|M|=|Y|−q\left|M\right|=\left|Y\right|-q. In other words, |M|+q=|Y|.\left|M\right|+q=\left|Y\right|.(71)The same argument (but applied toXXandppinstead ofYYandqq) yields |M|+p=|X|.\left|M\right|+p=\left|X\right|.(72)Hence, (69) becomes |S|=|M|+p⏟=|X|(by (72))+q=|X|+q.\left|S\right|=\underbrace{\left|M\right|+p}_{\begin{subarray}{c}=\left|X\right|\\ \text{(by (\ref{pf.HKMT.proof-using-gm.4}))}\end{subarray}}+\,q=\left|X\right|+q.(73) Now, from (71), we obtain |M|+q\displaystyle\left|M\right|+q=|Y|\displaystyle=\left|Y\right|≥|N(X∩S)|+|S|⏟=|X|+q(by (73))−|X∩S|(by (68))\displaystyle\geq\left|N\left(X\cap S\right)\right|+\underbrace{\left|S\right|}_{\begin{subarray}{c}=\left|X\right|+q\\ \text{(by (\ref{pf.HKMT.proof-using-gm.4b}))}\end{subarray}}-\left|X\cap S\right|\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.HKMT.proof-using-gm.1})}\right)=|N⁡(X∩S)|+|X|+q−|X∩S|\displaystyle=\left|N\left(X\cap S\right)\right|+\left|X\right|+q-\left|X\cap S\right|=|N⁡(X∩S)|+|X|−|X∩S|+q.\displaystyle=\left|N\left(X\cap S\right)\right|+\left|X\right|-\left|X\cap S\right|+q.Cancellingqq, we obtain |M|≥|N⁡(X∩S)|+|X|−|X∩S|.\left|M\right|\geq\left|N\left(X\cap S\right)\right|+\left|X\right|-\left|X\cap S\right|.(74)Thus, we have found a matchingMMofGGand a subsetUUofXX(namely,U=X∩SU=X\cap S) such that|M|≥|N⁡(U)|+|X|−|U|\left|M\right|\geq\left|N\left(U\right)\right|+\left|X\right|-\left|U\right|. This proves the Hall–König matching theorem (once again). New proof of Hall’s Marriage Theorem:Proceed as in the proof of the Hall–König matching theorem that we just gave. But now assume that our bipartite graph(G,X,Y)\left(G,X,Y\right)satisfies the Hall condition (i.e., we have|N⁡(A)|≥|A|\left|N\left(A\right)\right|\geq\left|A\right|for each subsetAAofXX). Hence, in particular,|N⁡(X∩S)|≥|X∩S|\left|N\left(X\cap S\right)\right|\geq\left|X\cap S\right|. Therefore, (74) becomes |M|≥|N⁡(X∩S)|⏟≥|X∩S|+|X|−|X∩S|≥|X|.\left|M\right|\geq\underbrace{\left|N\left(X\cap S\right)\right|}_{\geq\left|X\cap S\right|}+\left|X\right|-\left|X\cap S\right|\geq\left|X\right|.Hence, Proposition8.18(e)shows that the matchingMMisXX-complete. Thus,GGhas anXX-complete matching (namely,MM). This proves Hall’s Marriage Theorem (once again).

Exercise 10.69.

Letccandrrbe two positive integers. LetTTbe a tournament with more thanrcr^{c}vertices. Each arc ofTTis colored with one of thecccolors1,2,…,c1,2,\ldots,c. Prove thatTThas a monochromatic path of lengthrr.

(A path is said to bemonochromaticif all its arcs have the same color.)

[Hint:Induct oncc, and apply Gallai-Milgram to a certain digraph in the induction step.]

10.3Path-missing sets

We move on to less well-trodden ground.

Menger’s theorem (one of the many) is from 1927; the Gallai–Milgram theorem is from 1960. One might think that everything that can be said about paths in graphs has been said long ago.

Apparently, this is not the case. In 2017, when trying to come up with a homework exercise for a previous iteration of this course, I was experimenting with paths in Python. Specifically, I was looking at digraphsD=(V,A,ψ)D=\left(V,A,\psi\right)with two distinct verticesssandttselected. Inspired by the arc-Menger theorems, I was looking at the subsetsBBofAAthat could be removed without disconnectingssfromtt(more precisely, without destroying all paths fromsstott). I noticed that the number of such subsetsBBseemed to be even wheneverDDhas a cycle or a “useless arc” (i.e., an arc that is used by no path fromsstott)162162162With one exception: IfA=∅A=\varnothing, then it is odd., and odd otherwise.

I could not prove this observation. Soon after, Joel Brewster Lewis and Lukas Katthän came up with a proof and multiple stronger results. The proofs can now be found in a joint preprint[GrKaLe21].

The first way to strengthen the observation is to replace the parity claim (i.e., the claim that the number is even or odd depending on cycles and useless arcs) by a stronger claim about an alternating sum. This is an instance of a general phenomenon, in which a statement of the form “the number of some class of things is even” can often be replaced by a stronger statement of the form “we can assign a plus or minus sign to each of these things, and then the total number of plus signs equals the total number of minus signs”. The stronger statement is as follows:

Theorem 10.71(Grinberg–Lewis–Katthän).

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be a multidigraph. Letssandttbe two distinct vertices ofDD. A subsetBBofAAwill be calledpath-missingifDDhas a path fromsstottthat does not use any of the arcs inBB(that is, a path fromsstottthat would not be destroyed if we remove all arcs inBBfromDD). (In the terminology of Definition10.3, this is the same as saying thatBBisnotanss-tt-arc-separator.)

Let𝐌\mathbf{M}be the set of all path-missing subsets ofAA.

(a)IfDDhas an arc that is not used by any path fromsstott(this is what we call a “useless arc”), then ∑B∈𝐌(−1)|B|=0\sum\limits_{B\in\mathbf{M}}\left(-1\right)^{\left|B\right|}=0(and thus|𝐌|\left|\mathbf{M}\right|is even).
(b)IfDDhas a cycle, then ∑B∈𝐌(−1)|B|=0\sum\limits_{B\in\mathbf{M}}\left(-1\right)^{\left|B\right|}=0(and thus|𝐌|\left|\mathbf{M}\right|is even).
(c)IfA=∅A=\varnothing, then ∑B∈𝐌(−1)|B|=0\sum\limits_{B\in\mathbf{M}}\left(-1\right)^{\left|B\right|}=0(and thus|𝐌|\left|\mathbf{M}\right|is even).
(d)In all other cases, we have ∑B∈𝐌(−1)|B|=(−1)|A|−|V′|,\sum\limits_{B\in\mathbf{M}}\left(-1\right)^{\left|B\right|}=\left(-1\right)^{\left|A\right|-\left|V^{\prime}\right|},whereV′V^{\prime}is the set of all vertices ofDDthat have outdegree>0>0(and thus|𝐌|\left|\mathbf{M}\right|is odd).

Example 10.72.

LetD=(V,A,ψ)D=\left(V,A,\psi\right)be the following digraph:

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{{}{{}}{}{}{{}}{{{\lx@inpgf@ignorespaces}}{{{\lx@inpgf@ignorespaces}}{\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{fill=#000000} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{49.96582pt}{59.75198pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 69.14 82.68)} \lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke-dasharray=none,stroke-dashoffset=0.0pt} \lxSVG@begingroup@{stroke-linecap=round} \lxSVG@begingroup@{stroke-linejoin=round} \lxSVG@drawpath@unclipped{M -4.21 5.32 C -3.44 2.13 -1.73 0.62 0 0 C -1.73 -0.62 -3.44 -2.13 -4.21 -5.32}{fill:none} \lxSVG@closescope \lxSVG@closescope }}{{\lx@inpgf@ignorespaces}}}}\lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{27.73006pt}{63.685pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 38.37 88.12)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ .Letssandttbe the vertices labelledssandtthere. Then,DDhas neither a cycle nor a “useless arc”, and its arc setAAis nonempty; thus, Theorem10.71(d)applies. The path-missing subsets ofAAare the three sets{a,b,c,d}\left\{a,b,c,d\right\},{c,e}\left\{c,e\right\}and{d,e,f}\left\{d,e,f\right\}as well as all their subsets (such as{b,c,d}\left\{b,c,d\right\}). In other words,

𝐌\displaystyle\mathbf{M}={all subsets of{a,b,c,d}}∪{all subsets of{c,e}}\displaystyle=\left\{\text{all subsets of }\left\{a,b,c,d\right\}\right\}\cup\left\{\text{all subsets of }\left\{c,e\right\}\right\}∪{all subsets of{d,e,f}}\displaystyle\ \ \ \ \ \ \ \ \ \ \cup\left\{\text{all subsets of }\left\{d,e,f\right\}\right\}={∅,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c},{b,d},\displaystyle=\{\varnothing,\ \left\{a\right\},\ \left\{b\right\},\ \left\{c\right\},\ \left\{d\right\},\ \left\{a,b\right\},\ \left\{a,c\right\},\ \left\{a,d\right\},\ \left\{b,c\right\},\ \left\{b,d\right\},{c,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d},\displaystyle\ \ \ \ \ \ \ \ \ \ \left\{c,d\right\},\ \left\{a,b,c\right\},\ \left\{a,b,d\right\},\ \left\{a,c,d\right\},\ \left\{b,c,d\right\},\ \left\{a,b,c,d\right\},{e},{c,e},{f},{d,e},{e,f},{d,f},{d,e,f}}.\displaystyle\ \ \ \ \ \ \ \ \ \ \left\{e\right\},\ \left\{c,e\right\},\ \left\{f\right\},\ \left\{d,e\right\},\ \left\{e,f\right\},\ \left\{d,f\right\},\ \left\{d,e,f\right\}\}.Hence, the sum∑B∈𝐌(−1)|B|\sum\limits_{B\in\mathbf{M}}\left(-1\right)^{\left|B\right|}has1111addends equal to−1-1and1212addends equal to11; thus, this sum equals to11. This is precisely the value(−1)|A|−|V′|=(−1)6−4=1\left(-1\right)^{\left|A\right|-\left|V^{\prime}\right|}=\left(-1\right)^{6-4}=1predicted by Theorem10.71(d).

Proof 10.73(Proof of Theorem10.71.).

See[GrKaLe21, Theorem 1.3](where𝐌\mathbf{M}is denoted byPM(D)\operatorname*{PM}\left(D\right), and where arcs are called “edges”). Of course, part(c)is obvious, and part(a)is easy (since inserting a useless arc into a setB∈𝐌B\in\mathbf{M}or removing it from a setB∈𝐌B\in\mathbf{M}always results in a set in𝐌\mathbf{M}). Parts(b)and(d)are the interesting ones. The proof in[GrKaLe21, Theorem 1.3]relies on a recursive argument (“deletion-contraction”) in which we pick an arc with sourcessand consider the two smaller digraphsD∖aD\setminus aandD╱aD\diagup aobtained (respectively) by deleting the arcaafromDDand by “contracting”aa“to a point”.

Further levels of strength can be reached by treating𝐌\mathbf{M}as a topological space. Indeed,𝐌\mathbf{M}is not just a random collection of sets of arcs, but actually asimplicial complex(since any subset of a path-missing subset ofAAis again path-missing). Simplicial complexes are known to be a combinatorial model for topological spaces, and in particular they have homology groups, homotopy types, etc.. Thus, in particular, we can ask ourselves how the topological space corresponding to the simplicial complex𝐌\mathbf{M}looks like. This, too, has been answered in[GrKaLe21, Theorem 1.3]: It is homotopic to a sphere or a ball (depending on the existence of cycles or “useless arcs”); its dimension can also be determined explicitly. (The sum∑B∈𝐌(−1)|B|\sum\limits_{B\in\mathbf{M}}\left(-1\right)^{\left|B\right|}discussed above is, of course, its reduced Euler characteristic.) Even stronger statements can be made usingdiscrete Morse theory(see again[GrKaLe21]for details).

10.4Elser’s sums

We now return to undirected (multi)graphs. Here is a result found by Veit Elser in 1984 ([Elser84, Lemma 1]), as a lemma for his work in statistical mechanics:163163163I have restated the result beyond recognition; see[Grinbe21, Remark 1.4]for why Theorem10.74actually implies[Elser84, Lemma 1].

Theorem 10.74(Elser’s theorem, in my version).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph with at least one edge. Fix a vertexv∈Vv\in V.

IfF⊆EF\subseteq E, then anFF-pathshall mean a path ofGGsuch that all edges of this path belong toFF. In other words, it means a path of the spanning subgraph(V,F,φ∣F)\left(V,F,\varphi\mid_{F}\right).

Ife∈Ee\in Eis an edge andF⊆EF\subseteq Eis a subset, then we say thatFFinfectseeif there exists anFF-path fromvvto some endpoint ofee. (The terminology is inspired by the idea that some infectious disease starts atvvand spreads along theFF-edges.)

(Note that if an edgeeecontains the vertexvv, thenanysubsetFFofEE(even the empty set) infectsee, because(v)\left(v\right)is a trivialFF-path fromvvtovv.)

Then,

∑F⊆Einfectsevery edgee∈E(−1)|F|=0.\sum\limits_{\begin{subarray}{c}F\subseteq E\text{ infects}\\ \text{every edge }e\in E\end{subarray}}\left(-1\right)^{\left|F\right|}=0.

Example 10.75.

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be the following graph:

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\lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{60.83853pt}{25.3907pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 84.18 35.13)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}}\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}}{}{{}} {{{{{}}{}{}{}{}{{}}}}}{}{{{{{}}{}{}{}{}{{}}}}}{{}}{}{}{}{}{}{{{}{}}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@begingroup@{stroke=#000000} \lxSVG@setlinewidth{\the\pgflinewidth}\lxSVG@begingroup@{stroke-width=1.2pt} \lx@inpgf@ignorespaces{}\lxSVG@stroke\lxSVG@drawpath@unclipped{M 65.83 0 L 11.74 0}{fill:none} \lx@inpgf@ignorespaces\hbox{\hbox{{\lxSVG@begingroup@{_scopebegin=1} {{}{}{{ {}{}}}{ {}{}} {{}{{\lx@inpgf@ignorespaces}}}{{}{\lx@inpgf@ignorespaces}}{}{{}{\lx@inpgf@ignorespaces}} {\lx@inpgf@ignorespaces }{{{{\lx@inpgf@ignorespaces}}\lxSVG@begingroup@{_scopebegin=1} \lxSVG@transformcm{1.0}{0.0}{0.0}{1.0}{25.52881pt}{-10.37744pt}\lxSVG@begingroup@{transform=matrix(1.0 0.0 0.0 1.0 35.32 -14.36)} \pgfsys@hbox{62}\lxSVG@closescope }}} \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope }}} \lxSVG@closescope \lxSVG@closescope {\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}{\lx@inpgf@ignorespaces}\hss}\lxSVG@discardpath\lxSVG@closescope \hss}}\lxSVG@closescope\endpgfpicture}}\ \ ,and letvvbe the vertex labelledvv. Then, the subsets ofEEthat infect every edge are

{1,2},{1,4},{3,4},{1,2,3},{1,3,4},{1,2,4},{2,3,4},{1,2,3,4}.\left\{1,2\right\},\ \ \left\{1,4\right\},\ \ \left\{3,4\right\},\ \ \left\{1,2,3\right\},\ \ \left\{1,3,4\right\},\ \ \left\{1,2,4\right\},\ \ \left\{2,3,4\right\},\ \ \left\{1,2,3,4\right\}.Thus,

∑F⊆Einfectsevery edgee∈E(−1)|F|\displaystyle\sum\limits_{\begin{subarray}{c}F\subseteq E\text{ infects}\\ \text{every edge }e\in E\end{subarray}}\left(-1\right)^{\left|F\right|}=(−1)2+(−1)2+(−1)2+(−1)3+(−1)3+(−1)3+(−1)3+(−1)4\displaystyle=\left(-1\right)^{2}+\left(-1\right)^{2}+\left(-1\right)^{2}+\left(-1\right)^{3}+\left(-1\right)^{3}+\left(-1\right)^{3}+\left(-1\right)^{3}+\left(-1\right)^{4}=0,\displaystyle=0,exactly as predicted by Theorem10.74.

Proof 10.77(Proof of Theorem10.74.).

Elser’s proof is somewhat complicated. I give a different proof in[Grinbe21, Theorem 1.2], which is elementary and nice if I may say so myself.

My proof should also be not very hard to discover, once you have the following hint: It suffices to prove the equality

∑F⊆Edoesnotinfectevery edgee∈E(−1)|F|=0\sum\limits_{\begin{subarray}{c}F\subseteq E\text{ does {not} infect}\\ \text{every edge }e\in E\end{subarray}}\left(-1\right)^{\left|F\right|}=0(because the total sum∑F⊆E(−1)|F|\sum\limits_{F\subseteq E}\left(-1\right)^{\left|F\right|}is known to be00). In order to prove this equality, we equip the setEEwith some total order (it doesn’t matter how; we can just rank the edges arbitrarily), and we make the following definition: IfF⊆EF\subseteq Eis a subset that doesnotinfect every edgee∈Ee\in E, then we letε⁡(F)\varepsilon\left(F\right)be the smallest (with respect to our chosen total order) edge that is not infected byFF. Now, you can show that ifF⊆EF\subseteq Eis a subset that doesnotinfect every edgee∈Ee\in E, then the set164164164The symbol△\bigtriangleupstands for the symmetric difference of two sets. Recall its definition: IfXXandYYare two sets, then theirsymmetric differenceX△YX\bigtriangleup Yis defined to be the set(X∪Y)∖(X∩Y)\displaystyle\left(X\cup Y\right)\setminus\left(X\cap Y\right)=(X∖Y)∪(Y∖X)\displaystyle=\left(X\setminus Y\right)\cup\left(Y\setminus X\right)={all elements that belong toexactlyone ofXandY}.\displaystyle=\left\{\text{all elements that belong to {exactly} one of }X\text{ and }Y\right\}.F′:=F△{ε⁡(F)}F^{\prime}:=F\bigtriangleup\left\{\varepsilon\left(F\right)\right\}(that is, the set obtained fromFFby insertingε⁡(F)\varepsilon\left(F\right)ifε⁡(F)∉F\varepsilon\left(F\right)\notin Fand by removingε⁡(F)\varepsilon\left(F\right)ifε⁡(F)∈F\varepsilon\left(F\right)\in F) has the same property (viz., it does not infect every edgee∈Ee\in E) and satisfiesε⁡(F′)=ε⁡(F)\varepsilon\left(F^{\prime}\right)=\varepsilon\left(F\right). This entails that the addends in the sum∑F⊆Edoesnotinfectevery edgee∈E(−1)|F|\sum\limits_{\begin{subarray}{c}F\subseteq E\text{ does {not} infect}\\ \text{every edge }e\in E\end{subarray}}\left(-1\right)^{\left|F\right|}cancel each other in pairs (namely, the addend for a given setFFcancels the addend for the setF′=F△{ε⁡(F)}F^{\prime}=F\bigtriangleup\left\{\varepsilon\left(F\right)\right\}), and thus the whole sum is00.

Elser’s theorem, too, can be generalized and strengthened. The strengthening is similar to what we did with Theorem10.71: We treat the set of all “non-pandemic-causing subsets” (i.e., of all subsetsF⊆EF\subseteq Ethatdon’tinfect every edge) as a simplicial complex (since a subset of a non-pandemic-causing subset is again non-pandemic-causing), and analyze this complex as a topological space. The claim of Theorem10.74then says that the reduced Euler characteristic of this space is00; but we can actually show that this space is contractible (i.e., homotopy-equivalent to a point). Even better, we can prove that the simplicial complex of all non-pandemic-causing subsets iscollapsible(a combinatorial property that is stronger than contractibility of the corresponding space). See[Grinbe21, §5]for definitions and proofs.

We can furthermore generalize the theorem. One way to do so is to replace our “patient zero”vvby a set of vertices. This leads to a much less trivial situation. The recent paper[DHLetc19]by Dorpalen-Barry, Hettle, Livingston, Martin, Nasr, Vega and Whitlatch proves some results and asks some questions (that are still open as of 2022).

A different direction in which Elser’s theorem can be generalized is more fundamental: It turns out that the theorem is not really about graphs and edges. Instead, there is a general structure that I call a “shade map”, which always leads to a certain sum being00. See[Grinbe21, §4]for the details of this generalization. I will not explain it here, but I will state one more particular case of it ([Grinbe21, Theorem 3.2]), which replaces edges by vertices throughout Theorem10.74:

Theorem 10.78(vertex-Elser’s theorem).

LetG=(V,E,φ)G=\left(V,E,\varphi\right)be a multigraph with at least two vertices. Fix a vertexv∈Vv\in V.

IfW⊆VW\subseteq V, then aWW-vertex-pathshall mean a path𝐩\mathbf{p}such that all intermediate vertices of𝐩\mathbf{p}belong toWW. (Recall that the “intermediate vertices of𝐩\mathbf{p}” mean all vertices of𝐩\mathbf{p}except for the starting and ending points of𝐩\mathbf{p}.) (Note that any path of length≤1\leq 1is automatically aWW-vertex-path, since it has no intermediate vertices.)

Ifw∈V∖{v}w\in V\setminus\left\{v\right\}is any vertex, andW⊆V∖{v}W\subseteq V\setminus\left\{v\right\}is any subset, then we say thatWWvertex-infectswwif there exists aWW-vertex-path fromvvtoww. (This is always true whenwwis a neighbor ofvv.)

Then,

∑W⊆V∖{v}vertex-infectsevery vertexw∈V∖{v}(−1)|W|=0.\sum\limits_{\begin{subarray}{c}W\subseteq V\setminus\left\{v\right\}\text{ vertex-infects}\\ \text{every vertex }w\in V\setminus\left\{v\right\}\end{subarray}}\left(-1\right)^{\left|W\right|}=0.

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An introduction to graph theory(Text for Math 530 in Spring 2022 at Drexel University)

Abstract

Contents

1Preface

1.1What is this?

1.1.1Remarks

1.2Notations

2Simple graphs

2.1Definitions

Definition 2.1.

Example 2.2.

Example 2.3.

Definition 2.4.

Example 2.5.

2.2Drawing graphs

Definition 2.6.

Example 2.7.

2.3A first fact: The Ramsey numberR⁡(3,3)=6R\left(3,3\right)=6

Proposition 2.8.

Definition 2.9.

Example 2.10.

Proof 2.11(Proof of Proposition2.8.).

Proposition 2.13.

Exercise 2.14.

2.4Degrees

2.4.1Definition and basic properties

Definition 2.15.

Proposition 2.16.

Proof 2.17.

Proposition 2.18(Euler 1736).

Proof 2.19.

Corollary 2.20(handshake lemma).

Proof 2.21.

Proposition 2.22.

Proof 2.23.

2.4.2Mantel’s theorem

Theorem 2.24(Mantel’s theorem).

Example 2.25.

Proof 2.26(Proof of Mantel’s theorem.).

Exercise 2.27.

Theorem 2.28(Turan’s theorem).

Exercise 2.29.

Exercise 2.31.

Exercise 2.33.

Exercise 2.34.

2.5Graph isomorphism

Definition 2.36.

Proposition 2.37.

Proposition 2.38.

Proposition 2.39.

Proposition 2.40.

Proof 2.41.

2.6Some families of graphs

2.6.1Complete and empty graphs

Definition 2.42.

2.6.2Path and cycle graphs

Definition 2.43.

Definition 2.44.

2.6.3Kneser graphs

Example 2.45.

2.7Subgraphs

Definition 2.46.

Proposition 2.47.

Proof 2.48.

Example 2.49.

Example 2.50.

Proposition 2.51.

Proof 2.52.

Example 2.54.

Exercise 2.55.

Exercise 2.56.

2.8Disjoint unions

Definition 2.58.

2.9Walks and paths

2.9.1Definitions

Definition 2.59.

Example 2.60.

Exercise 2.61.

2.9.2Composing/concatenating and reversing walks

Proposition 2.62.