Graphs and hypergraphs (North-Holland mathematical library)

GRAPHS A N D HYPERGRAPHS North-Holland Mathematical Library Board of Advisory Editors: M. Artin, H. Bass, J. Eells, ...

Author: Claude Berge

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GRAPHS A N D HYPERGRAPHS

North-Holland Mathematical Library Board of Advisory Editors:

M. Artin, H. Bass, J. Eells, W. Feit, P. J. Freyd, F. W. Gehring, H. Halberstam, L. V. Hormander, M. Kac, J. H. B. Kemperman, H. A. Lauwerier, W. A. J. Luxemburg, F. P. Peterson, I. M. Singer and A. C. Zaanen

VOLUME 6

1973

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM * LONDON AMERICAN ELSEVIER PUBLISHING COMPANY, INC. - NEW YORK

Graphs and Hyper gr aphs CLAUDE BERGE University of Paris

Translated by Edward Minieka

1973

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM . LONDON AMERICAN ELSEVIER PUBLISHING COMPANY, INC. - NEW YORK

0 NORTH-HOLLAND PUBLISHING COMPANY - 1973 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyrighf owner

Library of Congress Catalog Card Number: 72-88288 North-Holland ISBN series: 0 7204 2450 X North-Holland ISBN volume: 0 7204 2453 4 American Elsevier ISBN: 0 444 I0399 6

Translation and revised edition of

GRAPHES ET HYPERGRAPHES ODUNOD, Paris 1970

Publishers:

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM NORTH-HOLLAND PUBLISHING COMPANY, LTD. - LONDON Sole Distributors for U.S.A. and Canada:

AMERICAN ELSEVIER PUBLISHING COMPANY, INC. 52 VANDERBILT AVENUE NEW YORK. N.Y. 10017

PRINTED IN GREAT BRITAIN

TO JEAN-MICHEL

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FOREWORD

Graph theory has had an unusual development. Problems involving graphs first appeared in the mathematical folklore as puzzles (e.g. Konigsberg bridge problem). Later, graphs appeared in electrical engineering (Kirchhof's Law), chemistry, psychology and economics before becoming a unified field of study. Today, graph theory is one of the most flourishing branches of modern algebra with wide applications to combinatorial problems and to classical algebraic problems (Group Theory, with Cayley, Ore, Frucht, Sabidussi, etc. ; Category Theory, with Pultr, Hedrlin, etc.). Graph theory as a separate entity has had its development shaped largely by operational researchers preoccupied with practical problems. It was with these practical problems in mind that we wrote our first book Thiorie des graphes et ses applications published by Dunod in January 1958. This text hoped to unify the various results then scattered through the literature. For this purpose, we emphasized two major areas. The first of these areas was the networkpow theory of Ford and Fulkerson which was beginning to transcend analytic techniques. This theory gave new proofs for more than a dozen graph theory results including some famous theorems by Konig and by Menger. The second area was the theory of alternating chains which started with Petersen sixty years earlier, but which appeared in optimization problems only in 1957. These two areas had many curious similarities; however, the integer linear programs that they solved did not overlap. Now, more than ever, we believe that these two areas should form the foundation of graph theory. The first mathematicians to work in graph theory (in particular the thriving Hungarian school with D. Konig, P. Erdos, P. Turin, T. Gallai, G. Hajbs, etc.) considered mainly undirected graphs, and this could lead students to believe that there are two theories-one for directed graphs and one for undirected graphs. This book is written with the viewpoint that there is only one kind of graph (directed) and only one theory for graphs. This is reasonable because a result for an undirected graph can be interpreted as a result for a directed graph in which the direction of the arcs does not matter. Convii

viii

FOREWORD

versely, a result for a directed graph can be interpreted for an undirected graph by replacing each edge of the undirected graph with two oppositely directed arcs with the same endpoints. Since 1957, research in graph theory has assumed astonishing proportions. Results have appeared from all over the world, and some of the conjectures of our first book have been solved notably by our students in Paris from 1959 to 1964 (in particular, the late Alain Ghouila-Houri, whose work frequently appears in this text) and by Soviet mathematicians (in particular, A. A. Zykov, V. G . Vizing, L. M. Vitaver, M. K. Goldberg, L. P. Varvak, etc. following the Russian edition of our first text). Thus, because of this embarrassment of riches, this book is vastly more extensive, but still cannot treat very specific applications.(l) The concept of a matroid due to H. Whitney and developed by W. Tutte has made possible an axiomatic study of cycles and trees. However, we cannot treat this algebraic aspect of graph theory too extensively without straying from our purpose. Similarly, new techniques have appeared for topological graphs, but these would also take us astray. Such strictly topological problems will be the subject of a future work.(a) On the other hand, this book intends to present a systematic study of the theory of hypergraphs. A hypergraph is defined to be a family of hyperedges which are sets of vertices of cardinality not necessarily 2 (as for graphs). Given a graph, a hypergraph can be defined by its cliques, or by its spanning trees or by its cycles. Thus, the theory of hypergraphs can generate simultaneously several results for graphs. The formulation of combinatorial problems in terms of hypergraphs often gives surprisingly simple results that will look very familiar to graph theorists. At the Balatonfured Conference (1969), P. Erdos and A. Hajnal asked us why we would use hypergraphs for problems that can be also formulated in terms of graphs. The answer is that by using hypergraphs, one deals with (l) There are several graph theory texts that emphasize operational research problems; in particular: C. Berge and A. Ghouila-Houri, Programmes, jeux, et rlseau de transport, Part 11. Dunod, Paris, 1962 (English edition, Methuen, London; Wiley, New York, 1965; German edition, Teubner, Leipzig, 1967; Spanish edition, Compania Edit. Continental, Mexico, 1965); L. R. Ford and D. R. Fulkerson, Ffows in Networks, Princeton Press, 1962 (French edition, GauthierVillars, 1967); R. G.Busacker and T. L. Saaty, Finire Graphs andNetworks, McGraw-Hill, 1965; A. Kaufman, Introduction a la combinatorique en vue des applications, Dunod, 1969 (English edition to appear); B. Roy, AIgebre nioderne et thPorie des graphes, Volume 1, Dunod, 1969; Volume 2, 1970; and, finally, A. A. Zykov, Graph Theory (in Russian), NAUKA Publishing House, Siberian Branch, Novosibirsk, 1969. (a) The topological aspects of graph theory will be treated separately in another book to include such topics as the planar representation of graphs, genus, thickness, crossing number of non-planar graphs, proof of the Heawood conjecture by Ringel and Youngs, Edrnonds’ methods, etc.

FOREWORD

ix

generalizations of familiar concepts. Thus, hypergraphs can be used to simplify as well as to generalize. This English edition contains some results that appeared too late for the original French edition, especially the Chvdtal existence theorem for hamiltonian cycles (Chapter 10) and the Lovdsz proof for the first perfect graph conjecture (Chapter 20). An index of all definitions is given at the end of the text so that the reader can pass over chapters without much loss of continuity. Theorems are appended with the name of their first discoverer and the year of discovery. Sometimes, an old or fundamental result is treated as a corollary, and the theorem from which it is derived is attributed to a recent author. This is done purely for didactic purposes and is not intended in any way to diminish the importance of results that have been generalized. A bibliography arranged according to chapters is also found at the end of the book. We first wish to thank Michel Las Vergnas and Jean-Claude Fournier who have made many notable and original contributions and alterations to this text, also to Pierre Rosenstiehl for the assistance given us during our weekly meetings. We wish to thank all those who have helped with suggestions, in particular, J. C. Bermond, J. A. Bondy, P. Camion, U. S. R. Murty, L. Lovdsz, J. M. Pla, and W. T. Tutte. C. Berge


TABLE OF CONTENTS PART ONE-GRAPHS CHAPTER 1. BASIC CONCEPTS

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3. List of symbols . . . . . . . . . . . . . . . . 1.Graphs

2. Basicdefinitions

CHAPTER 2. CYCLOMATIC NUMBER 1. Cycles and cocycles . . . . . . . 2. Cycles in a planar graph . . . . .

. . . . . . . . . . . . . . . .

CHAPTER 3 . TREES AND ARBORESCENCES 1. Trees and cotrees . . . . . . . . . . . . 2. Strongly connected graphs and graphs without circuits 3. Arborescences . . . . . . . . . . . . 4 . Injective. functional and semi-functional graphs . . 5 . Counting trees . . . . . . . . . . . . CHAPTER 4

.

. . . . . . . . . . . . . . . .

. . . .

. . . . . . . . . . . . . . . . . . . . . . . . . .

3. Centres and radii of a quasi-strongly connected graph . . . . . 4. Diameter of a strongly connected graph . . . . . . . . . 5 . Counting paths . . . . . . . . . . . . . . . .

.

9

12 17

24 28

32 36 42

PATHS. CENTRES AND DIAMETERS

1. The path problem . . 2. The shortest path problem

CHAPTER 5

3 5

55 59 61

66 74

FLOW PROBLEMS

1. The maximum flow problem . . . . 2. The compatible flow problem . . . . 3 . An algebraic study of flows and tensions . 4. The maximum tension problem . . .

. . . . . . . . . . . . . . . . . . . . . . . .

76 86 89

. . . . . . . . 95

CHAPTER 6. DEGREES AND DEMI-DEGREES 1. Existence of a p-graph with given demi-degrees . . . . . 2. Existence of a p-graph without loops with given demi-degrees 3. Existence of a simple graph with given degrees . . . . . xi

. .

102

. . 109 . . 115

xii

TABLE OF CONTENTS

CHAPTER 7. MATCHINGS 1 . The maximum matching problem . 2. The minimum covering problem . 3 . Matchings in bipartite graphs . . 4. An extension of the Konig theorem 5 . Counting perfect matchings . .

. . . . . . . . . . . .

.

122 129 131 141 142

. . . . . . . . . . . . . . . . . . . . . . . . . . .

150 153 155

. . . . . . . . . . .

164 175 181

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . . . . . .

.

CHAPTER 8. C-MATCHINGS

.

1 The maximum c-matching problem . 2. Transfers . . . . . . . . . 3 . Maximum cardinality of a c-matching

.

CHAPTER 9

CONNECTIVITY

1 . h-Connected graphs . . . . 2. Articulation vertices and blocks 3 . k-Edge-connected graphs . .

. . . . . . . . . . . . . . . . . . . . . .

CHAPTER 10. HAMILTONIAN CYCLES

1 . Hamiltonian paths and circuits . . . . 2. Hamiltonian paths in complete graphs . . 3. Existence theorems for hamiltonian circuits . 4 . Existence theorems for hamiltonian cycles . 5 . Hamilton-connected graphs . . . . . 6. Hamiltonian cycles in planar graphs (abstract)

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

186 192 195 204 217 223

CHAPTER 1 1 . COVERING EDGES WITH CHAINS 1 . Eulerian cycles . . . . . . 2 . Covering edges with disjoint chains 3 . Counting eulerian circuits . . .

. . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

228 232 239

CHAPTER 12. CHROMATIC INDEX 1 . Edge colourings . . . . . . . . 2. The Vizing theorem and related results . 3 . Edge colourings of planar graphs (abstract)

. . . . . . . . . . . . . . . . . . . . . . . .

248 254 267

xiii

TABLE OF CONTENTS

CHAPTER 13. STABILITY NUMBER 1 . Maximum stable sets . . . . . . . 2. The Turin theorem and related results . . 3 . a-Critical graphs . . . . . . . . . 4. Critical vertices and critical edges . . . . 5 Stability number and vertex coverings by paths

.

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

272 278 285 297 298_

CHAPTER 14. KERNELS AND GRUNDY FUNCTIONS

. .

1 Absorption number 2. Kernels . . . 3 Grundy functions 4. Nimgames . .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

303 307 312 318

CHAPTER 15 . CHROMATIC NUMBER 1. Vertex colourings . . . . . . . . 2. y-Critical graphs . . . . . . . . . 3 . The Haj6s theorem . . . . . . . . 4. Chromatic polynomials . . . . . . . 5 . Vertex colourings of planar graphs (abstract)

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

325 338 350 352. 355

1 . Perfect graphs . . . . . . . . . . . . 2. Comparability graphs . . . . . . . . . . 3 . Triangulated graphs . . . . . . . . . . . 4. i-Triangulated graphs . . . . . . . . . . 5 . Interval graphs . . . . . . . . . . . . 6 . Cartesian product and Cartesian sum of simple graphs

. . . . . .

. . . . . .

. . . . . .

. . . . . .

360 363 368 369 371 376

. . . .

. . . .

. . . .

. . . .

389 391 396 400

. . . . . . . . . . . . . . . . . . . . . . . .

414 420

CHAPTER 16. PERFECT GRAPHS

PART TWO-HYPERGRAPHS CHAPTER 17 . HYPERGRAPHS AND THEIR DUALS 1. Hypergraphs . . . . . . . . 2. Cycles in a hypergraph . . . . . 3 . Conformal hypergraphs . . . . 4. Representative graph of a hypergraph

. . . .

. . . .

. . . .

. . . .

. . . .

CHAPTER 18. TRANSVERSALS 1. Matchings and c-matchings 2. Transversal number . . .

TABLE OF CONTENTS

XiV

CHAPTER 19. CHROMATIC NUMBER O F A HYPERGRAPH

.

1 Stability number and chromatic number of a hypergraph 2. Cliques of a hypergraph . . . . . . . . . 3 . Good colourings of the edges of a graph . . . . . 4 . Generalizations of the chromatic number of a graph .

. . . .

. . . . . . . . . . . .

428 432 440 443

CHAPTER 20. BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

. .

1 Strong chromatic number 2. Balanced hypergraphs . 3 Unimodular hypergraphs 4. Stochastic functions .

. . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . .

448 450 463 469

CHAPTER 21. MATROIDS

. Matroid on aset . . . . . . The Rado theorem and related results . Image of a matroid . . . . . . . Minimum weight basis . . . . .

1 2. 3 4

REFERENCES

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . . . . . . . . . . . . . . .

INDEX OF DEFINITIONS

. . . . . . . . . . . . .

476 481

486 493 498 523

PART ONE

Graphs


CHAPTER 1

Basic Concepts

1. Graphs Intuitively speaking, a graph is a set of points, and a set of arrows, with each arrow joining one point to another. The points are called the vertices of the graph, and the arrows are called the arcs of the graph. The set of vertices of a graph is generally denoted by X , and the set of arcs of a graph is generally denoted by U. For example, in the graph in Fig. 1.1,

X = { a , b , c , d } , U = { 1,2,3,4,5,6,7,8,9, lo}. Arc 9, which goes from vertex c to vertex d, is said to be of the form (c, d ) , for short one may write 9 = (c, d ) . Arc 2, which goes from vertex a to vertex b, may similarly be written as 2 = (a, b). Note that arcs 3 and 4 have the same form as arc 2, but should not be confused with arc 2. In this book, we shall only consider finite sets X and U. Note that the position of the vertices in the drawing of the graph is not important: only the way in which the vertices are joined by arcs is important. A graph is completely determined by its vertices and by the family of its arcs. Formally, a graph G is defined to be a pair ( X , U ) , where (1) X i s a set { xl, x z , ..., x, } of elements called vertices, and (2) U is a family ( u l , u z , ...,u,) of elements of the Cartesian product X x X , called arcs. This family will often be denoted by the set U = { 1, 2, ..., m } of its indices. An element ( x , y ) of X x X can appear more than once in this family. A graph in which no element of X x Xappears more than p times is called a p-graph. The number of vertices in a graph is called the order of the graph. An arc of G of the form (x, x ) is called a loop. For an arc u = ( x , y ) , vertex x is called its initial endpoint, and vertex y is called its terminal endpoint. Vertex y is called a successor of vertex x if there is an arc with y as its initial endpoint and x as its termina1.endpoint. The set of all successors of x is denoted by

mx). 3

4

GRAPHS

Similarly, vertex y is called a predecessor of vertex x if there is an arc of the form (y, x). The set of all predecessors of vertex x is denoted by

l-m: The set of all neighbours of x is denoted by rG(x) = r,f(x) u l?i(x).

a

x5 C

Fig. 1.1. A 3-graph of order 4

Fig. 1.2. A 1-graph of order 6

Note that r,+ is a correspondence from X to X that associates with each x E X a subsetr&(x)of X . It is possible that T,+(x)= 12/ (the empty set). If r,(x) = 0,x is called an isolated vertex. For A c X,let rG(A)

=

u

rG(a),

aeA

If x E rG(A),x $ A, then x is said to be adjacent to set A . For p = 1, a p-graph is called a I-graph. The arcs of a I-graph are all distinct elements of the Cartesian product X x X.In this case,

U c X x X ,

IUl=m.

A I-graph G = (X,U)is completely defined by X and the correspondence

r = r;. Hence, G can be denoted by (X,r).

In a graph G = (X,U),each arc u, = (x, y) determines a continuous line joining x and y. Such a line, without any specification of its direction, is called an edge, and is denoted by e, = [x, y]. The family (el,ea, ..., em)of the edges of G is denoted by its set of indices E = { 1, 2, ..., m } .

BASIC CONCEPTS

5

If the directions of the arrows in a graph are not specified, it is convenient, for conceptual reasons, to deal with the pair (X,E ) , rather than the pair (A', U ) . Such a pair (X,E ) is called a multigraph (or undirected graph). A multigraph is called a simple graph if: (1) it has no loops, (2) no more than one edge joins any two vertices. In a simple graph, E denotes a subset of P2(X),the set of all subsets of X with cardinality 2.

Fig. 1.3. Multigraph

Fig. 1.4. Complete graph Ks

Fig. 1.5. Complete bipartite graph K3,

Graphs and multigraphs often appear under other names: sociograms (psychology), simplexes (topology), electrical networks, organizational charts, communication networks, family trees, etc. It is often surprising to learn that these diverse disciplines use the same theorems. The primary purpose of graph theory was to provide a mathematical tool that can be used in all these disciplines. It would be convenient to say that there are two theories and two kinds of graphs: directed and undirected. This is not true. All graphs are directed, but sometimes the direction need not be specified. Results f o r directed graphs can be applied to a multigraph G =' ( X , E ) by replacing G with a directed graph G * that has two oppositely directed arcs corresponding to each edge in G. Similarly, results for multigraphs can be applied to a directed graph G = (X,U ) after removing the direction from each arc in G. 2. Basic definitions Adjacent arcs, adjacent edges. Two arcs (or two edges) are called adjacent if they have at least one endpoint in common.

6

GRAPHS

Multiplicity. The multiplicity of a pair x, y is defined to be the number of arcs with initial endpoint x and terminal endpoint y. Denote this number by m$(x, y), and let mc(x,y ) =

d b ,x ) ,

rnc(x, y ) = m b ( x , Y)

+~

( xY) , .

If x # y, then mG(x,y ) denotes the number of arcs with both x and y as endpoints. If x = y , then mG(xy y ) equals twice the number of loops attached to vertex x. If A and B are two disjoint subsets of X,let rn,+(A,B)=

( U / U E U , U

r n , ( ~ , B ) = rncf(A,B )

=(x,y),xEA,yEB),

+ m,+(B,A ) .

Arc incident to a vertex. If a vertex x is the initial endpoint of an arc u, which is not a loop, the arc u is said to be incident out of rertex x . In graph G, the number of arcs that are incident out of x plus the number of loops attached to x is denoted by d,+(x)and is called the outer demi-degree of x . An arc incident into vertex x and the inner demi-degree d;(x) are defined similarly.

Degree. The degree of vertex x is the number of arcs with x as an endpoint each loop being counted twice. The degree of x is denoted by dG(x)= d,+(x) d;(x). If in a graph each vertex has the same degree, this graph is said to be regular.

+

Arc incident out of a set A c X. If the initial endpoint of an arc u belongs to A , and if the terminal endpoint of arc u does not belong to A , then u is said to be incident out of A , and we write u E @ + ( A ) . Similarly, we define an arc incident into A , and the set o - ( A ) . Finally, the set of arcs incident to A is denoted by w(A) = w+(A)u @ - ( A ) .

Symmetric graph. If m,+(x,y ) = m:(x, y ) for all x , y E X,the graph G is said to be symmetric. A I-graph G = (X,U)is symmetric if, and only if, (X,Y)EU

*

(Y,x)EU.

Anti-symmetric graph. If for each pair (x, y ) E X x X, mG+(xyY)+ 6 ( X , Y ) G 1

Y

7

BASIC CONCEPTS

then the graph G is said to be anti-symmetric. A I-graph G = ( X , U ) is antisymmetric if, and only if, (X,Y)EU

(Y,X)$V.

An anti-symmetric I-graph without its direction is a simple graph. Complete graph. A graph G is said to be complete if d x , Y ) = mc+(x, Y >

+ m&,

Y) 2 1

for all x, y E X,such that x # y. A I-graph is complete if, and only if, (X,Y)$U

(Y,X)EU.

A simple, complete graph on n vertices is called an n-clique, and is often denoted by K,,. See Fig. 1.4. Bipartite graph. A graph is bipartite if its vertices can be partitioned into two sets XI and X , such that no two vertices in the same set are adjacent. This graph may be written as G = ( X I , X,,U ) . Complete bipartite graph. If for all x1 E XIand for all x, E X,, we have mG(xl,x,) 3 1, then graph G = (XI, X,,V ) is said to be a complete bipartite graph. A simple, complete bipartite graph with I XII = p , and I,'A I = q is often denoted by Kp,,. Subgraphof G generated by A c X . The subgraph of G generated by A is the graph with A as its vertex set and with all the arcs in G that have both their endpoints in A . If G = (X,r)is a 1-graph, then the subgraph generated by A is the I-graph GA = ( A , rA) where

I',(x) = r ( x ) n A

(x E A)

.

Partial graph of G generated by Y c U. This is the graph ( X , V )whose vertex set is X and whose arc set is V. In other words, it is graph G without the arcs U - V. Partial subgraph of G. A partial subgraph of G is the subgraph of a partial graph of G. For example, if G is the graph of all roads in the United States, the set of all 4-lane roads is a partial graph of G; the set of all roads in Illinois is a subgraph, and the set of all 4-lane roads in Illinois is a partial subgraph. Chain of length q > 0. A chain is a sequence p = ( u ~u,, , ..., u,) of arcs of G such that each arc in the sequence has one endpoint in common with its predecessor in the sequence and its other endpoint in common with its successor in the sequence. The number of arcs in the sequence is the length of

8

GRAPHS

chain p. A chain that does not encounter the same vertex twice is called elementary. A chain that does not use the same arc twice is called simple. Path of length q > 0. Apath of length q is a chain p = ( u l , u 2 , ..., u i , ..., uq) in which the terminal endpoint of arc ui is the initial endpoint of arc u i + l for all i < q. For a 1-graph, a path is completely determined by the sequence of vertices xl, x z , ... that it encounters. Hence, we often write p =

((XI,

x2), (xz, x 3 ) ,

...) = 1x1, x2, .’., X k , % + l l = Pb1, X k + l l .

.Vertex

x1 is called the initial endpoint and vertex xk+ is called the terminal endpoint of path p. Similarly, for a simple graph, a chain p with endpoints x and y is determined by the sequence of its vertices, and we may write

Cycle. A cycle is a chain such that (1) no arc appears twice in the sequence, and (2) the two endpoints of the chain are the same vertex. Pseudo-cycle. A pseudo-cycle is a chain p = (ul,u2, ...,u,) whose two endpoints are the same vertex and whose arcs are not necessarily distinct. Circuit. A circuit is a cycle p = ( u l , u2, ..., u,) such that for all i < q the terminal endpoint of ut is the initial endpoint of ui + 1.

Connected graph. A connected graph is a graph that contains a chain p[x, y ] for each pair x , y of distinct vertices. Connected component of a graph. Clearly, the relation [x = y, or x # y and there exists a chain in G connecting x and y ] denoted by x = y is an equivalence relation because (1) x E x (2) x = y (3) x 3 y , y

*

=z

y=x

*

x

=z

(reflexivity) (symmetry) (transitivity)

The classes of this equivalence relation partition X into connected subgraphs of G called the connected componenfs. For example, the graph in Fig. 1.2 possesses two connected components. Articulation set. For a connected graph, a set A of vertices is called an articulation set (or a cufset) if the subgraph of G generated by X - A is not connected; the term “cutset” will not be used here, in order to avoid confusion with another kind of “cutsets” defined in the theory of transportation networks (and used in Chapter 5).

9

BASIC CONCEPTS

For example, { a, c } and { c } are two articulation sets of the graph in Fig. 1.1 ; vertex c is also called an articulation certex (or a cut-oertex). Stable set. A set S of vertices is called a stable set if no arc joins two distinct vertices in S ; for example, { b, d } is a stable set of the graph in Fig. 1.1. Matrix associated with a graph G. If G has vertices x l , x z , .. ., x,, let

+

i

aj = m~ (xi, X J ; The matrix ((a:))is called the matrix associated with G. For example, the matrix associated with the graph in Fig. 1.2 is x1

x2

x3

x4

0

1

0

0

0

0

1

1

1

1

0

0

0

0

0

0

0

0

0

0

1

0

0 . 0

0

0

0

0

0

0

0

0

0

0

1

1

x5

x6

+ ((a:))*is called the adjacency matrix.

The matrix ((a;))

3. List of symbols Set of all real numbers. w Set of all non-negative integers. N z Set of all integers. la Empty set. of set A . (i.e. number of elements) Cardinality IAl Set of all x such that ... { x/ ... } a is an element of set A . aeA a is not an element of set A . a$A AUB Union of sets A and B. Intersection of sets A and B. A n B A -B A less B (the elements of A that are not in B ) . A EB Set A is contained in set B (possibly A = B ) . Set A is not contained in set B. AckB Cartesian product of A and B (the set of all pairs (a, 6)where A x B a E A and b E B). Image of element a in correspondence r. Image of set A in correspondence r, or UaeA T(a) (r(Qr) = 0). r(4

.

10

GRAPHS

m) r-

Transitive closure of correspondence r. Inverse correspondence of correspondence

l(A)

r%) B,,,(A) (1) = (2) (1) e (2)

P!

Binomial coefficient.

n!

n

Multinomial coefficient.

= q (mod. k)

log P

I:[ [;I

x/~(x)~YY).

Set of all subsets of set A . Set -of all subsets of cardinality k . Set of all non-empty subsets of cardinality Q k. Property (1) implies property (2). Property (1) is equivalent to property (2).

-%4> Sk(A)

p

=(

r, or

*

((a:)) Det ((a!))

Integer p is equal to q modulo k (the remainder from the division of p by k is equal to the remainder from the division of q by k). Logarithm of p . Integer part of p / q. Smallest integer greater than or equal to p / q.

Matrix whose entry in the i-th row andj-th column is a:. Determinant.

For a graph G, Set of all neighbours of vertex x . Set of all successors (resp. predecessors) of vertex x . Degree of vertex x. dG(4 d,+(x),d;(x) Outer demi-degree, inner demi-degree of vertex x . Number of edges between sets A and B. mG(A,B ) . m$(A, B ) , m;(A, B ) Number of arcs going from A to B (resp. from B to A ) . Portion of the chain p between vertices x and y . A x , Yl Set of all arcs having exactly one endpoint in A . 44 o + ( A ) , o - ( A ) Set of all arcs with only their initial endpoint in A (resp., with only their terminal endpoint in A). TO(X)

r,+(x), r ; ( x )

For a family Y of sets, a member S E Y is defined to be a nrinimal set if it a member S E Y is defined to be does not contain any other member of 9; a minimum set if its cardinality has the minimum value. A maximal set and a maximum set are defined similarly.

11

BASIC CONCEPTS

EXERCISES 1. Show that if G is a simple graph with n vertices and p connected components, the maximum possible number of edges in G is 1 +-P)(n-P+

I).

2. Show that a simple graph with n vertices and more than +(n connected.

- 1) (n - 2) edges is

Chapter 2

Cyclomatic Number

1. Cycles and cocycles In a graph G = ( X , U),a cycle is a sequence of arcs p = ( 49

U2 9

* * ' Y

u,)

such that (1) each arc uk, where 1 < k < q, has one endpoint in common with the preceding arc U k - 1 , and the other end point in common with the succeeding arc uk+l (i.e., this sequence is a chain), (2) the sequence does not use the same arc twice, (3) the initial vertex and terminal vertex of the chain are the same. An elementary cycle is a cycle in which, in addition, (4) no vertex is encountered more than once (except, of course, the initial vertex which is also the terminal vertex). Given a cycle p, we denote by p + the set of all arcs in p that are in the direction that the cycle is traversed, and we denote by p- the set of all the other arcs in p. If the arcs in G are numbered 1,2, ..., m, then cycle p is defined by a vector

Henceforth, a cycle and its vector p will be used interchangeably, and when we say that cycle p is the sum of cycles p' p2,we will mean the vector sum.

+

Property 1. A cycle is the sum of elementary cycles that are pairwise arcdisjoint.

This is evident because as we traverse, p an elementary cycle is defined each time we return to a vertex. 12

13

CYCLOMAnC NUMBER

Property 2. A cycle is elementary i f , and only i f , it is a minimal cycle, i.e., no other cycle is properly contained in it.

The proof is obvious. If A is,a non-empty subset of X , * + ( A ) denotes the set of arcs that have only their initial endpoint in A and * - ( A ) denotes the set of arcs that have only their terminal endpoint in A . Let *(A) = W + ( A ) u * - ( A ) .

A cocycle is defined to be a non-empty set of arcs of the form o ( A ) , partitioned into two sets * + ( A ) and * - ( A ) . Corresponding to each cocycle, there is a vector 0

with *i=

=

(01

9

[+

0 1 -1

0 2

, *.',*,)

if if if

9

i$w(A), ~EO+(A), i~w-(A).

A cocycle may be identified by its vector a. A cocycle is called elementary if it is the set of arcs joining two connected subgraphs A l and A z such that A , 3 A , z Izf A , n A , = $3 A, U A , =

c,

where C is a connected component of the graph. A cocircuit is defined to be a cocycle o ( A ) in which all arcs are directed in the same direction, i.e. into set A , or out of set A . Property 3. A cocycle is the sum of elementary cocycles that are pairwise arcdisjoint.

Let o be a cocycle of the form o ( A ) , and let Al, A z , ..., Ak be the different connected components of the subgraph generated by A . Then

+

+ +

o ( A ) = 44,) u)(A,) W(Ak), and the cocycles w(A,), o ( A z ) , ..., o ( A , ) are pairwise disjoint. It remains to show that o ( A , ) is the sum of elementary disjoint cocycles. If C is the connected component that contains A,, and if the subgraph generated by C - Ai has connected components C1,Cz, ..., then o ( A , ) = - O(C,) - o(C2)-

*-.

14

GRAPHS

where -o(C,)is an elementary cocycle since o(C,) joins the connected subgraphs C , and Ai U Cz U C,....Furthermore, -o(C,),-o(Cz),... are pairwise arc-disjoint. - Q.E.D.

Property 4. A cocycle is elementary if, andonly if, it is a minimal cocycle (i.e. no other cocycle is properly contained in the cocycle). Let w(A) be a minimal cocycle. Therefore, A is contained in a connected component C. Let A l , A z , ..., A k be the connected components of the subgraph generated by C - A . If k 2 2, the vector - o ( A , ) is a cocycle properly contained in w(A). But this is impossible, and therefore, k = 1, and o is an elementary cocycle. Conversely, let o be an elementary cocycle that joins two connected subgraphs A , and A,. If we remove some, but not all, of the arcs from w, then we no longer have a cocycle. Therefore o is a minimal cocycle. Q.E.D. Arc Colouring Lemma (Minty [1960]).Consider agraph with arcs 1,2, ..., m. Colour arc I black, and arbitrarily colour the remaining arcs red, black or green. Exactly one of the following conditions holds (1) there is an elementary cycle confaining arc I and only red and black arcs with the property that all black arcs in the cycle have the same direct ion) (2) there is an elementary cocycle containing arc I and only green and black arcs, with rhe property that all black arcs in the cocycle have the same direction. Successively label the vertices of the graph using the following iterative procedure (1) Let arc 1 = (b, a). Label vertex a, (2) If vertex x is labelled, and vertex y is unlabelled, then label y if (a) there is a black arc (x, y ) , or (b) there is a red arc ( x , y ) or ( y , x). When the labelling procedure stops, exactly one of the following two cases occurs : CASE1: Vertex b has been labelled. The vertices used by the procedure to label b from a constitute an elementary cycle of red and black arcs with all black arcs having the same direction. (Thus, there cannot exist a cocycle of black and green arcs with all black arcs in the same direction.) This cycle is the sum of disjoint elementary cycles, one of which contains arc 1. CASE2: Vertex b has not been labelled. Let A denote the set of all labelled

CYCLOMAnC NUMBER

I5

vertices. Note that w(A) contains only black arcs directed into A or green arcs. Thus, there exists a cocycle o ( A ) of green and black arcs containing ajc 1 with all black arcs directed into A . (Thus there cannot exist a cycle of redand black arcs with all black arcs having the same direction). This cocycle isthe sum of disjoint elementary cocycles, one of which contains arc 1. Corollary. Each arc belongs either to an elementary circuit or to an elementary cocircuit, but no arc belongs to both. This is shown by applying the lemma with all arcs coloured black. The cycles pl,p2, ..., pk are said to be dependent if there exists a vector equation of the form

rl p1

+ r2 p2 + + rk pk = o ,

where rlr r2, ..., rk are real numbers, not all zero. If the cycles are not dependent, they are said to be independent. A cycle basis is defined to be a set { pl, p2, ..., pk 1 of independent elementary cycles such that any cycle p can be written as

p = rl p'

+ r2 p2 + + rk pk , *..

where rl, r2, ..., r k are real numbers. Clearly, k equals the dimension of the subspace of R" generated by the cycles and therefore does not depend on the choice of the basis. This constant k is called the cyclomatic number of G, and is denoted by v(G). A cocyc/e basis { wl, 02, ..., o1} is defined similarly, and its cardinality I is called the cocyclomatic number of G and is denoted by A(G).

EXAMPLE.Consider the graph G in Fig. 2.1 ; its elementary cycles are;

a

= [abca], p' = (1,6,2) = [abca], pz = (1,6,3) P3 = (293) = [ma], p4 = ( 1 , 4 , 5 , 2 ) = [abdca], = [acdb], p5 = ( 6 , 5,4) p6 = ( 1 , 4 , 5 , 3 ) = [abdcu].

Od C

Fig. 2.1

These cycles are not independent, since we have, for example: p' - pz

+ p3 = 0 .

The cycles pl, p2, p3 form a cycle basis, and therefore, v(C) = 3. A cocycle o ( A ) can be written as a sequence of arcs ( f il, f i2, ...) where

16

GRAPHS

each arc i of the sequence is preceded by a + sign if i E w + ( A )or by a - sign if i E o - ( A ) , it can also be denoted by { A }. For the graph in Fig. 2.1, the elementary cocycles are: = { a } = (+ 1 , + 2, - 3) , u2={ab} =(+6,+2,-3,+4), w3 = { U C } = (- 6 , + 1, + 9, 0 4 = { a b c } = (+ 4, + 5 ) , as = { a b d } = (+ 6 , + 2, - 3 , - 5 ) a6 = ( a d } = (- 6 , 1 , - 4 ) .

0'

+

3

Obviously, these cocycles are not independent. To form a basis, one could take, for example, w', m4 and w5; hence A(G) = 3. Theorem 1. Let G be a graph with n certices, m arcs andp connected components. The cardinality of a cycle basis is v(G) = m - n p . The cardinality of a cocycle basis is A(G) = n - p .

+

1. There exist n - p elementary cocycles. Suppose first that the graph is connected ( p = 1). Successively form n - 1 independent cocycles w(A,), w(A,), ..., m(A,,-,) in the following way: (a) Take an arbitrary vertex a, and let A , = { a , }. The cocycle o ( A , ) contains an elementary cocycle. Let [a,, a,J be an edge of this elementary cocycle such that a, E A l , %#A, * (b) Let A , = A , u { a 2 } . The cocycle w(A,) contains an elementary cocycle. Let [x, a,] be an edge in this elementary cocycle such that

X E A ~ , a39Az. (c) Let A 3 = A , u { a 3 } , and repeat the process until n - 1 elementary cocycles have been defined. These cocycles are independent because each contains an arc not contained in any of the others. If the graph is not connected (p > l), let C, C, ted components. Then, there exist

(I c1I - 1)

..., C, denote its connec-

+ (I c2I - 1) + + (I C, I - 1) = n - p

independent elementary cocycles.

+

2. There exist m - n p independent elementary cycles. Let v(G) = m - n p , and construct a sequence Go, G,, ..., G, = G of partial graphs. Graph Go consists of the isolated vertices of G. Each G, is obtained from its predecessor G i - , by the addition of an arc i of G - Gi-,.

+

17

CYCLOMATIC NUMBER

Initially, v(G,) then

=

0, and there are no cycles. If arc i forms a new cycle piy

+ 1,

v(GJ = v(Gi-J

since p remains unchanged and m is increased by 1. If arc i does not form a new cycle, then v(GJ = v(Gi- 1)

9

since p decreases by 1 and m increases by 1. Upon termination, v(G) = m - n + p cycles ptl, piz, ..., p * k have been defined. There is no vector of the form r1 pi'

+ r2 p**+ + rk pi* = 0, 9..

with some rk # 0 because cycle p'k contains arc ik which is not contained in any of the other cycles. Thus the pi are z(G) independent cycles.

3. There cannot exist more than v(G) = m - n + p independent cycles, and there cannot exist more than A(G) = n - p independent cocycles. Consider in R" the vector space M generated by the cycles and the vector space SZ generated by the cocycles. If p is a cycle and if o = w(A) is a cocycle, their scalar product m

equals zero because

(P, 44)= (P,

c "("1) c

aeA

=

aeA

(P,

4.)) = 0 .

Thus M and 0 are orthogonal subspaces of R", and their dimensions must satisfy dim M

+ dim 52 < m .

From Parts 1 and 2, dim M

+ dim 52 3 v(G) + A(G) = m .

Therefore equality holds throughout, and dim M = v(G) dim 152 = I.(G).

Q.E.D. 2. Cycles in planar graphs

Graph G is said to be planar if it is possible to represent the graph on a plane in which the vertices are distinct points, the arcs are simple curves

18

GRAPHS

and no two arcs cross one another. A representation of G that satisfies the above requirements on a plane is called a topological planar graph. Two topological graphs that can be made to coincide by an elastic deformation of the plane are considered to be the same.

EXAMPLE 1. A convex polyhedron in 3-dimensional space defines a simple graph : its “corners” (O-dimensional faces) are the vertices and its “sides” (l-dimensional faces) are the edges of the graph. It has been shown (Steinitz [1922]) that a simple graph G can represent a convex polyhedron in R3if, and only if, G is a connected planar graph that cannot be disconnected by the removal of less than three vertices. EXAMPLE 2. Problem of three factories and three utilities (Fig. 2.2). Three factories, a, b, and c rely upon underground supply lines for their water from point d, their gas from point e, and their electricity from pointf. Is it possible to arrange the three factories and the three utility stations so that no supply lines cross one another except at their endpoints? It can be shown that 8 supply lines can always be placed but that the 9th supply line must cross at least one other supply line. Thus K3,3is not planar.

Fig. 2.2

Let G be a topological planar graph. A face of G is defined to be a region of the plane bounded by arcs such that any two points in a region can be connected by a continuous curve that meets no arcs or vertices. Let 2 denote the set of all faces. The boundary of a face z is the set of all arcs that touch face z. Faces z and z‘ are said to be adjacent if their boundaries contain a common arc. (If two faces touch one another only at a vertex, they are not adjacent .) The contour of a face z is defined to be an elementary cycle formed with the edges of the boundary of z that contains in its interior the face z. Note that

(’YCLOMATIC NUMBER

19

there is exactly one unhounrled face and it has no contour. All the other faces are bounded and have exactly one contour.

EXAMPLE.A geographic map corresponds to a topological planar multigraph whose edges are the borders between countries. This graph has no isthmus. and each of its vertices has degree 2 3. A given face may be adjacent to another face along several different edges. Note in Fig. 2.3 that faces g and d have a common vertex but are not adjacent.

Fig. 2.3

Theorem 2. In a topological planar graph G, the contours qf the different bounded faces constitute a cycle basis. Clearly, the theorem is true if G has only 2 bounded faces. If the theorem is true for all graphs withf - 1 bounded faces, we shall show that the theorem is also true for a topological planar graph withf’bounded faces. If not all of the contours are edge-disjoint cycles, then the result is evidently true. Suppose that when arc i is removed, a graph G’ with f - 1 bounded faces is formed. By hypothesis, the contours of this graph G’ are a fundamental basis of independent cycles. If arc i is returned to the graph, a new finite face is formed. Its contour is a cycle independent of the cycles of G’ because it contains an arc not present in any cycle of G‘. Thus the addition of a n arc cannot increase the cyclomatic number by more than 1, and the bounded faces of G determine a cycle basis. Q.E.D. Corollary 1. If a connected topological planar graph has n rertices, 171 arcs and f faces, then

n -m

+f = 2

(Euler’s Formula) .

The number of bounded faces equals the cyclomatic number v(C). Thus

20

GRAPHS

f = v(G)+ 1

= (m

-n

i- 1)

+ 1 =m -n + 2,

and the corollary follows. Corollary 2. A simple planar graph G has a i’ertex x of degree d&) 6 5. Suppose that G is connected. Otherwise, each connedted component will be considered separately. Since G is a simple graph, each face is bounded by at least three distinct edges. Consider the bipartite graph formed by the set A of vertices representing the faces of G and the set B of vertices representing the the edges of G. Place an arc from a E A to b E Beach time face a is incident to edge b. (This graph is called the face-edge incidence graph.) Clearly, the number of arcs is < 2 m and B 3f. Thus

2m

f 2. The following properties are equicalent (and each characterizes a tree): (1) H is connected and has no cycles (2) H has n - 1 arcs and has no cycles, ( 3 ) H is connected and contains exactly n - 1 arcs, (4) H has no cycles, and i f a n arc is added to H , exactIy one cycle is created, (5) H is connected, and if any arc is remooed, the remaining graph is not connected, (6) Every pair of rertices of H is Connected by one and only one chain. (1) 3 (2) I f p denotes the number of connected components, m denotes the number of arcs and v(H) denotes the cyclomatic number (see Ch. 2): then (1) implies p=1, v(H)=m-n+p=O. Thus, m = n - p = n - 1. ( 2 ) 3 ( 3 ) Since v ( H ) = 0, In = n - 1, it follows that p=v(H)-m+n=

1.

Thus H is connected.

(3)

3

(4) Sincep

=

1, m

=

n - 1 , it follows that

v(H) = m

(4)

-

-n + p

=0.

Thus H contains no cycles. In other words, if an arc is added, the cyclomatic number becomes equal to 1, and there is exactly one cycle in the new graph. (5) If H were not connected, then two vertices, say a and 6, would not 24

TREES AND ARBORESCENCES

25

be connected, and an arc (a, b) could be added without creating a cycle, which contradicts (4). Thus p = 1, v ( H ) = 0, and therefore m=n-I. If an arc is removed, we obtain a graph H‘ with m‘ = n’ 2 v(H‘) = 0 . Hence p’ = v(H’) - m‘ n’ = 2

-

+

(5)

3

and H’ is not connected. (6) For any two vertices a and b there is a chain connecting them (since H i s connected). This chain is unique (otherwise the removal of an arc which belongs only to the second chain would not disconnect the graph).

(6) + (1) Clearly, if H had a cycle, at least one pair of vertices would be joined by two distinct chains, which contradicts (6). Q.E.D. Theorem 2. A vertex is called “pendant” if it is adjacent to exactly one other vertex. A tree of order n > 2 has at least two pendant vertices. Let H be a tree that has only 0 or 1 pendant vertices. Consider a traveller who traverses the edges of the graph starting from a pendant vertex (if there is one). If he does not permit himself to use the same edge twice, he cannot go to the same vertex twice (since H has no cycles). If he arrives at a vertex x , he can always depart using a new edge (since x is not a pendant vertex). Thus the trip lasts indefinitely, and this is impossible since His finite. Q.E.D.

Theorem 3. A graph G

=

( X , U )has a partial graph that is a tree if, and only

if, G is connected. I f G is not connected, no partial graph of G is connected. Therefore G cannot have a partial graph that is a tree. If G is connected, look for an arc whose removal does not disconnect the graph. If no such arc exists, G is a tree by virtue of property (5). If such an arc exists, remove it and look for another such arc, etc. ... When no more arcs can be removed, the remaining graph is a tree whose vertex set is X. Q.E.D. The tree obtained from G as above is called a spanning tree, and Theorem 3 yields a simple algorithm to construct a spanning tree of a connected graph. A spanning tree can also be constructed in the following way:

26

GRAPHS

Consider any arc uo. Find an arc u1 that does not form a cycle with uo. Then find an arc u2 that does not form a cycle with { uo, u1 }, etc. ... When the procedure cannot continue, a spanning tree has been obtained by property (4). Theorem 4. Let G be a connectedgraph, let H be a spanning tree of G, and let uibe an arc of G not in tree H . If arc ui is added to H , it creates a cycle p i by virtue of property (4). The diferent cycles piform a cycle basis of G, called the “basis associated uith tree H”. The cycles pi are independent since every one of them contains an arc not contained in any of the others. Moreover, the number of pi equals: m(G)

- m(H) =

m - (n - 1) = m

-n

+p

= v(G)

.

By Theorem ( I , Ch. 2), it follows that the pi form a cycle basis of G. Q.E.D.

n

H

G Fig. 3.1

Remark. This theorem yields a simple algorithm to construct a cycle basis of a connected graph G. If G is not connected, each connected component has to be treated separately. Given a connected graph G = ( X , U ) , Theorem 1 shows that a partial graph H = ( X , V ) is a spanning tree if it contains no elementary cycles and if upon the addition of any arc in U - V, the graph contains an elementary cycle. Similarly, we shall say that a partial graph ( X , W ) of G is a cotree if it con-


21

tains no elementary cocycles of G and if upon the addition of any arc in U - W , it does contain an elementary cocycle of G.

Theorem 5. Let G tion of U :

=

( X , U ) be a connected graph, and let ( V , W ) be a parti-

vv w =u,

Vn W=@.

A necessary andsufficient condition,for( X , W ) to be a cotree is that (X,V )be a tree. 1. SufJiciency. If ( X , V ) is a tree, we shall show that ( X , W ) is a cotree.

W contains no cycles of G. Suppose that W contains an elementary cocycle @ ( A ) of G. Then no chain in the tree (X,V ) connects A and X - A , which is a contradiction.

lf c' E V , the set of arcs W U { r } contains a cocycle of G. Clearly, (X,V { t' )) has two connected components A and B. Therefore, w(A) is an elementary cocycle of G contained in W u { /'}. 2. Necessity. If (X,W ) is a cotree, we shall s'how that (X,V ) is a tree by using the Arc Colouring Lemma (Chap. 2). V contains no cycles. Let c' E V. Colour arc black. Colour the arcs in V - { u } red, and colour the arcs in W green. Since ( X , W ) is a cotree, G has a cocycle of black and green arcs that contains arc c. Thus there cannot be a cycle of red and black arcs containing arc u. Since c was selec'ted arbitrarily, there cannot be any cycle in V. 21

If w E W, the set V u { IV } contains a cycle. Colour arc w black. Colour the arcs in V red, and colour the arcs in W - { w } green. Since G has no cocycle of black and green arcs containing arc w,there is a cycle of black and red arcs containing arc w . This proves that (A', V ) is a spanning tree of G. Q.E.D. Theorem 6 . Let G = (X,U ) be a connected graph, let F = (X,W ) be a cotree, and let u, be an arc of G not in F. l f u l is added to F, it creates exactly one cocycle a',and the different cocycles o'forni a cocycle basis of G.. Clearly, if in the graph ( X , U - W )an arc uiis removed, exactly two connected components A and B are formed, and @(A) = mi. The cocycles wi are independent since each of them contains an arc not contained in any of the others. The number of cocycles w' equals the number of edges in the tree ( X , U - W ) which equals

28

GRAPHS

n-I=A(G). From Theorem (1, Ch. 2), it follows that the mi form a cocycle basis. Q.E.D. This theorem yields a simple algorithm to construct a cocycle basis. 2. Strongly connected graphs and graphs without circuits

Consider a connected graph G = ( X , U). A path of length 0 is defined to be any sequence [XI consisting of a single vertex x E X . For x E X and y E X , let the relation x = y signify that there is a path pl[x,y ] going from x to y and also a path p J y , x ] going from y to x. This relation is an equivalence relation, i.e. it satisfies the following three properties: X E X

x=y

*

x=y,

y = z

for all x, y s x ,

*

x=z.

The sets of the form

A(x,) = ( X / X € x, x

= xg 1

partition X and are called the strongly connected components of G. A graph is said to be .strongly connected, if for all x , y E X , there exists a path pl[x,y ] and a path p J y , XI. In other words, graph G is strongly connected if it has only one strongly connected component.

Theorem 7. I f G is a connected graph with at least one arc, thefollowing conditions are equivalent: (1) G is strongly connected, ( 2 ) Every arc lies on a circuit, ( 3 ) G contains no cocircuits. (1)

3

(2)

a

(3)

3

(2) Let ( x , y ) be an arc of G; since there is a path from y to x , arc (x,y ) is contained in a circuit of G. (3) If G had a cocircuit that contains arc (x,y ) , then G cannot have a circuit containing this arc by the Arc Colouring Lemma with all arcs coloured black. This contradicts (2). (1) Let G be a connected graph without cocircuits. We shall assume that G is not strongly connected and produce a contradiction. Since G is not strongly connected, it has more than one strongly connected component. Since G is connected, there exist two distinct strongly connected components that are joined by an arc

29


(a, 6). Arc (a, b) is not contained in any circuit because otherwise a and b would be in the same strongly connected component. By the Arc Colouring Lemma, arc (a, b) is contained in some cocircuit. This contradicts (3).

Q.E.D. Theorem 8. If G is a graph with at least one arc, the following conditions are equicalent : (1) G is a graph without circuits, ( 2 ) Each arc is contained in a cocircuit. The proof is immediate, Theorem 9. I f G is a strongly connectedgraph of order n, then G has a cycle basis of v(G) circuits. To prove Theorem 9, it is sufficient to show that v(C) independent circuits can be found. This is obviously true if G has order < 2. Assume that this is true for all graphs of order less than n > 2. We shall show that this also is true for a graph of order n. Choose from the circuits of length > 1 a circuit p = ( u l , u2, ..., uk) of minimum length. No arc joins two non-consecutive vertices of this circuit but there may exist arcs parallel to the arcs of circuit p. Replace all the vertices of p by a single vertex a'. Replace each arc incident to p but different from u l , u2, ..., uk by an arc with the same index incident to a'. The new graph G' has order n' = n - k f 1 with m' = m - k arcs. Graph G' is strongly connected and, by virtue of the induction hypothesis, has a family of v(G') = q independent circuits pi, p;, 2. If this is true for all graphs with n - 1 vertices, we shall show that it is also true for a graph G with n vertices. Since G contains no circuits, there is at least one vertex b such that the length of the longest path from b equals 1, and there exists a vertex a without successors such that (b, a) E U. Consider the graph G’ obtained from G by deleting the arcs from b to a and by replacing a and b by a single vertex a’. We shall show first that graph G‘ contains no circuits. If a circuit p’ were present in G’, it would necessarily pass through a’ and have length > 1. Let d be the vertex that follows a’ in this circuit. The cycle p in G induced by p’ contains arc (a, b) (because G has no circuits). Hence (6, d ) E U and there exists a path of length > 1 from b to a, which is a contradiction. Thus G‘ is connected, has n - 1 vertices, has no circuits. By the induction hypothesis, G‘ has n - 2 independent cocircuits o ’ ( A ; ) , w‘(Ah), ..., c ~ ’ ( A k - and ~ ) we may assume a‘ E A ; , A ; ,

...,

.

Each of these cocircuits of G’ induces a cocircuit @(Ai) in G. Since the vector o ( A i ) has the same coordinates as the vector w’(A;),the w(A,) are linearly independent vectors. The vectors o ( a ) , o ( A l ) , o ( A , ) , ..., m(A,-,) are also linearly independent because o ( a ) contains the arc (b, a) that is not contained in any of the other cocircuits. Thus n - 1 independent cocircuits have been found. Q.E.D. Let G = (A’,U ) be a strongly connected graph without loops and with more than one vertex. For each vertex x, there is a path from it and a path going into it; therefore there exist at least two arcs incident to x. A vertex that has more than two arcs incident to it is called a node. Otherwise, it is called an anti-node. A path whose only nodes are its endpoints is called a branch. A strongly connected graph without loops that has exactly one node is called a rosace. A rosace has a very simple structure since each branch leaves from and returns to the only node. A graph G is said to be minimally connected if it is strongly connected and the removal of any arc destroys the strongly connected property. Clearly, a rosace is a minimally connected graph. Furthermore, each minimally connected graph is a 1-graph without loops. For a graph G = (A’,U ) , the contraction of a set A of vertices is the operation defined by replacing A by a single vertex a and by replacing each arc

31


going into A (resp. out of A ) by an arc with the same index going into a (resp. out of a).

Lemma 1. Let G be a minimally connected graph. Let A be a set of vertices that generates a strongly connected subgraph of G. Then the contraction of A yields a minimally connected graph. 1. We shall show first that the contraction of A yields a 1-graph. If this were not the case, there would exist a vertex x 4 A and two vertices a, a‘ E A such that (x, a), (x,a’) E U (or, with (a, x), (a‘, x) E U but this would not change the proof). If one of these arcs is removed, the graph remains strongly connected. Thus, G is not minimally connected, which is a contradiction.

2. We shall show now that the contraction of A yields a graph G‘ that is minimally connected. Clearly, graph G’ is strongly connected. If an arc u is removed, the remaining graph is not strongly connected, since the graph ( X , U - { 14 >)is not strongly connected. Q.E.D. Lemma 2. Let G be a minimally connectedgraph, and let G’ be the minimally conrlected graph obtained by the contraction of an elementary circuit of G. Then v(G) = v(G’)

+ 1.

Let p be the elementary circuit to be contracted. It is of length k > 1 and has no chords (otherwise G would not be minimally connected). Let n‘ and m‘ respectively denote the number of vertices and arcs in G’. Then

+

v ( G ) = rn’ - n’ 1 = ( m - k) - ( n - k = m - n = v(G) - 1 .

+ 1) + 1

Q.E.D. Theorem 11. I f G is a minimally connectedgraph of order n 2 2, then G has at least two anti-nodes. Since 1 X 1 > 1, G contains at least one circuit, and v(G) 2 1. If v(G) = 1, the result is true because G is an elementary circuit. We shall assume that the result is true for graphs with a cyclomatic number < k and show that it is also true for a graph G with v(G) = k > 1. CASE1. G has no circuits of length 2 3. Then any two adjacent vertices of G are necessarily joined in both directions. The simple graph H that has the same vertices as G with two vertices joined by an edge if, and only if, they are adjacent in G, is connected (because G is connected) and has no cycles (because G is minimally connected). Thus, H is a tree of order 2 2, and from Theorem 2, H has two pendant vertices x and y . The vertices x and y are both anti-nodes i n graph G and this proves the result.

32

GRAPHS

CASE2. G has a circuit p of length

2 3. This circuit has no chord (a chord is an arc joining two non-consecutive vertices of a circuit i if p had a chord, G would not be minimally connedted). Since v(G) 2 2, there exists a vertex of

G that is not contained in p. The graph G' obtained from G by the contraction of p has order > 1, and from Lemma 2, v(G') = v(G) - 1. Because of the induction hypothesis, G' possesses two anti-nodes x and y . I f one of these anti-nodes, say x, is the contracted image of p, then p contains an anti-node z of G (because its length is 2 3). Thus G has at least two anti-nodes, y and z. If neither of the vertices x and y is the contracted image of p, then G has at least two anti-nodes, x and y. Q.E.D. Corollary 1. Let G be a minimally connected graph that is not an elent~ntary circuit. Then there exists a branch whose anti-nodesform a non-empty set A such that the subgraph Gx - A is strongly connected. Construct from G a graph i7 whose vertices are the nodes of G and whose arcs are the, branches of G. Graph G is strongly connected, but it is not minimally connected (because it has no anti-nodes). Thus, an arc can be removed from without destroying the strong connectivity, This arc of G is necessarily a branch of G of length > 1, because G is minimally connected. Q.E.D.

Corollary 2. I f G is a strongly connected graph without loops having at least one node, then there exists a branch whose arcs and anti-nodes can be removed without destroying the strong connectioity. The proof is similar to the proof of Corollary 1.

Theorem 12. If G = ( X , U ) is a graph, the graph C' obtained.from G by contracting each strongly connected component contains no circuits. The proof is immediate. 3. Arborescences In a graph G = (X,U ) , a vertex a is called a root if all the vertices of G can be reached by paths starting from a. A graph does not always have a root. A graph G is said to be quasi-strongly connected if for each pair of vertices x, y , there exists a vertex z ( x , y ) from which there is a path to x and a path to y . A strongly connected graph is quasi-strongly connected because we can

33

TREES A N D ARBORESCENCES

let z ( x , y ) = x ; the converse is not true. A quasi-strongly connected graph is connected. Finally, an arborescence is defined as a tree that has a root. For example, the family tree of the male descendants of King Henry IV is an arborescence whose root is King R n r y IV.

Fig. 3.2: Arborescence

Lemma. A necessary and suficient condition that a graph G root is that G be quasi-strongly connected.

=

( X , U ) have a

Clearly, if G has a root, then G is quasi-strongly connected. Conversely, suppose G is quasi-strongly connected, and consider its vertices xl, x2,...,x,. There exists a vertex z2 from which there is a path to x1 and a path to x2. There exists a vertex z3 from which there is a path to z 2 and a path to x3, etc. Also, there exists a vertex z , from which there is a path to znV1and a path to x,. Clearly, vertex z, is a root of G. Q.E.D. Theorem 13. Let H be a graph of order n > 1. The following properties are equivalent (and each characterizes an arborescence) : H is a quasi-strongly connected graph without cycles, H is a quasi-strongly connected graph and has n - 1 arcs, H is a tree having a root a , There exists a certex a such that each other aertex is connected with it by one path f r o m a, and only one, H is quasi-strongly connected and this property is destroyed i f any arc is remoced f r o m H , H is quasi-strongly connected and has a certex a such that d,(n) = 0 , d&)

= 1

(x # a),

34

GRAPHS

(7) H has no cycles and contains a vertex a such that d,(Q) = 0

(x # a ) . d&) = 1 (1) 3 ( 2 ) From property (I), H is connected and without cycles. Thus H is a tree. Therefore, H has IZ - 1 arcs. (2) 3 (3) From property (2), H is connected and has n - I arcs. Thus H i s a tree. From the lemma, H has a root a. (3)

=I

(4) The root a of tree H has the desired property.

(4)

* (5) Suppose that the quasi-strongly connected property is not destroyed when an arc ( x , y ) is removed. Then, there exist two elementary paths

(5)

-

[ z , c i , c 2 , . . . , x l and [ z , d l , d 2 , + - . , ~ 1 that do not use arc ( x , y ) . Thus there are two paths in graph G from z to y , and there are two paths from a to y . This contradicts property (4). (6) From the lemma, graph H has a root a because it is quasi-strongly connected. Thus &(XI

z

1

(x # a ) .

If a vertex x satisfies d;(x) > 1, there exist two distinct arcs u, u E o - ( x ) and, therefore, there are two distinct paths from a to x. If arc u is removed, the graph still has a root at a and therefore remains quasi-strongly connected, which contradicts (5). Thus d&) = 1 (x # a ) . Finally, there cannot exist an arc incident into a because the graph obtained from H by removing this arc has a as a root and is quasistrongly connected, which contradicts (5).

(6)

3

(7) The number of arcs in H equals n

C

dK(xj) = n

- 1.

j=1

Since H i s connected and has n - 1 arcs, it is a tree, and contains no cycles. (7)

3

( 1 ) Starting from a vertex b # a, travel through the graph traversing

the arcs against their direction. No vertex is encountered twice because H has no cycles. I f a vertex x # a is encountered, the trip will

35


continue because d;(x) = 1. Therefore, the trip can only end at vertex a. Thus a is a root, and H is quasi-strongly connected. Q.E.D.

Corollary. A graph G has a partial graph that is an arborescence i f , and only

if, G is quasi-strongly connected. If G is not quasi-strongly connected, no partial graph is an arborescence. Conversely, if G is quasi-strongly connected, we can successively delete all the arcs whose removal does not destroy the quasi-strongly connected property. When no such arcs exist, the graph is an arborescence by virtue of Theorem 13, property (5). Q.E.D. The following theorem deals with simple graphs and is a constructive reformulation of a result of P. Camion [1968]. Theorem 14 (Crestin [1969]). Let G = (X, E ) be a simple connectedgraph, and let x1 E X. It is possible to direct all the edges of E so that the obtained graph Go = ( X , U ) has a spanning tree H such that: 1. H is an arborescerzce with root x l , 2. The cycles associated with tree H are circuits, 3. The only elementary circuits of Go are the cycles associated with tree H. Construct a sequence x l , x2, ... of distinct vertices as follows: Given the partial sequence x l , x 2 , ..., x i , find the vertex x j whose index j is as large as possible such that (1)

(2)

1<j a, (because p would pass through a, twice and would not be elementary). This contradicts that:

a4 = a , > a 2 . Q.E.D. Remark. The last part of this proof yields easily a result of Chaty [1966]: A strongly connected graph Go has exactly v(G,) elernentary circuits, i f , and 0nI-y if, there exists in Go a spanning tree H such that the elementary cycles associated with H a r e circuits.

4. Injective, functional and semi-functional graphs

The concept of an arborescence can be generalized in the following way: A graph G is said to be injectiue if d;(x) < 1 for all vertices x. If G is injective,

37


then G is a 1-graph and may be written as G dence is an injective correspondence, i.e.

=

( X , r).Then the correspon-

*

r(x)nr(y)=@. A graph G is said to be afuncfional if d;(x) < 1 for each vertex x. If G is a functional, then G = (X,r)is a I-graph, and the correspondence r is a function 9 defined o n X . For example, if X is a set of states, and if ~ ( xdenotes ) the unique state that follows x in a deterministic process, then the pair (X, cp) is a functional graph. Finally, a 1-graph G = (X,r) is said to be a senzi-functional if x # y

r ( x ) n Tb)# Qr

=.

r ( x ) = l7.~).

A functional graph is semi-functional. An injective graph G semi-functional, because

m)n r69 z fa

=>

x =y

*

=

( X , r) is

r ( x ) = r(y>.

t

A

Fig. 3.5. Functional graph

Fig. 3.4. Injective graph

Property 1. A 1-graph G = ( X , r ) is functional if, and only $, its inverse H = ( X , r - I) is injective. The proof follows since d;(.u) = d;(x).

r)

Property 2. A I-graph G = ( X , is semi-functional if, and only if, its inverse H = ( X , r-l)is semi-functional,

Suppose that G is semi-functional; let y and y’ be two vertices of G such that r - l ( y ) n r - ’ ( y ’ ) # 0 and let x,, E T - l ( y ) n r - ’ ( y ’ ) . We have

x ~ r - * ( y ) r=r 3

y~T(x)

=>

r ( x ) = r(x0)

T(x) n I‘(xo) # @

*

Y’ET(X)

*

a XEr-ICy’).

38

Therefore,

GRAPHS

r-' ( y ) c r- '(y'), and equality holds. Thus, r-'(y) n r-'@') z =. r-l(y)= r-lb'),

Hence, graph H is semi-functional.

Q.E.D. If G = (A', U ) is a graph, its adjoint G * is defined to be a 1-graph whose vertices ul,u,, ..., u, represent the arcs of G and which has an arc from ui to u j if the terminal endpoint of the arc in G corresponding to ui is the initial endpoint of the arc corresponding to u,. A path in G that uses all the arcs corresponds to a path in G* that uses all the vertices. Thus one can relate properties on the arcs of G to properties on the vertices of G*. There are many characterizations of adjoints. (See Heuchenne [1964].) The simplest is given by the following theorem:

Theorem 15. A 1-graph H is the adjoint of a graph i f , and only is, H is semifunctional. 1. If H = ( U , f ) is the adjoint of a graph G functional, since

=

(A', U ) , then H is semi-

r(u) n r(u') # 0 implies that in G the terminal endpoints of arcs u and u' coincide, and consequently, T ( u ) = I-(#'). 2. Let H = (U, r)be a semi-functional graph, and consider the family % of subsets of U of the form C(u) = {

r/f(r)= T(u)}, or C ,

=

{ c/T(P) = @}

The sets Co,Cl, C,, ..., C, of W form a partition of U . Consider a partition 9 of U formed by the sets

Di = T ( C J

( i = 1, 2, ...,q)

Construct a graph G with vertices x,, xl, x2, ..., x , , ~ and k arcs from xt to xi denoted by utl, ut z,..., u,, if

Di n C, = {uil, ui2, ..., uik}. Since 59 and B are two partitions of U, each vertex of H is represented by exactly one arc of G. Clearly, in G, the terminal endpoint of arc u coincides with the initial endpoint of arc u if, and only if u E f ( u ) . Therefore, H = G*. Q.E.D.

39


Theorem 16. An injectice 1-graph G = ( X , U ) is connected $ and only ih G is quasi-strongly connected. 1. Recall that a graph G is quasi-strongly connected if for each pair x, y E X, there is a vertex z whose set of descendants { z } u r(2)u r Z ( z )u ..* contains both x and y. If G is quasi-strongly connected, it is clearly connected.

2. Let G be injective and connected; then each pair x, y is joined by an elementary chain p[x,Y1 = LX, First, suppose that (x, a,) E L'; then

a2,

...Y

E

X , where x # y ,

ak,y] *

(x, u * (ak,Y)E u because, otherwise, there exists an a, with &(a,) > 1. Thus the vertex z(x,y ) = x has both x and y for descendants. Now suppose that ( y , a 3 E U ; then (Y,ak)EU

=>

(Q1,X)EU.

Thus vertex z(x, y ) = y has both x and y for descendants. Finally, suppose (al, x) E U and (ak,y ) E U ; then there exists at least one vertex a, in the chain p [ x , y ] such that (ai, a,- 1) E u, (ai, a(+1) E u * Since G is injective, p[a,, x ] and p[ai, y ] are paths, and z(x, y ) x and y for descendants. This shows that G is quasi-strongly connected.

=

ai has both Q.E.D.

Corollary. Afunctional I-graph G = ( X , r)is connected if, and onZy if, the inverse 1-graph ( X , r - ') is quasi-strongly connected. The proof is obvious.

Theorem 17. A necessary and suficient condition that the edges of a simple graph G = (A', E ) can he directed to form an injectire (resp.functional) graph is that each connected component of G contains at most one cycle. 1. Necessity. Let H be an injective graph; each cycle is a circuit (otherwise, there would be two arcs leaving the same vertex). No two distinct circuits can have a common vertex because then two arcs would enter the same vertex. On the other hand no two circuits in the same connected component can be without a common vertex, because each arc incident to the circuit is directed out of the circuit (and there is no vertex z that has descendants i n both circuits).

40

GRAPHS

Thus there is at most one cycle in each connected component.

2. Suficiency. Let G = ( X , E ) be a simple graph with at most one cycle in each connected component. Direct its edges in the following way: If a connected component of G contains no cycles, it is a tree, and its arcs can be directed to form an arborescence. If a connected component of G has only one cycle, direct first the edges of this cycle so that it becomes a circuit. Then, by contracting into a single vertex xo all the vertices in the cycle, the connected component becomes a tree: direct the edges of this tree to form an arborescence with root xo. Clearly, the directed edges induce in G an injective graph: Q.E.D. Theorem 18. Let cp be a mapping from a subset of X into X.Each connected component C, ofthefunctional graph G = ( X , q) is the union of two connected components D, and Dl(,, of the functional graph H = ( X , q2)L Moreotler, fi D , = Dm(x)rthen D , is a connected component of H with a circuit of odd length. Let C, denote the connected component of G that contains x. 1. If y E C,, there exist from Theorem 16 integers p and q with q z q ( y )= cp2p(x)or cp2q(y) = q2p+’(x). Let g(x) = rp2(x).We may write

either gq(y) = gp(x) or gq(y) = g p [ q ( x ) ] . Hence Y

Conversely, y

E

D,

E0 ,

v Ddx) *

U De(,) implies that

qp(x) = cpq(y),and therefore y

E

C.,

2. If D, = D+(,), there exist integers p and q with = gq[cp(x)]

or, equivalently, q2P(x)

=

yP+’(X).

This proves the existence in G of two paths p [ x , z] and v[x,z ] , one of which is odd and the other even. Hence, there exists a cycle of odd length. Since G is functional, this cycle is a circuit, and its vertices are the vertices of a circuit in graph H = ( X , q2). Q.E.D.

APPLICATION (Rufus Isaacs). Find a real valued function q ( x ) on R such that cp[cp(x)] = ax + b, where a, b E [w.

41


If a > 0, Menger found:

This function cp is the required solution, since

If a < 0, the problem is more difficult. For example, take a = - 1 and b = 0. From the preceding theorem, we know that a connected component of the graph G = ( X , q) is the union of two distinct connected components of H , for example, D, and D,,, if x E [2 k , 2 k 1 J or D, and D x - l if x E [2 k + 1, 2 k 21. This immediately gives the graph of the function (shown by the dotted lines in Fig. 3.6).

+

+

Fig.

Fig. 3.7. Function p such that (p2(x) = - x

42

GRAPHS

5. Counting trees

Before presenting results about the number of different trees in a graph, we shall state some properties of multinomials coefficients :

By convention, these coefficients will be equal to 0 if we do not have n l , n,, ..., np 2 0 and nl n2 + n, = n. Proposition 1. Let X be a set of n distinct objects. Let n,, n 2 , ..., np be nonn2 n, = n. The number of ways to negative integers such that nl place the n objects intop boxes X,, X,,..., X,,containing n, ,n,, ..., npobjects respectively, is

+ +

+ + +

n

n!

(n l , n,, ..., n p)= nl ! n, ! ... n p ! The set X, can be chosen in

(3( iznl)

different ways. Suppose the set X, is

chosen, then the set X, can be chosen in

different ways, etc. Hence,

the required number is

=

n! ( n - nl) ! (n - nl - n,) ! np ! n l ! ( n - n l ) ! n , ! ( n - nl - n,)! n 3 ! ( n - nl - n, - n 3 ! ) " ' n p !

-

n! nl ! n2 ! n3 ! ... np !*

-

Proposition 2 (multinomial formula). Giren'p real numbers we have (a1

+ a, + *.- + a,).

= nl,n2,

z(

n

...,n , , ~ n~

Consider real variables a f , where 1 product

nZ7

(at).' ***)

(a,Y ... (Up).?

np

< i < n, 1 < j < p ,

and form the

43


Given the integers n , , n 2 , ..., np whose sum is n, consider in the above polynomial a monomial of the form il

iz

(a, a1

... a inl l ) (a2II a,j z ... a2jn ... (a,

fl

2)

kl

n,

... a,

kn p).

This monomial corresponds uniquely to an arrangement of the set N = { 1, 2, ..., n } into the boxes N , , N , , ..., N,,, where

IN1I=n,,IN2I =nz,*..,IN,I=n,. By proposition 1, the total number of such monomials is therefore

n n! n l , n 2 , ..., np> = nl ! n , ! ... n,!

'

If we put a? = a? =

= a; = a,

for all i, we obtain the desired formula.

+.

Q.E.D. Proposition 3.

n

n - 1 n i , n 2 , ..., n i - 1 , ni - 1 , n i + l , ..., np

Clearly, (a,

+ a, + + up)" = (a, + a2 + -.. + up)(a, + a, + + up)"-' , - a .

and the general term of this polynomial is

(nl, n 2 , ..., n p1 n

a;1

a? =

... a:=

c i/,l#o

.i(

.

nl,

n-l

..., ni -

1,

1

..., np

a;l

... a;'- ... a: . Q.E.D.

We are now ready to consider the problem of counting the number of ways to choose a set E c P , ( X ) such that the simple graph ( X , E ) is a tree. , Theorem 19. Let T ( n ; dl , d,, ..., d,) denote the number of distinct trees H with vertices xl,xz, ..., x, and with degrees dH(xl) = dl. dH(xz)= &, ..., d H ( x n )= dn. Then

T ( n ; d l , d 2 , ..., d,) =

n-2 d l - 1,dZ - 1 ,

>

..., dn - 1

1. Clearly, the sum of the degrees is twice the number of edges. From Theorem 1, the sum of the degrees for a tree is 2 (n - 1). Thus, T # 0 only if

44

GRAPHS n

C (di - I) = 2(n - 1) - n = n - 2 .

i= 1

Without loss of generality, we may suppose that dl 2 d2 2 ... 2 d,; since the above equality implies that d, = 1, vertex x, is pendant in the tree.

2. We shall show that T ( n - 1 ; d l , d z , ... , di - 1, ..., d,-l)

T ( n ; d l , d z , ..., d,) = ijdj 3 2

.

Let C, be the set of the trees H with vertices xl,x,, ..., x, and degrees dH(x,:)= d k , such that the pendant vertex x, is joined to xi.If di 2 2, then

I @ i I = T(n - 1 ; d,, d2, ..., di - 1 , ..., d , - l ) . Since the set of all trees is the union of the sets V, for di 2 2, the above equality follows.

3. The theorem is true for n is true for n - 1. Then T(n ; d l , d z , ..., d,) =

C

ijdj 3 2

=

2. Assume that n 2 3 and that the theorem

T ( n - 1 ; d l , d2, ..., di - 1 ,

..., d,-l)

=

n-2 = (d, - 1 , d 2 - 1, ..., d,

-1 Q.E.D.

Corollary 1 (Cayley [1897]). The number of direrent trees with vertices x l , x2,. .., x, is nn-2. Using Proposition 2, the number o f trees equals

Corollary 2 (Clarke [1958]). The number of diferent trees H with vertices

x l , x 2 , ..., x, and with dH(xl) = k is

45


The desired number equals

(by setting all variables equal to 1 in the multinomial formula). Corollary 3 (Moon [1967]). Let G = ( X , E ) be a simple complete graph of order n. Let (XI , X2,..., X,) be a partition of X , and let

m,

w2,

..., H, = (x,,jE,)I, H~ = (Ily H , = be pairwise disjoint trees of orders I Xt I = n, . The number of spanning trees of G that have H I , H2, ..., H p as subgraphs is ..., Hp) = n , nz ... np nP-'.

T ( H l , H,,

If each set Xt were contracted to a unique vertex a t , then, the number of trees

w with

dg(ai) =

(i = 1 , 2,

di

..., p )

is

1.

P - 2

- 1, ..., d , - 1 exactly (nJdl (nJd2 ...

( d , - 1, d, To each tree R correspond trees H of graph G. Hence TW,Y H , ,

.*.Y

(~z,)~P

different spanning

Hp) =

= nl n,

... n,(nl

+ n2 + + np)P-2. *.*

Q.E.D. Corollary 4 (Cayley [1889]). The number of forests with rertices xl, x 2 , ..., x, and with p connected components such that xl, x 2 , ..., x, belong to p

different trees is T'(n ;p ) = pnn-P-'

.

46

GRAPHS

Let C be the set of trees H on the vertices x o , xl, x,, ..., x, such that dH(xo)= p . From Corollary 2,

If P c { 1,2, ..., n } and I P I = p , let VPdenote the set of trees in V such that, for all i E P,the vertex x, is joined to x,. Then

Hence

Therefore,

T'(n ; p ) =

.

(n - l ) ! P ! ( n - P I ! ,,n-p = n! ( p - 1) ! ( n - p ) !

p,tn-~-l

.

Q.E.D Let X = { xl, x,, ..., x,} be a set of n vertices, and let E c P,(x) be a set of q edges that join pairs of vertices in X. We now propose to calculate the number T ( X , E ) of different tPees on X that do not contain any edge of E. Let ( X , F ) be a graph with n vertices, q edges and p connected components with, respectively, n,, n2, ..., np vertices. Let v(F) =

( ni n *, ... n p

if graph (X,F ) contains a cycle, otherwise.

Theorem 20 (Temperley [1964]). The number of different frees on sef X that do not contain any edge in E is T ( X , E ) = nn-'

c- v(F)($)IF1

F c E

If e E E, Iet A , denote the set of all trees that contain edge e. Let F

c

E. If

(X,F ) has no cycles and hasp connected components, Corollary 3 to Theorem 19 states that the number of different trees that contain all the edges of F is

41


If (X,F ) contains a cycle, the above formula is still valid, since bath sides of the equality are 0. From Sylvester's Formula, we have

Q.E.D. Corollary 1 (Weinberg [1958]). / f E is a set of qpairwise disjoint edges, then

T ( X , E ) = nn-2( 1 - ; ) q . In this case, if F

c E, then v(F) = 2 I F ' ,

and

Q.E.D. Corollary 2 (O'Neil [1963]). I f E is a set of q edges, all with a common endpoint xl, then

(

T ( X , E ) = nn-' 1 - ;)q-l(l

-+).

The proof follows, since we have

C

F c E

v(F)(+)lFi

=

k=O

k=O

(k + l)(:)(+)'

=

k - 1=0

Q.E.D. Corollary 3 Let S c X with I S I = s. /s E is the set of edges that join all the possible pairs of i7ertices in S (i.e. ( S , E ) is a complete graph), theti

(

T ( X , E ) = nn-2 1 - -

.

48

GRAPHS

Let Fp denote the family of subsets F c E such that (S, F ) is acyclic and has p connected components. Then

F c E

F E 9,

p= 1

For P C S, I P I = p , and F E F,,,consider the triples (S, F, P) such that the graph (S, F ) has a vertex of P in each connected component. From Theorem 19, Corollary 4,

1 { ( S , F, P)/ FE .Fp} I =

pss-p-'

.

Therefore

Consequently,

c

V(F)(+y

- p= 1

FI

(+)s-p(;)pss-P-l

s- 1

=

s-p

s

c (y)( p

p - 1= o

-1

- ):

- J'(1

s-1

Q.E.D. Corollary 4 (Scoin [1962]). rfgruph (X,E ) is the union of two disjoint cornplete graphs ( S , V ) and ( T , W ) with I S I = s and I T 1 = t , then

T ( X ,E ) = s ' - 1 From Theorem 19, it follows that

--.

p-1.

T ( X , I/ u W ) - T ( X , V ) T ( X , W ) ,n

-2

n.

-2

,n-2

'

Therefore, by Corollary 3,

Q.E.D.

49


Corollary 5 (Moon [1967]). r f E is a set of m - 1 edges that forms an elementary chain on a set Y of m rertices, then T ( X ,E ) =

ri! f

+ p - l ) ( -nl)m-p

(m

m -P

p=l

If F c E determines a graph ( Y , F ) with p connected components, then IFI=IYI-p=rn-p.

If m , , m2, ..., m pare respectively the numbers of vertices of these connected m, = m. For I F I = m - p , there are components, then m, ma * * * as many graphs ( Y , F ) as there are ways to choose positive integers m,, ma, ..., m, that sum to m. Thus

+

F c E

+ +

p= 1

FcE nr-p

c...

ml m,

... m p .

mt,mZ,... 2 0 m l t m l t tm,=m

The last summation equals the Coefficient of xm in the expansion of (x

+ 2 x2 + 3 x3 + -.)" = XP(1 - x)-ZP ,

From the binomial formula, this coefficient equals (-

1)"-p

- 2 A - 2 ~ - 1 ) ( - 2 p - 2 ) ...(- 2 p - ( m - p (m - PI!

- (m + p

- 1) ( m

+ p - 2) ... ( 2 p + 1) 2 p

( m - PI!

=(

m

I)) -

+

P -1 m - p Q.E.D.

We now turn to the problem of counting the partial subgraphs of a given graph G that are arborescences. Let A = ((a:))be the matrix associated with graph G, where a: = mG+(xi,x i )

denotes the number of arcs in G from xi to x,. Let D matrix defined by d;

=

mG+(X - { x i > , x i )

if if

=

((dj))be the diagonal

i # j i =j .

50

GRAPHS

The matrix D - A

((dj

=

D - A =

- a:)) can be written as

- a:

- a:

...

- a;

- a;

...

3

- an

Caf i#n

The determinant of this matrix equals 0 because the sum in each row is 0. The minor obtained by removing the first row and first column of this matrix is denoted by

Ai =

- a:

- a:

...

-a:

Cat, i+n

Lemma. Let G = ( X , U ) be a graph with m = n - 1 arcs and no loops. Then G is an arborescence wirh root x1 iL and only iL A l = 1. Otherwise A , = 0. 1. If G is an arborescence with root xl, then d: is equal to d; (xi) = 1 for i = 2, 3, ..., n. Index the vertices so that the indices increase along any path (this is possible since G is an arborescence). Then

11

- a:

0 A, =

1

0

0 .

.

lo

.

.

.

.

.

.

.

a

2 - u4

...

-a:

...

1

-

a,’

I

...

. . . . . . . . . . . . . . . . . . . . . . .

0

0

...

1 1

2. We shall now show by induction on n that if G is a graph with n vertices, rn = n - 1 arcs and with A , # 0, then G is an arborescence with root xl Each vertex x k , where k # 1, is the terminal endpoint of at least one arc of G (because, otherwise, the k-th column vector of A , is the zero vector and d l = 0). Since m = n - 1, the inner demi-degrees (see Chapter 1) satisfy the equalities:

.

51


&(XI)

7

0

CiG(xk) = 1

( k = 2, 3 , ..., n)

.

Hence, by Property (6) of Theorem 13, it is sufficient to show that G is connected in order to show that G is an arborescence. Suppose that G is not connected. Then A , can be decomposed into two square matrices B' and B" in the following way :

From equation (i), we see that the vertices x, with s E S and the vertex x1 generate a subgraph G ' with m' = n' - 1. The vertices xt with t E T and the vertex x1 generate a subgraph G"with m" = nu - 1. Since Det (B') # 0 and Det (B") # 0, we know by the induction hypothesis that G and G" are arborescences with root xl. This contradicts the assumption that G is not connected. Q.E.D.

Theorem 21 (Tutte 119481). Let G = ( X , U)be a graph, and let x1 E X . The number of partial subgraphs of G that are arborescences with root x, equals a:

I ..............................

where a: = mo+(x,,x,). Without loss of generality, we may assume that G has no loops. Note that = A,(a,, a3, ..., a,,) is a linear function of the last n - 1 column vectors of the matrix ((a;)),i.e.,

A,

Al(a;

+ a;, a 3 , ..., a,) = dl(a;, a,, ..., a,) + dl(a;, a 3 , ..., a,)

A,(&, a 3 , ... , a,)

=

Adl(%, a,, ... , a,).

Denote by ek = (O,O, ..., l , O , ..., 0) the n-vector with its k-th coordinate equal to 1 and with all other n - 1 coordinates equal to 0. Then

52

GRAPHS

C

=

k2 sk3

a:' ak,) ... a: Al(ek,, ek3,..., e k n ) .

>..

From the lemma, A l(ek2,eks, ...,ekn)equals 0 or 1. For each term in the sum, the partial graph defined by

a, = ek2, a3 = ekn, ... , a,, = ekn, has no loops (because G has no loops) and is formed from the n ( x k 2 , x2),

(xk,,

**-)

x3)3

- 1 arcs

(xk,, x n ) .

Finally, from the lemma, we know that dl(ek,, eka,..., ekn)= 1 if, and only if, the graph is an arborescence with root xl. This completes the proof. Q.E.D.

Corollary. ZfG = (X,E ) is a simple graph, the number of spanning trees in G is equal to the minor (which is independent of the coeffjcients of the principle diagonal) of the square matrix ((bi)) of order n, where =

bf

dG(xi)

= - 1

=o

if

i=j

if if

i #'j i# j

and and

[xl,xj] E E [xi, x,] 4 E .

Let G * be the graph obtained from G by replacing each edge by two oppositely directed arcs. A spanning tree of G corresponds uniquely to an arborescence of G* rooted at x1 (say). Therefore, by Theorem 21, the number of spanning trees in G equals

Al=i

b:

'

b:

1

:

1

b,3

d&3)

.

f

. .

b;

b;

...

i i_

d,(x,,)

Q.E.D. EXAMPLE. Consider the graph G in Fig. 3.8. As shown in Fig. 3.9, the number of spanning trees in G equals

A, =

i -:

= 16.

-1 -1

-1

31


53

.f, Fig. 3.8

1

t

w*

Fig. 3.9

EXERCISES 1. Let m and n be two integers.

(1) State a necessary and sufficient condition for the existence of a strongly connected 1-graph with n vertices and nz edges. (2) Show that, for all strongly connected 1-graphs with n vertices and rn edges, the lower bound on the number of edges whose removal can destroy the strong connectivity

is

[:I.

2. Show that if G = (A’, E) is a simple graph such that each edge is contained in some elementary cycle, then the anti-symmetric graph G’ constructed from G as in Theorem 14 is strongly connected; show that G’has only m n 1 = v(G) elementary cycles, and

- +

54

GRAPHS

that a graph without circuits can be Constructed from G' by reversing the direction of a subset of its arcs with cardinality min { rn - n

+ 1, n - 1 } .

(Chaty [1968]) 3. Show that the number of trees on n vertices that have exactly k pendant vertices is It s;$, where S; denotes the Stirling number of the second kind.

I'

( R h y i [1959]) 4. Consider a set X = {xl,xa, ..., x,,} and the group S,,of permutations on X. A set T o f transpositions [x,, x,] defines a simple graph ( X , T ) . Show that: (1) A set Tof n 1 transpositions generates the symmetric group S,, if and only if the graph (X, T)is a tree. (2) (Den&) I f f is a circular permutation of degree n, then the number of ways of writingfas a product of n - 1 transpositions equals nn-a.(Use Corollary 2, Theorem 19; for a detailed proof, see C. Berge, Principles of Cornbinarorics, Academic Press, New York, 1971, p. 143.)

-

5 . A circulation tree of a connected graph G = (X,U)is defined as a spanning tree such that each associated cycle is a circuit of G. Show that a strongly connected graph with n vertices and rn arcs has a circulation tree if, and only if, the total number of elementary circuits is rn n 1. (Chaty [19711)

- +

CHAPTER 4

Paths, Centres and Diameters

1. The path problem

The path problem is the following: Find (as quickly as possible) a path from a given uertex a to a giuen vertex b in a 1-graph G = ( X , V ) .The chain problem is similarly defined for a simple graph G = (X, E ) . Note that the chain problem becomes a path problem in the I-graph G * = (X,U) obtained from G = ( X , E ) by replacing each edge in E by two oppositely directed arcs. EXAMPLE. Problem of the hunter, wolf and Brussels sprout. A hunter, a wolf and a Brussels sprout arrive simultaneously at a river bank. The ferry boat is too small to take more than one passenger (in addition to the boatman) at the same time. For obvious reasons, the boatman cannot leave the hunter and the wolf alone together, nor can he leave the wolf and the Brussels sprout alone together. How should he arrange their passage across the river? This well-known problem can be solved mentally by considering only a small number of states. Nonetheless, it is a typical example of the path problem. A graph of the various states can be constructed, and a path must be found from state a (the hunter H , wolf W, Brussels sprout S and boatman Bare all on the right bank) to state b (all are on the left bank). One solution to the problem is shown in Fig. 4.1.

b

*

H WSB

HWB

WSB

HSB

WB

aHS

W

W

S

0

Fig. 4.1

For more complicated cases, several systematic algorithms have been proposed. If the graph is already known, it is always possible to find all the elementary paths starting from vertex a by constructing all the different arbo55

56

GRAPHS

rescences rooted at Q. Such an arborescence for the above example is shown in Fig. 4.2. H WB

Fig. 4.2

By inspection we see that there are two paths from a to b. A formal statement of an algorithm for finding all elementary paths from a was given by B. Roy [1960], and others. Of more interest are localalgorithms applicable when the entire graph is not known. Ideally, a local algorithm will not trace through the entire graph. Let G = ( X , U ) be a graph, and let a E X . If u l , u2, ..., u, are the arcs of G, define an alphabet whose letters are u l , u2, ..., u,, - u l , - u2, ..., u, (positive and negative letters). A word is a sequence of letters, written in the form

+

-

u1

If

~

i

= ~

+ + + ui + 02

*.*

+

u1+1

+

+ .'. + U k .

- u i , the word +

~

01

+ 02 + + U i - 1 + u i + 2 + + .*'

***

uk

is called a redziction of the preceding word. is called the trajectory of graph G when : A word p k = ul + u2 (1) v1 is an arc with a as its initiakendpoint, and tracing through the arcs corresponding to the consecutive letters of a word (in the direction of the arc if the letter is positive, and in the opposite direction if the letter is negative) defines a chain pk. (2) If U E U , then p k contains the letters + u and - u at most one time each. (3) If the letter - u is in the word, then the letter + u precedes - u in the sequence. (4) After word p k has been reduced as much as possible, the reduced word contains no more negative letters (and, therefore, the reduced word jlkdefines an elementary path starting at a). The TrCmaux Algorithm given below is a rapid, local algorithm to construct a trajectory terminating at vertex b.

+ +

ilk

PATHS, CENTRES AND DIAMETERS

57

TrCmaux algorithm We shall construct successively a sequence of the trajectories pl,p2, ., .. Each time an arc is used or a vertex is encountered, it is labelled. The algorithm consists of three rules: RULE1. Label vertex a. If 14 = (a, x ) is an arc with initial endpoint a, put + u. (In other words, we advance along arc u and label it.)

p1 =

+

RULE2. Let p i = u1 v2 21, be a trajectory terminating at x with vt > 0. If vertex x has not been previously encountered, and if x is the initial endpoint of an arc ui+ = ( x , y), let p'+' = pi ui+ and label arc ui+l and vertex y. Otherwise, let p i + l = pi - cir and label arc ci again. + . * a +

+

RULE3. Let pi = u1 +-.+ vi be a trajectory terminating at x with u, < 0. If x is the initial endpoint of an arc u , , ~= (x, y ) $ pi, let p f c l = p i + u ~ + ~ . Otherwise, let pi+' = pi - uj where uj is the last letter of the reduced word

6;. TrCmaux's theorem. When the above algorithm terminates, the terminal trajectory p* has the property that .for each arc u in a path starting from a, pk contains the letters u and - u exactly once.

+

1. When the procedure ends, the trajectory has returned to a, and all the arcs incident out of a have been labelled (by Rule 3).

2. We shall show that all arcs u labelled by the procedure have been traversed in reverse. In fact, if, each time the procedure reaches a labelled vertex via an unlabelled arc, we detach the arc from its terminal endpoint before traversing it in reverse (Rule 2), the graph of labelled arcs would become an arborescence rooted at a (by Theorem 13, Ch. 3, Property (6)). Hence, if this arborescence is explored through the same trajectories, all its arcs will have been traversed in reverse. 3. We shall show that if there exists an elementary path from a, say

then the arc up has been labelled by the procedure. has not been labelled (because, otherwise the arc If not, then vertex preceding vertex a p - l in the trajectory has not been traversed in reverse by either Rule 2 or Rule 3, which contradicts Part 2 of this proof). Therefore, arc has not been labelled. For the same reason, arc u p - 2 has not been labelled, and arc u1 has not been labelled. But this contradicts Part 1 of this proof. Q.E.D.

58

GRAPHS

Remark 1. With T r h a u x ' s Algorithm, a path from a to b will be found by purely local methods, if such a path exists. In fact, the procedure never labels the same arc more than twice-once in each direction. This givesa bound on the number of steps in the algorithm. Remark 2. If a chain between two vertices in graph G = ( X , U ) is sought, the algorithm can be applied to the symmetric graph G* constructed from G by replacing each edge in G by two oppositely directed arcs, To avoid the possibility of using an edge of G four times, we can, after labelling an arc in G for the first time, remove the oppositely directed arc from the graph. This can also simplify the procedure. Remark 3. For a graph G = ( X , U ) , it is easy to see that the Trkmaux Algorithm described above can be used to construct a maximal arborescence rooted at a given vertex. This arborescence is defined by the set of all labelled arcs which have been traversed in reverse by using only Rule 2.

Algorithm (P. Rosenstiehl. J. C. Bermond)

P. Rosenstiehl [1966] noted that the TrCmaux Algorithm is a special case of a class of algorithms described by Tarry [1895]. He also devised another labelling procedure, called the algorithm for new arcs, which is described below. We shall successively construct the trajectories p l , p2, .. . by the following rules: RULE1. If u = (a, x ) is an arc whose initial endpoint is a, let p1 = In other words, advance along arc u and label it.

+ u.

RULE2. Let pi = r1 t'[ be a trajectory with x as terminal vertex. If x is the initial endpoint of an unlabelled arc u ' + ~= ( x , y ) , let p ' + l = I*'+ u i + l . Label the arc u ~ , ~If.there does not exist an unlabelled arc with initial endpoint x , and if the reduced word jic is not null, let p'+l = pi - u, where u is the last letter of the reduced word. Theorem. When the above algorithm terminates, the terminal trajectory pk has the property that for each arc u in a path from a, p k contains the letters u and - u exactly once. (The proof is identical to the proof of the TrCmaux theorem.) +.a-+

+

In other words, a path from a to 6, if it exists, will be found by purely local methods and without labelling the same arc more than twice. Note that if G is

PATHS. CENTRES AND DIAMETERS

59

an arborescence rooted at a, the algorithm for new arcs and TrCmaux’s Algorithm are equivalent. Algorithm for a planar graph.

If a chain is sought between two vertices a and b in a planar graph G, and if both a and b are on the unbounded face, we have the mazeproblem. A traveller, who tries to get out of the maze by following the passageways, can find a chain between a and by the following rule: At a,junction, always take the passageways on the extreme right. This algorithm always locates the exit to the labyrinth without traversing any passageway more than twice.

2. The shortest path problem The shortest path problem is the following: Consider a graph G , and for each arc u, a number I(u) 2 0, called the length of u. Find an elementary path from a to b that minimizes

44 =

c Ku)

UEll

EXAMPLE. Find the shortest route on a map from city a to city b. To do this, construct a graph G by representing each road between two localities on the map by oppositely directed arcs in the graph; let the road’s mileage correspond to the arc’s length. Thus, this example reduces to a shortest path problem. (Similarly, we might search for the fastest or most economical journey.) Shortest path algorithm (Dantzig [1960]) Let the starting vertex be denoted by a , ; we shall determine, for each vertex x, the length r(x) of the shortest path from a, to x by the following rules: (1) To begin with, let t ( a l ) = 0. Function t is therefore defined on the set A 1 = { a1 1. (2) Suppose that on the k-th step, the function t has been defined on the set Ak = { a,, a 2 , ..., ak }. For each vertex a, E A,, select a vertex b, E X - A k such that (a,, b,) E U and such that the length l(aj, b,) is minimum. Find a vertex a4 E Ak such that t(aq)

+ l(a,, b,) = min { l ( a j ) + [(aj, b j ) } . i

Then, put Ak+l

=

”{

f(hJ= t(a,) +

bg

1 7

bq) .

60

GRAPHS

It remains to show that t(b,) is the length of the shortest path to b,, and that this path passes through a,. Assume that r(a,) is the length of the shortest path to a j , for all a, E A k . All paths that leave A k have a length 2 t(a,) /(aq,6,) = t(b,). To reach b, one must surely pass through A k , since a, E A k a Therefore t (b,) is the length of the shortest path from a to b. Q.E.D. Note that each time a vertex 6, is selected, all the arcs with terminal endpoint b, may be deleted. Remark. Suppose that the vertex b, that is associated with a, E Ak can be found without difficulty. Then the selection of the (k + I)-th vertex requires only k comparisons. The maximum number of comparisons needed for a graph with n vertices is 1 1 2 + ... ( n - 1) = - n(n - 1 ) . 2

+

+

+

This bound can be improved if we search simultaneously for the length of the shortest path from bl to b.

Fig. 4.3

EXAMPLE.Consider the graph in Fig. 4.3. The number next to each arc u is its length I(u). 1 . t(a) = 0. Compare ah = (0 1) and QC = (0 2). Choose ah. 2. t(h) = 1. Compare he = (1 + 3), hd = ( I + 3) and ac = (0 2). Choose ac. 3. t ( c ) = 2. Compare he = (1 3), hd = (1 3) and cf = (2 + 2). Choose he, hd and cf. 4. t(e) = t ( d ) = t(f) = 4. Compare eg = (4 3), dg = (4 3) and f g = (4 2). Choose,fg. 5 . t ( g ) = 6. Compare gb = (6 1) and f b = (4 4). Choose gb. 6. t(b) = 7. Thus a shortest path from a to b is acfgb.

+

+

+

+

+

+

+

+ +

+

The next theorem follows immediately from the above algorithm: Theorem 1. I f G = (X,U ) is a graph with root a, and with a length I(u) 2 0 for all u E U , then there exists in G a spanning tree H = ( X , U ) which is an


61

arborescence with root a, and such that each path in H is a shortest path from a in G. Clearly, the different arcs chosen by the algorithm form an arborescence H (since H has no circuits and is connected). Analogous algorithm for symmetric graphs For a symmetric graph, a simple algorithm to find a shortest path is to represent the arcs of the graph by strings of the appropriate length and to let the vertices be represented by knots that tie the arcs together. To find a shortest path between knots a and b, simply pull knots a and b apart. The taut strings between a and b will represent the shortest path.

3. Centres and radii of quasi-strongly connected graphs Consider a 1-graph G = ( X , U ) and two vertices x and y of G. The directed distance d(x, y ) is defined to be the length of the shortest path from x to y . (If no such path exists, let d(x, y ) = co.)The associated number e(x) of a vertex x is defined to be e(x) = max d ( x , y )

.

YCX

Y Z X

A traveller at vertex x can reach any other vertex in e(x) or less steps. A centre of G is defined to be a vertex xo with the smallest associated number. The associated number e( xo) of vertex xo is called the radius of G, and is denoted by p ( G ) . These concepts are important in telecommunications. A communication network may be represented by a graph (not necessarily symmetric), and a centre of the graph represents an optimal site for a transmitting station. Proposition 1. The directed distances d(x, y ) satisfy (1)

d(x, x ) = 0

d ( x , Y ) + dCv, 2) .> d ( x , z ) . (2) If the graph is symmetric, Hpe have also

(3)

4x9 Y ) = 4 Y , 4 .

The proof is immediate. If the graph is symmetric, the function d(x, y ) satisfies (l), (2), and (3) and is a distance in the topological sense. Recall (Ch. 3, Q 3) that a “root” of a graph is a vertex xo such that for each vertex y , there is a path from xo toy. Also, recall that a graph has a root if, and only if, it is quasi-strongly connected.

62

GRAPHS

Clearly, a centre has a finite associated number if, and only if the graph has a root, and therefore, if, and only if, the graph is quasi-strongly connected. Henceforth, we shall assume that graph G is quasi-strongly connected.

Proposition 2. I f x , and y o are two centres of graph G, then they both belong to the same strongly connected component. The proof is obvious.

Theorem 2. I f G is a 1-graph of order n, without loops, andsuch that max d,+(x) = p > 1 , then its radius p(G) satisfies

+

If p(G) = co,the theorem is obvious. Suppose that p(G) < co. Let xo be a centre of G. The number of vertices with a directed distance of 1 from xo is < p . The number of vertices with a directed distance of 2 from xo is 6 p2. Thus

or n(p - 1)

+ 1 < pP+l,

and hence, log(np - n

+ 1) < (p + 1) l o g p .

This gives the formula.

Q.E.D.

Lemma. Let G = ( X , U ) be a strongly connected graph of order n, and let a be a root of G. Consider a spanning arborescerzcd H = (X,V ) of G with root a such that each path of H is a shortest path of G. Let B denote the set of all terminal vertices in arborescence H. Then

I Ble(a)2 n -

1.

Equality holds if; and only if; H consists of I B I paths of length e(a) starting from a without common certices (except a).

The vertices of H (except a ) can be placed on e(a) horizontal lines such that = i is placed on the i-th horizontal line.

a vertex x with d(a, x )

63


If b E B, let X ( a , b) denote the set of all vertices (exctpt vertex a) in the path of H that goes from a to b. Then,

Theorem 3 (Goldberg [1965]). vertices and m arcs, then

If

Q.E.D. G is a strongly connected l-graph with n

where [r]*denotes the smallest integer 2 r. For all m and all n, there exists a strongly connected l-graph with n vertices and m arcs such that the aboue equation holds with equality. 1. Let a be a centre of G,and let H be a spanning arborescence with root a as defined in the lemma. Since G is strongly connected, each pendant vertex b E B in arborescence H is the initial endpoint of an arc in G - H. Since the number of arcs in G - H is m(G - H ) = m(G) - m ( H ) = m - n 1, by Theorem ( 1 , Ch. 3), we have 181 < m - n 1. Then, from the lemma,

+ +

This yields the required inequality. 2. We shall construct a strongly connected graph G with n vertices and m arcs, with radius

rn-n+l This graph G will, in fact, be a rosace with centre a (see Fig. 4.4). It consists of m - n 1 circuits (the “branches” of the rosace) having a common vertex a. Let

+

Thus, we can write: n - 1 = (m - n

+ 1 ) ( q - 1) + r ;

0 0. Hence, G is strongly connected with n vertices, and p ( G ) = q. Q.E.D.

Fig. 4.4. Rosace with centre u, n = 15, m = 18, p = 4.

We shall now study the properties of the centres of a complete 1-graph. Theorem 4. If G = ( X , r) is a complete I-graph, each vertex xo such that

I r ( x o ) - { xo 1 j is a cenfre, and p ( G )

I

- { x 1I

= max ~ ( x ) XEX

< 2.

If

1

1

max ~ ( x ) {x } = I X1 - 1, X

the theorem is true, and p ( G ) = 1. Otherwise, consider a vertex xo for which I T(x,) - { xo } I is maximum. Since e(x) 2 2 for all x , we have only to show that e(xo) = 2 . Suppose that e(xo) > 2. Then there exists a vertex y # xo that cannot be

Fig. 4.5

65


reached from xo by any path of length 1 or length 2. Since y .$ T ( x o ) ,then xoE T ( y ) . Moreover, if z is a vertex of T(xo)- { xo}, tben y .$ T(z), since, otherwise, there exists a path of length 2 from xo to y . Then z E T ( y ) , and z E T ( y ) - { }. Hence, T(xfJ) - { xo 1 = rcu> - Y j * Since xoE T ( y ) - { y }, the above inclusion is strict, and consequently,

Q.E.D.

which is clearly a contradiction.

Theorem 5 (Maghout [1962]). If G = ( X , U ) is a complete I-graph M + t h radius 2, then,for each y E X there exists a centre xosuch that (xo,y ) E U .

Let y E X. Since the radius of the graph is greater than 1 , there exists a vertex x 1 # y such that (Y, 4 4 u * Hence, (xl, y ) E U. If x 1 is a centre, the theorem follows. Otherwise, there exists a vertex x 2 # x l , y , with (xl, x 2 ) .$ U and ( y , x,) .$ U. Hence, (xz, x1) E

u,

(X2,Y) E

u.

If x 2 is a centre, the theorem follows; otherwise, there exists a vertex x3 # xl, x 2 ,y such that (x2’x3)#

u,

(x17x3)#

u,

(y,xj)$u .

Hence (X3,X2)EU,

(X3tX1)Eu,

(x3,y)EU*

If x g is a centre, the theorem follows; otherwise, there exists a vertex x g , etc. At least one vertex xk located by this procedure is a centre. Otherwise,

TC(Y) = x - { y } . Therefore a centre of the subgraph generated by X - { y } is also a centre of graph G , and by Theorem 4, a centre would have been located during the procedure. Thus, this vertex xk is the required centre. Q.E.D. Corollary. A complete 1-graph G with radius 2 has at least 3 centres. Let C ‘ be a complete, anti-symmetric graph obtained from C by removing some of its arcs. This new graph G’ has a centre, y1 and, by Theorem 5, there

66

GRAPHS

exists another centre y 2 with ( y z ,y,) E U.Also, there exists a centre y , with ( y 3 ,y z )E U . Since the graph G’is anti-symmetric, Y3 f

Y1

-

Thus, G has at least three distinct centres y l , yz and y , .

Q.E.D. 4. Diameter of a strongly connected graph

The diameter 6(G) of a graph G is the maximum of the directed distances, 1.e. : S(G) = max d ( x , y) ,

,\

I

00 m

m = 5 , p = 4 , 6 = 4.

Rosace m = 8,p

m

m =9,p = 2,6

Rosace =

1,6 = 2.

= 6 , p = 3 , 6 = 4.

m =6,p

=

6 , p = 2 , 6 = 4.

=4

Rosace = 3,6 =

m=7,p=2,6=4.

Fig. 4.6

4.

m

m

=

7 , p = 2,6 = 3.

= 20, p

-;L

1 , 6 = 1.

PATHS, CENTRES A N D DIAMETERS

67

The diameter is finite if, and only if, G is strongly connected, In this section, we shall assume that G is always strongly connected. Also, without loss of generality, we may assume in this section that G is a I-gr$ph without loops. EXAMPLE. In Fig. 4.6, each graph has 5 vertices and is strongly connected. The number of arcs is denoted by m, the radius by p, and the'diameter by 6. A circle is drawn around each centre. A square is drawn around each vertex x such that e(x) = 6(G). Note that for n = 5, the equation 6 + m 2 9 is always satisfied. Only the complete symmetric graph has a diameter equal to 1. Since the graph is finite, then clearly, 6 < 00, and if the graph has a centre, 6 2 p. In a graph representing the avenues of communication between the various members of an organization, the diameter 6 represents the maximum number of times that a message must be relayed before it reaches its destination. The problem of constructing strongly connected graphs with n vertices a6d m arcs whose diameter is as large as possible (or as small as possible) was considered by Bratton [1955]. The problem of maximum diameter graphs has been solved by Ghouila-Houri [1960], and the problem of minimum diameter graphs has been solved by Goldberg [1966]. Before proceeding, note that if G is a strongly connected graph without loops and with n vertices and m arcs, the numbers n and m cannot be chosen arbitrarily: If n > 1, G has at least one cycle, hence, the cyclomatic number u(G) = m - n 1 > 1, and consequently, we have m >, n . From Theorem (9, Ch. 3), we know that equality holds only if G is an elementary circuit. Furthermore, since the number of arcs in G cannot be greater than the number of arcs in a symmetric complete graph with n vertices, we have

+

rn

< n(n - 1 ) .

Theorem 6 (Goldberg [1966]). I f G is a strongly connected I-graph without loops and with n vertices and m arcs, and if G is not an elementary circuit, then

6(G) 2

[ m2(n- n-+ ')l

*,

Furthermore, this is the best possible result.

If G is not an elementary circuit, then, from Theorem (9, Ch. 3),

rn

- n + 1 = v(G) 2 2 .

68

GRAPHS

Consider a strongly connected graph G with cyclomatic number v(G) 2 2; we shall show first that v(G)6(G) 2 2 (n - I) . A trail of G is defined to be a n y elementary path p = [x,, x,, ..., xs] such that for i = 0, 1, ..., s - 1, only one arc of G is incident out of x,. Let xo denote the initial vertex of a longest trail of G,,say: Po = [xo,

--.>

%I*

Let s be the length of p,. Construct an arborescence H rooted at xo as defined in the lemma to Theorem 3. Clearly, H begins with path p,, and its first node after xo is x,. Let B denote the set of terminal vertices of arborescence H.

+

1. We shall show that I B I 1 < v(G). The number of arcs in G - H is v(C). Since each vertex b E B is the initial endpoint of at least one arc of G (because G is strongly connected), we have v(G) 2 1 B 1. If v(G) = I B 1, then each vertex of B-is the initial endpoint of only one arc of G, and there exists a vertex b, E B such that (bl, x,) E U. Clearly, since I B I = v(G) B 2, the vertex 6 , is not on p o = [x,, xl, ...,x,], and [b,, x o , xl, ..., x,] would be a trail of length s + 1. This contradicts the maximality of p o . Hence, v(G) > 1 B 1. 2. We shall show that s 2 3 6(G) implies n - 1 < 3 6(G) v(G). Let v(G) = v, and let 6 ( G ) = 6. If we denote the associated number of a vertex x in the arborescence H by e,,(x), then, from the lemma to Theorem 3,

n-l,<s+e,(x,)IBJ,<s+(S-s)IBI= 6 = s I B J - s(l BI - 1) < Sl B I - ?(I BI

- 1) =

6 6 = -2( I B I + 1 ) 6 z . v . 6 . 3. We shall show that s < - implies 2

6 1’

n - 1 2 for z E B3. Therefore t(zo) =

+

Finally, we have

n

< s + 1 + s l B l + (6 - 2 s ) I B3 I < s + 1 + s l B l + (6 - 2 s ) y1’ .

Hence,

n-l 1 andn, > 1 w i t h j - i > 1, then the value offcan be increased either by taking n; = ni - 1 and n; = n j 1, or by taking ni = n, 1 and n; = n, - 1, since

+

f ( n , , ...)n, k 1,

= f ( n o , ...)n,)

..., n j T 1, ... n,) )

+

=

+ 1 k (nt-1 + + ni,l) n1

T (nj-1

It follows that there exists an index k, with 1

i Z k }>.,=1. i # k + l

+ n, + n j + l ) .

< k es.A set X of residents belong to various clubs C1,C,, ..., C, (which are not necessarily disjoint subsets of X) and t o various political parties PI,Pz,..., P, (which are disjoint subsets of A'). Each club must choose one of its members to represent it, and no person can represent more than one club, no matter how many clubs he belongs to. HOWshould one choose a system of distinct representatives A = { al, a,, ..., a, }, such that the numbers of representatives belonging to each party Pisatisfies bj

MY, x)

*

I f sink b is labelled b y this procedure, we shaN show that rhe arcflow q1 can be augmented, i.e., Hie shall construct a new flow cp' such that cp; > q,. Let

P

=

b,ai, U 2 , ..., Uk, bj

be a chain from a to b in which each vertex ai+l.has been labelled from its predecessor a,. (1) If the edge [a,,a,,,] is directed from a, to a,,,, then we have q(ai, ai+d
ci

'pi
0, there exists an arc i with d,(cpi)> 0. Suppose for example that d,(cp,) > 0, and that cpl < b,. Then, sequentially, label the vertices according to the following rules: RULE1. Label vertex a, the terminal endpoint of arc 1 with the index

+ 1.

RULE2. If x is labelled and y is not labelled, label y with the indexj if < cp

(x, y ) is arc j and if cp,

RULE3. If x is labelled and y is not labelled, label y with the index - j if ( y , x) is a r c j and if 'p, > b,. If vertex 6, the initial endpoint of arc 1, is labelled, then a new flow cp' such that d(cp') < d(q) can be constructed by using the method of the Ford and

88

GRAPHS

Fulkerson algorithm. If the initial endpoint of arc 1 cannot be labelled, then the set A of labelled vertices satisfy a E A and b $ A . Since (pl (A)

iEW

ci-

-(A)

1

bi.

LEO - ( A )

From the lemma, no compatible flow exists, and the problem has no solution. This algorithm establishes the following result :

Compatible flow theorem (Hoffman [1958]). For a graph G with arc numbers b, and c, such that - 00 < b, < cl < + co for all i, a necessary and suficient condition that there exists aJow cp with bt < 'pi < ct for all i is that, for each set A c X ,

The necessity of the condition follows from the lemma. The condition is sufficient because each flow cp with d(cp) > 0 can be improved by using the preceding algorithm, until we obtain a flow cp' with d ( q ) = 0. Q.E.D. The following algorithm is in general more efficient.

Second compatible flow algorithm Let G be a graph with vertices x l , x2, ..., x , and with arc numbers b f , ci. The following rules construct a flow cp in G such that

bi

< cpi f

ci

'

( i = 1 , 2 ,..., m).

RULE1. Construct a transportation network R' from G by adding a source a, asink b, a return arc (6, a), and the various source and sink arcs.

RULE2. If b(x, y ) 2 0, let the capacity of arc ( x , y ) in network R' equal CYX,

Y ) = 4x9 y> - 4 x 9 Y> ,

and create a sink arc ( x , b) with capacity c'(x, b) = b(x, y )

.

Also, create a source arc (a, y ) with capacity

m,v) = W , Y )

*

RULE3. If b(x, y ) < 0, let the capacity of arc ( x , y ) in network R' equal

89

FLOW PROBLEMS

CYX,

Y ) = 4 x 9 u> -

m,Y>,

and create a source arc (a, x ) with capacity c’(a, x) =

- b(x, y ) .

Also, create a sink arc ( y , b) with capacity CYY,

b) =

- 0 ,v) -

We shall now show that a compatibleflow exists in G i f , and only i f , in transportation network R’, there exists a maximumjow cp‘ that saturates all source arcs and sink arcs. If we let q ( x , y ) = cp’(x,y ) + b(x, y ) , then clearly cp is a flow in G because for each vertex x in G,

C

bm+W

pi

- C

pi =

iE(D-(X)

C

( ~ f bi) ++

IErO+(X)

bi 3 0

Flow cp is compatible in G, because

0 < pi

< C; - bi

(i=1,2

,..., m),

and consequently,

b,

(i = 1 , 2,

< 401 < ci

..., m)

Conversely, it is easily seen that each compatible flow cp in G corresponds in R’ to a flow cp’ that saturates all source and sink arcs. This completes the proof. Q.E.D.

3. An algebraic study of flows and tensions Each flow considered above is a vector in Z“.Flows could also be considered in any ring R, with s , t ~ R OE R SER s , t ~ R

=-

s + t ~ R , (zero element), * --sER, => s.tER.

The space Z” is not a “vector space” on Z, because Z is not a “field”, but a “module” on Z, and

90

s, t E b ,

E

n,s E n m

-

GRAPHS

=

s

+ t = (sl + t l , ...)s, + tm)E Z",

is = ( h 1 , ..., Asm)E n r n .

Consequently, the set @ of all flows in graph G constitutes a subiiiodule of

Z",i.e., we have: cp'+$€@,

=>

'P',cp2E@ S€h,cp€@

,

*

SCpE@.

Theorem 3. Let G = ( X , U ) be a connected graph. Let H = ( X , V ) be a spanning tree of G. Denote the arcs of U - V by 1, 2 , ..., k, and denote the associafed cycles of'H by p', p2, ..., pk. A J o w cp is uniquely defined from the i d u e s cpl, cp2, ..., (Pk by 9 =

p1

+

'p2

p2

+

f

(Pk

pk.

Consider the vector k

This vector cp is a flow, since it is a linear combination of flows. Clearly cp takes only zero values outside of tree H . Let W c Vbe the set of arcs ifor whichcp, # 0. We shall show that W = 0 i.e., each connected component C of the partial graph (X,W ) reduces to a single vertex. If not, C is a tree, and from Theorem (2, Ch. 3), C necessarily contains a pendant vertex a. If, for example, the pendant arc is incident into a , then

c

O#

pi=

c

cp;=o,

I E m'(a)

i€m-(.)

which is a contradiction.

Q.E.D. Corollary. A rrecessary and suflcient condition that a rector cp he a J o w is that it is of the form Cp =

$1

p1

+

$2

p2 f ' * '

+

sk

pk 3

where sl,s,, ..., sk E Z and pl,p2, ..., pk are elementary cycles.

Since a cycle is a flow, this linear combination of cycles is a flow. The converse follows immediately from Theorem 3, because graph G may be assumed to be connected (otherwise, each connected component could be considered separately). Q.E.D. In particular, Theorem 3 shows that a cycle p can be obtained by the addi-

91

FLOW PROBLEMS

tion of all the cycles p*whose "out of tree" arc is used by p, In this addition, a cycle is preceded by a sign if the cycle is in the direction of p. Otherwise, the cycle is preceded by a - sign. Hence, Theorem 3 reduces the number of unknowns for the determination of a flow from 112 to m - n + 1.

+

Theorem 4. A necessary and su#icient condition that a rector cp be a flow uith no negatire components is that it is of the form cp = s1 p1

+

where sl,s,, ..., s k E Z,sl, s,, ..., s k

s2

p2

+ + sk p k "*

7

> 0 and pl, p2, ..., p k are circuits.

Clearly, a vector cp of the indicated form is a flow 2 0. Conversely, consider a non-zero flow cp 2 0, and let C be the graph obtained from G by removing all arcs i with 'pi = 0. Graph C contains no cocircuits, and therefore by the arc colouring Lemma, it contains at least one circuit pl. Let s, > 0 be the smallest flow in an arc of pl. The vector cpl = cp - s1 p1 is a flow 2 0 that has more zero.components than cp. If cp' is not a zero vector, repeat this process, etc. ... Finally, a zero flow of the form cpk

= cp - s1 p1 - s2 p2

- ... - s k p k = O ,

is obtained with sl,s2, ..., sk E Z and sl,s2, ..., sk 2 0.

Q.E.D. A tension (or potential difference) is defined to be a vector 8 = O,,J E Z" such that, for each elementary cycle p ,

(el,O,, ...,

This equality can be restated by saying that the .scalar product ( p, 0 ) = p i di is zero, Let 0 denote the set of all tensions. Note that 0 is a submodule of Z", i.e.,

01,02E@ S€Z,0€0

=. *

O1

+02€@,

*SO€@.

Theorem 5. A cector 0 = (01,02, ..., 0,) is a tension and only if, there exists a function t ( x ) defned on the certex set X it.ith ralues in Z such that, f o r each arc i = (a, h), 8, = t(b) - t(a). The function t ( x ) is called a potential attached to tension 0. 1. If 8 is a vector defined by a function t ( x ) , then consider the cycle p = (il, i,,

..., ik)

92

GRAPHS

that successively encounters the vertices a,6, c, Pi,

..., z. Then, we may write

= t(b) - t(a)

Oil

. . . . . .=. . t(4 . . . .-. . 0 . . .) .

Pi,.Bi,

Pi,

ei, = t(a) - t ( z ) .

Adding the above equations yields

c ei- c ei=o.

isp+

isp-

2. It is easy to calculate successively the coefficients t ( x ) for a given tension 8, by the following rules: RULE1. Take any vertex xo, label xo and set 0. RULE2. If x is labelled and y is unlabelled, and if i t(x,) =

= (x, y )

is an arc, put

rb) = t ( x ) + oi .

If i

= (y, x) is

an arc, put

tb)= t

( ~) ei

.

In this way, all the vertices of a connected graph will be labelled. (If the graph is not connected, each connected component can be treated separately.) Each coefficient is uniquely defined by this process. Otherwise, there would exist two chains p1 and p2 from xo to x such that # ,

and consequently, (pl

- p 2 , e ) # 0.

Since p1 - p2 is a flow, it is a linear combination of elementary cycles by virtue of Theorem 3. Therefore, there exists an elementary cycle p such that ( L O >

+ 0,

which contradicts the definition of a tension.

Q.E.D. Consequence. This theorem clearly shows that a cocycle w(A) is a tension, since we may let t(x) =

Hence, for an arc i = (a, b),

0 1

if if

XEA x$A.

93

FLOW PROBLEMS

I

t(b) - t(a) =

+1 -1 0

if if if

I

~EO+(A) i E o - ( A ) = o,(A) i$o(A)

.

Theorem 6. Let G = (X,U ) be a connected graph. Let H = (A’, V ) be a spanning tree of G with arcs 1, 2, ..., I, let 02,02, ..., m1 be the cocycles associated with H. A tension 8 = ( G , , e2, ..., 6,) is uniquely defined from the values GI, G2, ..., O1 by:

+ e2 m2 + + 8, .

e = el

+.-

The vector 8’ = 0 -

el o1

- e2 w 2 - ... - e1 0

1

is a tension that has zero value on each arc of the tree H. If 8‘ corresponds to a potential t’(x), then

t’(xl) = t‘(x2) = ... = t’(x,,). Consequently, 8’

=

0.

Q.E.D. Corollary. A necessary and suficient condition that a vector 0 be a tension is that it is of the form

e = s1 0 1 + s2 m2 + ..I + sk m k ,

where sl,s2, ...,sk E Z,and d,02, ._., okare elementary cocycles. Clearly, each linear combination of elementary cocycles is a tension, and conversely, each tension is a linearly combination of elementary cocycles, from Theorem 6. Q.E.D. Theorem 6 shows that a tension

e = (el, 02, ..., em), can be determined from A(G)

=

n

- 1 unknowns.

Theorem 7. A necessary and suficient condition thai 8 be a tension 2 0 is that

e = s1 0 1 + s,

0 2

+ -..+

-k,

where sl, s2, ..., s k E Z, sl,s2, ...,s k 2 0 and ol, w2, ..., wk are elementary cocircuits. Clearly, any linear combination of cocircuits with all coefficients 2 0 is a tension 2 0.

94

GRAPHS

Conversely, consider a tension 8 # 0, 8 > 0. We shall show that there exists an elementary cocircuit o1and an s1 > 0 such that the vector

e - s1 o1 has more zero components than the vector 8. Let i = 1 be an arc with el=min(8i/8i#0, 1 ~ i ~ n z ) . Put 0, = s1 > 0. Colour black all arcs i such that 0, > 0. Colour red all arcs i such that 0, = 0. There cannot exist a red and black elementary cycle with all black arcs in the same direction that contains arc 1. Thus, by virtue of the Arc Colouring Lemma, arc 1 is contained in a black elementary cocycle o' with all arcs in the same direction. Thus, o1is a cocircuit, and the vector

6 - s,

o1

has more zero components than the vector 6. This vector is also a tension > 0. If 6 - s1 o1# 0, this reduction process can be repeated until a tension - .-. - s k U k = 0. 8 - s1 d - s2

is obtained.

Q.E.D. Theorem 8. A vector cp E Em is apow if, and only if, it is orthogonal to each vector of 0.A rector 0 E Em is a tension if, and only if, it is orthogonal to each vector of 0. (Hence, 0 and @ are two orthogonal submodules of Em.) 1. We shall show that if cp E @ and

@ € 0,then

i.e., m

21 v i e i = ( c P , =~ )0 .

i=

For each elementary cycle p, (p,Q> From Theorem 3, cp is of the form

Thus,

=o.

6 and cp are orthogonal,

95

FLOW PROBLEMS

2. Let cp be a vector such that (cp, 0 ) = 0 for all 0 E 0.Vector cp is a flow, because if we take 0 = o ( x ) for some vertex x , then

C

2

'pi-

isw+(x)

~ i = ( ~ ( x ) , ~ ) = o -

iEa-(x)

I,.

We shall successively reduce the quantity m

d(e) =

C1 di(eJ

*

i=

If d(8) = 0, the tension 8 is compatible, and the procedure stops. Otherwise, there exists an arc i with d,(ei) > 0; let 1 = @,a) and suppose that

-=

el kl . d,(e,) = k , - o l , To construct a tension 8’ such that d(8’) < d(B), successively label the vertices of G in the following way: RULE1. Label vertex a, the terminal endpoint of arc 1, with the index

+ 1.

RULE2. If x is labelled and y is unlabelled, label y with the index + i if (x,y ) = i and if Bi < ki . RULE3. If x is labelled ar,d y is unlabelled, label y with the icdex - i if

( y , x) = i and if 0, b li .

If vertex b, the initial endpoint of arc 1, cannot be labelled by this procedure, then the tension 8 can be improved. In fact, the set A of labelled vertices satisfies UEA, b#A.

+ 1, and 6’ compatible, because The tension 8’ = 8 - o ( A ) satisfies 0; = i E o + ( A )implies Bi > k, and 0; 2 k,. Similarly, i E o - ( A ) implies 0, < and 0; < l i . If this labelling procedure labels vertex b, then there exists a chain v = [a, a,,

a2,

ee.7

4

9

in which each vertex has been labelled from the preceding one. Thus iEvf i E vP =

< k,, ei 2 I,.

0,

40,b] + [b, a] is a cycle. Since Ol < k l , we have

98

GRAPHS

From the preceding lemma, it follows that no compatible tension exists. The algorithm yields the following result:

Compatible tension theorem (Ghouila-Houri [1960]). Giren a graph G and numbers k, and li where - co < k, < I, < + cc for i = 1,2, ..., m, a necessary and suficient condition that there exists a tension 0 = (el,&, ..., 0,) with k, < Oi < I, for all i, is that for each cycle p,

C+lii e p

C

ki>O*

lop-

The necessary part of the theorem follows from the preceding lemma. The condition is sufficient because, then, each tension 0 with d(0) > 0 can be improved using the preceding algorithm until we obtain tension 8’ with d(8’) = 0. Q.E.D.

Corollary 1 (Roy, [1962]). There exists a tension 8 uith 0, 2 kifor all i, K and only if, for each circuit p,

+ OD for all i. There exists a tension 8 suih that Oi < I, for all i, if, and only i f ,

The proof is achieved by taking 1,

=

Corollary 2. f o r each circuit p,

CI,20. iEl(

Corollary 3. A vector Y

=

( Y , Y , ... Y,) 3

9

9

is called a subtension if there exists a tension 8 such that yi < 9, for all i. A necessary and sufJicient condition that a rector y is a subtension is that (Cp,Y) 0. 1. If y satisfies the above inequality, then by taking for cp a circuit p, we obtain

99

FLOW PROBLEMS

From Corollary 1, y is therefore a subtension. 2. Conversely, let y be a subtension. If cp E @, cp 2 0, then m

m

Furthermore, equality holds if, and only if, ‘pi

>o

yi =

ei.

Clearly, if ( cp, y ) = 0, then y is a tension on the partial graph generated by the arcs i such that qpi> 0. Finally, if y is a tension on the partial graph generated by the arcs i with ‘pi > 0, then

C

YiVi=O,

Vi’O

and, consequently,

(cP,Y>=

C

W’O

yiqi+

C

yiqi=O*

v1=0

Q.E.D. Maximum tension algorithm We shall construct a tension that maximizes the value 8, of the tension in arc 1 = (b, a). Starting with any compatible tension 8, achieves the labelling procedure for the compatible tension Problem. If vertex.b cannot be labelled, the set A of labelled vertices satisfies a E A , h $ A , and the tension 8’ = 8 - w(A) satis1. Thus, the value of the tension in arc 1 can be improved. fies 0; = 0, If vertex b can be labelled, we shall show that the value 0, of the tension 8 in arc 1 is maximum.

+

For each compatible tension 8 and for each cycle p that uses arc I along its direction, we have

If the algorithm labels vertex b, then there exists a chain v[a,b] such that i e v+ => Oi = k i , i e v* 6, = l i .

100

GRAPHS

Hence, for the cycle p

=

+ [b, a ] , we have

v[a, b]

From inequality (l), we see that 0, has reached its maximum value. Theorem 9. For a graph G with arcs numbers k , and I,, the maximum calue of a compatible tension in arc I is

m a x e l = min e

(i$-

li

C

-

P

k).

iEP+

1 Ep+

iZ1

If 8 is a compatible tension with

it can be improved by using the above algorithm. The theorem follows.

Q.E.D.

EXERCISES 1. Let

.T = (TI, Tz, ..., Tn) be a partition of a finite set X , and let

Y

sz,...,Sm)

= (Sl ,

be a family of subsets of X . Show that if every union of k of the Ti contains at most k of the S, (for k = 1, 2, ..., n), then there exist indices i,, ia, ..., i, such that T t p n S p# 0

p = 1 , 2,...,m .

for

Hint: This result is easily shown by constructing the appropriate transportation network. 2. Let

Y = (TI, Tz,..., Tn) be a partition of a finite set X , and let Y

= (Sl, SZ,

...,Sm)

be a family of subsets, and let c,, ca, ..., cn be positive integers. Show how to construct a set of representatives A = { a 1 , a * ,..., a m } . such that al

E

S,,a2 E Sz, ..., etc., and such that IAnTjISq

Dour

j=1,2,

..., n

101

FLOW PROBLEMS

By reducing this problem to a flow problem, show that such a set of representatives exists if

for all I c { 1, 2, ..., m } and all J

C

{ 1, 2,

..., n}.

3. Let

9-= (Tl, T2, ..., T n ) be a partition of a finite set X , and let

Y = (Sl,S2, ...,Sm) be a family of subsets, and let b l , b z , ..., b, be positive integers. Show how to construct a set of representatives A

with al

E

=

{ a l , a z ,..., a m } ,

S1,u2 E Sz,..., and such that IAnTjIgbj

for

j=l,2

,..., n .

Show that such a set of representatives exists if

for all I c { 1, 2,

..., m } and for all J

c { 1, 2,

..., n }.

4. Use the Compatible Flow Theorem to prove the following!

Let R be a transportation network with source a and sink b. Associated with each source arc i E w + ( u ) are two numbers b, and c, such that 0 < b, < c,. Associated with each sink arc i E w - ( b ) are two numbers b; and c; such that 0 < b; < c;. A necessary and sufficient condition that there exists a flow 'p satisfying b~ < qt < C I bJ < pj < c)

for for

iE

w+(a),

j E w-(b) ,

is that both of the following conditions hold: (1) There exists a flow 'pl with (p:

3 br

pf

< cJ

for for

iEw+(a)

for for

i e w+(a) j e w-(b) .

j E

w-(b).

(2) There exists a flow 'p2with 2

< CI

2

2 b;

qi

(pj

CHAPTER 6

Degrees and Demi-Degrees

1. Existence of a p-graph with given demi-degrees

For a graph G = ( X , U ) , the outer demi-degree d $ ( x ) of a vertex x is defined to be the number of arcs having x as their initial endpoint, i.e.

Similarly, the inner demi-degree d ; ( x ) o f a vertex x to be the number of arcs having x as their terminal endpoint, i.e. &(x) =

c mE(x, Y ) .

Y E X

Finally, we define the degree dG(x) of a vertex x is defined to be the integer d,(x) = d G + ( X )

+ dG(X) .

Thus, a loop at vertex x increases the degree of vertex x by 2.

A p-graph is a graph with m,+(x,y ) 6 p for all vertices x, y. Given integers r l r r 2 , ..., r , , sl,s2,..., s,, we may ask if there exists a p-graph G with vertices xl, x2, ..., xn such that dC+(xk)= rk

(k = 1 , 2 , ..., n)

di(Xk) =

(k = 1, 2 ,

sk

n)

In this case, the pairs (rk, sk) are said to constitute the demi-degrees of a p graph. Theorem 1. Let (rl, sl), (r2,s2), ..., (r,,, s,) be pairs of integers with si 2

The pairs

( r k , sk)

n

s2

2 *.. 2 S " .

constitute the denii-degrees of a p-graph if, and only (fi k

102

103

DEGREES AND DEMI-DEGREES n

Construct a transportation network R with vertices x l , x 2 , ..., x,, ,TI,,T2,

..., X,, and with a source a and a sink b. Join vertices x, and f , by an arc with capacity c(x,, Xj) = p . Join vertices a and x, by an arc with capacity c(a, x i ) = r,. Join vertices 2, and b by an arc with capacity c(F,, b e s,. Any flow that saturates the source and sink arcs of R defines a p-graph (X,U ) having demi-degrees ( r k , s,). Conversely, each p-graph (X,U ) with demi-degrees ( r k ,sk) defines a flow in R that saturates the source and sink arcs. From Theorem (2, Ch. 5), a necessary and sufficient condition for the existence of such a flow is that condition (2) hold and that, for each set B = { Fil, Xiz, ..., Xi* ), the total flow that can enter B is greater than the total demand at B, i.e., -

-

-

F(xi,, xi2, ..., xi*)

-

-

> d ( x i , , x i r , ..., Xi,)

(1

< il

< i2
si, + si2 + ... + sir

i=

(1 < il < i2

r2 > - * * ’ > r, be a n-tuple of integers. Corresponding to this n-tuple, associate the sequence (r:, r t , ...) where r z denotes the number of.r, that are greater than or equal to the integer k. Hence, r: > 1; 2 r: > * - To visualize the numbers r z , we can construct a diagram, called the Ferrers diagram, as in Fig. 6.1. Hatch the first r l squares in the i-th column in the positive quadrant. Then it is easily seen that rT equals the number of hatched squares in thej-th row in the positive quadrant. By counting, in two different ways, the number of hatched squares, we obtain: n

Cri= Cr:. i= 1

k>l

The sequences (Ti) and (r;) are called conjugates. Using these definitions, several corollaries follow from Theorem 1.

104

GRAPHS

j’k - - -

Fig. 6.1. Ferrers diagram for the sequence (r,) = (5, 4, 2, 2, 1, 1). Here, Cr:) = (6,4,2,2, 1)

Corollary 1. Given n pairs

(Ti,

s1

sJ of integers such that

2

... 2 s,,

s2

there exists a p-graph with d,C(x,) = r i , d;(xt) = s1for i only if,

c

=

1,2,

..., n if,

and

k

r: 2

i>1

c sI

(k = 1 , 2 , ...,n

- 1)

j=1

j=1

By counting in two different ways the number of hatched squares in the first pk rows of the associated Ferrers diagram for (rl , r 2 , ..., r,) we find that It

C min { ri, p k } = i=1

pk i=1

r:

.

Corollary 2 (Ryser [1957]; Gale [1957]). Consider npairs ( T i . such that s1

2

s2

There exists a I-graph G with d$ (xJ k

2 . * - 2 s,. =

ri and &(xJ

= st

if, and only if,

k

r: 2

j= 1

sI

Q.E. D. st) of integers

(k = 1 , 2 , ..., n

- 1)

105

DEGREES AND DEMI-DEGREES

This follows immediately from Corollary 1.

Corollary 3. Giuen a sequence (rl , r 2 , ..., r,,), there exists a 1-graph G with dG+(x,)= r, and d;(xi) = 1for all i if, and only if, t r i = n . i= 1

This condition is necessary because if such a graph Gexists, then n

n

c ri c =

i=l

=

&(Xi)

n

i= 1

Conversely, assume that the condition holds; by Theorem 1, we have only to show that, for all k < n, n

2 min { k, ri } 2 k .

i= 1

If there exists an ri 2 k , the above inequality is satisfied. Otherwise, each rl is less than k, and consequently, n

n

i= 1

i= 1

C min { k , ri } =

C

n 2 k

ri

.

Thus there exists a graph G with the desired properties.

Q.E.D. We shall now present necessary and sufficient conditions for the pairs ( r l , rl), ( r 2 , r2), ..., (rn,r,,) to constitute the demi-degrees of a symmetric pgraph, i.e. a p-graph G with d ( x , Y> = m a y , x>

(x, Y E X )

*

Theorem 2. Let Go be a symmetric graph such that each odd cycle contains a vertex with one or more loops, and let r l , r2, ..., r,, be positive integers. If Go has apartialgraph H with d i ( x i ) = d;;(xi) = rifor each vertex x i , then Go has a symmetric partial graph G with d i ( x i ) = d G ( ~ i= ) ri for all i. We shall assume that the theorem is true for any graph Go of order < n, and we shall show that the theorem is true for a graph Go of order n. Given a graph H, we shall construct from H the symmetric graph G. 1. Suppose first that there exists a vertex xo such that

m&o, Y> = m a y , xo>

(Y EX).

106

GRAPHS

Consider the subgraph Go of order tz - 1 obtained from Go by removing with given demi-degrees Fi = vertex xo. Graph Go has a partial graph ri - mH+(x,,xi), which can be transformed to a symmetric graph G with demi-degrees Ji (by the induction hypothesis). This yields graph G. 2. We may now suppose that for each vertex x of Go,there exists a vertex y adjacent to x with

-

m,+(x,Y> > d ( Y , x> Since d$(x) = d;(x), there exists a vertex z adjacent to x with

m&, z ) < m,+(z,x). Clearly x , y and z are distinct vertices. 3. Choose any vertex xl, and let x2 be any vertex such that

m 2 x l , x2> >

.

m&2

XI)

Let x3 be any vertex such that

, x3) > 4 x 3 xz> In this way, a ,sequence xl,x2, x3, ... is defined and, since the graph is m&2

9

finite, an elementary cycle =

[xp

9

x p + k - l ~x p + k

xp+l,

=

xpl

will be formed. 4. If cycle ,u is even, transform H to H ' by removing an arc between each of the pairs: (xp, x p + l ) ,

(xp+2, xp+3),

(xp+k-2,

as.9

Xp+k-l)

and adding an arc between each of the pairs: (xp+2,

xp+A ( x p + 4 ,

xp+3),

.-.,(Xp+k,

Xp+k-l).

This does not alter the demi-degrees, and produces a graph H ' with

5. If cycle ,u is odd, and if H has a loop at one of its vertices, say x, then transform H into H ' by removing an arc between each pair ( x p , x p ) , ( x p + l , Xp+2)9 ( x p + 3 , x p + 4 ) 9

...) ( x p + k - 2 ,

xp+k-l),

and adding an arc between each pair (XP+l,

x p ) , ( x p + 3 ,x p + 2 ) ,

*",

( X P + k , Xp+k--l)*

This does not alter the demi-degrees, and the graph H ' satisfies (1).

107

DEGREES A N D DEMI-DEGREES

6. Finally, if cycle p is odd and if H has no loops attached to p , then there exists a vertex of p, say x,, that is incident with a loop in Go, Transform H into H ’ by adding an arc between each pair ( x p , xp)? ( x p + 2 3 x p + l > ? ( x p + 4 ,

xp+3),

..., ( X p + k - l ,

Xp+k-2),

and removing an arc between each pair (xp? x p + l > ,

(xp+Z,

xp+3),

.--)( X p + k - l ,

1 Xp+k)*

Again, this does not alter the demi-degrees, and graph H’ satisfies (1). After repeating this procedure a finite number of times, a symmetric graph G is obtained. Q.E.D. Corollary. Given integers rl 2 r2 2 .*.2 r,, a necessary and suflcient condition that there exists a symmetric p-graph with &(xi) = & ( x i ) = ri for all i, is that

The result follows by applying Theorem 2 to the p-graph G,, with vertices x l , x 2 , ..., x, a n d p arcs going from xi to x, for all i and allJ, and then, invoking Corollary 1 to Theorem I . The following consequence of Theorem 1 is used to characterize tournaments.

Theorem 3. (Landau [1953]; Moon [1963]). There exists a coniplefe antisymmetric 1 -graph with outer demi-degrees rl

< r2 < -..< r,,

il; and only il;

I where

I:(

ri = i= 1

(3

denotes the binomial coefficient

P! 4!(P-4)!’

The condition is necessary because, in a complete anti-symmetric I-graph the number of edges joining the vertices of the set { x1 ,x 2 , ..., xk } is less than or equal to the number of arcs leaving the vertices of the set.

108

GRAPHS

The condition is suficient. To prove sufficiency, let si

= (n

- 1) - ri

(i = 1 , 2,

..., n) .

Thus, s1

p

2

s2

2

s3

2

*..

2

s,.

1. First we shall show that the conditions 6f Theorem 1 are satisfied for 1. Note that

= n

n

C sI = n(n - 1) - C ri = n(n - 1) - n(n

- 1) - n(n - 1) =

2

2 Furthermore, by letting rk denote the number of r, that are < k, i=1

i= 1

C ri.

i= 1

Note that, for any integers k and t ,

2[(:)+(!j’-k(r-l)]=r(t-

l)+k(k-l)-2kt+2k =

(k

- t ) 2 + ( k - r) z o

,

Thus,

k

2 k(n - 1) -

C ri = i=1

c k

si

y

i=l

and the conditions of Theorem 1 are satisfied.

2. From Part 1, there exists a 1-graph G & ( x i ) = ri , & ( x i ) = si

=

( X , U ) such that

(i = 1,2,

..., n) .

If this 1-graph G has neither loops nor two oppositely directed arcs joining the same pair of vertices (call such arcs multiple edges), then G is a complete anti-symmetric 1-graph because the number of arcs is case, the theorem has been verified.

xr=lri

=


109

Otherwise, we shall alter G and decrease the number of loops and multiple edges without changing its demi-degrees. Suppose that at vertex xi there are p(xi) loops and q(xi) multiple edges. The number of vertices that are not adjacent to X I is

1

1

(n - 1) - T G ( X J =(n - 1) - p i i-si -q(xJ-2 p(x*)]=q(xJ +2 P(Xi) . Colour red each multiple edge and loop, and insert a green edge between each pair of non-adjacent vertices; then the number of red edges incident to x equals the number of green edges incident to x. Suppose we travel through the coloured edges of the graph without using two edges of the same colour in succession and without using the same edge twice. Then, after arrival a t a vertex x , we can always leave x except perhaps if x is the initial vertex of the tour. In other words, if there exists red edges, then there exists a cycle

bl, Y 2 ,

Y11 > with alternately red and green edges such that **-)

Y2k3

u, 4. u d v 3 , Y 4 ) , dv4, Y 3 ) E u , Oll,YZ),dYZ,Y1)E

02,

Y3), (Y39

Y2)

s....................

@Zk,

Y l ) , @I

9

Y2k)

$

u-

Thus G may be altered by removing arcs ( y 2 , y l ) ,(y4, y3), etc... and by adding arcs ( y 2 ,y3),( y 4 ,y5), ..., ( y Z ky, l ) , without changing any demidegree. This process decreases the number of red edges. It can be repeated until no more red edges are present. Q.E.D.

2. Existence of a p-graph without loops and with given demi-degrees Consider the following problem: Given a graph Go = ( X , U),construct a partial graph H with gitlen demidegrees d; (x) and d; (x). For A c X and B c X,let m& ( A , B ) denote the number of arcs in Go whose initial endpoint is in A and whose terminal endpoint is in B. If an integer ri is associated with each xi E X then, for each A c X , let

Theorem 4. For a graph Gowith certices x1 , x2 , ..., x,, and integers ri, si, for i = 1, 2, ..., n, a necessary and suflcient condition that Go hare a partial subgraph H with

110

GRAPHS

i= 1

i= 1

_ -

Consider a bipartite transportation network R = ( X , X , U ) with vertex s e t s X = { x , , x , ,..., x , , ) a n d X = ( x , , f , ,..., x-,,},withasourceaanda sink 6, and with arcs ( 4 9

Xi)

(a, xi)

(F,,b)

(for 1 (for 1

with capacity mc',(xi,xi) with capacity r, with capacity s, .

< i < n) < j < n)

If condition (2) is satisfied, the desired partial graph H exists if, and only if, network R has a maximum flow that saturates the sink arcs; from Theorem (2, Ch. 5), this is equivalent to

(Ac X),

F(A) a d(A)

(1')

where F ( z ) is the maximum amount of flow that can enter

Z,i.e.

n

W) = i C min { T i , d 0 ( x i , 1)}, = 1 and d ( Z ) is the total demand of set 2,i.e.

x-

d(A) = XJ

E

SJ

=

s(2) .

A

Clearly, condition (1') is equivalent to condition (1).

Q.E.D.

Corollary 1. Given pairs of integers ( r l , s,), ( r 2 , s2), ..., (r,,, sn), a necessary and sufficient condition f o r ;he pairs to constitute the demi-degrees of a p-graph H without loops, is that n

(1)

C1 min { ri, P I A -

i= n

{ x i }

I)

( A c X)

S(A)

n

We apply Theorem 4 to a complete symmetric p-graph Go without loops. Clearly,

I

m&(xt, A ) = P A - { x i

The result follows.

1I

*

Q.E.D.


111

These conditions simplify considerably when rl s1

> r2 2 2 r, > s2 z . * z s,. ***

Consider a sequence ( r l , r 2 , ..., r,) of positive integers such that rl 2 r2

Z

.-*

> r,.

Let Fk denote the number of indices i such that i < I$ and ri 2 k - 1 plus the number of indices i such that i > k and ri z k . The sequence (rk) is called the corrected conjugate of sequence ( r J . The numbers r, can be visualized on the corrected Ferrers diagram (Fig. 6.2), formed by dividing the positive quadrant into three parts: hatched, dotted or empty. All squares on the principal diagonal are dotted. In the i-th column, the first r, squares not on the diagonal are hatched. All other squares are empty.

t

hFig. 6.2. Corrected Ferrers diagram for the sequence (7, 5, 5 , 5 , 5 , 3, 2, 1). The corrected conjugate sequence is (7, 6 , 5, 4, 4, 5 , 1, 1)

Clearly, the number of hatched squares in the k-th row of the diagram equals Fk, and therefore r i = C rk.

1

iZ1

k3l

112

GRAPHS

Corollary 2. Let (r,, si) be pairs of integers with

> > s2 >

2 rn ,

rl 2 r2 s1

***

> s,.

Let (Fi) be the corrected conjugate of ( T i ) . There exists a 1-graph H without loops with d i ( x i ) = ri and d;; ( x i ) = si for all i, and only if, (1) n

C-

Ti=

i31

C si. i= 1

1. If such a I-graph exists, then n

C1 min { ri, 1 A

i=

- {xi } 1 2 s ( 4

(A

= XI.

By taking A = { x l , x z , ..., XI, >,condition (1) follows. Condition (2) obviously holds. 2: Conversely, suppose that conditions (1) and (2) are satisfied. For any set A of cardinality k, n

c min { ri, I A -

i= 1

min{r,,k- 1)

= xl€A

+ C

{Xi

1

> = k

k

min{ri,k)

xiex-A

From Corollary 1, there exists a graph H without loops with dH+(xi)= ri and d;(x,) = si for all i. Q.E.D. We shall now consider necessary and sufficient conditions for the pairs ( r l , rl), ( r 2 , r2), ..., (r,, r,) to constitute the demi-degrees of a symmetric p graph without loops, i.e. a p-graph with: mcf(x, Y> = m a y , XI

(x, Y E X ) .

First, we shall prove the following very general result : Theorem 5 (Fulkerson, Hoffman, McAndrew [1965]). Let Go = ( X , V )be a symmetric graph of order n without loops, such that any two rertex-disjoint elementary cycles of odd length are joined by an edge. Let r l , r z , ..., r, be integers whose sum is eren. If Gohas a partial graph H with d i (xi) = d; ( x i ) = ri for all i, then Go has a symmetric partial graph G with

113


(3) If the numbersf(x, y ) are all even, then the graph G defined by 1 mc+(x,Y > = +,

Y)

is the required graph. Otherwise,-fl(x, y ) determines a non-empty set El c Y 2 ( X )defined by

EI = { [x, YI / 1x3 YI E 9 2 ( f l , f i ( x , Y )

1

(mod. 2)

}*

Let H , = ( X , El) denote the simple graph having El as its edge set. Note that in H , all degrees are even, because

dHI(xk= )

C

fl(x,, y )

= 2 rk

=0

(mod. 2)

.

Y E rH,(xk)

2. We shall now successively construct a sequencef,,f,, ...,f, of functions of two variables that satisfy (l), (2), and (3) and define (as above) sets of edges

E2 2 2 Eq and simple graphs H I , H,, ..., H p . When we encounter a set E, El

2

=

(i.e., a functionf,(x, y ) with even values that satisfy conditions (l), (2), and (3)), we shall obtain the required graph G, defined by

1

Y).

mc+(x, Y ) = #’

3. If the simple graph H1 has an edge, then it has a cycle (because all its degrees are even). Let p denote such a cycle. If cycle p is even, define a function f,(x, y ) , that differs from fi(x,y ) only on the edges of p, by adding alternately 1 and - 1 to f l ( x , y ) while traversing cycle p. Since cycle p is even, f i ( x , y ) also satisfies conditions (l), (2) and (3), and determines a new set of edges E, = El - p. The simple graph H z = ( X , E,)

+

114

GRAPHS

defined in this way has only even degrees. If H1 contains an even cycle p', we can similarly define H3 = (X,E3), etc. Repeat this process until a simple graph H , = (X, E,) without any even cycles has been found. 4. If the graph H, has an edge, it contains a cycle p (because all its degrees are even). This cycle p is necessarily elementary (because otherwise the edges of p would contain an even cycle which contradicts the definition of H,) and odd. Let p =

[a17 a 2 r

a2k+l3 a2k+2

=

'11

be this elementary odd cycle of H,.Then 2k+l

c f,(ai, a i + l )= 1

(mod. 2)

i= 1

and

n

(mod. 2)

= x r i = O

.

i= 1

This shows thatf, has at least one odd value on an edge not in p, and that H , contains an edge not in p. Thus H , has another odd elementary cycle v # p , and v has no common vertices with p (since, otherwise, they would form an even cycle). From the assumptions on Go, these two cycles are joined by an arc of Go. For example, let

I

p = a2, v = Cbl, 6 2 , (a1 Y b , ) E u

a2k+ 1,

.*.,b,+,b l l t

1

Edge [al b,], which is not in H,, satisfies f,(a,, b,)

=0

(mod. 2)

.

,

Iff&, b,) = 0, the function f,+ is obtained by lettingfp+,(a,, b,) = 2, and by adding alternatively - 1 and 1 to the edges [u,, a,], ..., [azk+ a,], and also to the edges [b, b,], ..., [b,t+ b,]. Iffp(al, 6,) # 0, let f,+l(al, b,) = fp(ul,6,) - 2 and add alternately 1 and - 1 to the edges [a,, a,], ..., [ a z k + ,all , and to the edges [b,, b,], ...,

+

+

[b21+17

bll*

Function f,+, also satisfies conditions (l), (2) and (3), and it yields


115

the simple graph H,+, = (X, E P + , ) ,where E,,, = E, - ( p u v). Since E is finite, we obtain in a finite number of steps a set Eq = 0. Q.E.D. Corollary. Given the integers p 2 I, r1 2 rz >...a r, 2 1 such that condition for the existence of a symmetric p-graph G without loops whose vertices satisfy dC+(x,)= d;(xi) = r.I 2 is that

2ri is ecen, a necessary and suficient n

The proof follows by applying Theorem 5 to a graph Go with vertices x l , x 2 , .. ., x, and with p arcs from xi to x j for each pair (xi,x j ) with xi # x j . 3. Existence of a simple graph with given degrees

This section describes necessary and sufficient conditions for a sequence of integers dl 2 d2 2 ... 2 d, to constitute the degrees of a simple graph. Theorem 6 (Erdos, Gallai [1960]). Let dl 2 d2 2 * * * >d, be a decreasing sequence of n integers with 2 di ecen, and let (di) denote its corrected conjugate sequence. The following conditions are equit'alent: (1) There exists a simple graph G n!hose tiertices x, satisfy dG(xi)= di; k

k

( k = 1 , 2 , ..., n ) ; (3)

C di G k(k - 1) + C i=1

min { k , d, }

( k = 1 , 2 ,..., n ) .

j=k+l

(1) A (2) Condition (1) implies the existence of a I-graph G* without loops such that d,+,(xi)= d,;(xi) = di, for all i; this implies (2), by Corollary 2, Theorem 4. (2) 5 (1) Condition (2) implies the existence of a I-graph H without loops such that d,f(xi) = dG(xi) = d i , from Corollary 2, Theorem 4. Since 2 diis even, this implies from Theorem 5 the existence of a symmetric 1-graph G* without loops such that d&(xi) = &*(xi) = di . (2)

3

(3) Consider the corrected Ferrers diagram for the sequence (di),and denote by xrC the number of empty squares in [0, k ] x [0, k ] ; condition (3) is equivalent to k

(3')

k

116

GRAPHS

Since ctk > 0, condition (2) implies condition (3’), which implies condition (3).

(3)

2

(2) Suppose condition (3’) is satisfied and there exists adinteger k with

We shall show that this results in a contradiction. Clearly k > 1. Since only one square in [0, 11 x [0, 11 is dotted, we have a1 = 0 and, from condition (3’), dl < dl a1 = dl. Let q be the largest integer such that d, 2 k - 1. Then q < k (since ctk > 0). Hence

+

1 di > ic= di= q(k - 1 ) + k

k

1

i= 1

c k

di

+

i=qt 1

c n

di.

i=l+l

Thus n

a

2d i > q ( k -

di.

1)+

i= 1

i=k+l

On the other hand, from condition (3), we have 4

n

i= 1

i=q+ 1

1 di < q(q - 1) + C

min { di, 4 ) G n

L

G q(q

- 1)

+ 1

i=q+ 1

min { di, q }

< q(q - 1) + ( k - 4 ) 4 +

+ 1 i=k+

1

di
s2 2 with p G q. There exists a simple bipartite graph G = (A’, Y, E ) on the sets > . a * >

... 2 s,,

117


X = { x l ,..., x p }

and

Y = { y l ,..., y , } ,

such that dG(xi)= ri dG(Yj) = sj

(i = 1 , 2,

,..,p ) ,

( j = 192, ...) 4 ) ,

and only i f i

(1 1 4

r: =

C

sj

j= 1

i2t

Clearly, a necessary and sufficient condition that there exists such a graph G is that there exists a I-graph whose demi-degrees are given by the pairs ( 0 , s,), (0. sz), ..., ( 0 , s,) ( r l ,s,+l>,( r z , s q + J 3..., ( r p ,sq+&

where Sqfl

= sq+2 =

..* = s q c p= 0 .

From Corollary 2, Theorem 1, such a l-graph exists if, and only if, we have both L

I,

1 ~ : aC i=

j= 1

1

sj

( k = 1 , 2 ,..., q + p ) , a n d

(2')

These conditions are equivalent to conditions (1) and (2) above.

Q.E.D. Theorem 8. The numbers dl , d 2 , ..., d,, constitute the degrees of a tree i f i and only i f , (1) di 2 1 ( i = 1 , 2 )..., n),

c di n

(2)

=

2(n - 1).

i= 1

From Theorem (19, Ch. 3), these conditions are equivalent to T ( n ; d , , d 2 , ..., d,) # 0 .

Q.E.D. Theorem 9. Let dl

> d2 >...> d,,

be a sequence of integers, n 2 2. A

118

GRAPHS

necessary and sufficient condition for the existence of a simple connected graph G with degrees d,(x,) = d,, is that

>1

d,

(1)

n

2 di is etren

(3)

i= 1 k

(4)

k

CdiG i= 1

Cdi i= 1

( k = 1 , 2 ,..., n ) .

Suppose these conditions are satisfied; then from conditions (3) and (4) there exists a simple graph with the given degrees. From (2), it has at least I I - 1 edges, and therefore, if this graph is not connected, it has a cycle (Theorem I , Ch. 2). Let [x,y ] be an edge of this cycle, and let [a, b] be an edge of a different connected component. This exists, from condition (1). If edges [x,y ] and [a, b] are replaced by two new edges [x, a] and [ y , b],the number o f connected components is reduced without changing any degrees. By repeating this operation as many times as needed, we obtain a connected graph. Concersely, suppose there exists a simple connected graph G with degrees dG(Xi)= d i . Then conditions (l), (3) and (4) are satisfied. Furthermire, if m denotes the number of edges in G, n

i= 1

n

d, =

dG(xi)= 2 m

> 2(n - 1)

i= 1

(since the cyclomatic number satisfies v(G) dition (2) is also satisfied.

=

m -n

+ 1 > 0). Thus conQ.E.D.

Recall that a graph of order n > 2 is defined to be 2-connected if it is connected and if it has no articulation vertex (a vertex whose removal disconnects the graph).

Theorem 10. Let dl 2 d2 d,, be a sequence of integers, n > 2. A necessary and sujicient condition for the existence of a simple 2-connected graph G with degrees dG(xi)= di is that > a * * >

(1)

d, 2 2

119


It

(3)

C di is ecen i= I

I

k

1. Assume that conditions (I), ( 2 ) , (3) and (4) are satisfied. Then there exists from Theorem 9 a simple connected graph G with d,(x,) = di for all i. If xk is an articulation point, the subgraph G‘ generated by X - { x k } has p’ 2 2 connected components. At least one of these connected components has a cycle, since

y(G’)= m’ -

11’

+ p’ = ni - d,

-

(n - 1)

+ p’ 2

Let [ y , z ] be an edge of this cycle, and let [ t , u] be an edge of another connected component of G’ (which exists from condition (1)). If the edges [ y ,z ] and [t, u] are removed, and two new edges [ y , f ] and [z, 211 are added, the degrees are not altered, but the number of connected components of G‘ is reduced. Furthermore, for each vertex x, this operation does not increase the number of connected components of the subgraph Gx+) (supposing that [ t , u] has been selected in a cycle if its component in G ‘ is not a tree). Repeating this operation as many times as needed, we obtain a graph H such that HX-(,, has only one connected component (for each vertex x). Graph His the required graph. 2. Concersely, the degrees of a 2-connected graph G satisfy the required conhas m - dl edges, n - 1 vertices, and ditions, because Gx-,,,,

Thus

Hence, condition (2) holds. Conditions (l), (3) and (4) clearly hold.

Q.E.D.

120

GRAPHS

EXERCISES 1. For a multigraph G without multiple edges (but possibly with loops) let ~ G ( x = ) d&) if vertex x has no loops. Let 6,(x) = dG(x)- 1 if vertex x has a loop. In other words, a loop increases by 1 (not 2) the “corrected degree” 6&). Show from Theorem 2 that for a sequence dl > dz 3 *.. > d., there exists a multigraph G (without multiple edges) with n vertices xl,x 2 , ...,xn,such that S,(X;) = d; for i = 1, 2, ..., n, if and only if k

k

d: 3 1=1

1dj

(k = I , 2,

...,n)

i=l

(Ramachandra Rao [1969]) 2. Show that the pairs (r,, sl)are the demi-degrees of an arborescence with root x1 if, and

only if, rl = 0

rr = 1

(i # I )

fsc=n--l. I= 1

3. Show that the pairs ( r , , sl)are the demi-degrees of a strongly connected functional graph if, and only if, (i=l,2,

rt=si=l

...,n ) .

4. Show that a sequence d l , d z , ..., d, constitutes the degrees of

if,

2 dl is even.

a multigraph if, and only (Hakimi [1962])

5. A simple graph G is said to be k-edge-connectedif it is not disconnected by the removal of less than k edges. Show that if n > 1 and k > 1, then a sequence d l , d z , ..., d. con-

stitutes the degrees of a k-edge-connected graph if, and only if, (1) the dl are the degrees of a simple graph, (J. Edmonds [1964]) (2) dl > k for all i. 6. Let dl 3 dz p - . B d, be a sequence of integers with 2 dl even. Show that there exists a multigraph G with multiplicityp without loops with vertices xl of degree dG(xI)= dl if, and only if,

14n

Hint:This can be shown from the Corollary to Theorem 5 . 7 . Let dl 2 dz 2 > d,, be a sequence of integers. From its corrected conjugate sequence (a,, dz, a,, ...). Show that di 3 d2 > d, 3-, or that there exists an integer k such that -

-

-

> dr >, ... B db , = dii + 1 dtbz > dktJ > .*. di

.

dnbl 8. Let dl 3 dz B - . be a decreasing sequence, and let (d,) be its corrected conjugate sequence. Consider an integer k s n, and let lo be the largest integer 1 such that the number of hatched or dotted squares in the I-th column of the corrected Ferrers diagram is 3 k.

121


For 1, k k, show that n

di i=l

=

min

=

131. ( i i l

1

d,+hlo-k.

i=lo+ 1

Let ok denote the number of empty squares in the square [O,k] x [0, k] of the corrected Ferrers diagram. For lo < k, show that

i= 1

9. Show that if G is a 3-connected graph with degrees d1

(1)

dk

c

(3)

< d, 0) is

Note that in the case n

=

2 q , the graph K Z q ,a clique with 2 q vertices,

is clearly a graph of order n with m

=

(‘2”)

edges and with a maximum

matching of cardinality q. In the second case, the graph formed by the union of a ( 2 q + 1)-clique

126

K,

GRAPHS

+

and a set S, - (,

,

+ 1)

of n - (2 q

+ 1) isolated vertices is clearly a graph

of order n with a maximum matching of cardinality q and with m = (2q2+ l) edges. Finally, in the third case, take a q-clique K, and a stable set Sn-,and join in all possible ways the vertices of Kq with the vertices of S,,-,. Clearly, this graph of order n has a maximum matching of cardinality q since n - q > q and has m =

(3+

q(n - q) edges.

We shall now show that the given numbers represent the maximum possible number of edges. For the first case (n = 2 q), this is evident. Suppose that n22q+l. Let S(Eo)denote the set of unsaturated vertices in- a maximum matching Eo where I Eo I = q. Since n > 2 q , we have S(Eo) # @. Let El denote the edges in matching Eo that have one endpoint adjacent to setleral vertices of S(Eo).By Theorem 1, the other endpoint of such an edge cannot be adjacent to S(Eo),because then there would exist an alternating chain between two distinct unsaturated vertices, and Eo would not be a maximum matching. Let E, = Eo - El,and let q1 = I El I and q, = I E, I. Thus, q1 + q2 = q. For i = 1, 2, let X,denote the set of endpoints of the edges of Ei. Thus,

X, n X, = 0,X, n S ( E , )

=

0,X ,

n S(E,) = @ .

1. Two edges of El cannot generate a 4-clique because then there would be an alternating chain joining two vertices of S(Eo).Thus, the number of edges of Gjoining two vertices of X, satisfies

2. The number of edges of G joining Xl and S(Eo)satisfies

GE,))6 ql(n - 2 q ) .

mG(X1,

3. Let [x,, y,] be an edge of E 2 . If neither x2 nor y , is adjacent to the set X i of vertices of Xl that are non-adjacent to S(Eo),then mG({ x2,

y2 },

x - x2) 6 2 41 + 2 < 3 41 + 2 .

If the edge [x,,y 2 ] has an endpoint x, adjacent to Xi,(see Fig. 7.2), its other endpoint y 2 is not adjacent to S(Eo).Similarly, endpoint y , cannot be adjacent to two vertices of Xi.Thus,

MATCHINGS

n z , ( x z , X - X2) - X2)

m ~ ( ~X 2 ,

< 29, + I , < 9, + 1 ,

and, hence,

'"Gt{ x2, Y2

1, X - X2) < 3 q1 + 2 .

Finally, we obtain

E2 Fig. 7.2

Consequently, m = m,(x,,

x - X2) + ~ G ( X XZ Z, ) + m G ( x 1,S(E,))+ w , ( x , ,

3 q z q I +2q2

If nG7 Hence m

5q+3 7

c

I f n >-5 q + 2 m

(:)

+i212)+

then n - 3 9

ql(n - 2 q ) + ( 2 : l ) - ( y )

+-

2

=

=

+qFGO.

(2 q + 1) . 3 3

+ UOZ

then - 9)

+ (4 - ql)

v j

- n)

x,)

-

1 2ql(q - ql),

128

GRAPHS

Combining the above inequalities for all cases yields,

Note that

is equivalent to

or

Q.E.D. Maximum matching algorithm Consider a simple graph G = ( X , E ) with a matching E,, and associate with it a 1-graph = (A', U ) , where (x, y ) E U if there exists a vertex such that [x, 21 E E - Eo and [z, y ] E E,. For each even alternating chain p in G,

H

Fig. 7.3

129

MATCHINGS

that starts from an unsaturated vertex, there corresponds i n mentary path ,ii For . example in Fig. 7.3, jl

= [ a , b , c , e , d l corresponds to

c,

= [ a , c,

c a unique ele-

4.

For each elementary path ji in there corresponds a unique even chain fl of G, but this chain is not necessarily simple. For example, ,= i[ a , c , d , b] corresponds to p = [ a , b , c , e , d , c , b] (edge bc appears twice). A path in will be called legal if it corresponds to a single chain in G. Otherwise, it will be called illegal. Hence, the matching problem reduces to finding a legal path in that connects an unsaturated vertex (e.g. vertex a) to the neighbours of another unsaturated point (e.g. vertices b or g). One could use known algorithms for finding all elementary paths in (see Ch. 4,§ 1) and then each path of G that is illegal could be eliminated. In Fig. 7.3, the chain ji = [a, c, e, g] yields the desired alternating chain p = [a,6, c, d, e , f , g , h] in G . As noted by Jack Edmonds ([1962], [1965]), it is not necessary to explore each elementary path of starting at vertex a in order to reach an unsaturated vertex by a legal path. Modifications of Edmonds’ algorithm have been suggested (C. Witzgall and C . T. Zahn [1965]; M. Balinski [1970]; B. Roy [1969]).

c

c

c

c

2. The minimum covering problem

Given a simple graph G = (X, E ) , a coaering is defined to be a family F c E such that each vertex x E Xis the endpoint of at least one edge of F. The problem of finding a minimum cardinality covering has many similarities to the maximum matching problem. The covering problem is a more general case of a problem known in logic as “Quine’s Problem”. EXAMPLE. In the fort shown below (Fig. 7.4), there is a tower at the endpoints

Fig. 7.4. Minimum covering of a graph (in dark lines)

130

GRAPHS

of each wa!l. A guard stationed at a wall can watch both towers a t the end of his wall. What is the minimum number of guards needed to watch all the towers? Since the minimum covering of the corresponding graph is 7 edges, it follows that 7 guards will be required.

Theorem 3 (Norman, Rabin [1959]). In a simple graph G = ( X , E ) of order n, a maximum matching Eo and a minimum covering Fo satisfy

I Eo I f I Fo I = n. Giuen a maximum matching E,, a minimum couering

F1 = Eo u { e, 1 Y E S ( E d 1 is obtained by adding to Eo,for each unsaturated certex y , an edge e, of G that is incident to y. Giren a minimum corering F,, a maximum matching El is obtained by remotling successir7elyfrom F, edges that are adjacent to an unremored edge.

If Eo is a maximum matching, the set F~ = E o u ( e , / y E % E o ) ) is clearly a covering, and

IF1 I

=

IEO

I + (n - 2 I E O I )

=n

- IEOI.

Furthermore, if Fo is a minimum covering, the set El obtained by the successive elimination of edges of Fo that are adjacent to an (unremoved) edge of Fo is a matching. Since in G, the edge of Fo do not form chains of length 3, each removed edge creates exactly one unsaturated vertex of E l . Hence,

I Fo I - IE, I = I X - W

l

)

I = n - 2 I El I

and IF01

Since 1 El

1 < 1 Eo 1, IF,

=n-

IEII.

it follows that

I = n - l E o l < n - IE, I

=

I FoI.

Thus, the covering Fl is also a minimum covering. Since I Fl I = I Fo 1, it follows that I El 1 = 1 Eo 1, and, consequently, the matching El is a maximum matching. 1 Fo 1 = n. Finally, 1 Eo 1 Q.E.D.

+

This theorem shows that the minimum covering problem reduces to the maximum matching problem, which we shall study in the next section.

MATCHINGS

131

3. Matchings in bipartite graphs A graph G is said to be bipartite if its vertex set can be partitioned into two classes such that no two adjacent vertices belong to the same class. Theorem 4. For a graph G, the following conditions are equiralent: (1) G is bipartite, (2) G possesses no elementary cycles of odd length, ( 3 ) G possesses no cycles of odd length.

(1) =- (2) because if G is bipartite, we can colour the vertices red and blue such that two adjacent vertices have different colours. If G has an elementary cycle of odd length, then the vertices of the cycle cannot alternate in colour.

(2) 3 (3) Suppose that G possesses no elementary cycles of odd length, but there exists a cycle p = [ x o ,xl, ..., x,, = x o ] of odd length. If there are two vertices xt and xk in cycle p such that j < k and x j = x k , then the cycle can be decomposed into two cycles p [ x , , x k ] and p [ x o ,x j ] + p[Xk, x o ] . Furthermore, one of these cycles has odd length (otherwise, p would have even length). Clearly, each time that the cycle p is decomposed in this way, an odd cycle remains. When this decomposition terminates, there will remain an odd elementary cycle, which contradicts (2).

(3)

3

(1) We shall show that a graph without odd cycles is bipartite. Suppose that the graph is connected (otherwise, each connected component could be considered separately). Successively, colour the vertices using the following rules : RULE1. Colour an arbitrary vertex a blue. RULE2. If vertex x is blue, colour red all vertices adjacent to x. If vertex y is red, colour blue all vertices adjacent to y. Since the graph is connected, each vertex is coloured. A vertex x cannot be coloured both red and blue, since then vertices x and a would be contained in a cycle of odd length. This colouring determines a partition of the vertices into two classes and G is bipartite. Q.E.D.

Henceforth, a bipartite graph with vertex sets X and Y and with edge set E will be denoted by G = (A', Y, E ) . For any A c X u Y, the set of vertices adjacent to set A is denoted by T,(A).

132

GRAPHS

Konig's theorem [1931]. For a bipartite graph G number of edges in a matching equals

-AI

min (I

A c X

=

( X , Y, E), the maximum

+ I r G ( A ) I)

Consider the transportation network with vertices X u Y and a source a and a sink b. Source a is joined to each y j E Y by an arc of capacity c(a, y j ) = 1. Sink b is joined to each xi E X by an arc of capacity c(xl, b) = 1. Finally, y, is joined to xi by an arc of capacity 1 if y, E Tc(x,).

Fig. 7.5

For a set A c X,the total demand of set A equals d ( A ) = I A I. The maximum amount of flow that can be sent into A equals F ( A ) = I T,(A) I. A flow in this network defines a matching in the graph in which xi and y j are matched if a unit of flow traverses arc ( y j ,x,). Conversely, each matching defines a flow. The cardinality of a maximum matching is, therefore, equal to the value of a maximum flow between a and b. By Theorem (2, Ch. 3, max I E ,

I = d ( x ) + min ( F ( A ) - d ( A ) ) = A c X

€0

= IX

I

+ min

(I T,(A) I - I A I ) =

A c X

=

min (I X - A I

+ TG(A)).

A c X

Q.E.D.

133

MATCHINGS

For a graph G, a transversal set T is defined to be a set of vertices such that each edge has at least one endpoint in T.An equivalent formulation of Konig’s Theorem is: Corollary 1. For a bipartite graph G, the maximum number of edges in a matching equals the minimum number of vertices in a transversal set. Let Eo be a maximum matching, and let To be a minimum transversal set. Clearly, I To 1 2 I Eo 1 since To contains at least one endpoint of each edge in Eo. Furthermore, for each A c X , the set T = ( X - A ) u T,(A) is a transversal set of G, and from Konig’s theorem,

I E, I

= min

(I

X -A

A C X

I + I T,(A) I ) 2 I To I .

Hence, I Eo 1 = I To I.

Q.E.D. For a graph G, a stable set S is defined to be a set of vertices such that no edge has two distinct endpoints in S ; another formulation of Konig’s theorem is :

Corollary 2. For a bipartite graph G, the maximum number of vertices in a stable set equals the minimum number of edges in a cotiering. If T c X denotes a transversal set, and if E,, denotes a matching, then, from Corollary 1, max I Eo I = min I T I . T

Eo

If T is a transversal set, its complement S = ( X u Y ) - T is a stable set. If S is a stable set, its complement is a transversal set. Thus, max 1 S

I = IX 1

S

+ I Y I - min 1TI. T

From Theorem 3, min 1 F 1 = I X

I + 1 Y 1 - max I Eo 1 .

F

Eo

Hence, maxIS( =minIFI. S

P

Q.E.D. I f a bipartite graph G = ( X , Y , E ) has a matching that saturates all the vertices of X , then we say that X can be matched into Y. If this matching also saturates all the vertices of Y, we say that X can be matched onto Y.

134

GRAPHS

The following theorem is an easy consequence of the Konig theorem. G

Theorem 5 (P. Hall [1934]; “Konig-Hall Theorem”). In a bipartite graph = (X,Y , E ) , X can be matched into Y if, and only if,

I

rG(A)

I 2 IA I

(A

x)

-

From the Konig Theorem, X can be matched into Y if, and only if,

I x I = max I E,, I = min (1 x - A I + I rG(AjI). A c X

EO

This is equivalent to

Q.E.D. Corollary 1. In a bipartite multigraph G = ( X , Y, E ) with

IXl=p, IYI=q, index the certices x , E X and y j E Y such that

< dG(XJ

dG(x1) < (1IG(xz)< dGb1)

> dG(Yl)

2

**’

9

2 dG(yq) .

A sujicient condition that X can be matched into Y is that q 2 p and dG(xl)> 0 and

i

1-1

k-1

do(&)

’2 4 d Y J 1-1

(k = 2,3, ....PI.

Consider two subsets A c X and B c Y with k and k - 1 elements, respectively. From the above inequality, it follows that k

2

k- 1

dG(Y) =

mG(X,

B, *

YCB

Thus, the number of edges leaving A is strictly greater than the number of edges entering B. Hence, T,(A) C$ B for each set B of k - 1 elements; consequently 1 T,(A) I > k - 1 = I A 1 1. Finally,

-

I rG(A) I 2 I A 1 Therefore,

(A

x>.

X can be matched into Y. Q.E.D.

135

MATCHINGS

Corollary 2. If, in a bipartite multigraph G

mind,(X) 2 max dG(y)

=

and

(X,Y, E ) , we hare

I Y 1 2 IX I ,

Y E Y

XEX

then X can be matched into Y. Let min d G ( x ) = d,, max dG(y) = d2 . YEY

X P X

Thus d, 2 d,, and by indexing the vertices as described above, &(XI)

+ dJx2) + + dG(Xk) 2 kd, > ( k - I)dl 2 (k - I)d2 2 1..

2 dG(Y,)

+ dG(Y2) + ... + dG(Yk-1)

(k

=

2 , 3 , ..., P> *

Therefore, X can be matched into Y. Q.E.D. Corollary 3. I f G = (X,Y , E ) is a bipartite multigraph with no isolated vertices and Hith I Y I 2 I X 1, and such that for some vertex x1 E X,

min d G ( x ) 2 max d,(y) , Y € Y

xfx, xex

then X can be matched into Y. We may suppose that x1 has minimum degree (otherwise, the proof is immediate). Then

+

+ + dG(xk)2 dG(xl)+ ( k - 1) dl =-

d G ( x l ) dG(xz)

> (k - 1) d2 2 d G ( ~ 1 )+ ... + dG0.k-

I).

Hence, X c a n be matched into Y.

Q.E.D. Corollary 4. In a bipartite multigraph G = (X,Y, E ) , there exists a matching that saturates all the certices with maximum degree.

First, suppose that there exists a bipartite multigraph -

a,

G =(X,Y, with G = (X,Y , E ) as a subgraph suppose X c of degree h = max d , ( z ) .

1,Y

c

y, and G is regular

zeXuY

Prom Corollary 2, 1 can be matched onto L in G, since This matching saturates each vertex in G of degree h.

I XI

=

I Fl.

136

GRAPHS

We shall construct this graph Denote these h replicas by

G

= ( X , Y ,E ) ,

c by taking h replicas of multigraph

G.

G’ = ( X ’ , Y‘, E ’ ) , G” = ( X ” , Y ” ,E ” ) , ...

..., ~ ( h - 1 )= ( X ( h - l ) ,

E(h-1)

y(h-1) 7

1.

Let X consist of X U X‘ U--VF h - l ) together with some additional vertices. Similarly, let F consist of Y u Y’ u . . . u Y ( h - l )together with some additional vertices. These additional vertices are determined as follows : If xi E X and &(xi) < h, create, in F, h - &(xi) additional vertices, and join ~ X ~ X”, E ..., xIh-’)€X ( h - l ) , the each of these vertices to xi E X , x l X’, analogues of x i . Repeat this construction if y j E Y and &(y,) < h. In this way, a multigraph having G as a subgraph is constructed. Q.E.D.

c

Note that this result allows us to give an affirmative answer to the Dating Problem (Example 4, 0 1). Corollary 5. In a bipartite graph G = ( X , Y, E ) , X can be matched into Y if, and only i f ,

IX-

fG(B)I

I T,(A) I. Let B = Y - T,(A). Since no vertex of B is adjacent to A ,

IA I

A c X - I-,@). Hence,

IX -

I

~ G ( B ) 2I I A I > TG(A)I = I Y - B I . Q.E.D.

This corollary is in fact a reformulation of the Konig-Hall theorem that will be needed later. The next result is a reformulation of Bernstein’s theorem.

Theorem 6. In a bipartite graph G = ( X , Y, E ) , a necessary and suficient

137

MATCHINGS

condition that rhere exists a matching that simultaneously saturates X and B c Y is that (1) X can be matched into Y, i.e.

I T,(S) I 2 I s I

(S

=X);

(2) B can be matched into X , i.e.

I w-)Ia I T I

( T cB ) .

Clearly, conditions (1) and (2) are necessary. We shall show that if there exists a matching Eo from B into A c X , and that if there exists a matching El from X into Y , then there exists a matching saturating both X and B. We shall now construct from El a matching E; in which the saturated vertices of X remain saturated, and an unsaturated vertex b E Bin El becomes saturated in E ; . Since b E S(Eo),b is the end-point of a chain p of the form: Ab,

ZI =

( e , , e2, e37 e4r

with

...I

e l , e 3 , ... E E, - El , e 2 ,e4, ... E E , - E, .

Fig. 7.6

Suppose that p is as long as possible. Then, the last vertex z of this chain belongs to Y (otherwise, z belongs to A , and z is an endpoint of an edge of El - Eo that can extend chain p). If z E Y , then z $ B (otherwise z is an endpoint of an edge of Eo - El that can extend chain p). Thus, z E Y - B. Thus, E; = El u (p n Eo) - (p n El)is a matching saturating b ; besides, each vertex of X u B that is saturated in El remains saturated in E ; . By repeating this procedure as many times as needed, a matching saturating both X and B can be obtained. Q.E.D.

138

GRAPHS

Corollary. In a bipartite graph G = ( X , Y, E), a necessary and sufJicient condition that there exists a matching simultaneously saturating X and B c Y is that ( S c X) .

min{ITc(S)I,IXI-IB-J;;(S)I)21SI

Using Corollary 5 to Theorem 5, the two conditions of the above theorem can be written as

I TG(S>1 2 I s I

(1)

(2')

IB-rG(S)I]= [O, m,(xi, ; Cm', m'l

~ j ) ]

if if if if

u

=

( a , xi)

u = ( y j , b) u = (xi, Y j ) u = (b,a ) .

Each flow in R that is compatible with these intervals determines a partial graph G' with I E' 1 = m' and with

1 < dG.(xi)< h' max { 0, d G ( y j )- h" 1 < d G , ( y j )< h'

max { 0 , d,(xl) - h"

The partial graph G " dGW(si)

=

G - G ' satisfies I E"

=

t i G , , ( ~ ' j )=

t/G(xi)

I = m" and

- d G , ( x i ) < h"

d,(yj) - dG,(yi)

< k"

Thus the required partition of the edges of G has been found. Conversely, such a partition determines in graph R a compatible flow within the permitted intervals. From the Compatible Flow Theorem (Ch. 5 , Q 2), a necessary and sufficient condition for the existence of a compatible flow in R is that

b(o-(~))< c(w+(S))

(S c X u Y u ( a , b

1).

It is left to the reader to verify that the above condition is equivalent to that of the lemma. Q.E.D.

G

Theorem 7. (Dulmage, Mendelsohn [1961]). For a bipartite multigraph = (A', Y, E ) with I E I = m and with maximum degree < h, let

a = m a x ( n ~ - r n , ( A , ~ ) - ( / 1 - 1 ) ( 1 ~ - ~+ I I Y-BI)}, Ac X Bc Y

p = min Ac X Bc Y

{ m G ( A ,B ) + I x - A 1

+ I Y - B I} .

Then,

Furthermore, for each integer m' such that IS < m' p, then there exist sets A c X and B c Y such that

n? > h(m,(A, B)

+ IX

-A1

+I

Y - B ] )2

~m,(A,B)+m,(X-A,Y)+m,(A, Y - B ) = m , which is a contradiction. Thus,

2. If

[ t] < c,then there exist sets A c X and B c Y such that m -<m-mG(A,B)-(hh

1 ) ( ~ x - - ~ ~ + ~ Y - B ~ ) .

Hence,

m -

-i >in,(A, B ) + ( h in

h - 1 h

2 --m,(A,

B)

1)

(I

X -A

+ -h- (-hh1

1

+ 1 Y - B 1)

IX - A

2

I + h I Y - B I) 2

h -1 > --m, h which is a contradiction. (3) If m' satisfies c < m' have

< p, then, for all A

c

X and for all B c Y, we

2 in - i n & , B ) - ( h - I ) (1 x - A I + I Y - B ~ ~ ~ ' ~ ~ ? , ( A , B ) + I x - A I + I Y - B I .

\ in'

Let h'

=

1, h"

=

I)

h - 1, m" = m - m'. The above inequalities become

m" - h" I X - A I - h" I Y - B I in' - h'l X - A I - h' I Y - B I

< in,(A, B) < in,(A, B ) .

From the lemma, G can be decomposed into two partial bipartite multigraphs G' = (A', E') G" = ( X , E - E ' ) , and with

I E' I = in', max d,.(z) < 1, max dG,,(z)Q h

-

1.

Clearly, E' is the required matching.

Q.E.D.

141

MATCHINOS

4. An extension of the Konig theorem

In this section, we shall consider a multigraph with the properties: (1) there are no loops, (2) if p and p’ are two odd cycles without a common vertex, there exist two adjacent vertices x E p and x’ E p’. Such a graph is called semi-bipartite (see Ch. 6 , Q 2). We shall now study conditions for which a semi-bipartite graph possesses a perfect matching (i.e. a matching that saturates all vertices). Lemma 1. Let G = (X, r)be a symmetric semi-bipartite I-graph with I X I even; there exists a perfect matching i f , and only i f , there exists a partial graph H of G with (x 1 The proof follows from Theorem (5, Ch. 6). dH+(X) =

&(x)

=

EX)

.

Lemma 2. Let G = (X, r) be a 2-graph; there exists a set of elementary circuits of G that partition X i f , and only i f ,

I f(4 I 2 IA I

(A

=

x,

Associate with G a bipartite graph GI = ( X , E ) obtained by taking two replicas X and of set X,and joining xi E X to 2, E by an edge if x, E T(x,). In graph G, set Xcan be partitioned into. circuits if, and only if, the bipartite graph GI has a perfect matching, i.e., if, and only if,

x

I

fG,(4

I = I T(4 I > I A I

x

(A

=XI. Q.E.D.

Theorem 8. A semi-bipartite graph G = (X,E ) possesses a perfect matching i f , and only i f , (1) I X I is even, (2) I rG(A) I A ( A Apply Lemmas 1 and 2 to the symmetric semi-bipartite 1-graph G* obtained from G by replacing each edge by two oppositely directed arcs. Corollary. I f G is a semi-bipartite regular multigraph of degree h that has an ecen number of certices, then G possesses a perfect matching. Make two replicas X and of the vertex set of G, and construct a bipartite multigraph H = (X,1,E ) with m,(x, j ) = mG(x,y). For each A c X,we have

1A1

=

mH

((A,

rH(A))

mH

(x,r H ( A ) )

=

I rG(A) I

*

142

GRAPHS

Thus I f , ( A ) 1 2 I A I and, from Theorem 8, the simple graph G' obtained from G by collecting the multiple edges has'a perfect matching, which is also a perfect matching in G. Q.E.D.

5. Counting perfect matchings For certain graphs (particularly planar graphs) simple methods are available to count the number of distinct matchings. These methods are the result of work done independently by Fisher [1961] and by Kasteleyn [1961]. Kasteleyn's proof has been simplified by RCnyi [1966] and by Pla [1970]. We shall use Pla's proof. First, let us review some definitions from matrix algebra. If A = ((a:))is a square matrix of order n, the determinant of A is written as

where

2 42)

1 (r

=&I)

... ...

is a permutation of degree n, and E(O) = + 1 or - 1 if the permutation is even or odd, respectively. The permanent of A is defined to be the number Perm A = C

a&)

... a&) ,

0

Proposition 1. Let G = (X,U ) be a I-graph with certices x l , x2, ..., x,, and let A = ((a;))be a square matrix of order n defined by a: = 1 i f ( x , , x,) E U, and af = 0 $(xi, x j )6 U. Then the permanent of A equals the number of pairwise disjoint systems of circuits that partition X . Each non-zero term of the expansion of the permanent corresponds to such a system of circuits, and conversely. Q.E.D. A skew-symmetric matrix B = ((bj)) can be associated with an antisymmetric l-graph G = ( X , U ) by letting + 1 if ( x i , x j ) e U , b'. = - 1 if ( x j , x i ) U ~, 0 otherwise. Matrix B is called the adjoint matrix of G. If B = ((bj)) is a skew-symmetric matrix of even order n = 2 k , the pfafian of B is defined to be Pf B = CE(O,) bi: bf: ... b2t-l ,

4

n

143

MATCHINGS

where n = [ i l ,i 2 ] .[i3, ia]...[i2k-l, i z k ] is a permutation of degree n that decomposes into k cycles of length 2 (Le., x(i) = j impliesj # i and x(j) = i ) , such that i l < i3 < < i2k-1 il < i,, i3 < i4, ..., izk,

-=

and where cn is the permutation

Proposition 2. r f G = ( X , U ) is an anti-symmetric I-graph of even order n = 2 k , and if B = ((bj))is its adjoint matrix, then 1 Pf B I is less than or equal to the number of perfect matchings in G . Furthermore, the number of perfect matchings in G equals I Pf B I if, and only if, each term in the expansion of Pf B has the same sign. Each term in the expansion of the pfaffian corresponds to a perfect matching, and conversely. If two non-zero terms have opposite signs, then they cancel out one another. Q.E.D. Note that the pfaffian is easily calculated using the following well-known theorem from linear algebra: Proposition 3. I f B is a square skew-symmetric matrix of order n, then Det B = (Pf B)*

=o

i f n is even , i f n is odd.

The proof can be found in most comprehensive texts on linear algebra. Consider an anti-symmetric 1-graph G = ( X , U )with a perfect matching Wo c U. If p is a cycle of G, let p + denote the arcs of p that are directed in the direction of travel through the cycle, and let I p + I denote the cardinality of this arc set. If, for a family M = (pi / i E I ) of cycles, the numbers I p t I are all odd, G is said to be net1 directed with M . Finally, cycle p is said to be alternable if there exists a perfect matching in which p is an alternating cycle.

Theorem 9 (Kasteleyn [1961]). Let G = ( X , U ) be an anti-symmetric 1-graph of eren order 2 k ; let B be its adjoint matrix, and let W, be a perfect matching in G. The follorving three statements are equivalent: (1) AN the non-zero term in the expansion of Pf B hace the same sign, (2) G is t%3elldirected for the family of its alternable cycles, (3) G is well directed with the family of the alternating cycles for Wo.

144

GRAPHS

If W, is the only perfect matching in G, the result follows since G has no other alternable cycle. Thus we may assume that G has several perfect matchings. (I)

3

(2) For example, let p = [x,, x2, ..., x Z p x,] , be an alternable cycle, and let Wand W' be two perfect matchings i n G such that

-

(W

W') u (W' - W ) = p .

Suppose that the set of arcs common to W and W' is W n W' = {

(XZP+l, X Z p + 2 ) ,

.-.,( X Z k - 1 3

X,d>

*

The terms corresponding to Wand W' in the expansion of Pf B are respectively 1

b43 ... bii-' b;;:; ... b:i-'

8 = E(O)

b2

0'

biP b: b;

and =

~(0')

... bf,Pfi ... b i i - '

,

where

61

=(:.

2 2P

...

3 2

...

2p+ 1 2 p 1

2P '

+

2 p -1

... ...

Note that E(U) ~(0') = + I , because 2 p - 2 transpositions are required to pass from o to d.Thus, from (I),

+1

=

6 0' = - bi b j

... b;;-'

b:" .

Consequently,

Ip'l

E

Ip-I

=1

(mod2),

and for the alternable cycle p, graph G is well directed.

(2) 3 (3) This follows because each alternating cycle of W, is an alternable cycle. (3)

3

(1) Let W be a perfect matching different from W , . From Corollary 1 to Theorem I , a sequence of perfect matchings W, , W , , ..., W, can be constructed so that for each i the arcs of

(Wi- Wi+,>u ( W i + l

- Wi)

define an elementary cycle pi that is alternating in W,. Thus, from (3), I p; I is odd.

145

MATCHINGS

As above, we can show that the terms O(Wo)and O(W) corresponding to matchings W, and W satisfy O(Wo)B(W ) = + 1. Consequently, all nonzero terms in the expansion of Pf B have the same sign. Q.E.D.

Theorem 10. If G = (A’, U ) is an anti-symmetric I-graph, and i f M = (pi1 i E I ) is a family of linearly independent cycles, then, by reversing thl direction of certain arcs, a 1-graph G‘ = (X,U‘) that is well directedfor M can be obtained. Let si = number of arcs of G directed in the direction of travel in p i , s‘ = number of arcs of G directed against the direction of travel in pi. Let = 1 if ujepi ‘ij = o if uj+pi. Finally, let = 1 if the direction of arc u, should remain unchanged to obtain graph G‘, zi = 0 otherwise.

i

Graph G’ will be well directed for M if, and only if. for each i, rn

s1

+ j1 cij z j 3’1 =1

(mod 2).

Hence, G’ is obtained from a system of equations of m variables zl, z 2 , ...,

z, in the field of integers modulo 2: (i E I )

.

j= 1

If family M consists of independent cycles, then this system consists of principal equations and, therefore, it has a solution (zl, z2, ..., z,) in the field of integers modulo 2. Q.E.D. For example, consider the non-planar graph i n Fig. 7.7 with the perfect matching W , shown in dark lines. The alternating cycles are linearly independent since each possesses a distinct edge (marked with a cross in Fig. 7.8). Thus, from Theorem 10, it is possible, by directing the edges, to obtain a graph G = ( X , U ) in which the alternating cycles are well directed. The adjoint matrix of graph G = ( X , U ) is:

146

GRAPHS

0

6

5

1

1

0

4

Fig. 7.7

- 1

1 - 1

1

0

1

-1 0

-1

0

From Theorem 9, the number of distinct perfect matchings is

I Pf B I

= JDi5t B = J 3 . 3 . 2 . 2 = 6

We shall show that this method always works for planar graphs. Theorem 11 (Kasteleyn [1961]). Let G = (A', U ) be a connected planar graph with a perfect matchirig Wo that is well directed for the family of contours of its bounded faces. (From Theorem 10, one such orientation always exists.) Then G is well oriented for the family of its alternating cycles. Let p be an alternating cycle for the perfect matching Wo. Cycle p surrounds an even number of vertices because the vertices in the interior p are matched together by W , . Let H be the subgraph of G generated by vertices situated on p or in the interior of p. Clearly, H i s planar and connected. Suppose H has n vertices, m arcs and f finite faces, v l , v 2 , ..., v,. Successively traverse the cycles v l , v 2 , .. ., v, along their direction and then traverse p against its direction. While doing this, count the total number 5 of arcs traversed along their direction. It is clear that each arc of H will be traversed once in each direction; thus, ( = m.

147

MATCHINGS

Next, by summing over the cycles, we have f

t

=

c I :v

i= 1

I+

(w- I p + I) ,

where I@) is the length of p. Since G is well directed for the cycles vl, we have m=f+Z(p)+Ip+I.

The numbers n, m andfsatisfy the Euler relation for planar graphs (see Corollary 1, Theorem 2, Ch. 2), f = m - n 1. Hence

+

IpfI--m+f+/(p)=n+

= 1 + (n - I(p)) = 1 ,

1 +/(p)

since n - l ( p ) is the number of vertices in the interior of p, which is even. Thus, G is well directed for p. Q.E.D. Remark. T o calculate the number of distinct perfect matchings in a simple planar graph G, it suffices to give the edges of G a suitable anti-symmetric orientation, to determine the adjoint matrix B and to calculate 4 D e t B.

EXERCISES 1. Suppose that G = ( X , E ) is a connected graph without isthmi (an edge whose removal disconnects the graph), and each vertex of G has degree 3. Consider a maximum matching Eo. Show that there exists a chain whose edges belong alternately to Ea and E - Eo that uses exactly once each edge of E - Eo and uses each edge of Eo twice. (P.Medgyessy [1950]) 2. In a bipartite graph G = (A', Y, E), let

6(A)

, A I - I Tc;(A)

1,

60 = max & ( A ) . AcX

Show that

+

(1) &At U A z ) ~ ( A n I Az) > 6 ( A i ) f & A d . (2) Using the above inequality, show that the family

d -{ A / & A ) - - n } satisfies Ai,AzCd

7

Ai

u

A2,

AinA2Ed.

(3) If So > 0, the set of vertices of X not saturated by at least one maximum matching is

3. Consider a bipartite graph G = (A', Y, E ) for which there exists a matching of X into Y. Show that there exists an xo E Xsuch that, for each y E r G ( x o ) ,at least one maximum matching uses the edge [ X O , yl.

148

GRAPHS

Hint: If I TJS)

I > I S I for each

S #

@’,

any xo E X can be chosen. (M. Hall [1948])

4. Deduce from the preceding exercise that if the minimum degree dG(x) for x E X is equal to k. then there exist at least k ! distinct maximum matchings. (M. Hall [1948]) 5. In a bipartite graph (X,Y, r)with I XI = I Y 1, let k = max{ d d x d ddxd * - * 4-d ~ ( x d -ddvd - d&d - *.* - d ~ b - d .]

+

k

+

Show.that there exist k disjoint matchings of X into Y. (0.Ore [1955]) 6. Showfhat in a graph G with n vertices and minimum degree k , a maximum matching V satisfies

I V : 3 min ( k ,

[-;I 1.

(P. Erdos, L. P6sa [19621) 7. Show that if V is a matching and T is a transversal set, then min I TI < 2 max I V I . Show that the equality holds if, and only if, the connected components of the graphs are all cliques of odd cardinality. 8. If k = min do(x), if G is connected, and if max I V I < n--+ 2 , then 3 mini TI < 2max 1 VI - k . (Erdos, Gallai [1961])

9. If a graph remains connected after any k - 1 of its vertices are removed, and if n-I maxl VI < , then k < max I VI,and 2 min I TI < 2max I V I - k (Erdos, Gallai [1961]) Show that this bound is attained by a graph G formed from a k-clique K8 and 1 > k 1 cliques K z , +~I, with each vertex of K,,, + being joined to each vertex of K k . 10. Show that 2 mi n I V I < m + m a x I V l . Also show by an induction on m that the equality holds if, and only if, the connected components of the graph are 2-cliques or 3-cliques. 11. Show that 4 min I TI < 2 n nr - max I V I . If G is a 2-clique, 3-clique, 4-clique or two triangles joined by one edge, then the equality holds. Are there other connected graphs for which this equality holds? (Erdos, Gallai [1961]) 12. Let a tree with diameter < 3 be called a double star. Let f ( G ) denote the minimum number of double stars needed to cover all the edges of a graph G with n vertices. (1) If r is the cardinality of a maximum matching of G, show that f(G)< 2 r . (2) Show thatf(G) < n - 2 r . ~

+

+

(3) Using (1) and (2), show that f(G)< (L. Lovisz [1968])

149

MATCHINCS

13. In a simple graph G = ( X , E ) , consider a family of sets %?= { c1, c2, . . . I c, 1 , with (1) I Ct I odd (i = 1, 2, ...,p ) ( 2 ) For each [ x , y ] E €, there exists an index i such that

.

min { I Ci I, { x , y 1, let c(CJ = max { k, 1 }, and let

I Ci n { x , y } I

If

I C, I < 2 k

+

=

1I} .

P

C(W =

1 C(C0. I=1

Show that for each matching Eo,

1 EOI C min c(W. Y

Also show that max l E o l = m i n c ( a , Eo

v

(J. Edmonds [1964]) 14. Let G = (X,Y, r )be a bipartite 1-graph without isolated vertices (where Tis acorrespondence from X onto Y ) . Show that the minimum number of arcs that must be added to G to make a strongly connected graph is max { I X 1, I Y I }. Hint: Use Corollary 1 to the Konig theorem.

CHAPTER 8

c -Matchings

1. The maximum c-matching problem

Consider a multigraph G = ( X , E ) with vertices xl, x z , ..., x, and a ntuple c = (cl, c2, ..., c,) of integers with ( i = 1, 2 , ..., n). 0 < ci < &(xi) The set E0 c E is called a c-matching if for each i, the set Eo(x,) of edges of Eo incident to x, satisfies

I EO(Xi) I G ci

'

A vertex x, is said to be saturated in the c-matching Eo if I Eo(x,)1 = c,. In this section, we shall study the problem of constructing a maximum c-matching. The maximum matching problem (Chap. 7) is a special case of the maximum c-matching problem for c = (1, 1, ..., 1). To each multigraph G, there corresponds a simple graph G defined as follows: For each xi E X , define two disjoint sets A i = ( a f / e E E ( x i ) } and B i = ( b : / k = 1 , 2 , . . . , d G ( x i ) - c i } Let the vertex set of G be the union of U;=l At and of U;=l B,. For each i, join each vertex of A , to each vertex of B , . For each edge c = [ x , ,xi] of G , construct an edge 2 = [a;, a;] in Theorem 1. A maximum matching i?, in graph that saturates U;=l B, induces a maximum c-matching Eo in graph G,and comersely. 1. Let Eobe a maximum matching of G. We may assume that Eo saturates each b: (by interchanging if necessary the edges of Eo and of E - Eo along a chain [b:, a;, u!] of length 2). Matching Eo in G defines a set of edges E, in G, and this set Eo satisfy

c.

I

I

€,(xi) < dG(xJ - I Bi I = dG(xi) - (dG(xi) Thus, Eo is a c-matching in G ; furthermore,

Ci)

= ci

151

C-MATCHINOS

2. Now consider a maximum c-matching El c E of graph G. Set El defines in G a matching Elthat saturates each b:, as shown in Fig. 8.1.

-E - Eo

b Fig. 8.1

Consequently,

Since Eo is a c-matching in G, I Eo I

< I El 1,

and therefore

Since Eo is a maximum matching, 1 El1 = I ED1. Consequently, I El I = Hence, Eo is a maximum c-matching of G, and El is a maximum matching of G. Q.E.D.

I Eo I.

Remark. This theorem demonstrates that a maximum c-matching can be constructed by determining the maximum matching Eo in graph G and then saturating each vertex b: by an interchange along alternating chains of length 2. Consider a multigraph G with a c-matching Eo. From G, construct a multigraph R(Eo) by adding to G a vertex xo called the origin that is joined to each vertex xi by cI edges, for all i. Multigraph R(Eo)is called a transfer net work.

152

GRAPHS

Let the edges of Eo be represented by dark lines, and let ci - I Eo(xi)I edges from xo to xi for all i be also represented by dark lines. All other edges of R(Eo)are represented by light lines. An alternating chain is defined as a chain of R(Eo)with edges alternately dark and light that repeats no edge. A transfer along an alternating chain p is defiiled as the interchange of the dark and light colouring along p. Theorem 2. (Berge [1958]). A c-matching Eo of a multigraph G = ( X , E ) is maximum if, and only if, there exists no alternating chain in R(Eo) that joins xo to itself and has dark initial and terminal edges. As in Theorem 1, construct a graph G corresponding to the multigraph G and a matching Eo in G corresponding to Eo that saturates each vertex in (J Bi. An alternating chain with the above properties in R(Eo)corresponds in G to an alternating chain that connects two distinct unsaturated vertices (if Eo is properly chosen), and vice versa. In graph G, such a chain exists if, and only if, the matching Eo in G is not maximum (Chapter 7, Theorem l), or, from Theorem 1, if, and only if, the c-matching Eo in G is not maximum. Q.E.D. Theorem 3. I f Eo and El are two maximum c-matchings of a multigraph G = ( X , E), then El can be obtained from Eo by transfers along alternating cycles of R(Eo)that are pairwise edge-disjoint (but not necessarily elementary). From Corollary 1 to Theorem 1 (Ch. 7), each maximum matching El in G can be obtained from the maximum matching Eo by a series of transfers along vertex disjoint alternating chains. Each of these chains is either an alternating elementary cycle or an even elementary chain starting at an unsaturated vertex. In either case, this alternating chain corresponds in R(Eo)to an alternating cycle that is not necessarily elementary. Q.E.D. Theorem 4. A “free edge” is defined to be any edge of G = ( X , E ) that is contained in some maximum c-matching but not in every maximum c-matching. An edge e is,free if, and only if, gicen a maximum c-matching Eo of G, e lies on an alternating cycle of R(Eo).

1. If e is contained in an alternating cycle of R(Eo),then e is evidently a free edge, since a transfer can be made along this alternating cycle. 2. Let e be a free edge, and suppose at first that e is contained in the maximum c-matching Eo. There exists a maximum c-matching El that does not contain e. Thus, from Theorem 3, e is contained in an alternating cycle of W O ) .

If e is not contained in Eo,then e is contained in a maximum c-matching

153

C-M ATCHINGS

El obtained from Eo by a transfer along an alternating cycle of R(ED).Again, e is contained in an alternating cycle of R(Eo). Q.E.D. 2. Transfers In this section we shall consider a simple graph G and show how to obtain from G all the graphs with the same degrees as G by a sequence of transfers of a particular form. Lemma. If G = ( X , E ) is a simple graph such that each even cycle of length > 4 has a chord that divides the cycle into two euen cycles. Let E, c E be a set of dark edges. Then any transfer along an alternating cycle p can be obtained by a sequence of transfers along alternating cycles of length 4.

If the length of cycle p equals 4, the result is trivial. If the result is true for cycles of length < 2 k , then we shall show that it is also true for an alternating cycle p of length 2 k , say c1 = [a,, a27

**.,

a11

02k,

-

Since this cycle has a chord of the form [a,, a,,,,,,], cycles : Pl =

bi,

~ 1 2=

[a,,

ai+1, **., ai+2p+1,arl

we have two even

Y

ai, ai+2p+1,a i + 2 p + 2 ,

e a . 3

a2kr

a13

*

Relative to E D ,only one of these two cycles is alternating, say p l . Since the length of p 1 is < 2 k, the transfer along p l can be accomplished by a sequence of transfers along alternating cycles of length 4: Eb = Eo - (eel n E,) u (p1 - E,) . E, Relative to ED,cycle p2 is an alternating cycle of length < 2 k. The transfer -+

-+ ES = EL - ( p 2 n Eb) u ( p 2 - EA) Eb can be accomplished by a sequence of transfers along alternating cycles of length 4. Thus, the transfer

E,

+

Eg = E ,

- (p n E,)

u (p - E,)

has been obtained. Q.E.D. Theorem 5. Let HD= ( X , Eo) and H I = ( X , El) be two simple graphs, with the same vertex set, such that

dHo ( x i ) = dH1( x i ) = d,

(i = 1, 2, ... , n)

.

154

GRAPHS

Let a, 6, C, d be any four rertices in X with ac E E,, bd E Eo, ad 4 Eo and bc 4 Eo. A “direct transfer on H,” is defined as an operation that removes edges ac and bd and adds edges ad and bc. Then, graph H 1 can be obtained from graph Ho by a sequence of direct transfers.

Clearly, Ho and H I are two partial graphs of a complete graph of the form G = ( X , E). In G each even cycle [ a l ,a,, ..., a,] has a chord of the form [a,,a, + 31. From the above lemma and from Theorem 3, Ho can be transformed into fZl by transfers along alternating cycles of length 4. These transfers are direct transfers. Q.E.D.

Theorem 6. Let Ho graphs such that

=

( X , Y , E,) and H ,

=

( X , Y, El)be two bipartite 1, 2,

&&xi) = d H , ( x i )= ri

(i

dHo(Yj) = dH,(yj) = s j

( j = 192,

=

Let xl, x , E X be two distinct certices, and let y,, y , vertices with

x,Y,EEo,

X*Y2EEo,

XIY24E0,

E

... , p ) ,

..., 4) Y be two distinct

X2Yl$EO.

A “bipartite transfer on H,” is dejned to be an operation that remores edges x1 y , and x2y , and adds edges x1 y , and x2y , Graph H1can be obtained from graph Ho by a sequence of bipartite transfers.

.

Clearly, H o and H I are partial graphs of a complete bipartite graph of the form G = (X,Y, E ) in which each x E X is joined to each y E Y. From the lemma and from Theorem 3, Ho can be transformed into H I by a sequence of transfers along alternating cycles of length 4. These transfers are bipartite transfers. Q.E.D.

Theorem 7. Let G

=

(X,U ) and G‘

=

( X , U ) be I-graphs such that

dG+(xi) = dG+,(xi)= ri

(i = I , 2, ..., n) .

&(xi) = d&(xi) = si

(i = 1, 2, ..., n ) .

Let a, b, x, y be vertices of X with a # 6, x # y , (a, x) E U, (6, y ) E U, (a, y ) 4 U, (b, x) 4 U. An “oriented transfer” is dejined to be an operation that removes arcs (a, x) and (6,y ) and adds arcs (a, y ) and (b, x). Graph G’ can be obtained from Graph G by a sequence of oriented transfers.

C-MATCHINGS

155

Graph G = (X,U ) corresponds to a bipartite graph H = ( X , X,E ) where [ x i ,Z,] E E if, and only if, ( x i , x,) E U. The proof follows when Theorem 6 is applied to the bipartite graph H. Q.E.D. Remurk. Consider a 1-graph G such that d,+(x) = d;(x) for each vertex x . Such a graph defines a permutation of degree n, and vice versa. Thus, Theorem 7 generalizes the well known theorem from algebra: Ecery permutation is a product of transpositions.

3. Maximum cardinality of a c-matching Consider a multigraph G = (A', E ) with a c-matching Eo c E, and the corresponding multigraph R(Eo)with dark and light edges (see Section 1). Recall that the origin xo is joined to some of the vertices of X by dark and light lines, the edges of Eo are dark, and the edges of Fo = E - Eo are light. Consider the chains starting at the xo and beginning with a dark edge. If there exists an alternating chain ,u starting at xo and going to X E X , then orient each edge in ,u toward x. It is possible that an edge will be given two opposite directions. If vertex x is the endpoint of a dark edge oriented toward x and is not the endpoint of any light edge oriented toward x , then vertex x is called dark. The set of all dark vertices is denoted by Xd. If vertex x is the endpoint of a light edge directed toward x and not the endpoint of any dark edge directed toward x , then vertex is called light. The set of all light vertices is denoted by X', If a vertex x is the endpoint of a light edge directed toward x and also the endpoint of a dark edge directed toward x , then x is called a mixed vertex. The set of all mixed vertices is denoted by X".Finally, if a vertex x is not the endpoint of any edge directed toward x, then vertex x is called inaccessible. The set of all inaccessible vertices is denoted by X'. Vertex xo, the origin of the network R(Eo), is so far unclassified: it will be defined to be right if there exist no alternating chains from xo to xo with dark edges at each endpoint; otherwise, xo is defined to be mixed. Each vertex of R(E,) is either light, dark, mixed or inaccessible. Theorem 8. Two dark certices can only be joined by a dark edge. Two light certices can only be joined by a light edge. A mixed i3ertex and an inaccessible oertex cannot be joined by an edge. A dark uertex and an inaccessible vertex can only be joined by a dark edge. A light uertex and an inaccessible vertex can only be joined by a light edge.

This follows immediately from the definitions.

156

GRAPHS

These results are summarized below : TvDe of vertex

I

Liaht

Dark

Mixed

I

Dark edge

Light

Inaccessible

I

Dark edge

Light edge

Mixed Inaccessible

Dark edge

Light edge

Consider the subgraph of R(Eo)generated by the set Xm.Denote the connected components of this graph by M , , M 2 ,.... Chain p is said to enter M I via edge [a, b] if p is of the form p = [ x o , a,, a 2 ,..., ak = a ,

6 , = b , b 2 , ..., b,]

with xo, a l , a,,

..., ak # M 1

bi,

..., b l E M 1 .

b2, b 3 ,

Lemma 1. Let x o 6 M I and x E M I . If an alternating chain p[xo,x]enters M I via an edge [a, b] and ferminafes with a dark (respecticely, light) edge, then there exists an alternating chain p'[xo, x ] entering M I ria [a, b] and terminating with a light (respecticely, dark) edge.

Fig. 8.2

If x is a mixed vertex, then there exists an alternating chain v [ x o ,x ] that terminates with a light edge. Since xo 6 M 1 and x E M I , there exist in v [ x o ,x ] edges with one endpoint in X - M I and the other endpoint in M1. Let [c, d] be the last edge of this type in v [ x o ,XI.

C-MATCHINGS

157

Finally, let y be the first vertex of ,u that is on v[d,x ] (such a vertex always exists). If p[b,y ] and v[d,y ] terminate with edges of the same type, the theorem is true because of the alternating chain : P'

= P[X,,

Yl

+vb,XI.

I f p[b,y ] and v[d,y ] terminate with edges of different types, then [a, 61 = [c, dl since, otherwise, p[x0,vl + V [ Y ,c ] would be a simple alternating chain, and c would be mixed, which contradicts c $ M,. Thus, the chain P' = P h , bl

+ v[d, X I

is alternating and satisfies the requirements of the theorem. Q.E.D.

Lemma 2. Let xo6 M , , x E M1 ; let [a, b] be an edge incident to M1 and directed into M I . There exists an alternating chain p [ x o ,x ] from xo to x that enters M , via edge [a, 61. Let Y be the set of vertices of M I that are accessible by an alternating chain entering via [a, b]. Let Z be the set of other vertices of M , . Since b E Y, we have Y # @. Suppose that Z # 0, then there exists an edge [ y , 21 with y E Y and z E Z. We shall show that this leads to a contradiction. From Lemma 1, y is accessible by two alternating chains ,ul[xo,y ] and ,uz[xo,y ] entering M , via edge [a, b] and terminating respectively with a dark and light edge. Thus, z is accessible to an alternating chain entering via [a, 61. This contradicts z E Z. Q.E.D.

Theorem 9. (Gallai [1950]).Let Eo be a set of dark edges in G,. Let M I be a component of the subgraph of R(Eo)generated by the mixed vertices. r f x o $ M , , there exists exactly one edge incident to M I and directed into M1. rfxo E M I , there exists no edge incident to M I and directed into M I .

1. If xo $ M I , let p be an alternating chain going from xo to M,. Let b be the first vertex of p in M , . The alternating chain p [ x o ,b] enters M , via an edge [a, b] that is incident to MI and is directed into M I . Let [c, d ] be an edge other than [a,b] that is also incident to M 1 and directed into M , . From Lemma 2, we know that d E M 1 is accessible by an alternating chain entering via [a, 61; from Lemma 1, we may assume that this chain terminates with an edge of a type different from [c, d ] . Thus c is a mixed vertex, which contradicts c $ MI. Thus [a, 61 is the only edge that is directed into M,.

158

GRAPHS

2. I f xo E M , , then we can return to the preceding case simply by adding vertices a, and b,, a dark edge [a,, b,] and a light edge [b,, x,]. Let a, replace xo as the origin. The only edge entering M 1 is [b,, xo].Thus there exists no edge of the original graph that enters M , . Q.E.D.

Theorem 10. Let Eo be a maximum matching, and let M , be a component of the subgraph generated by the mixed vertices in the multigraph R(E,). There is a dark edge incident to M 1 and directed exclusively into M,. All other edges incident to M , are light and exclusively directed out of M I . 1. Since the matching Eo is maximum, x, is light, and x , $ M , . From

Theorem 9, there exists exactly one edge directed into M,. Denote this edge by [c, X I , where c $ M I , x E M I . Suppose [c, x ] is light: since x is mixed, there exists an alternating chain p from xo to x that terminates with a dark edge. Chain p necessarily uses edge [c, x ] (which is the only possible entrance edge to Ml), followed by a dark edge incident to x and terminates with another dark edge incident to x. But this is impossible, since there is only one dark edge incident to x. Thus, edge [c, x ] is dark. 2. Let [ y , b] be another edge incident to M , with y E M 1 and b 6 M,. From Theorem 9, [ y , b] is necessarily directed out of M , . Furthermore, [ y , b] cannot be dark because y, being a mixed vertex, can be reached by an alternating chain terminating with a dark edge. Q.E.D.

Corollary 1. If Eo is a maximum matching, and if M 1 is a component of the subgraph generated by X", then I M , I 2 3 and 1 M , I is odd. Let [c, x ] be a dark edge incident to M1 with x E M,. Then, I M 1 1 2 3, because, otherwise, x cannot be reached by an alternating chain terminating with a light edge. Furthermore, since M1 consists of vertex x and pairs o f vertices joined by dark edges, I M 1 I is odd. Q.E.D.

Corollary 2. I f Eo is a maximum matching, each non-mixed certex that is adjacent to a mixed vertex is a light vertex. I f [c,x ] is a dark edge incident to a component M , with c 6 M1, and M , , then c $ X i , c $ X d and c $ X". Consequently, c E X ' . If [a, b] is a light edge incident to M 1 with a $ M , , b E M,, then similarly, a 4 X', a $ Xd,and a $ X". Thus, a E XI. Q.E.D.

x

E

159

C-MATCHINGS

Corollary 3. vertex is light.

Eo is a maximum matching, each vertex adjacent to a dark

Let a E X d . If [x,a] is a dark edge, then x 4 X i(since x is the endpoint of a directed edge); x 4 X d (because, otherwise, a would be mixed), and x 4 Xm (from Corollary 2). Thus x E X'. If [x,a ] is a light edge, then x 4 X i , x $ X d , and x 4 X" (from Corollary 2). Thus, x E X'. Q.E.D. Theorem 11. Let Eo be a maximum matching and let Il be a connected component of the subgraph generated by the inaccessible vertices. Each edge incident to I, is light and undirected. Each vertex x $ I , that is adjacent to a vertex of Il is light. rfgraph G has no isolated vertices, then I I, I is even and 2 2. 1. If [c, x ] is an edge incident to Il with x E I,, then c $ Xm,c 4 Xi,and c 6 X d (from Corollary 3). Thus, c E X', and [c, x] is right. 2. Since Z, does not contain xo nor unsaturated vertices, it contains only pairs of vertices joined by dark edges. Thus, 1 Il I is even and 2 2. Q.E.D.

Theorem 12. (Berge, [1958]).Given a siniple connected graph G and a subset S of the vertices, let p,(S) denote the number of components of odd order in the subgraph generated by X - S. The number of unsaturated vertices in a maximum matching is

5

= max (pi(S) sc

x

-I S

I)

*

1. Consider a set S c X , and let C1, C,, ..., C, be the components of odd order in the subgraph generated by X - S. If a component ck has no unsaturated vertices, there is at least one dark edge going from c k to a vertex s k e S because I c k I is odd. Two distinct components c k correspond to two distinct vertices sk. Thus, no, the number of unsaturated vertices, satisfies

p,(S) - no < (number of

c k

without unsaturated vertices)

< 1 S 1,

Hence, Pi(S)

- I S I G no

( S c x)

2. We shall show that S can be chosen so that p,(S) - S = no (this establishes the theorem). Let Eo be a maximum matching, and let X' be the set of light vertices. From Theorem 10 and Corollaries 1, 2, and 3, pi(X') = (number of components in G y )

+ I Xd I .

160

GRAPHS

Furthermore, the dark edges of G define a bijection between the set X' and the components M , that contains no unsaturated vertices and no dark saturated vertices. Thus

I X 1I

= (number of components in

Gp) + 1 X d I - no.

Hence

no = p i ( X r ) - I X'

I.

Q.E.D. Corollary 1. In a simple connected graph G = ( X , E ) of order n, the number of edges in a maximum matching equals 1

- (n

2

-0,

where

The proof is immediate from Theorem 12.

Corollary 2. (Tutte [1947]). A necessary and suficient condition for a connected graph to possess a perfect matching is that PAS) G I S I The proof is immediate.

(S c X )

The following result generalizes the Petersen theorem [18911 for regular graphs of degree 3. Theorem 13. Let G = ( X , E ) be a connected multigraph that is regular of degree h, with an etlen number of vertices, without loops, and with (ScX; S#@,X). mG(S,X-S)>h-l Then G possesses a perfect matching. Furthermore, each edge of G is free (i.e. each edge of G belongs to at least one perfect matching but does not belong to all perfect matchings). 1. By using Corollary 2, we shall show that there exists a perfect matching. Let S be a non-empty subset of X , where S # X , and let C,,C2,... be the connected components of odd order in the subgraph generated by X - S. By hypothesis, m,(S, C,) 2 h - 1. However, m,(S, C,)> h - 1 , because, otherwise, the number of edges in C, would equal 1

1

mG(Cl,C , ) = 2 [ h I C, I - ( h - l)] = 2[h(l C , I - 1)

+ 11

161

C-MATCHINGS

which is not an integer, since I C, I is odd. Thus, mG(S,

c1) > h

Hence,

h

IsI2

mG(S,

x-s) 2 m,(s, c1 u c,?u "*) = 1 mG(S, ck) 2 hp,(S). k

Thus, pi(S) < I S 1 for all S # X, S # 0. This inequality remains valid when S = X because p i ( X ) = 0, or when S = 0 because p i ( @ ) = 0 (since the graph has an even number of vertices and is connected). Thus, from Corollary 2, there exists a perfect matching Eo.

2. To show that.an edge e E E is free, it is sufficient from Theorem 4 to show that e is on an alternating cycle relative to a perfect matching Eo. We may assume that e E E - Eo (because if e E Eo and if some edge of E - Eo that is adjacent to e is on an alternating cycle, then edge e will also be contained in an alternating cycle). We shall suppose that e is a light edge of G that does not belong to any alternating cycle, and we shall show that this leads to a contradiction. Contract the endpoints of edge e into a single vertex xo and define a transfer network R(Eo) with origin xo. Since edge e appears in no alternating cycle, vertex xo is light. Assume that there exist mixed vertices; let C , be a connected component of the subgraph of R(Eo) generated by the mixed vertices. Clearly, xo 6 C,. Since I C, I is odd from Corollary 1 to Theorem 10, we know from Part 1 of this proof that mG(X

- c1

9

c1)

>, h

a

Furthermore, from Theorem 10, there is a single dark edge of R(Eo) that leaves C , , and thus there are at least h - I light edges that leave C , . Now consider the network R(Eo)obtained from R(E0)by contracting each component C, of mixed vertices. Thus, C , becomes a dark vertex c,, and no edge has been oriented in two directions. From Theorem 10, c1 is incident to a dark edge directed into c1 and to light edges each directed out of cl. Let X d and 8' respectively denote the sets of dark vertices and light vertices in network R(Eo).Let d:(x) denote the number of light edges of R(Eo) directed out of x. Consequently, d ; ( x ) >, h - 1

(x E 2 ) .

Besides, the number d;(x) of light edges directed into x satisfies d;(x)

< d,(x)

- 1=

h

-1

(X

EX' - { x0 }) .

162

GRAPHS

Since the theorem is obvious for h

=

1, we may assume that h > 1 ; thus,

By counting in two different ways the number of dark edges of R(Eo) that have exactly one direction, we obtain Since we have obtained two incompatible relations, the proof is achieved. Q.E.D. Corollary (Errera [1922]). I f G = (X, E ) i; a simple connectedgraph regular of degree 3 such that all the isthmi of G are on the same elementary chain, then G possesses a perfect matching. Recall that an “isthmus” is an edge [a, b] whose removal disconnects the graph. In a connected graph, the removal of an isthmus creates exactly two Connected components. Since 3 I X I = 2 1 E 1, the number of vertices of G is even. Consider the different connected components created by the removal of all of the isthmi. From the hypothesis, each of these connected components is joined to the rest of the graph G by one or two isthmi. Suppose a connected component Ci is joined to the rest of the graph by two isthmi [x, c] and [y, c’], where c, c’ E Ciand x, y $ C , . Consider the graph G, obtained from graph Gc, by joining vertices c and c‘. Graph G, is connected, regular of degree 3 and has no isthmi. Therefore, from Theorem 13, graph G, possesses a perfect matching that contains edge [c, c‘]. This matching corresponds to a matching Eo(Ct)of graph Gc, that saturates all vertices except c and c’. Suppose a connected component Djis joined to the rest of the graph by only one edge [d, x ] with d~ D,,x $ D,.Consider the graph Hiobtained from G,, by removing vertex d and by joining the vertices dl and d2 of D that are adjacent to vertex d in graph G. From Theorem 13, graph H , has a perfect matching that does not contain edge [d,, 41, and that corresponds in graph GD,to a matching Eo(D,) that saturates every vertex except d E D,. Thus, a perfect matching for G can be formed from the union of sets U Eo(C,),U Eo(D,), and the set of isthmi of graph G. Q.E.D.

I63

C-MATCHINGS

EXERCISE 1. Show the following result: Given a multigraph G = ( X , E ) , there exists a partial multigraph H with dH(xt)= d(i) if, and only if, for each pair S,T of disjoint sets, the number q(S, T )of components C of the subgraph of G generated by X S - T with rnc(C, T ) d ( C ) odd satisfies

-

q(S, T ) < d(S)- &T)

Hint: Apply Theorem 5 to the graph

+ dC x - s ( T ) .

+

G constructed from G as shown in Theorem 1. (W. T. Tutte [1954])

CHAPTER 9

Connectivity

1. &Connected graphs

The connectiviry ti(G) of a connected graph G is defined to be the minimum number of vertices whose removal disconnects G or reduces G to a single vertex. If G is not a clique, there exist two non-adjacent vertices a and 6 ; thus, X - { a, b } is a set whose removal disconnects G, and, consequently, K(G)

I= n - 2 .

On the other hand, if G is the n-clique K,,, we have K(K,) = n - 1 .

Graph G is said to be h-connected if its connectivity K(G) is 2 h. An articulation set of G is any set of vertices of G whose removal disconnects G. Let d denote the family of all articulation sets of G. A graph G # K,, is hconnected if, and only if,

K(G) = min I A I 2 h . A€ d

Thus G is h-connected i f , and only i f , (1) h < n - 1, and (2) there is no articulation set A of G with I A

I = h - 1.

Theorem 1. For a simple connected graph G, K ( G ) < min d,(x) . XEX

If G has n vertices, then K(G) < n - 1. Let xo be a vertex of minimum degree h, and suppose that h < K(G).The set Tc(xo)has cardinality h < n - 1. Thus G is not a clique, and K ( G ) = min I A A E J ~

I < I T,(x,) I64

1 = h < K(G),

165

CONNECTIVITY

which is a contradiction.

Q.E.D.

Theorem 2 (Harary [1962]). For each m, n with 0 Q n - 1

< in
h.

Thus, h < n - 1. Furthermore, G has no articulation set A with I A I < h because, otherwise, we could choose two vertices a and b in two distinct components of the subgraph generated by X - A ; each chain joining a and b passes through set A , but there cannot be h distinct vertices in A . Thus G is h-connected. Q.E.D.

Corollary 1. If G is h-connected, the partial graph obtained by remocing an edge is (h - 1)-connected. If two vertices can be joined by h vertex-disjoint chains in G, they can be joined by at least h - 1 vertex-disjoint chains in the new graph. Q.E.D.

Corollary 2. If G is h-connected, the subgraph obtained by removing a vertex is (h - I)-connected. If two vertices of the new graph can be joined by h vertex-disjoint chains in G, then they can be joined by at least h - 1 vertex-disjoint chains in the new graph. Q.E.D.

C0N N ECTI V I TY

169

Corollary 3. Let G be a simple h-connected graph. Let B = { b, ,b,, .. ., bh } be a set of vertices with I B I = h. I f a E X - B, there exist h vertex-disjoint elementary chains pi[a,b,]joining a and B. We may assume that none of the b, are adjacent to a because, if k vertices of B are adjacent to a, then h can be replaced by h - k in the theorem. Let graph G' be the graph obtained from graph G by adding a vertex z and joining it to each vertex b, E B. We shall show that graph G ' is h-connected. Let S be a subset of X with 1 S I < h - 1;we have then to show that the subgraph generated by X u { z } - S is connected. This is clearly true if z E S, since Gx-sis connected. If z $ S, then, since I S I < h, one of the bi does not belong to S, and since this bi is adjacent to z , the subgraph under consideration is connected. From the lemma, there exist, then, h vertex-disjoint chains in G ' between a and z. These chains induce in C the required chains p,[a, b,].

Q.E.D. Corollary 4. Let G be a simple h-connectedgraph with h 2 2. An elementary cycle passes through an arbitrary set of two edges el and e, and h - 2 vertices a13 a29 .-.)ah-2. 1. If h = 2, the graph G' obtained from G by adding vertices a and b in the middle of edges e, and e 2 , respectively, is again 2-connected because no vertex of G' can disconnect G'. From the Menger theorem, there exist two vertex-disjoint chains in G' that join a and b. These chains determine an elementary cycle passing through edges e , and e,, and the proof is achieved. 2. Suppose h > 2 and suppose that the theorem is true for all k-connected graphs with k < h. We shall show that it is also true for a given hconnected graph G. We may assume that a,, a,, ..., uh-, are not the endpoints of either el or e, since then the theorem would follow from the induction hypothesis. The subgraph obtained from G by removing vertex ah-, is (h - 1)connected, from Corollary 2. Hence by the induction hypothesis, there is a cycle p o passing through e l , e 2 ,a,, a,, ..., a h - 3 . Let B be the vertex set of cycle po. Clearly, I B I 2 h. From Corollary 3, vertex a,,-, and set B can be joined by h vertex-disjoint chains. Suppose that each of these chains encounters B at only one vertex (i.e. does not return to B). Denote these chains by

170

GRAPHS

Suppose, for example, that cycle p o encounters vertices b,, b,, ..., b,E B

in this order. Among the h segments of p o determined by two consecutive vertices of the sequence b,, b,, ..., bh,b,, there is at least one segment that does not contain in its interior any of the h - 1 elements a,, a2,..., ah -3, el, e,. For example, let po[bl,b,] be this segment. The cycle required for the proof is then :

C(Z[ah-2,621 + P o [ ~ z6h-J , + I*o[bh,1711 - pI[ah-2,b,]. Q.E.D.

Corollary 5 (Dirac [1960]).If G is a simple h-connected graph with h 2 2, then there is an elementary cycle passing through h arbitrary vertices. Select any set of h vertices a,, a2, ..., ah. Consider an edge [ah- ,, x] and an edge [ a h , y ] .From Corollary 4, there exists an elementary cycle passing through these two edges and through vertices a,, a,, ..., Therefore, there exists an elementary cycle containing vertices a,, a,, ...,a,, , Q.E.D. Theorem. (Halin [1969]). I f G is a simple h-connected graph, either there exists an edge such that the partial graph obtained by remoaing this edge is also h-connected or there exists a vertex of degree h. The proof, which is omitted, uses Theorem 3. For the cases h the theorem was first proved by Las Vergnas [1968].

=

3 and 4,

We shall now study conditions on the degrees of a graph that imply h-connectivity. The following result is a slight generalization of a result of Bondy [1969].

Theorem 4. Let G be a simple graph with n vertices x,,x,, . .., x, such that

&(xi) G dG(x2) 6 < dG(X,) = d . Let il < iz < be the sequence of indices i such that dG(xi)< i. For the I-th index of this sequence, let 1

=

[f c d,(x,,,] k=l

.

Then, the number p of connected components of G satisjies

Let c1

< cz < -.
I because, if p 6 I, the proposed inequality is obvious. Graph G contains at least c1 vertices of degree < c1 - 1, and c1 c2 c k vertices of degree < c k - 1. Hence,

+ +

a*.+

< ck -

dG(X,,+cl+...+Ck)

+ + + ck ( k = 192, P) numbers cl + c2 + .-.+ c k are all different. < c1

c2

*“

*

For k = I , 2, . . . , p , the ik, and so Hence, c1 c2 ck

+ +-+

Thus, from the definition of q(f), 1

1

dG(xiJ

2 lq(l) *

k=l

I, we have d,(x,,) 2 q(I). Hence, n 2 Iq(0 + ( p - 1 - 0 q ( 0 + d

Furthermore, k >

+ P,

and thus

n

+ q(l) - d q(1)

+1

’

which implies that

Q.E.D. EXAMPLE.Consider the graph G in Fig. 9.1 with the following degree sequence; 1, 1 , 3, 3, 4, 4, 4, 4,4. The vertices xi with &(xi) < i are circled.

For

I = 1, we have q(1) = 1, and pG[9+;-4~=3.

For 1 = 2, we have q(2) = 2, and

( [

p 6 max 2,

+

-

“1)

=2.

172

GRAPHS

Thus, Theorem 4 shows that G cannot have more than 2 components.

Fig. 9.1

Corollary 1 (Bondy [1969]). Let G be a simple graph with n uertices x l , x 2 , ...,x, such that dG(X1)

< dG(Xz) < < dG(Xn) = d . * * a

Ih for some integer q, dc(xk) > k

(k = 1 , 2,

..., 4).

then the number p of connected components of G satisfies

n+q-d p g

q + l

-

There exists an index i with dG(xi)< i because dG(x,) < n ; let il be the smallest such index. We have q(1) = dG(Xi,) 2 dG(XJ 2 4 . Hence

The proof follows by letting 1 = 1 in Theorem 4.

Q.E.D. The graph in Fig. 9.1 shows that Corollary 1 is weaker than Theorem 4. Corollary 2. Let G be a simple graph with n certices x l , x2, ..., x, such that dG(x1)

< dG(xz) < < dG(Xn) = d . *.*

IfdG(xk) > k for all k ,< n - d - I , then G is connected.

173

CONNECTIVITY

In Corollary 1 , let q=n-d-1.

We have

n +q - d=(q and, from Corollary 1,

+ d + 1) +

- d < 2(q

+ I),

Thus graph G is connected.

Q.E.D. Corollary 3 (Bondy [1969]). Let G be a simple graph with n vertices xl, x2, ..., x, such that &(xi) < dc(xz) < ... < dG(x,). I f f o r some integer h < n dG(xk)2 k f h - 1

(k

< n - dG(x,-h+l) - 1)

then G is h-connected.

Consider a set A C X with cardinality h - 1. We shall show that A cannot be an articulation set. Index the vertices XIof G’ = G X e A so that dG.(x;) < dG,(x;)
dG(xk) - 1 A I

2k

+ h - 1 - ( h - 1) = k .

Corollary 2 shows that Gx-a is connected. Since this is proved for each set A c X with cardinality h - 1, graph G is h-connected.

Q.E.D. Corollary 4 (Chartrand, Kapoor, Kronk [1968]). Let G be a simple graph of order n such that,,for some integer h < n, the,following two conditions hold: n - 11 (1) for each k < - , the number of certices with degree < k + h - 1 2 is less than k ;

174

GRAPHS

(2) the number of certices with degree
-and k < n - d G ( x , - , + , ) - 1 cannot be 2 satisfied simultaneously because this would imply that

and hence

which is a contradiction. Thus, if k

2- 1(p)

+ 2 + 21 / ( p ) + 2 - 4 = l ( p ) . -

For example, let v, be this cycle vi. Then, KV,)

< 51 K p ) + 2 < QP)

*

Consequently, v1 generates a clique, and there exists in p at least one odd chord. From Part (1) of the proof, p generates a clique. If the two chords do not cross one another, suppose, for example, ak E [a,, a,], and aj E [a,, a,], where k < j (see Fig. 9.5). Consider the cycle = lal,

ail

$- p[ai, akl

+ [%, aj] -k p [ a j ,

*

180

GRAPHS

Since cycle v is shorter than cycle p, it generates a clique, and cycle p contains an odd chord. From part (1) of the proof, p generates a clique. Hence, in all cases, p generates a clique. This contradicts the definition of p. Q.E.D.

Lemma 2. Ifeach even cycle has at least two chords, then an odd elementary cycle with at least one chord generates a clique. It is sufficient to show that if p = [a,, a,, ..., a,, a,] is an odd cycle and if a, - and [a,, a, + are also chords. It is

,]

[al, at]is one of its chords, then [a,, evident that one of the cycles ~1

= ~ [ a lail ,

+ [ai, all

or

~2

ail

=

+ ~ [ a ia11 ,

is even. Suppose p1 is even. From Lemma 1, we know that p, generates a clique and this implies that [a,, ai - E E. Since [a,, a, - ,] E E, consider the cycle

P’ =

at-11

+ p[ai-l,

011

*

This cycle is even, and so we know from Lemma 1 that [a,, a i + , ]E E. Q.E.D.

Theorem 7. (Dirac [1960]).If each e w n elementary cycle has at least two chords, then each block is either a clique or an odd cycle without chords. Let B be a block of G. If B is not a clique, then there exist two non-adjacent vertices b, and b, in B. Since G, is 2-connected, there is a cycle p o that contains 6 , and b,. This cycle p o is odd and has no chords (since, otherwise, 6 , and b, would be adjacent, from Lemmas 1 and 2). Let Bo c B denote the set of vertices on p,,. Suppose first that Bo # B. Let a E B - Bo. From Corollary 3 to Theorem 3, there are two vertex-disjoint elementary chains p,[a, x] and p,[a, y ] joining a and Bo, that contain no vertices of B, except x and y. One of the cycles plra, XI

+ p o [ x ,Yi - P2[a,Yi

& [ a ? XI

- Pob, X I

-

P2[Q,Y1

is even. Since this cycle generates a clique (from Lemma l), vertices x and y are adjacent. Since p , has no chords, x and y are consecutive vertices of the cycle. Therefore, the even cycle contains b, and b,. Thus b, and h, are adjacent, which is impossible. Thus Bo = B is an odd cycle without chords. Q.E.D.

181

CONNECTIVITY

3. k-Edge-connected graphs In this section we shall study, for a general graph G = ( X , U ) , the cardinality of a minimum cut between two vertices a and b, i.e. the number: cc+(u, b)

= min mc+(A,x - A) , A cX aEA

beX-A

where mb(A, B ) denotes the number of arcs going from A to B. We shall also study, for a multigraph G = (A’, E ) , the number: c,(a, b)

= min m,(A, X

- A) .

AcX asA beX-A

Theorem 8. 1. Let G = ( X , U)be a graph; then c; (a, b) equals the maximum number of arc-disjoint paths from a to b. 2. Let G = ( X , E ) be a multigraph; then c,(a, b) equals the maximum number of edge-disjoint chains from a to b. 1. If G = ( X , U)is a graph, form from G a transportation network R with a as a source CI and b as a sink. Let each arc U E U have capacity c(u) = 1, and add a return arc (b, a) = uo with infinite capacity. In network R, the maximum flow problem (Ch. 5, 5 1) is to find a flow cp, satisfying 0 < q(u) < 1 for all u in CJ, that maximizes the flow value in the return arc u,,. Theorem (1, Ch. 5 ) shows that

max rp(uo) = min r n , + ( ~X, 9

- A).

Ac X As0 A+b

The left side of this equality equals the maximum number of arc-disjoint paths from a to b in graph G. The right side is, by definition, c,+(a, b). This proves Part 1 of the theorem.

2. If G = ( X , E ) is a multigraph, consider the graph G* obtained from G by replacing each edge of G by two oppositely directed arcs. The maximum number of elementary chains in G between a and b with no common edge equals the maximum number of elementary paths in G* from a to b with no common arc. From part 1, this number equals &(a, b) = cJa, b) . Q.E.D.

182

GRAPHS

Nash-Williams Lemma. If G = ( X , E ) is a multigraph, it is always possible to construct a set E' of new edges that matches the certices of odd degree, and is such that graph G' = (A', E ' ) satisfies

Ajb

for all a E X , b E X . Let Y denote the set of vertices of odd degree, and let 2 denote the set of vertices of even degree; then

Thus 1 Y 1 is an even number, and it is always possible to match together the vertices of Y. The proof of this lemma is long, and the reader is referred to the original presentation (Nash-Williams [1960]).

A connected multigraph G is said to be k-edge-connected if it cannot be disconnected by the removal of less than k edges, i.e., if and only if c,(x, y ) 2 k for all x, y E X , x # y . A multigraph is I-edge-connected if, and only if, it is connected. A multigraph is 2-edge-connected if, and only if, it is connected and has no isthmus. The least k such that G is k-edge-connected is called the edge-connectivity. Lemma. In a connected multigraph G, an edge is an isthmus if, and only if, no elementary cycle of G contains this edge. If [a, b] is not an isthmus, the graph remains connected after the removal of [a,b] and there is an elementary chain p[a, b] of G that does not contain [a, b]. Thus, [a,b] together with p[a, b] form the required elementary cycle. Conversely, i f edge [a, b ] lies on a cycle, then the removal of [a, b] does not disconnect the graph, and so [a, b] cannot be an isthmus. Q.E.D.

Theorem 9. A connected multigraph is 2-edge-connected $, and only if: euery edge lies on a cycle. The proof follows from the lemma. Theorem 10 (Robbins [1941]). Given a simpre graph G, its edges can be and only if, G is 2-edgedirected to.form a strongly connected 1-graph H connected.

v,

1. If such a graph H exists, then G is 2-edge-connected because the removal of an edge cannot disconnect H and, consequently, cannot disconnect G.

I83

CONNECTIVITY

2. Conversely, if G is 2-edge-connected, each edge of G lies on an elementary cycle. Let A , denote the set of vertices on a first elementary cycle of G. Direct the edges of this cycle to form a circuit, and arbitrarily direct all the edges with both endpoints in Al. The subgraph generated by A , is strongly connected. If X = A,, the proof is achieved. If X - Al # 0, there is a vertex a2 4 A , that is adjacent to a vertex a, E A , (because the graph is connected). By hypothesis, edge [al,az]is on an elementary cycle that contains a chain leaving A , at vertex a, and returning to A , at some vertex b , , no edge of which is already directed. Direct this chain so that it becomes a path. Let A z denote set A , augmented by the vertices of this path. Arbitrarily direct all undirected edges with both endpoints in A z . The subgraph generated by A 2 is strongly connected. If X # A 2 , this procedure can be repeated, as many times as needed, to obtain a strongly connected graph H. Q.E.D.

Applications of this result to traffic problems are easily seen. If the streets of a city are represented by a graph, then the streets can be directed for one-way traffic without cutting off traffic between any two points, if, and only if, the graph is 2-edge-connected. The above theorem can be generalized: Theorem 11 (Nash-Williams [1960]). Theedgesof a multigraph G can be directed to form a graph H = ( X , U ) with

=

(X,E )

We may assume that multigraph G is connected; otherwise, each connected component could be considered separately.

CASE1. Each certex of G has euen degree. In this case, it is easy to see that there exists a cycle p that uses each edge exactly once. (This result, due to Euler, is one of the oldest results in graph theory; a proof is presented in Chapter 1 1, $1.) If the edges are directed along the direction of travel in p, then a graph H i s A # X,we have obtained. For each A c X,A #

a,

because a traveller through p exits set A as many times as he enters set A . Thus, rnG(A,

x - A ) = rn,+(A,x - A ) + m,(A, x - A ) = 2m,+(A,X - A ) .

184

GRAPHS

Consequently,

c i ( x , y) = min r n , f ( ~x,

- A ) = -1 min. r n , ( ~ , x - A ) = 31 c,(x,

A3x AlY

y> ,

A3x

A+Y

which establishes the theorem for this case.

CASE2. There exist uertices of odd degree. The proof for this case depends upon the Nash-Williams Lemma given above. Add to G a set E' of edges that matches the vertices of odd degree, as in the lemma. Let G G' = ( X , E u E'). Since each vertex of graph G + G' has even degree, its edges can be directed as shown in Case 1 to form a graph H H' = ( X , U u V ' ) where H = ( X , U ) is graph G = ( X , E ) with its edges directed, and H' = ( X , U ' ) is graph G' = ( X , E ' ) with its edges directed. If A is a set of vertices corresponding to a minimum cut in H between two given vertices a and 6, then

+

+

+ ( u , b) cH

=

m,+(A,x - A ) = rn,f+,.(A, x - A )

- rn,+,(A,X

- A).

From Case 1 and the Nash-Williams Lemma, we have

=

[ p1G ( a ,b ) ] . Q.E.D.

Note that if G is a simple 2-edge-connected graph, then CdX,

v) 2 2

(x, Y E

x ; x # u) .

Consequently there exists a graph H , with the same edges as G, such that c&,

y)

>1

(x, Y

EX; x # Y).

In other words, His strongly connected. This is another proof for Theorem 10.

CONNECTIVITY

185

EXERCISES 1. Show that a k-connected graph is k-edge-connected. 2. Show that a connected graph is 2-connected if, and only if, for every three vertices a, b, x , there exists an elementary chain p[u, b ] that contains x. 3. Let G be a simple regular graph of degree 3. Show that G is k-connected if, and only if, G is k-edge-connected. 4. Tutte has shown that: A graph G is 3-connected if, and only if, G is a “wheel” (an elementary cycle p and a vertex xo joined to each vertex in p) or can be formed from a wheel by a sequence of operations of the two following types: (1) the addition of a new edge, (2) the replacement of a vertex x of degree > 4 by two adjacent vertices x’ and x” of degrees 2 3 such that a neighbour of x in the original graph is a neighbour of exactly one of x‘ and x u in the new graph. Verify this result with some simple examples. (Tutte [1961]) 5. A simple 2-connected graph is defined to be rriininially 2-connected if the partial graph resulting from the removal of any edge is not 2-connected. If G is minimally 2-connected, show that: ( I ) No edge is the chord of a cycle. (2) If G # K:,, then G contains no triangles. (3) If [ x , y ] is an edge and z is an articulation vertex of G - [ x , y ] , then z lies on every cycle that contains [x, y ] . (4) Each chain that contains [x. y ] and z contains a vertex of degree 2 in its interior. ( 5 ) If G # K3, there is a cycle of G that contains two non-adjacent vertices of degree 2. (6) The set of all the vertices of degree 2 is an articulation set. ( M . Plummer [1968])

CHAPTER 10

Hamiltonian Cycles

1. Hamiltonian paths and circuits

In a graph G = (A', U),a hamiltonianpath is defined to be a path that meets every vertex exactly once. Similarly, a hamiltonian circuit is defined to be a circuit that passes through every vertex exactly once. In a simple graph G = (A', E ) , a hamiltonian chain and a hamiltonian cycle are defined similarly.

EXAMPLE1. Voyage around the world (Hamilton). Consider 20 cities ..., t , represented by the vertices of a regular dodecahedron (polyhedron with twelve pentagonal faces and 20 vertices). How can we travel to every city exactly once and return home using only the edges of the dodecahedron? In other words, can we find a hamiltonian cycle in the graph in Fig. 10.1? Hamilton solved this problem as follows. a When the traveller arrives at the endpoint of j an edge, he has the choice of taking the edge to S his right (denote this choice by R), or the edge r to his left (denote this choice by L), or by staying where he is (denote this choice by 1). In k the obvious way, define a product of these operations: for example, L2R denotes the Fig. 10.1 operation of going left twice and then right once. Finally, two operations are defined to be equal if after starting from the same vertex they terminate at the same vertex. Note that the product is not commutative (for example LR # RL), but it is associative (for example (LL)R = L(LR)). For the graph in Fig. 10.1, a, b, c,

@

,

R5 RL2 R LR2 L RL3 R LR3 L

= L' = I = LRL =

RLR

= L2

= 186

R2.

,

HAMILTONIAN CYCLES

187

Hence, 1 = R5 = R2 R3 = (LR3L ) R3 = (LR3)" = [L(LR3L ) RI2 = = (L2 R3 LR)" = [L2(LR3L ) RLR]" = [L3R3 LRLR]" = =LLLRRRLRLRLLLRRRLRLR. This sequence contains twenty operations and contains no partial sequence that equals 1. Therefore, it represents a hamiltonian cycle. Another hamiltonian cycle is obtained by reversing the sequence. It is left to the reader to verify that no other cycles exist. Note that the voyage around the world can begin (and terminate) at any of the twenty cities. Hamilton solved the problem with the additional constraint that the first five cities to be visited had to be a, 6, c, d, e in that order. He found four solutions, starting with edge [a, b ] : RLRLRLLLRRRLRLRLLLRR RLRLLLRRRLRLRLLLRRRL RLRLRRRLLLRLRLRRRLLL RLRRRLLLRLRLRRRLLLRL EXAMPLE 2. Open voyage around the world. Suppose that the traveller in Example 1 need not return home. In this case, each trip corresponds to a hamilton chain in-the graph in Fig. 10.1, and the number of distinct trips available to the traveller increases greatly. All the trips starting with an R are indicated below, All other possible trips can be obtained by interchanging the operations L and R in the sequences given below. RRRLRLRLLLRRRLRLRL RLRLLLRRRLRLRLLLRR RLRRRLLLRLRRRLLLRL RLLLRRRLRLRLLLRRRL RLRLLRRRLRLLLRRLRL RRRLLLRLRLRRRLLLRL RLRRRLLLRLRLRRRLLL RLRLRRRLLLRLRLRRRL RRRLRRLRLRLRLLRLLL RRLRLRLLLRRRLRLRLL RRRLLRLRRRLLLRLRLR RLRLRLLLRRRLRLRLLL RRRLRRLRLRLLLRRRLR RLLLRLRLRRRLLLRLRL RRLLLRLRLRRRLLLRLR RLRLRLLLRRRLRLLRRR RLLLRLRLLRRLRLRRRL RRRLLLRLRLRLRRLRRR RLRRRLLLRLRLRRLRRR RRRLRRLRLRLRLLLRRR RLRLLLRRLRLLRRRLRL RLRLRRRLLLRRLRLLLR RLLLRLRRLLLRRRLRLR RRRLLLRLRLRLRRRLLL RLLLRLRLLRRRLRLLLR RLLLRLRRRLLRLRLLLR RLLLRLRRRLLLRLRRRL

188

GRAPHS

EXAMPLE 3. Knight’s journey (Euler). How can a knight be moved on a chessboard so that he visits each square exactly once? This problem of finding a hamiltonian chain has interested many mathematicians: e.g. Euler, de Moivre, Vandermonde, etc. Many methods have been proposed. One method that seems to work in practice is the following: Move the knight to the square from which he will control the smallest number of unvisited squares.

Fig. 10.2. Examples of the knight’s journey

Another method is to move the knight on only half of the chessboard and then to have him repeat this pattern on the other half (see Fig. 10.2). This method depends on the special structure of the chessboard and does not work in more general situations. EXAMPLE 4. The problem of Mr. No. Mr. No, a mythical Japanese detective, lives in the upper left square of a chessboard. He wishes to visit Mr. Go who lives in the lower right square of the chessboard. Mr. No can move to any adjacent square but cannot move diagonally. Is it possible for him to visit each square of the chessboard exactly once en route to Mr. Go? Clearly, the problem of Mr. No is to find a hamiltonian chain between two

189

HAMILTONIAN CYCLES

given vertices. To visit each square exactly once, Mr. No must make 63 moves and 63 colour changes. Clearly, after 63 colour changes Mr. No stops on a square with a colour different from the colour of his home square. Since Mr. No and Mr. Go have home squares of the same colour, no such trip is possible.

Theorem 1. The number h(G) of hamiltonian paths in a 1-graph G = (X,U) has the same parity as the number of hamiltonian paths in the complementary 1-graph c = (X,X x X - U). We shall show that if G and G are two complementary 1-graphs, then h(G) = h(G) modulo 2. Suppose that the vertices of G are indexed from 1 to n. Given a subset V of the arcs of G, let h(V) denote the number of arrangements (il, iz, ..., in) of (1,2, ...,n)suchthateacharcof Visoftheform(ik,ik+l).Note that ifh(V) is not zero, then IVIGn-1, and the arcs of V form a family of pairwise disjoint paths. Furthermore, if h(V) # 0 and if I V I < n - 1, the connected components of the partial graph ( X , V ) consist of r disjoint paths, where r > 1, and consequently, h( V ) = r ! = 0. Hence, h(V)$O

*

(VI=n-l.

Now note that h(G) equals the number of arrangements ( i l , iz, ..., in) such that no arc of G is of the form ( i k , ik+l). Thus,

n ! - h(G) =

c

vc

h(V) il

c

h(Y)

+ .-.+

(-Qk-I

c h(V) +

**a

V C IJ

Therefore

h ( 5 ) = n ! - h(G) =

C

VCU I VI=n-l

Hence, h(c)= h(G) modulo 2.

h(V) = h(G) . Q.E.D.

Theorem 2 (C. A. B. Smith [1946]). 111 a simple regular graph of degree 3. the number of hamiltonian cycles that contain a given edge is even. Suppose’ that there exists a hamiltonian cycle (otherwise, the result is 3 trivially true). This cycle is of even length (since there are - n edges, which 2 implies that n is even). Thus the edges of the cycle can alternately be coloured with two colours c1 and /3 and the edges not on the cycle can be coloured y.

190

GRAPHS

Thus, there exists a partition { E,, E 8 , E, } of the edge set into three perfect matchings. Such a partition will be called here a 3-colouration of the edges. A 3-colouration of the edges with three colours a, fi, y determines a vector pa, = (pl,p2, ..., p") with p i = 1 if edge e, is c( or fi, and pi = 0 otherwise. Let p E M if vector p defines a family of vertex-disjoint even cycles that use all vertices. Each 3-colouration { E,, E,, E,} determines three vectors pa87 p a y , BY M , with (mod. 2). Pub + P U , + PbLpV 3 0 Furthermore, if a vector p E M consists of a family of k(p) pairwise disjoint cycles, then it corresponds to 2k(u)-1distinct 3-colourations. Summing the above identity over all 3-colourations { E,, E,, E, } gives

C

2k(p)-1

=o

(mod. 2).

p=o

(mod. 2).

p

lc=M

Hence,

C p

I k(d=

1

Since this sum is over all the hamiltonian cycles of G, the number of distinct hamiltonian cycles that contain a given edge is even. Q.E.D. Corollary 1. I f a simple regular graph of degree 3 has a hamiltonian cycle, it has at least three hamiltonian cycles.

Let e be an edge of the hamiltonian cycle. At least two distinct hamiltonian cycles p, and p2 contain e. Let x denote the first vertex at which these two cycles diverge after passing through e. If [x, y] is an edge of p, and not of p2, then at least two hamiltonian cycles pass through [x, y ] , and they are both different from p z . Q.E.D. Corollary 2 (N. J. A. Sloane [1969]). I f a graph G = ( X , E ) has two hamiltonian cycles without common edges, it has at least three hamiltonian cycles. Let p1 = [ x , , x z , ..., x,, x,] be a hamiltonian cycle of G , and let p2 be another hamiltonian cycle of G that does not contain any edge of pl. CASE1. Suppose n is even. A regular graph of degree 3 can be formed with all the edges of p 1 and some edges of p z . Corollary I established that this graph has three hamiltonian cycles. Therefore, the result is true for G. CASE2. Suppose n is odd and [xt, x t + J E E for all i. Then, there exists a third hamiltonian cycle, namely: p3

= [XI,X " ,

x2,

%,x4,

' S . 2

&-l, X l l

*

191

HAMILTONIAN CYCLES

CASE3. Suppose n is odd and [x,, xz] 4 E. Suppose that p2

=

[xi,

xj,xkl,

xkz7

* * * Y

x k , - , ~ xi]

i,j f

*

Form a new graph G by adding edge [x,. xz] and by removing vertex x, and all edges incident to xl. Graph has a hamiltonian cycle F1 = [xz,x3, ..., x,, xz]. The edges of cycle ,iil and the edges [xi,

Xkll,

[xki,

xkj],

Cxk4, x k s l ,

xi]

[xk,-j,

Y

form a regular graph of degree 3 that has two hamiltonian cycles that use edge [x,, x,] by Theorem 2. Therefore G possesses two hamiltonian cycles distinct from p z . Q.E.D. Theorem 3 (Boshk [1967]). A simple regular bipartite graph of degree 3 has an even number of hamiltonian cycles. Let G = ( X , Y, E ) be a regular bipartite graph of degree 3 with 1 X 1 = 1 Y 1 = 22’ and let h(G) denote the number of hamiltonian cycles in G. For n < 6 , the theorem is true since the complete bipartite graph K 3 , , contains exactly 6 hamiltonian cycles. Assume that the theorem is true for graphs of order < n ; we shall show that it is also true for a graph G = ( X , Y, E ) of order n > 6. Since G contains no triangles, an edge [x,,y,] is adjacent to four edges [xl,yzl, Ixl,y31, [yll~21,[Y,, x31such that the vertices xl, x z I x3,Y 1, y z , ~3 are all distinct. Let G’ and G” denote the two graphs obtained from G by the transformations shown in Fig. 10.3. C

.

G‘

G”

Y2

u n

Y3

x3

‘>( x3

Y3

Fig. 10.3

Let h,, h; and hl denote the number of distinct hamiltonian cycles in the graphs G, G’, and G ”using the vertices x, , xz, x3,y , ,y z , y , as shown in Fig. 10.4. From the induction hypothesis,

+ h; + h i + hi + hi = 0 h(G”) = h’; + hi + h’j + hi + h’; E 0 h(G’) = hi

(mod. 2) (mod. 2)

.

192

GRAPHS

Furthermore, it is evident that hi = h i , hz = h ; ,

hs = h:, Hence, h(G) = hi

h6

= h;,

hj

= hj,

hi = h i ,

h,+ =

hi

h; = h ; .

+ hz f + h, + h~ -k = h'; hi + hi + hi 4- h; + h; = h'; + h; + hi + h i + h'; + h; + (hi + h;) + (hi + h;)

=

h3

h;

+

h;

=0

h6

(mod. 2 ) .

Therefore, the number of hamiltonian cycles in G is even.

Q.E.D.

2. Hamiltonian paths in complete graphs

Theorem 4 (Camion [1959]). IfC = (A', f)is a strongly connected, complete 1-graph, then G has a hamiltonian circuit.

HAMILTONIAN CYCLES

193

Let p = [a,, a,, ..., ah- 1, ah,ah+, = a,] be a circuit of maximum length h. Suppose that circuit p does not encounter some vertex b. Then, u i Er ( b ) 3 b #T(ai-l) * ui-l E T ( ~ ) bET(a3 3 ai+1 #T(b) * b eT(at+l) U

Fig. 10.5

Therefore, the vertices not lying on circuit ,u can be divided into two classes B, and B, as follows: If b E B , , then each arc joining b and p is directed towards p . If b E B,, then each arc joining b and p is directed towards 6 . By hypothesis, B, u B2 # $3. Since G is strongly connected, B1 # $3, B, # $3,and there exists an arc from Bz to B,. Denote this arc by ( b 2 ,b,), where b, E Bz and b, E B,. Thus the circuit [al,a,, ...,a h ,b ,b,, all is longer than p, which contradicts the maximality of p . Q.E.D. Corollary. Let G = ( X , r) be a complete 1-graph. There exists a t’ertex xo such that for each vertex y # xo there is a path from xo to y . Such a vertex xo is calleda “root” of G. Each root is the initial endpoint of a harniltonianpath of G.

,

1. The existence of a root of G was established in Theorem (4, Ch. 4), since any centre of G is a root. 2. Let xo be a root of G. Consider the graph G’ obtained from graph G by adding a vertex z, an arc (z, xo), and the arc ( x , z ) for every x # x o . By Theorem 4, graph G’, which is complete and strongly connected, has a hamiltonian circuit, which corresponds in G to a hamiltonian path starting from x o . Q.E.D. Algorithm. This corollary provides a very simple algorithm to construct

a hamiltonian path in a complete 1-graph G = ( X , r)without loops:

194

GRAPHS

Let a, denote the vertex with the largest outer demi-degree d&J. Vertex a, is a centre (Theorem 4, Ch. 4). In the subgraph generated by F(al) - { a1 }, let a2 denote the vertex with the largest outer demi-degree. In the subgraph generated by r(a,) - { a , , a2 }, let a3 denote the vertex with the largest outer demi-degree, etc. Then [a,, a z , ...,] is a hamiltonian path. Theorem 5. I f G = (X,r)is a complete, anti-symmetric, transitive 1-graph, then G has exactIy one hamiltonian path. A hamiltonian path must start from a root. Since G is transitive, vertex x , is a root if, and only if,

r(x,)u(x,}=x. There is always a root in G, and since G is anti-symmetric, the above equation shows that the root is unique. Thus, the first vertex of a hamiltonian path must necessarily be this root xl. By the same argument, the second vertex of the hamiltonian path is the unique root of the subgraph generated by X - { x1 }, etc. ..., and the hamiltonian path is unique. Q.E.D. Theorem 6 (Ridei [1934]). I f C is a complete, anti-symmetric I-graph, then the number h(G) of distinct hamiltonian paths in G is odd. We shall show that if G is a complete, anti-symmetric graph, then reversing the direction of an arc does not change the parity of h(G). Since graph G can be obtained from a complete, anti-symmetric, transitive graph by making successive reversals of arc directions, this fact will establish the theorem. Let (a, b) be an arc of G , and let G ' be the graph obtained from G by reversing the direction of (a, 6). Let G, be the graph obtained from G by adding arc (b, a). Let G2 be the graph obtained from G by removing arc (a, b). We want to show that

h(G) G

= h(G')

G'

Fig. 10.6

(mod. 2 ) .

195

HAMILTONIAN CYCLES

Since graph G, is anti-symmetric, each hamiltonian path of G, (taken in its reverse direction) defines a hamiltonian path of GI,the complementary l-graph of G,, and vice versa. Therefore (from Theorem l), we have h(G2) = h(??~)= h(Gl)

(mod. 2)

.

Let h,(ab), h,(ba), h,(O) denote respectively the number of hamiltonian paths of G, that contain arc ab, arc ba, neither arc ab nor ba. Then,

hl(0) = h(G2) = h(G,) = h,(O) Thus, h,(ab)

= h,(ba) modulo 2, h(G) = h,(O)

+ h,(ab) + hl(ba) .

and consequently,

+ hl(ab) = hl(0) + hl(ba) = h(G’).

This shows that h(G) = h(G’) modulo 2, and completes the proof. Q.E.D.

3. Existence theorems for hamiltonian circuits

Clearly, the greater the demi-degrees d,+(x)and d;(x) of a 1-graph G are, the greater are the chances that G has a hamiltonian circuit. In this section, we present a theorem of Ghouila-Houri [1960] that gives conditions for the existence of a hamiltonian circuit in terms of the demi-degrees of the graph. First, a lemma is needed:

Lemma. Consider a circuit with m tlertices, each labelled with either a cross or a circle. suppose it contains exactly no # 0 circles and exactly n, # 0 sequences of q consecutive crosses. Then no+nl<m-q+l. Suppose that there are p maximal sequences of crosses of cardinality 2 q. Each such sequence is framed by two circles. By hypothesis, p 2 1. Let c denote the total number of crosses. Let mi denote the total number of sequences of q crosses contained in the i-th maximal sequence. Then the length of this maximal sequence is q a i - 1. Hence c 2 (q

+ al

+ - 1) + ( q + a,

- 1)

+ -.. + (q + ap - 1) 2 2 q + (a1 + a2 + ..* + ap) - 1 .

Hence, no+nl=m-c

+

P

Ca,<m-q+l. i= 1

Q.E.D.

196

GRAPHS

Theorem 7 (Ghouila-Houri [1960]). Let G be a strongly connected 1-graph for each vertex x, d,+(x) d,(x) z n , then G has a hamiltonian circuit. Let n 2 3. Suppose that the result is true for any graph of order < n, and that there exists a graph G = ( X , r)of order I X I = n for which the theorem fails. We shall show that this leads to a contradiction.

of order n and uithout loops.

v,

+

By hypothesis,

pm1+ pm1

- 1x1 > 0 ( X E X ) . (i) Let p = [ x o , x l , x,, ..., x , , - ~ ,x,] be an elementary circuit of G of maximum length. Since G is strongly connected, h 2 2, and since there are no hamiltonian circuits, m < n. Let Xo = { xo, x1, x2, x,-11. Let X I , X,, ..., X , denote the strongly connected components of the subgraph generated by X - X,. 1. We shalI show that each subgraph G,, , G,, , ...,Gx,contains a hamiltonian circuit. Let X E Xi, 1 < i 6 p . Then, f o r k < m, ..a,

xk

ri(x)

x&+1 6

fi(x)

(Otherwise, the circuit p could be lengthened.) Hence,

1~

( xn)x0 I Q I X , I -

I r c fn~x0 1 .

ForyEX,,j#i,O,

Y Erdx) * Y $T,+(x) (because X, is a strongly connected component different from Xi). Thus, for each j 2 0, j # i, rii(x) n X , Q I X j I - r , + ( x ) n X j 1 *

I

1

I

xo Fig.'l0.7

197

HAMILTONIAN CYCLES

Combining this result with (i) yields

1 T,+(x)n xi I + I

2

I - I Xi I >

nXi

G(X)

-

C (I

G ( x ) nX j

j#i

I

+I~

( xn )

xiI - I X , I) 2 o

Since this result is valid for each x E Xi, the subgraph Gxl has a hamiltonian circuit by virtue of the inductive hypothesis. 2. We shall show that there exists an X i , i # 0, such that and

r,+(Xo) n X i # fa

G ( X o ) n Xi # fa. It suffices to show that there exists an Xithat is joined in both directions to X,. Since graph G is strongly connected, there exists at least one X,such that

mxo, x,# 0.

Consider a vertex x directions to X,,then

+ X,

f-l

U

X , . If xk E X,,and if no Xiis joined in both

rJ(xk)

=>

x $ri(xk).

Hence,

I T,+(x,)

n ( X - xo - X,)

IG

< Ix -xO -xjI - I r i ( x k ) n (x-xO

If y

E

X,,then for each x + X, u X,, xErJ(Y)

*

-xj)I.

x$ri(Y)

Hence,

Im

y)

(7 ( X

- xo - X,)

Finally, for each y (ii)

I G(Y) n

(10

2

-

Ix,

E

I< I + ~ ( y n)(x, u x,) I 2

I

I

U X , ~- (X - x0 - X,)n ~ , + ( y )-

I (x- x0 - X,)

nri(y)

1 + I x - x0 - xiI 2 I x0 u X , I .

Furthermore, since G is strongly connected, there exists a path from X , to

X, of the form [z,, zl,..., z ~ - z,], ~ , with zl,z2,..., z t d l $ X o u X,.Since X,n T;(X,) = 0, this path has length > 1. Consider the subgraph generated by X, u X,together with the arc (z,, zt). Since this graph is strongly connected and has fewer vertices than G, inequality (2) shows that it contains a hamiltonian circuit. This circuit necessarily contains arc (z, , zt). By replacing arc (z,, zt) with the path z,, zl,. .., zt , we obtain a circuit of G that is longer than p, which is a contradiction.

198

GRAPHS

Hence, there is a component Xi with the required property. Denote this component by X , . 3. Here we shall show that each y E X , satisfies T,+(y)n Xo # @ and G ( Y ) n xo z 0. We may assume 1 X , I > 1, because, if I X , I = 1, this is obviously true. Let yo be a vertex of X , that does not satisfy, for example, G(Y0)

n xo f:

a.

Let [ y o , yl, ..., yp-l, y o ] be a hamiltonian circuit of XI with length q, such a circuit exists.) Lets be the smallest index such that r ; ( y g ) n Xo # 0.From the definition of X , , such an s exists, and s # 0. 1

< q < m. (From part (l),

1 .'!,

Xm-YO -1

--4--

XO

YO

s-1

XI Fig. 10.8

Since graph G has no circuit of length > m, xk E T;(y,) implies that x ~ + ~ , XktZ, x k t q 4 ri(yS-l). Thus,

I G~Y,-A n x0I Q m - 4 .

For i # 0, 1,

~ ~ ~ ( y s - l ) n x i ~ + ~ ~ -~ (1 y~S ~ l1) n ~ x0i ~ since there are no double edges between y S - , and Xi. Hence,

199

HAMILTONIAN CYCLES

which contradicts (i). Hence r a y ) n x0 z By a similar argument,

o

G ( y ) n .U, # la 4. We shall show that for each y , E XI,

1 GAYJ

n

(M).

(iEI).

xoI + I r2(yS-dn xo I < m

-q

+1.

In fact, X0 is a circuit of m vertices that can be labelled with circles and crosses in the following way: If x, E ~ Z ( Y ~ -label ~ ) ,x j with a circle.Otherwise, label x, with a cross. We have seen in Part 3 that x k E ri (y,) implies the existence of a sequence Xk-Cl, X k f 2 ,

Xk+q

of q crosses. From 3, there are no # 0 circles and n, # 0 sequences of q crosses. From the lemma,

I G(YJ

n

xo I + I rc+(yS-l) n x0 I < nl + no < m - q + 1

5. We shall show that there exists a vertex y, E Xl with

I MYJ

n

xo I + I C ( y s - l ) n x0 ] z

m

-4

+2.

From inequality (i), each vertex y E X , satisfies

I G ( Y ) n xo I + I m y ) n xo I a Z I Xo I - (I r,'cv> n Xl I + I Gcv)

- I xi I) - C (I ~ C + ( Y ) n xj I + I T ~ ( . Yn)x j I - I Xj I) 2 j+o,1

z

m - [(4

A', I

- 1) + (q - 1) - q] - o = m

-4

+2 .

By counting in two different ways the number of arcs joining X,, and XI,we obtain

Therefore, there exists at least one ys that satisfies the above inequality. This contradicts Part 4. Q.E.D. Corollary 1. I f C = (A',r) is a 1-graph without loops much that &(x) dD(x) 2 n - 1 (XEX) then G has a hamiltonian path.

+

9

200

GRAPHS

Add to G a vertex xo and join it to each other vertex by two oppositely directed arcs. The new graph G' is strongly connected, and d&(x)

+ d,.(x)

3 n-1

+ 2 = n'

(x E X ' ) ,

Graph G' has a hamiltonian circuit .from Theorem 7. Thus graph G has a hamiltonian path. Q.E.D. Corollary 2. If G is a complete I-graph, then G has a harniltonian path. (This result follows also from the corollary to Theorem 4.)

Corollary 3. Let G = ( X , r) be a strongly connected I-graph of order n without loops. If the graph remains strongly connected afier the removal of any vertex, and if d i ( x ) -t d,(x) 2 n

+1

(XEX),

then for each pair a, b of distinct vertices, there exists in G a hamiltonian path with endpoints a and b.

Consider the graph G' obtained from G by contracting { a, b } into a single vertex c and by letting T$(c) = T:(b) - { a } , f&)

=

f,-(a) - { b } .

Graph G' is strongly connected because for x # a, b graph G has a path from x to a that avoids b, and a path from b to x that avoids a. Thus, G ' has a path from x to c, and a path from c to x. These properties hold also for the graph G" obtained from G by contracting

- 1 + I TJb) Thus at least one of the following two inequalities is satisfied:

c )

1 + I I--&) I 2 I X' I ,

G ( C )

I + I TG,(C) I 2 I X ' 1 .

Im I

-1

20 1

HAMILTONIAN CYCLES

Suppose, for example, that the first inequality is satisfied. Then graph G ‘ satisfies dZ,(x)

+ d&X)

2 I x I - 1 = I X’ I

(x EX’)

.

Thus, from Theorem 7, G‘ has a hamiltonian circuit that corresponds in G to a hamiltonian path between a and b. Q.E.D.

Corollary 4 (Nash-Williams [1969]). Zf G loops, of order n, such that

=

( X , r) is a I-graph without

then G has a hamiltonian circuit.

From the Ghouila-Houri theorem, it sufKces to show that G is strongly connected. More precisely, we shall show that a 1-graph G without loops with

is strongly connected. Suppose that G were not strongly connected. Then there exist in G several strongly connected components Xl, X,, ..., X,,and their contraction yields a graph without circuits (Theorem 12, Ch. 3). There exists at least one connected component XI with &(XI, X - Xl) = 0, and one connected component X, with m $ ( X - X,, X z ) = 0. Suppose, for example, that I XI I 2 I X z I. If xo E X z , then

This contradicts the hypothesis.

Q.E.D. Corollary 5 (Bermond [1970]). .ISG = ( X , ZJ is a 1-graph of order n without loops such that

where k is an integer, 0 < k < n - 1, then each elementary path of length k is contained in a hamiltonian circuit.

Let p o

= [ao,a,,

..., a], be a path of length k in G. Let A = {ao,a,,

..., a k } .

202

GRAPHS

Construct a 1-graph Gofrom G by removing A , adding an auxiliary vertex a, adding an arc ( x , a) for each x E X - A with x E r;(ao), and adding an arc (a, y ) for each y E X - A with y E r z ( a k ) . It suffices to show that Gohas a hamiltonian circuit. Graph Go has no = n - k vertices. For a vertex x # a, d:o(x) = dGo(X] =

&(X)

rn;(x, A

- &(X,

A

- { a, }) 2 &(x) - k , - { ak }) 2 &(X) - k .

Furthermore,

dio(a) = d;(ak) - m i (ak, A - { ak }) 2 d:(Uk) dGo(a) = dF(ak) - mF(ak, A - { ak }) 2 d,(ak) Thus, for each vertex x of graph Go,

and

d&)

2

-k -k

9

n $ .

Hence, from Corollary 4, Go has a hamiltonian circuit, that corresponds in G to a hamiltonian circuit containing ,uo. Q.E.D. Corollary 6. I f G

=

( X , r)is a 1-graph without loops of order n such that

then for each pair a, b of distinct vertices there exists a hamiltonian path from a to b. If a E T(b),let G’ = G; otherwise let G’ = G + (b, a). Graph G’ satisfies the above conditions. Therefore, from Corollary 5 , G ” has a hamiltonian circuit containing (b, a) that corresponds in G to a hamiltonian path from a to b. Q.E.D. The following two conjectures are due to Nash-Williams [1969]. Conjecture 1. If G is a 1-graph without loops of order n > 5, such that for all x E X , n n d,‘(x) Z and dG(x) 2 2 then G has two arc disjoint arc-disjoint hamiltonian circuits.

HAMILTONIAN CYCLES

203

If G is a 1-graph without loops of order n 2 5 such that for all (x) < k is < k, and the number of with d g ( y ) < k is < k , then G has a hamiltonian circuit.

Conjecture 2.

n the number of vertices x with d$ k < ?, certices y

When sufficient conditions for the existence of a hamiltonian circuit are difficult to find, necessary conditions can be found by using the concept of a dissection. A dissection of graph G is defined to be a set of elementary paths of G such that each vertex of the graph is contained in exactly one path. Call a path closed if it also defines a circuit, otherwise call it open. The value of a dissection is defined to be the number of open paths in the dissection. The value is always less than or equal to the number n of vertices since a single vertex is a path of length 0. A hamiltonian path is a dissection of value 1. A hamiltonian circuit is a dissection of value 0.

Dissection theorem. For a 1-graph G = ( X , f), the minimum calue of a dissection equals a,, = rnax (i s I - I r ( ~ I).) S C X

Consider a dissection a = (d, a2, ..., a”, B1, B2, ..., Bq) of value q, where CI‘ is a closed path, and B’ is an open path. Let A‘ denote the set of all vertices a‘, and let B’ denote the set of all vertices in PI. Make two copies Xand Xof X , and form the bipartite graph H = ( X , 1, E) where ( x i , 2,) E E if, and only if, x, E f ( x , ) . Each circuit CI‘ defines uniquely in H a matching of I A‘ I edges, and each open path defines in H a matching of I B j I - 1 edges. Therefore the dissection defines a matching Eo of cardinality IEOI = C I A ‘ l + C ( I B j I - 1 ) . i

i

Thus n = ~ I A ’ I + ~ I B ’ I = ~ I A+ixI ( I B ’ I - l ) + q = I E o 1 + 4 . i

i

i

i

This correspondence between the matchings and the dissections is a bijection. Therefore, from the Konig theorem (Chapter 7), the minimum value of a dissection equals n - max I Eo I = n - (n - 6,) = 6, . This gives the required formula. Q.E.D. Corollary 1. I f C has a hamiltonian circuit, then 6 , = 0. The proof is immediate.

204

Corollary 2.

GRAPHS

IJ’ G has a hamiltonian path, then 0 < do 6

1.

The proof is immediate. Corollary 3. IJ’ a I-graph has no circuits, then a necessary and suficient condition for the existence of a haniiltonian path is that 6, = 1. The proof is immediate.

4. Existence theorems for hamiltonian cycles

Without loss of generality, we shall assume throughout this section that G is a simple graph. The main existence theorem for a hamiltonian cycle (P6sa [1962]) was a generalization of a previous result of Dirac [1952]. A new proof for P6sa’s theorem, due to Nash-Williams, permitted J. A. Bondy [1969] to give a stronger result. Finally, Chvital [ 19721 has proved an even stronger theorem, that is in some sense the best possible. The following theorem is an extension of Chvital’s theorem that has been modified to generalize a result of Kronk [19691. Theorem 8. Let G = ( X , E ) be a simple graph of order n with degrees dl < d2 < ... < d,. Let q be an integer, 0 < q < 11 - 3. I f , for erery k with q < k < 3 (n + q), the following condition holds:

(A)

dk-q < k

* dn-k 2 n

-k +q

then for each subset F of edges with I F I = q such that the connected components of ( X , F ) are elementary chains, there exists a hamiltonian cycle of G that contains F. Furthermore, this result is the best possible in the following sense: each sequence of degrees that does not satisfy condition (A) is majorized by a sequence of degrees of a graph that does not haoe the desired property. 1. We shall first show that condition (A) implies: dl > q. Suppose that dl 6 q, and let k q
k-q *

*

1Sk12k-q

I SA-k+q-l I

0*

[~i,ynl$E

[yl, ~ 2 ..., , yi, Y n , Y n - 1 , ..*> Yi+l, ~ would be a hamiltonian cycle containing F.

1 1

206

GRAPHS

4. Let k = dG(y,). We shall show that

Since dl > q, the first inequality of (D) holds. From (C), it follows that ~G(Y,>= ( X - {YnH

-(Yi/i~Il.

Hence,

< ( n - 1) - [ ~ G ( Y-I )41 -

&(yn)

Thus ~ G ( Y ,+ ) dG(yn)G n

(El

+ q - 1.

Finally,

5. From (C), we know that, for each i E I, [Yi,Ynl # E

and ~G(Y,)+ d ~ ( ~ k, then G has a hamiltonian cycle.

2

Since k

n-1. k , the number of vertices of degree < k is

2

< k. From Corollary 7, G contains a hamiltonian cycle. Q.E.D. Corollary 9 (Erdos, Gallai [1959]). Let G be a simple graph of order n 2 3.

Let x1 be the vertex of minimum degree. IfdG(xl) 2 2, and i f d G ( x )2 ?for all 2 x # x l , then G has a hatniltonian cycle.

If the vertices are indexed as in Corollary 1, we have dG(xl) > 1. For n n 1 < k < 1, we have d,(x,) 2 - > k. From Corollary 8, there exists a hamil2 2 tonian cycle. Q.E.D. Corollary 10 (Ore [1961]).Let G = (A', E ) be a simplegraph of order n 2 3, such that 44x1 + d&) < n 5 [x,Y ] E E -

Then G has a hamiltonian cycle. n Let k From Corollary 2, it suffices to show that the set 2' { x / x E X,d,(x) < k } has cardinality < k. an, 61, at-,, bt-1, ...)bz, a11 would be a hamiltonian cycle containing F. Thus TG(bl)c X - { a i - , / i € Z } and so

dG(bd

< n - I 1 < n - dG(al)+ 4 ,

or

dAad + dAb1) < n + q . (B) 3. Letj(a,) = s, k(b,) = t , so that a, = xs,b, = y t . If i E I, it follows from Part 2 that [ai- b,] $ E, and, therefore, Hence Thus there are at least I I I vertices x withj(x) s 2

1I I

< s, and, therefore,

2 dc(al)- q .

Hence

dG(X8) ,i s

+q.

dG(Yd G t

+4

By the same argument, *

Since G satisfies condition (A), and since [x,, y t ]$ E, it follows that

ddx,)

+ d&t)

3n

+q + 1.

This contradicts (B).

Q.E.D.

Corollary 1 (Bondy [1969]). Let G = (X,Y, E ) be a simple bipartite graph with 1 X I = 1 Y I = n 2 2. r f the vertices of X and of Y are indexed with increasing degrees, and if

d~(xi>< j , MY& dk

&(xi)

+ d&d

2n

then G has a hamiltonian cycle, The corollary is established by setting q = 0 in Theorem 11.

+ 1,

216

I

GRAPHS

Corollary 2. Let G = (A', Y, E ) be a simple bipartite graph with I X I = Y I = n > 2, and let q be an integer, where 0 Q q < n, such that (1) for each integer j Q 4 , the set S, = { x / x E X , d G ( x )< j ) has +

2 cardinality < j - q, (2) for each integer k < -t- ' , t h e set Tk = { y / y € Y,dG(y)6 k } has 2 cardinality < k - q. Then for each set F of q edges that forms vertex disjoint elementary chains, there is a hamiltonian cycle that contains F. This follows easily from Theorem 1 1 . Corollary 3 (Moon, Moser [1963]). Let G = ( X , Y, E ) be a simple bipartite graph with I XI = I Y I = n 2 2. r f l s k 1 < kand I Tk I < k foreachinteger n k < * Then G has a hamiltonian cycle. 2 The corollary is established by letting q = 0 in Corollary 2.

-

Corollary 4. Let G = (X,Y, E ) be a simple bipartite graph with I X I = n 1 Y 1 = n 2 2. r f each certex z E X v Y satis$es dG(z) > , except for two 2 vertices, x1 E X with dG(xl) 2 2, and y , E Y with dG(y,) Z 2, then G has a hamiltonian cycle.

If k = 1, then I f k > I andk

I sk I = 0 < k ; I T k I n 0. Since the covering M’ is minimum, each of these vertices is the endpoint of exactly one path of

M’. 5. Thus Vertex a, is the endpoint of a chain p’[ac,zt] E M’.Suppose that x E $[ai, zJ. Let y: denote the next to the last vertex encountered in the partial chain ,u’[u,, XI.Since vertex y: is necessarily a b,, it follows that y: is the endpoint of a chain p’[y:, z:] E M’. If x E p ’ b t , z:], define similarly a vertex y:, etc. In this way, a finite sequence 0

ai=Yi,

1 2 Yi, Yi,***

is determined for each i < k. It is easy to see that yf’ = yl implies p and i = j . Furthermore, the set Y of all yf (for all i and all p ) satisfies

{

/i - , the number k = minxex dG(x) can be written as

k=[i]+r

,

r>O ,

where

Let x, be a vertex of degree k 2 2 r. We shall show that the subgraph generated by T,(xl) contains a matching of cardinality r. In fact, if this subgraph contains a matching of only r - 1 edges bl,yzl, Iy3, ~ 4 1 ,..., L V Z-~3, Yzr - 21, vertex YZ,- 1 is not adjacent to ~ G ( X I ) {yl,..- 3 Yzr - 21. Thus,

< (2 r - 2) + n - k = =2r -2

+n

-[;I-

r= n-

[;I+).-

2
0 1 C, - C, I = q - &(a) + 1 2 h - dc(a) + 1 > 0. Thus, let a E C, - cb,/3 [a, allo with colour u.

E

Cb - C,. Let a, be vertex joined to a by an edge

2. We shall show that C,, 2 C,, - C., The connected components of the partial graph G(a, B) generated by colours a and p consist of either isolated vertices or alternating even cycles coloured u and 8, or alternating open chains coloured a and B. Since a E C, - cb, p E cb - C,, the connected component of G(a, p) that contains vertex a is neither an isolated vertex nor an alternating cycle. Therefore, it is an open chain with endpoint a that contains a l . Furthermore, b is the last vertex of this alternating chain because, otherwise, by interchanging colours a and B along this chain, colour a could be removed from set C, and [a, blO could be recoloured a. Thus p E C, for each E c b - C,. Hence

ca,I3 Cb - c,.

262

GRAPHS

3. We shall show that C,, 3 C, - C,.Suppose that this is not so. Let tl' E C, - C, and a' $ Gal. Let a2 be the vertex joined to a by an edge with colour a'. We see, as in Part 2, that the connected component of the partial graph

0

Fig. 12.4

G(a', j?) that contains a is an alternating chain with endpoints a and b. Thus the connected component of G(a', j?) that contains a, is an alternating open chain p [ a l ,z ] that contains neither a nor 6. Clearly, z # a , , because j? E C,, . Colours a' and j? can be interchanged along chain p [ a l , z]. Then [a, allo can be recoloured with j?, and [a, bI0 can be recoloured with a, which contradicts q(G) = q 1. Hence

+

c,,

3 Cb

- c,.

4. From Parts 2 and 3, cu,

= (cu

-

cb)

(cb

-

c4)

*

Hence dG(al)

=

Ico, I > I

c4

- cb I + I cb

2 (4 - dG(b) + 1)

- ca I 2

+ (4 - dG(a) + 1)

Thus

4

1

+ < 2 ( d G ( a l ) + dG(a) + dG(b))

Y

which contradicts the definition of q. Q.E.D. Corollary (Shannon [1949]). Let G be a multigraph without loops and wirh maximum degree h. Then

The proof follows immediately from Theorem 7.

263

CHROMATIC INDEX

Remark I . The above corollary can also be deduced from Vizing's theorem. Let h > 0, and suppose there exists a multigraph G with maximum degree < h such that q(G) = 32h 1 . By removing some edges if necessary, we may assume that G is critical with respect to this property, i.e. if we remove

[-1 +

from G any edge, the resulting graph has chromatic index

G contains two distinct vertices a and b with

because, otherwise, m,(x, y )

, ID1 I If A , , , #

I A2 I 2 I B2 I 2

*..

2 I Bq I 2 I A q + l

Bq #

a ; Bq+l= 0 .

a

a

121, we have IAI'

which contradicts that 1 A

I

IBI.

I 9

276

GRAPHS

If A , , , = 0, then set B, reduces to a single vertex (that is not pendant in G,i,"BJ. Thus, I A , I > I B, I, and

c 1

c a

I A I = i= 1 141> i=1 I 4 1 = I B I In both cases, a contradiction results. Q.E.D.

Lemma 2. Let G be a tree of order n, and let ( A , B ) be a bicolouring of its vertices such that I A I = I B I or I A I = I B I + 1. Then, there is an alternating sequence (a,, 6 , ,a,, b 2 , ...,) that uses each vertex of G exactly once. 1. Clearly, the lemma is true for n = 1 and n = 2. 2. If the lemma is valid for n = 2 k, we shall show that it is also valid for n = 2 k + 1. Since 1 A I = I B I + 1 > 1 B I, there exists by Lemma 1 a , A . For G x + ), there is an alternating sequence pendant vertex Q ~ + in ( a , , b,, .. ., bk) that uses all vertices, by the induction hypothesis. Hence ( a , , b l , ..., b k ,a k + , )is the required alternating sequence. 3. A similar argument shows that if the lemma is valid for n = 2 k 1, then it is valid for n = 2 k + 2. Q.E.D. Theorem 2. A stable set B is maximum i f , and only if, there is no maximal alternating sequence of odd length. 1. If such an alternating sequence a exists, then B is not a maximum stable set, since B' = ( B - a) U (a - B ) is a stable set with greater cardinality. 2. Let A be a maximum stable set, and let B be a stable set with I B I < 1 A I. We shall show that there exists a maximal alternating sequence of odd length relative to B. Let Bo = B - A , and let A . = A - B. Hence, I BO I < I A . 1. Let A , V B,, A 2 U B,, ,.., Ak U Bk be the connected components of subgraph GAOUBO, where A , c A o , B, c Bo for i = 1, 2, ..., k. Thus,

+

Hence, there exists an index i with I Bi I < I A, I. Let i = 1 be this index. 1 = I A , 1, a spanning tree of the connected subgraph G , i l V B l If I B1 I has a bicolouring ( A l , B,). By Lemma 2, its vertices constitute a maximal odd alternating sequence a relative to B,. Clearly, sequence CT is also for G an alternating sequence relative to B ; furthermore, sequence 0 is a maximal alternating sequence because if b E B - 6,either b E B - A = Bo and b is not adjacent to any a, € 6 ,or b E B n A and b is not adjacent to any a, E CT (since A is a stable set).

+

277

STABILITY NUMBER

+ +

If I B1 I 1 c I A , I, then pendant vertices can be removed from the spanning tree of G,,,,, until the remaining tree has a bicolouring ( A ; , B,) with I B1 I 1 = I A ; I, by Lemma 1. Then, as before, its vertices constitute the required sequence u. Q.E.D. meorem 3. A vertex of G is deJined to be "free" if it belongs to at least one maximum stable set but not to all maximum stable sets. Let B be a stable set. A vertex x is free and only is, there is an alternating sequence u = (a,, b, , a2, ..., aq, b,) relative to B that contains x and such that * I ' , ( b ) n { a , , a , ,..., a,] = bEB-a

a.

1. If such a sequence u exists and if x E 0 , let

A = X-B, B' = ( B - 0) u (0 - B ) , A' = ( A - a) u (0 - A ) .

B' is also a stable set, and since 1 B' 1 = I B I, B' is a maximum stable set. If x E B, then x $ B'; if x $ B, then x E B'. Thus, vertex x is free.

2. Let B be a maximum stable set that contains x , and let A be a maximum stable set that does not contain x. Let Bo = B - A. Let A . = A - B. Clearly, x E Bo. In the subgraph generated by A , U Bo, let A, u B1 denote the connected component with A1 c A0 , B , c BO, X E B ~ . If I B, I < I A, I, it is easy to see that there exists in G a maximal alternating sequence of odd length relative to B (as in the preceding theorem). This contradicts that B is a maximum stable set. Similarly, 1 B1 1 > I A , I contradicts that A is a maximum stable set. Thus

141= 1 . 4 1 1 . Consider a spanning tree of the connected graph generated by A, u B,. From Lemma 2, its vertices constitute an alternating sequence (T = (a,, b l , a2, ..., b,) of even length in graph G,,,,,. Sequence u is also an alternating sequence of even length in graph G. Furthermore, if b E B - a, either b E B - A = B,, which implies that vertex b is non-adjacent to { a,, a2, ..., u q } , or b E B n A , and b is nonadjacent to { a,, as, ..., aq} since A is a stable set. Thus, sequence o is the required alternating sequence. Q.E.D.

278

GRAPHS

2. The Turh theorem and related results No good algorithm is known for determining the stability number of a graph; however, there are several known bounds for a(G) in terms of various other invariants. Theorem 4 (J. C. Meyer [1972]). Let G = ( X , E ) be a simple graph of order n with vertices X I , x,, ..., x , such that 1 Q d~(x1Q ) ~ G ( x , )Q Jf for an integer p, 2 Q p Q n,

< do(&) .

dc(xn) + + ~ G ( X , - , + 2) Q n - P , then each stable set with fewer than p oertices is contained in a stable set with p vertices. The theorem will be proved by an induction on p. If p = 2, and if dG(xn)Q n - 2, then dG(x)< n - 1 for all x E X, and each vertex belongs to a stable set with two vertices. Thus the result is true. Let p 2 2. We shall assume that the result is true for p, and we shall show that it is also true for p + 1. Suppose that dG(xn) ... d G ( ~ n - P < n - p - 1 . Then, a fortiori, dG(x,) .-. dG(xn-,+,) Q n - p - 1 < n - P . Hence, from the induction hypothesis, a stable set S with fewer than p vertices is contained in a stable set So with p vertices. Let So = { y l , ..., y,). Then, dG(y,) -.. dG(yp)Q dG(xn) -.. d O ( ~ n - P +Ql )n - p - 1 . Therefore, l~,(So)I < n - p = I X - S o l . Consequently, there exists in X - Soat least one vertex that is not adjacent 1 vertices. to So.Therefore, S is also contained in some stable set with p Q.E.D.

+ +

+ +

+ +

+ +

+

Corollary (Berge [1960]). In a simple graph G with order n and maximum

degree h, each,maximal stable set has cardinality Let p =

n

-

at least.

[ h : ll*

*. Clearly

+

+ +

dG(xn) dG(xn - 1 ) ... d ~ ( x n - p+2) Q h(p The definition of p implies that h(p - 1) < n - p.

- 1)

219

STABILITY NUMBER

Thus 4AXn)

+ ddxn - 1) + ... + &(xn

- p + 2)

6 n -P

and the corollary follows from Theorem 4.

Q.E.D. EXAMPLE. There are 99 bridge players. No player will play with someone whom he does not know. Let h be the maximum number of players unknown to the same player. For what values of h can a bridge table (four players) be formed ? Let the vertices of G represent the players, and let the edges of G join pairs of players who do not know one another. If h = 32, then

and no table of four is possible. On the contrary, if It = 31, then a table can be formed. In fact, given any three players who know one another, a fourth can be found to play with them. Remark 1. Clearly, the above corollary implies that

For each n and for each h < n, we shall show that this inequality is the best possible. 1. We shall show first that if a and b are two integers, then

*

a-1

We can write:

a=qb+r, If r

=

0, then a - 1

=

~ < r 0, r < b, then a - 1 = qb

[;I

*=

+ (b - l), 0 < b - 1 < 6. Therefore, q =

I-[

a - 1

+1

+ ( r - l), 0 < r - 1 < 6. Therefore,

q + l = [ a +- 1J + l .

2. Let

Since n - 1 = ( p - l)(h -t- 1) + r, 0 tices can be formed from a clique with r

2, define graph G,,.k by adding to graph G,,-k+z,2a set of k isolated vertices. Then,

[2

1 m(Gn,,) = n i ( G , , - k + 2 , z = ) -(n - k

-2

+ 2 - 2 ) ( n - k + 2)]

Again, equality holds. The inequalities in Part 1 can be used to show the uniqueness of graph G,,.k . Q.E.D.

Corollary.

If G is a simple graph with n ilertices and m edges, then P*(G) < I I + 1 - ,, 1 + 2 ni

From the inequality of Theorem 1, we have (n - k)2 + 2(n - k ) - 2 111 2 0 . Since n - k 2 0, then

or

< ?I + 1 -,1 + 2 m . ~-

k

Q.E.D. 2. Kernels

For a 1-graph G = ( X , r),a set S = Xis defined to be a kernel if it is both stable and absorbant, i.e. if (1) (2)

S

+

x $S

+

XE

T(x)n S = T(x) n S #

0 0

(stable) (absorbant).

308

GRAPHS

Not every graph has a kernel (see graph in Fig. 14.3), and if a graph possesses a kernel, this kernel is not necessarily unique (see graph in Fig. 14.4). In this section we shall present existence and uniqueness theorems.

Fig. 14.3. Graph without kernels

Fig. 14.4. Graph with two kernels

EXAMPLE 1 (Von Neumann, Morgenstern [1944]). The concept of a kernel was first presented (under the name solution) in game theory. Suppose that n players, denoted by (l), (2), ..., (n), can discuss together to select a point x from a set X (the “situations”). If player (i) prefers situation a to situation 6, we shall write a > *b. The individual preferences might not be compatible, and, consequently, it is necessary to introduce the concept of effecticepreference. The situation a is said to be effectirelypreferred to b, or a > b, if there is a set of players who prefer a to b and who are a11 together capable of enforcing their preference for a. However, effective preference is not transitive, i.e., a > b and h > c do not necessarily imply that a > c. Consider the 1-graph (A’, r),where T(x) denotes the set of situations effectively preferred to x. Let S be a kernel of the graph (if one exists). Von Neumann and Morgenstern suggested that the selection be confined to the elements of S. Since S is stable, no situation of S is effectively preferred to another situation of S. Since S is absorbant, for every situation x q! S, there is a situation in S that is effectively preferred to x , so that x can be immediately discarded. EXAMPLE 2. Basis of axioms. Consider a “theory”, i.e., a set of propositions a, 6, c, ..., that we shall represent by vertices; add arc (a, b) if proposition b implies proposition a. The resulting graph G = ( X , U ) is transitive, i.e.: ( a , C)E u . (U,b)E u, ( 6 , C)E u A basis ofaxioms of the theory is a set B of propositions (called “axioms”) such that: (1) each proposition not i n B follows from one of the axioms, (2) no axiom follows from another axiom. The problem of finding a basis of axioms reduces to finding a kernel of G. It will be shown later that each transitive graph has a kernel.

309

KERNELS AND GRUNDY FUNCTIONS

Proposition 1. A necessary and suficient condition for a set S c X to be a kernel of a 1-graph G = ( X , r)is that its characteristicfunction cps(x)satisfies cp,(x) = 1

- max cp,(Y)

*

YE r(x)

Recall that the characteristic function cps(x)of set S is defined by cp,(x>

( == O

if if

XES, x$S.

If T ( x ) = 0,we define max cp,(y) = 0 . rw

YE

1 . Let S be a kernel. Since S is stable. we have cp,(x) = 1

=s

XE

Since S is absorbant, we have cp,(x) = 0

e-

max cp,(y) = 0 .

S

x$ S

-

Y E

rw

max cp,(y) = 1 .

IE r w

Combining these, the required formula follows.

2. Let cp,(x) be the characteristic function of a set S which satisfies the formula; then we have XE S =s cp,(x) = 1 =$ max cp,(y) = 0 =. T(x)n S = @ Y Er w x$S

*

cp,(x) = 0 Y

max cp,(y) = 1 rw

=$

T(x)n S #

0.

Thus, S is a kernel.

Q.E.D. Proposition 2. If S is a kernel, then S is a maximal stable set and a minimal absorbant set. Let S be the kernel of a I-graph G = (A', r).I f a 6 S, the set S u { a } cannot be stable because r ( a ) n S # fa. I f b E s, the set T = S - { b } cannot be absorbant because b 6 T and T(b) n T = 0. Q.E.D. Theorem 2. If G = ( X , r) is a symmetric I-graph, then G has a kernel. Furthermore, a set S c X is a kernel 5f, and only S is a maximal stable set. I . Clearly, a maximal stable set S of G is absorbant (otherwise, there would exist a vertex x 6 S non-adjacent to S , and S could not be a maximal stable set). Thus, S is a kernel.

310

GRAPHS

2. Conversely, if S is a kernel of a symmetric graph G, then S is a maximal stable set (because, otherwise, S would not be absorbant). Q.E.D.

Algorithms to construct all kernels of a graph have been presented by Roy [1970] and Rudeanu [1966]. Theorem 3. r f G = (X,r) is a transitiue I-graph, each minimal absorbant set has cardinality /I(G). Furthermore, a set S c X i s a kernel if, and only i f , S is a minimal absorbant set Let G

=

( X , r) be a transitive 1-graph, i.e.: yEr(X),

.Erg)

9

ZE T(X).

Consider the strongly connected components of G . If C is a component with m:(C, X - C ) = 0 (i.e., no arcs leave C ) , C will be called a terminal component. Since the graph obtained from G by contracting each strongly connected component contains no circuits, G possesses terminal components. Let C1, C 2 ,..., Cq be the terminal components of G. Each of these is a complete symmetric subgraph, because G is transitive. 1. If A is a minimal absorbant set, then A contains at least one vertex from and each x E C, satisfies each terminal component. Otherwise, A n C, = 0, x#A,

T(x)nA=

0,

which contradicts that A is absorbant. Let a, E A n Cifor each i. Consider the set A’

=

a , , a 2 , ..., a4 1 .

A’ is also an absorbant set because each x $ A’ is the initial endpoint of an arc leading into A‘ (because of transitivity). Hence A’ = A , and each minimal absorbant set is also a minimum absorbant set. 2. If S is a kernel, then S is a minimal absorbant set by Proposition 2. Conversely, if A = { a,, a2,. . ., aq } is a minimal absorbant set, then A is stable because no arc leaves a terminal component; therefore, A is a kernel. Q.E.D. Corollary 1. A transitice I-graph has a kernel, and ull of its kernels have the same cardinality.

Consequently, in Example 2, all the axiom bases have the same cardinality.


Corollary 2. In a graph G

=

311

( X , U ) , there is a set B c X such that

(1) no paths join trvo distinct t'ertices of B, (2) each vertex x 6B is the initial endpoint of a path leading into B.

The corollary follows by applying Corollary 1 to the transitive I-graph

( X , P ) obtained by creating an arc (x, y ) if there is a path from x to y in G. Q.E.D. Theorem 4. A 1-graph M'ithout circuits possesses a kernel, and this kernel is unique. Given a 1-graph G

=

( X , f) without circuits, consider the sets

X ( 0 ) = ( X / X E X , f ( x ) = fzr} X ( l ) = ( x / x # X ( O ) , m)C X < O ) > X(2) = { x / x $ X ( O ) u X(t), f ( x ) c X ( 0 ) u X ( 1 ) } , etc... These sets are pairwise disjoint, and x E X ( k ) if, and only if, the longest path from x has length k. Since G contains no circuits, the sets X ( k ) form a partition of X . A characteristic function qs of a kernel Sc a n be successively defined on the sets X(O), X(1), X ( 2 ) , etc. by the equality of Proposition 1. Furthermore, cpS(x)is uniquely defined for each x. Q.E.D. The following theorem due to Richardson shows that a graph without odd circuits possesses a kernel. The original proof of this result was long and involved. The less complicated proof presented here is due to Victor Neumann. A semi-kernel of a 1-graph G = ( X , r) is defined to be a non-empty stable set S c X such that each x adjacent to S has at least one successor in S. Lemma. if for each non-empty subset A c X , the subgraph GAhas a semikernel, then G has a kernel. Let S be a maximal semi-kernel of C , and let A = X - S - fG(S). If A = fzr, then S is a kernel. If A # fzr, then GAhas a semi-kernel T. Sets S and T are nonadjacent; thus, S u T is stable. Each x $ S v T that is adjacent to S u T has a successor in S u T. Thus, S u T is a semi-kernel of G, which contradicts the maximality of S. Q.E.D. Theorem 5 (Richardson [1953]). If C circuits, then G has at least one kernel.

=

( X , r) is a I-graph without odd

312

GRAPHS

From the lemma, it suffices to show that G possesses a semi-kernel. We may assume that G is connected. Let XI be a strongly connected component of G with I'(X,) XI.If 1 Xl 1 = 1, then X1is a semi-kernel. If 1 X , I > 1, and if xo E X , , let x be a vertex in XI distinct from x o . Then, all paths p [ x o , x] remain in the interior of XI ; and they have the same parity (since, otherwise, an odd circuit could be formed with a path p[x, x,,]). Let S denote the set of all x E XI such that all paths p[xo, x ] are even. Set S is stable. If z E Xis adjacent to S, then each successor of z belongs to S. Thus, S is a semi-kernel. Q.E.D. Note that the graph in Fig. 14.3 has no kernel: therefore it contains an odd circuit. Similarly, the graph in Fig. 14.4 does not have a unique kernel, therefore it contains a circuit.

3. Grundy functions Consider a 1-graph G = ( X , r)without loops. A non-negative integer function g(x) is called a Grundyfunction on G if for every vertex x, g(x) is the smallest non-negative integer which does not belong to the set { g(y) / y E f ( x ) }. This concept originated with P. M. Grundy El9391 for graphs without circuits. It was extended to 1-graphs by Berge and Schutzenberger [1956]. The Grundy function can also be defined as a function g(x) such that (1) g(x) = k > 0 implies that for each j < k , there is a y E T ( x ) u+th g(y>= J , (2) g(x) = k implies rhat each y E T(2) satisfies g(y) # k. A pseudo-Grundyfunction is defined to be a function that satisfies property (1) above and (2') g(x) = 0 impIies that each y E T ( x ) satisfies g ( y ) # 0. We shall see in the following development that a Grundy function (or a pseudo-Grundy function) determines a kernel of a graph. Remark 1 . Some graphs have no Grundy function. Some graphs have more than one Grundy function. (See Fig. 14.5 where the value of a Grundy function is written next to each vertex.) 0

--1 --

1

0

0

0

07-T 1

Fig. 14.5

313

KERNELS AND G R U N D Y FUNCTIONS

Remark 2. If G has a Grundy function g(x), then G has a kernel, since the set S = ( x / x E X , g(x)=O} satisfies simultaneously : (1)

XE

S

=-

g(x) = 0

*

T(x)n S = 0 ,

min g(y) > 0 rw

YE

(2)

x$ S

*

g(x)

>0

=>

min g(y) = 0 YE

3

T(x)nS #

0.

r(x)

The converse is not true. It is left to the reader to verify that the graph G in Fig. 14.6 has a kernel { d } but has no Grundy function. Theorem 6. I f G is a 1-graph such that each subgraph has a kernel, then G possesses a Grundy .function. Let G be such a graph, and let So be a kernel of G. Let S, be a kernel of GI = Gx-so ; let S, be a kernel of G, = Gx-csovsl,,etc. The sets S, form a partition of X . Let g ( x ) be an integer defined by: g(x) = k

X E & .

0

We shall show that g(x) is a Grundy function of G. 1. Let g(x) = k ; we shall show that for each j < k, there exists a vertex T ( x ) such that g(y) = j . Since x E S,, and k > j , vertex x is present in graph G,. Since x $ S,, there is a vertex y E S, such that y E T ( x ) ; thus, there is in T ( x ) a vertex y with g(y) = j .

y

E

2. Let g(x) = k ; there is no vertex y then S, would not be stable.

E

T ( x ) such that g ( y ) = k , because

Thus g(x) is a Grundy function.

Q.E.D. Corollary 1. A symmetric graph possesses a Grundy function. This follows from Theorem 2. Corollary 2. A transititle graph possesses a Grundy function. This follows from Theorem 3. Corollary 3. A graph without odd circuits possesses a Grundy function. This follows from Theorem 5.

314

GRAPHS

Theorem 7 (Grundy [1939]). A graph G without circuits possesses a unique Grundy function g(x). Moreoter, .for each rertex x , g(x) does not exceed the length of ?lie longes? path from x. As in Theorem 4, consider the sets:

a}

X(0) = { X / X € X , T ( X ) = X(1) = ( X / X # X ( O ) , T(x) = X(O)> X(2) = ( x / x $ X ( O )u X(1),T(x) c X(0)u X(l)} etc. Clearly, these sets partition X,and x E X(k) if, and only if, the longest path from x has length k. The values g ( x ) can be successively defined on the sets X(O), X(l), etc. If x E X(O), let g(x) = 0. If x E X(1), let g(x) = 1. If for each y E X ( k ) , the value g ( y ) is uniquely defined and satisfies g ( y ) < k, then for x E X ( k l), the value g ( x ) is uniquely defined and g(x) < k I. Q.E.D.

+

+

Before demonstrating the fundamental properties of Grundy functions, it is necessary to define the Cartesian sum of I-graphs: r,),..., G, = (X,,r,) be 1-graphs, and let Let G1 = (X,,rl),G, = (X2, P = { 1, 2, .. ., p }. The normal product of these I-graphs is defined to the 1-graph G = G1.G, ... G, with vertex set

. . x = x, x x2 x .'. x x, = I7 xi ieP

and with correspondence

The Cartesian sum of these graphs is defined to be the 1-graph G = + G, with vertex set X = Xiand with correspondence G, G,

+

n,,

+

r(x1,x2,*.-,xp)=

u

({xl>x

***

x { x i - ~ ~ x G ( x J x { x i + l > xx { x p > ) *

ieP

Finally, the Cartesian product of these graphs is defined to be the 1-graph

G = G, x G, x with vertex set X

=

x

G,

niEP Xi and with correspondence

315


mG

H

h

II

I

QY

d

G+H ox

Gx H

b.v

nx

hx

Fig. 14.6

EXAMPLE. Consider two machines, and let X , denote the set of possible states for the first machine. For xl, x l E X , , let x i E f , ( x , ) if state x i can follow state x,. Thus, the first machine defines a graph G, = (XI, rl). Similarly, the second machine defines a graph G, = ( X , , f2). Let ( x l , x,) describe the state x1 of the first machine and the state x, of the second machine. If the operator works both machines simultaneously, graph G1 x G, represents the possible changes of situations. If the operator works only one machine at a time, graph G, + G, represents the possible changes of situations. If the operator can work either one machine or two machines simultaneously, graph G, .G, represents the possible changes of situations. Remark. The connectivity of a Cartesian product of I-graphs has been studied by McAndrews [1963], who has shown the following result: If G and H are two strongly connected graphs respectirely with the sets M ( G ) and M ( H ) of circuits, then the Cartesian product G x H is strongly

316

GRAPHS

connected g, and only i f , the lengths of the circuits in M ( G ) V M ( H ) are relatiilely prime. Similar properties for the Cartesian sum have been studied by Aberth [1964],who has shown that: Graph G H is strongly connected i f , and only i f , both G and Hare strongly connected. Other properties of G H a n d G x H have been studied by Picard [I9701 and Pultr [1970].

+

+

Proposition 1. If 1-graph G = ( X , r)possesses a kernel S,and i f I-graph H = ( Y, A ) possesses a kernel T, then the normal product G. H has the Cartesian product S x T CIS a kernel. It is left to the reader to verify the stability and absorption of set S x T.

Does the Cartesian product of two graphs, each with a kernel, also have a kernel? The example in Fig. 14.6, due to C. Y . Chao [1963],provides a counter example: graphs G and H respectively have kernels { d } and { y }, but graph G + H does not possess a kernel, because such a kernel would necessarily contain vertex dy, (since r ( d , y ) = 0 )and two of the three vertices ax, bx, or cx, which contradicts stability. Later, it will be shown that if both graphs G and H possess a Grundy function, then graph G H has a kernel. Before demonstrating these results some preliminary developments are required. The binary expansion of an integer p is a sequence (p', p2,...,p k ) o f digits such that

+

p =pl

+ 2p2 + 4p3 + + 2 ***

k - y

Its binary form is k

P=P P

k-1

2

*'*PP

1 '

For example:

Decimal Binary form form

-

-

0 1

2 3 4 5

= = = = = =

0 = 1 = 10 = 11 = 100 = 101 =

Binary expansion

Decimal form

-

-

(0) (1) (0,1) (1,l) ( O , O , 1) (1,0, 1)

6 7 8

Binary form

Binary expansion

-

=

= =

9

=

10 11

= =

110 111 1000 1001 1010 1011

(0, 1, 1) (1,1,1) ( O , O , O , 1) ( l , O , O , 1) = (0,1 , 0 , 1) = (1,1,0,1)

= = = =

317


Let [nIc2,denote the integer n modulo 2 (which is the remainder of n divided by 2). The 2ko-1 + kalc 2k-1

k31

=

0, because,

qk

k 2. Let Go be the subgraph obtained by the removal of vertex xo. Clearly, we have

(1)

Y(G) < Y(G0) + 1 *

(2)

y(a

< y(G,) + 1

*

1. If in both (1) and (2) equality holds, then dG(xO)

2

ddxo) 2 y(G0).

Hence, y(Go)

Hence y(G)

+ y(Go) < &(xo) + &(xo)

= n - 1.

+ y(G) < (n - 1) + 2 = n + 1 .

2. If either (1) or (2) is not satisfied with equality, then, by the induction hypothesis, y(G)

+ y(G) < y(Go) + y(co)+ 1 < n + 1 .

331

CHROMATIC NUMBER

Hence, the theorem is valid in both cases.") Q.E.D. Remark. This bound is the best possible : A graph of type T,(n,p ) is defined to be a graph with n vertices firmed from a stable set S, with p vertices and a with n - p + 1 vertices such that clique Kn-p+l

I K"--,+, n s, I = 1 -

Fig. 15.4

If graph G is of type TI, then y(G) = n - p

+ 1,

Y G ) = P. Similarly, a graph of type T,(n,p) is defined to be a graph with n vertices formed from a cycle C, of length 5 without chords, a stable set S, with p < n - 5 vertices, and a (n - p - 5)-clique K,,-,- 5 , such that these three sets are disjoint and (1) each vertex C, is adjacent to each vertex of Kn-,,-,, (2) no vertex of C, is adjacent to a vertex of S,. If G is a graph of type T,, then (') A stronger inequality has been obtained by R. P. Gupta [1968]. A quasi-colouring wirh k colours is defined to be a mappingf : X+ (al,aa , ak) such that i # j implies the existence of two adjacent vertices x and y with f ( x ) = u, and f ( y ) = aI.A colouring with a minimum number of colours is a quasi-colouring because, otherwise, two colours a ' and a, are not adjacent and the vertices with any one of these two colours could be coloured by using only one colour. Let #(F) denote the maximum number of colours needed for a quasi-colouring of G. Thus, y(C) < I)(@. Gupta has shown that * a * ,

+

Y(G) v(G) < n Clearly, this result is stronger than Theorem 2.

+ 1.

3 32

GRAPHS

y(G) = ( n

- 5) + 3 = n - p - 2 ,

-p

y(E) = p + 3 y(5) = n + 1 . 0

Hence

y(G)+

K2

G E T2(11, 4)

Fig. 15.5

The bound given in Theorem 2 is attained by graphs of type TIand. T2. Furthermore, H. J. Finck [I9661 has shown that only the graphs of types TI and T, attain this bound.

Corollary. I f G is the complementary graph of a simple graph G of order n, then

By Theorem 2, we have 4 Y(G)y(G) G (Y(G) -

y(G)’ + 4 Y(G) y(G) G (w) + Y(G)’G (n + I)* .

Q.E.D. Remark. This bound is the best possible. If n is odd, then each graph G of type

satisfies

If n is even, then each G of type

333

CHROMATIC NUMBER

satisfies

Theorem 3. If G is a simple graph with n vertices and m edges, then y(G)

n2 >7 n -2m

Let y(G) = q. Let S1, S2, ..., S, denbte the sets of vertices of colour a l , respectively. The adjacency matrix of graph G has the form:

a z , ..., a,,,

s,

-I

I

s 2

...

lo

___-

0 Let n, = I Si I. Let No denote the number of 0 entries in this matrix, and let Nl denote the number of 1 entries in this matrix. Then,

Nl = 2 m ,

This follows from the Cauchy-Schwartz inequality I x I I y with x = ( n l , n2, ..., n,) and y = (1, 1, ..., 1). Thus, the total number of entries in the matrix satisfies:

n2 = N ,

n” + N1 2 2 m + -

Hence,

(n2 - 2 m) q 2 n2 .

4

1

2

I ( x, y ) 1,

334

GRAPHS

Hence the required formula.

Q.E.D. n2 Remark. In order to have equality, it is necessary that No = - and that 4 I

IXl.IYI = I(X,Y>l,

which implies that ( n l , n 2 , ..., n,) is proportional to (1, 1, ..., 1). Thus, equality holds only if

n , = n2 =

= n,

’

and if the matrix has only 1 entries outside of the blocks S, x S , , i.e., the graph is formed from q stable sets S , , S2,..., S, of the same cardinality with two vertices joined together if, anh only if, they belorg to distinct S,.

Theorem 4. (Roy [1967], Gallai [1968]). If G = ( X , E ) is a simple graph with y(G) = q, then, for each orientation of its edges, there exists an elementary path with q uertices. Furthermore, there exists an orientationfor which there is no path with more than q uertices. 1. Suppose that G = (X, r) is a 1-graph. Let t ( p ) denote the number of vertices encountered by an elementary path p (i.e., the length of path p plus one). We shall show that max, t ( p ) 2 y(G) where the maximum is taken over all elementary paths p of G. More precisely, we shall show that the vertices of G can be coloured with max, t(p) colours. By Corollary 2 to Theorem (3, Ch. 14), there exists a set Sfsuch that each vertex of G can be reached by a path starting from S:, and no two vertices of Sy are connected by a path. Let

s,”= r(s:)- s: ; s: = r(s;)- (s:u S,”) . ...................... Clearly, these sets partition X. Let (X, A ) denote the partial graph generated by the arcs ( x , y ) of G such that X€S,O, y E S : + l .

If, for some index i, there exist two vertices a E S, and b E S, with b E r ( a ) , then define a partition (Si,S:, S ; , ...) of X, where

335

CHROMATIC NUMBER

s; s:

=

q

=

sp - { b } ,

f . j = 1 , 2,..., i -

1)

s:+1= s;+lu{bL s;+2

=

sP+z ”

u d(&)) .....................

s;+3= $+3

-

Repeat the process until all sets in the partition are stable. The initial partition (S:, S,”,S,”, ...) has the property that for each p and for each x E S,O, there exists an elementary path p x terminating at x with t(p,) = p . All the successive partitions have this property because if (Sf, Sg, S,”, .. .) satisfies the property, then (Sf+ l, S,k S,”+l, ...) also maintains this property for all vertices that do not change class (and also for the other vertices, obviously). Thus, a partition into h stable sets with the desired property can be obtained. Hence, +

y(G)

Gh

< max W . P

2. This result is the best possible: let the vertices of a simple graph G be coloured with q = y(G) colours, a l , aZ, ..., a,, and let each edge [x, y ] be directed from x to y if x is coloured a, and y is coloured aj with i < j ; then it is evident that no path of the resulting graph contains more than q vertices. Q.E.D.

Corollary. Let G be a simple graph coloured with q = y ( G ) colours al,az , Then there exists an elementary chain that encounters consecutively the q colours at in this order.

..., a,.

Direct edge [x,y ] from x to y if x is coloured a t , y is coloured with a j , and i < j , From Theorem 4, there exists an elementary path containing exactly y(G) vertices. Clearly, this path has the required property. Q.E.D. It follows from Theorem 4 that a complete graph contains a hamiltonian path.

Theorem 5. Let ( S , ,Sz, .. , S,) be a q-colouring (not necessarily minimum) of a simple graph G, and let d , = max d G ( x ) . XESk

336

GRAPHS

Then, y(G) B max min { k

, dk + 1 } .

k, 3, with y(G) = h + 1, and that contains no (h + I)-clique. We shall show that this leads to a contradiction.

1. Since vertices that are not essential to the above properties can be removed, we may assume that G is minimal with respect to these properties. Let xo E X . The subgraph Go generated by X - { xo ) contains no (h 1)clique, and therefore y(Go) < h 1. Hence, y(Go) = h. This implies that dG(xo)B h because, otherwise, one of the h colours used to colour Gocould 1. be used to colour x o , which contradicts y(G) = h Thus, dG(xo)= h. We may assume that, if y,,y,, ...,y , denote the h vertices adjacent to xo,they are coloured with the colours ul, a2, . ., a,, respectively in a giuen h-colouring of Go.

+

+

+

.

2. Let C(a,, aj) be the subgraph of Go generated by the vertices with colours at or a, in the h-colouring of Go.Vertices y , and y , belong to the same connected componenr of G(a,, a,) because, otherwise, after interchanging colours ai: and a, in the connected component containing y,, xo could be coloured a [ , which contradicts y(G) = h + 1. 3. We shall show that the connected component of G(ui,a,) that contains y, and y,,is an elementary chain p[y,, y,] going from y, to y,. Vertex y r is adjacent to only one vertex with colour a, (otherwise, since dG(yi)Q h, vertex y, could be recoloured with a colour c ( k , k # i,j, and vertex xo could be coloured a!). Consider a chain in G ( q , a,) from y , to y,. We shall show that this chain is

unique. Let x be the first vertex of this chain with dGcar, a,)(x)> 2. I f x has colour a,, for example, then there are three vertices with colour a, adjacent to x ; since d,(x) ,< h, vertex x can be recoloured a*, k # i,j. This would disconnect y , and y, and contradict Part 2. 4. We shall show that the two chains pb,, y,] and p [ y l ,yk] that constitute components G(a,, aJ) and G(at, a,) cannot contain a common vertex z # y l .

338

GRAPHS

If such a vertex z existed, then it would have colour at and would be adjacent to two vertices with colour a, and to two vertices with colour aka Thus, h 2 dc(z) 2 4, and there is a fourth colour aI# a*, a,, a,, to recolour z. This would disconnect yi and y , and contradict Part 2.

+

5. Since G does not contain any (h 1)-cliques, there exist in Tc(xo)two non-adjacent vertices, say y , and y,. Consider the connected component p b l , y z ] of G(al, a,) and the connected component p ' [ y l , y 3 ] of G(a,, a&. Let x be the first vertex after y , in chain p [ y , , y,]. Since y, and y, are nonadjacent, x # y,. If colours a, and a3 are interchanged in the component of G(a,, as) that contains y, and y,, then vertex y1 is recoloured a3 and vertexy3 is recoloured a l . Furthermore, the new component H,, with colours a1 and a, that contains y, satisfies HI2

= P [ X , Y2l

Y

, and p ' [ y , , y3] have no common vertex since from Part 4, chains ~ [ y ,y,] (except rl). On the other hand, the component H23 with colours a, and a3 that contains y , satisfies H23

= &,

XI

9

since x has colour a2 and is adjacent to y,. Hence, x is a vertex common to the connected components H12and H Z 3 .This contradicts Part 4 and completes the proof. Q.E.D. The next section will present extensions of the Brooks theorem. 2. y-Critical graphs

A simple graph G is defined to be y-critical if for each vertex xo the subgraph Go generated by X - { xo } has chromatic number y(Go) < y(G). Note that in this case, y(G,) = y(G) - 1 because if y(G) = q 1 and y(Go) c q, then Gocan be coloured with q - 1 colours and vertex xo can be coloured with a q-th colour, which contradicts that y(G) > q. We shall now study the properties of y-critical graphs.

+

Property 1. A graph G with y(G) = q 1.

y(G) = q

+

+ 1 has a y-critical subgraph

with

If G is not y-critical, there is a vertex xo whose removal does not decrease the chromatic number. If the subgraph Go generated by X - { xo 1 is not y-critical, then again there is a vertex xl whose removal from Go does not

339

CHROMAnC NUMBER

decrease the chromatic number, etc. Eventually, this process locates a y-critical subgraph. Q.E.D.

+

Property 2. If G is a simple y-critical graph with y(G') = q 1, then d,(x) 2 q.for all x . If dG(xo)< q, the subgraph Go generated by X - { x o } can be coloured with q colours. At least one of these q colours is not adjacent to vertex x ; thus, x can be coloured with this colour, which contradicts y(G) = q 1. Q.E.D.

+

Pmperty 3. A y-critical graph is connected. The proof is obvious. Property 4. A y-critical graph contains no articulation set that is a clique. Let G be a y-critical graph with y(G) = q + I , and let A be a clique that is an articulation set of G. Since G is y-critical,

-

I A I = ?(GA) < 4 Let B, , B2,B3, . .. be the connected components of the subgraph generated by X - A , and let B; = B, V A , B2 = B; v A , .., be the corresponding pieces. Since G is y-critical, each piece can be coloured with q colours. Thus, G can be coloured with q colours, which contradicts y(G) = q 1. Q.E.D.

+

Property 5. A y-critical graph has no articulation points, This follows from property 4 since an articulation point is a 1-clique.

+

Property 6. If G is a y-critical graph with y(G) = q 1, and if A = { a, b } is an articulation set of G, then there are exactly two pieces B; and B, relatice to this articulation set and =

Y(W

= 4.

Clearly, a piece B' relative to A can be coloured with q colours. Three cases must be considered : (1) B' cannot be coloured with q colours so that a and b have the same colour, (2) B' cannot be coloured with q colours so that a and b have different colours, (3) There is a colouring of B' in which a and b have the same colour, and there is also a colouring of B' in which a and b have different colours. Case (3) cannot apply to a piece B' because the corresponding component B can be removed without changing the chromatic number of G. If Case (1)

340

GRAPHS

applies to B ‘ , then y(B‘) = q because if y(B’) = q - 1, vertices a and b can be coloured with a q-th colour, which contradicts that Case (1) applies to B’. Similarly, if Case (2) applies to B’, then y(B’) = q. Since y(G) = q 1, there is a piece satisfying Case (1) and a piece satiifying Case (2). Since G is y-critical, there is only one piece of each type.

+

Q.E.D.

(2 pieces)

(5 pieces or less)

(3 pieces or less)

@ (3 pieces or less)

@ (2 pieces)

@ (5 pieces or less)

Fig. 15.6. Various types of articulation sets in a y-critical graph

Property 7. I f G is y-critical with y(G) = q + 1 2 4, and if A = { a, b, c } is an articulation set, then one of the following cases occurs: I . G, contains only one edge and A has at tnost three pieces, each with chromatic number q. 11. G, contains exactly two edges, and A has at most two pieces, each with chromatic number q. 111. G, contains no edges, and A has at mostfire pieces, each with chromatic number q or q - 1.

We shall show only Case I. For example, suppose that a and b are the adjacent vertices in A . If piece B‘ is coloured with colours 1,2, ...,q, then vertices a, b, c may be coloured 121, or 122, or 123. However, not all of these

341

CHROMATIC NUMBER

three colourings are possible for B' because G is y-critical; it follows that y(B') = q (because if y(B') < q - 1, then using a q-th colour, all three of these colourings would be possible). If two of these colourings are possible for piece B;, then there is a piece B; for which only the third colouring is possible (and there are only two pieces since G is y-critical). If no piece admits two of the three colourings, then there exist three pieces BI, B;, Bj that respectively admit one of each of the three colourings. Q.E.D. Property 8. A y-critical graph G = ( X , E ) with y(G) disconnected by the removal of q - 1 edges, i.e.,

=

q

+I

cannot be

mG(A,X-A)aq ( A c X ,A # 5 2 / , X ) . We shall show that a contradiction results if there are two nonempty sets A and B of vertices G with A U B = X , A A B = 0 , mG(A,B) Q - 1 . Since G is y-critical, the subgraph GAis q-colourable; consider a q-colouring function for G A ,i.e., a mappingf(a) from A into { 1, 2, ...,q 1 such that

X,YEA, b , Y l E E =. f ( x ) Z S b ) . Let a,, a,, ...,a, be the vertices of A adjacent to B; choose function f so that (i = 1 , 2 , ...,r ) . f(aJ 6 i Let e,, e,, ., ,, ep denote the edges that join A and B, with indices chosen so that e, = [ a i , x ] , e, = [ a j , y ] , i < j =. s < t . Let g' be a q-colouring function for the subgraph G B .We shall show that by a sequence of transformations, this q-colouring of GB can become com1. If edge patible with the q-colouring of G A ,which contradicts y(G) = q e, is of the form [a,, b,], define a q-colouring g2(y) of G, by

+

= g'(y)

= g'(bJ

if

g'(v> f 2, g'(b3

if

g'(y) = g'@J

if

g'(y) = 2

.

Thus Now consider e,

=

[a,, b,], and similarly, define a q-colouring g3((y)by

342

GRAPHS

Thus,

This process can be continued, and since p edge ep gives a q-colouring gP+ with

=

rnc(A, B)

< q - 1, the last

gPf1(b,)> f ( a , ) *

ram btl

+

Thus G can be coloured with q colours, which contradicts y(G) = q 1. Q.E.D. Consider a y-critical graph G with y(G) = q G satisfies

min &(x)

q

+ 1; from Property 2, graph

.

xsx

To study the structure of G, we shall consider the set M =

{X

/ X € X ,

d&) = q ) ,

and study the structure of the subgraph G , generated by M.

Lemma 1. Let p = [a,, a , , a 2 , ..., ak-,, a,] be a cycle of G , and let f be a (q 1)-colouring of G with only certex a, having the (q + I)-st colour; then there exists a (q + 1)-colouringg of G ivitli only vertex a, having the (q + I)-st colour such that:

+

g(x) =fW d u o ) =f(a,) da1) = f ( a z >

( X 4 d

=f(a3)

..............

Let C Yctz, ~ , ..., a q + , be the colours in colouringf, where f(ao) = a g + l . Since dG(ai)= q, each colour a l , x Z , ..., Q, appears exactly Once in ~ G ( U , ) . Hence, interchanging the colours of a, and of a, gives a (q + 1)-colouring with only vertex a, having the (q 1)-st colour. This procedure can be re-

+

CHROMATIC NUMBER

343

peated around cycle p until vertex a, has colour a,,,. This produces the required colouring g . Q.E.D. Lemma 2. Let p = [a,, a,, ..., a,-,, a,] be a cycle of GM with no chord incident to certex a,; then p is an odd cycle.

+

1. Since graph G is ( q 1)-critical, there exists a (q + 1)-colouringfof G with only vertex a, having the (q + I)-st colour a,,,. Since dc(ao) = q, the set Tc(a,) - { a,, a k - , } contains exactly q - 2 vertices, and none of these vertices lies on p . Each of these vertices take a different colour in colouring f; let ul,a 2 , ...,a, - be these colours. 2. We shall show that only the colours and a, are assigned to vertices a,, a2, ..., a k - 1 in colouringf. If this is not true, then there exists in cycle p a vertex ai, i > 0, with colour u , , j # q - 1, q. After repeating i - 1 times the operation of Lemma and with 1, a colouring g is obtained with only vertex a. having colour g(x)

=f(4

(X#P)

d u o ) = f (a01 d a l ) = f (a3 = aj g(a2) = f (ai+ 1)

..............

d a k - 1)

= f ( a i - 1)

*

But, then, the set rc(ao) has two vertices with colour a, in colouring g. Thus, a, can be recoloured with a colour other than which contradicts y(G) = 1.

+

3. In colouring f, all vertices of cycle p, except a,, have either colour or u,. Since vertices a, and a k - l belong to T,(a,) and cannot have the same colour, cycle p is necessarily odd. Q.E.D.

aq-,

Theorem 7 . Let G be a y-critical graph with y(C) = q + 1; then dc(x) 2 q for all X , and each block of the subgraph GMgenerated by set M = { x x E X, dc(x) = q } is either a clique or an odd cycle without chords.

From Lemma 2, each even cycle of graph G M has at least two chords. Thus, from Theorem (7, Ch. 9), G M has the required form. Q.E.D. Remark. If q equals

h = max d,(x) , xsx

344

GRAPHS

the Brooks theorem follows from Theorem 7: Let G be a graph with max dC(x) = h

y(G) = h

and

+ 1.

Consider the y-critical subgraph G‘ of G with y(G’) = h + 1 (which exists by Property 1). Since dG.(x) 2 h (Property 2), and since dG.(x) < dG(x)< h, graph G’ is regular of degree It. In other words, the vertex set X’ of graph G’ is equal to the set M = { x / x E X’; d&) = h } . Furthermore, G’ is connected and has no articulation points (Properties 3 and 5). Since GL = G’ is a block, Theorem 7 shows that G’ is either a (h 1)clique or an odd cycle without chords. Hence graph G has either a connected component that is a (h I)-clique, or, if h = 2, an odd cycle without chords. The Brooks theorem follows.

+

+

The next theorem is also an extension of the Brooks theorem.

+ 1 2 4.

Lemma. Let G be a y-critical graph with y(G) = q clique C = { c l , c 2 , ..., cq } nith q tiertices such that (dG(ci)

- 4) G 4 -

G contains a

9

i=l

then in a q-colouring of X - C, the sets Ai = r G ( C i )

-c

contain a common colour. Suppose that d,(c,)

< dG(C2) < ..* G d G ( C q ) .

Since G is y-critical, then by Property 2, for at most q - 3 of the c i , then dG(cl)

=

dGG(c.2)

=

dG(c1)

dG(c3)

=

Z q. Since d,(c,)

-q 2 1

4.

Thus I A l I = 1. Let a1 be the colour of the unique vertex of A l in the q-colouring g of X - C. Suppose that there is a set A,, that does not contain colour al. (If several such sets exist, take index io to be as small as possible.) We shall now show that this is impossible, i.e., clique C can be coloured 1. with the q colours used for X - C, which will contradict y(G) = q Let a l , a z , ..., aq be the q colours of g ( X - C). Define the q-colouring g(x) of C successively in the following way: - let g(ci,) be colour a,;

+

345

CHROMATIC NUMBER

q # io. let g(c,) be any colour different from the colours of A , u

-if

{ Go 1;

-if

q - 1 # i,,, let g(c,-l) be a n y colour different from the colours of Aq-1 U { ci,, ~9 1; -etc. ... Note that, f o r j < q, a

a

Thus, (q-j+1) (lAjl-l)q

- 11.

347

CHROMATIC NUMBER

We have d,,(x)

+ 1 < d,(x)

(x

ET) ,

(because each vertex x E T is adjacent to X - T, since X - T is a maximal stable set.) Hence S' c S. In fact, S' is a proper subset of S since xoE S and xo$9.Hence,

C

xos'

[&,(x)

- (4 - I)] G

C

(dc(x)

- 4)

(dG(X)

- q) - 1
+ 1 (i = ..., PI. '

1 9 2 9

Therefore, a maximum stable set S, in Gcf,(,) satisfies

( Set So =

Up,l

I Si I = 4 G C J { a

} c Si c

+1

9

ci u { a } .

S, is stable in G,and P

'I

So I =

c

i= 1

Wc,)

+1

*

Moreover, since a partition V of G into cliques can be formed with A and with the B(G,,) cliques of a minimum partition of C,, for i = 1, 2, . . . , p , we have

363

PERFECT GRAPHS

Hence, a(G) CASE 2.

=

B(G).

Upzl A, = A . Then, for all i, u Al)

=

a(GCl)

.

Otherwise, there would exist a stable set S, c A , u C, with

I

Si

I

=

~ ( G c ,+ ) 13

and there would exist a vertex a E A , that is non-adjacent to some maximum stable set of Ci, which contradicts the definition of A t . Hence,

Hence, a(G)

=

B(G).

Q.E.D. The next three sections treat particular classes of a-perfect and y-petfect graphs. 2. Comparability graphs

A simple graph G = ( X , E ) is called a comparability graph if it is possible to direct its edges so that the resulting 1-graph (A', U ) satisfies: (x,y ) € (x,Y ) E

u, (.v, z ) € u u

=>

(x, z ) E u cV,x)4U

(transitivity) (ant i-symmetry).

Clearly, a bipartite graph is a comparability graph. Furthermore, each subgraph of a comparability graph is a comparability graph.

A

The graph in Fig. 16.1 is not a comparability graph because edges ab, bc, cd, be, ce can be directed appropriately, but then it is impossible to direct .f edge eJcorrectly.

a

/

Fig. 16.1

364

GRAPHS

Theorem 4. Every comparability graph is a-perfect.

Consider a I-graph G = ( X , U ) whose arcs represent an order relation. From Corollary 2, to Theorem (14, Ch. 13), a(G) equals the smallest number of elementary paths that partition the vertex set. Because of transitivity, each path generates a clique of G , and, by the corollary to Theorem (4, Ch. 10) each clique is the vertex set of some elementary path of G. Hence, m(G) = O(G). Q.E.D.

Theorem 5. Each cotriparability graph is y-pe$ect. It suffices to show that if G = ( X , U ) is the I-graph of an order relation, then y(G) = w(G). Let t ( x ) denote the length of the longest path from x plus one. Since G has no circuits, t ( x ) < co. If max t ( x ) = k, then there exists a k-clique. There I)-cliques (because this clique would contain a path passing exist no (k through all its vertices, and the longest path contains only k vertices). Thus

+

o(G) = k

.

Consider k colours denoted by 1,2, ..., k , and colour each vertex x with colour t ( x ) . Two adjacent vertices cannot have the same colour, because if there is an arc directed from x toy, then t ( x ) > t(y). Thus Y(G)d k

.

Since y ( C ) b o(G) = k, we have

y(G) = k = o ( G ) . Q.E.D.

Theorem 6. Let G = ( X , U ) be a transitive I-graph, and let E denote the set of all puirs of a4acent certices. Then, the simple graph ( X , E ) is a comparability graph. We shall remove from G all the loops and one arc in each multiple edge, so that the resulting I-graph (A', V ) satisfies the transitivity property. Let xl, x 2 , ..., x, be the vertices of G , and let.x, = x j if either i = j or f j (Xi,X j ) E ( x , , Xi) E u . From the transitivity of G , the relation = is an equivalence. Define set V as follows : (1) If x, and x, are in the same equivalence class, and if i > j , let ( x i ,x j ) E V, ( 2 ) If xi and x, are in different equivalence classes, and if ( x i ,x,) E U, let 7

(XI,X I ) E

v.

7

3 65

PERFECT GRAPHS

Hence, we have:

*

(x, z ) E V

=>

( y , X) $ V

(transitivity) (anti-symmetry). Q.E.D. Theorem 7 (Ghouila-Houri [1962]). A relation > is said to be a semi-order relation if nie have

(x,y ) E V , ( y , z ) E V (X,Y)E v

a > b, b a>b

>c

* *

a > c or c not b > a

>a

(semi-transitivity) (anti-symmetry)

A simple graph can be directed so that its arcs represent a semi-order relation

5f, and only if, it is a coniparability graph. It suffices to show that i f G = ( X , U ) is a I-graph whose arcs represent a semi-order relation, then there exists an orientation of the edges of G that represent an order relation. Assume that the theorem is true for all graphs of order < n ; we shall show that it is valid for a 1-graph G = ( X , U ) of order n whose arcs represent a semi-order. First, note that if three distinct vertices a, b, and c satisfy

( a , b) E U , ( b , c>E u, ( c , a>E u , then each other vertex is adjacent with either 0 or 2, or 3 of these vertices. Suppose that G is not transitive; then there exist three vertices x l , x2, x3 satisfying (XI

2

x*)E

u,

( x 2 , x3) E

u,

(x3, X I ) E

u

*

Two cases must be considered:

CASE1. Each itertex x # xl, x 2 ,x3 adjacent with one of these three vertices is adjacent with all three.

Remove vertices x2 and xg and direct the edges of the resulting subgraph so that it is transitive. (This is possible from the induction hypothesis.)

A

Fig. 16.2

366

GRAPHS

Then, direct each edge of the form [x,x,] or [x,x3],where x # xl, x 2 , x3, in the same direction as [x, x l ] ,and direct transitively the triangle formed by x,, x,, x3. Thus, the obtained graph is transitive. CASE2. There exists at least one i?ertexa # x, , x,, xBthat is adjacent with exactly two of these vertices, say, x2 and x3.

Let A be the set of vertices y that satisfy C ~ , X Z ) E Uand

(x,,y)~U.

Thus, x, E A. Furthermore, since a is not joined to x l , then a E A. We shall show that i f x 4 A, then either x is adjacent with aI1 the pertices of A, or x is not adjacent with set A.

Fig. 16.3

Let b E A , and suppose that x is joined to a but not to b. By considering the triangle formed by a, x 2 ,x 3 , we see that x is joined to at least two of these three vertices and thus to x2 or x s . By considering the triangle formed by b, x, , x 3 , we see that x is necessarily joined to x, and x3. Since x is not adjacent with b, it follows that

( x , x2) E U

and

( x g ,x)

E

U.

This contradicts x $ A . Now, direct transitively the subgraph generated by A and the subgraph obtained by removing from G the vertices of A other than x1 (this is possible from the induction hypothesis), then direct each' edge of the form [x,a], with x $ A and a E A , in the same direction as [x,x,]. The resulting graph is transitive. Q.E.D. A conjecture due to A. Hoffman can now be proved:

Theorem 8 (Gilmore, Hoffman [1964];Ghouila-Houri, [1962]). A necessary and suficient conditionfor a simple graph G = (X,E ) to be a comparability

367

PERFECT GRAPHS

graph, is that for each pseudo-cycle [a,, a,, ..., a,, , ,, a,] of odd length, there exists an edge of the form [a,, a,,,] (where the addition is modulo 29 1). A pseudo-cycle is a sequence of vertices starting and ending with the same vertex such that any two consecutive vertices are adjacent. For example, graph G in Fig. 16.1 contains a pseudo-cycle

+

[a, b , c , d, c , e , f , e, b , a1

of length 9, but no edges of the form indicated. Thus, this graph is not a comparability graph. 1. Necessity. If the graph ( X , U ) of an order relation contains an odd pseudo-cycle [a,, a 2 , ..., a,, + a,] without any edge of the form [acJa, + J , then when this pseudo-cycle is traversed, arcs are alternately encountered along and against their direction. But, this is impossible if the pseudo-cycle is odd.

2. Sufficiency. Consider a simple graph G the simple graph H = ( Y , F ) defined by (l)

(

-

Y b,Y’lEF Y E

c>

=

( X , E ) , and associate with G

y = (a, b), a , b E X , (a, ~ ] E E { Y J ’ } ={(a,b),(b,c)}, [a,cI#E.

It is easy to show that the hypothesis implies that H contains no odd cycles, and therefore, by Theorem (4, Ch. 7), H is bipartite. Thus, the vertices of H can be partitioned into two classes Yo and Y - Y o ,with

Y,Y‘E y o Y , Y ’ E Y - yo

=. =>

b,Y’l#f;, Iv,Y‘l#F.

Direct each edge [a, b] of G from a to b (and write a > b) if (a, b) E Yo: From (l), note that in graph H the vertices y = (a, 6) and y’ = (b, a) are adjacent (because [a, a] $ E, i.e. G has no loops); thus, these vertices belong to two distinct classes, and each edge of G has received exactly one direction. Furthermore, we have =>

b>c

[

(a, b) E yo

or ( b , c> E yo

]

*

=>

[(a, b ) , ( b ,C

[a,c]~E

-

) ] U

=>

[

c>a.

368

Therefore, the relation comparability graph.

GRAPHS

> is a semi-order relation, and by Theorem 7, G is a

Q.E.D. Other characterizations of comparability graphs have been discovered by Gallai [1967].

3. Triangulated graphs A graph G is defined to be a triangulated graph if each cycle of length > 3 possesses a chord (i.e. an edge joining two non-consecutive vertices of the cycle). The concept of triangulated graphs is due to Hajnal and Surhyi [1958]. A subgraph of a triangulated graph is also a triangulated graph because, otherwise, it would have a cycle of length > 3 without chords, and G would also have a cycle of length > 3 without chords

1. A tree is a triangulated graph. EXAMPLE

EXAMPLE 2. A cactus with only cycles of length 3 is triangulated because it contains no cycles of length > 3.

3. Graph L(G) is defined to be the graph in which each vertex EXAMPLE Zi represents an edge ei of G and with two vertices joined together if, and only if, they represent adjacent edges in G. If G is a cactus with only cycles of length 3, then the graph L(G) is triangulated: Otherwise, L(G) contains a cycle [Fl,F,, ..., g k , F1] without chords and with k > 3, and this cycle corresponds in G to a cycle ( e l , e , , e 3 , ..., ek, e l ) of length > 3, which contradicts the definition of G . The structure of triangulated graphs can be clarified by the following theorem, due essentially to Hajnal and Sursinyi [1958].

Theorem 9. If G is a triangulated graph, then each minimal articulation set of G is a clique.

Let A be a minimal articulation set of G. Removing A creates several connected components C, C', C", .... Each vertex a E A is joined to each of these components. (Otherwise, A - { a } would be an articulation set of G, which contradicts the minimality of A.) Let a , and a, be two vertices in A . There exists a chain p = [a,, cl, c, ..., c,, a,], where c1,

cz,

...)cp E c .

369

PERFECT GRAPHS

Assume that p is a chain of this type with minimum length. There also exists a chain p’ = [az,c;, c;, ..., cb, all, where c;,c;, ..., C ; E C

Assume that /I’ is a chain of this type with minimum length. The cycle p p’ has no chords of the following types:

+

- [a,, cil

( i # 1)

- [ci, cj]

(i # j )

- Ca,,cil

(i #

- Cci,

P)

- [a,, cjl

( j # 1)

- [ci,

(i # j )

- [a,,

,

because C and C‘ are two distinct connected components of Gx-a

c~I c;]

1

because p would not be the shortest chain

41 (i #

because p’ would not be the shortest chain.

4)

+

Since the graph is triangulated, cycle p p’, which has a length of at least 4, possesses a chord. This chord must necessarily be [a,, az]. Thus, any two vertices of A are adjacent, and A is a clique.

Q.E.D. Corollary 1 (Berge [1960]). A triangulated graph is y-perfect. Clearly, this is true for graphs with 1, 2, or 3 vertices. If this is true for all graphs with < n vertices, we shall show that it is true for a graph G with n vertices. Suppose that G is neither disconnected nor a clique (because then the proof would follow immediately). From Theorem 9, G contains an articulation set A that is a clique. Each piece relative to A is y-perfect by the induction hypothesis. Thus, from Theorem 2, G is y-perfect. Q.E.D.

Corollary 2 (Hajnal, Surinyi [1958]). A triangulated graph is a-perfPct. The proof is identical to the proof for Corollary 1, with Theorem 3 replacing Theorem 2. 4. i-Triangulated graphs A graph G is defined to be i-triangulated if each odd cycle p of length > 3 has a set of chords E(p) that form with p a planar graph whose unbounded

310

GRAPHS

face is the exterior of p, and whose bounded faces are all triangles. This concept is due to Gallai [1962]. The following theorem gives several characterizations of i-triangulated graphs.

Theorem 10. For a graph G,the following properties are equivalent: (1) G is i-triangulated, (2) For each elementary cycle p of odd length in G, and for each edge e of p, there exists a vertex of p that forms a triangle with edge e, (3) Each elementary cycle of G of odd length k has k - 3 chords that do not cross one another, (4) Each elementary cycle of G of odd length greater than 3 has at least two chords that do not cross one another, (5) If two distinct vertices x and y on an elementary cycle p of odd length are non-adjacent,, then p contains two other vertices z and i (distinct from x and y ) that are adjacent and such that the four uertices x, z, y , t are encountered on 11 in this order, ( 6 ) For each elementary cycle p = [x,,

x1, x21

.-.,X2qr X o l

of odd length and for each index i such that [ x i - 1 , x i + 1 ]$ E, there exists an index j # i - 1, i, i 1, such that [xi, x,] E E.

+

The proofs for (1) immediate.

(6)

(1)

3

-

=>

(2), (2)

=$-

(4), ( 3 )

=>

(4), (1)

3

(5), (5) + (6) are

(4) Let p = [x,, xl, x l , ..., xlp, x,] be a cycle in G of odd length > 3. If there is an edge [xi,x,] E E w i t h j # i - 1, i + 1, i - 2, i + 2, then one of two cycles determined by this edge and by p is of odd length > 3 and has a chord; since this chord does not cross chord [x,, x,], condition (4) follows. On the other hand, if each chord is of the form [x,, X , + ~ ]condition , (4) also follows. (3) Let p be a cycle of odd length n. Let m be the number of chords of E(p). Let f denote the number of triangles in this triangulation of p. From Euler’s formula, we have n - (n m) ( f 1) = 2. On the other hand, we have 3 f + n = 2(m n) . Hence, m = n -3.

+ + + +

(4)

=>

(1) Let p be a cycle of odd length, and let E,, denote a maximum set of chords that do not cross one another. The edges of p and of E,,

371

PERFECT GRAPHS

form a planar graph. Each bounded face of this planar graph is either a triangle or an even face. Since p is odd, at least one of the faces is a triangle. If not all of the bounded faces are triangles, then an even face can be found adjacent to a triangle. By removing the edge that separates them, an odd face having at least five sides is obtained. Since this face has at least two chords that do not cross one another, this contradicts that E,, is maximum. Q.E.D. Gallai [1962] proved : Eiiery i-triangulated graph is a-perfect. The original proof was very involved, but later, L. Surhyi [1968] presented a simpler proof. In fact, Gallai proved a stronger result: I f G is a connected i-trianguluted graph, at least one of the foliowing three conditions is satisfied:

( I ) G contains an articulation set that is a clique, (11) There exists a set A c X such that G A is bipartite and Gx-,, is com-

plete, and such that

aEA, b E X - A

3

[a,b]EE,

(111) The relation ‘‘x = y or [x, y ] c$ E” is an equit3alencerelation.

Other classes of perfect graphs have been investigated, and the following results have been proved : (1) I f each cycle of odd length > 5 has two chords that cross one another, then the graph is a-perfect (Olaru; in Sachs [1970]). (2) If each odd cycle contains an edge such that each maximal clique that contains this edge contains three certices of the cycle, then the graph is a-perfect (Berge, Las Vergnas [1970]). The proof for this result in a different form is presented in Chapter 20. 5. Interval graphs

Consider a family &’ = ( A , , A 2 , .. ., A,) of intervals on a line. The representative graph of &‘ is defined to be a simple graph G in which each vertex a, corresponds to an interval A , , and with two vertices joined together if, and only if, the two corresponding intervals intersect. Such a graph is also called an intercalgraph. G. Haj6s 1957 and N. Wiener were the first to study interval graphs, and so far two topological characterizations have been found; the first is due to Lekkerkerker and Boland [1962], and the other is due to Gilmore and Hoffman [1964]. For bipartite interval graphs, see also Kotzig [1963].

372

GRAPHS

EXAMPLE 1. Each student visits the university library once a day, and at the end of the day he submits the list of the names of the students met in the library while he was there. The problem is to find the order in which the students entered the library. Construct a graph G in which each vertex represents a student, with two vertices joined together if, and only if, the corresponding students were present in the library at the same time. This graph is an interval graph, because it represents the intervals of time during which the students were present in the library. The theorems of this section will give all the possible solutions. EXAMPLE 2. In genetics, tests can be performed to determine if two chromosomes overlap one another, and the problem is to prove or disprove that a set of chromosomes are linked together in linear order. Construct the graph G whose edges are the pairs of overlapping chromosomes; if this graph is not an interval graph, it follows that the chromosomes cannot be linked in linear order. EXAMPLE 3. Consider the following problem that occurs in psychology: Given a finite number of points x, , x2, , ., x, on a line and an infinite family s2 of intervals, two points x, and x j are said to be indistinguishable if there is an interval in family Q that contains both of them. The indistinguishable pairs determine a graph, and we may ask for the characteristic properties of these graphs. In fact, such a graph represents a family of intervals I,, Z,, ..., Z, , where interval ZIcorresponds to point xland is defined by

.

I , = [xt,

+~~]~U{W,/O~ER,O~~X~}.

If xi < x j , point x, and point x, are indistinguishable if, and only if, xI E I t , which is equivalent to It n Zj # 0. Theorem 11. Every interval graph G is triangulated. Suppose that there is a cycle [al, a,, ..., a,, a,] without chords. Let At be the interval corresponding to vertex al, since interval A k does not overlap with interval A k - 2 , the initial endpoints of the A* constitute a monotone sequence; and therefore, A , cannot overlap with A , , which contradicts that [a,, a,] is an edge of G. Q.E.D.

Remark. The converse is not true: Graph G in Fig. 16.4 is triangulated, but we shall show that G is not an interval graph. Clearly, the intervals A , , A , , A , are pairwise disjoint and may be placed in this order on the line. But, then, interval B, that intersects intervals A , and

PERFECT GRAPHS

373

"2

Fig. 16.4

A3 must also intersect interval A2, which contradicts that vertices a2 and b3 are non-adjacent.

Corollary. Every interval graph G is a-perfect and y-perfect. Since G is triangulated, G is a-perfect (Corollary 2 to Theorem 9), and yperfect (Corollary .1 to Theorem 9). Q.E.D.

APPLICATION 1 (Gallai).

If&'

is afinitefamily of intervals on a line, and if

k is the maximum possible number ofpairwise disjoint intervals in -c9, then there exist k points on the line such that each interval contains at least one of these points. Let graph G represent these intervals. Each clique in G corresponds to a family of intervals having one point in common, by the Helly theorem.") The result follows. APPLICATION 2. If& is a family of intervals on a line, and k is the maxintum number of interiials that together have a non-empty intersection, then the intervals can be coloured with k colours such that no two intervals with the same colour intersect. The first application can be related to Example 1. The minimum number of photographs of the library that are needed so that each student is photographed at least once equals the maximum number of *students who were pairwise not present together in the library.

We shall now study characterizations of interval graphs. Lemma 1. If G is an intercal graph, then its complementary graph comparability graph.

is a

(l) Helly's theorem for intervals can be stated as follows: If a family of intervals does not contain two disjoint intervals, then all the intervals have a common point. A simple proof is given in Chapter 17, Section 3, Example 1.

3 74

GRAPHS

If graph G represents a family of intervals d,two vertices x and y are ' . linked together in G if, and only if, they represent disjoint intervals of a Direct edge [ x , y ] from x to y if interval y is to the right of interval x on the line. This produces a I-graph ( X , V ) such that:

* *

(X,Y)EU

( x ,Y ) E u, (v,4 E Thus,

u

(Y,X)#U. (X,

4E u.

C is a comparability graph. Q.E.D.

The following lemmas treat a simple graph G with the following properties:

(1) Euery cycle of length 4 has a chord, ( 2 ) the elementary graph C is a comparability graph. Let G = (A', E ) be a graph with these properties. Let V denote the family of the maximal cliques of G. Let C = ( X , U) denote the complementary graph of G, assuming that the edges of G are directed transitively.

c

Lemma 2. Let C1,C2E %'; there exists an arc in that joins together set C1 and set C,. Furthermore, all the circs in G that join Cland C, hare the same direct ion. If C, and C2are two distinct maximal cliques of G , then G contains two non-adjacent vertices a E C, and c E C2.(Otherwise, C1U C2 would be a clique.) For example, let (a, c ) E U. Let bd be another edge of C with b E C1, d E C2 and a # 6. If c = d, then edge bd has the same direction as edge ac (from C , to C,) because, otherwise, G would not be a comparability graph. If c # d, then either ad or bc is an edge of (since, otherwise, the cycle [a, d, c, 6, a] of G would contain no chords). Without loss of generality, let this edge be ad. Then, (a, d ) E U. Hence,

(6, d)E u . Thus, in both cases, edge bd is directed from C1to C2,as is edge ac. Q.E.D.

c,

c2

@::+-;:@ h

- - - -- - - - - - - Fig. 16.5

-- --- Arc EdgeofofGG

3 75

PERFECT GRAPHS

Lemma 3. Let H be a I-graph whose vertices represent the cliques of $? and with an arc from C, to C, if there exist two certices a E C, and c E C, in G with (a, c ) E U. Then, H is a complete, transitive I-graph.

By Lemma 2, H i s complete and anti-symmetric. It remains to show that if (C,, C,) is an arc of H and (C,, C,) is an arc of H , then (C,, C), is an arc of H. We shall suppose that (C,, C,) is not an arc of H (and we shall show that this leads to a contradiction). Then, (C,, C,) is an arc of H .

Edge of G

Arc of

G

Fig. 16.6

For example, suppose that: a,bEC1, c , d € C 2 , e,fEC3 (b, c) E

u,

(d,4 E

u, (f,4 E u .

I n this'case, ad is an edge of because, otherwise, the cliques { a, d } and C, would contradict Lemma 2. Thus, from Lemma 2, (a, d ) E U. Since U is a transitive relation, (a, e) E U . But, then, (A a) E U and (a, e) E U implies that (f,e) E U. This contradicts that both vertices e and f belong to C,. Q.E.D. Theorem 12. (Gilmore, Hoffman [1964]). A simple graph G is an interval graph if, and only if, the following two conditions hold:

(1) every cycle of length 4 has a chord, (2) the complementary graph C is a comparability graph. Necessity. Condition (1) is necessary because a graph that represents a family of intervals is triangulated (Theorem 11). Condition (2) is necessary from Lemma 1. Suficiency. Let G = ( X , E ) be a simple graph that satisfies conditions (1) and (2). Let k? = { C,, C2,..., C, }be the family of the maximal cliques of G. As in Lemma 3, form the I-graph H . Graph H is complete, anti-symmetric

376

GRAPHS

and transitive. From Theorem (5, Ch. lo), H contains a unique hamiltonian path p. Suppose that the cliques of%' are indexed so that p = [ C , , C,, . .., C,]. Consequently, (Ci, C,) is an arc of H if, and only if, i < j . Note that G represents the sets

z,

=

{ i/

ciE w, ci 3 x )

9

because two vertices x and y are joined together in G if, and only if, I,nI,#

@.

To show that I, is an interval, it suffices to show that

::q

==.

XEC,.

p p ) and (y, x) E U (since r < 4); this gives the required contradiction. Q.E.D. 6. Cartesian product and Cartesian sum of simple graphs

In Chapter 14, three operations were defined for 1-graphs: normal product, Cartesian product, and Cartesian sum. If G = ( X , E ) and H = ( Y , F ) are two simple graphs, the Cartesian sum of graphs G and H is defined to be a simple graph G H whose vertex set is X x Y, and with two vertices xy and x'y' joined together if, and only if,

+

either x

= x', [y, y ' ] E

F, or [x, x'] E E, y

=

y' .

311

PERSECT GRAPHS

Let the Kronecker delta be defined on graph G as follows: &(x, x’) = 1 if x # x‘, and &(x, x’) = 0, otherwise. The number of edges that join vertices xy and x’y’ in graph G + H can be written as ~zG+H(xY,

x ’ v ’ ) = &Ax, x‘) m d y , Y’)

+ b ( y , Y’) mc(x, x‘) .

The Cartesian product of graphs G and H is defined to be a simple graph G x H whose vertex set is X x Y, and with two vertices xy and xfi’ joined together if, and only if [x, x‘] E E and [y,y’] E F. The number of edges joining these two vertices in graph G x H can be written as

m, x H ( V 7 x’ r’>= 4 x 9 x’) ’%f(YY

r’).

The nornialprodtrct (or simply, product) of graphs G and H is defined to be a simple graph G.H whose vertex set is X x Y, and with two vertices xy and x’y’ joined together if and only if, either

x = x’,

[ y , y’] E F

3

or [X,X‘]EE,

or [ x ,X ’ I E E Y

y=y’7

[v,4”l E F

*

The number of edges joining these two vertices can be written as nzG.H(X,

y> = = mG(X, x’) mHb, y’)

+ dG(x, x‘) nrH(Y, Y’) + mG(x, x’) 611(y,y’)

*

Note that these operations are commutative. If the definitions are extended to more than two graphs, it can be shown that the operations are associative and distributive (C. Picard [1970]). These definitions were first introduced to study the chromatic number and the stability number. The relationships between these operations and the main fundamental numbers are described in this section.

Proposition 1. Let G and H be two graphs. Then

o(G

+ H ) = max { o(G),o ( H ) ) .

Let C1, Cz, ..., C, be the maximal cliques of G, and let D1,Dz,..., D,be the maximal cliques of H ; then the maximal cliques of G H are the sets C, x { y , } and the sets { xi } x D,. Hence,

+

o(G + H) = max{ I { xi 1 x D j I , I C, x { y j } I }

= max { 4 G ) , OW)}.

Q.E.D.

3 78

GRAPHS

Proposition 2 (Vizing [ 19631, Aberth [ 19641). Let G and H be two graphs. Then Y(G

+ H> = m a { Y ( G ) 7 Y ( H ) ) .

1. Let r = max { j ( G ) , y ( H ) }; colour G and H with r colours 0, 1, 2, r Let g(x) = k if vertex x is coloured k. For each xy E X x Y, let

- 1.

+

Consequently, g(xy) defines a colouring of G H because if xy and x‘y’ are adjacent and have the same colour, then either x = x’ and vertices y and y‘ are adjacent in H, or y = y’ and vertices x and x‘ are adjacent in G. In the first case,

A4 + g

o

= A4 + gD’>

(mod. r )

and g ( y ) = g(y’), which is impossible because [ y ,y’] E F. A similar result follows for the second case. Hence, G H is r-colourable.

+

+

+

2. Graph G H cannot be (r - I)-colourable, because, graph G H contains a subgraph isomorphic to G and a subgraph isomorphic to H . Hence, y(G H ) = r. Q.E.D.

+

G + H

Y

Y-

P-

H

aI

I

a

h

Fig. 16.8

Proposition 3. Let G and H be two graphs. Then a(G

+ H ) 2 a(G) a(H) .

I

c

d

X

PERFECT GRAPHS

319

If S is a maximum stable set of G, and if T is a maximum stable set of H, then the set S x T is a maximum stable set of G x H. Hence, a(G

+ H> 2

IS x T ! = a(G)a(H) Q.E.D.

In Fig. 16.8, the vertices of a maximum stable set are circled, and it is easily seen that

a(G

+ H ) = 4 > a(G)a(H) = 3 .

Proposition 4. Ifgraphs G and H respectively have orders n(G)and n(H),then a(G

+ H ) < min { a(@ n ( H ) , a(H) n(G)} . +

Let So be a maximum stable set of G H. Its intersection with set X x { y , } cannot have more than a(G) vertices; hence, 1 So I ,< Y I a(G). Similarly, I So I < I X 1 a(H). The formula follows. Q.E.D. Proposition 5. Ifgraphs G and H respectively haw order n(G) and n(H), then

B(G

+ H) < min { n(G) O(H), n ( H ) ' X G ) } .

Let $9 = (C,,C,,..., C,) be a minimum partition of G. The sets Cix { y, are cliques of G H and cover X x Y. Thus,

+

O(G + H )

ICx

DI

=

w(G) o ( H ) .

.

Conversely, if Co is a maximum clique of G H, let the projection of Co on X be C, and let the projection of Co on Y be D. Since C and D are cliques in G and H , the set C x D is a clique in G . H. Hence, C, = C x D (since Co is a maximal clique). Thus, w(G.H) = I Co I = I C I x

I D I < o ( G )o ( H ) .

The required equality follows.

Q.E.D. Proposition 9. Let G and H be two graphs. Then cr(G.H) 2 cc(G) a ( H ) . If S and Tare maximum stable sets respectively in G and H , then

I SI

= .(G),

I TI

= U(H).

The Cartesian product S x T is a stable set in G . H , and, consequently, we have o((G.H) 2 I S x TI = I SI x I TI = a(G)cc(H). An upper bound for a(G .H ) will be given in Chapter 19, 92.

Q.E.D.

Proposition 10. O(G.H) < O(G) O(H) .

Let (Cl, C2,..., C,) be a minimum partition into cliques of graph G. Let (Dl, D2,..., DJ be a minimum partition into cliques of graph H

382

GRAPHS

Ingraph G.H, set C, x D,is a clique, for i = 1,2, ..., p , and j = 1,2,..., q. The cliques C,x D,partition graph G.H. Therefore, O(G.H)

< p q = O(G) 8 ( H ) . Q.E.D.

APPLICATION (Shannon

[1956]). Consider a transmitter that can emit five

signals, a, b, c, d, e, and a receiver that can interpret each of these signals in two different ways. Signal a can be interpreted as either p or q, signal b can be interpreted as either q or r, etc., as shown in Fig. 16.9. What is the maximum number of signals that can be used for a code so that there is no possible confusion on reception? This problem reduces to finding a maximum stable set S of a graph G shown in Fig. 16.10, in which two vertices are adjacent if and only if they represent two signals that the receiver can confuse. For example, we can take S = ( a , c } , since graph G in Fig. 16.10 has stability number a(G) = 2.

S

u(G)=2

e Fig. 16.9

Fig, 16.10

C.G=G2

b

~

t

b

c

a(G2)=5

Fig. 16.11

d

e

383

PERFECT GRAPHS

Instead of single letter words, we could use a code of two letter words, provided that no two letter words of this code can lead to confusion on reception. Thus, the letters a and c which cannot be confused can form the code: aa, ac, ca, cc which has a vocabulary of (a(G))2 = 4 words. But an even richer code is: aa, bc, ce, db, ed. It is easily seen that no two of these words can be confused by the receiver. This gives a vocabulary of 5 words. Note that the words xy and x'y' can be confused if, and only if, these two words are adjacent vertices in the normal product G G = GZ.A code consisting of 2-letter words has a maximum vocabulary of a(G2) words. More generally, the maximum possible vocabulary for a code of k letter words is the stability number of the product.

.

Gk = G.G.G ... G . k

With this as motivation, the capacity of graph G (or, "zero-error capacity") is defined to be the number c(G) = sup

73).

The capacity of the graph G in Fig. 16.10 is known to be between 2 and 3; however, its exact capacity remains unknown. Furthermore, Ljubich [19641 has shown that IC4a(G)"tends to c(G)when k tends to infinity.") An important problem is the characterization of the graphs whose code is improved when longer words are used. We shall prove the following result, due to Shannon : I f a graph G satisfies a(G) = B(G), then the code cannot be improved by using longer words. From Propositions 9 and 10, we have

.

Fekete's theorem states that if a sequence of numbers al, as,. . is sub-additive (ie., urn+,,"u,+un), then

4n n

+

inf n

an

-.

"

If we let a, = -log a(G"), then a sub-additive sequence is formed because a(Gmtn) 3 a(Gm)a(Gn), log a(Gm+n)2 log a ( P ) + log a(Gn).

Thus,

and

384

GRAPHS

Hence,

Thus, a(G)

=

B(G) implies that c(C) =

and the code cannot be improved..

Q.E.D. In particular, when each signal is determined by its modulation frequency, two signals can be confused if, and only if, the corresponding frequency

intervals of the two signals intersect (“linear noise”). In this case, graph G represents a family of intervals, and therefore, by the corollary to Theorem 11, u(G) = B(G). Hence, ifnoise is linear, then a code cannot be improved by using longer words.

EXERCISES 1. A graph G is defined to be vertex-critical a-imperfect if it is not a-perfect, but if the removal of any vertex makes the resulting graph a-perfect. Show that such a graph G satisfies the following properties: ( I ) for each vertex, there exists a maximum stable set that does not contain this vertex, (2) for each vertex, there exists a minimum partition of G into cliques such that one of the cliques contains only this vertex. E. Olaru has shown that each vertex x belongs t o a cycle of length 3 5 such that all the chords of the cycle are incident to x. (Sachs [ 19701) 2. Graph G is defined t o be edge-critical a-imperfect if G is not a-perfect but if the graph G e is a-perfect for each edge e of G.

-

Show that such a graph is a-critical (see Chapter 13). Show that a connected graph G is edge-critical a-imperfect, if, and only if, C is an odd cycle of length 2 5 without chords. Hint: Use Corollary 3 to Theorem (6, Ch. 13).

c,

3. Let C, be a cycle of length 7, and let G = be its complementary graph. Show that (1) G has no odd cycles of length > 3 without chords, (2) G is vertex-critical a-imperfect. 4. Consider a simple graph G = (X,E) with the following property: if x l r x z , x 3 , x4 are four distinct vertices with [ X I , xz] E E, [ x z , x 3 ] E E, [x3, xq] E €, then [xlr x3] E E, or [ x z , x4]E E. Show that graph G contains a vertex a that is adjacent to every other vertex of the graph. Show (by induction on n) that the complementary graph of G is a comparability graph. (Wolk [1965]) 5. Show that a simple graph G is the transitive closure of an arborescence if, and only if, it satisfies the two following properties:

385

PERFECT GRAPHS

(1) G is connected, (2) Ifxl, xa, x3, x4are four distinct vertices with [xl, xp]E €, [xl, x31 E E, ( ~ 3xr] , E €, then [xl, X J E € or [xa,x4JE E. Hint: The sufficiency can be proved by induction on n, and by using Exercise 4.

(Wolk [1965]) 6. Show that if q = w(G), then NG

+ 4)

= n(C).

7. If p ( G ) denotes the number of connected components of graph G, show that p(G

+ HI

= p(G)dH).

8. Show that if G and H are two simple graphs with hamiltonian cycles, then graph

+

H has a hainiltonian cycle. Show that if both G and H have eulerian cycles, then graph G cycle. G

+ H has an eulerian (Aberth [1964])


PART TWO

Hypergraphs


CHAPTER 17

Hypergraphs and Their Duals

1. Hypergraphs Let X = { xl, x,, ..., x,} be a finite set, and let 8 = ( E L /i E Z ) be afamily of subsets of X.The family € is said to be a hypergraph on X if (1)

Ei

#

la

(iEI)

UEi=X ief

The couple H = (X,S)is called a hypergraph.. I X I = n is called the order of this hypergraph. The elements xl, x,, ..., x, are called the vertices and the sets E l , E,, ..., Em are called the edges. A hypergraph is shown in Fig. 17.1. An edge Ei with I El I > 2 is drawn as a curve encircling all the vertices of El. An edge ELwith I Ei I = 2, is drawn as a curve connecting its two vertices. An edge Ei with I Ei I 3 1, is drawn as a loop as in a graph. If the edges Ei are all distinct, the hypergraph is said to be simple, and 8 is a set of non-empty subsets of X.If I Ei I = 2 for all i, and if the hypergraph H is simple, then H is a simple graph without isolated vertices. The main purpose of the theory of hypergraphs is to generalize results from graph theory. In a hypergraph, two vertices are said to be aa’jacent if there is an edge El that contains both of these vertices. Two edges are said to be adjacent if their intersection is not empty. The incidence matrix of hypergraph H = ( X , 8)is a matrix ((a:)) with m rows that represent the edges of H and n columns that represent the vertices of H , such that

4

[

= 1 =0

if if

xjeEI xi$&.

Each (0, 1)-matrix is the incidence matrix of a hypergraph if no row or column contains only zeros. To each hypergraph H = (X;E l , E,, ..., Em)there corresponds a hyper389

390

HYPERGRAPHS

graph H* = ( E ; Xl, X 2 , ..., X,) whose vertices are points e l , e 2 , ..., em (that respectively represent E l , E,, ..., Em)and whose edges are sets XI, X,, ..., X,, (that respectively represent x l , xz, ...,x,), where, for allj,

X,

= {ei/i
0, the k-section of hypergraph H = ( X , 8)is defined to be the couple H(,) = ( X , b,,,) formed by X and the set &(k)

=

{F/F

c

x ; 1 G I F I < k ; F c Ei

for some E , E 8

}.

Proposition 2. If H is a hypergraph with rank function r(S), the k-section H(,) of H is a simple hypergraph with rank function r(,)(S) where r(k)(S) = min

{ k , r ( ~1.)

HYPERGRAPHS AND THEIR DUALS

391

The proof is immediate. Thus, the 2-section H,,, of a hypergraph H i s a graph with the same vertices as H and with a loop attached to each vertex. Let denote this 2-section without its loops. Clearly, ( H ) z is a simple graph. 2. Cycles in a hypergraph

In a hypergraph H = ( X , S),a chain of length q is defined to be a sequence (x,, E l , x z , Ez,..., E,, x q + d such that (1) xl, x2, ..., x, are all distinct vertices of H,

(2) E l , Ez,..., E, are all distinct edges of H, (3) X k , X k + l E Ek for k = 1, 2, ..., 4. If q > 1 and x,, , = xl, then this chain is called a cycle oflength q. If H is a graph, these definitions correspond to those for an elementary chain and an elementary cycle (except that a loop is not considered as a cycle in the theory of hypergraphs).

Fig. 17.1. Connected hypergraph (A’:E l , E,, E 3 , E,, Es, Ee) of order 8 with a E, I - 1) = 2 1 2 2 1 -k 0 = 8. unique cycle and with

c(I

+ + + +

If there is a chain in the hypergraph that starts at vertex a and terminates at vertex b. then we shall write a = b. Proposition 3. The relation a = b is an equicalence relation, whose classes are called the “connected components” of the hypergraph. If Clis a connected component that intersects edge E, then C, contains E. It suffices to show that this relation is an equivalence (1) a F a, because there exists a chain of length 1 beginning and terminating at vertex a, (2) a = b implies b = a,

392

HYPERGRAPHS

(3) a = b, b = c implies a = c, because if there exists a chain of length p from a to b and a chain of length q from b to c, then the edges of these chains form a chain of length < p + q from a to c. The theorem follows. Q.E.D.

Proposition 4. If H is a hypergraph with n certices, ~ ? edges i and p connected components, it contains no cycles i f , and only $ m

Consider the bipartite graph G ( H )whose vertices represent the vertices and edges of H , where the vertex representing x, is joined to the vertex representing Ei if, and only if, x, E Ei. Graph G ( H ) has p connected components, m + n vertices, and zr=lI Ei I edges. Hypergraph H contains no cycles if, and only if, G ( H ) is a forest, i.e. if m

or m

Q.E.D.

Proposition 5. If H = (X,8)is a hypergraph with G: = (E, / i E I), then H has no cycles $ and only if,

1 . If H contains a cycle ( a l , E l , a , , E,, { 1, 2, ..., q } , we have

..., E4, al), then, by letting Q

=

and the inequality of the proposition cannot hold. 2. If H = (A’,8)contains no cycles, then a partial family (El / i €1)also contains no cycles. If this family forms a hypergraph with p connected components, then by Proposition 4,

Q.E.D.

393

HYPERGRAPHS A N D THEIR DUAL3

Proposition 6 . A hypergraph ( X , 8) with p connected components and n wrtices has a unique cycle if, and only if,

c (IEJ-l)=n-p+l. m

i= 1

The bipartite graph G ( H ) defined in Proposition 4 has p connected components, nr + n vertices, and IfI Ei1 edges. It has exactly one cycle if, and only if, its cyclomatic number is ni

Q.E.D. Theorem 1 (LovAsz [1965]). r f a hypergraph H = (X,8)of order n with p connected components contains no cycles of length 2 3, and no loops, and if

I Ein E j 1

2. Assume that it is not true for a hypergraph H with n vertices, and we shall show that this leads to a contradiction.

2. Hypergraph H is connected because, otherwise,

for each connected component C,, and by summing over k, the inequality is obtained.

3. We shall show that H contains no vertex whose removal disconnects H (“articulation vertex”). If such a vertex xo existed, and if its removal created two connected components C, and C 2 ,then by letting D1= C, u { xo }, we would have:

C

(I Ei I

- 2)

< I D,I - 1

3

EicDi

Since H has no loops, there is no edge contained in both D , and

394

HYPERGRAPHS

D2.Furthermore, each edge intersecting D , (resp. D,)is entirely contained in D, (resp. DJ.Thus,

x,

4. We shall show that no vertex x1 belongs to onlyone edge. If x1 E El and 4 Ei for i # 1, and if I El I 2 3, then the hypergraph

-

(El { x1 1, E2Y ... E m ) contains no cycles of length 2 3 ; then, by the induction hypothesis, Y

m

2 (IEJ

- 2) - 1 < (n

- 1) - 1 ,

i= 1

which implies that the inequality of the theorem holds. If I EiI = 2, the same result is obtained by considering the hypergraph (E,, E3, ..., Em). 5. We shall show that I Ein E, 1 = 0 or 2 for Ei# E,. Suppose that El n E2 = { xo}. Since I Ei I 2 2, consider a vertex a E El with a # x o , and a vertex b E E2 with b # xo. Since xo is the only vertex of E, n E,, it follows that a # b. Since xo is not an articulation vertex from Part 3, there exists a chain (a, E ; ,

4, E ; , ..., x;,

E;, b)

connecting a and b that does not use vertex xo. If there existed two indices i a n d j such that E,' = El and E; cycle (XO,E!,Xf+l,EI+l,

...,E[i,X

=

E2, then the

0)

+

would be of length 2 3 (because i 1 # j ) . If there existed an index i such that E; = El and no index j such that E,' = E,, then the cycle

(~g,Ef,x;+l,E;+ - .~. ,,E i ,b,Ez,Xo) would be of length 2 3. If there existed no index i such that E( = El or E; = E,, then the cycle (aYE;,x;Y *-*YE;,byJ%,Xo,E1,a) would be of length 2 3.

395

HYPERGRAPHS A N D THEIR DUALS

In all cases, there is a cycle of length 2 3 in H , which contradicts the hypothesis. 6. Consider the family 9 = { F / F = E n E’, E, E’ E 8,E n E‘ # 0 } where each set contains exactly two vertices. We shall show that 9is a partition of X . Clearly, U F E F= ~ X because, from Part 4, each vertex x E Xis covered by at least two edges Ei and E, E 8,and E, n E, E 9. Furthermore, let F and F‘ be two different sets of 9with

F = El n E z ,

F’

=

E; nE ; .

If there exists a vertex xo E F n F’, we can write: F={xo,xl},

F ’ = { x ~ , x ; } , xi # x ; *

If El = E ; , hypergraph H contains a cycle (xo, E,, xl, E l , x i , EL, xo) of length 3, which contradicts the hypothesis. If El # EI ,E4 ,and if E2 # E; , E & let E l n E; =

{.yo,

a, }

E , n EL

=

{ x o , a, } .

Clearly, a, # a, because, otherwise, El n E2 = Ei n E i . Therefore, we may assume that, say, a, # x i . Thus, there is a cycle (xo, E l , al, E ; , xi, €;, xo) of length 3, .which contradicts the hypothesis.

7. Construct a bipadte graph G, whose vertices represent the m edges of 8 in one class and the sets of 9 in the other class, and where the vertex representing Ei E d and the vertex representing F? E F are joined if, and only if, E, n F, # 0. Note that, in this case, F, = Ei because, otherwise, F, = E, n E; with E,, E; # E,, and therefore Ei n Ej and F, could not be two different classes of 9. Thus, the number of edges in graph G is

m+n From Part 6, the number of vertices in G is n(G) = 7 . 8. Note that graph G contains no elementary cycle [F,, El,F,, E,, ..., E k , F,], because then k > 4, and if we select one x, in each Fi,the cycle ( x , , E l , x,, E,, ..., xl) would be of length 2 3, which contradicts the hypothesis. From Proposition 4, graph G satisfies

E,

m(G) < n(G) - 1 .

396

HYPERGRAPHS

Hence.

and

I

Q.E.D.

In particular, Theorem 1 shows that if H is a uniform hypergraph of rank 3 without cycles of length 2 3, then

m denote the family of subsets A , of X such that the subgraph G,, is also a tree. We shall show that this family satisfies the Helly property. Consider three vertices a, b, c of G. Let P be the set of indices p such that A , contains two of these three vertices. If the three vertices belong to the same elementary chain of G, then one of the vertices, say 6, is between the other two in the chain. Thus, b E npPPqp. If the three vertices do not belong to the same elementary chain, then there exists a vertex z of G that separates any two of them. Thus,

400

HYPERGRAPHS

n

# @. Thus, from the Corollary to Theorem 3, the Ai In both cases, satisfy the Helly property.

4. Representative graph of a hypergraph Let H = ( E ;Xl ,X2, ..., X,) be a hypergraph with n edges. The representative graph of H is defined to be the simple graph L ( H ) of order n whose vertices xl, x2, ..., x, respectively represent the edges XI, X,, ..., X , of H and with vertices xi and x j joined by an edge if, and only if, XI n X , # 0. For each simple graph G, there exists a hypergraph H such that G = L(H). More precisely, we have the following: Proposition 1. Let G be a simple graph with vertex set X , and let ( E l , E2,..., Em)be a family of subsets of X with the following properties: (1) each ELis a clique of graph G, (2) each vertex and each edge of G is covered by at least one E l . Then, G is the representative graph of the dual H of the hypergraph

H* = ( X ; E l , E z ,

..., Em).

Conversely, if G is the representative graph of a hypergraph

xi,xz,.-*,xJ, = (X.; E l , E2, ..., Em)satisfiesproperties H=(E;

then the dual H*

(1) and (2).

G = L ( H ) means that vertices xi and x, are adjacent if, and only if, there exists in H a vertex e, E Xt n X,. This is also equivalent to saying that there exists an edge Ek in H * that contains both xt and x,, or to saying that G

= (H*)Z.

Clearly, this can be expressed as H* satisfies (1) and (2).

Q.E.D.

For example, consider the graph G shown in Fig. 17.2. Graph G is the representative graph for each of the following hypergraphs: (1) the multigraph H , (the Et are the maximal cliques of G), (2) the hypergraph H2 (the Et are the edges of G), (3) the simple bipartite graph H 3 , etc. Let Q(G) denote the minimum order of the hypergraphs H for which G = L(H). For graph G in Fig. 17.2, we have R(G) = 2 because G = L(Hl). The following proposition shows that the determination of Q(G) reduces to the determination of a chromatic number.

401


G

d

e

Fig. 17.2

B

E

H , = (A. B. C, D. E )

Order

=

6

Fig. 17.3

Proposition 2. Let G be a simple graph with vertices xl, x2,...,x,,,none of which is isolated. Let G be a graph, each of whose vertices represents an edge of G, with two certices corresponding to edges [a,b] and [ x , y ] of G joined together if, and only i f , { a, 6, x,y } is not a clique in G. The minimum order Q(G) of the hypergraphs for which G is a representative graph equals the chromatic number

?)(GI.

402

HYPERGRAPHS

The graph G corresponding to graph G in Fig. 17.2 is shown in Fig. 17.4. bc

Cd

de Fig. 17.4

G

ce

1. We shall show that each q-colouring s2,..., s,) of G with q = y(C) colours yields a hypergraph ( E ; X , , X2,..., X,) of order q for which

(sl,

graph G is the representative graph. The set Sl of vertices of C with the i-th colour is a stable set. If [a, b] is an edge of G belonging to Si,then vertex a is either identical or joined to each of the endpoints of an edge of S,. Therefore, the endpoints of the edges in generate a clique Eiof G. The hypergraph ( X ; E l , E,, ..., E,) has the property that each edge and each vertex of G are covered by at least one of the E l . From Proposition 1, G is the representative graph of the dual hypergraph ( E ; X I , X2,..., X,) which has order I E I = q. Hence,

s,

Q(G)

G Y (GI

*

2. We shall show that a hypergraph H = ( E ; X,, X,, ..., X,,) of order q = Q(G) for which G = L ( H ) can determine q-colouring ( S , , S2, ..., 3,) of the vertices of G. Let A k denote the set of vertices of H that belong to exactly k of the sets XI. Then, = I E I=

I ~1 I + I AZ I + I ~3 I + ... .

} where e E To each To each vertex e E Al , associate the I-clique { and X,(e),associate the vertex e E A,, that belongs to exactly two sets x,(,) 1 of G. To each vertex e E A3 that belongs to exactly three 2-clique { xlCe), X,(e),X k ( e )associate , the 3-clique { of the sets x , ( ~ X) ,k ( e ) } of graph G, etc. Since H i s of minimum order, we have A' = 0. In this way, a family ( E l , E,,..., E,) of q-cliques in G is defined. Clearly, each edge [ x , , x,] of G is covered by at least one of these cliques (because Xin X , contains a vertex of E ) . Let 3' denote the set of edges of G contained in clique E l . Let S2 denote the set of edges of G covered by clique E2 that

403


s3

are not covered by El. Let denote the set of edges of G covered by clique E3 that are not covered by either El or E,, etc. The family ($, sa,..., S,) clearly is a q-colouring of the vertices of G. Hence, y(c)< Q(G), and SO Y(G)= Q(G). Q.E.D.

Proposition 3. Let G be a simple graph without isolated vertices and without triangles and with m edges. The minimum order of the hypergraphs H for which G = L ( H ) is R(G) = m. Graph defined as in Proposition 2 is an m-clique. Therefore, the required minimum order is R(G) = m. Q.E.D. Proposition 4 (Erdos, Goodman, P6sa [1966]). If G is a simple graph of order n without isolated vertices, then

For each n, this bound is the best possible. 1. From Theorem (5, Ch. l l ) , the edges and vertices of G can be covered by a family of 2-cliques and 3-cliques (El, E,, ..., &) where k < n2 From

[ - I.

Proposition 1, G is the representative graph of the dual of hypergraph ( X ; El, E,, ..., E J . Since this dual has order k , the minimum required order is

< k < [ %] . 2

R(G)

2. We shall show that for each n, there exists a graph G of order n such that

If n = 2 k is even, let G be the complete-bipartite graph Kk,k. Since graph G contains neither isolated vertices nor triangles, Proposition 3 can be applied to G. Thus,

R(K,,,)

=

k2

n2

=4 =

[ -3n42

If n = 2 k + 1 is odd, then let G be the complete-bipartite graph Kk,k+l. From Proposition 3, 2 1 1 - 1 n+l n2 I Q ( K k , k + i ) = k ( k + 1) = 4 2 .2 - 4 Q.E.D.

-[?I.

404

HYPERGRAPHS

The following proposition characterizes the representative graphs of the maximal cliques of a graph. Proposition 5 (Roberts, Spencer [1969]). A simplegraph G is ihe representative graph of ihe maximal cliques of some graph i f , and only i f , ihere exists in G a family of cliques (Ei / i E I ) such that (1) each edge of G is covered by an Ef, (2) (Ei i E I ) satisfies the Helly property.

Clearly, we may assume that G contains no isolated vertices. 1. Let G be a representative graph of the maximal cliques Xl , X , , ..., A',, of some graph. These maximal cliques determine a conformal hypergraph H = ( E ; X I , X,, ..., X J , where n

€ = (J

xi.

i=1

Let E l , E,, ... denote the edges of the dual hypergraph H * . By Proposition 1, these are cliques of G and satisfy condition (1); since H is conformal, they also satisfy condition (2), by Theorem 3. 2. Conversely, consider a family of cliques (Ei / i E I ) in G that satisfy conditions (1) and (2). Cliques E,' = { xi } for i = 1, 2, ..., n can be added to this family without violating conditions (1) and (2). Clearly, H" = ( X ; E l , E,, ..., E;, E;, ..., EJ is a hypergraph. Let H = ( E ;Xi,

9

XJ

be its dual. Condition (1) implies that G is a representative graph of H by Proposition 1. Condition (2) implies that H is a conformal hypergraph by Theorem 3. Moreover, edge Xi contains vertex e; because xi E Ei = { x, }, and this is the only edge of H that contains vertex e ; . Hence, the edges of H are all maximal edges. Thus, H is the family of maximal cliques of graph (H),. Q.E.D. To show that graph G in Fig. 17.2is the representative graph of the maximal family of cliques of a graph (namely, H3), it sufficesto consider El = { a, 6, c } and E, = { c, d, e } as the family of cliques satisfying conditions (1) and (2). The following result which is a new generalization of a result of Krauz, characterizes the representative graph of the k-cliques of a graph.

Proposition 6. A simple graph is the representative graph of the k-cliques of


405

Some graph i f , and only if, there exists a family (El / i E I ) of cliques in G such that (1) each edge of G is covered by an E,. (2) each vertex of G belongs to exactly k of the E,, ( 3 ) each partial family ( E j j E J ) formed froni k of the cliques E, that are pairwise intersecting, has an intersection of cardinality one.

1. Let G be the representative graph of the family V k = (XI, X , , ..., X,,) of k-cliques of a graph (E, r);assume that each vertex or edge that does not belong to a k-clique has been removed from (E, I-). Family%, determines a uniform hypergraph H = ( E ; XI, X,, ..., Xk) of rank k, and

G

= (H*),

Let E l , E,, ... denote the edges of the dual hypergraph H*. Since G = (H*),, these edges are cliques of G and satisfy condition (1). They also satisfy condition (2) because an edge of Hcontains k vertices, and thus a vertex of H * belongs to k edges. Condition (3) is also satisfied because if k vertices e, of H are pairwise adjacent, then there exists a unique edge of H that contains all k vertices (since H is conformal).

2. Conversely, let (Et / i E I) be a family of cliques of G that satisfy conditions (l), (2) and (3), and let H * = ( X ; E l , E,, ..,, En) be the hypergraph whose edges are these sets. By condition ( l ) , G = ( H * ) 2 . By condition (2), each edge of the dual hypergraph (H*)* = H contains exactly k vertices. Finally, the k-cliques of graph ( H ) , are the edges of H, and by condition (3), graph ( H ) , contains no other k-cliques. Thus, G is the representative graph of the k-cliques of graph (H)Z. Q.E.D. Corollary (Krauz [1943]). A simple graph G is the representative graph of some simple graph H $ and only if, G contains a family of cliques (E, / i E I ) such that (1') each edge of G belongs to exactly one El, (2') each vertex of G belongs to exactly two E1. Let k = 2 in Proposition 6 ; then conditions (1) and (3) imply condition (1'), and vice versa, and condition (2) becomes condition (2').

Q.E.D. Note that this corollary gives only a global characterization of an L(H). We shall now present a local characterization that is easier to verify; first some new definitions are required.

406

HYPERGRAPHS

G1

f

A

a. x

Y

Fig. 17.5

A triangle of a graph G = (A’, E ) is defined to be normal if each vertex of the graph is adjacent to an even number of vertices of the triangle. Otherwise, the triangle is defined to be specia!. Note that a normal triangle is a maximal clique of G.

Lemma. If G possesses two special triangles abc and abd with [c, d ] 6 E, then G contains as a subgraph at least one of the graphs shown in Fig. 17.5. We shall consider separately the following cases. CASE1. There exists a vertex of G that is adjacent to an odd number of vertices of triangle abc and to an odd number of certices of triangle abd. Let x be this vertex. Two subcases must be considered : CASE1’. Vertex x is adjacent to exactly one vertex of each triangle. I f x is adjacent to a, then subgraph GI is obtained with vertices x, c, d, a. I f x is adjacent to c and to d, then G2 is obtained. CASE1”. Vertex x is adjacent to all the vertices of one of the two triangles. Then, x must be adjacent to a, b, c, d, and G3is obtained. CASE2. Each certex of G is adjacent to an even nuniber of vertices in either one of the triangles abc or abd. Let x be a vertex of G adjacent to an odd number of vertices of abc; let y be a vertex adjacent to an odd number of vertices of abd. Thus, x # y.


401

First note that x I S not adjacent to both c and d (because, otherwise, x would be adjacent to a, b, c, d and then the conditions for Case 2 would not be satisfied). If x is not adjacent to c, then x is adjacent to exactly one of the vertices a or b, and consequently, to d. Thus, three subcases must be considered. I. x is adjacent to a, b, c but not to d, 11. x is adjacent to c, but not to a, 6, d, 111. x is adjacent to a, d, but not to b, c. By considering the conditions on x and y nd whether or not x and y are adjacent, one of the nine graphs of Fig. 17.5 can be obtained. The details are left to the reader. Q.E.D. Theorem 4 (Beineke [1968]). For a simple graph G, the following conditions are equicalent: (1) G is the representative graph of a simple graph, (2) G contains none of the subgraphs shown in Fig. 17.5, (3) thefamily 5f of all the maximal cliques of G that are not normal triangles

corers each certex at most twice, and corers each edge at most once, and G contains ti0 subgraph of the type G, , (4) there exists a family B of cliques such that each rertex belongs to exactly two cliques of B and each edge belongs to exactly one clique of PP. (1) * (2) From the Krauz theorem, there exists a partition of the edges of G into cliques such that each vertex of G is covered by at most two cliques. If G contains a subgraph of the form G,,then this partition induces in G, a partition of the edges of G1with the same property. It is easily verified that none of the graphs G, of Fig. 17.5 admits such a partition. (2)

(3)

-

(3) If two maximal cliques C and D which are not normal triangles have a common edge [a, b],then there exist two vertices c E C and d E D that are not adjacent. If C # { a, b, c }, then the triangle abc is special. I f C = { a , b, c } , then the triangle abc is also special. From the lemma, the special triangles abc and abd with [c, d ] E, yield that there exists a subgraph of G of one of the types shown in in Fig. 17.5. This contradicts condition (2). (4) Let % be the family of all the maximal cliques that are not normal

triangles. We shall successively add new cliques to this family to form family 9.

408

HYF'ERGRAPHS

Let [a, b] be an edge that is not covered by a clique of V. This edge necessarily belongs to a normal triangle abc. No clique of %' covers only one vertex of this triangle (because, otherwise, the triangle would be special). Similarly, no clique of V covers the three vertices of this triangle. Two cases must be considered: CASE1. No vertex of triangle abc is covered by a clique of %'.There are two subcases to consider. CASE1'. Triangle abc intersects with no other normal triangle. Form 9 by adding the cliques { a, b, c }, { a }, { b } and { c } (or, by adding the cliques { a, b 1, { a, c 1, { b, c 3). CASE1'. Triangle abc has a common edge with another normal triangle (and with only one, because GIis forbidden). Let abd be this new triangle. Noother normal triangle meets abed (because then it would intersect each triangle at two vertices). Triangle abd does not meet any clique of V (because then a or b would be covered by a clique of V). Form B by adding the cliques { a , b, c}, { a, b, a>,{ c 1, { d ) .

Case 2"

Case 1" Fig, 17.6

CASE2. A vertex of triangle abc is covered by a clique C E W. In this case, exactly two of the vertices are covered by clique C (because, otherwise, the triangle abc would not be normal). Without loss of generality, let b and c be these two vertices. CASE2'. No other clique of V meets triangle abc. Form 9 by adding the cliques { a, b } and { a, c }. CASE2". One other clique C' of %' meets triangle abc. In this case, C' meets the triangle at two vertices (otherwise, abc would not be normal), and these two vertices must be u and c from the definition of edge [a, b]. No other

409

HYPERGRAPHS AND THEIR.DUALS

clique ofV meets triangle abc. Thus, form B by adding the clique { a, b }. See Fig. 17.6.

(4) * (1) by the corollary to Proposition 6. Q.E.D. 1

H

568

78

Fig. 17.7

Remark. Note that Theorem 4 presents an efficient method to determine if a graph G is the representative graph of some simple graph. An example is given in Fig. 17.7. The graph H represented by G can be constructed from the family 5f of the maximal cliques of G that are not normal triangles (and which are circled by a continuous line in Fig. 17.7). In graph G, the only normal triangles are 234,456, and 678, which are not to be included in family V. Family B can be formed from 9? by adding the 2-clique { 7 , 8 } and the I-clique { 1 }. Hence, the required graph H , shown in Fig. 17.7, is obtained.

Proposition 7 . A simple graph G is the representatiue graph of some multigraph without triangles and without loops if, and only i f , each vertex of G belongs to at most two maximal cliques. 1. Suppose that G = L ( H ) is the representative graph of multigraph H. A vertex x of G represents an edge [c, d ] of H , and T,(x) U { x } represents the set of edges of H incident to c or to d. Let C, denote the set of vertices corresponding to the edges incident to c, and let 0,denote the set of vertices corresponding to the edges incident to d. Then T,(x) u { x }

=

c, u D, ,

and C, and D, are two cliques of G. Since H contains no triangles, the sets C, - D, and D, - C, are not joined by an edge.

410

HYPERGRAPHS

Let C be a maximal clique in G that contains x . Then, C c C, u 0,.If y E C and y E C, - D,, then each vertex y’ E C that is adjacent to y belongs to C, ;hence, C c C,, and it follows from the maximality of C that C = C,. Therefore, each maximal clique that contains x equals either C, or D,, and consequently, there are at most two maximal cliques that contain x . 2. Conversely, let X’ denote the set of vertices in G that belong to exactly one maximal clique, and let X”denote the set of vertices in G that belong to exactly two maximal cliques. Assume that X = X’ u X”. Consider the hypergraph H * = ( X ; E l , E,, ..., { x i }, { xk }, ...) whose edges are the maximal cliques of G, and the cliques { xi 1 for all xi E X’. Consider the hypergraph H = ( E ; X,,X,,..., X,) dual of H*. Clearly, graph G is the representative graph of H . Hypergraph H i s a multigraph without loops, because each vertex x, of H* belongs to exactly two distinct edges. Suppose that H contains a triangle el e2 e,; let x , y , z be the vertices of H * that represent respectively edges [ e l , e,], [e,, e,], [e,, e l ] . Then,

xEE1 n E 2 ; x#E3

Y E En~E3 ; Y #El

Z E En~El ; z#E2

-

Let C be a maximal clique of G that contains triangle xyz. Then,

c z E,,E,,E3 Consequently, vertex x belongs to three distinct maximal cliques E l , E2, C, which is a contradiction. Q.E.D.

Proposition 8. A simple graph G is the representatice graph of some bipartite multigraph if, and only if, the two following conditions hold: (1) each tlertex of G belorigs to at most two maximal cliques, ( 2 ) each odd elementary cycle of G contains two sides of a triangle. 1 . If G = L ( H ) is the representative graph of a bipartite multigraph H, then condition (1) follows from Proposition 7. Condition (2) also holds, because if [xl, x2, ..., x , , x,] is an elementary odd cycle of G that does not contain two sides of a triangle, then p > 3, and the corresponding edges X , , X,, ..., X,, X I of multigraph H also constitute an odd elementary cycle. This contradicts that H is bipartite.

2. Conversely, consider a graph G that satisfies conditions (1) and (2). From condition (l), there exists a multigraph H = ( E ; Xl, X2,..., X,,) without triangles such that G = L ( H ) . It remains to show that multigraph H contains no odd elementary cycle [ e , , e 2 , ..., e,, e l ] with p > 3. Suppose that H contains such a cycle. Let Xi be an edge of multigraph H

41 1


that joins e, and e,,, in the cycle. The sequence [x,, x l , x,, ..., x,] is an elementary odd cycle of G. From (2), there exists a triangle formed from three consecutive vertices of this cycle, say x l , x,, x3. In multigraph H, edges X , , X,, X3 are pairwise adjacent and, since H contains no triangles, they have a common endpoint. But this contradicts that these three edges lie on an elementary cycle. Q.E.D. This theorem shows why graph G shown in Fig. 17.2 is the representative graph of a bipartite multigraph H 3 .

Remark. Consider a graph G1 that is a triangle, and a graph G, that is three edges with a common endpoint. Clearly, L(Gl) and L(G,) are both isomorphic to a triangle. H. Whitney [1932] has shown: Zf two connected simple graphs have isomorphic representative graphs, then they are isomorphic, except for G, and G,. This result has been extended to hypergraphs by Berge and Rado [1972]. Two hypergraphs H = ( X ; E l , E2,..., Em) and H ’ = ( Y ; F,, F,, ..., F,) are said to be isomorphic if they have the same number m of edges, and if there exists a bijection cp: X - t Y and a permutation 71 on M = { 1,2, ..., m } such that (i = 1, 2, ..., m) . cp(El) = F,,,, H and H’ are said to be strongly isomorphic if there exists a bijection cp: X - t Y such that

(i = 1 , 2, ...,m). cp(E,) = Fl In this case, we write H r H’. Clearly, two simple graphs (El / i E M ) and (6 / i E M ) have the same representative graph if, and only if, (E,, E,) (El, F,) for all i,j . The main results for hypergraphs are the following:

I. For m > 1, there exist two uniJorm hypergraphs K,(M) = (E, / i E M ) and L,(M) = (F, / i E M ) which are not strongly isomorphic and which satisfy

( E , / i E M - { k } ) g ( F , / i E M - {k})

EM)

The vertices of K,(M) arepoints x,, for all J c M with I J I = m modulo 2, and its edges are the sets El = { x, / i E J }for all i < m. The vertices of L,(M) are points y , for all J c M with I J I = m + 1 modulo 2, and its edges are the sets 6 = {y,/iEJ}. 11. Let p be an integer, p > 1, p < m ; let H = ( X ; El,E2, ..., Em) and H‘ = ( Y ;F1, Fa, ..., F,) be two hypergraphs such that f o r each set J c M with cardinalityp, thepartial hypergraphs (Ei/ i E J ) and (F, / i E J ) arestrongly

412

HYPERGRAPHS

isomorphic. Then, Hand H' are strongly isomorphic, except if there exists three sets A c X , B c Y, Z c A4 with I Z I = p and ( A n Ei / i €11)

( B n Fi / i E Z) E' Lp(Z),

Kp(Z),

or vice versa.

For the proofs, see Berge, Rado [1972].

EXERCISES 1. A hypergraph (X,8)is said to be hereditary if A€&,

B c A , B # O

3

BE&'.

Let 2 be the set of all subsets of 8 of the kind { E l l ,E l z ,..., El, } with Ell c Eta c C €in.

Show that the derived hypergraph (8, X ) is conformal. (In algebraic topology, a hereditary hypergraph is an abstract simplicia1 complex, and the derived hypergraph is its first barycentric subdivision.) 2. Let el,e2, ..., em be points on a line, and let XI, X z , ..., X,,, be intervals such that no interval is contained in another. Show that the dual hypergraph H* of H = (XI, Xz,..., X,,) is also a family of intervals. Also, show how to place the points x l , x z , ..., x,, and the intervals E l , E z , ..., Em on a line such that el E X j if, and only if, x j E E,. '

3. Let G be a graph such that the maximum cardinality w(c) of the cliques of G is < 3. Show that G is the representative graph of the maximal cliques of some graph if, and only if, the maximal cliques of G satisfy the Helly property. Show that G is the representative graph of the maximal cliques of some graph if, and only if, G contains no partial subgraph with six vertices a, b, c, d, e, f a n d with edges (Roberts, Spencer [1969]) ab, ac, bc, bd, be, ce, cf, de, ef. 4. Calculate the minimum cardinality d(n) of a set A such that each graph of order n

is the representative graph of n distinct subsets of A . Show by induction on n that

d(2) = 2 ,

d(3) = 3 , d(n) =

[TI

if

n 2 4. (Erdos, Goodman, P6sa [1966])

5. Consider a family of non-empty sets of the form

( x / x = ( x * , x 2 ,...,x * ) E W ; a *
1. The stability number of H is defined to be the maximum cardinality of a stable set of H. The chromatic number x ( H ) is defined to be the smallest number of colours needed to colour the vertices of H such that no edge Ei of H with 1 EiI > 1 has all its vertices with the same colour. This concept was introduced by Erdos and Hajnal [1966]. A q-colouring is defined to be a partition of X into q stable sets

s1, s,

*-*,

s q

9

each corresponding to a colour. A hypergraph for which there exists a qcolouring is said to be q-colourable. ci

Fig. 19.1 Projective plane with 7 points

EXAMPLE. The projective plane with 7 vertices forms a uniform hypergraph H of rank 3 with 7 edges aef, adg, abc, fdb, fgc, ced, beg. See Fig. 19.1. Set { a, c, e, g } is a maximum stable set. Its chromatic number is x ( H ) = 3, and a 3-colouring of H is shown in Fig. 19.1. 428

429

CHROMATIC NUMBER OF A HYPERGRAPH

Proposition 1 (Erdos, Hajnal, [1966]). I f H is a uniform hypergraph of rank 3 with n vertices that satisfy

IEinEII < 1

(i#j),

then each maximal stable set S satisfies

Let S be a maximal stable set with s vertices. Since S is maximal, each X - S belongs to an edge Ex such that Ex - { x } c S. From the hypothesis, xE

IE,nE,.I G 1

( x , y ~ X S - ;x f y ) .

Thus, the sets Ex n Shave cardinality 2 and are pairwise different; hence

It is easily seen that this1 implies s2

[JG].

Q.E.D. Proposition 2. Let H be a hypergraph of order n with chromatic nitmber x ( H ) and with stability number ,B(H); then X(H)B(H) 2 n 9 x(H) + B ( H ) G + 1* Consider a q-colouring (Sl, S,, ..., S,) of H with q Then,

c I Si I G 4BW)

=

x(H) colours.

4

n =

i= 1

= X(H)B(H) Y

and the first inequality follows. Let S be a maximum stable set of H. Colour the vertices of S with a first colour, and colour each of the remaining vertices with n - B(H) additional colours. Hence x ( H ) G [. -P(H)] and the second inequality follows.

+1 Q.E.D.

Given a hypergrap,h H = (X, &), the degree of a vertex x is defined to be the maximum number of edges different from { x } that form a partial family (E,/ j E J ) with EinE, = { x }

( i , j ~ J i;# j ) .

430

HYPERGRAPHS

Let the degree of x be denoted by d,(x). Clearly, d,(x) = 0 if, and only if, the only edge that contains x is { x } . The following theorem shows that the relationship between the chromatic number of a hypergraph and its degrees is analogous to the relationship between the chromatic number of a graph and its degrees:

Theorem 1 (Tornescu [ 19683). Let (S,, S,, .. ., S,) be q-colouring of H , and let dk = max d,(x) . X€sk

Then, x(H)

< max min { k, dk + 1 } . kbq

1. We shall show that there exists an r-colouring (Si,Sh, ..., S:) of H with I

min ( k,r )

( S; is a maximal stable set in X -

U Sir . i < k

Clearly, if S, is not a maximal stable set in X , then vertices can be added to S, until a maximal stable set S, is obtained. If S, - S; is not a maximal stable set in X - S ; , then vertices can be added to it until a maximal stable set 5'; is formed, etc. This gives the required colouring of H. 2. Let x $ UlklS!, and l e t j < k ; then from the maximality of S; there exists an edge E; with

Since x E E;, for all j

< k , the family ( E ; , E i , ...,Ell.) satisfies E;nE;={x}

(i#j;i,jk.

~H(x)

-

3. Let i(x)-denote the index of the set Si that contains x . From part 2, i(x) 2

By letting k

=

i(x)

k

+1

- 1, it follows that i(x) < d&)

+1

dH(x)> k . ( x E X) .

431


< k (from part l), and i(a) < max i ( x ) < max (d&) + 1)

Let a E Sk,then i(a)

d, + 1 ,

I

XSsk

XSsk

Hence, x(H)

< max i(x) < max min { k, dk + 1 } , k XPX

Q.E.D. Corollary 1. I f H is a hypergraph with x(H) = q + 1, and fi the hypergraph .‘to has chromatic number x(Ho) = q, then

Ho obtained by removing vertex d&O) 2 4.

S2, ..., S,) be a q-colouring of Ho = Hx-( x,, ). If dH(xo)< q, then Let (S1, by considering the (q I)-colouring ( S , , S,,..., S,, { xo I), it follows that

+

x ( H ) < max min { k , dk k < q + l

which contradicts x ( H ) = q

+1}

Q q,

+ 1. Q.E.D.

Corollary 2. Ifg is a positive integer such that

I { x / x E X , dH(x) 2 4 ) I < 4 , then

x(H)

< 4.

Index the vertices such that

d&i) 2 ddxz) 2 d&)

2

**.

Z d,(xn).

Then, for k > q,

+ 1) < q .

min{k,dH(xk)

Clearly, this inequality also holds for k < q. For the n-colouring ({ x1 }, { x, }, ..., { x,, }), Theorem 1 yields

x(H)

< max min { k, df,(xk) + 1 ) Q q . k4n

Q.E.D. Corollary 3. If H is a hypergraph of maximum degree do, then ~

. x ( H ) < do

-I- 1 .

The proof follows from Corollary 2 with q

=

dl

+ 1.

APPLICATION. The following theorem is due to T. S. Motzkin [1968]:

432

HYPERGRAPHS

For a simple graph G with maximum degree h, it is possible to colour the vertices with

[;] + 1 colows such that no cycle has only one colour.

To show this, let H = (X,8) be the hypergraph in which X i s the set of vertices in G that belong to at least one cycle and in which d is the family of the sets of vertices that are contained in the same cycle. h The maximum degree of hypergraph H is do < 2 and the Motzkin

[

1,

theorem follows from Corollary 3. We can also state: The edges of a simple graph G can be coloured with q colours such that each cycle has more than one colour when q = max min

{ &(a), d&)

1.

C4,bI E E

To show this result, consider the hypergraph H whose vertices are the edges of G that belong to at least one cycle. The degree in H of an edge ab of G equals

dH(ab)< min{ dG(a)- 1, dG(b)- 1 ) Q q

- 1,

and the result follows from Corollary 3.

2. Cliques of a hypergraph Let H = (X, 8)be a hypergraph of rank h, and let r < h. A set A c X is defined to be a clique of rank r if either I A I < r, or I A I 2 r and each subset of A with cardinality r is contained in at least one edge of H . Clearly, each subset of a clique of rank r is also a clique of rank r. Let w,(H) denote the maximum cardinality of a clique of rank r in H. Since r 6 h, and since an edge E1 with 1 Ei = h is a clique of rank r, then

I

w,(H)

> I Ei 1 = r ( H ) .

If the vertex set X of hypergraph H i s a clique of rank r, then hypergraph H i s said to be r-complete. If H is a uniform hypergraph of rank h, then a “clique of rank h” is often called a clique, and each set with less than h vertices is called a triaial clique. Remark 1. Sauer [1971] has proved the following generalization of the Turan theorem:


Consider a set X with I X classes A , with

Let H,,, that

I = n and apartition (A,, A 2 , ..., A,)

433

of X into p

,= (X,8)be a hypergraph in which & is the set of subsets E of X such (El=h (,EnAil 6 1

(idp).

Then, H,,, is the only simple uniform hypergraph with n(H) = n, r(H) = h, o,(H) < p , and with the maximum number of edges. Remark 2. Let m(n, h, r ) denote the smallest possible number of edges in a uniform hypergraph of rank h and order n that is r-complete. If E l , E2, ..., Em are the edges of such a hypergraph, then

ij

l=l

Y,(EO

=

Y,(W

Y

and, hence,

Therefore,

Furthermore, equality holds if, and only if, there exists a uniform hypergraph of rank h and order n such that each subset of X with cardinality r is contained in exactly one edge of H. The hypergraphs with these properties are studied in finite geometries and are called Steiner systems.

Remark 3. Chvdtal [I9711 has shown how the numbers m(n,h, r ) are related to the Turdn numbers. Let n, k , b be integers such that 1 < k < b < n, and let T(n, k,6 ) denote the smallest value of t n such that there exists a uniform hypergraph of rank k and order n with m edges whose stability number is smaller than b. The Turdn theorem (Ch. 13, $2) determines the value of T(n, 2,b). The number T(n, k , b) is called the Turbn number in n, k , 6. If a hypergraph H = (Et / i E Z) has a stability number smaller than b, then each set B with 1 B I = b is non-stable and contains some edge E i . Thus,

434

HYPERGRAPHS

an (n - b)-set X - B can always be covered by X - E, for some i, and hence, the hypergraph ( X - Ei / i E I ) of rank n - k is (n - b)-complete. Therefore, T(n, k , b) = m(n, n - k , n - b) . From Remark 2, it follows that

T(n, k, b) 2

(

b)

(n --b

=

-

n! (n - b) !(b - k ) ! (n - k ) ! b !(n - b) !

n! k!(b-k). q)(b)-l. k !(n - k ) ! b! k k

This inequality, which was also discovered by Katona, Nemetz and Simonovits, yields a bound for the stability number P(H) of a hypergraph H with n vertices and m edges such that minter I Et [ = k. If H is uniform of rank k, then

and, therefore,

If H is not uniform, this inequality is satisfied a fortiori.“)

Theorem 2. Let H be a simple uniform hypergraph of rank h, and let K he a clique with k vertices, k 2 h. Then each vertex of K has degree

and the chromatic number of the subhypergraph generated by K is

,

+

x ( H R ) = do 1. 1. The degree of a vertex x E K is the maximum number of pairwise disjoint subsets of K with h - 1 elements that do not contain x. Thus,

For asummaryof previous results for T(n, k, b), see V. ChvatalI19711. More recently J. Spencer [1972] found an improved lower bound for T(n, k, b) for large values of n. He proved that if n l b k ( k - l ) - l , then n” k - ’ . T(n, k , b) 2

43 5


2. A minimum q-colouring of Hkis obtained from a partition ( S ,7

s, ..*,S,) 7

such that ISiI = h - 1 IS,l G h - 1

(i= 1,2,...,q-

1).

Therefore,

From that part 1, it follows that

k

* h-1

=do

Furthermore, from Corollary 3 to Theorem 1, x(HJ X(HK) = d,

+1.

< do + 1. Hence,

+ 1. Q.E.D.

Remark I . Theorem 2 shows that the bound given by Corollary 3 to Theorem 1 is the best possible: If H is a uniform hypergraph of rank h and maximum degree do, then x ( H ) < do 1, and equality holds for each H that contains a clique whose vertices have degree do. However, x ( H ) = do 1 does not imply the existence of such a clique, as shown by the hypergraph H in Fig. 19.2 due to L. Lovrisz. The edges of H are abc,, abc,, abc,, c, c2 c,, acb,, acb,, acb, , b, b, b, , bca, , bca, , bca, , a, a, a, .

+

+

Fig. 19.2

436

HYPERGRAPHS

Hypergraph H is uniform of rank 3 and contains only vertices with degree do = 2. The chromatic ndmber of H is 3 = do f 1. However, H contains no cliques of degree do = 2. The following result due to Lov6sz [1965] generalizes Theorem (6, Ch. 15). If H is a connected uniform hypergraph of rank h > 1 with maximum degree do > 1 , and i f I U E EEX I < do(h - 1) 1 f o r each x, then x ( H ) = do 1 holds only in the two following cases: (1) h = do = 2, and H is an odd cycle, or (2) H contains a clique of degree do.

+

+

Remark 2. If H is a uniform hypergraph of rank h with a maximum clique Kof cardinality k,then from Theorem 2, x ( H )is bounded below by a function of h and k. However, x ( H ) is not bounded above by a function of h and k. In fact, Erdos and Hajnal [1966] have shown that for each q > 1, there exists a uniform hypergraph H of rank h with x ( H ) = q that contains no cliques of order > h. More precisely, If h, q, lare integers > 1, there exists a uniform hypergraph H(h, q, 1) of rank h with chromatic number q and without cycles of length < 1. A constructive proof has been given by Lov6sz [1968]. The following is a well known result:

Theorem 3 (Ramsey [1930]). Consider three integers p, q, h with p , q 2 h. Thete exists a$nite integer R,(p, q ) such that each uniform hypergraph of rank h with n vertices, n 2 R,(p, q), contains a clique with p vertices or a stable set with q oertices. The “Rarnsey number of rank h in p and q” is dejned to be the smallest number R,(p, q) with this property. We shall show by induction on h that if I X 1 is sufficiently large, then for any partition of y h ( X ) into two cIasses 6 and S,there exists a set A c X with 1 A I = p , Y k ( A ) c 8,or there exists a set B c X with

I B I = 4 , g,,(B) c 9 .

1 . Clearly, the theorem is true for h = 1 , because if 1 X I = p + q - 1 , and if X is partitioned into two classes, with no p elements of X in the first class, then there exist at least q elements of X in the second class. Therefore, (1)

RdP, 4) = P

+4- 1

*

We shall assume that the theorem is true for ranks less than h, and we shall show that it is true for the rank h > 1. 2. Clearly, the theorem is true for p = h because if I X 1 >, q and if X contains no h-set in 8,then each h-set belongs to 9. Therefore, (2) &(h, 4) = 4 *

437


By the same argument, the theorem is true for q = h, and (2‘)

h)

Rh@,

p

*

+ q = 2 h, we shall prove it by induction

Since the theorem is true for p o n p q.

+

3. Let p, q be two integers withp

+ q > 2 h ; we shall show that

Rhb, q) Q Rh-l[Rh(P -

(3)

=F

Y

41, R h b , 4 - l)]

+

*

Let Pf =

Rh(P

- 1, 41,

q’ = Rh(P,

- 1)

7

and consider a set X with cardinality n > Rh- l(p‘, 4’) + 1. It suffices to show that for any partition (8,P) of Y,,(X), there exists either a set A with [ A I = p and Ph(A) t 8, or a set B with 1 B I = q and Ph(B) c 9. Let a E X,and let X‘ = X - { a }. For E‘ E g h - I ( x ‘ ) , let

1-’;: Since 1 X ’

I

if if

E’ u { a } E 6“ E‘u ( a } E F .

2 Rh-l(p’,q’), there exists (for example) a set A‘

I A’ I = p‘, Since [ A’ I = p‘

=

gh-l(A‘)

t

t

X’with

8’.

R,(p - 1, q), there exists either a set B” c A‘ with

I B” I = 4

ph(p) C F (in which case the above proposition is true), or a set A” c A‘ with 9

I A” I = p -

1,

Ph(A”) c 8 .

In this case, set A” u { a } satisfies

I

IA”u { a } =p,

’ g h ( A ” u { a } ) c 6“

9

and the above proposition is true. Q.E.D.

Corollary. Let p , q

> 2, and let G = (X,E ) be a simple graph of order p + 4 * .( p - 1 *)-

Then there exists in G either a clique of cardinality p or a stable set of cardinality 4.

438

HYPERGRAPHS

From Theorem 3, it suffices to show that

This is true for p = 2, since from (2)

(;)=

4 = R,(2,4)

*

Suppose that the inequality is valid for each R,(p’, q’) withp’ then from (1) and (3), it follows that

u p , 4) < R,(P

- 1,q) + M

4

P - 1+4 p - 2

P ,4

-’)+(P+;-

+ q’

- t + l ‘ Thus there exists a colouring of degree f i n

[=]h + 1 colours.

I t +

11

*

colours, i.e.

Q.E.D. For t = 0, this corollary reduces to Corollary 2 to Theorem (5, Ch. 15). For t = 1, this result was proved by Gerencstr [1965].

*

*

*

A second generalization of the chromatic number of a graph has been studied by G . Chartrand, D. P. Geller, S. Hedetniemi [1968]. Let f k ) ( G ) denote the smallest number of colours needed to colour the vertices of a simple graph G such that no elementary chain with length k is monocoloured For k = 1, then $l)(G) is the chromatic number y(G). For k = 2, then, from the preceding corollary,

No‘e that f k ) ( G )is also the chromatic number of a uniform hypergraph of rank k 1, whose edges are the sets of vertices that lie on an elementary chain of length k in G. Chartrand, Geller and Hedetniemi [1968] have shown that if G is a simpIe graph, and i f the Iength of the longest elementary chain in G is I, 2 < I < k, then

+

446

HYPERGRAPHS

A third generalization of the chromatic number of a graph can be defined by replacing "chain" with "path" i n the definition of z")(G).

*

*

*

A fourth generalization of the chromatic number of a graph has been studied by S . Hedetniemi [1970]. Let zk(G) denote the smallest number of colours needed to colour the vertices of graph G such that no connected subgraph of order k has only one colour. Thus,

XZW)

= y(G)

9

Clearly, z k ( G )is also the chromatic number of a hypergraph defined in the obvious way.")

*

*

*

Another generalization of chromatic number has been given by H. Sachs and M. Schauble 119671. For a simple graph G = ( X , E ) and a number k 2 2, a coloiiring by k-cliques is defined to be a partition (XI, X,, ..., X,) of X such that no k-clique is contained entirely in one class of the partition (i-e., no k-clique has only one colour). Let zk(G)denote the smallest integerp for which there exists such a partition in p classes. Thus, the chromatic number of G (in the usual sense) is x2(G). Consider the hypergraph H formed from the k-cliques of graph G (which was characterized in Proposition 6, Ch. 17). Clearly,

K W ) = X,(G)

'

Sachs and Schauble have discovered an inductive construction that yields the following result: For each p 2 1 and for each k 2 2, there exists a graph G = G ( p , k) with the following properties:

(I) G contains no (k

+ I)-cliques,

(2) ZAG) = P,

( 3 ) for each colouring by k-cliques that uses p colours, there exist at least p pairwise disjoint ( k - I)-cliques that have only one colour. For k = 2, this construction gives a graph without triangles and with chromatic number p for each integer p (Blanche Descartes [1947]; Zykov [1949]; Mycielski [1955]; Erdos, Rado [1960]). Zn particular, Hedetniemi [1970] has shown that if G is a planar graph, then it is possible to colour its vertices in 4 colours such that none of the 4 subgraphs generated by the vertices of the same colour is connected. Furthermore, if G contains no triangles, then 2 colours are sufficient.


447

EXERCISES 1. Using the argument for Proposition 1, Section 1, show that if a uniform hypergraph H with rank h 2 3 satisfies

1-4n E j l

< h-2

(i # j ) ,

then the stability number k of hypergraph H satisfies

2. Show that a hypergraph H with chromatic number x ( H ) = q contains a chain of length q - 1. (Tomescu [19691)

3. Consider a hypergraph with n vertices and with edges E, such that (1) I EfI = 3 for all i, (2) for a, b E ,'A a # b, there exists exactly one edge that contains both a and b. Such a hypergraph is called a Steiner system of order n. For example, the projective plane with 7 points (Fig. 19.1) is a Steiner system of order 7. Show that for n 3 7, then x ( H ) 2 3. It is known that a Steiner system exists if, and only if, n I 1 modulo 6, or if n = 3 modulo 6. If n I 3 modulo 6, a Steiner system H of order n with x ( H ) = 3 can be constructed by a well known method due to M. Hall (see, Combinarorial Theory, GinnBlaisdell, Toronto, 1967, Theorem 15.3.2). If n = 1 modulo 6, a Steiner system H with x ( H ) = 3 can be constructed by a method due to A. Rosa [1970]. 4. Given an n x ti chessboard, define the Queen's hypergraph H,O as the hypergraph whose vertices are the squares of the chessboard and whose each edge E, is the set of squares that a queen at square x controls (including square x). Similarly, define the King's hypergraph H Z , Rook's hypergraph H,", and Bishop's hypergraph H!. Show that x(H,f') = ,y(Hf) = x(H,?) = x(H,f) = 2 .

5. In the h-dimensional space Rh, let be a family of R h + l ( p ,q) compact convex sets such that each subfamily of q members, contains h + I members whose intersection is non-empty. Show that there exist p sets of 5 ' whose intersection is not empty. Hint: Use Helly's theorem: If V: is a family of p 2 h 1 convex compact sets in Rh, and if the intersection of any h 1 of them is non-empty, then the intersection of the family W is not empty. (Berge, Espaces Topologiques, Paris [1962]).

+

+

6. In the plane, let S be a finite set of points with 1 S I 3 R 4 ( p , 5), such that no three points are on a line. Show that there exist p points of S that are the vertices of a convex polygon. Hint: Show that (1) if I S I = 5, then there exist 4 points of S that form a convex quadrilateral. (2) if all the quadrilaterals formed from a set o f p points are convex, then thesep points are the vertices of a convex polygon. (Erdos, Szekeres [1935]).

CHAPTER 20

Balanced Hypergraphs and Unimodular Hypergraphs

1. Strong chromatic number

S) be a hypergraph with rank function r(S). A set S is Let H = (X, defined to be strongly stable if r(S) = 1, i.e. if (i = 1 , 2 ,

ISnE, I d 1

..., m).

Note that each strongly stable set is also a stable set. The strong stability number a(H) of hypergraph H is defined to be the maximum number of vertices in a strongly stable set. Clearly,

a(H) = a ( ( W 2 ) . The covering number p ( H ) of H is defined to be the smallest number of edges of H that cover all the vertices of H. Clearly, p ( H ) 2 13((H)2),the smallest number of cliques needed to cover graph (I&. Furthermore, if H is a conformal hypergraph, then p ( H ) = O((H)&. A strong q-colouring of hypergraph H is defined to be a q-colouring of the vertices of H such that no two vertices contained in the same edge have the same colour. Clearly, a strong q-colouring is a partition of X into q strongly stable sets. The strong chromatic number y(H) of H is defined to be the smallest integer q for which there exists a strong q-colouring. Clearly, y(H) 2 r(X). Hypergraph His said to be y-perfect if y(HA) = r(A) for all A = X. Proposition. Euery y-perfect hypergraph is conformal.

Let H = ( X , 8)be a y-perfect hypergraph. To show that H is conformal, it suffices to show that each clique C of graph (H)2 is contained in an edge of H. Since hypergraph H is y-perfect,

Ic I = W

C )

= r(C),

and, consequently, there exists an edge of H that contains C. Q.E.D. 448

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

449

Remark. Let V denote the family of maximal cliques of graph G . If graph G can be characterized by a property of family ??, it is often interesting to relate the chromatic number of G to properties of hypergraph H = ( X , 59) since y(G) = y(H). The relationship between the rank of a hypergraph and its strong stability number is described in the following results. Theorem 1. Let H

=

( X , b) be a hypergraph with rank r(A). Then

To demonstrate the first inequality, suppose that S is a maximum strongly stable set. Then, r(S) = 1, and

To demonstrate the second inequality, let k be the smallest number of edges that cover X,and denote these edges by E l , E,, ..., Eli. Thus,

+ + I A n Ek I < kr(A) . Therefore, and

and, finally,

[ -1

A H ) = k 2 max I A I A#@ dA)

*

A cX

Corollary.

If H is a y-perfect hypergraph, then

ACX

Suppose A is a non-empty subset of X . Let q = r(A) = y(HA). If ( A l , A 2 , ..., A4) is a strong q-colouring of H A , then

I A I = 1 A1 1 + ' * *

+ 1 A, 1 < qL%(H)= r(A) cr(H).

450

HYPERGRAPHS

Hence,

and

a ( H ) L max I A I A# 0 r(A) From Theorem 1, it follows that

[--]

*.

a ( H ) = max I A I A#O r(A)

[-3

*.

Q.E.D. Remark. If hypergraph H is a graph, then the inequalities of Theorem 1 are the best possible, since the expression

can attain the value of cr(H) as well as the value of p ( H ) . For example, consider the graph GI with 6 vertices shown in Fig. 20.1.

Fig. 20.1

We have : a(Gl) = 3 = ko(Gl) < p(G,) = 4 .

Let G2be the graph obtained from G, by removing the isolated vertex. For graph G 2 , we have: a(G,) = 2 < ko(G,) = 3 = p(GJ

.

2. Balanced hypergraphs A hypergraph H is said to be balanced if every odd cycle (al, E l , a2, E2, ..., E2p+1, a l ) has an edge Ei that contains at least three vertices a, of the cycle. For example, a family of intervals of points on a line


45 1

is a balanced hypergraph. Clearly, a multigraph is balanced if, and only if, the multigraph is bipartite.

Proposition 1. Let H be a balanced hypergraph, then ecery partial hypergraph H' is balanced. If H' had an odd cycle with no edge containing three of its vertices, then this sequence would be also an odd cycle of H with no edge containing three of its vertices. This contradicts that H i s balanced. Q.E.D.

Proposition 2. Let H be a balanced hypergraph, then every subhypergraph Hs is balanced.

If Hs had an odd cycle with no edge containing three of its vertices, then this sequence would define for H an odd cycle with no edge containing three of its vertices. Q.E.D. Proposition 3. Let H = (Et I i E I ) be a balanced hypergraph, and let Eo be = (E, u Eo 1 i E I ) is also balanced.

a set. Then the hypergraph H'

Suppose that H' has an odd cycle p with no edge containing three of its vertices. Let this cycle be denoted by p = ( x i , E;, x;, Eh, ..., xk, EL, x i ) .

At least one vertex xi lies in Eo (otherwise, p would define an unbalanced cycle of H ) . Since k 2 3, p has an edge E; w i t h j # i - 1 and j # i, and E; contains x;, x ; + and ~ xi. This contradicts the definition of p. Q.E.D.

Proposition 4. Let H = ( E l , E 2 , . .., Em)be a balanced hypergraph, and let x,, $ U Ei. Then the hypergraph (El u { xo }, E,, ..., Em)is balanced. The proof is immediate.

Proposition 5. Let H = ( X , 6 ) be a balanced hypergraph, and let x1 E X . The hypergraph H' = (XI,a') obtained f r o m H by adding a new vertex xi and by setting X' = x u { x i } if x1$ E, ~'{:zu{x~'} if xlEE, is also balanced. The proof is immediate.

452

HYPERGRAPHS

Proposition 6. Let H Then its dual H* = (e,,

= (x, ,

..., x, ;E l , ..., Em)be a balanced hypergraph.

..., em;XI, ..., X,)

is also a balanced hypergraph.

By definition,

X , = { e i / i < m,E,

3

x,}.

Consider an odd cycle p = ( e l , XI, e 2 , ...,e2p+ xaP+ e l ) in H* ; it corresponds to an odd cycle (xl,E,, x 3 , ..., E z P c l ,X 2 p + l ,E l , xl) in H. Since H i s balanced, some E,contains three of the x,. Therefore some vertex e, of p belongs to three of the X,;or, equivalently, some X, contains three of the e,. Thus, cycle p is balanced. Q.E.D. Proposition I. If H = (Ei / i E I) is a balanced hypergraph, and if H’ = (Ei / i E J ) is a partial hypergraph such that i, j E J implies El n E, # 0,then

nE

,

Z

~

ieJ

(In other words, the edges of a balanced hypergraph satisfv Helly’s property.) The proof is an induction on the number of edges in H‘. The proposition is true for every subfamily with two edges. Assume that it is true for every subfamily with less than p edges; we shall prove it for a subfamily (E, / i E J ) = ( E ; , E6, ..., Ek) wherep B 3. By the induction hypothesis, for every k < p , there exists a vertex a, with

a, E

n E,’ .

t#k

We may assume that the a, are all distinct (otherwise Consider the sequence p = (a,, E;, a3,E ; , a,, E4, 4

n E ; # @). .

If two of the sets €I, EL, El are equal, for instance if €; = €;, then a, belongs to n t G I E ; and , the proof follows. If no two of the sets E; , E;, E; are equal, then p is an odd cycle, and one of these edges, say E; , contains { a , , a2,u3 }. Hence,

n E,.

IEJ

Q.E.D. Propositions 6 and 7 show that a balanced hypergraph is conformal. Theorem 2. A hypergraph H = (X,6 ) is balanced S c X , the subhypergraph Hs is bicolourable.

and only i f , for every


If every subhypergraph of H cause otherwise, there exists an Ef containing three of the a,, and graph Hs with x(HJ > 2, which

453

is bicolourable, then H is balanced beodd cycle (a,, El, a2, ..., E p , al) with no S = { a,, a2, ..., up } generates a subhyperis a contradiction.

Conversely, let H be a balanced hypergraph that is not bicolourable, with minimum order n = I X I. We shall show that this yields a contradiction.

1. We shall show that each vertex xo belongs to at least two different edges of H with exactly two elements. The subhypergraph Ho generated by X - { xo } is balanced (Proposition 2) and has order n - 1. Therefore it has a bicolouring (Sf,Si).If xo did not belong to an edge with exactly two elements, then (Sp u { xo }, Si) would be a bicolouring of H, which contradicts the definition of H . If xo belongs to only one edge with two elements, and if its other endpoint lies in Sp, then (Sf, SX u { xo )) would be a bicolouring of H. Hence, xo belongs to at least two edges with two elements, say [x0, y ] and [xo, z ] with z # y. 2. Denote by 9 the family of edges with exactly two elements. Consider Since G is balanced (Proposition I), G is a bipartite graph. graph G = ( X , 9). Consider a connected component C of G o . Since G has order 2 3 (from part l), there exists at least one vertex x1 in C which is not an articulation point.

3. Consider the subhypergraph Hi generated by X - { x1 }. It is balanced and has order n - 1. Therefore H I admits a bicolouring (&, SJ, and all the vertices adjacent to x1 in G have the same colour. Let S, be the set of vertices with this colour. Then, (Sl, S, u { x1 }) is a bicolouring of H because each edge of H with two elements is bicoloured (since it is an edge of G), and each edge of H with more than two elements is also bicoloured (because its intersection with X - { x1 } is bicoloured). This contradicts the assumption that H does not have a bicolouring. Q.E.D. Theorem 3. For a balanced hypergraph H = (E, / i E I ) , let k = mi ti,,, I Ei There exist k transversals of H that partition the vertex set X .

I.

Let (Sl, S,, ..., S,) be a partition of X into k classes, and let k(i) be the number of classes which meet edge Ei.If k(i) = k for every i, then X is partitioned into k transversals. If k(i) < k for an index i = io, then k(io) c I E,, 1; hence, there exist indices p and q such that: I S p n E i , I 2 2,

454

HYPERGRAPHS

and IS,nE,,,I = O .

The subhypergraph H' generated by S, u S, is balanced. Thus, by Theorem 2, H' has a bicolouring ( S ; , Si). Let 5'; = S, for j # p , q. The partition (Sl, S;l, ..., SL) determines new coefficients k'(i) with k'(io) = k(i,) 1

+

k'(i) 2 k(i)

( i # io).

By repeating this transformation as many times as needed, we obtain a new partition (Si,S ; , ..., SL) with k'(i) = k for every i. Clearly, this is the required partition of X . Q.E.D.

Theorem 4. A hypergraph H is balanced f, and only i f , y(H') = r(H') for eoery partial subhypergraph H' of H . 1. Let H be a hypergraph, and suppose that the above equality holds for every partial subhypergraph H' of H . If hypergraph H is not balanced, then there exists an unbalanced cycle, say, =

(01,El,

a2, E2,

.-.)a k ,

Ek,

a

The partial subhypergraph H' = (S, 8') defined by S = { a l , a2, ..., ak] and ..., Ek n S ) is a graph consisting of an odd cycle. Hence y(H') = 3, r(H') = 2, which contradicts that y(H') = r(H').

b' = (El n S, Ezn S,

2. Let H be a balanced hypergraph of rank h ; we have y ( H ) 2 h since h different colours are needed to colour a maximum edge; therefore, it suffices to show that there exists a strong h-colouring of H. Consider a hypergraph H' = ( X ' , 6 ' ) obtained from H by adding a set A , with 1 A , I = h - 1 E, I elements for every i and by letting:

X'=

XU

u A' i €1

and €'= ( E , u A i / i E Z ) .

Clearly, H' is a uniform hypergraph of rank h, and by Proposition 4, H' is balanced. By Theorem 3 applied to H', there exists a partition ( T i , T;, ..., 7';) of X' into h transversals. Since I 7';n E, 1 < 1 for every i and everyj, the sets S, = T; n X are the classes of a strong h-colouring of H. Q.E.D.

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

Corollary. If H is a balanced hypergraph with rank h, and if hl then there exists a partition (A', , X,)of X such that

455

+ h, = h,

r(Xl) = h, r ( X 2 ) = h,. Let (Sl, S,, ..., S,) be a strong h-colouring of H. Let Xl

=

u

s,

l 1. Assume that the proposition is valid for each normal hypergraph Ill with v ( H l ) < q. We shall show that it is also valid for a normal hypergraph H = ( X , 8)with v ( H ) = q. I f there exists a vertex x E X with v(H - 8,) < q, then there exists a transversal T of H - &,‘ with q - 1 vertices, and T u {XI is a transversal of H with q vertices. Hence, the proposition is valid. Therefore, we may assume that

v(H - 8x)= q (X E X). For each xi E X,consider a maximum matching gi of H - B,, . Let n = I X I. Consider the hypergraph Ho = P1+ g2+ ... + S,,formed with all the edges of the Pi for i = 1,2, ..., n. (Thus, hypergraph Ho contains exactly nq edges.) Since each vertex x, is not covered by Fitit follows that S(Ho) < n .

460

HYPERGRAPHS

Moreover, in a colouring of the nq edges of H , , the same colour occurs at most q times, and consequently, there are at least n different colours. Hence, dH0) 2 n >

wfo)

However, from the lemma, Ho is a normal hypergraph, which implies that q(H,,) = 6(Ho). This contradiction achieves the proof. Suficihncy. Consider a hypergraph H = ( X , 8) such that each partial hypergraph H’ of H satisfies v(H‘) = T(H’). It suffices to show that 6 (H) = q(H)* Consider a hypergraph R whose vertices correspond to the matchings in H, and whose edge E, is the set of all the matchings in H that contain edge E1.A subfamily H ’ = (Ej / j E J ) has a non-empty intersection if, and only if,

El n Elf # 12/ ( j , j ’ €4 (because this implies that T ( H ’ ) = v ( H ’ ) = 1). This is equivalent to

Ej n El, = 0

( j , j’ E J ) .

In other words, h sets of the family (€, / j E J ) have a common vertex in H if, and only if, (E, / j E J ) is a matching in 17. Thus, (1)

q(R)= T ( H ) ,

(2)

6(H) = v(R).

Moreover, the sets of a family (El / j E J ) have a common vertex 6, if, and only if, for each j E J , edge E, belongs to matching go,that is, if, and only if, (El / j E J ) is a matching in H . Thus,

(3)

q(H) =

(4)

6(R)= v ( H ) ,

7(m,

By applying (1) and (4) to a partial hypergraph H ‘ of H and to the corresponding partial hypergraph R‘ of R, it follows that

q(R’)= 6(R’). Thus, R is a normal hypergraph, and from part I , from ( 2 ) and (3), it follows that

v(R) = s(R).Therefore,

6(H) = q ( H ) .

Q.E.D. Fournier and Las Vergnas [i972] have also shown that a normal hypergraph is bicolourable. More precisely, if each odd c.vcIe of H contains three edges with a non-empty intersection, then H is bicolottrable.


46 1

Theorem 7 has been used by Lovsisz to prove that a graph is a-perfect if, and only if, it is y-perfect. More precisely:

Theorem 8 (LovQsz ([ 19721). Let G = ( X , E ) be a graph such that a(G,) = O(Gs) for each S c X ; then y(G) = w(G). Construct a hypergraph whose vertices are the cliques of G and whose edge X , is the set of all cliques of G that contain vertex xi E X . Clearly, graph G is a representative graph of the edges of H . Thus, (1)

v(H) = 4G)

(2)

q ( H ) = Y(G) *

9

Furthermore, from its definition, hypergraph H satisfies the Helly property. Therefore,

(3)

6(H) = w(G),

(4)

T ( H ) = O(G)

.

Each subgraph G, of G corresponds to a partial hypergraph H’ of H , and this correspondence is bijective. Applying (1) and (4) to G, and to H‘, it follows that v ( H ’ ) = 7 ( H ’ ) . Therefore, from Theorem 8, H is normal. Hence, q ( H ) = 6 ( H ) , and from (2) and (3), we have y(G) = w(G). Q.E.D. We shall now use the above properties of balanced hypergraphs to determine a sufficient condition for a hypergraph to contain k pairwise disjoint transversals. This condition was first shown for k = 2 by Lovasz [1968] and then extended to the general case independently by Lovhsz [1970] and Las Vergnas [1970]. First, a lemma is needed. Lemma. Let k 2 2. Let G = ( X , Y ,

r ) be a

simple bipartite graph such

that

I T(S)I

2 (k

- 1)

1 SI +

1

( S C X, S # @).

If no partial

graph of G (other than G itself) satisfies this condition, then each certex x in X has degree k.

First note that by letting S = { x } , the above condition implies that d,(x) 2 k for every vertex x.

Let d denote the family of all sets A c X such that 1T(A))=(k-l)IA\+l.

462

HYPERGRAPHS

1. We shall show that if A , A’ ES? and if A n A’ #

0, then A n A’ ~

d .

We have :

I T ( A u A’) 1 + I T ( A n A ’ ) I G I T ( A )u r ( A ’ ) 1 + I T ( A ) n r(A’)I = IW ) I + I W ’I ) =(k-l)IA]+l+(k-l)(A’(+l = ( k - I ) ( A U A ’ (+ 1

+(k- 1 ) I A n A ’ I

+ 1.

Moreover, since G satisfies the condition of the lemma, and since A n A’ #

gf,we also have: (T(’4UA’)I 2 ( k - 1 ) I A u A ’ )

+1

[ T ( A n A ’ ) [2 ( k - l ) [ ~ n + ~ 1’. [ Therefore, equality must hold in these two inequalities. Hence, A n A’ ES?.

2. Consider a vertex a in X , and let T(a) = { b l , b2, ..., 6, ). Let G, = ( X , Y, r,)denote the partial graph of G obtained by removing edge [a, b,] There exists a subset At of X such that

I Tf(4) I < (k - 1) 14 I + 1 Hence,

A, 3 a ; Let A

=

At E&’;

6, $ T,(A,).

A l n A 2 n ... n A,. From Part 1 , A

E&.

T(A - { a }) n r(a)= @

From above,

.

Therefore, =

+

+

I T ( A - { a > )I 2 k ( k - 1) I A (k - 1) I A I 1 = I T ( A ) I .

I r(A)I = &(a)

+

- { a >I

Consequently, equality holds in this inequality, and &(a) = k .

Q.E.D. Theorem 9 (Lovsisz [1970], Las Vergnas [1970]). Let k 2 2, and let H (El / i E I ) be a hypergraph such that

I

u

Ei~2(k--l)IJl+l

(Jl=I,Jf0).

1EJ

Then hypergraph H contains k pairwise disjoint traversals.

=

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

463

Consider a family H' = (El / i E I) such that

E,'

I 1u €J

C

E(

E*'IZ(k-l)lJI+1

(i E Z) (JCZ,J#I),

and that is minimal with respect to the order relation H' < H" defined by: E; c E ; for all i E Z. Consider the bipartite graph C,. = (I, X , r) where r(i)= El. Since this bipartite graph satisfies the conditions of the lemma, I E ; I = k for every i E I. Moreover, for each J c I, J # 0,

From Proposition (5, Ch. 17, $2), this implies that hypergraph H' contains no cycles. Therefore, H' is balanced, and, from Theorem 3, H' contains k painvise disjoint transversals. These sets are also transversals of H. Q.E.D.

3. Unimodular hypergraphs This section treats a class of hypergraphs which, like balanced hypergraphs, generalizes the concept of a bipartite graph. These hypergraphs which are important in integer linear programming give sharper results. First note that the concept of a q-colouring of a graph can be extended to hypergraphs in many different ways. (1) Aq-colouringof a hypergraph H = (Ei/ i E I ) is a partition (Sl,S,, S,) of its vertices into q classes such that iEZ, j 1 * Et Q S , .

...,

The smallest number of classes is the chroriiatic number x(H). See Chapter 19. (2) A strong q-colouring is a partition (Sl, S,, ..., S,) into q classes such that I E , n S , I G 1 ( i ~ z j, d 4). The smallest number of classes is the strong chromatic number y ( H ) . See Section 1. (3) An equitable q-colotiring is a partition (Sl, S,, ..., S,) into q classes such that for each i E Z and for each j , j' d q, - 1 d IE,nS,I

- IElnSi.I G 1 .

The smallest number q 2 2 for which there exists an equitable q-colouring is defined to be the equitable chromatic number k ( H ) of hypergraph H.

464

HYPERGRAPHS

If H is a graph, then these three definitions coincide with the ordinary definition of the chromatic number of a graph. The reader can verify that the hypergraph H defined by the projective plane with 7 points (Fig. 19.1) satisfies

x(H) = 3 ,

y(H) = 7 ,

k(H) = 7 .

The reader can also verify that the hypergraph H with 9 edges in Fig. 20.2 satisfies x(H) = 2 , y(H) = 4 , k(H) = 3 . Proposition 1. For each hypergraph H, X(H) G k ( H )

Y(H)

-

Clearly, each strong q-colouring of H is also an equitable q-colouring of H . Similarly, each equitable q-colouring of H is also a q-colouring of H. The proposition follows. Q.E.D. Proposition 2. If there exists an equitable bicolouring of H , and if GH = ( I , X,r) denotes the bipartite graph of the vertex-edge incidences in H , then for each partial hypergraph H' of H , the graph G,. is either non-eulerian, or eulerian with the number of edges equal to a multiple of 4.

The bipartite graph GH is defined by T(i) = E t . A graph is said to be eulerian if it has only even degrees. If H has an equitable bicolouring, then clearly H' also has an equitable bicolouring. Let (Sl, S,) be such a bicolouring. For k = 1, 2, let Fk denote the set of edges of G H d that are incident to a vertex x E S,. If Gw is eulerian, then each vertex i E I is incident to an equal number of edges of Fl and F,. Thus, I Fl

I = I F, I. Moreover,

I Fl I = 2

dCH' (x).

XESl

is even. Therefore, the number of edges in G H , is I Fl which is a multiple of 4.

1

+ I F2 I = 2 I Fl I,

Q.E.D. If H is a graph, or if H is the dual of a graph, it is easy to show that H contains an equitable bicolouring $ and only %,for each partial hypergraph H' of H, the graph G ., is either non-eulerian or eulerian with the number of edges equal to a niultiple of 4.

BALANCED HYPERGRAPHS AND UNlMODULAR HYPERGRAPHS

465

This result is similar to a theorem of Camion [1965] for totally unimodular matrices. A hypergraph H = ( X , 8)is defined to be unimodular if for each S c X , the subhypergraph Hs admits an equitable bicolouring.

EXAMPLE 1. Clearly, a graph is unimodular if, and only if, the graph is bipartite. Thus, the concept of a unimodular hypergraph is an extension of the concept of a bipartite graph. 2. Consider a set X of points on a line, and a family 8 = EXAMPLE ( E l , E 2 , ..., Em)of non-empty intervals whose union is X . The hypergraph H = ( X , 8)is unimodular because for each S c X an equitable bicolouring of Hsis obtained by alternately colouring red and blue the points of S as they appear in order on the line. Proposition 3. If H is a unimodular hypergraph, each partial sirbhypergraph of H is also imirnodulur:

This follows from the definition of unimodularity. Proposition 4. A unimodular hypergraph is balanced.

Suppose that hypergraph H is unimodular but not balanced. Then H contains an odd cycle ( a l , E l , a2, E,, ..., Ep,a,) of which no edge contains three of the a,. Thus, the subhypergraph H A generated by A = {a,, a,, ..., a,} does not have an equitable bicolouring, which contradicts the unimodularity

of H. Q.E.D. H i

Fig. 20.2

The converse is not true. Consider the hypergraph H in Fig. 20.2. Each odd cycle of H contains three vertices of edge E l . Hence, H is a balanced hypergraph. However, H is not unimodular. Hypergraph H has a unique

466

HYPERGRAPHS

bicolouring of its vertices. (This can be seen by colouring vertex a red, then successively colouring in red or blue vertices b, c, d, e, S, g, h, i, j . ) This bicolouring is not equitable for edge E l . Theorem 10. r f H is a hypergraph without odd cycles, then H is unimodular. Since each subhypergraph of H contains no odd cycles, it suffices to show that a hypergraph H = ( X , 8)without odd cycles has an equitable bicolouring. There exists a function x,(t) such that Ei can be written as { x l ( l ) ,xl(2),

...)xdrt) 1 ,

where

r,

=

I El I .

Let 9, be the set of pairs xi( l)xi(2), xi(3)x,(4), etc. Consider the graph G = (X,a),where G = U 4. We shall show that graph G cannot have an odd cycle. Suppose that G contains odd cycles. Let p = [a,, a,, ..., a,] denote an odd cycle of G with minimum length. Cycle p is elementary; suppose that it contains two disWithout loss of generality, let [a,,a,,,] joint edges in the same class 4. and [a,, a, + ,] be these two edges. Transform 4 by replacing these two edges by edges [as,a,,,] and [a,, a , + , ] . Replace p by the sequence either ( a l , a , , ..., a,, a , , , , ..., a , ) or (a,,,, a , , , , ..., a,, a , , , ) that has odd length. Repeat this process as many times as possible. Upon termination, an odd sequence is obtained that determines an odd cycle in hypergraph H , which is a contradiction. Since graph G contains no odd cycles, there is a bicolouring (S, , S,) of its vertices. Clearly, (S, , S,) is also an equitable bicolouring of H . Q.E.D.

Corollary. A hypergraph H = (Ei I i E I ) contains no odd cycles if, and only if, each hypergraph H' = (El/ i E I ) with E( c Elfor all i E Isatisjies k(H') < 2. The condition is necessary because hypergraph H' contains no odd cycles. The condition is sufficient because an odd cycle of Zf (if it exists) would induce a hypergraph H' that is a graph consisting of an odd cycle. Hence, k ( H ' ) = 3, which is a contradiction. Q.E.D. The following theorem and its corollary are direct extensions to unimodular hypergraphs of a property initially discovered by D. de Werra [I9701 for the dual G* of a bipartite graph: Theorem 11. A unimodular hypergraph H has an equitable q-colouring for each positive inteaer Q 2 2.


467

For q = 2, the theorem is clearly true. For q > 2, consider a unimodular hypergraph H = (E, / i E I ) , and consider a partition ( S , , S,, ..., S,) of its vertices into q classes. For eachj, k 6 q, let ejk(i) = 1 sj e ( i ) = max

Ei

,.k

I - I SICn Ei 1 el,&)

.

Clearly, e(i) Z 0. If e ( i ) for all i E Z, the partition is an equitable q-colouring of hypergraph H and vice versa. Suppose that there exists an index io with e(io) 2 2, and let r and s be two distinct indices with er,s(io)= e(io). Then, for allj,

I ss n Eio I 6 I S, n Eio I

-

6 I S r n Ei, I Since H is unimodular, the subhypergraph H' of H generated by the set Sru S, has an equitable bicolouring ( S ; , S:). Let S; = S, for j # r, s. The partition (Si,SL, ..., Si) defines as before coefficients e;,(i), and f o r j # r, s, we have : e 2 i 0 ) < e,,(io) - 1 6 e(io) - 1 e:,(io) 6 e,,(i0) - 1 6 e(io) - 1 e:,(io) 6 erj(io) - 1 < e(io) - 1 e;,(io) 6 e,s(io)- 1 ' 6 e(io) - 1 e:,(io) 6 1 6 e(io)- 1 . Since eb,,(io) = ep,q(io)for p , q # r, s, the number of pairs ( j , k ) with e;,(io) 6 e(io) - 1 is greater than the number of pairs ( j , k ) with e,k(io) < e(io) - 1. Moreover, for all i # i o , e;,(i> 6 e(i). By repeating this transformation, a partition with e(i) 6 1 for all i is finally obtained. This partition is an equitable q-colouring of H. Q.E.D. Corollary. I f H = ( E J i E Z ) is a unimodular hypergraph, and if k = mini,, I EiI, then there exists a partition ( T , , T,, ..., T,) of the vertex set X of H into k transversals such that

Note that there always exists a partition of X into k transversals from 'Theorem 3 (since a unimodular hypergraph is balanced). Moreover, from Theorem 1 1 , hypergraph H has an equitable k-colouring ( T I ,Tz,-.*,.Tk), and

468

HYPERGRAPHS

Furthermore, since k = min 1 Ei1, each T,is a transversal of H. Q.E.D.

APPLICATION (de Werra [1970]). If G = (X,E ) is a bipartite multigraph and if q > 1, then the edges of C can be partitioned in q partial graphs GI, Gz,..., G, such that for each j < q and for each x E X ,

It suffices to apply Theorem 11 to G*. Let A = ((a:))be a matrix with m rows and n columns. Matrix A is defined to be totally unimodular if the determinant of each square submatrix of A equals + 1, 0 or - 1. If A is unimodular, then clearly each coefficient aJ of A must be + 1, 0 or - 1, since a: is a minor of order 1. We shall state without proof two fundamental properties of unimodular matrices.

Theorem of Hoffman and Kruskal [1956]. A H n x m matrix A is totally unimodular if, and only if, f o r any integer m-rectors b and b' and for any integer n-vectors a and a', each face of the polytope { X / X E

R";a

< x < a';b C Ax 6 6')

contains an integer point. It follows that each vertex of this polyhedron has integer coordinates.

Theorem of Ghouila-ilouri [1962]. An n x m matrix A = ((a:)) is totally unimodular f, and only f, each set J c { 1,2, ..., n } can be divided into two disjoint sets J1 and Jz such that

1

2 a: - C fEJI

af

Ic

1

( i ~ m ) .

j EJz

A consequence of this theorem is:

Property 1. A hypergraph H is unimodular if, and only if, its incidence matrix ((a;))is totally unimodular. Property 2. If H is a unimodular hypergraph, then its dual H* is also a unimodular hypergraph.

This follows from Property 1 since the incidence matrix of H * is the transpose of the incidence matrix of H .


469

4. Stochastic functions Let H = (X,6 ) be a hypergraph. A real valued functionf(x) defined on X is said to be a stochastic function associated with H if

0 I FI - F2 I

*

478

HYPERGRAPHS

There exist two matchings Vl and V, such that F, = A n SCV,). F, = A n S(V,), At least one connected component of the partial graph generated by

(V1 - V z ) u ( V z is an open elementary chain the other endpoint b in Fl Therefore,

- V,) = p[a, b] with an endpoint

11

- F,,

a in Fz - Fl and from the lemma to Theorem (1, Ch. 7).

v; = ( V , - P ) u ( V 2 n 14 is a matching that saturates all the vertices in Fl u { a }. This contradicts the maximality of Fl, and completes the proof. EXAMPLE 6. Consider a family ( A , / j E Q ) of subsets of a set E such that: A(Q) = U A j = E . j aQ

A subset T = { t l , t,, ..., tk } of E is called a partial transcersal if there exists an injection j ( i ) from { I , 2, ..., k } into Q = { 1, 2, t . ., q } such that ( i = 1 , 2, ..., k ) . ti € A j ( i , We shall show that the family of partial transversals is a matroid on E with rank r(S) = I Q I min (I A ( J ) n S I - I J I ) .

+

JcQ

This matroid is called the transversal matroid of the family ( A , / j E Q). Consider the bipartite graph (Q, E, r)where r is defined by:

W = Aj

( j e P) *

From Example 5, we know that the sets of vertices that are contained in S( V ) for some matching Vdefine a matroid. The family of partial transversals is a subhypergraph of this matroid. Therefore, the family of partial transversals is also a matroid. From the Konig theorem (Ch. 7, g 3), the rank-function of this matroid is:

r ( S ) = min J C

Q

(I

Q-J

I

+ 1 r(J)n S I) = q + min (I A ( J ) n S I - I J I] . J C Q

EXAMPLE 7. If (C, , C,, ..., C,) is a partition of a set E into p classes, and if cl, c z , ..., c, are integers with 0 6 c1 < I C, I, then the family 9 = { F I F c E , F # @ , IFnC,I 1. Then, there would exist two distinct vertices 6' and b" in B,. Let B; = B , - { b' )

B; = B ,

- { b" }

B!=B!'=B,

if

j # 1 .

483

MATROIDS

From the minimality of g,there necessarily exist two subsets I, J of Q such that: @’(I)) < I I I + k r(B”(J)) < I J I k - q . Hence, @(I)) r(B”(J)) < I I I I J I 2(k - q) - 2 .

+

+

+

+

Moreover,

B’(I) u W ( J ) = B(I u J ) , B‘(I) n B”(J) =I B(Z n J - { 1 1). Combining these results with Proposition 4 yields r(B’(I)) r(B”(J))2 r(B’(Z) u B”(J)) r(B’(I)n B”(J)) 3 r(B(I u J ) ) r(B(I n J - { 1 })) 2 zu J I ~nJ - { 1 } 1 + 2(k - q) 3 I1I + I J I + 2(k - 4 ) - 1 which is a contradiction. Thus, we have shown that 28 is of the form (( bi } / i E Q). If we let B= {bi/iEQ}, then from (1) it follows that r ( B ) = r(B(Q)) 2 I Q I + k - q = k . Therefore, there exists a subset K of Q with I K I = k and an independent set F = { b, / i E K } contained in B, such that:

+

+I

+

+

9

biEAi

(iEK).

Q.E.D.

The well-known Rado theorem follows immediately from this result: Theorem 4 (Rado [1942]). If A4 = (E, .F) is a matroid, then a f a m i 4 at = ( A l , AP, ..., A,) of subsets of E hare an independent set of distinct representatices if, and only $, ( J c Q) . r(A(J)) 2 I J I The proof follows by letting k = q in Theorem 3.

Q.E.D.

Corollary 1. Two families .d= ( A , , A , , ..., A,) and 93 = ( B l , B,, ..., BJ hace a common set of distinct representatives $ and only if,

IA(J)nB(K)) I J I + I Kl - 4 (J, K c Q ) . Consider the transversal matroid M of family 9 (see Example 6).The rank of this matroid is r ( S ) = q + min (I B ( K ) n S I - I K I). K= Q

484

HYPERGRAPHS

From Theorem 4,there exists a transversal set of .dthat is independent in M if, and only if, for all J c Q,

Q.E.D. Corollary 2. If V = (Cl, C z ,..., C,) is a partition of a set E, and i f c l , c2, ..., c, are integers with 0 < ci < I Ci1 for all i, then a faniily s4 = ( A , , A 2 , . .., A,) has a set T of distinct representatives wfth I T n C, 1 < ci f o r all i, $ and only i f ,

Consider the matroid M formed from the sets F c E with I F n CiI for all i (see example 7), whose rank is

< c,

D

r(S) =

1 min { c i , I S n CiI ] . I= 1

There exists a set of distinct representatives of .d that is independent in M if, and only if, for all J c Q, P

r(A(J)) =

C

min { ci, I A ( J ) n Ci I } 3 I J

I.

i= 1

Q.E.D. The Corollary to Theorem (6, Ch. 7), can be restated in terms of sets of distinct representatives as follows: A necessary and suficient condition .for a set B c E to be contained in some set of distinct representatives of a faniily d = ( A l , A 2 , ..., A,) is that min{IA(J)uBI, q - I B - A ( J ) ( } > I J I ( J cQ ) . We shall extend this result to matroids, but first a lemma is needed. Lemma. Let A4 = (E, 9) be a tnatroid of rank r. Let B E 9, and let q 2 I B I. Then, the family

FGe,q = { F / F c E , F u B E F, I F u B I ,< 4 } is a matroid on E with rank

} - IB - S I. E, and let Sobe a subset of S that belongs to F&.If F satisfies the rGeJS)= min { r(S u B ) , q

Let S c two following conditions:

(1)

F EFB,q

(2)

S o c F c S

485

MATROIDS

then, clearly F satisfies I F I d min { r(S u B) , q ] - I B - S I Thus, it remains to show that equality holds for some F.

Fig. 21.1

Set B u So is independent in M and, consequently, is contained in an independent set F' of B u S with cardinality IF') = r ( S u B ) . Let F" be an independent set such that B u So c F" c F' , I F" I = min { r(S u B ) , q ] . Clearly, set F = F'

A

I FI

S satisfies conditions (1) and (2), and its cardinality is = min{ r ( S u B ) , q } - I B

- SI. Q.E.D.

Theorem 5 (Las Vergnas [1969]). Let M = (E, F)be a matroid with rank r. Let B E F , and let q 2 1 B 1. Then, a family &' = ( A l , A , , ..., A,) ofsubsets of E hace a set T of distinct represenfatiliessuch that T E 9and T 3 B i f , and only if, min{r(A(J)uB),q}-lA(J)-

BI>,IJI

(JcQ).

Consider the family

FB,q = { F / F c E , F u B E 9, I F uB I d 4 1. By the lemma, this family is a matroid on E. If there exists a set Tof distinct representatives of&' with T E .Fand T then I T 1 = q. Therefore, T E FB,q

3

B,

*

Conversely, if there exists a set T of distinct representatives of d without T E FB, , then T u B E F , I T u B l d q , I TI = 4 ,

486

HYPERGRAPHS

and, consequently,

TEB, T 2B. Then, from the Rado theorem, there exists a set T that satisfies the above requirements if, and only if, the rank rB,qof the matroid ( E , SBvQ) satisfies rs,,(A(J))2 I J

I

(J

c

Q)

or

min(r(A(J)uB),q) -

IB-A(J)[

2

IJI. Q.E.D.

3. Image of a matroid Many of the ideas to be presented in this section originated in the work of Edmonds and can be found in Edmonds, [1965], [1970]. Let M = (E, F)be a matroid on E = { e l , e,, ..., en,}, and let cp be a mapping of E onto a set E. Consider the family -

9 = (cp(F) / F E 9). Since cp is a mapping onto E, the pair R = (E, 8 )is clearly a hypergraph, called the image of M . Theorem 6 (Nasi-Williams 119661). Zf R = (E, 9) is the image of a matroid M = (E, 3)by a mappitig cp of E onto E, theti R is a matroid, and its rank is:

I.(E) = min (r(q-' (A))+ I E - A I) . ACE 1. We shall show that

-inax - IF I F € f

Clearly, max

+ I E - A 1).

= min ( r ( c p - ' ( ~ ) ) ACE

I F I is the greatest integer k such that the family

(v-'(G), (P-*(ez), *..>v-'(L)) has a partial set of distinct representatives that is independent in M and has cardinality k. From Theorem 3, this is the greatest integer k such that min (r(q-'(A))+

IE

-

Al)2 k .

ACE

Hence, max I F I = inin ( r ( q - ' ( i ) )+ I E AC

21).

E

2. To complete the proof, it suffices to show that the image of M by cp is a matroid.

481

MATROIDS

Consider a mapping such that

of E = { el, e,, ..., em } onto

'p

E

=

{ Z2,&,

-

..., em1

d e l ) = Z2 'p(ei) =

Zi

for i # 1

,

Mapping cp is said to be an elementary contraction. Since each mapping is a product of elementary contractions, it suffices to show that the image of matroid M by an elementary contraction cp is a matroid. We shall Consider an independent set Fo E .F such that Fo is maximal in 9. show that Fo is a maximum set. From part 1, it suffices to show that there exists a set c E such that

lF0 I

= r(cp-'(A))

+I E -A/.

In order to simplify, let Eo = E - { e l , e2 }. Since Fo is maximal in 9, we may assume that Fo is a maximum set of F. We may also assume that Fo contains both el and e, because, otherwise, Fowould be maximum, since

1 Fo I

= I F,

I = r(E) = r(qD-l(E)) + I E - Zo1 .

Three cases must be considered: CASE1. r(Eo) = r(E). Since

G ' ( E - { el 1) G r(E) 9 it follows that r(E - { el )) = r(E). Since Fo contains both el and e2, then @o)

I F~ - { el 1 1 = r(E) - I < r(E - { el }) . Consequently, there exists a maximum independent set FA that does not contain el and that satisfies: -

Fb

Thus,

-

3

F,

- { e, } = F o .

Fo = F i , and

I Fo I

=

I %I

= I F;

I = r(E) = r(cp-'(E))

+ I E - Eo I .

Hence, the required equality is proved. CASE2. r(Eo) = r ( E ) - 1. Then, each maximum independent set contains either e, or e2. Moreover, I Fo n Eo I = I Fo I 2 = r(E) - 2 < r(E,).

-

Thus, there exists a vertex a E Eo such that (Fon Eo) U { a } is an independent set with cardinality r(Eo) = r ( E ) - 1. Let FA be a maximum independent set that contains this set. Since FA is maximum, it contains el or e2.

488

HYPERGRAPHS

Without loss of generality, we may assume that

>.

Fb = (Fo n E o ) u ( a , e ,

Fo.Therefore, Fo = FA,and 1 F b I = I F, 1 = r(E) = r(cp-'(E)) + I E - E I .

Clearly, FJ2

Hence, the required equality is proved. CASE3. r(Eo) = r ( E ) - 2. Then, each maximum independent set contains both el and e,. Consequently, Fo 2 F, and

I Fo I = I F , 1 - 1 = r(E) - 1 = r(E,) + 1 = r(p-'(&)) + 1 E - E, I Hence, in all cases, Fo is a maximum set of 9. c

Q.E.D. If H 1 = ( X , Sl),H 2 = ( X , S2),..., H P = ( X , 8")are hypergraphs on a set X,their join is defined to be the hypergraph H

=

H 1 v H a v - v H P = (A',

P

V

&'),

1=1

where : P

v

8' = { E'

LJ

E2 v

*--

u E P / E' €8'' ; E 2 € E 2 ; ... : EP E P} .

i= 1

Clearly, H is a hypergraph on .'A

Theorem 7. If (E, Sl), (E, P) ..., ,(E, 9") are matroids whose ranks are respectively rl, r2, ..., rp, then their,join hypergraph is a matroid, and its rank is -

r(E) = min (r'(A) + ... + rp(A)+ I E - A

I).

AcE

Make p identical copies E l , E 2 , ..., E P o f set E. Let ef, denote the element of El that corresponds to ek E E. Consider the mapping cp of Uf=lE' into E that maps e: E E' onto e,. Clearly, M = (U E', V F j ) is a matroid, and its rank is r(U E,) = r l ( E ' ) r2(E2) ... r P ( E P ) .

+

+ +

a,

From Theorem 6, the image of this matroid by cp is a matroid and R is clearly the join Vfml9'Moreover, . the rank of the joint matroid is

F(E)

=

min ( r ( v - ' ( A ) )

+ I E - A I)

ACE

Q.E.D.

489

MATROlDS

Corollary 1 (Edmonds [1968], Nash-Williams [1968]). For a matroid M = ( E , F),the cocering number p(M), i.e. the minimurii number of independent sets that are needed to cotrer E is equal to: p ( ~ =) max

[---I

IAl

*.

r(A)

ACE A # 0

By definition, p ( M ) is the smallest integer k such that the join matroid V M V *.. V M of k identical copies of matroid M has rank I E I. From Theorem 7, p ( M ) equals the smallest value of an integer k such that M

min (kr(A)

+ I E - A I) = I E I ,

ACE

or, equivalently, such that min ( k r ( ~-) I A

I) = 0.

A cE

This is equivalent to: kr(A) = I A

I20

( A c E),

or

k > 'r ( A )

a>.

(A c E , A #

Thus, p ( ~ =) max

[-dI A ) * IAI

ACE A # r'

'

Q.E.D. Corollary 2. For a matroid M = (E, F), the maximum number k , of pairwise disjoint maximal independent sets is

r(A)fr(E)

Clearly, ko equals the greatest integer k such that the joint matroid of k identical copies of matroid M has rank kr(E), where min (kr(A) + 1 E - A

I)

= kr(E)

.

ACE

This is equivalent to rnin (kr(A) - kr(E)

+ I E - A I) = 0,

ACE

or to

-

k(r(E) r ( A ) ) < I E - A I This yields the required equality.

(A c E). Q.E.D.

490

HYPERGRAPHS

Corollary 3. Consider a matroid M of integers szrch that

=

(E, S)and a sequence k l , k,,

..., k,

a

Let k: denore the number of k, that are 2 j . Then, set E can be partitioned into q independent sets Fl , F2,..., F,, with I Ff I = k, for all i, if, and only if,

C

k;<JE-AJ

(ACE).

j>r(A)

Consider the ki-section M ( k l )composed of the family .F(ki)

=

{ F / F E 9, IFI

< ki 1 .

It is a matroid, and this matroid has rank r'(A) = min { r(A), k, }. There exist q sets that satisfy the conditions of the corollary if, and only if, the join matroid M = Vp= M(kl)has rank I E I. This is equivalent to

which is equivalent to 4

1

4

min(r(A),ki}+IE-A121EJ=

C i= 1

i=l

k;

ki=

(ACE)

j>O

or to (Ac E).

Q.E.D. The corollary follows. A!

Corollary 4. Let H M be a hypergraph consisting of the circuits of a rnatroid = (E, 9) of rank r. Then the chromatic number of H M is

A#

0

A set F c E i s independent if, and only if, it contains no circuits. Therefore, a partition ( A l , A 2 , ..., A,) is a colouring of hypergraph H M if, and only if, sets A l , A , , ..., A , are independent. Thus, z ( H M )= p ( M ) , and the corollary follows from Corollary 1. Q.E.D. The above results easily yield several theorems for graphs that were proved earlier by direct but very involved methods.

49 1

MATROIDS

APPLICATION 1 (Theorem of Tutte [1961]). The edge set of a simple connected graph G = (X,E ) contains k pairwise disjoint spanning trees i f , and only i f , for each partition 9 of X,the number mc(P)of edges that join two distinct classes of the partition satisfies

2 k(l 1 - 1 ) . 1. If there exist k spanning trees H I , H , , ..., Hkthat are pairwise edgedisjoint in G , then for each partition B of the vertices, filG(9)

m H , ( 9 )2 191 - 1

( i = 1 , 2 ,...,

k).

Therefore, k

mG(9>

2

mH,(9L) I=1

2 k(l 9 I - 1)

9

which is the required condition. 2. Suppose that the condition of the theorem is satisfied. Consider the where 9is the family of forests F l , F,, ..., F, of graph matroid M = ( E , 9) G. Let r(A) denote the rank of M . If A C E defines a partial graph of G with p connected components that constitute a partition 9'= (XI, X , , ..., X,) of X,then r(E) - r ( A ) = ( n - 1) - (n - p ) = p - 1 = 191 - I , Therefore, from the condition of the theorem, it follows that

1E - A I

> m G ( 9 )> k(l 9 I - 1)

= k(r(E) - r ( A ) ) .

Hence,

Thus, from Corollary 2 to Theorem 7, there exist k disjoint spanning trees in G. Q.E.D. APPLICATION 2 (Theorem of Nash-Williams [1964]). The edges of a siiizple graph G = (X,E ) can be coloured with k colours sirch that no cycle has all its edges with the same colour i f , and only if, for each set A c X , the number mG(A,A ) of edges with both endpoints in A satisfies

mG(A, A ) < k(l A 1 - I ) . In other words, the chromatic number of the hypergraph Gc formed consisting of the cycles of G is

492

HYPERGRAPHS

1. Suppose that the edges are coloured with k colours 1,2, ..., k. Let mi@, A ) denote the number of edges with colour i that have both endpoints in A . These edges form a forest. Therefore, nr,(A, A ) < I A I - 1. Hence,

m,(A, A )

=

rn,(A, A )

+ ... + n1,(A, A ) G k(l A I - 1) ,

and the result follows. 2. Conversely, suppose that the condition of the theorem is satisfied. Conformed by the family of forests of graph G. Let r desider the matroid ( E , 9) note the rank of this matroid. If the partial graph ( X , F) of G generated by F c E has exactly p connected components (XI, Fl),(X2, F2),..., (Xp,Fp) that are not isolated vertices, then kr(Fi) - I Fi I 2 k(l

XiI

- 1) -

rnG(Xi,X i ) 2 0 ,

Hence, P

kr(F) - I F

1

=

1 (kr(Fi) - 1 F iI) 2 0, i= I

and

k2max[!$]*. FCE F+ 0

Thus, from Corollary 4, it is possible to colour the edges of G in k colours such that no cycle has only one colour.

Q.E.D. APPLICATION 3. f l G is a simple graph with maximum degree h, then it is

"ll*+

possible to colour its edges with edges with the same colour.

1 colours such that no cycle has all its

Let G = ( X , E ) be such a graph, and let A = A - { a } . Thus

C

X . If I A

I

> 1. For a E A ,

let

From the Nash-Williams theorem (Application 2), it follows that

493

MATROIDS

G I*+

Therefore, it is possible to colour the edges of G with -

1 colours such

that no cycle of G has all its edges with the same colour.

Q.E.D. 4. Minimum weight basis

A maximal independent set of a matroid M is called a basis. Associate with each element e of matroid M a positive number p(e) called the weight of e. The ueight of a basis F o f M is defined to be P(F) =

c P(4.

esF

We shall consider the problem of determining a minimum weight basis of a matroid.

EXAMPLE 1. Minimum tree in a weighted graph. Let G = ( X , E ) be a connected simple graph. To each edge e associate a “length” p(e) > 0. The minimum tree problem is to find a spanning tree H = ( X , F ) of G whose total length c EP

is as small as possible. This problem due to G. Choquet [1938] and J. B. Kruskal [1956] has numerous applications. For example, n cities in the Benelux countries must be connected by canals, and the total length of the canals must be minimized (and no branching is allowed except at the cities). Since the system of canals must be connected and since its length must be minimum, no cycles are allowed. Thus this problem becomes a minimum tree problem. The minimum tree problem reduces to finding a minimum weight basis for formed by the forests of G. (See Example 3, Q 1.) the matroid (E, 9)

EXAMPLE 2. The following problem studied by Chein [1970] also reduces to a minimum tree problem: Consider a connected simple graph G = ( X , E ) with a positive length y ( e ) associated with each edge e, and a set S c X . Find a tree in G (not necessarily spanning) that contains all the vertices of S and that has a minimum total length. When all lengths are equal to 1, this problem can be solved by treating separately each of the connected components X , , X,, ..., X , of the subgraph Gs, that respectively have spanning trees (XI, F J , ( X 2 ,F J , ..., ( X p ,F,), and by finding a spanning tree F, of minimum weight in the graph obtained from GbecontractingX,, X , , ..., X,.Then,F, u F , u . - u F,istherequiredtree.

494

HYPERGRAPHS

Let F = ( e l , e,, ..., e,) and F’ = ( e i , ek, ..., e:) be two ordered bases of the elements of these r-tuples are indexed such that: matroid M = (E, 9); p(el) 6 p(e2> 6

... < p ( e , ) ,

< p(e;) 6

G p(e:)-.

p(e;)

Let F

< F’ if, and only if, there exists an index k < r such that

i

p(eJ = p(e;)

for

i < k

p ( 4 < pk;) . This order relation < on the set of bases is called a lexicographic ordering: i f the “weight” of a letter is its rank in the alphabet, then the relation F < F‘ between two words F and F’ indicates that word F appears before word F’ in the dictionary.

M

Proposition 1. If F = ( e l , e,, ..., e,) is an ordered basis of a rnatroid = (E, .F),then the following three conditions are equiralent : (1) F is a minimal basis vith respect to the lexicographic ordering, c(, ( 2 ) each ordered basis F‘ = (e;, eh, ..., ek), satisfies p(e,) < p(e,’)for all i, ( 3 ) F is a minimum weight basis.

(1) 3 (2). Otherwise, let F be a minimal basis, and let F’ be a basis that does not satisfy (2). Let k be the first index such that p(eJ > p(ek) A = { e,,e,,

..., e k - * , e ; , e > , ..., e ; } ,

has rank r(A) > k, and consequently there exists an index i set ( e , , e 2 ,..., e , - , , e J ) = F , is an independent set of cardinality k. Moreover, since i

p ( 4 ) < p(ek). contains Fo with I B I = r.

< k such that the

< k, we have:

p(el) 6

Let B be the basis that contradicts the minimality of F.

Thus, B

< F,

which

(2) 3 (3). This is obvious.

(3) => (1). Let F be a minimal basis with respect to the lexicographic ordering, and let F’ be a minimum weight basis. Since (1) implies (2), basis F satisfies (2), and (i = 1, 2, ..., r ) . p(eJ < p(e9 Since F’ has minimum weight, it follows that p(ei) = p(e{)

(i = 1, 2, ..., r ) .

495

MATROIDS

Thus F' < F ; since F is minimal with respect to the ordering ci, it follows that F' is minimal with respect t o the lexicographic ordering a. Q.E.D. Algorithm 1. To construct a minimum weight basis, first choose an element el of minimum weight, and let Fl = { el }. Then choose any element e2 with e, 4 Fl and Fl U { e2 } E 9, that has minimum weight; let F2 = Fl u { e2 } etc. After r steps, a basis F, is determined. Since F, is minimal with respect to the lexicographic ordering, it is a minimum weight basis by Proposition 1. Dual algorithm. If F i s a basis of matroid A4 = (E,.F) then , its complement E - F is called a cobasis. Clearly, the cobases of M are the bases of some matroid M* called the dual matroid of M . If M is a matroid whose bases are the spanning trees of a connected graph, then the cobases of M are the cotrees of this graph. Instead ofconstructing a minimum weight basis for matroid M with weights p(e), we could construct the maximum weight cobasis or construct a minimum weight basis for the dual matroid M * with weights p*(e) = p o - p(e), where p o is a suficiently large positive number such that p*(e) > 0 (e E E ) . T o construct a maximum weight cobasis for M, first remove from E an element e, of maximum weight. Let E, = E - { e l }. Then, remove an element e2 E Elthat has maximum weight. Set E2 = El - { e, } also has rank r. After m - r steps, a maximum weight cobasis E and a minimum weight basis have been found.

For the special case of constructing a minimum length spanning tree of a graph G, the above algorithms can be improved. First, we need the following proposition : Proposition 2 (Rosenstiehl [1967]). Let G = ( X , U ) be a connected graph with a length p(u) associated uwth each arc u. Suppose that the lengths are all diyerent. Then, the tree of miniminn length is unique. A necessary and suficient condition that arc u belongs to this tree ( X , V ) is that there exists an elementary cocycle o in G such that p(u) = min p(u)

.

UECD

Another necessary and suficient condition that arc u belongs to tree ( X , V ) is that f o r each cycle p,

496

HYPERGRAPHS

1. The minimum tree is unique. Otherwise, if V = ( u , , u,, ..., u,) and W = ( w , , wZ,..., w;) were both minimum trees, then from Proposition 1, (i = 1, 2, ..., r ) p(ui) < p(wJ < d o i ) This contradicts the assumption that the arc weights are all different. 2. If arc u belongs to the minimum tree (X,V ) , then u separates two sets A and X - A , clearly, p(u) = min p ( u ) . u E m(A)

(Otherwise, there would exist an arc u' # u in the minimum tree chosen by the algorithm 1 to link sets A and X - A . ) Therefore, there exists a cocycle w(A) that satisfies (1). 3. If arc L' belongs to a cocycle w that satisfies (l), then u also belongs to an elementary cocycle w(A) that satisfies (l), and u would be the first edge of w(A) to be chosen by Algorithm 1 to construct the minimum spanning tree V. Therefore, t i E V. The proof of condition (2) is left to the reader. Algorithm 2 (Kruskal [1956], Sollin [1962]). A sequence V,, V,, ... of forests can be successively formed in the following way: (I) Let V , = forest ( X , V,) consist of n isolated vertices. (11) If forest ( X , V,) has an isolated vertex x, then join x to its closest neighbour; i.e., if u i C l is the arc of minimum length incident to x, then let

a;

V i + l = V,U { I ' i + l 1 . (111) If forest ( X , V,) has no isolated vertices and is not connected, then

join a connected component to its closest connected component. (IV) If ( X , V,) is connected, then it is a spanning tree and, clearly, this tree has minimum length by Proposition 2. Algorithm 3 (P. Rosenstiehl [1967]). Starting with a cocycle basis { w1,w2, ..., w k } remove from each of these cocycles its minimum weight arc. Then, proceed as in Algorithm 2. Dual algorithm. Starting with a cycle basis { p l , pz, . .., p' } (for example, if G is planar, the contours of the bounded faces), remove successively from each cycle an arc with non-maximum weight. Continue until no cycles remain.

EXERCISES 1. Show that the set 9 of circuits of a matroid satisfy the following: (1) S' c S, S E Y implies S' = S (2) S, S' E 9, S # S, a E S n S' implies that there exists an So E Y with So c S U S' - {a}.

497

MATROIDS

Conversely, show that if a family of sets satisfies (1) and (2), then the family is the set of circuits of some rnatroid. 2. Let Kn denote the complete graph with n vertices, and let K: denote the hypergraph consisting of the cycles of K , . Show that the Nash-Williams theorem implies that

Show how to construct a colouring of the edges of graph K,, in no cycle has all its vertices with the same colour. 3. Let K,, , denote the complete bipartite graph. Show that

4. Let M be a matroid on set E. Show that if 9 is the family of its bases then { E - B / B E } is the set of bases of matroid M . 5. Let G = ( X , P) be a 1-graph, and let Bo be a subset of X with cardinality k. Let d = ( P(x) u { x } / x E X - So}. For a set B C X with cardinality k, show that the following conditions are equivalent : (1) Sets B and Bo can be joined together by k paths that are pairwise vertex disjoint. (2) The set X - B is a set of distinct representatives of family d (see Section 2). 6. Use Exercises 4 and 5 to show the following result: Let G = (X,T)be a 1-graph, and let Bo C X. Then the sets B c Xsuch that I B I = I Bo I and such that there exist 1 Bo I pairwise vertex-disjoint paths joining B and Bocon-

stitute the bases of a matroid on X.

(H. Perfect, Applications of Menger’s Theorem, Math. Anal. Appl., 22, 1968, pp. 96-111.)

7. Let G = (X,T)be a transitive 1-graph, and let S be the family of subsets Fof X such that F 1 ( X - F ) 3 F. Show that P is a matroid on X. 8. Let G be a multigraph. Let 9 be the family of subsets of the edge set that do not contain an even elementary cycle and that d o not contain two edge-disjoint odd elementary (A. Duchamp, Thesis, Caen [1971]) cycles. Show that S is a matroid on E.

9. In the (directed) graph G = (X,U ) , let P be the family of subsets Uo of U = ..., urn} that do not contain an “anti-circuit” (a simple even cycle whose arcs are alternatively directed). Show that 9 is a rnatroid on U.Determine its rank. (A. Duchamp) Hint: Consider the graph H = ( Y , V) with Y = { a,b }, and with m arcs v1, v 2 , ..., urn directed from a to 6. Consider the “product” G 8 H, i.e. the graph on X x Y with an arc from ( x , y) to (x’, y‘) denoted by wk, if (x, x’) = uK and ( y , y’) = Uk. Clearly, each anti-circuit of G is a cycle of G 8 H, and each cycle of G 8 His an anticircuit of G. 10. Show that if (E, F)and (E, 9’) are two rnatroids on E whose ranks are respectively r and r’. then max I S I = [r(A) r’(E - A)] . { ulr ua,

ss9n-F’

y2

+

(This is the “Matroid Intersection Theorem” of J. Edmonds).

References

CHAPTER 2 : Cyclomatic Number G. DEMOUCRON, Y. MALGRANGE, R. PERTUISET, Graphes planaires. Revue FranGuise de Rech. Opkrat., 30, 1964, 33-47. T. GALLAI,u b e r regulare Kettengruppen. Acta Math. Ac. Sc. Nungar., 10, 1959, 227-240. A. GHOUILA-HOURI, Flots et tensions duns un graphe, Thesis, UniversitC de Paris, 1964. K. KURATOWSKI, Sur le problkme des courbes gauches en topologie. Fund. Math., 15, 1930, 271-283. A. LEMPEL,S . EVEN,I. CEDERBAUM, “An Algorithm for Planarity Testing of Graphs”. ThCorie des Graphes. Rome I. C. C. (P.Rosenstiehl, ed.), Dunod Paris, 1967, 215-232. S. MACLANE,A Combinatorial Condition for Planar Graphs. Fund. Math. 28, 1937, 22-32. G. J . MINTY,Monotone Networks. Proc. Roy. SOC., A 257, 1960, 194-208. R. PELLET,Initiation a la thCorie des graphes, vocabulaire descriptif. Entreprise Moderne $Edition, Paris, 1966. H . SAcns, Einfuhrung in die Theorie der endlichen Craphen, Teubner, Leipzig, 1970. E. STEINITZ, Polyeder und Raumeinteilungen. Enzykl. Math. Wiss., 3, 1922, 1-139. W. T. TUTTE,A Class of Abelian Groups. Canad. J. Math., 8, 1956, 13-28. W. T. TUTTE, How to Draw a Graph. Pror. London Math. SOC.,13, 1963,743-767. H . WHITNEY,Planar Graphs. Fund. Math., 21, 1933, 73-84. A. A. ZYKOV,Theory of Finite Graphs (Russian), Nauka, Novosibirsk, 1969.

CHAPTER 3 : Trees and Arborescences AARDENNE-EHRENFEST, N. G. DE BRUIJN,Circuits and Trees in Oriented Linear Graphs. Simon Steuin, 28, 1951, 203-217. C. BERGE,Principles of’ Conibinatorics, Academic Press, New York, London, 1971. ‘‘ Matrices and Trees”. Economic Activity Analysis, R. BOTT,J. P. MAYBERRY, Wiley, New York, 1954, 391-407. A. CAYLEY, A Theorem on Trees. Quart. J. Math. 23, 1889, 376-378. P. CAMION, Modules unimodulaires. J. Cornbinat. Theory, 4, 1968, 301-362. G. CHATY,Graphes fortement connexes c-minimaux. C. R. Ac. Sc. Paris, 266, 1968, 907-909. T.

VAN

498

REFERENCES

499

G. CHATY,Graphes fortement connexes c-minimaux et graphes sans circuit co-minimaux. J . Combinat. Theory, 10, B, 1971, 237-244. G. CHOQUET, Etude de certains rCseaux de route. C . R. Ac. Sc. Paris, 206, 1938, 310-311. L. E. CLARKE, On Cayley’s Formula for Counting Trees. Proc. Cambridge Phil. SOC.,59, 1963, 509-517. J. P. CRESTIN,Un algorithme d’orientation des ar2tes d’un graphe connexe fini. MimCographe du SCminaire sur les Probltmes Combinatoires, I. H. P., 1969. p. J. KELLY,A Congruence Theorem for Trees. Pacific J . of Math. 7, 1957, 961968. J. B. KRUSKAL, On the Shortest Spanning Subtree of a Graph. Proc. Am. Math. SOC.,71, 1956, 48-50. B. LECLERC, P. ROSENSTIEHL, “ Arbres”. Structures algkbriques et combinatoires, chapter 1, Unesco 1970. V. V. MENON,On the Existence of Trees with Given Degrees. Sankhya, 26, 1964, 63-68. J. W. MOON, “Enumerating Labelled Trees”. Graph Theory and Theoretical Physics, (F. Harary) Academic Press, London, New York, 1967, 261-271. J. W. MOON,On the Second Moment of the Complexity of a Graph. Mathematika, 11, 1964, 95-98. L. NESBESK~, Algebraic Properties of Trees. Acta Uniu. Carolinae Philologica Monograph, 25, 1969. P. V. O’NEIL,The Number of Trees in a Certain Network. Notices Amer. Math. SOC., 10, 1963, 569. C. F. PICARD, Graphes et questionnaires. Gauthier-Villars, Paris, 1972. C. DE POLIGNAC, Theorie des ramifications. Bull. SOC.Marl?. France, 8, 1880, 120. C. RAMANUJACHARYULU, Trees and Tree-equivalent Graphs. Canad. J. Math. 17, 1965,731-733. A. RENYJ,Some Remarks on the Theory of Trees. Magy. Tudom. Akad. Mat. Kut. Intkz. Kolz., 4, 1959, 73-85. P. ROSENSTIEHL, ‘‘L‘arbre minimum d’un graphe”. Thkorie des Craphes, Rome 1. C. C. (P. Rosenstiehl, ed.), Dunod, Paris, 1967, 357-368. P. ROSENSTIEHL, Graph Problems Solved by Finite Automate Network, Abstract, Calgary International Conf. 1969. B. ROY, Cheminement et connexitk duns les graphes. Thesis, Universitd de Paris, 1962. H.I. SCOINS,The Number of Trees with Nodes of Alternate Parity. Proc. Camb. Phil. SOC.,58, 1962, 12-16. J. SEDLACEK, Regular Graphs and their Spanning Trees. Res. paper 97, Calgary, 1970. H. N. V. TEMPERLEY, On the Mutual Cancellation of Cluster Integrals. Proc. Phys. SOC., 83, 1964, 3-16. W. T. TUTTE,The Dissection of Equilateral Triangles into Equilateral Triangles, Proc. Cambridge Phil. SOC.,44, 1948, 203-217. H. M. TRENT,A Note on the Enumeration and Listing of All Possible Trees in a Connected Linear Graph. Proc. Nut. Ac. Sciences, 40, 1954, 1004-1007. L. WEINBERG, Number of Trees in a Graph. Proc. IRE, 46, 1958, 1954-1955.

500

REFERENCES

CHAPTER 4 : Path Problems S. B. AKERS,On the Construction of (d, k)-graphs. IE E E Transactions on Electronic Computers, 19, 1965, 488. M. BALINSKI,On the Structure of Convex Polyhedra. Pac. J. of Math., 11, 1961, 431-434. M. BECKMANN, C. B. MCGUIRE,C. B. WINSTEN,Studies in the Economics of Transportation. Yale University Press. New Haven, 1956. R. BELLMAN, The Theory of Dynamic Programming. Bull. A . M. S., 60,1954,503. B. BOLLOBAS, “Graphs with Given Diameter”. Theory of Graphs (Tihany), Acad. Press 1968, 29-36. M. BORILLO,Quelques remarques sur la determination de I’algorithme optimal pour la recherche du plus court chemin. Revue FranGaise de Rech. OpPrat., 33, 1964, 385-388. J. BOSAK,A. KOTZIG,3. ZNAM.“ O n some Metric Problems in Graph Theory”, in Beitruge zur Graphentheorie (Sachs, Voss, Walther, ed.), Leipzig, 1968, 33-39. J. BoshK, A. KOTZIG,3. ZNAM,Strongly Geodetic Graphs. J. of Combinat. Theory, 5, 1968, 170-176. D. BRATTON,Efficient Communication Networks. Cowles Comm. Disc. Paper, 2119, 1955. G. B. DANTZIG,On the Shortest Route Through a Network. Manag. Sc., 6, 1960, 187-189. J. C. DERNIAME, C. PAIR,Problemes de cheminement duns les graphes, Monographie @),Dunod, 1971. E. W. DIJKSTRA, A Note on Two Problems in Connection with Graphs. Numerische Muthematik, 1, 1959, 269-271. S. E. DREYFUS, An Appraisal of Some Shortest-path algorithms. Opens. Res., 17, 1969, 395-412. P. ERDOS,H. SACHS,Regulare Graphen gegebener Taillenweite mit minimaler Knotenzahl, Wiss. Z . Uniii. Halle. Math., 12, 1963, 251-258. A. GHOUILA-HOURI, Diamktre maximal d’un graphe fortement connexe. C. R. Ac. Sc. Paris 250, 1960, 4 254-4 256. M. K. GOLDBERG, The Radius of a Graph (Russian). Uspehi Mat. Nairk 20, 1965, no 5. M. K. GOLDBERG, The Diameter of a Strongly Connected Graph (Russian). Doklady, 170, 4, 1966. B. GRUNBAUM, Convex Polytopes. Wiley, New York, 1967. C. HEUCHENNE, Sur une certaine correspondance entre graphes. Bull. SOC.Roy. Sc. Li2ge, 33, 1964, 743-753. A. HOFFMAN, R. SINGLETON, On Moore Graphs with Diameters 2 and 3. I. B. M. J . of Res., 4, 1960, 5, 497-504. C. JORDAN,Sur les assemblages de lignes. J. Reine Angew. Math. 70, 1869, 185190. A. KAUFMAN and Y . MALGRANGE, Recherche des chemins et circuits hamiltoniens d’un graphe. Revue Frangaise de Rech. OpPrat., 26, 1963, 61-73. V. KLEE,Diameters of Polyhedral Graphs. Canad. J. Math., 16, 1964, 602-614. E. LUCAS,RPcrPations mathPmatiques, 1, Blanchard, Paris, 1921. K. MAGHOUT,Thesis, University of Paris, 1962.

REFERENCES

501

U. S. R. MURTY,On Some Extremal Graphs. Acta Math. Acad. Sc. Hungaricae, 19, 1968, 69-74. U. S. R. MURTY,Extremal Non-separable Graphs of Diameter Two. Uniu. of Waterloo mimeograph, 1968. U. S. R. MURTY,K. VIJAYAN, On Accessibility in Graphs. Sankhya, A 26, 1965, 270-302. C. PAIR, “Sur des algorithmes pour les probl6mes de cheminements dans les graphes finis”. ThPorie des Graphes, Rome I. C. C. (P. Rosenstiehl, ed.), Dunod, Paris, 1967, 271-300. R. RADNER,A. TRITTER,Communication in Networks. Cowles Comm. paper, 2098. 1954. P. ROSENSTIEHL, Existence d‘automates finis capables de s’accorder bien qu’arbitrairement connectks et nombreux. I. C. C. Bulletin, 1966, 5 , 245-261. B. ROY,Contribution de la Theorie des Graphes a l’btude de certains problkmes lineaires. C . R. Acad. Sc. Paris, 248, 1959, 2437. B. ROY, Cheminement et connexitt duns les graphes. Thesis, University of Paris, 1962. B. ROY, Algebre moderne et Thkorie des Graphes. Dunod, Paris, 1969. H. SACHS,Uber selbstkomplementare Graphen, Publ. Math. Debrecen, 9, 1962, 21C288. G.TARRY,Le probleme des labyrinthes. Nouvelles Annales de Math., 14, 1895, 187-189. V. G.VIZING,The Number of Edges in a Graph with Given Radius (Russian). Doklady, 173, 1967, no 6. ZNAM,Decomposition of the Complete Directed Graph into Factors with Given Diameters. Abstract Calgary Conference, 1969.

s.

CHAPTER 5 : Flow Problems C. BERGE,Sur la dbficience d’un rbseau infini. C. R. Acad. Sciences, 245, 1957, 1206-1 207. C. BERGE,Sur l’bquivalence du probltme du transport gknCralisC et du probleme des reseaux. C. R . Ac. Sciences, 251, 1960, 324-325. C. BERGE,Les problemes de flot et de tension. Cahiers du Centre d’Etudes de Rech. OpPrat., Bruxelles, 3, 1961, 69-83. C . BERGEand A. GHOUILA-HOURI, Programmes, Jeux et RPseaux de Transport Dunod, Paris, 1962. (English translation: Methuen, London; Wiley, New York, 1965.German translation : Teubner, Leipzig, 1967.-Spanish translation: Compania ed. Continental, Mexico, 1965.) R. FAURE, Eltments de Recherche Opkrationnelle. Gauthier-Villars; Paris, 1968. L. R. FORD and D. R. FULKERSON, Maximal Flow Through a Network. Canadian J. of Math., 8, 1956, 399. L. R. FORD and D. R. FULKERSON, Dynamic Network Flow. Rand Corp. Paper, P967, 1956. L. R. FORDand D. R. FULKERSON, A Network Flow Feasibility Theorem and Combinatorial Applications. Canad. J. Math., 11, 1959, 440-450. L. R. FORD and D. R. FULKERSON, Flows in Networks. Princeton Press, Princeton, 1962 (French translation by J. C. Arinal, Gauthier-Villars, Paris, 1967).

502

REFERENCES

D. GALE,A Theorem on Flows in Networks. Pacific J. Math., 7 , 1957, 1073. G. DE GHELLINCK, Aspects de la notion de dualite en ThCorie des Graphes. Cahiers du Centre d’Etudes de Rech. OpCrat., Bruxelles, 3, 1961, 94-102. A. GHOUILA-HOURI, Sur l’existence d’un flot ou d’une tension prenant ses valeurs dans un groupe abklien. C . R. Ac. Sciences, 250, 1960, 3931-3932. A. GHOUILA-HOURI, “Recherche du flot maximum dans certains riseaux lorsqu’on impose une condition de bouclage”. Proc. Secondlnt. Conf. Op. Res.. London, 1960, 156-157. A. GHOUILA-HOURI, Flots el tensions dans i m graphe. Ann. Ec. Normale 81, 1964, 267-339. A. J. HOFFMAN (in C. BERGE,ThPorie des Graphes, Paris, 1958, p. 80). J. C. HERZ,Cours (mimeo), Lille, 1967. C. RAMANUJACHARYULU, Transport Networks with Pairs of Homologous Arcs. Cahiers Centre I?t. Rech. Operat. Bruxelles, 5, 1963, 203-210. C. RAMANUJACHARYULU, Trees and Tree Equivalent Graphs. Canad. J. Math., 170, 1965, 731-733. B. ROY,Thesis, Metra. 1962. B. ROY,Alg2bre moderne et Thiorie des Graphes, I, Dunod, Paris, 1969; 11, Paris, 1970. P. SLEPIAN, Mathematical Foundations of Network Analysis. Springer Tracts, No. 16, Springer Verlag, Berlin, Heidelberg, New York, 1968. L. WEINBERG, A Survey of Linear Graphs; Fundamentals and Applications to Network Theory. Matrix Tensor Quarterly, 14, 1963-1964, 103-1 15. CHAPTER 6 : Degrees and Demidegrees C. BERGE,Espaces topofogiques, p. 193. Dunod, Paris, 1966 (p. 184, English translation, Oliver and Boyd, 1963). L. W. BEINEKE, F. HARARY, Local Restrictions for Various Classes of Directed Graphs. J. London Math. SOC.,40, 1965, 87-95. Jack EDMONDS, Existence of k-edge Connected Ordinary Graphs with Prescribed Degrees. J. Res. Nut. Bur. Standards, B 68, 1964, 73-74. P. ERDOS, T. GALLAI,Graphen mit Punkten vorgeschriebenen Grades. Mat. Lapok, 11, 1960, 264-274. D. R. FULKERSON, Zero-one Matrices with Zero Trace. Pacific J. Math., 10, 1960, 831-836. D. R. FULKERSON, A. J. HOFFMAN, M. H. MCANDREW, Some Properties of Graphs with Multiple Edges. Canad. J. Math., 17, 1965, 166-177. D. GALE,A Theorem on Flows in Networks. Pacific J. Math., 7 , 1957, 1073-1082. S. L. HAKIMI, On Realizability of a Set of Integers as Degrees of the Vertices of a Linear Graph, I. J. ofthe SIAM, 10, 1962, 496-506. G. H. HARDY, J. E. LITTLEWOOD, G. POLYA, Inequalities. Cambridge, 1952. H. G. LANDAU, On Dominance Relations. Bull. Math. Biophys. 15, 1953, 143-148. V. V. MENON,On the Existence of a Tree with Given Degrees. Sankhya, 26, 1964, 63-68. J. W. MOON,An Extension of Landau’s Theorem on Tournaments. Pac. J. Math., 13, 1963, 1343-1345. R. F. MUIRHEAD, Some Methods Applicable to Identities and Inequalities of

REFERENCES

503

Symmetric Algebraic Functions of n Letters. Proc. Edinburgh Math. SOC., 21, 1903, 144-150. A. RAMACHANDRA RAO, Some Extremal Problems and Characterizations in the Theory of Graphs. Thesis, Indian Stat. Inst., Calcutta, 1969. S . B. RAO, A. RAMACHANDRA RAO, Existence of 3-connected Graphs with Prescribed Degrees. Pac. J . Math., 33, 1970, 203-207. H. J. RYSER,Combinatorial Properties of Matrices of Zeros and Ones. Canad. J. Math., 9, 1957, 371-377. H. J. RYSER,Matrices of Zeros and Ones. Bull. A. M . S., 66, 1960, 442464. CHAPTER 7 : Matchings

M. BALINSKI, On Maximum Matching, Minimum Coverings and Their Connections. J . Combinat. Theory (to appear). C. BERGE,Two Theorems in Graph Theory. Proc. Nut. Ac. Sciences, U. S . A., 43, 1957, 842. C . BERGE,Sur le couplage maximum d’un graphe. C. R. Acad. Sciences, 247, 1958, 258-259. C. BERGE,“Alternating chains”, in: Graph Theory and Computing (R. C. Read) Acad. Press, 1972. A. L. DULMAGE and N. S . MENDELSOHN, Some Graphical Properties of Matrices with Non-negatiue Entries. Mimeo. Univ. Alberta, 1961. Jack EDMONDS, Covers and Packings in a Family of Sets. BciII. A . M . S., 68, 1962, 494499. Jack EDMONDS, Paths, Trees and Flowers. Canad. J . Math., 17, 1965, 449-467. P. ERDOS,T. GALLAI,On Maximal Paths and Circuits of Graphs, Acta Math. Acad. Sc. Hungar., 10, 1959, 337-356. P. ERDOS, L. PosA, On the Maximal Number of Disjoint Circuits in a Graph. Publ. Math. Debrecen, 9, 1962, 3-12. P. ERDOS,T. GALLAI, On the Minimal Number of Vertices Representing the Edges of a Graph. Piibl. Math. Mag. Tud. Ak., 6, 1961, 181-203. M. E. FISHER, Statistical Mechanics of Dimers on a Plane Lattice. Phys. R., 124, 1961, 278-286. J. FOLKMAN, D. R. FULKERSON, “Edge Colouring in Bipartite Graphs”. Conibinatorial Mathematics and their Applications (Bose, Dowling ed.), Univ. of N.C. Press, Chapel Hill, 1969. R. L. GRAHAM, L. H. HARPER, Some Results on Matching in Bipartite Graphs, SIAM., J . 17, 1969, 1017-1022. R. P. GUPTA,Studies in the Theory of Graphs. Tata Institute of Fund. Res., Bombay, 1967, Thesis, Calcutta, 1967. R. P. GUPTA,“ A Decomposition Theorem for Bipartite Graphs”. Th&orie des Graphes Rome I. C. C., (P. Rosenstiehl, ed.), Dunod, Paris, 1967. M. HALL,Distinct Representatives of Subsets. Bull. Am. Math. SOC.,54, 1948, 922-926. P. HALL,On Representations of Subsets. J. London Math. Soc., 10, 1934, 26-30. P. R. HALMOS and H. E. VAUGHAN, The Marriage Problem. Am. J . of Math., 72, 1950, 214-215. A. J. HOFFMAN, H. W. KUHN,Systems of Distinct Representatives and Linear Programming. Am. Math. Monthly, 63, 1956, 455466.

504

REFERENCES

P. W. KASTELEYN, The Statistics of Dimers on a Lattice. I The Number of Dimer Arrangements on a Quadratic Lattice. Physica, 27, 1961, 1209-1225. P. W. KASTELEYN, “Graph Theory and Crystal Physics”. Graph Theory and Theoretical Physics

Graphs and hypergraphs (North-Holland mathematical library)

Graphs and Hypergraphs

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