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I NTRODUCTION TO GRAPH

AND HYPERGRAPH THEORY

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I NTRODUCTION TO GRAPH

AND HYPERGRAPH THEORY

V ITALY I. V OLOSHIN

Nova Science Publishers, Inc.New York

c© 2009 by Nova Science Publishers, Inc.

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Library of Congress Cataloging-in-Publication DataVoloshin, Vitaly I. (Vitaly Ivanovich), 1954-

Introduction to graph and hypergraph theory / Vitaly I. Voloshin.p. cm.

Includes index.ISBN978-1-61470-112-5 (eBook)

1. Graph theory. 2. Hypergraphs. I. Title.QA166.V649 2009511’.5–dc22

2008047206

Published by Nova Science Publishers, Inc.✜ New York

To Julian, Olesea and Georgeta for unlimited love and support

The Essence of Mathematics is in its generalizations,

The Beauty of Mathematics is in its ideas,

The Power of Mathematics is in its absolute truth...

Contents

Preface xi

I G RAPHS 1

1 Basic Definitions and Concepts 51.1. Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2. Graph Modeling Applications . . . . . . . . . . . . . . . . . . . . . .. . 81.3. Graph Representations . . . . . . . . . . . . . . . . . . . . . . . . . . .. 121.4. Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .151.5. Basic Graph Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.6. Basic Graph Operations . . . . . . . . . . . . . . . . . . . . . . . . . . .. 251.7. Basic Subgraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.8. Separation and Connectivity . . . . . . . . . . . . . . . . . . . . . .. . . 34

2 Trees and Bipartite Graphs 392.1. Trees and Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.2. Trees and Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3. Minimum Spanning Tree . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.4. Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45

3 Chordal Graphs 513.1. Preliminary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.2. Separators and Simplicial Vertices . . . . . . . . . . . . . . . .. . . . . . 523.3. Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.4. Distances in Chordal Graphs . . . . . . . . . . . . . . . . . . . . . . .. . 593.5. Quasi-triangulated Graphs . . . . . . . . . . . . . . . . . . . . . . .. . . 62

4 Planar Graphs 674.1. Plane and Planar Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . .674.2. Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.3. K5 andK3,3 Are not Planar Graphs . . . . . . . . . . . . . . . . . . . . . . 714.4. Kuratowski’s Theorem and Planarity Testing . . . . . . . . .. . . . . . . . 734.5. Plane Triangulations and Dual Graphs . . . . . . . . . . . . . . .. . . . . 76

vii

viii Contents

5 Graph Coloring 795.1. Preliminary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.2. Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . .. 805.3. Structure of Colorings . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 835.4. Chromatic Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . .895.5. Coloring Chordal Graphs . . . . . . . . . . . . . . . . . . . . . . . . . .. 955.6. Coloring Planar Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1025.7. Perfect Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1085.8. Edge Coloring and Vizing’s Theorem . . . . . . . . . . . . . . . . .. . . 1125.9. Upper Chromatic Index . . . . . . . . . . . . . . . . . . . . . . . . . . . .116

6 Traversals and Flows 1236.1. Eulerian Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1236.2. Hamiltonian Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1256.3. Network Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

II H YPERGRAPHS 131

7 Basic Hypergraph Concepts 1357.1. Preliminary Definitions . . . . . . . . . . . . . . . . . . . . . . . . . .. . 1357.2. Incidence and Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1397.3. Basic Hypergraph Classes . . . . . . . . . . . . . . . . . . . . . . . . .. 1447.4. Basic Hypergraph Operations . . . . . . . . . . . . . . . . . . . . . .. . . 1467.5. Subhypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1517.6. Conformality and Helly Property . . . . . . . . . . . . . . . . . . .. . . . 154

8 Hypertrees and Chordal Hypergraphs 1618.1. Hypertrees and Chordal Conformal Hypergraphs . . . . . . .. . . . . . . 1618.2. Algorithms on Hypertrees . . . . . . . . . . . . . . . . . . . . . . . . .. . 1688.3. Cyclomatic Number of a Hypergraph . . . . . . . . . . . . . . . . . .. . . 174

9 Some Other Remarkable Hypergraph Classes 1819.1. Balanced Hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . .1819.2. Interval Hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1839.3. Normal Hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1859.4. Planar Hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .187

10 Hypergraph Coloring 19310.1. Basic Kinds of Classic Hypergraph Coloring . . . . . . . . .. . . . . . . . 19310.2. Greedy Algorithm for the Lower Chromatic Number . . . . .. . . . . . . 19710.3. Basic Definitions of Mixed Hypergraph Coloring . . . . . .. . . . . . . . 20110.4. Greedy Algorithm for the Upper Chromatic Number . . . . .. . . . . . . 20710.5. Splitting-Contraction Algorithm . . . . . . . . . . . . . . . .. . . . . . . 21310.6. Uncolorability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 21910.7. Unique Colorability . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 227

Contents ix

10.8. Perfection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23610.9. Chromatic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24410.10.Coloring Planar Hypergraphs . . . . . . . . . . . . . . . . . . . .. . . . . 254

11 Modeling with Hypergraphs 26311.1. List Colorings without Lists . . . . . . . . . . . . . . . . . . . . .. . . . 26311.2. Resource Allocation . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 264

12 Appendix 26712.1. What Is Mathematical Induction . . . . . . . . . . . . . . . . . . .. . . . 26712.2. Graph Theory Algorithms and Their Complexity . . . . . . .. . . . . . . 26912.3. Answers and Hints to Selected Exercises . . . . . . . . . . . .. . . . . . . 27012.4. Glossary of Additional Concepts . . . . . . . . . . . . . . . . . .. . . . . 275

References 279

Index 281

Preface

Graph Theory is an important area of contemporary mathematics with many applicationsin computer science, genetics, chemistry, engineering, industry, business and in social sci-ences. It is a young science invented and developing for solving challenging problems of“computerized” society for which traditional areas of mathematics such as algebra or cal-culus are powerless.

This book is for math and computer science majors, for students and representativesof many other disciplines (like bioinformatics, for example) taking the courses in graphtheory, discrete mathematics, data structures, algorithms. It is also for anyone who wants tounderstand the basics of graph theory, or just is curious. Noprevious knowledge in graphtheory or any other significant mathematics is required. Thevery basic facts from set theory,proof techniques and algorithms are sufficient to understand it; but even those are explainedin the text.

Structurally, the text is divided into two parts where Part II is the generalization of PartI. The first part discusses the key concepts of graph theory with emphasis on trees, bipartitegraphs, cycles, chordal graphs, planar graphs and graph coloring. The second part consid-ers generalizations of Part I and discusses hypertrees, bipartite hypergraphs, hypercycles,chordal hypergraphs, planar hypergraphs and hypergraph coloring. There is an interactionbetween the parts and within the parts to show how ideas of generalizations work. Themain point is to exhibit the ways of generalizations and interactions of mathematical con-cepts from the very simple to the most advanced.

The reader is conducted from the simplest examples, definitions and concepts step bystep towards an understanding of a few most fundamental facts in the field. When writing Ipursued the following goals:

• to make it as readable as possible;

• to choose the most instructive (not complex!) theorems and algorithms;

• to exhibit sequential generalization of concepts and ideas;

• to show an interaction between the sections and chapters forthe sake of integrity;

• clearly expose the essence and core of graph and hypergraph theory, including hyper-graph duality and hypergraph coloring;

• in Part I, to prepare the reader for understanding Part II;

• in Part II, to use the knowledge from Part I.

xii Vitaly I. Voloshin

Hypergraphs model practical situations in different sciences in a much more generalsetting than graphs do. In addition, they help to find optimalsolutions for many new opti-mization problems. While vertices represent the elements of a set (as in graphs), the hyper-edges represent subsets of any cardinality (not just 2 as in graphs), or, even more generally,arbitrary statements about arbitrary subsets.

One of the features of this text is theduality of hypergraphs. There are the only twoplayers in graph theory: vertices and edges. In dual hypergraphs, they just swap the roles.This fundamental concept is missing in graph theory (and in its introductory teaching) be-cause dual graphs are not properly graphs, they generally represent hypergraphs. However,as Part II shows, the duality is a very powerful tool in understanding, simplifying and uni-fying many combinatorial relations; it is basically a look at the same structure from theopposite (vertices versus edges) point of view.Teaching and applying graph theory withouthypergraphsdoes not allow to use duality; it is like teaching graphs without their comple-ments. Among the goals of the text, one is to fill up this gap.

Part I may be used on undergraduate level for one semester introductory course, PartII may be used as a text or supplement for senior and graduate students. Some chaptersor sections from Part II may be used on undergraduate level for most advanced students asprojects in undergraduate research to report on departmental seminars. The book includesmany examples, figures and algorithms; each section ends with a set of exercises and a setof computer projects. The answers and hints to selected exercises are provided at the endof the book. The material has been tested in class during morethan 20 years of teachingexperience of the author. Math majors will pay more attention to theorems and proofs,computer science majors will work more with the concepts, algorithms and computations,and representatives of other sciences will find models and ideas for solutions of optimizationproblems in their fields.

On the contents,four core areas of graph theory have been chosen: bipartite graphs,chordal graphs, planar graphs and graph coloring. The text exhibits the survey of basic re-sults and their generalizations to hypergraphs in these areas. Bipartite graphs, planar graphsand graph coloring were the source, the origin of graph theory. Chordal graphs, discoveredmuch later, have a very special place in the entire theory: itis the best playground forintroduction to graphs and hypergraphs. The fact is that many unrelated (!) fundamentalparameters introduced for general graphs (like, for example, related to degrees, or com-plements, or colorings), achieve their optimal values on chordal graphs. There are manyrelations of chordal graphs to trees, but only the language of hypergraphs allows to showthat chordal conformal hypergraphs are dual to hypertrees.This is usually very impressiveand unexpected to the reader since it is sufficient simply to transpose the incidence matrixof a hypertree to obtain a chordal hypergraph. It explicitlyshows the strength of hypergraphtheory.

At last graph coloring, generalized to hypergraphs, allowsto consider the colorability,upper chromatic number, hypergraph perfection, the gaps inthe chromatic spectrum, etc.Such concepts grew up from graph coloring and essentially represent thegraph coloringunfolding. Several basic results from mixed hypergraph coloring, taken, adapted and up-dated from research monograph [6], will lead to unforeseen discoveries in Chapter 10; theydemonstrate the power of generalizations. All this reflectsthe fact that for the last twodecades, a significant number of new fundamental ideas, results and publications have led

Preface xiii

to the situation where hypergraph theory in general, and hypergraph coloring in particular,are taking a new shape. The theory has a great future since it continues to generate newresearch problems that never arose before.

The bookfills up the gapin educational materials on graphsand hypergraphs; itcon-tributes to diversityof books and textbooks in the field; it isto understand the basic ideasof Graph and Hypergraph Theory.That is why many more special and advanced topics likeDirected Graph Theory, Extremal Graph theory, TopologicalGraph Theory, Ramsey The-ory, Random Graphs, Sperner Theory, Block Designs and many other topics (which usuallyhave entirely devoted books) are not covered. The reader, however, will be well prepared tounderstand and to begin to work in any of these directions.

There are several pedagogical methods consistently used throughout the text:

• when formulating a new definition or concept, a formulation and examples of thenegation often follow;

• when formulating a new theorem, the cases and examples when it does not work oftenfollow;

• the names are given to many special graphs and hypergraphs; they are used insteadof drawings thereafter;

• the same examples of graphs and hypergraphs are used for manydifferent compu-tational problems; as the opposite, the same problems and algorithms are used fordifferent examples of graphs and hypergraphs;

• structurization of more complex proofs is made in order to ease the understanding ofa few basic steps;

• detailed proofs of some long theorems are omitted and only ideas or sketch of theproofs are provided;

• contradictory facts or statements are used to call a surprise and make the readingsimply interesting;

• an idea is explained first and then the details follow;

• since the comparison is crucial for understanding, opposite versions of some conceptsare provided and the respective graphs, hypergraphs and algorithms are compared;

• since the visualization is the feature of graph theory, “look ⇒ read⇒ look” - rule isimplicitly applied;

• the main goal of exercises is to test understanding of concepts and theorems, and themain goal of computer projects is to train in programming forscientific computationsin graph and hypergraph theory;

• bold type is used for definitions and paragraph headings,italic type is used to call aspecial attention, and symbol� is used to indicate the end of proofs.

xiv Vitaly I. Voloshin

After all, the ultimate goal of the book is to popularize graphs and hypergraphs.

Acknowledgements: I am grateful to Troy University for repeated support of this project.I also thank the students of Troy University who took the course Introduction to GraphTheory and helped in polishing the text.

Vitaly I. Voloshin

Troy UniversityTroy, [email protected]

November 18, 2008

Part I

GRAPHS

1

“...Graph Theory begins with “Graph”

and

ends with “Theory”...

Chapter 1

Basic Definitions and Concepts

“Pictures speak louder than words...”

“– Give me the definition first...”

1.1. Fundamentals

b

b b

bb

Figure 1.1. This is a graph.

An example of a graph is shown in Figure 1.1. The most simple and least strictdef-inition of a graph is the following: a graph is a set of points and lines connecting somepairs of the points. Mathematicians name and number everything: in graph theory, pointsare calledvertices, and lines are callededges. So, the graph in Figure 1.1 consists of fivevertices and seven edges.

Throughout the book, we use the standard notation: upper case lettersA,B, . . . ,X, Y,Z for sets (all sets are finite), lower case lettersa,b, . . . ,e, . . . ,x,y,z for the elements ofa set and curly braces{,} for listing the elements of a set. It is convenient to assign indicesif we have many elements of the same type. A finite set is a list of its elements; no element

6 Vitaly I. Voloshin

b

b b

bb

x1

x2

x3x4

x5

e1

e2

e3

e4

e5

e6

e7

Figure 1.2. GraphG = (X,E).

is repeated and the order of elements in the list does not matter (we read the lists from leftto right). The number of elements in a setA is denoted by|A|, and the empty set is denotedby /0. If a set contains other sets as elements, then it is called afamily ; in a family, elementsmay be repeated but order still does not matter.

In a graph, the set of vertices is denoted byX and is written asX={x1, x2, ..., xn} wherexi is thei-th vertex andn is the number of vertices. The set of edges is denoted byE and iswritten asE = {e1,e2, . . . ,em} whereei is thei-th edge andm is the number of edges.Eachedge ei is identified by the pair of respective vertices which are connected by ei . It remainsto “invent” the last letter to denote the entire graph:G. Now we are ready to present theformal definition of a graph:

Definition 1.1.1 A graph G is a set X of vertices together with a set E of edges. It iswritten as

G = (X,E).

Figure 1.2 presents the same graph shown in Figure 1.1 using the definition and agree-ments above. It hasn = 5 vertices andm = 7 edges. We writeG = (X,E) whereX = {x1,x2,x3,x4,x5} andE = {e1,e2,e3,e4,e5, e6,e7}. Since each edge is a pair of ver-tices we writee1 = {x1,x2}, e2 = {x2,x3}, e3 = {x3,x4}, e4 = {x4,x5}, e5 = {x5,x1},e6 = {x2,x5}, e7 = {x2,x4}, and thereforeE = {{x1,x2},{x2,x3}, {x3,x4},{x4,x5}, {x5,x1},{x2,x5},{x2,x4}}. SinceE is a set of sets it is a family.

If two vertices are connected by an edge, then they are calledadjacent, otherwise theyare calleddisjoint. For example, verticesx1 andx2 are adjacent, but verticesx1 andx3 aredisjoint. For a given vertexx, the number of all vertices adjacent to it is calleddegreeof thevertexx, denoted byd(x). In our example,d(x1) = 2, d(x2) = 4, and so on. The maximumdegree over all vertices is called themaximum degree ofG, denoted by∆(G). For graphG, see Figure 1.2,∆(G) = d(x2) = 4.

Basic Definitions and Concepts 7

The adjacent vertices are sometimes calledneighborsof each other, and all the neigh-bors of a given vertexx are called theneighborhood of x. The neighborhood ofx is de-noted byN(x). In our graph, for example,N(x1) = {x2,x5}. Evidently, the degree of avertex is the cardinality (the number of elements) of its neighborhood:d(x1) = |N(x1)|= 2,d(x2) = |N(x2)| = 4, and so on.

For a graphG, if we count the degree of each vertex and arrange these degrees in nondecreasing order, then we obtain a sequence called thedegree sequenceof G. The degreesequence forG, see Figure 1.2, is: (2,2,3,3,4).

Two edges are said to beadjacent if they have a vertex in common anddisjoint other-wise. In Figure 1.2 edgese1 ande2 are adjacent, and edgese1 ande3 are disjoint. If a vertexx belongs to an edgee, then we say that they areincident to each other. In the exampleabove, edgee1 is incident to vertexx1 and is not incident to vertexx5 and so on. As one cansee, adjacency is referred to the elements of the same type and incidence is referred to theelements of different types.

Sometimes the vertex set of a graphG is denoted byV(G) and the edge set byE(G).So, generally any graphG = (V(G),E(G)). The number of vertices is usually denotedby n = n(G), and the number of edges bym = m(G). The set of edges containing avertex x is denoted byE(x). In our example,V(G) = X = {x1,x2,x3,x4,x5}, |X| = n = 5,E(G) = E = {{x1,x2},{x2,x3},{x3,x4}, {x4,x5}, {x5,x1}, {x2,x5},{x2,x4}}, |E|= m= 7,E(x1) = {e1,e5}, and so on.

Proposition 1.1.1 (Degree equality)For any graph G= (X,E), the following equalityholds:

Σni=1d(xi) = 2m. (1.1)

Proof. Indeed, if we sum the degrees of all the vertices then each edge is counted twicebecause it has two ends. �

Applying formula (1.1) to graphG, see Figure 1.2, gives:

2+4+2+3+3= 14= 2×7.

Proposition 1.1.2 In any graph G= (X,E), there are two vertices with the same degree.

Proof. If a graphG hasn vertices, then the degree sequence hasn integer numbers. Forany vertexx, the minimum number of neighbors is 0, and the maximum numberof neigh-bors isn−1, therefore 0≤ d(x) ≤ n−1.

Assume that alln numbers are different. Then they must take all values on the intervalof integers from 0 ton−1. 0 means that there is a vertex with no neighbors, andn−1 meansthat there is a vertex adjacent to all the rest of vertices. This cannot occur simultaneously.Therefore, there must be two vertices with the same degree. �

In graphG, see Figure 1.2, for example,d(x1)=d(x3)=2, andd(x4) = d(x5) = 3.


Exercises 1.1.

b

b

b b

b b b

b

bbb

bb

b

b

b

b

G1 G2 G3

Figure 1.3.

1. For each of the graphs in Figure 1.3, find the number of verticesn and the number of edgesm.

2. For each of the graphs in Figure 1.3, name or number the vertices and for each pair of verticesshow if they are adjacent or not.

3. For each of the graphs in Figure 1.3, name the edges and for each pair of edges show if theyare adjacent or not.

4. In each of the graphs in Figure 1.3, for each pair of vertices and edges show whether they areincident.

5. In each of the graphs in Figure 1.3, find degree and neighborhood of every vertex and degreesequence.

6. Apply degree equality (Proposition 1.1.1) to each of the graphsG1,G2 andG3.

7. In each of the graphs in Figure 1.3, find the vertices of the same degree.

8. For each of the graphs in Figure 1.3, find the maximum degree∆(G).

1.2. Graph Modeling Applications

Consider a simple instructive problem. Suppose we have a chemical plant that producesfive chemical compounds A, B, C, D, and E which must be stored instorage areas. It isknown however, that chemical A combined with chemical B might explode, so they mustnot be stored in the same storage area. The same occurs if chemical A is combined withE, chemical B is combined with C or E, chemical C is combined with D, and chemical Dis combined with E. What is the minimum number of storage areas and how the chemicalcompounds should be stored to avoid any explosion hazards?

Let us “translate” the problem from the wording above into the language of GraphTheory. Denote chemicals A, B, C, D and E respectively by lettersx1, x2, x3, x4 andx5 anddraw five vertices with the namesx1, x2, x3, x4 andx5 in the plane (put them, for example,in some imaginary circle clockwise withx1 on the top). Now draw the edges: read theproblem again and connect by an edge every pair of vertices corresponding to the chemicalswhich are explosive if combined. Thus, since A and B might explode if combined, connectcorresponding verticesx1 andx2 with an edge; in the same way, connectx1 with x5, connect


b

b b

bb

x1

x2

x3x4

x5

storage area 1

storage area 2

storage area 3


x2 with x3 andx5, connectx3 with x4 and connectx4 with x5. The graph obtained is shownin Figure 1.4. It is visually reflecting the relations between the chemicals. Now the problemcan be mathematically formulated in the following way: how can we partition the vertexsetX = {x1,x2,x3,x4,x5} into the smallest number of parts, i.e. the subsets, in such awaythat no subset contains adjacent vertices? Each such subsetwill be considered as the subsetof chemicals that can be safely stored in the same storage area.

Looking at Figure 1.4, we can evidently partitionX into five subsets:{x1}, {x2}, {x3},{x4}, {x5}; this partition is feasible, i.e. good, but not optimal. We can partitionX intofour subsets:{x1,x3}, {x2}, {x4}, {x5}, what is better but still not optimal. At last, wecan partitionX into three subsets:{x1,x4}, {x2}, {x3,x5}, see Figure 1.4 (the partition isshown by dotted closed curves), and show that it is optimal, i. e. minimal. Indeed, visuallyone can see that any partition ofX into two subsets leaves edges in one of the subsets andtherefore is not good. Strict mathematical proof would be the following: consider triangleformed by verticesx1,x2, andx5; any partition of it into two parts (even if one part is empty)leaves an edge in one of the parts. So, two storage areas is nota possible solution, and threestorage areas are sufficient and represent the optimal solution of the problem. Looking atthe picture, the reader can easily find at least one another optimal solution, say{x1,x3},{x2,x4}, and{x5}. How many and which optimal solutions do exist?

We will see in Chapter 5 and other chapters how such and many other problems whichask not only for the optimal but for all possible solutions, can efficiently be solved. In allapplications, like in the example above, visualization is the key feature. Using vertices,edges and their meanings, one apply graphs to depict situations in different sciences. Dif-ferent problems, first formulated in ordinary language, as in the example above, are thentranslated into the language of Graph Theory and solved mathematically. Mathematiciansthen provide algorithms for finding optimal solutions whichare implemented in a software


by computer engineers and passed back to the respective businesses for use in the industry.We next provide a series of situations from different sciences which can successfully

be modeled by graphs.In the next chapters we will consider many mathematical problemswhich have a respective meaning if applied to graphs as models. In each of the examplesbelow, the reader can easily draw a respective graph with a few vertices and edges keepingin mind their meaning.

Mathematics:

• the vertices are natural numbers from 1 to 100; two vertices are adjacent if the re-spective numbers have a common divisor different from 1;

• the vertices are intervals on the real line; two vertices areadjacent if the respectiveintervals intersect;

• the vertices are alln-dimensional vectors with binary coordinates (each componentis either 0 or 1); two vertices are adjacent if the respectivevectors differ in preciselyone component.

Computer science:

• vertices are computers in a network; two vertices are adjacent if the respective com-puters are linked together by telecommunications circuits;

• vertices are processors in parallel architectures; two vertices are adjacent if the re-spective processors have a direct link;

• vertices are files in a data base; two vertices are adjacent ifthe respective files cannotbe opened simultaneously;

• vertices are all web pages in the world; two vertices are adjacent if the respective webpages are connected by any hypertext link;

• vertices are the image fragments in image segmentation in computer vision; two ver-tices are adjacent if the respective fragments are related.

Genetics:

• the vertices are fragments of a DNA sequence; two vertices are adjacent if the respec-tive fragments overlap;

• the vertices are species; two vertices are adjacent if the respective species have acommon hereditary property.

Chemistry:

• the vertices are atoms in a molecule; two vertices are adjacent if the respective atomshave a bond;


• the vertices are chemical compounds used by a chemical factory; two vertices areadjacent if the respective compounds are explosive when combined.

Engineering:

• the vertices are junction points of an electric circuit; twovertices are adjacent if therespective junction points are connected by a wire.

Economics:

• the vertices are all companies in the world; two vertices areadjacent if the respectivecompanies are the suppliers for each other.

Healthcare:

• the vertices are drugs; two vertices are adjacent if the combination of the respectivedrugs is lethal.

Sociology:

• the vertices are employees in a company; two vertices are adjacent if the respectivepeople are in conflict;

• the vertices are the people in a town; two vertices are adjacent if the respective peopleare friends.

Broadcasting:

• the vertices are radio transmitters in a region; two vertices are adjacent if the respec-tive transmitters interfere.

Geographical maps:

• the vertices are cities; two vertices are adjacent if the respective cities are connectedby a highway.

Generally, graphs represent the simplest visual models of systems: any system is a set ofelements together with a set of relations between the elements. In the most general settinghowever, the relations are expressed by statements about any subsets rather than just pairsof elements.

Exercises 1.2.

1. Suppose there are the following intervals on the real line: [0,3], [4,9], [2.7,5], [5,7], [2,4.3],[1,4], [10,11] and[0,12]. Draw a graph where vertices represent the intervals and twoverticesare adjacent if and only if the respective intervals intersect (the “intersection” or “interval”graph).


2. There are following eight possible sequences of length three consisting of 0 and 1: 000, 001,010, 011, 100, 101, 110, 111. Draw a graph where the vertices represent the sequences andtwo vertices are adjacent if and only if the respective sequences differ in precisely one digit.Why is this graph called “cube”?

3. Make the list of all your friends. Draw a graph where the vertices are your friends andtwo vertices are adjacent if and only if the respective friends are friends themselves (the“friendship” graph).

4. For Florida, Alabama, Georgia, Mississippi, South Carolina, Tennessee, Kentucky, Virginiaand California construct a graph where the vertices are these states and two vertices are adja-cent if and only if the respective states have a common border.

5. There are four workers A, B, C and D and five jobs 1, 2, 3, 4, and5. Worker A can do jobs1 and 2, worker B can do jobs 1, 4 and 5, worker C can do jobs 2, 3, and 4, and worker Dcan do job 5. Draw a graph where the vertices are the workers and jobs and two vertices areadjacent if and only if they correspond to a worker and a job that the worker can do.

6. Think about five different ways you can drive from home to the school. Draw a graph wherethe vertices are your home, the school and all street crossings on your way; two vertices areadjacent if and only if the respective crossings, home and the school are consecutive on yourway.

7. There are three houses and three wells. Draw a graph where the vertices are houses and wellsand connect each house with each well by a curve representingan edge. Is it possible to drawthe graph in such a way that the curves do not intersect in the plane at points other than ahouse or a well?

1.3. Graph Representations

For the last 50 years, Computer Science became a major provider of problems to GraphTheory and, simultaneously, it became a major consumer of the solutions to such problems.In practical applications, graphs involved have not five, and even not ten, or twenty, buthundreds and thousands of vertices and edges. It is not possible for human being to solveany problem with such a huge number of elements just using thedescription or even drawinga graph. Computers are used for such tasks, and there are several special ways to storegraphs in computer memory.

Adjacency lists. Let us consider graphG shown in Figure 1.5. For every vertexx, forma list of all of its neighbors. The set of all such lists is called theadjacency list. In G, theneighbors are:

• For x1 : x2,x5,

• For x2 : x1,x3,x4,x5,

• For x3 : x2,x4,

• For x4 : x2,x3,x5,

• For x5 : x1,x2,x4.


b

bb

b bx1

x2

x3

x4x5

e1 e2

e3

e4

e5

e6 e7


The adjacency list, denoted byL(G), is:

L(G) = {{x2,x5},{x1,x3,x4,x5},{x2,x4},

{x2,x3,x5},{x1,x2,x4}}.

Adjacency matrix. It is a matrix (rectangular table from letters and/or numbers) whichhas one row and one column for each vertex. If vertexxi is adjacent to vertexx j , then(i, j)-entry (element at the intersection ofith row and jth column) in the matrix is 1, otherwise itis 0. In fact, adjacency matrix is a square matrix.

For graphG, see Figure 1.5, the adjacency matrix denoted byA(G) is:

A(G) =

0 1 0 0 11 0 1 1 10 1 0 1 00 1 1 0 11 1 0 1 0

.

One can compareA with the adjacency lists: in every row, the 1’s indicate the respectiveneighbors from the lists and vice versa.

If we write (from left to right) the rows as columns (1st row as1st column, 2nd row as2nd columns and so on), then the columns become rows, and we obtain another matrix,A′

which in this particular case is the same asA. Such an operation is called atransposition ofthe matrix. SinceA= A′, this matrix is calledsymmetric. One can think about transpositionas of a rotation of the matrix about its diagonal, imaginary line in space connecting upperleft and lower right entries.


Incidence matrix. It is a matrix which has one row for each vertex and one column foreach edge of a graph. If vertexxi is incident to edgeej , then the(i, j)-entry in the matrixis 1, otherwise it is 0. For our graphG, the incidence matrix denoted byI(G) has 5 rowscorresponding to the vertices and 7 columns corresponding to the edges:

I(G) =

1 0 0 0 1 0 01 1 0 0 0 1 10 1 1 0 0 0 00 0 1 1 0 0 10 0 0 1 1 1 0

.

As one can see, every column has precisely two 1’s; their rowspoint out on the verticeswhich are connected by the respective edge. If we transpose this matrix (write rows sequen-tially as columns), we obtain another matrix which is different fromI . We will se in Part IIhow important the transposition of the incidence matrix is.

Edge lists.One can describe graph by giving just the list of all of its edges. For graphG, the edge list, denoted byJ(G) is the following:

J(G) = {{x1,x2},{x2,x3},{x3,x4},

{x4,x5},{x1,x5},{x2,x5},{x2,x4}}.

Important comment. If we compare Figure 1.2 and Figure 1.5, then we observe thatthey represent the same graphG; the only difference is in the positions of vertices in theplane. It is the feature of graph theory that the same graph may be drawn in many differentways. Not only the vertices may have different positions, the edges may be drawn as curvesconnecting the same pairs. As far as the names of vertices andedges remain the same, weaccept the agreement that it is the same graph because it has the same mathematical model.

At this moment it is important to understand that all three drawings Figure 1.1, Fig-ure 1.2 and Figure 1.5, the description of pairG = (X,E), adjacency listL(G), adjacencymatrix A(G), incidence matrixI(G), and edge listJ(G), all are different representations ofthe same concept called graph and denoted by just one letterG. Inverse, having any of thesedescriptions one can draw a graph and/or construct any otherrepresentation as well.

In practice, depending on the problem, some representations are more suitable than theothers.

Exercises 1.3.

1. For graphsG1,G2 andG3, see Figure 1.6, construct an adjacency list.

2. For graphsG1,G2 andG3, construct an adjacency matrix.

3. For graphsG1,G2 andG3, construct an incidence matrix.

4. For graphsG1,G2 andG3, construct an edge list.

5. Write down an arbitrary adjacency list, adjacency matrix, incidence matrix, edge listand draw the respective graph.


b

b

b b

b b b

b

bbb

bb

b

b

b

b

G1 G2 G31

2

3

4

a

bc

d

Figure 1.6.

Computer Projects 1.3.Write a program with the following input and output.

1. Given an adjacency list, find the adjacency matrix.

2. Given an adjacency list, find the incidence matrix.

3. Given an adjacency list, find the edge list.

4. Given an adjacency matrix, find the adjacency list.

5. Given an adjacency matrix, find the incidence matrix.

6. Given an adjacency matrix, find the edge list.

7. Given an incidence matrix, find the adjacency list.

8. Given an incidence matrix, find the adjacency matrix.

9. Given an incidence matrix, find the edge list.

10. Given an edge list, find the adjacency list.

11. Given an edge list, find the adjacency matrix.

12. Given an edge list, find the incidence matrix.

1.4. Generalizations

When degenerated cases may occur.In some cases, especially from theoretical viewpoint, it is convenient to consider an edge connecting a vertex to itself. Such edges arecalled loops. It may also happen when some vertex has no neighbors, i.e. itsdegree is 0.These vertices are said to beisolated. Loops and isolated vertices are shown in Figure 1.7.

When repeated edges occur.In some areas of applications, there are many connectionsbetween some of the points (for example, in geographic maps), so in graph models, theremay be many edges connecting the same pair of vertices. Such edges are calledparallelor multiple . The number of repetitions of an edge is itsmultiplicity . Graphs which admitmultiple edges are calledmultigraphs. Multiple edges are shown in Figure 1.7.

When direction/order is important. Until now we made no distinction in writing thesame sets in different ways. Usually, we read from left to right, and the following twowritings for the same set are equivalent:{a,b} = {b,a}. However, there are situations inreal life when an order is crucially important. For example,a road (no direction specified)is not the same as a one-way street (direction should be strictly observed). To reflect suchsituations, one introduce an order which becomes important. In writing ordered sets instead


b

loop b

b

b

b

isolated vertices inE4

b b

b

isolated vertex in a graph

b b

multiple edges

b b

arcs

Figure 1.7. Generalizations.

of curly braces{,}, one use parentheses(,). So now(a,b) 6= (b,a). In a graph, an orderedpair of vertices is called anarc. If (x,y) is an arc, thenx is called theinitial vertex andy is called theterminal vertex. A graph in which all edges are ordered pairs is calledthedirected graph, or digraph. Even loops may be ordered, see Figure 1.7.

Comparing edges and arcs one can accept the following point of view: an edge of agraph is equivalent to two arcs going in opposite directions, see Figure 1.8.

b b b b=

Figure 1.8.

Graphs in which order is not important are calledundirected graphs. Figure 1.8 impliesthat directed graphs represent more general structures than undirected graphs. Or, equiva-lently, the undirected graphs represent a special case of directed graphs, namely, when eachedge is replaced by a pair of arcs going in opposite directions. Undirected graphs with-out loops and multiple edges are calledsimple graphs or simplygraphs. Usually, unlessotherwise stated, one consider simple graphs.

Graph representations can accordingly be adjusted. For undirected multigraphs, adja-cency listL may contain multiple elements and empty sets, adjacency matrix A may havezero rows and zero columns and integer numbers different from 1 (=multiplicity), incidencematrix I may have zero rows and repeated columns, and edge listJ may have repeated ele-ments. For directed graphs, adjacency listL will not change, adjacency matrixA may not besymmetric (the order may be “row→ column”), incidence matrixI will have -1’ for initialvertices, and edge listJ will become ordered.

An example of a directed multigraphG with a loop is shown in Figure 1.9.The adjacency list ofG is:

L(G) = { /0,{2},{2,4},{1,1,3}};


b

b

b

b 1

23

4

a

b

cd

e fG

Figure 1.9.

the adjacency matrixA is:

A(G) =

0 0 0 00 1 0 00 1 0 12 0 1 0

;

the incidence matrixI is:

I(G) =

1 1 0 0 0 00 0 1 l 0 00 0 −1 0 1 −1

−1 −1 0 0 −1 1

;

and, at last, the edge listJ is:

J(G) = {(4,1),(4,1),(3,2),(2,2),(4,3),(3,4)}.

Notice that the columns inI(G) are in the order of arcsa,b,c,d, e, f , and “l ” in fourthcolumn and second row indicates that arcd is the loop at vertex 2.

Some graph modeling examples where the order is important are:

• the vertices are street crossings in the city map; crossingsxi andx j form an arc(xi ,x j)if the traffic is allowed in the direction fromxi to x j ;

• the vertices are all companies in the world; companiesxi andx j form an arc(xi ,x j) ifcompanyxi is a supplier for the companyx j ;

• the vertices are all web pages in the Internet; the web pagesxi andx j form an arc(xi ,x j) if there is a hypertext link from pagexi to pagex j ;

• the vertices are folders of a folder system in a computer; foldersxi andx j form an arc(xi ,x j) if folder xi contains folderx j as a subfolder;

• the vertices are states of a discrete system; statesxi andx j form an arc(xi ,x j) if theprobability of transition from statexi to statex j is positive;

• the vertices are all your predecessors; predecessorsxi andx j form an arc(xi ,x j) ifpredecessorxi is the parent of predecessorx j ;


• the vertices are students in a classroom; studentsxi and x j form an arc(xi ,x j) ifstudentxi likes studentx j .

Exercises 1.4.For graphs in Figures 1.3 and 1.6, replace each edge with an arc and construct the adjacency list

L, adjacency matrixA, incidence matrixI , and edge list (i.e. arc list)J.

Computer Projects 1.4.Repeat Computer Projects 1.3. for: a) undirected multigraphs; b) directed multigraphs.

1.5. Basic Graph Classes

Empty graphs. Any graph must have at least one vertex. In other words, we do not acceptgraphs without vertices, though in hypergraph theory we will show how this case may betreated. A graph may have no edges at all; such graphs are called empty, denoted byEn

wheren is the number of vertices. GraphE4 is depicted in Figure 1.7. If a graph is notempty, then it has at least one edge. Notice that an empty graph is not the same as theempty set which is always denoted by/0.

b b b b b

x1 x5x2 x3 x4

P5

Figure 1.10. PathP5.

Paths. A graph in which all vertices can be numbered (ordered from left to right)x1,x2, . . . , xn in such a way that there is precisely one edge connecting every two consecutivevertices and there are no other edges, is called apath.

The number of edges in a path is itslength. A path onn vertices is denoted byPn.Evidently, in anyPn, the number of edgesm = n− 1. Any edge itself is a pathP2. Theverticesx1 andxn both have degree 1 in a pathPn; we say that the pathconnectsverticesx1

andxn. Generally, any path connecting verticesx andy is called(x,y)-path. In Figure 1.10,pathP5 of length 4 connects verticesx1 andx5, or is (x1,x5)-path.

b b

b

b

b

b

b

b

b

b

G1 G2

Figure 1.11. Connected graphG1 and disconnected graphG2.

Connected graphs.A graph is calledconnectedif in it any two vertices are connectedby some path; otherwise it is calleddisconnected.It means that in a disconnected graph


there always exists a pair of vertices having no path connecting them. Any disconnectedgraph is a union of two or more connected graphs; each such connected graph is then calledaconnected componentof the original graph.

For example, see Figure 1.11,G1 is a connected graph, butG2 is a disconnected graphhaving two connected components. Any isolated vertex is a connected component. Gener-ally, an empty graphEn hasn components. If each component of a graphG represents thesame graph, sayG′, and ifG hask connected components, then we write:G = kG′.

Cycles.A connected graph in which every vertex has degree 2 is calledacycle, (some-times “simple cycle”) denoted byCn wheren is the number of vertices.

b

C1

b b

C2b b

b

C3

b b

bb

C4

b

b b

bb

C5

b b b

bbb

C6

Figure 1.12. CyclesC1,C2,C3,C4,C5, andC6.

If n is an even number, thenCn is calledeven cycle. If n is odd, thenCn is odd cycle.In Cn, the number of edges coincides with the number of vertices and it is called thelengthof the cycle, see Figure 1.12.C1 represents an edge connecting the vertex with itself, it isa loop. C2 represents two parallel edges connecting the same pair of vertices. The cycleC3 is calledtriangle. In simple graphs, if there are cycles, then they must have length atleast 3. If a graph is not a cycle, then either it has a vertex ofdegree other than 2, or it isdisconnected.

b

b

b

bb

b

Figure 1.13. WheelW6.

Wheels. If for any cycleCk,k ≥ 3, we add a new vertex and connect it to each of the


vertices ofCk, then the graph obtained is called awheel, denoted byWk+1. The wheelW6 isshown in Figure 1.13.

Complete graphs. A graph in whicheverypair of vertices is an edge, is calledcom-plete, denoted byKn where as usually,n is the number of vertices. It is called completebecause we cannot add any new edge to it and obtain a simple graph. For everyn≥ 1, thedegree of each vertex inKn is n−1, and the number of edges is:

b

K1

b b

K2 b b

b

K3

b b

bb

K4b

b b

bbK5

b b

b

bb

b

K6

Figure 1.14. Complete graphsK1,K2,K3,K4,K5, andK6.

m=

(

n2

)

=n(n−1)

2.

This formula can easily be obtained by counting and adding degrees of every vertex, i.e.applying Proposition 1.1.1: havingn vertices of degreen−1 each, we obtain the numbern(n−1) = 2m. For example, the number of edges inK6 is: m= 6(6−1)/2 = 15. If a graphis not complete, then it has at least two vertices which are not adjacent.

Complete graphsK1,K2,K3,K4,K5, andK6 are shown in Figure 1.14. Notice thatK2 =P2, K3 = C3, andK4 = W4.

Trees. A connected graph which has no cycles is called atree. Usually any tree onnvertices is denoted byTn. In contrast toPn, Kn, Cn andWn there are many distinct trees onnvertices.Pn is a special (simplest) case of a tree. We will have the concept of isomorphismto distinguish graphs.

b b b b b bb

b

b

b

b

b

Figure 1.15. Trees.

Figure 1.15 shows three examples of trees, among which the 1st and the 3rd “are the same”just because they both areP4, and the 2nd is “different” from them. There are no other trees


b

b

b

b

b

X1 X2

G

b

b b

b

b

bb

b

K1,3 C4

Figure 1.16. Bipartite graphs.

on 4 vertices. In any tree onn vertices, the number of edgesm= n−1. Disconnected graphwithout cycles is called aforest. Evidently, in a forest every component is a tree.

The last example shows that a graph may be drawn in many different ways; differentdrawings are like different views of the same object. It is not important how a graph isdrawn; it is important which vertices are adjacent and whichare not. As we have seen, notevery intersection of edges in a drawing is a vertex. But every vertex lies on the intersectionof the respective edges.

Bipartite graphs. A graph G = (X,E) is calledbipartite if its vertex setX can bepartitioned into two disjoint setsX1 and X2, called parts, in such a way that every edgeconnects vertices from different sets, see Figure 1.16. It means that there are no edgesinside X1 and there are no edges insideX2. In other words, for a bipartite graph, in itsdrawing, it is possible to color the vertices using just two colors (say blue and red) in sucha way that adjacent vertices have different colors. If a graph is not bipartite, then in anypartition of its vertex set into two subsets, at least one of the subsets contains at least oneedge. Trees and even cycles are bipartite graphs; odd cyclesare not. The smallest simplegraph which is not bipartite is triangleC3. Notice that any graph which contains a triangleor an odd cycle cannot be bipartite. IfG = (X,E) is a bipartite graph, it is convenient towrite it asG= (X1,X2;E) whereX1 is its left part andX2 is its right part (as is the first graphin Figure 1.16).

A complete bipartite graph is a bipartite graph in whichevery vertexfrom partX1 isadjacent toevery vertexfrom partX2. It is called complete because it is not possible to adda new edge to it and obtain another bipartite graph. If in a complete bipartite graph|X1| = rand |X2| = s, then the graph itself is denoted byKr,s. The number of edges inKr,s clearlyequalsrs.

Among the examples of bipartite graphs shown in Figure 1.16,the first is not complete,the second isK1,3, and the third isC4. Observe that bipartitionX1,X2 is explicitly shown forgraphG and it is not shown forK1,3 andC4; this fact exhibits the difference between “canbe partitioned” (as in definition) and “is partitioned”.

Regular graphs.A graph in which every vertex has the same degreek is calledregularof degreek or k-regular. Empty graphEn is 0-regular, cycleCn is 2-regular, andKn is(n−1)-regular graph. The 3-regular graphs are called thecubic graphs.

It is easy to construct cubic graphs. For example, one can take two cycles of thesame length, draw one inside another and connect the respective vertices by edges called


b

b b

b

b

b b

b

Cube

b

b

bb

b

b

b

bb

b

Petersen graph

Figure 1.17. Cubic graphs.

“spokes”. If we do that withC3 we obtain a graph calledprism; if we do that withC4 weobtain a graph calledcube. If we do that withC5 by drawing inner cycle differently, weobtain a famous graph called thePetersen graph, see Figure 1.17.

b

b b

b b

b bb

b

b b

b

G1 G2 G3

1

2

4

3

1

2

4

3

1

2

4

3

Figure 1.18. Isomorphic graphs.

Isomorphic graphs. How can we compare different graphs? Consider graphG1 inFigure 1.18. Let us re-draw it aside by placing vertex 4 inside triangle formed by vertices1, 2 and 3, and preserving the edges. We obtain graphG2. Is G2 different fromG1, orthat is the same graph? Let us re-drawG1 again by placing the vertices on a square as inG1 but now replacing the segments of straight lines (representing the edges) by arbitrarycurves. We obtain graphG3. Is G3 different fromG1, or that is the same graph? On onehand, of course, three graphs are pairwise different because the vertices are different pointsin the plane. On the other hand, they all have the same adjacency matrix, i.e., the samemathematical model. We can say that graphsG1, G2 andG3 are “essentially the same”. Buthere is an important point: graphsG1, G2 andG3 have the same adjacency matrix becausethe vertices have the same names; if we choose a different numbering of vertices in any ofthe graphs, we obtain a different adjacency matrix. In practical applications, the names ofvertices are not given at all; so, how to recognize if two graphs are “essentially the same”?

Mathematical definition for two graphs “to be essentially the same graph” is expressed


in the concept of “isomorphism” (Greek: “iso” = equal, and “morphe” = shape). Two simplegraphsG1 = (X1,E1) andG2 = (X2,E2) are calledisomorphic if there exists a one-to-onecorrespondence between vertex setsX1 andX2 such thatany two vertices are adjacent inG1

if and only if their images in the correspondence are adjacent in G2. Any such one-to-onecorrespondence is called anisomorphism. If G1 andG2 are isomorphic, then we say thatG1 is isomorphic toG2, andG2 is isomorphic toG1, and we writeG1

∼= G2.Suppose|X1| = |X2| = n. Generally, how many one-to-one correspondences do exist?

The first vertex ofX1 can be mapped into any ofn vertices ofX2, so we haven possibilitiesfor it. Once the first vertex is mapped, the second vertex hasn− 1 possibilities, then thethird vertex hasn−2 possibilities and so on. Total number of possibilities isn(n−1)(n−2) · · ·3 · 2 · 1 = n!. If in at least one of thesen! cases there is a complete “coincidence”of G1 andG2, then they are isomorphic. Otherwise they are not. So, two graphs are notisomorphic, if no matter how we map the vertices of one into the vertices of another, therewill always be a pair of vertices which are adjacent in one graph and disjoint in another. Insuch case the graphs are really different because we can never match them.

We can think about the number of isomorphisms to be 0 (when graphs are not isomor-phic), and 1, 2, . . .n! when they are isomorphic. In Figure 1.15, for example, the first andthe third trees are isomorphic, they both representP4. There are two different choices tomap end vertices, then the rest of mapping is determined univocal. So, we can observe thatthere are two isomorphisms between any two pathsPn. On the other hand, the second treeis not isomorphic toP4. GraphsEn andKn haven! isomorphisms, graphsCn have 2n and soon.

b

b

b b b b b

b

T1

b

b

b b b b b

b

T2

Figure 1.19. Non isomorphic trees.

To be isomorphic, graphs must have the same number of vertices, the same number ofedges, the same number of vertices of each degree. However, graphs may have the samenumber of vertices and edges and even degrees and still not tobe isomorphic, see examplein Figure 1.19. Both treesT1 andT2 have eight vertices, seven edges, four vertices of degree1, two vertices of degree 2 and two vertices of degree 3. However, in T1 the vertices ofdegree 3 are adjacent but inT2 the vertices of degree 3 are disjoint. This observation andcommon sense tells us that there will be no match in all 8!=40320 one-to-one correspon-dences. Generally, to prove that graphs are not isomorphic,instead of considering alln!mappings, it is sufficient to find some property in one of them and show that it is missingin another. To show that graphs are isomorphic, it is sufficient to exhibit that very sameone-to-one correspondence from the definition.

Next example, see Figure 1.20, shows two graphsG1 andG2 which appear to be iso-


b b b

b b b1 2 3

4 5 6

b b

b

bb

b

a

b

c

d

e

f

G1 G2

Figure 1.20. Isomorphic graphs.

morphic. In fact, the figure represents two “very different”drawings of the same graph. Themapping (one-to-one correspondence) of the vertices ofG1 into the vertices ofG2 denotedby σ is the following:

σ =

(

1 2 3 4 5 6a b c d e f

)

.

One can check manually that every two vertices are adjacent in G1 if and only if therespective vertices are adjacent inG2. For example, vertices 1 and 2 are disjoint inG1, soare the corresponding verticesa andb in G2. Vertices 1 and 4 are adjacent inG1, so are thecorresponding verticesa andd in G2, and so on foreach pair of the vertices. The numberof all such comparisons is

(62

)

= 15 which is much less than 6!= 720, the number of allone-to-one correspondences.

In determining a graph class, one proceed in the following way: first, observe somegraph property, then investigate all graphs having that property. There are hundreds ofgraph classes that have been investigated. In the simplest classes such asEn, Kn, Cn, Wn,etc, the lower index always shows the number of vertices.

Exercises 1.5.

1. ConstructL(G) for each graphG drawn in this section.

2. ConstructA(G) for each graphG drawn in this section.

3. ConstructI(G) for each graphG drawn in this section.

4. ConstructJ(G) for each graphG drawn in this section.

5. Find out and explain which graphs in Figure 1.21 are bipartite.

6. For each pair of integer numbersr ands, 1≤ r,s≤ 5, draw a complete bipartite graphKr,s.

7. Explain whyKr,s andKs,r are isomorphic.

8. Find out and explain which pairs of graphs in Figure 1.21 are isomorphic and which are not.

9. Compare graph representations for two isomorphic graphsG1 andG2 in Figure 1.20. Whenare they identical?

10. Explain when two isomorphic graphs have the same adjacency matrix and the same incidencematrix.


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Figure 1.21.

Computer Projects 1.5.Write a program with the following input and output.

1. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a path.

2. Given any ofL(G),A(G), I(G),J(G), recognize ifG is connected.

3. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a cycle.

4. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a complete graph.

5. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a tree.

6. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a wheel.

7. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a complete bipartite graph.

8. Given any ofL(G),A(G), I(G),J(G), recognize ifG is a bipartite graph.

9. Given any ofL(G),A(G), I(G),J(G), recognize ifG is regular.

10. Given any ofL(G1),A(G1), I(G1), or J(G1), any ofL(G2), A(G2), I(G2), or J(G2) and aone-to-one correspondenceσ between the vertices ofG1 andG2. Recognize ifσ realizes anisomorphism betweenG1 andG2.

1.6. Basic Graph Operations

In many proofs and algorithms, one often apply graph operations that allow to obtain onegraph from another. We now consider some of them.

Deletion of a vertex.Let us have a graphG = (X,E) and a vertexx∈ X. A deletionofx from G is the removing ofx from setX and removing fromE all edges ofG that containx. Recall thatE(x) denotes the set of edges containing vertexx in graphG. If X1 = X−{x},andE1 = E−E(x), then deletion ofx from G results in obtaining the graphG1 = (X1,E1),


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G

x

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G1

deletex

Figure 1.22. Deletion ofx from G.

see Figure 1.22. We write this operation asG1 = G− x. In G1, we can choose and deleteanother vertex to obtain a graphG2 and so on; sequential deletion of vertices results in asequence of graphs.

We may want to delete an entire subset of vertices; it is equivalent to a sequentialdeletion of the respective vertices in any order. Sometimesdeletion of vertices is calledstrong deletion because the vertices are removed from a graph together with all incidentedges.

Weak deletion of a vertex.Again, let us have a graphG = (X,E) and a vertexx∈ X.Weak deletionof x from G is removing ofx from setX. Now the setE(x) remains in thegraph but it loses the vertexx. The only exception is ifE(x) contains loops: we assume theloops disappear. All other edges fromE(x) become loops and remain in the graph obtained.

If the very same vertex is weakly deleted fromG, see Figure 1.23, we obtain a differentgraphG1. We also write this operation asG1 = G−x with understanding that the meaningof deletion (strong or weak) is clear from the context. In ourexample, see Figure 1.23,E(x) = {e1,e2,e3} and all these edges become loops inG1.

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b b

b

G

x

b b

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G1

deletex weaklye1e2

e3

e1 e2

e3

Figure 1.23. Weak deletion ofx from G.

Deletion of an edge.It is the simplest operation of deletion: we just remove an edgefrom the list of edges. All the rest remains unchanged, see Figure 1.24. Sometimes thedeletion of an edge is called aweak deletion.

Strong deletion of an edge.It is the removing of an edge from the list of edges andweak deletion of both of its vertices from the graph obtained. Strong deletion of the edgee3 is shown in Figure 1.24. In both cases we write:G1 = G−e3.

In graph theory, the term “deletion of a vertex” by default ismeant as “strong deletionof a vertex”, and the term “deletion of an edge” by default is meant as “weak deletion of


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deletee3 stronglye1e2

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deletee3 weaklye1e2

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Figure 1.24. Strong and weak deletions ofe3 from G.

an edge”. However, we are making this distinction here to fit to the general hypergraphapproach. In Part II (Section 7.4.), we will see that strong deletions of vertices is the sameas strong deletions of edges, and weak deletion of vertices is the same as weak deletion ofedges in dual hypergraphs.

Contraction of an edge.Let us have a graphG = (X,E) and an edgee= {x,y} ∈ E.Contraction of edgeeconsists in the following two steps, see Figure 1.25:

1. Identifying verticesx andy in a new vertex calledxy and removinge from E;

2. All edges ofG having one end atx or y will have this end atxy with other endunchanged.

b bex yb

xy

N(x) N(y) N(x) N(y)

contracte

G G1

Figure 1.25. Contraction ofe in G.


If G is a simple graph andN(x)∩N(y) 6= /0, then each vertex fromN(x)∩N(y) will beconnected with vertexxy by two edges, i.e. graphG1 will have multiple edges. IfG is notsimple and has loops atx or y, the loops remain inG1 at vertexxy.

At last, if G is not simple and had another edgee′ connectingx andy, thene′ becomesa loop at vertexxy. So, contraction of an edge indeed means contraction of thatedge up toone point. An example of contraction is shown in Figure 1.26.

b b b

b bex y

b b b

b

G G1

xy

contracte

Figure 1.26.G1 = G ·e.

A graphG1 is contractible to a graphG2 if G2 may be obtained fromG1 by a sequenceof contractions of the edges. For example, anyCk,k≥ 4 is contractible toC3 = K3, any treewith at least two vertices is contractible toK2, anyKn is contractible toKm if n ≥ m andso on. On the other hand, noCk can be contracted toK4, no tree can be contracted toK3.The maximum value ofn such that a graphG is contractible toKn is called theHadwigernumber of the graphG, denoted byη(G).

Taking the complement.Consider a simple graphG = (X,E) with |X|= n. Thecom-plement ofG denoted byG is the graph on the same vertex setX in which two vertices areadjacent if and only if they are disjoint in the original graph G. In other words,G = (X,E′)whereE′ is such an edge set thatE∪E′ forms the edge set ofKn. This is whyG is called thecomplement ofG. Evidently,G = G, En = Kn andKn = En. Notice thatP4 is isomorphicto P4, C5 is isomorphic toC5, andKr,s = {Kr ,Ks}. It may happen thatG is connected graphbut G is not, see Figure 1.27.

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complement

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G G

Figure 1.27.G andG.

Sequential application of operations. Any of the basic operations above (except tak-ing complement) may be applied sequentially many times whatresults in a sequence ofgraphs. Since we consider finite graphs, there will always bethe last graph in this sequence.We then can consider the sequence of inverse operations and reconstruct the original graphfrom the last graph. This method is very common in graph theory and is used in proofs bymathematical induction.


Exercises 1.6.

1. Implement a weak sequential (in any order) deletion of vertices ofC4, W5, K5, K3,3 andPetersen graph.

2. Implement a strong sequential (in any order) deletion of vertices ofC4, W5, K5, K3,3 andPetersen graph.

3. Implement a weak sequential (in any order) deletion of edges ofC4, W5, K5, K3,3 and Petersengraph.

4. Implement a strong sequential (in any order) deletion of edges ofC4,W5, K5, K3,3 and Petersengraph; at each step, weakly remove the loops.

5. Implement a sequential (in any order) contraction of edges ofC4, W5, K5, K3,3 and Petersengraph; at each step, weakly remove multiple edges.

6. Find the Hadwiger number of any tree,C5, W5, K2,3.

7. Construct the complement ofC3,C4, C5,C6, K5, K3,5, Petersen graph, cube,P7, any tree on 6vertices,E5, 2C3.

8. Show thatC5∼= C5.

9. Show thatC6 is isomorphic to a prism.

10. How are the degree sequences ofG andG related ?

Computer Projects 1.6. Write a program for the following operations.

1. Given any graph representation of a graph, delete a vertexstrongly.

2. Given any graph representation of a graph, delete a vertexweakly.

3. Given any graph representation of a graph, delete an edge strongly.

4. Given any graph representation of a graph, delete an edge weakly.

5. Given any graph representation of a graph, construct the same representation for the comple-ment.

6. Given any graph representation, implement exercises 1-5with drawing on the screen a graphobtained at each intermediate step. Disregard multiple edges and loops.

1.7. Basic Subgraphs

Graphs as combinatorial structures may have many differentproperties. The propertiesare very often determined by the presence or absence of some specific substructures calledsubgraphs. We next consider some of them.

Subgraphs. Let us have a graphG = (X,E). Any graphG′ = (X′,E′) is called asubgraph of G if and only if X′ ⊆ X, andE′ ⊆ E. In such case, we writeG′ ⊆ G. SinceE′

contains only the elements ofE, both ends of any edge fromE′ must be inX′. Therefore,G′

can be obtained fromG by strong deletion of verticesX−X′ (sequentially in any order or atonce) and further weak deletion of remaining edgesE−E′ (sequentially in any order or atonce). In Figure 1.28, bothG1 andG2 are subgraphs ofG: G1 is obtained by strong deletionof vertexx1 and weak deletion of edge{x3,x5}, andG2 is obtained by strong deletion of


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G G1 G2

Figure 1.28. GraphG, subgraphG1 and induced subgraphG2.

x2 andx4. It is evident that the order in which the vertices and edges are deleted is notimportant.

Induced subgraphs.A graphG′ = (X′,E′) is called aninduced subgraphof a graphG= (X,E) if X′ ⊆X and all edges ofG having both ends inX′ form edge setE′. Sometimeswe say thatG′ is a subgraphinduced by X′. Induced subgraphG′ may be obtained fromG by just strong deletion of verticesX−X′ (sequentially in any order or at once). Inducedsubgraph is a special case of subgraph. A subgraph is not induced if at least one edge ofG with both ends inX′, is missing. In a graphG, it is convenient to denote the subgraphinduced by a setY ⊆ X by GY.

In Figure 1.28,G2 is an induced subgraph ofG but G1 is not. One can see thatG con-tains three subgraphs isomorphic toK3, two subgraphs isomorphic toC4 and one subgraphisomorphic toC5. Among these subgraphs, onlyK3’s are induced.

Cycles.In a graph, any subgraph representing a sequence of verticessuch that every twoconsecutive vertices are connected by an edge, and, the firstand the last vertices coincide,is called thecycle. The number of edges in a cycle is called itslength. A cycle is calledodd or even if its length is respectively odd or even. Cycles may be induced or not; notevery induced cycle is isomorphic to the simple cycleCk for somek≥ 3. For example, thesequence of verticesx2,x3,x4,x5,x2, see Figure 1.28, forms the cycle of length 4, which isisomorphic toC4 (represented by graphG1), however, if we consider an induced subgraph,then we need to add the edge{x3,x5} to G1. A cycle which has all the vertices different(except the first and the last) is calledsimple. For example, the cyclex1,x2,x3,x4,x5,x1 issimple, but the cyclex1,x2,x5,x3,x4,x5,x1 is not simple because the vertexx5 is used twice.Though the length of the last cycle is 6, it is not isomorphic toC6 by the same reason. Noticethat any odd cycle, if it is not simple, then it can be split into two cycles one of which is oddand another is even. This implies that if a graph has an odd cycle, then it has a simple oddcycle.

Cliques. Since graphs have vertices, they always contain subgraphs isomorphic to someof K1, K2, K3,. . . , Kn. They are calledcliques. If we have a subgraph isomorphic toKr

which is not contained in a subgraph isomorphic toKr+1, then we say thatKr is themaximalby inclusion complete subgraph, or, equivalently, themaximal clique. Being maximalby inclusion, different cliques may have different size, i.e. the number of vertices. Thelargest size of a clique among all the cliques of a graphG is called theclique number of G


denoted byω(G). Simply,ω(G) is the maximum number of pairwise adjacent vertices. Forany graphG, 1≤ ω(G) ≤ n.

For example, see Figure 1.28, verticesx2 andx3 induceK2 in graphG but it is not amaximal clique since it is not maximal; it is contained inK3 induced by verticesx2,x3,x5

which is maximal by inclusion. On the other hand, the same verticesx2 andx3 in graphG1

form a maximal clique. We conclude thatω(G) = 3 andω(G1) = ω(G2) = 2.Independent sets. Similarly to complete subgraphs, any graph contains subgraphs

isomorphic to some ofE1, E2, . . . , En. In a graphG, a subset of vertices which inducesa subgraphEk is called thestable set,or independent set. There are also maximal byinclusion and not maximal by inclusion stable sets. The largest size of a stable in a graphG is called thestability number , denoted byα(G). Simply,α(G) is the maximum numberof pairwise disjoint vertices. For any graphG, 1≤ α(G) ≤ n.

Observe thatω(G) = α(G) and α(G) = ω(G) because when taking the complementevery complete subgraph becomes a stable set and vice versa.

For graphG, see Figure 1.28, verticesx1 andx3 form a stable set, verticesx2 andx4

form another stable set. Both are maximal by inclusion, andα(G) = 2.

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fG1 G2

Figure 1.29. Maximal cliques and stable sets.

One more example is shown in Figure 1.29. InG1, vertices 1 and 2 form a clique, andvertices 2, 3, and 4 form a clique,ω(G1) = 3. In G2, verticesa,b, andd form a maximal byinclusion stable set, and verticesa,b,e, and f also form a maximal by inclusion stable set;the last being maximal is maximum, soα(G2) = 4.

Important comment. In graph theory, given any property, it is common to use termmaximal (minimal) in the sense ofmaximal (minimal) by inclusion and termmaximum(minimum) in the sense oflargest (smallest) over all maximal (minimal). The differenceis similar to the difference between the local and global maximum (minimum) of a function.

Transversals (vertex covers). In a graphG = (X,E), a subset of verticesT ⊆ X iscalled atransversal (vertex cover)if its complementX \T is an independent set. It meansthat every edge ofG has at least one end inT. The minimum cardinality of a transversalof a graphG is called thetransversal number and denoted byτ(G). It follows from thedefinitions above that for any graphG,

α(G)+ τ(G) = |X|.

Spanning subgraphs.Let us have a graphG = (X,E), |X| = n. Any subgraphG′ ⊆ Gsuch thatG′ = (X,E′) is called aspanning subgraph.Thus spanning subgraphs have thesame vertex set as the graph itself. Spanning subgraph whichis a tree is calledspanning


tree. If a graph has a spanning tree, then it is connected because any tree is a connectedgraph.

In Figure 1.29, if we weakly delete edges{1,2} and{2,3}, then we obtain a spanningtree ofG1.

Matchings. In a simple graphG, a subgraph in which every vertex has degree 1, iscalled amatching. Every matching simply represents a collection of edges which haveno common vertices, i.e., which are pairwise disjoint. Aperfect matching is a matchingwhich is a spanning subgraph. Themaximum size of a matching(over all matchings) isdenoted byν(G). If a graphG has a perfect matching, then clearly,n is an even number andν = n/2.

In graphG2, see Figure 1.29, edge{c,d} forms a maximal by inclusion matching.However, a maximum matching is formed for example by edges{a,c} and {d,e}, andν(G2) = 2.

Since at least one vertex from every edge of any matching mustbelong to any transver-sal, the cardinality of any transversal is at least the cardinality of any matching; i.e, for anygraphG,

τ(G) ≥ ν(G).

Therefore, if we have a matching and a transversal of the samecardinality, then bothare optimal.

Factors. Let G be a simple graph. Ak-factor of G is a spanning subgraphin whichevery vertex has degreek. In this way, 1-factor represents a perfect matching, 2-factor is acycle or a collection of disjoint cycles, and so on.

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K4Figure 1.30. Factors.

Factors not always exist, sometimes there are many distinctk-factors. Complete graphshave the largest number of factors. Examples of factors in a graphG (which is a prism) areshown in Figure 1.30. Regular edges show a 1-factor which is aperfect matching; dashededges show a 2-factor consisting of two connected components beingC3 each. There is onemore 2-factor which is connected and represented by cycleC6. The graph itself is a 3-factor.

Partition of the edges of a graph intok-factors is calledk-factorization. The 1-factorization ofK4 is shown in Figure 1.30.

Graph minors. A graphG′ is aminor of a graphG, if G′ can be obtained fromG by asequence of any vertex deletions, edge deletions and edge contractions.


Exercises 1.7.

b b

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Figure 1.31.

1. For graph in Figure 1.31, write down the list of all cliques, the list of all maximal cliques, andthe list of maximum cliques. Begin with all 1-vertex cliques, then all 2-vertex cliques, and soon.

2. Write down the list of all cliques ofP6,C3,C4,C5,W4, W5, K2,3, cube and Petersen graph.Begin with all 1-vertex cliques, then all 2-vertex cliques,and so on.

3. Write down the list of all maximal cliques ofP6,C3,C4,C5, W4,W5,K2,3, cube and Petersengraph.

4. Write down the list of all maximum cliques ofP6,C3,C4,C5, W4,W5, K2,3, cube and Petersengraph.

5. Using the concept of a tree, suggest a way to construct the list of all maximal independentsets in a graphG.

6. Write down the list of all independent sets ofP6,C3,C4,C5, W4,W5,K2,3, cube and Petersengraph. Begin with all 1-vertex independent sets, then all 2-vertex independent sets, and soon.

7. Write down the list of all maximal independent sets ofP6, C3, C4,C5, W4, W5,K2,3, cube andPetersen graph.

8. Write down the list of all maximum independent sets ofP6,C3,C4, C5, W4, W5,K2,3, cube andPetersen graph.

9. Implement exercises 2-6 for the complements of the same graphs.

10. Find the longest path inC7, cube,W9, Petersen graph.

11. Find the shortest and the longest cycles inK4,4, cube, Petersen graph.

12. Find a spanning tree in cube,K6,9, Petersen graph.

13. Findω(G) for every graph drawn in this section and in Figure 1.21.

14. Findα(G) for every graph drawn in this section and in Figure 1.21.

15. Findτ(G) for every graph drawn in this section and in Figure 1.21.

16. Findν(G) for every graph drawn in this section and in Figure 1.21.

17. Findα,ω,τ, andν of cube, Petersen graph,K999, K1000,C20, C21,W99, W100, Pn, Kn, Cn,Wn,for all integersn≥ 2.

18. Find all non isomorphic minors ofW5.


Computer Projects 1.7. Using a convenient graph representation, write a program for thefollowing algorithmic problems.

1. Given a graphG, a subset of vertices and a subset of edges. Check if the subsets form asubgraph ofG.

2. Given a graphG and a subset of vertices. Output the subgraph induced by the subset.

3. Given a graphG and a subset of vertices. Check if the subset induces a clique.

4. Given a graphG and a subset of vertices. Check if the subset is an independent set of vertices.

5. Given a graphG and a clique. Check if the clique is maximal.

6. Given graphG and an independent set. Check if the set is maximal.

7. Input: Petersen graph. Output: the collection of all maximal independent subsets.

8. Given a graphG and a subset of vertices. Check if the subset is a transversal.

9. Given a graphG and a list of edges. Check if the edges form a matching inG.

10. Given a graphG and a list of edges. Check if the edges form ak-factor inG.

1.8. Separation and Connectivity

Recall that while deleting vertices we always mean strong deletion. LetG = (X,E) be asimple connected graph andx,y∈X. If verticesx andy are not adjacent, deleting the setX−{x,y} from G leaves only verticesxany, i.e. a disconnected graph. Any setS⊆X of verticeswhich after deletion fromG leaves a disconnected graph is called aseparatoror vertex cut.A subgraph induced by some separator is also calledseparator. Among connected graphs,only complete graphs do not have separators. For disconnected graphs, a separator is anyset of vertices deleting of which increases the number of connected components. So, ifS isa separator inG, then there are at least two vertices, sayx andy which are separated byS;in this case,S is called(x,y)-separator. The meaning of a separator is that all(x,y)-pathspass through it.

Separators may contain other subsets which are also separators. A separator whichdoes not contain any other separator as a proper subset, is called aminimal separator. Asusually, minimality is meant by inclusion; different minimal separators may have differentsize. The minimum over all sizes of all minimal separators iscalled theconnectivity ofG and denoted byκ(G). There is only one exception here, namely graphKn. Since it hasno separators, it is convenient to put by definitionκ(Kn) = n− 1. So, for any incompleteconnected graphG, κ(G) is the size of the smallest separator.

There is one more important concept related to separation. AgraphG is calledk-connectedif connectivity κ(G) ≥ k. So, all graphs are 0-connected, connected graphs are1-connected, connected graphs havingκ(G) ≥ 2 are 2-connected and so on. Generally, ifa graphG is k-connected, then it is(k−1)-connected,(k−2)-connected, and so on. Butthis implication is not working in the opposite way: if a graph isk-connected, it may be not(k+1)-connected. The inclusion of connectivity classes is shownin Figure 1.32.


All graphs

Connected graphs

2-connected graphs

3-connected graphs

k-connected graphs

Figure 1.32. Connectivity classes.

Figure 1.33 shows a graphG that has many separators. Set{2,3,5, 6} is a (1,4)-separator though it is not a minimal separator. Subsets{2,6} and {2,5} also represent(1,4)-separators both being minimal. It is seen from the figure that for this graphκ(G) = 2.

GraphG in Figure 1.33 is 2-connected. Therefore, it is 1-connectedand 0-connected.But it is not 3-connected and it is notk-connected for anyk≥ 3.

b b b

b b b1 2 3

456

G

Figure 1.33. 2-connected graphG.

Suppose we delete a separatorS from a connected graphG and obtain two connectedcomponents induced by vertex setsX1 andX2. The two subgraphs induced byX1∪S andX2∪S are calledderived subgraphsof the graphG with respect to separatorS. We willdenote them byGX1∪S andGX2∪S, or simply byG1 andG2 respectively, see Figure 1.34.Notice that bothX1 andX2 are not empty sets, and none of the vertices fromX1 is adjacentto any of the vertices fromX2.

Proposition 1.8.1 In a connected graph G, if S is a minimal separator, then each vertex ofS has neighbors in both X1 and X2.

Proof. If somex∈ Shas no vertex inX1 adjacent to it, thenS−{x} is also a separator of


X1 S X2

G1

G2

Figure 1.34. Derived subgraphs: general scheme.

G what contradicts the minimality ofS. �

For example in Figure 1.35, vertices 3, 5 and 6 in separator{2,3, 5,6} do not haveneighbors in both components because the separator is not minimal; however, vertices 2and 5 form a minimal separator and both have neighbors in eachof two components. Thetwo connected components produced by this separator are induced by vertex sets{3,4} and{1,6}. The two derived subgraphs produced by separator{2,5} are induced by vertex sets{1,2,5,6} and{2,3,4,5}. They are shown asG1 andG2.

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456

G

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b b

b b

b b

b

6 5

1 2

G1 G2

5 4

32

Figure 1.35. Derived subgraphsG1 andG2.

The separation scheme above and Proposition 1.8.1 easily generalize for the case ifdeletion ofS leaves any numberk ≥ 2 connected components induced byX1,X2, . . . ,Xk.The derived subgraphs then are:G1 = GX1∪S, G2 = GX2∪S, . . . , andGk = GXk∪S.

Another idea is that one can delete edges from a connected graph and obtain a discon-nected graph. Such subsets of edges are callededge-separatorsor edge-cuts. If an edgeitself is a separator, it is called abridge.

What do we need this for? The answer is that many important properties of graphs


can be successfully investigated by using derived subgraphs. Derived subgraphs have lessvertices than original graph what opens the way for mathematical induction.

We end the section with formulation of the important Menger’s Theorem:

Theorem 1.8.1 (Menger, 1927)In a connected graph G, for two non adjacent vertices xand y, the minimum number of vertices in an(x,y)-separator equals the maximum numberof (internally) vertex disjoint(x,y)-paths.

Idea behind the proof: to disconnect verticesx andy, one need to destroy every(x,y)-path. �

Exercises 1.8.

b

b

b b

b bb

b

1

2

34

5

67 8

G

Figure 1.36.

1. For every pair of disjoint vertices of graphG, see Figure 1.36, find a separator.

2. For every pair of disjoint vertices of graphG, see Figure 1.36, find a minimal separator.

3. For graphG in Figure 1.36, find the connectivityκ(G).

4. For which integerk≥ 0, graphG in Figure 1.36 isk-connected?

5. For separator{1,3,4,7,8}, construct the derived subgraphs.

6. For separator{1,8,3,4,6}, construct the derived subgraphs.

7. For separator{5,6,7,8,2}, construct the derived subgraphs

8. Apply Proposition 1.8.1 for separator{1,8,3}.

9. Find an edge-separator for graphG in Figure 1.36.

10. For graphG in Figure 1.36 find a minimal (1,4)-separator.

11. For graphG in Figure 1.36 find a minimal (1,4)-edge-separator.

12. What is the smallest number of edges that disconnect graph G in Figure 1.36?

13. Apply Menger’s Theorem to vertices 3 and 6 in graphG, see Figure 1.36.


Computer Projects 1.8. Using an appropriate graph representation, write a programfor thefollowing algorithmic problems.

1. For a given subset of vertices in a graphG, find out if the subset is a separator.

2. For a separator of a graphG, determine if the separator is minimal.

3. Find the connectivity of a graphG.

4. For a given subset of edges in a graphG, find out if the subset is an edge-separator.

Chapter 2

Trees and Bipartite Graphs

“– Why are these graphs called “trees”?

2.1. Trees and Cycles

Theorem 2.1.1 For a simple graph G, the following statements are equivalent:

1. G is a tree;

2. G is connected and m(G) = n(G)−1;

3. G has no cycles and m(G) = n(G)−1;

4. There is a unique path connecting any two vertices of G;

5. G has no cycles and connecting any of its two nonadjacent vertices by an edge resultsin precisely one cycle.

Proof. 1. ⇒ 2. G is connected by the definition of tree. Prove the equality by inductionon n. For n = 1,2 the statement is trivial. Assumen(G) > 2. SinceG is a tree it has apendant vertex; strongly delete it and obtain a treeG1. ForG1 by the induction hypothesis,m(G1) = n(G1)−1. Now return deleted vertex with an edge and reconstructG. Evidently,m(G) = m(G1)+1, andn(G) = n(G1)+1, and the implication follows.

2. ⇒ 3. Prove thatG has no cycles. By contradiction, suppose thatG has a cycle.Weakly delete an edgee= {x,y} of this cycle;G−e is connected since(x,y)-path remains.If G−econtains cycles, repeat the procedure until graph obtainedcontains no cycles. Sinceit is connected, it is a tree, and by 1.⇒ 2. m= n−1, a contradiction to the fact that wedeleted at least one edge.

3.⇒ 4. Let G have no cycles andm= n−1. It may have many components. If it hask > 1 componentsG1,G2, . . . ,Gk, everyGi is a tree. By 1. ⇒ 2. mi = ni − 1 for eachGi.Thereforem(G) = m1 + m2 + · · ·+ mk = (n1 − 1)+ (n2 − 1)+ · · ·+ (nk − 1) = n− k. Sowe obtainm= n−1 = n− k which impliesk = 1 andG is connected. It means that thereis a path between any pair of vertices. If there are two different paths connecting a pair ofvertices, then there is a cycle what contradicts to condition 3. Therefore, 4. holds.


4.⇒ 5. G cannot have cycles because in any cycle any two vertices are connected bytwo different paths. Letx,y be two nonadjacent vertices. Hence there is a unique(x,y)-path. Therefore adding the edge{x,y} to G produces a cycle. This cycle is unique becauseotherwise, if we obtain two cyclesCk andCl , the subgraph(Ck∪Cl)−{x,y} forms the thirdcycle, a contradiction to 4.

5.⇒ 1. G has no cycles. If it is not a tree (i.e. it is disconnected), then connect twovertices from different components by an edge. We obtain no cycles, a contradiction. �

Let G be a simple graph havingk connected components. The numberΛ(G) = m(G)−n(G) + k is called thecyclomatic number of G. If G is connected, thenk = 1 andΛ =m−n+ 1 = m− (n−1). Theorem 2.1.1 in fact states thatn−1 is the minimum numberof edges for a graph to be connected, or, equivalently, to be atree. Therefore, the numberm−(n−1) shows how many extra edges graphG has. Starting from any spanning tree, onecan sequentially add remainingΛ edges to reconstructG. Every such edge forms preciselyone cycle with spanning tree. These cycles are calledelementary and edges are calledchordswith respect to the spanning tree.

If G is disconnected, we apply the same reasoning to each component and replace “tree”with “forest”. In other words, the cyclomatic number indicates “how far” graphG is fromthe forest. That is why it is called the “cyclomatic number”.

Indeed, the following holds:

Corollary 2.1.1 Λ(G) = 0 if and only if G is a forest.

b

b

b

b

b

G

Figure 2.1. Spanning tree, chords and cycles.

An example of a graph and its cyclomatic number is shown in Figure 2.1. The solidedges form a spanning tree, the dashed edges show the chords.Correspondingly,Λ(G) =m−n+1= 7−5+1 = 3. One can see that every chord forms exactly one elementary cyclewith the spanning tree. The total number of cycles however, is greater thanΛ. There areadditional cycles that can be expressed as combinations of elementary cycles.

Generally, a graph may have many spanning trees, and therefore many different sets ofelementary cycles; important however is that any cycle can be expressed as a combinationof elementary cycles and the number of elementary cycles is always the same, namely equalto the cyclomatic number.

Let us agree that two spanning trees ofKn are considered different if they are formedby different sets of edges; in fact, some of them may be isomorphic.

Theorem 2.1.2 (Cayley’s Formula, 1889)The number of spanning trees in graph Kn,n≥ 1, equals nn−2.

Trees and Bipartite Graphs 41

Since any tree is a connected graph, with the agreement aboveCayley’s formula impliesthat there are totalnn−2 trees for everyn. For example, ifn = 3, then there are three“different” trees which in fact all are isomorphic to each other.

Exercises 2.1.

1. Find the cyclomatic number ofEn, Cn, Kn, Wn, prism, cube and the Petersen graph.

2. Find all spanning trees for graphG in Figure 2.1.

3. Describe the procedure for finding all spanning trees inC6, K4, W6.

4. Describe the procedure for finding all spanning trees in cube and prism.


1. Generate all spanning trees in prism, cube and the Petersen graph.

2. Generate all spanning trees inKn, Wn, n≥ 5.

2.2. Trees and Distance

GraphEn is calledtrivial . Unless stated otherwise, we consider nontrivial graphs. It meansthat any graph has at least two vertices and at least one edge.

Let G = (X,E) be a graph,x,y∈ X. Thedistancefrom x to y denoted byd(x,y) is thelength of the shortest(x,y)-path. If there is no such path inG, thend(x,y) = ∞; evidently,in this caseG is disconnected andx andy are in different components. Any segment of ashortest path is a shortest path itself. The distance between x and a set of verticesY ⊆ X isdefined asd(x,Y) = miny∈Y d(x,y). It is the shortest distance betweenx and any vertex ofY.

For allx,y,z∈ X, the following properties of distances hold:

1. d(x,y) ≥ 0, andd(x,x) = 0;

2. d(x,y) = d(y,x);

3. d(x,y) ≤ d(x,z)+d(z,y) (triangle inequality).

The diameter of G denoted bydiam(G) is maxx,y∈X d(x,y); in other words it is thedistance between the farthest vertices. For connected graphs, diameter is a positive integernumber. An(x,y)-path for whichd(x,y) = diam(G) is called adiametral path. There maybe many diametral paths in a graph. LetN∞(x) denote the set of farthest vertices from vertexx. It means that ify ∈ N∞(x) andz 6∈ N∞(x), thend(x,z) < d(x,y). The distance betweenvertexx and setN∞(x) is called theeccentricity of x. Thecenter of G is a set of verticesof minimum eccentricity. Theradius of G is equal to eccentricity of any vertex from thecenter. At last, a vertex of degree 1 is calledpendant vertexor leaf.

Theorem 2.2.1 Let x∈ X be an arbitrary vertex in a tree T . Then every vertex y∈ N∞(x)is pendant.


b b b b b

b

x yz

z′

Figure 2.2.

Proof. Let y∈ N∞(x). Consider(x,y)-path and a vertexz on this path adjacent toy, seeFigure 2.2. By contradiction, assume there is another vertex z′ adjacent toy. The shortest(x,z′)-path cannot usey because it would be longer thand(x,y). Therefore, it must usevertexzor any other vertex from(x,y)-path. In any case we obtain a cycle (at least triangle)what contradicts thatT is a tree. �

Corollary 2.2.1 Any tree has at least two pendant vertices.

Proof. Considerx,y such thatd(x,y) = diam(T). Evidently, x ∈ N∞(y) andy ∈ N∞(x),hence by Theorem 2.2.1 bothx andy are pendant. �

Theorem 2.2.2 (Jordan, 1869)The center of a tree is either K1 or K2.

Proof. Notice that strong deletion of any pendant vertex from a treeT leaves all distancesbetween remaining vertices unchanged. Delete all pendant vertices fromT, obtain treeT1.Since we delete all pendant vertices, by Theorem 2.2.1, the eccentricity of each vertex inT1

is less than the eccentricity of the same vertex inT by 1. Therefore vertices with minimumeccentricity inT andT1 are the same. Repeat the procedure as many time as possible. Weobtain a sequence of treesT1, T2, T3, . . . , the last beingK2 or K1 what is the center. �

The theorem explicitly presents an algorithm how to find the center of a tree. One canprove thatdiam(T) can be found in the following two steps: 1) start at any vertexzand findanyx∈ N∞(z); 2) repeat the procedure forx and find anyy∈ N∞(x): diam(T) = d(x,y).

Exercises 2.2.

1. For the tree in Figure 2.3, find the diameter, radius and center.

2. For each vertexx in the tree in Figure 2.3, findN∞(x) and the eccentricity.


1. Given any tree, find the diameter.

2. Given any tree, find the center.

3. Given any tree and a vertexx, find N∞(x).


b b b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

bb

b b

b

b

Figure 2.3.

2.3. Minimum Spanning Tree

Sometimes, as in this section, we use lettersD for an edge set andD for its elementsbecause we need letterE for another purpose. A graphG = (X,D) is calledweightedif each edgeD ∈ D is assigned a positive real numberw(D) called theweight of edgeD. Usually, in many practical applications, the weight represents a distance, time, cost,capacity, resistance, probability, etc. Consider any spanning treeT = (X,E) of G. Theweight w(T) of treeT is the sum of weights of all edges ofT. Different spanning trees mayhave different weights. The problem of finding a spanning tree of minimum (maximum)weight is called theminimum (maximum) spanning tree problem. It is one of the fewoptimization problems that allows an efficient algorithm for any graph.

Algorithm 2.3.1 Finding minimum spanning tree (Kruskal’s algorithm)INPUT: A connected weighted graph G.OUTPUT: A tree T of minimum weight.

1. Order the edges of G in increasing (non-decreasing) orderof their weights and set Tto be an empty graph.

2. Add the first edge from the ordering to T .

3. Consider the next edge in the ordering. If it produces a cycle in T with alreadyincluded edges, skip it. Otherwise, include it in T .

4. Repeat step 3. until T is connected.

5. Output T .

Theorem 2.3.1 (Kruskal, 1956)For any connected weighted graph G, Algorithm2.3.1constructs a spanning tree of minimum weight.

Proof. First of all, Algorithm produces a tree because it does not create cycles and can-not end whenT is disconnected. SinceG is connected, some edges between connectedcomponents ofT exist and at least one must be included inT.


SupposeT is not an optimal spanning tree, i.e. there exist a spanning treeT∗ 6= T ofminimum weight such thatw(T∗) < w(T). Let D be the first edge in the ordering that waschosen forT and is missing inT∗. Adding D to T∗ creates a unique cycle. SinceT hasno cycles, this cycle contains an edgeD′ which is not inT. Construct the spanning treeT∗ + D−D′. Now notice thatw(D) ≤ w(D′), and thereforew(T∗ + D−D′) = w(T∗) +w(D)−w(D′) ≤ w(T∗). We obtain a new spanning tree of the minimum weight which hasone edge more in common withT. Repeating this procedure as many times as necessary,we eventually obtainT what proves that it is an optimal spanning tree. �

Algorithm and Theorem work if we need to find a spanning tree ofmaximum weight;we just proceed in inverse ordering of the edges.

Exercises 2.3.

b

b

b b

b

b b

b2

3

1

41

10

1

23 9

10

11

Figure 2.4. Weighted graphG with integer weights.

1. Choose any spanning tree in graphG, see Figure 2.4, and compute its weight.

2. In graphG in Figure 2.4, find the minimum spanning tree in two differentways: a) intuitively;b) by using Kruskal’s algorithm.

3. In graphG in Figure 2.4, find the maximum spanning tree in two differentways: a) intuitively;b) by using Kruskal’s algorithm.

4. Given weighted complete graphKn with all weights equalw > 0, how many minimumweighted trees does it have?


1. Given a weighted graphG, find a spanning tree and compute its weight.

2. Given a weighted graphG and a spanning tree; check if the tree is of maximum (minimum)weight.

3. Given a weighted graphG, implement the Kruskal’s algorithm.

4. Given a weighted graphG, find the maximum spanning tree.


2.4. Bipartite Graphs

Theorem 2.4.1 (Konig, 1936) A graph G is bipartite if and only if it does not have oddcycles.

Proof. ⇒ Let G be a bipartite graph with partsA andB, andCk be any cycle in it with avertexx∈ A. TraverseCk starting atx in any direction. SinceG is bipartite, each time wealternate partsA andB. Since we end atx, k is even.

⇐ Let G = (X,E) be a connected nontrivial (n≥ 2) graph without odd cycles; chooseanyx∈ X. Denote{x} by N0, neighborhoodN(x) by N1, vertices at distance 2 fromx byN2, vertices at distance 3 fromx by N3 and so on, The last set in this sequence isN∞(x)denoted byNk. We obtain the following partition ofX:

X = N0∪N1∪N2∪N3∪ . . .∪Nk.

Observe thatNi ∩Nj = /0, i 6= j. Moreover, any edge ofG connects vertices either fromthe sameNi or from consecutive sets.

x x′

y

z

bb

bb

b

b

N0 N1 Ni N∞

Figure 2.5.

Now form a bipartition ofG by placing inA all sets with even indices and placing inB all sets with odd indices, soX = A∪B. It remains to prove that there are no edges ofGinside any of setsNi. By contrary, assume there is an edgee= {y,z}, such thaty,z∈ Ni forsomei > 0, see Figure 2.5. Consider the shortest(x,y)-path and(x,z)-path. They may havecommon vertices other thanx. Recall that every segment of a shortest path is also a shortestpath. Letx′ be the last common vertex counting fromx. Then(x′,y)-path and(x′,z)-pathhave the same length, sayl , and no common vertices other thanx′. Therefore, these pathsand edgee form a cycleC2l+1, a contradiction. �

The idea of the proof of Konig’s theorem allows to suggest a simple algorithm which notonly gives the possibility to recognize bipartite graphs but is also used for finding solutionsof many other problems.

Algorithm 2.4.1 (Breadth-first search)

INPUT: A graph G= (X,E)OUTPUT: A labeling of vertices of G by the numbers 0,1,2,. . .

1. Start at arbitrary unmarked vertex and mark it with 0; set i= 0;

2. Mark by i+1 all vertices which are not marked and adjacent to vertices marked i;


3. If there are unmarked vertices adjacent to marked vertices, set i:= i + 1 and go tostep 2;

4. End.

Clearly, each marked vertex ofG has a mark (label)i which equals the distance to orfrom the initial vertex. To check whetherG is bipartite it is sufficient to add one morestep: verify that no two vertices with equal labels are adjacent. If two different verticeswith the same label induce an edge, then there is an odd cycle,and by Theorem 2.4.1,Gis not a bipartite graph. IfG is connected, then Algorithm 2.4.1 marks all the vertices. IfG is not connected, then some vertices remain unmarked; one can run it again starting atany unmarked vertex. The number of re-runs will coincide with the number of connectedcomponents ofG.

To find the distance, i.e. the length of the shortest path between any two vertices ofG, itis sufficient to run the algorithm starting at any of these vertices, and the label of the secondvertex will be the distance.

If G is a directed graph, then Algorithm 2.4.1 may be used to find the set of vertices thatcan be reached from a given vertexx of G.

As the opposite to the breadth-first algorithm, there is another way of search in a graphwhich is called thedepth-first search algorithm. We next describe the idea of the algo-rithm.

Assume we need to find a spanning tree in a connected graphG. Choose a vertex anddeclare it visited. Choose any unvisited vertex adjacent tothe last visited vertex, declareit “current” and add connecting edge to the spanning tree. Ifthere are edges connectingcurrent vertex with other visited vertices, declare them asback edges. Choose anotherunvisited vertex adjacent to the last visited vertex and repeat this procedure as long aspossible. We get stuck at a vertex which has no unvisited neighbors. At this point, wereturn back to the vertex from which the current vertex was visited (this step is calledbacktracking) and look for another unvisited neighbor. If there are such,we visit oneand repeat the procedure. If there are none, we return one more step back and repeat theprocedure again. We add to the spanning tree one edge at a timewhen visiting a new vertex;the back edges are not added because they form cycles with theedges of the tree. Eventuallyalgorithm stops at the very first visited vertex. That vertexis called theroot of the spanningtree. As the breadth-first search, the depth-first search algorithm can be applied to solve anumber of different search problems.

We now continue the discussion of bipartite graphs. LetG = (X,E) be a graph. Theneighborhood N(S) of a subsetS⊆ X is the union of all neighborhoods of the verticesfrom SminusS itself, i.e.,

N(S) = ∪x∈SN(x)\S.

If G = (X,Y;E) is a bipartite graph, then the neighborhood of any subset ofX is inY, and the neighborhood of any subset ofY is in X. We will use the notationNG(S) toemphasize that the neighborhood is considered in a graphG.


Recall that a matching is a set of pairwise disjoint edges. Wesay that amatchingcovers a set of verticesif each vertex from the set is incident to an edge of the matching.

Theorem 2.4.2 (Hall, 1935)A bipartite graph G= (X,Y;E) has a matching that covers Xif and only if for every subset S⊆ X,

|NG(S)| ≥ |S|.

Proof. ⇒ If G = (X,Y;E) has a matching that coversX, then evidently, any subsetS⊆ Xhas at least|S| neighbors inY, i.e.,|NG(S)| ≥ |S|.

⇐ Let G = (X,Y;E) be a bipartite graph. We prove the sufficiency of the theorem byinduction on|X|. If |X| = 1, then by the definition of bipartite graph, the single vertexx ofX has at least one neighbory in Y, thus the edgexy is the required matching.

Let now |X| > 1 and the theorem be true for all bipartite graphs with the first part on< |X| vertices. There are two cases to consider.

Case 1:for any subsetS⊂ X, S 6= X,

|S| < |NG(S)|. (2.1)

Consider an arbitrary edgexy of G. Delete strongly verticesx andy from G and denotethe bipartite graph obtained byG′ = (X′,Y′;E′). Evidently, |X′| = |X| − 1 < |X|. Sinceprecisely one vertex is deleted fromY, the inequality (2.1) implies that inG′ for any subsetS′ ⊆ X′, the following holds:

|S′| ≤ |NG(S′)|.

By the induction hypothesis, there exists a matching inG′ that coversX′. Add the edgexy to it and obtain a matching that coversX in graphG.

Case 2:there exists a subsetS0 ⊂ X, S0 6= X, such that

|S0| = |NG(S0)|. (2.2)

In Figure 2.6, which illustrates this case, the respective sets are shown by ellipses. We nowsplit the vertex set ofG into two subsets:

A = S0∪NG(S0) and B = (X∪Y)\A

and consider two induced bipartite subgraphs,GA andGB.In GA, for any subsetS⊆ S0, we have thatNGA(S) = NG(S). Therefore,|S| ≤ |NGA(S)|.

By the induction hypothesis,GA has a matching that coversS0.In GB, for any subsetS⊆ X−S0, we have that

NGB(S) = NG(S)\NG(S0),

see Figure 2.6. Therefore,

|S0|+ |S| = |S0∪S| ≤ |NG(S0∪S)| = |NG(S0)|+ |NGB(S)|.


X Y

S0 NG(S0)

S

X−S0

NG(S)

GA

GB

Figure 2.6.

Since|S0| = |NG(S0)| by (2.2), we obtain that|S| ≤ |NGB(S)|. Hence by the induction hy-pothesis, graphGB has a matching that coversX−S0.

Combining matchings ofGA andGB we obtain a matching ofG that coversX.�

Corollary 2.4.1 Every regular bipartite graph has a perfect matching.

Proof. Let G = (X,Y;E) be ak-regular (k ≥ 1) bipartite graph. Counting the degrees ofvertices inX and inY leads to the equalityk|X| = k|Y| what implies|X| = |Y|. It meansthat every matching that coversX, also coversY.

Let S⊆X. SinceG is k-regular, the numberi of edges fromSto NG(S) is i = k|S|. Sinceeach vertex fromNG(S) has degreek, i ≤ k|NG(S)|; therefore, for anyS⊆ X, |NG(S)| ≥ |S|.By Theorem 2.4.2,G has a perfect matching. �

Recall thatτ(G) is the cardinality of a minimum transversal of a graphG, i.e. theminimum number of vertices that “touch” all edges. As we mentioned in Section 1.7., forany graphG, τ(G) ≥ ν(G).

Theorem 2.4.3 (Konig, 1931) For any bipartite graph G,

τ(G) = ν(G).


A

B

S NG1(S)

X Y

G1

G2

Figure 2.7.

Proof. Let G = (X,Y;E) be a bipartite graph andT ⊆ (X∪Y) be a minimum transversalof G. We will construct a matching of sizeτ(G) = |T| what proves the theorem.

Let X∩T = A andY∩T = B, andG1 = GA∪(Y\B) andG2 = GB∪(X\A), see Figure 2.7.SinceA∪B is a transversal, there are no edges betweenY \B andX \A. For eachS⊆ A,considerNG1(S). If |NG1(S)| < |S|, then, becauseNG1(S) “touches” all edges incident toSthat are not “touched” byB, we could replaceSbyNG1(S) and obtain a smaller transversal ofG thanT. SinceT is a minimum transversal, this is impossible, and therefore, |NG1(S)| ≥ |S|for any subsetS⊆ A. By Theorem 2.4.2, graphG1 has a matching that coversA. Applyingthe same reasoning to graphG2, we obtain a matching inG2 that coversB. Since graphsG1 andG2 have disjoint vertex sets, we combine these two matchings inone matching ofgraphG that has|A|+ |B|= |T| = τ(G) edges. Henceτ(G) = ν(G). �

Exercises 2.4.

1. Run the breadth-first search algorithm for prism, cube andPetersen graph starting at an arbi-trary vertex.

2. For integersm,n≥ 1, formulate the conditions whenKm,n has a perfect matching.

3. Find all perfect matchings in the cube.

4. For integersm,n≥ 1, find τ(Km,n) andν(Km,n).

5. Run the breadth-first search algorithm for graph in Figure2.8 to determine if it is bipartite.

6. For graph in Figure 2.8, find the values ofτ andν and the respective transversal and matching.


1. Given an arbitrary graphG, run the breadth-first search algorithm to recognize ifG is bipar-tite.


b b b b b b b

b b b b b b

b b b b b b

b b b b b b b

Figure 2.8.

2. Given an arbitrary graphG, run the breadth-first search algorithm to recognize ifG is con-nected.

3. Given an arbitrary graphG, run the breadth-first search algorithm to recognize ifG is a tree.

4. Repeat projects 1-3 using the depth-first search algorithm.

5. Given a graphG and a subset of vertices, determine if the subset is a transversal.

6. Given a graphG and a subset of edges, determine if the subset is a matching.

Chapter 3

Chordal Graphs

“Smart people find shorter ways, i.e. the chords in common ways...”

3.1. Preliminary

If a connected graphG is not a tree, then it has cycles. Some cycles may have two non-consecutive vertices which are adjacent inG. The edge connecting them is achord, or adiagonal of the cycle. GraphG is calledchordal if everycycle of length≥ 4 has a chord.Since cyclesCk, k ≥ 4, have no chords as separate graphs, chordal graphs cannot containthem as induced subgraphs. In other words, if a chordal graphcontainsCk, k≥ 4, then noneof them is induced. Notice that cyclesC1,C2, andC3 do not have non-consecutive verticesand therefore cannot have chords; it explains why “chordality” begins with the cycles oflength≥ 4.

A chord splits a cycle into two smaller cycles; if graph is chordal and at least one of thecycles is not a triangle, then it has another chord, and so on.Eventually, every cycle is splitinto a number of triangles. That is why chordal graphs are also known as “triangulated” or“rigid circuit” graphs.

As follows from the definition, the smallest graph which is not chordal isC4. Figure 3.1shows three graphs among whichG1 andG3 are chordal andG2 is not. ForG1, one canmanually check that all cycles of length≥ 4 have chords;G3 is a tree and has no cycle atall; G2 contains an induced cycleC4 shown by dashed edges.

Every graph is either chordal, or not. If a graph is not chordal, then it contains aninduced cycleCk,k≥ 4. Since trees contain no cycles, they all are chordal graphs.

A vertex is calledsimplicial if all of its neighbors are pairwise adjacent, i.e., the neigh-bors induce a complete subgraph. It follows that a vertex is not simplicial if it has twodisjoint neighbors.

In Figure 3.1, simplicial vertices are labeled by “s”. We will see that chordal graphsalways have simplicial vertices. Moreover, as trees are thespecial case of chordal graphs,in the same way pendant vertices represent the special case of simplicial vertices.

Proposition 3.1.1 In a chordal graph, every induced subgraph is chordal.


b b

b

b

b

b b

b

b

b

b

b

b

b bG1 G2 G3

s

s

s

s s

s

Figure 3.1.

Proof. If in a chordal graph an induced subgraph is not chordal, thenit contains an in-duced cycleCk,k ≥ 4. Evidently, Ck is an induced subgraph in the original graph, whatcontradicts that it is chordal. �

The above property is important because it allows to prove many results about chordalgraphs by induction: if we delete any vertex from a chordal graph, we obtain a chordalgraph again.

3.2. Separators and Simplicial Vertices

“Separate and dominate – how old is that?”

Theorem 3.2.1 (Minimal separator theorem) A graph is chordal if and only if every min-imal separator is a clique.

Proof. ⇒ Let G= (X,E) be a chordal graph with a minimal separatorS. Assume we havetwo derived subgraphsG1 = GX1∪S andG2 = GX2∪S, see Figure 3.2.

X1 S X2

b

b

bb

bb

x

y

Figure 3.2. Minimal separator in a chordal graph.

We need to prove thatGS is a clique. Consider any pair of verticesx,y∈ S. SinceS isa minimal (by inclusion) separator, each ofx,y has neighbors in bothX1 andX2. Choosea shortest(x,y)-path throughX1; its length is at least 2. Next choose a shortest(x,y)-paththroughX2; its length is also at least 2. Combining these two(x,y)-paths we obtain a cycle

Chordal Graphs 53

Ck,k ≥ 4. Since there are no edges betweenX1 andX2 andG is chordal, verticesx andymust be adjacent. Sincex andy are arbitrary vertices ofS, all vertices ofS are pairwiseadjacent, i.e.,GS is a clique.

In the case when there are more than two derived subgraphs, itis sufficient to considerany two of them.

⇐ Let G = (X,E) be a graph in which every minimal separator is a clique. By contra-diction, assumeG is not chordal. Then it contains an induced subgraphCk, k≥ 4. Considertwo nonadjacent vertices ofCk, say,x andy. Ck is formed by two different(x,y)-paths.Sincek≥ 4, each such path contains at least one internal vertex. SinceCk is induced, inter-nal vertices of the first path are not adjacent to the internalvertices of the second path. Nownotice that any minimal(x,y)-separator contains at least one internal vertex from each path,and therefore contains two nonadjacent vertices, a contradiction. �

Theorem 3.2.2 (Dirac theorem, 1961)Every noncomplete connected chordal graph con-tains at least two simplicial vertices which are not adjacent.

Proof. Let G = (X,E) be a connected chordal graph which is not a clique; hence|X| =n≥ 2. We prove the statement by induction onn. The theorem is evident forn= 3. Assumethe theorem is true for all noncomplete connected chordal graphs on less thann vertices andprove it forG.

SinceG is not a complete graph, there are two vertices, sayx and y, which are notadjacent. Hence vertex setX −{x,y} is a separator. Choose any minimal separator fromthis set and denote it byS. By Theorem 3.2.1,GS is a clique.

Suppose deleting ofS from G leaves two connected components,GX1 and GX2; letG1 = GX1∪S andG2 = GX2∪S be two derived subgraphs with respect toS, see Figure 3.2.GraphsG1 andG2 both are chordal as subgraphs ofG.

Consider subgraphG1. If it is a clique, then every vertex fromX1 being simplicial inG1

is simplicial inG because there are no edges betweenX1 andX2. SupposeG1 is not a clique.It is connected becauseGX1 is connected andGS is a clique. It has< n vertices becauseX2 6= /0. By the induction hypothesis,G1 contains at least two simplicial vertices which arenot adjacent. SinceGS is a clique, both simplicial vertices cannot be inS. Therefore, oneof them is inX1. Again, since there are no edges betweenX1 andX2, the simplicial vertexfrom X1 is a simplicial vertex inG.

Consider subgraphG2 and apply the same reasoning. We obtain that inG there are twosimplicial vertices which are not adjacent.

If deletion ofS from G leavesk ≥ 3 connected components, apply the same reasoningto each component; evidently, in such caseG will contain k simplicial vertices which arepairwise not adjacent. �

Observe that the statement of Dirac theorem may be extended to any complete graphKn, n ≥ 2, where all vertices are simplicial, with the exception that all they are pairwiseadjacent.

Corollary 3.2.1 Every nontrivial tree contains two pendant vertices.

Proof. Any tree being a chordal graph by Theorem 3.2.2 must have two simplicial verticeswhich in this case are pendant. �


Let G1 = (X,E) be a chordal graph; it contains a simplicial vertex, denote it by x1.Deletex1 from G1 and denote the graph obtained byG2. GraphG2 is a subgraph ofG1

and therefore is chordal. It contains a simplicial vertex, denote it byx2. Deletex2 from G2,obtain another chordal graphG3, which has a simplicial vertexx3; delete it and continue thisprocedure on. At every step we obtain a chordal graph and apply Theorem 3.2.2. The laststep in this procedure will occur when we delete the last vertex xn and arrive toGn+1 = /0.We obtain the ordering of verticesx1,x2, . . . ,xn. Denote it byσ, soσ = (x1,x2, . . . ,xn). Theorderingσ is called asimplicial elimination ordering or a perfectelimination ordering.The procedure of deleting the vertices in orderingσ is called thesimplicial decomposition.Its main feature is that every vertexxi is a simplicial vertex in graphGi induced by verticesxi ,xi+1, . . . ,xn andGi+1 = Gi −xi, i = 1,2, . . . ,n.

It follows from the observations above that any chordal graph has an orderingσ, or,equivalently, a simplicial decomposition. But if a graph has a simplicial elimination order-ing, is it chordal then? Suppose it is not; then it contains aninducedCk,k≥ 4. If a simplicialdecomposition exists, then sooner or later, the first vertexof Ck must appear in it. However,no vertex ofCk is simplicial in any induced subgraph, a contradiction.

Summarizing the observations above, we arrive to the following conclusion:

Theorem 3.2.3 (Simplicial decomposition theorem)A graph G is chordal if and only ifit has a simplicial elimination ordering.

Figure 3.3 shows graphG1 and its simplicial elimination ordering. Respectively, onecan construct a sequence of graphsG2, G3, . . . ,G8,G9 = /0 by sequential deletion of simpli-cial vertices in order 1, 2, 3,. . . , 8, soσ = (1,2,3,4,5,6,7,8). Observe that there are onlytwo simplicial vertices inG1: 1 and 2. However, each of the remaining vertices becomessimplicial at some step.

b

b b b

b

bbb

1 2

34

5

6

7 8

G1

Figure 3.3.G1 and simplicial elimination orderingσ = (1, 2, 3, 4,5, 6, 7,8).

The power of simplicial elimination ordering is that in determining if a graph is chordal,it is not necessary to investigate all subsets of sizes 4, 5, 6, . . . , and check if every subsetdoes not induceCk; instead, it is sufficient to find at least one simplicial elimination orderingor show that none exist. In other words, searching for a simplicial elimination orderingreplaces an exhaustive search for recognition of chordal graphs.

For example, consider graphG1, see Figure 3.3. If we use only the definition of chordalgraph, then we have to check

(84

)

subsets of size 4 for the case if any inducesC4. Thenrepeat the procedure forC5, it would give us

(85

)

subsets. Then we would have to consider(8

6

)

subsets forC6,(8

7

)

subsets forC8 and(8

8

)

subsets forC8. Visualization ofG could save

Chordal Graphs 55

some steps but only in such small examples. However, using the last theorems, instead ofthat long procedure, we just look for the simplicial elimination ordering.

Non-chordal graphs may have simplicial vertices but if we start simplicial decompo-sition, sooner or later we get stuck because in obtained graph no vertex is simplicial. InFigure 3.1, graphG2 is not chordal and simplicial decomposition gets stuck after removingof the unique simplicial vertex.

Theorem 3.2.4 In a chordal graph, any vertex may be the last vertex in some simplicialelimination ordering.

Proof. Indeed, letG be a chordal graph andx be any vertex. SinceG has at least twosimplicial vertices at any step of simplicial decomposition, we can avoid deletingx at everystep, except the last one. �

For example, in Figure 3.3 vertex 1 being the first in orderingσ will be the last inorderingσ′ = (2,3,8,7,6,5,4,1).

In a chordal graph, if any vertexz may terminate someσ, then any vertexy adjacentto z may be the last but one. Then any vertexx forming a triangle withz andy may bethe third from the end. If we develop this procedure further,then if graph is chordal, wecould reconstruct an ordering inverse to simplicial elimination ordering. The main point isthat we can start at any vertex. This idea is at the base of the most efficient algorithm forrecognizing chordal graphs, the so called “Maximum Cardinality Search”.

Theorem 3.2.5 If x is a simplicial vertex in a graph G= (X,E), then there exists a maxi-mum independent set S⊆ X such that x∈ S.

Proof. Sincex is simplicial, every maximum independent setS′ contains precisely onevertex from set{x} ∪N(x) because otherwise the setS′ ∪{x} would be independent andhave greater cardinality. If there is a vertexy∈ N(x) such thaty∈ S′, then we replace it byx and putS= S′ \{y}∪{x}. �

Notice that the theorem above holds for any graph, not necessarily chordal. However,the existence of a simplicial elimination ordering allows to suggest a simple algorithm forfinding the independence number and the respective maximum stable set of a chordal graph:

Algorithm 3.2.1 (Finding maximum stable set of a chordal graph)INPUT: A chordal graph G= (X,E)OUTPUT: A maximum independent set S⊆ X with |S| = α(G)

1. Set S= /0.

2. Find a simplicial vertex x, delete strongly{x}∪N(x) and include x in S;

3. If at least one vertex remains, repeat step 2.

4. End.


If we run the algorithm for graphG shown in Figure 3.3, then on the first step, vertex1 is included inS, and set{1,4,5} is strongly deleted; on the second step, vertex 2 isincluded inS, and set{2,3,8} is strongly deleted; at last, on the third step, vertex 6 isincluded inS, and vertices 6 and 7 are deleted. There are no more vertices left; we end andconclude that one of the maximum stable sets of graphG is S= {1,2,6}, andα(G) = 3.Respectively, vertices{3,4,5,7,8} form a minimum transversal andτ(G) = 5. Dependingon the simplicial vertices chosen at each step, there are other maximum stable sets andminimum transversals. Notice that Algorithm 3.2.1 also works for graphs which are notchordal.

Further, observe that deleted sets induce cliques; one can prove that they form a min-imum number of cliques thatcover X, namely: the cliques induced by{1,4,5}, {2,3,8}and{6,7} have the property that each vertex ofG belongs to one of them, and the numberof such cliques is the minimum.

In general, theclique cover number θ(G) of a graphG is the minimum number ofcliques in graphG such that each vertex belongs to precisely one clique. We saythat thecliquescover the vertex set ofG. As we will see, clique coverings play an important rolein graph theory. So, for the graph in Figure 3.3,θ(G) = 3.

Exercises 3.2.

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b b

bb

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b

b b b

b

b

b

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G1 G2 G3

Figure 3.4.

1. In each of the graphs in Figure 3.4 determine which vertices are simplicial and which are not.

2. For each of the graphs in Figure 3.4 determine if it is chordal or not.

3. Which of the graphs in Figure 3.4 has a minimal separator which is not a clique?

4. For chordal graphs in Figure 3.4, find all simplicial elimination orderings.

5. In each chordal graph shown in Figure 3.4, choose a vertex at random and find a simplicialelimination ordering in which the vertex is the last.

6. Run Algorithm 3.2.1 for each of the graphs in Figure 3.4.

7. For each of the graphs in Figure 3.4 find the minimum clique cover; compare the number ofcliques with the independence number.

Chordal Graphs 57

8. What is the minimum number of edges to be added to the cube (prism,Wn, Petersen graph)to make it chordal?

9. What is the minimum number of edges to be deleted from the cube (prism,Wn, Petersengraph) to make it chordal?


1. Given a graphG and a vertexx, determine ifx is a simplicial vertex.

2. Given a graphG, find out if it contains simplicial vertices.

3. Given a graphG, recognize if it is chordal.

4. Given a chordal graphG, find the maximum independence number and minimum cliquecover.

3.3. Degrees

Degree of a vertex is the number of its neighbors; it is a very general term defined forall graphs. The vertices of minimum degree play an importantrole in many optimizationproblems. Sometimes one investigate the maximum possible value of minimum degreeover all subgraphs in a given graph. When we say “the degree ofa vertex in a subgraph”we mean the number of its neighbors in the subgraph only.

It appears that chordal graphs have a special place if we investigate the degrees. For agraphG= (X,E), let us define the following parameter, known asSzekeres-Wilf number:

M(G) = maxX′⊆Xminx∈G′d(x).

As it follows from the definition, to findM(G) one need to consider all induced sub-graphsG′ of the graphG, count the minimum degree in each of them and find the maximumvalue over all of them. However, there is a simple procedure for findingM(G). It consistsin decomposition ofG by sequential elimination of vertices of minimum degree; maximumvaluet over all these minimums coincides withM(G).

Indeed, on one hand,M(G) cannot be less thant since t is the minimum degree injust one subgraph from the sequence of graphs. On the other hand, if M(G) > t, thenthere is an induced subgraphG′ ⊆ G which has minimum degree greater thant. But inthe decomposition by minimum degrees, the first vertex ofG′ appears at some stepi, seeFigure 3.5; its degree cannot exceedt by definition oft. So, we concludeM(G) = t.

Generally, notice that the degree of any vertex in subgraphG′ cannot exceed the degreeof the same vertex inGi , and moreover in the original graphG.

Proposition 3.3.1 For any graph G, M(G) ≥ ω(G)−1.

Proof. Indeed, by definition,ω(G) is the maximum size (number of vertices) of a cliquein G. Therefore, there is at least one induced complete subgraphof G with minimum degreeω(G)− 1. SinceM(G) is the maximum of minimum degrees over all induced subgraphs,the inequality follows. �


b G′ Gi

Figure 3.5.

As we see, the valueω(G)−1 in fact is the lower bound forM(G) for all graphs. Thenext theorem shows that chordal graphs realize the absoluteminimum forM(G):

Theorem 3.3.1 The following statements are equivalent:1) M(G′) = ω(G′)−1 for each induced subgraph G′ ⊆ G;2) G is chordal.

Proof. 1)⇒ 2) Let M(G′) = ω(G′)−1 for each induced subgraphG′ ⊆ G and supposeGis not chordal. ThenG contains an induced cycleCk of length≥ 4. ButM(Ck) = 2= ω(Ck),a contradiction.

2) ⇒ 1) Let G be a chordal graph. Since every subgraph of a chordal graph isalsochordal, without loss of generality, prove the equality forG. Note thatM(G) ≥ ω(G)−1.As chordal graph,G has a simplicial decomposition. Let the highest degree of a simplicialvertex in such decomposition bet. Then the size of maximum clique inG is ω(G) = t +1.The simplicial vertex of degreet is not necessarily a vertex of minimum degree, thereforewe haveM(G) ≤ t = ω(G)−1. HenceM(G) = ω(G)−1. �

One can see in the conclusion that for chordal graphs, both simplicial decompositionand minimum degree decomposition give the same maximum value of degree equal toω(G)−1. In special case, whenG is a tree, both simplicial decomposition and minimumdegree decomposition coincide. In general case, the situation is much less attractive. Forexample, for complete bipartite graphKn,n with n≥ 2, we obtainM(Kn,n) = n, ω(Kn,n) = 2,and evidently, there are no simplicial vertices.

Exercises 3.3.

1. For each of the graphs in Figure 3.6, find the value ofM(G) and compare withω(G)−1.

2. Determine which of the graphs in Figure 3.6 is chordal and which is not. For a non chordalgraph, find a minimal induced subgraphG′ such thatM(G′) > ω(G′)−1.

3. DetermineM(Cn), M(Kn), M(Wn), and the value of Szekeres-Wilf number of the cube, prismand Petersen graph.

4. Construct a graphG with arbitrarily large difference

M(G)− (ω(G)−1).

5. Prove that for anyk-regular graphG, M(G) = k.

6. Find the Szekeres-Wilf number for the complements of graphs from 3.

Chordal Graphs 59

b

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G1 G2 G3

Figure 3.6.


1. Given a graphG, find a vertex of minimum degree.

2. Given a graphG, find M(G).

3. Given a graphG and a subset of vertices; find the minimum vertex degree in thesubgraphinduced by the subset.

3.4. Distances in Chordal Graphs

Next theorem is a generalization of Theorem 2.2.1:

Theorem 3.4.1 If G is a chordal graph, then for every vertex x, the set N∞(x) of farthestvertices contains a vertex which is simplicial in G.

Proof. Without loss of generality, consider a connected nontrivial chordal graphG =(X,E). Prove the theorem by induction onn. The statement is evident forn = 2. Assume itholds for all chordal graphs on< n vertices, and|X| = n.

Let Ni be the set of vertices at distancei from x. We obtain the following partition ofX:

X = N0∪N1∪N2∪ ·· ·∪Nk−1∪Nk

whereNk = N∞(x), see Figure 3.7.

x

b

b

N0 N1 Ni Nk = N∞Nk−1

y

Figure 3.7.


If k = 1, thenN∞(x) = X −{x}. Chordal graphG− x has a simplicial vertex. Anysimplicial vertex inG− x is simplicial in G becausex is adjacent to all other vertices.Therefore in this case,N∞(x) has a vertex simplicial inG.

Let nowk ≥ 2. The setNk−1 evidently is a separator inG. It contains a minimal sepa-rator, and subgraphGNk has at least one connected component. Without loss of generality,supposeNk−1 is a minimal separator and, moreover,GNk has just one connected component.

By Theorem 3.2,GNk−1 is a clique. IfGNk−1∪Nk is a clique, then every vertex fromNk

is simplicial in G. If GNk−1∪Nk is not a clique, then as chordal graph, it contains at leasttwo simplicial vertices which are not adjacent. They both cannot be inNk−1 becauseNk−1

induces a clique. Hence at least one of them is inNk. It remains to observe, see Figure 3.7,that every vertexy∈ Nk which is simplicial inGNk−1∪Nk, is simplicial inG. �

b

b

b

b

x y

x′ y′G

Figure 3.8.

Next corollary generalizes both Theorem 3.2.2 and Corollary 2.2.1.

Corollary 3.4.1 Any connected nontrivial chordal graph has at least two simplicial ver-tices.

Proof. Let G = (X,E) be a connected nontrivial chordal graph. Consider verticesx andysuch thatd(x,y) = diam(G), see Figure 3.8. By Theorem 3.4.1, there is a vertexy′ ∈ N∞(x)which is simplicial inG. Obviously,d(x,y′) = d(x,y) = diam(G). By the same reason, forvertexy′, there is a simplicial vertexx′ ∈ N∞(y′) such thatd(y′,x′) = d(y′,x) = diam(G). �

b

b b b

b

bbb1 542

10

3

9 8 7

6

11

12G

b

b

b

b

Figure 3.9.

Next example (Figure 3.9) shows how it looks in a graphG. It is easy to see thatdiam(G) = 4. Let us start at vertex 2. So,N0 = {2}. ThenN1(2) = {1,12,10,3}. Next wefind N2(2) = {11,4,8,9}. At last,N3(2) = N∞(2) = {5,6,7}. Observe that vertices 5 and 7are simplicial.

Chordal Graphs 61

We now run the same procedure starting at, say, 5.N0(5) = {5}, N1(5) = {4,6},N2(5) = {7,8,9,3}, N3(5) = {2,10}, N4(5) = N∞(5) = {1,12,11}. We find two simplicialvertices, 11 and 5 such thatd(11,5) = diam(G) = 4. There are two diametral (11,5)-paths:11-10-3-4-5 and 11-10-9-4-5.

Notice that simplicial decomposition of a chordal graph does not change any distancein graphs obtained at each step; this is true even if graph is not chordal but has simplicialvertices.

Last example shows that in chordal graphs,N∞(x) may contain vertices which are notsimplicial. In Figure 3.9,N3(2) = N∞(2) = {5,6,7} where vertex 6 is not simplicial. There-fore Theorem 2.2.2 cannot be generalized to chordal graphs.There are other methods forfinding the center in chordal graphs.

Exercises 3.4.

b b b b

b

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b b

b b

b

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b

b

G

Figure 3.10.

1. In chordal graphG, see Figure 3.10, for each vertexx, find N∞(x) and a simplicial vertex inN∞(x).

2. Find the diameter, radius and center of the graph in Figure3.10.

3. In chordal graphG, see Figure 3.10, find a pair of simplicial vertices which arethe ends of adiametral path.

4. In chordal graphG, see Figure 3.10, find a diametral path with both ends not being simplicialvertices.

5. For chordal graphG in Figure 3.10, find a simplicial decomposition ordering; reconstructGin inverse ordering and track the change of the diameter, radius and center.


1. Given a chordal graphG and a vertexx, find all simplicial vertices fromN∞(x).

2. Given a chordal graphG find a simplicial decomposition ordering; reconstructG in inverseordering and track the change of the diameter, radius and center.


3. Given a chordal graphG, find a diametral path.

4. Given a chordal graphG, find diameter, radius and center.

3.5. Quasi-triangulated Graphs

Not only chordal graphs have many interesting properties, but they serve as an importantbase for further generalizations in graph and hypergraph theory. This section represents anexample of such generalization and its application to cyclic structure of graphs and theircomplements.

In a graphG, a vertex is calledweakly cyclic if it belongs to no inducedCk,k ≥ 4.Evidently, in any graph, simplicial vertices belong to no inducedCk,k ≥ 4; so they areweakly cyclic. Not only simplicial vertices are weakly cyclic. In chordal graphs, all verticesare weakly cyclic. The concept of weakly cyclic vertex is more general than the concept ofsimplicial vertex.

A graphG is calledlatticed if each vertex belongs to some inducedCk,k≥ 4 andsomeinducedCl , l ≥ 4. Latticed graphs are invariant with respect to taking the complement in thesense that both do not contain weakly cyclic vertices. One can think about latticed graphsas of graphs with all vertices being “strongly” cyclic. All cyclesCk,k≥ 6 are latticed: eachvertex belongs toCk andC4, see Figure 3.11.

b b

b b

b

b

b

b

Figure 3.11.

In a graphG, a vertexx is calledco-simplicial if it is simplicial in the complementG.That means its non-neighbors form an independent set of vertices, or, equivalently, no edgeentirely lies outside the neighborhoodN(x).

Definition 3.5.1 A graph G is calledquasi-triangulated if it has a decomposition by se-quential elimination of vertices that at each step, are simplicial or co-simplicial.

Quasi-triangulated graphs are invariant when taking the complement: ifG is quasi-triangulated, then its complementG is also quasi-triangulated.

For example,C4 is a quasi-triangulated graph: every vertex is co-simplicial; when wedelete any vertex, the remaining graph is chordal. GraphC4 is chordal, with all verticesweakly cyclic.

Chordal Graphs 63

CycleC5 is not quasi-triangulated because it has no simplicial or co-simplicial vertices.Moreover, it is latticed because evidently,C5 andC5 are isomorphic.

Consider all graphs that can be decomposed by sequential elimination of vertices whichat each step are weakly cyclic in graph or its complement. Howthey are related to quasi-triangulated graphs? Surprisingly, these two classes of graphs coincide. To prove this, weneed the following

Lemma 3.5.1 If G is a graph and x is a vertex that belongs to no induced Ck,k ≥ 4, thenevery minimal separator S in the neighborhood N(x) is a clique.

b

bb

b b

x

y zS

Y

Figure 3.12.

Proof. Let G be a graph andx be a vertex that belongs to no inducedCk,k ≥ 4. LetYinduce a component ofG−S that does not containx, see Figure 3.12. Choose any twoverticesy,z∈ S. SinceS is a minimal separator, each ofy andz has a neighbor inY. SinceY induces a connected component, there is a shortest path of length at least two joiningyto z throughY. This path together withx forms an inducedCk,k≥ 4, a contradiction to ourassumption onx. Therefore,y andzmust be adjacent, i.e.S induces a clique. �

Theorem 3.5.1 (Characterization theorem, Gorgos, 1984)For any graph G the follow-ing statements are equivalent:

(i) G is quasi-triangulated.

(ii) G can be decomposed by sequential elimination of weakly cyclic vertices in graph orits complement.

(iii) G does not contain latticed subgraphs.

Proof. If G is quasi-triangulated, then it can be decomposed by sequential elimination ofweakly cyclic vertices in graph or its complement because every simplicial vertex is weaklycyclic in graph and every co-simplicial vertex is weakly cyclic in the complement. In turn,if G can be decomposed by sequential elimination of weakly cyclic vertices in graph or itscomplement, then it does not contain latticed subgraphs because they are not decomposableby definition. So, (i) implies (ii), and (ii) implies (iii). Therefore, we only need to provethat (iii) implies (i).


We prove (iii)⇒(i) by induction on the number of verticesn(G). Let G = (X,E) be agraph satisfying (iii). IfG contains a simplicial or co-simplicial vertex, then we delete it, theobtained graph has< n vertices, we use the induction hypothesis and prove the implication.Therefore, assumeG contains no simplicial vertex and no co-simplicial vertex.The proofis split into the following steps.

Step 1. Proof thatG is a connected graph.SupposeG is disconnected. If at least oneconnected component is a chordal graph, then it has a simplicial vertex what was alreadyexcluded. If no one component is chordal, then each containsan inducedCk,k≥ 4. Chooseany two of them as induced subgraphs, say,Ck,k≥ 4 andCl , l ≥ 4. Considered together theyform a latticed subgraph because each vertex belongs to somecycle (Ck,k≥ 4 orCl , l ≥ 4)and to the complement ofC4, see Figure 3.13. Therefore,G is a connected graph.

b

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b

b

b

b

b

b

C4

Figure 3.13.

Step 2. Finding subgraphG′ and vertex y. SinceG is not a latticed graph, one ofG or G contains a weakly cyclic vertex. Without loss of generality, supposeG contains aweakly cyclic vertex. Denote the set of all weakly cyclic vertices byX′. Hence,X′ 6= /0.

Let G′ = G−X′. SubgraphG′ has at least one vertex, because otherwiseX′ = X, allvertices ofG are weakly cyclic, i.e.G is chordal and contains a simplicial vertex.

SinceX′ 6= /0 we conclude thatn(G′) < n(G). Evidently,G′ does not contain latticedsubgraphs. By the induction hypothesis,G′ contains a simplicial or co-simplicial vertex;denote it byy.

Since every vertex ofG′ lies in someCk,k ≥ 4, y is co-simplicial. We will prove thaty is adjacent to all vertices ofX′: this will imply y is co-simplicial in the initial graphG, acontradiction.

Step 3. Proof that y is co-simplicial in G. Let x be any vertex inX′. SinceG isconnected andx is not co-simplicial, there is a nonempty setSof vertices inN(x) that is aminimal separator ofG, see Figure 3.14 (x is not shown). By Lemma 3.5.1,S is a clique.Let G1,G2 be induced subgraphs ofG such thatG = G1 ∪G2,G1 ∩G2 = S, and there isno edge betweenG1−SandG2−S. If at least one ofG1 or G2 is chordal, then there is asimplicial vertex inG1−S, or G2−S, and thus it is a simplicial vertex inG, a contradiction.

Therefore,G1 contains aCk,k≥ 4 andG2 contains aCl , l ≥ 4. SinceS is a clique, oneedge, saye1, of the first cycle lies completely inG1−S. Similarly, there is an edge, saye2,of the second cycle that lies completely inG2−S. All endpoints ofe1,e2 are inG′. Sincey is co-simplicial inG′, all of its non-neighbors fromG′ are pairwise disjoint. Ify lies in

Chordal Graphs 65

b

b

b

b

b

b

b

b

bbe1

e2

S

y

G1

G2

Figure 3.14.

G1−S, then both ends ofe2 are adjacent non-neighbors. Ify lies in G2−S, then both endsof e1 are adjacent non-neighbors. Therefore, the only possibility for y is to be inS. ButS⊆ N(x), and we obtain thaty is adjacent to vertexx. Sincex was chosen arbitrary fromX′, y is adjacent to all vertices fromX′ and therefore is co-simplicial for the entire graphG.This final contradiction proves the theorem. �

Exercises 3.5.

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G1 G2 G3

Figure 3.15.

1. Which of the graphs in Figure 3.15 is quasi-triangulated and which is not?

2. In a graph which is not quasi-triangulated in Figure 3.15,find a latticed subgraph.

3. Construct a quasi-triangulated graph having only one vertex which is simplicial or co-simplicial.

4. For each of graphsG1,G2, andG3 in Figure 3.15, find a minimum clique cover and a maxi-mum independent set.


5. Which of the graphsCn, Kn, Wn, n≥ 3, cube, prism and Petersen graph is quasi-triangulatedand which is not?


1. Given a graphG, recognize if it is quasi-triangulated.

2. Given a graphG, recognize ifG is quasi-triangulated.

3. Given a quasi-triangulated graphG, find a maximum independent set and a minimum cliquecover.

4. Given a quasi-triangulated graphG, find a maximum independent set and a minimum cliquecover of the complementG.

Chapter 4

Planar Graphs

“Planarity is the cause of car collisions...”

4.1. Plane and Planar Graphs

Omitting some topological details we say that a continuous curve in the plane which con-nects two points (called the first and the last) and has no intersection with itself is called aJordan curve. A Jordan curve isclosedif the first and the last points coincide. We will usethe following

Theorem 4.1.1 (Jordan Curve Theorem)A closed Jordan curve L partitions the planeinto precisely two regions, bounded and unbounded, each having L as boundary.

It is clear that the unbounded region contains the infinite point. A region isconnectedinthe sense that any pair of its points can be connected by a Jordan curve which lies inside theregion. In drawing graphs, the vertices are represented by points in the plane, and the edgesare represented by Jordan curves connecting the respectivepoints. Evidently, a segment ofa straight line is the simplest case of a Jordan curve. If two Jordan curves intersect at a pointdifferent from the first and the last for each of them, then we say that the respective edgescross (intersect), or, alternatively, we have anedge-intersectionor crossing.

A planar graph is a graph thatcan be drawnin the plane without crossings of theedges. If a planar graphis drawn in the plane without intersections, then such drawing iscalled aplane graph.

Sometimes a plane graph is calledplane embeddingof a planar graph. Plane graphdivides the plane into connected regions calledfaces. So each face is bounded by somecycle. The number of vertices of such cycle is called thesize of the face.For a plane graphG, we will denote the number of faces byf (G) or simply by f .

Any planar graph may have several plane embeddings. Figure 4.1 shows three drawingsof the same graphK4; only the first two of them are plane graphs. One can see that inthefirst drawing, facef1 is bounded by the cycle 1-2-4-1, while in the second by the cycle 1-2-3-1 and so on. The number of faces in both embeddings is 4; there is always one face whichis not bounded. It is called theunbounded face. The unbounded face in both embeddingsof K4 is denoted byf4.


b b

b

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f1 f2

f3

f4

b b

bb

1

2 3

4

f2

f3

f4

f1

b b

bb

1

2 3

4

Figure 4.1. Three different pictures ofK4: only the first two are plane graphs.

Having the third drawing ofK4 with one crossing, we could re-draw it to obtain one ofthe first two plane embeddings. However, it is not possible todo that for every graph. Gen-erally, graphs may have different plane embeddings with different number of crossings. Theminimum number of crossings over all possible drawings is called thecrossing numberofa graph. Clearly, for planar graphs the crossing number is 0.

Exercises 4.1.

b

b

bb

b

b

b

b

b

b

b

G1 G2

Figure 4.2.

1. For each of the graphsG1 andG2 in Figure 4.2, find the number of crossings. AreG1 andG2

plane graphs?

2. For each of the graphsG1 andG2 in Figure 4.2, find a plane embedding, denote the faces andfind the size of each face; find the cycle which forms the unbounded face.

3. AreG1 andG2 planar graphs?

4. Connect the two lower vertices in graphG1 (Figure 4.2) by an edge and, if possible, find aplane embedding of the obtained graph.

5. Connect the two upper vertices in graphG2 (Figure 4.2) by an edge and, if possible, find aplane embedding of the obtained graph.

Planar Graphs 69

6. Find at least two different plane embeddings of the prism,cube,K2,3 andWn (n≥ 4).

7. Find a drawing of the Petersen graph with the smallest number of edge-intersections.


1. Given a drawing of a graph in the plane (vertices = points, edges = straight line segments)find out if it is a plane graph.

2. Given a drawing of a graph in the plane (vertices = points, edges = arc segments) find out ifit is a plane graph.

3. Given a drawing of a graph in the plane (vertices = points, edges = straight line segments)find the number of edge-intersections.

4.2. Euler’s Formula

Theorem 4.2.1 (Euler, 1750)If G is a connected plane graph with n vertices, m edges andf faces, then

n−m+ f = 2.

Proof. Let T be a spanning tree ofG. Evidently, m(T) = n− 1, f (T) = 1, son(T)−m(T)+ f (T) = n− (n−1)+1= 2, and the formula holds forT.

Now add sequentially the remaining edges ofG to T: each such adding increases thenumber of edges and the number of faces by 1. Since in the formula m and f are of theopposite signs, the equality holds forG itself. �

Corollary 4.2.1 If a graph G is planar, then all of its plane embeddings have the samenumber of faces equal to m−n+2.

Proof. Indeed, sincen(G) andm(G) are constant, Euler’s formula implies thatf (G) =m−n+2. �

Corollary 4.2.2 If G is a connected planar graph without parallel edges, then

m(G) ≤ 3n(G)−6.

Proof. Consider a plane embedding ofG. Since there are no parallel edges, the size ofeach face is at least 3. Count the edges around each face. The minimum number that wecan obtain is 3f . In fact, each edge is counted twice, so we obtain 2m. Therefore, 3f ≤ 2m.From Euler’s formula,f = 2−n+m. Hence, 3(2−n+m) ≤ 2mwhat givesm≤ 3n−6. �

Corollary 4.2.3 Every connected planar graph without parallel edges contains a vertex ofdegree at most 5.


bb

bb

1

2

1’

2’

N

Figure 4.3. Stereographic projection.

Proof. If the graph contains at most 6 vertices, then every vertex has degree at most 5.Therefore, it remains to consider the case when the graph hasat least 7 vertices. Supposeeach vertex has degree at least 6. Counting edges around eachvertex results in 6n ≤ 2m,or, equivalently, 3n ≤ m. Combining this inequality with Corollary 4.2.2 we obtain thefollowing: 3n≤ m≤ 3n−6, which leads to contradiction 0≤−6. �

For any plane graphGand every its bounded facef , there exists such a plane embeddingof G that facef becomes unbounded. This can be shown using the so calledstereographicprojection.

Suppose we have a plane embedding ofG. Put a sphere tangent to the plane and mapthe plane onto the sphere “to the north pole” as shown in Figure 4.3. In this way we obtainan image ofG on the surface of the sphere. We now rotate the sphere in such away that thedesired face contains the north pole. From the new north pole, we then project the sphereback onto the plane. A new plane embedding ofG is obtained; the face that was boundedin the first embedding, becomes unbounded in this new plane embedding. Therefore, thefollowing proposition holds.

Proposition 4.2.1 A graph can be drawn on the sphere without intersections of edges ifand only if it is planar.

The statement above leads to the following observation about polyhedra (3-dimensionalfigures bounded by intersections of planes):

Corollary 4.2.4 If a convex polyhedron has n vertices, m edges and f faces, then n−m+f = 2.

Proof. Indeed, having such a polyhedron, place it into a sphere and project it out ontothat sphere. The vertices and edges of the polyhedron form anembedding of a graph onthe sphere. We then use stereographic projection to projectit onto the plane and obtain aplane embedding of the respective graph. The vertices, edges and faces of such embeddingcorrespond to the vertices, edges and faces of the initial polyhedron. �

Planar Graphs 71

For example, consider the cube:n = 8,m= 12, f = 6 and evidently, the formula holds.

Exercises 4.2.

b

b

b

bb

b

b

b

bb b b

b

b b

Figure 4.4.

1. Draw a plane embedding of the prism, cube,K2,3, each of the graphs in Figure 4.2 and verifythe Euler’s formula.

2. For each of the drawings in 1., check the inequality of Corollary 4.2.2.

3. Using the stereographic projection, show how to obtain from each other all three embeddingsof the same graph in Figure 4.4.

4. Check the inequality of Corollary 4.2.2 for prism, cube,Kn, Wn, Km,n (m≥ 1,n≥ 3), and thePetersen graph.

5. Draw a plane graph on 6 vertices with all the vertices belonging to the unbounded face; re-draw it in such a way that all the vertices belong to a bounded face.


1. Given a plane graphG and a facef , construct a plane embedding ofG with f being theunbounded face.

2. Given a planar graphG and its drawing in the plane having one crossing. Construct aplaneembedding ofG.

3. Given a planar graphG, a vertexx and a plane embedding ofG− x. Construct a planeembedding ofG.

4.3. K5 and K3,3 Are not Planar Graphs

Theorem 4.3.1 K5 and K3,3 are not planar graphs.

Proof. SupposeK5 is planar. Let us try to draw it in the plane, see Figure 4.5. Since ithas cycle 1-2-3, any plane drawing must have this cycle. The cycle partitions the plane intotwo faces. Vertex 4 may be embedded inside or outside of this cycle. In both cases, since it


b

b

bb

b

1

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34

5

b

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1

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?no face

for 5

?no face

for 5

b b

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1

23

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1

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3 4

4 inside

4 outsideK5

b

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3 inside

3 outside

?no face

for 6

?no face

for 6

1 4

5 2

1 4

5 2

1 4

5 2

3

3

K3,3

Figure 4.5.K5 andK3,3 are not planar.

is adjacent to vertices 1, 2, and 3, the inside or outside faceis split into three faces of size3 each. So, all the faces in the plane have size 3. It remains tofind the face for vertex 5.However, since vertex 5 has degree 4, there is no face for it (see Figure 4.5).

If we take vertex 5 instead of 4, and embed it inside or outsideof the cycle 1-2-3, wewill not be able to find a face for vertex 4. All cases are exhausted; it implies thatK5 cannotbe drawn in the plane without crossings of the edges, i.e., itis not planar.

Planar Graphs 73

Now supposeK3,3 is planar, and vertex setX = {1,2,3,4,5,6}. SinceK3,3 is bipartite,without loss of generality assume that vertices 1, 2, and 3 form the first part, and vertices 4,5, and 6 form the second part, see Figure 4.5.

Let us try to drawK3,3 without crossings in the plane. Since it has cycle 1-4-2-5, anyplane drawing must have this cycle, so start with it. The plane is now partitioned into twofaces. Vertex 3 may be drawn in inside or outside face. Since it is adjacent to vertices 4 and5, each case leads to three faces of size 4 each. Suppose we draw vertex 3 in inside face.So, there are three faces to draw vertex 6 which must be adjacent to 1, 2, and 3. If we placeit in face 1-4-3-5, then it cannot be connected with 2. If we place it in face 2-4-3-5, then itcannot be connected with 1. If we place it in outer face 1-4-2-5, then it cannot be connectedwith vertex 3. So, there is no face to place vertex 6. The case when vertex 3 is drawn inoutside face is considered similarly.

If, instead of vertex 3 we proceed with vertex 6 and embed it inside or outside the cycle1-2-4-5, by similar reasoning we arrive to the conclusion that there is no face for vertex 3.

All possible cases are considered; it implies thatK3,3 cannot be drawn in the planewithout crossings of the edges, i.e., it is not planar as well. �

Exercises 4.3.

1. Find the values ofk for whichK5 andK3,3 arek-connected.

2. Using graphK3,3 disprove the following statement: letG be a graph andS be a minimalseparator having at least three vertices with derived subgraphsG1 andG2. ThenG is planarif and only if bothG1 andG2 are planar.

3. Given a graphG and a 2-vertex separatorK2 producing planar derived subgraphsG1 andG2.Using plane embeddings ofG1 andG2, construct a plane embedding ofG.

4. Check the inequality of Corollary 4.2.2 for bothK5 andK3,3.

5. Prove that any graphG containingK5 or K3,3 as any subgraph (induced or not) is not planar.

6. Find the crossing number ofK5 andK3,3.


1. Given a 4-regular graphG, determine if it containsK5 as a subgraph.

2. Given a 3-regular graphG, determine if it containsK3,3 as a subgraph.

4.4. Kuratowski’s Theorem and Planarity Testing

If a graph is planar, then every its subgraph is evidently planar. If a graph contains a non-planar subgraph, then the graph itself is not planar. Theorem 4.3.1 implies that any graphcontainingK5 or K3,3 as subgraphs is not planar. But if a graph is not planar, does it containK5 or K3,3 as subgraphs? The answer is that it may not containK5 or K3,3 as subgraphs, butit must contain subgraphs closely related toK5 or K3,3.

Let us define two graphsG1 andG2 to behomeomorphic if both can be obtained fromsome graphG3 by replacing some edges with some paths. In other words, we simply draw


(“put”) additional vertices on some edges ofG3. It is clear that any such replacing does notchange the planarity ofG3: if G3 is planar, then bothG1 andG2 are planar and vice versa.For example, any two cyclesCk andCl , with k, l ≥ 3 are homeomorphic because they bothcan be obtained fromC3 by such replacing. Observe that any graph homeomorphic toK5 orK3,3 is not planar even if it does not containK5 andK3,3 as subgraphs of any type.

The next theorem is one of the fundamental theorems in graph theory. It shows thatgraphs that are homeomorphic toK5 andK3,3 represent the unique cause of non-planarity.

Theorem 4.4.1 (Kuratowski, 1930)A graph is planar if and only if it does not containsubgraphs homeomorphic to K5 or K3,3.

Proof. We omit the complete proof of this theorem because it is long and involves manyadditional results. It can be found in several extended texts, such as e.g. [3, 4, 5, 7]. Theidea of the proof can be described using the following steps.

1. Observe that if a graphG is planar, then for any edge ofG, there exists an embeddingsuch that the edge belongs to the unbounded face. This embedding can be found bythe stereographic projection.

2. Suppose a graphG contains a separator with two vertices, sayx,y, such that thederived subgraphs areG1 andG2. Let G′

i = Gi ∪{x,y}. Using step 1, one prove thenthat if G is not planar, then at least one ofG′

1,G′2 is not planar.

3. The last implies that any minimal non-planar graph must be3-connected. In otherwords, it is sufficient to consider further 3-connected graphs only.

4. At this point one use the lemma stating that any 3-connected graph on≥ 5 verticescontains an edge whose contraction does not change its 3-connectivity.

5. Next one prove the following lemma. In a graphG, contract an edge and obtain agraphG′. The lemma states that ifG′ contains subgraphs homeomorphic toK5 orK3,3, then the initial graphG also contains subgraphs homeomorphic toK5 or K3,3.

6. Final step: one prove that ifG is a 3-connected graph without subgraphs homeomor-phic toK5 or K3,3, then it has a plane embedding. The proof is by induction on thenumber of vertices. InG, contract an edge (guaranteed by step 4) to obtain a graphG′ which has less vertices. By step 4, it is 3-connected. By step5, G′ does not con-tain subgraphs homeomorphic toK5 or K3,3. Therefore, by the induction hypothesis,there exist a plane embedding ofG′. The final argument consists in considering afew possible cases how the initial graphG can be reconstructed fromG′. In each ofthese cases one show that it is possible to construct a plane embedding ofG unless itcontains subgraphs homeomorphic toK5 or K3,3. �

Planarity testing algorithm. There are many algorithms for determining if a graph isplanar. Surprisingly, they do not use search for subgraphs homeomorphic toK5 or K3,3.Next we informally describe the basic idea of such an algorithm.

Planar Graphs 75

Consider a graphG and a subgraphG′ ⊆ G (not necessarily induced). If we deletethe vertices ofG′ from G, we obtain a number (may be 0) of connected components. Afragment of G with respect toG′ is one of the following:

1) an edge ofG which is not inG′ but connects two vertices ofG′;2) a connected component ofG−V(G′) together with edges connecting it toG′ (vertices

of attachment fromG′ included).

Now, letG′ be a plane graph (algorithm usually starts with a cycle). Find all fragmentsof G with respect toG′. For each fragmentA, determine a set of facesF(A) that contain allvertices of attachment. IfF(A) = /0 for someA, thenG is not planar. If|F(A)|= 1 for someA, then selectA for the next step. If|F(A)| > 1 for all A, then select any fragmentA.

Next, chose any path connecting two vertices of attachment of the selected fragmentA.Embed the path inside a face fromF(A). Call the resulting plane graphG′ and repeat theprocedure., i.e., find a new set of fragments, for each of themfind all admissible faces, andso on. If we arrive to the initial graphG, thenG is planar. One can prove that the algorithmworks correctly, i.e., ifG is planar, then it finds a plane embedding, otherwise it stopsatsome subgraphG′ 6= G.

Exercises 4.4.

b b

b

b

bb

b bb

b

b b

bb

bb

b

b

b

G1 G2

Figure 4.6.

1. For which values ofm≥ 1 andn≥ 3, graphsKn, Km,n, Wn are planar?

2. Count the number of crossings in each of the graphs in Figure 4.6.

3. Which of the graphs in Figure 4.6 is planar and which is not?

4. In Petersen graph, find a subgraph homeomorphic toK3,3.

5. What is the minimum number of vertices (edges) that must bedeleted from Petersen graph tomake it planar?

6. Prove that replacing an edge by multiple edges does not change the planarity of a graph.



1. Given a plane graphG and an edge, construct an embedding ofG with the edge being on theunbounded face.

2. Given a graphG, determine if it is homeomorphic toK5.

3. Given a graphG, determine if it is homeomorphic toK3,3.

4. Given a graphG andn≥ 4, determine ifG is homeomorphic toCn.

5. Given a graphG andn≥ 4, determine ifG is homeomorphic toKn.

6. Given a graphG andn≥ 4, determine ifG is homeomorphic toWn.

7. Given a graphG, determine if it is homeomorphic to a cube.

8. Given a graphG, determine if it is homeomorphic to a prism.

9. Given a graphG, determine if it is planar.

4.5. Plane Triangulations and Dual Graphs

Plane triangulations. A simple connected plane graph is calledplane triangulation ifevery its face, including unbounded, represents a triangle(i.e. has size 3). A simple planar(plane) graph is calledmaximal planar (plane) graph if adding any new edge to it resultsin a non-planar graph.

b

b

bb

b

Figure 4.7. Plane triangulation.

One can prove that these two concepts are equivalent, namelya graph is a plane trian-gulation if and only if it is a maximal plane graph. It is easy to determine the number ofedges in a plane triangulation:

Theorem 4.5.1 For any plane triangulation, m= 3n−6.

Proof. Indeed, if we count the edges around each face, we obtain the equality 3f = 2m.Substitution off from Euler’s formula results inm= 3n−6. �

Every plane graph is a spanning subgraph of some plane triangulation; the latter can beobtained by adding edges to a given plane graph. Plane triangulations are important because

Planar Graphs 77

sometimes it is sufficient to prove results for them to conclude that the results hold for allplanar graphs.

An example of a graphG which is a plane triangulation is shown in Figure 4.7. Onecan see that all faces including unbounded face are triangles. G hasn = 5 vertices andm = 3n− 6 = 15− 6 = 9 edges. Thus any plane graph on five vertices is a subgraph ofgraphG.

Dual graphs. For any plane graphG one can construct another plane graph denoted byG∗ and called thedual of G. The rules are the following:

1. In each face ofG choose a point which becomes a vertex ofG∗.

2. For each edge ofG separating facesfi and f j construct an edge ofG∗ connectingvertices fi and f j .

An example of a graphG and its dualG∗ is shown in Figure 4.8. The edges ofG aredrawn by solid lines while the edges ofG∗ are drawn by dashed curves. SinceG had twofaces,G∗ has two vertices. Every edge ofG is crossed by the corresponding edge ofG∗. Asone can see, the dual to simple graphG is not a simple graph, in particular, the separatingedge ofG corresponds to a loop ofG∗.

b

b

b

b bb b

G

G∗

Figure 4.8. Plane graphG and its dualG∗.

It is evident thatn(G∗) = f , m(G∗) = m and f (G∗) = n. The last follows from theobservation that all faces around each vertex inG are consecutively connected by the edgesof G∗ and thus produce a face ofG∗.

Theorem 4.5.2 If G is a plane connected graph, then(G∗)∗ is isomorphic to G.

Proof. Evidently, one can reconstruct graphG from the plane embedding ofG∗ using thesame rules. �

One can show however, that different embeddings of the same planar graphG may havenon-isomorphic duals.


Exercises 4.5.

1. InK5, delete an edge, draw a plane embedding which is a triangulation and construct the dual.

2. In K3,3, delete an edge, draw a plane embedding which is a triangulation and construct thedual.

3. Construct the dual to a prism.

4. Construct the dual to a cube.

5. Construct the dual toWn,n≥ 4.


1. Given a plane graphG, drawG∗.

2. Given a plane graph, complete it to a triangulation.

Chapter 5

Graph Coloring

“Warnings: do not use in painting; do not give any preferenceto any of the colors...”

5.1. Preliminary

Coloring theory started with the problem of coloring the countries of a map in such a waythat no two countries that have a common border receive the same color. If we denotethe countries by points in the plane and connect each pair of points that correspond tocountries with a common border by a curve, we obtain a planar graph. The celebratedFourColor Problem asks if every planar graph can be colored with 4 colors. It seems to havebeen mentioned for the first time in writing in an 1852 letter from A. De Morgan to W.R.Hamilton. Nobody thought at that time that it was the beginning of a new theory. The first“proof” was given by Kempe in 1879. It stood for more than 10 years until Heawood in1890 found a mistake. Heawood proved that five colors are enough to color any map. TheFour Color Problem became one of the most famous problems in discrete mathematics ofthe 20th century. Besides colorings it stimulated many other areas of graph theory.

Generally, coloring theory is the theory about conflicts: adjacent vertices in a graphalways must have distinct colors, i.e. they are in a permanent conflict. If we have a “good”coloring, then we respect all the conflicts. If we have a “bad”coloring, then we have a pairof adjacent vertices colored with the same color. This lookslike having a geographic mapwhere some two countries having common border are colored with the same color. Graphsare used to depict “what is in conflict with what”, and colors are used to denote the state ofa vertex. So, more precisely, coloring theory is the theory of “partitioning the sets havinginternal unreconcilable conflicts” because we will only count “good” colorings.

In day by day life, perhaps the most common and simple application of coloring is intraffic lights. People who invented traffic lights did not even realize how smart they were:they were first to observe that it was not important what sign (stop, or drive) to put; it wasonly important that whatever they put should be in differentstates at any moment of timefor two given streets which intersect. Of course, as in many practical applications of math,they borrowed “red” and “green” from maritime rules, added “yellow” for an intermediatestate, and put that on the most visible position. But mathematical model only represents thecoloring of graphK2.


It may look surprising but in graph coloring it does not matter which color is “blue” orwhich is “red”; it only matters how many different colors areavailable. Instead of reallycoloring the vertices of a graph, we just label them by numbers 1,2, . . . , λ whereλ is thenumber of available colors. In this sense, graph coloring isthe most color-blind subject.

In the most general setting, a color of a vertex may be thoughtof as a “state of a point”,or even more generally, a “statement about anything”.

5.2. Definitions and Examples

Let G = (X,E) be a simple graph and{1,2, . . . ,λ} be theset of available colors. Anylabeling of vertices of graphG by the numbers from{1,2, . . . ,λ} is called acoloring. Eachvertex is assigned precisely one color.

Definition 5.2.1 A coloring is calledproper if adjacent vertices havedifferent colors.

Not every coloring is proper. In a proper coloring, verticesof the same color induce anindependent set. If we change the color for at least one vertex, we obtain another coloring.It may be proper or not. The number of all proper colorings of agraphG with at mostλcolors is denoted byP(G,λ). We will see thatP(G,λ) is a polynomial inλ and thereforewe call it thechromatic polynomial. When we need to underline the number of availablecolors, we say that we consider aproper λ-coloring. Notice that the number of really usedcolors in a properλ-coloring may be strictly less thanλ; but it can never be greater thanλ. Often, when we say just “coloring” we mean proper coloring.The minimum number ofcolors over all proper colorings is called thechromatic number of a graphG, denoted byχ(G). Evidently, the maximum number of colors that can be used in aproperλ-coloring ismin{n(G),λ}.

Since there are no colorings with less thanχ(G) colors, P(G,λ) = 0 for all integervalues ofλ such that 1≤ λ ≤ χ(G)−1. Since any coloring withχ(G) colors is at the sametime a properλ-coloring for anyλ ≥ χ(G), we conclude thatP(G,λ) ≥ 1 for any integerλ≥ χ(G). Sinceλ is the number of colors, in our discussions it is always an integer variable.

Consider an example. LetG = (X,E) where X = {x,y,z}, and E = {{x,y},{y,z}}, see Figure 5.1. There is no coloring with one color, soχ(G) > 1. On the otherhand, it is easy to see a coloring with two colors, say 1-2-1, so χ(G) = 2. Suppose the setof available colors is{1,2,3}, consequently,λ = 3. Having totally three colors and keep-ing in mind thatχ(G) = 2, we obtain the following possibilities for the numberi of reallyused colors:i = 2 andi = 3. If we use only 2 colors, then we have to chose them from{1,2,3}. Thus we have the choices: colors 1,2, colors 2,3, and colors1,3. The total numberof choices equals

(λi

)

=(3

2

)

= 3. For the first choice, the colorings are 1-2-1 and 2-1-2, seethe first two colorings in Figure 5.1, columni = 2. For the second choice, the colorings are2-3-2 and 3-2-3. For the last choice, the colorings are 1-3-1and 3-1-3. Notice that in thiscase the colorings come in pairs, and in each pair, we just permute the colors.

If i = 3, we observe that any coloring will be proper because all thevertices are ofdifferent colors. To construct all colorings, fix color 1 forx and obtain two colorings bypermuting colors 2 and 3 ony andz: 1-2-3 and 1-3-2. Do the same for color 2: we have

Graph Coloring 81

b b b

G

x y z

b b b

b b b

b b b

b b b

b b b

b b b

1 2 1

2 1 2

2 3 2

3 2 3

1 3 1

3 1 3

b b b

b b b

b b b

b b b

b b b

b b b

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1i = 2 i = 3

Figure 5.1. GraphG and 12 its proper 3-colorings: 6 strict 2-colorings and 6 strict 3-colorings.

colorings 2-1-3, 2-3-1. At last, for color 3 we obtain the colorings 3-1-2 and 3-2-1, seeFigure 5.1.

Summarizing all the cases we arrive to the conclusion thatP(G,3) = 12. What aboutP(G,λ) for an arbitrary value ofλ? Should we consider all possibilities by exhaustivesearch? And what aboutP(G,λ) for any graph? Is there any formula? We will showthat there are general procedures for computation ofP(G,λ) for an arbitrary graphG.

Before proceeding to the next section, let us consider one more concept. LetG= (X,E)be a graph, andi be the number of used colors. So, each ofi colors is in use. Such propercolorings are called thestrict i-colorings. Each stricti-coloring partitions the vertex setXinto i nonempty subsets, calledcellswhere each cell is the set of vertices of the same color.Suchpartitions are calledfeasibleand cells are calledcolor classes. In other words, in afeasible partition ofX into i cells, adjacent vertices belong to different cells and eachcellis an independent subset of vertices. Notice that proper colorings and feasible partitionsare different concepts: colorings are labelings of vertices, partitions are divisions ofX intononempty subsets. However, they are closely related. They are even more close than itappears.

In contrast withλ which has no upper limit, the numberi of really used colors satisfiesthe following inequality: 1≤ i ≤ n(G). Let r i(G) be the number of feasible partitions ofG into i cells. The vector

R(G) = (r1, r2, . . . , rn)

is called thechromatic spectrum of G.Since by definitionχ(G) is the smallest number of colors in a proper coloring, the

chromatic spectrum, in fact, always has the following form:

R(G) = (0,0, . . . ,0, rχ, rχ+1, . . . , rn).


In our example, see Figure 5.1, wheni = 2, we have six proper colorings. Each ofthese colorings is a proper 3-coloring and simultaneously,a strict 2-coloring. For all cases,there is only one feasible partition ofX: X1 = {x,z}(first cell) andX2 = {y} (second cell).Therefore,r2(G) = 1.

Wheni = 3, we have other six proper colorings. Each of these colorings is a proper 3-coloring and simultaneously, a strict 3-coloring. All the colorings generate only one feasiblepartition with cellsX1 = {x}, X2 = {y}, andX3 = {z}. Hencer3(G) = 1. Since evidently,r1(G) = 0, we obtain that the chromatic spectrum ofG is:

R(G) = (0,1,1).

The chromatic spectrumR(G) is calledcontinuous (gap-free)if it does not containzeroes between positive components. Otherwise it is calledbroken (has gaps).

Theorem 5.2.1 For any graph G, the chromatic spectrum R(G) is continuous.

Proof. As we noticed,R(G)= (0,0, . . . ,0, rχ, rχ+1, . . . , rn). By definition,rχ > 0. Considerany strict coloring ofG with χ colors and choose any color class which has> 1 vertices.Split it into two non-empty subsets; we obtain another feasible partition usingχ + 1 cellswhat impliesrχ+1 > 0. Repeating this procedure further for any other cell having> 1 ver-tices, conclude thatrχ+2 > 0, rχ+3 > 0, . . .. We cannot continue the splitting when all cellshave precisely one vertex each, and that corresponds torn = 1 > 0. �

We will see in Part II that this fundamental property does nothold in a more generalcase of hypergraph coloring.

There are many results about the bounds on the chromatic number. We end the sectionwith the following observations. The Konig theorem (Theorem 2.4.1) about bipartite graphsmay now be reformulated in terms of colorings:

Theorem 5.2.2 For a graph G,χ(G) ≤ 2 if and only if it has no odd cycles.

An important theorem relating maximum degree∆(G) and chromatic numberχ(G) ofa graphG is

Theorem 5.2.3 (Brooks, 1941)If G is a connected graph different from a clique and anodd cycle, thenχ(G) ≤ ∆(G).

Exercises 5.2.

1. Find the chromatic number ofEn, Kn, Km,n, treeTn, C2n, C2n+1, Wn, prism, cube and Petersengraph.

2. In graphG, see Figure 5.1, weakly delete an edge and find all feasible partitions,P(G′,3) andR(G′).

3. For all graphs from 1. starting with any proper coloring bythe minimum number of colors,show that the chromatic spectrum is gap free.

Graph Coloring 83

b b

b

b b

b

b

b

b

b

111 1 2

1 10 3

2

11363

Figure 5.2.

4. Figure 5.2 exhibits a graphG and a proper coloring. Find a respective feasible partitionandvalues ofλ for which the coloring is a properλ-coloring. Find a value ofi for which thecoloring is a stricti-coloring.

5. Beginning with the feasible partition intoi = 6 cells for the graph in Figure 5.2, construct asequence of feasible partitions intoi +1, i +2, . . . ,n = 10 cells.

6. For the graph in Figure 5.2, find a bound on the chromatic number by applying Brooks The-orem.

7. For the graph in Figure 5.2, find the exact value of the chromatic number and the correspond-ing optimal coloring.

8. For the graph in Figure 5.2, starting with optimalχ-coloring, construct a sequence of feasiblepartitions intoχ, χ +1, χ +2, . . . ,n = 10 cells.

Computer Projects 5.2.Using an appropriate graph representation, write a programfor thefollowing algorithmic problems.

1. Given a graphG and a coloring, check if the coloring is proper.

2. Given a graphG and a proper coloring, output the respective feasible partition.

3. Given a graphG, using a generator of random numbers, generate a proper coloring.

5.3. Structure of Colorings

Let us haveλ ≥ 1 colors and consider graphKn. Evidently, ifλ < n, then there are no propercolorings ofKn because all the vertices must be colored pairwise differently. Suppose nowλ ≥ n. Any assignment ofn different colors to the vertices ofKn results in a proper coloring,soχ(Kn) = n. What aboutP(Kn,λ)?

Let us start coloringKn. We haveλ possibilities to color the first vertex. Then, forthe second vertex, we have(λ − 1) possibilities. For the third vertex there are(λ − 2)possibilities. Any color used once, cannot be used again. Because of the symmetry ofKn,the order of such coloring procedure does not matter. If we continue it until the last vertexis colored, we arrive to the conclusion that

P(Kn,λ) = λ(λ−1)(λ−2) · · · (λ−n+1).


b

λ

b b

λ(λ−1) b b

b

λ(λ−1)(λ−2)

b b

bb

λ(λ−1)(λ−2)(λ−3)

Figure 5.3. Complete graphsK1,K2,K3,K4, and their chromatic polynomials.

Figure 5.3 shows examples of the chromatic polynomials of the simplest complete graphs:The productλ(λ − 1)(λ − 2) · · · (λ − n+ 1) is denoted byλ(n) and sometimes called thefalling factorial .

Let us have now a graphG = (X,E) which is not aKn. How to computeP(G,λ)? SinceG is not Kn, it has two nonadjacent vertices, sayx andy. All proper colorings ofG withλ colors are split into two classes: whenx andy have different colors and whenx andyhave the same color. All properλ-colorings ofG whenx andy have different colors are theproperλ-colorings of the graphG1 = G∪{x,y}. All proper λ-colorings ofG whenx andy have the same color are the properλ-colorings of the graphG2 = G1 · {x,y} (recall thatG ·edenotes the graph obtained by contraction of an edgee, see Section 1.6., Figure 1.25).Therefore,

P(G,λ) = P(G1,λ)+P(G2,λ). (5.1)

Notice that compared toG, graphG1 has the same vertices and edges except a newlyadded edge{x,y}. In its turn, graphG2 has the same edge set and the same vertex set exceptthat x andy are replaced be the new vertexxy. GraphG1 has more edges, and graphG2

has less vertices thanG. It is important to observe that graphG2 may have multiple edges.Namely, if verticesx andy have common neighbors inG, then contraction of edge{x,y}leads to multiple edges. Since in the definition of proper coloring multiple edges play norole (only adjacency is important), we replace every multiple edge in graphG2 by a singleedge. So, without any loss of generality for colorings, we can assume that graphG2 is alsosimple.

Figure 5.4 shows how we can depict the equality above as the equality of graph draw-ings. In this way, graphG1 stands for “connection” and graphG2 stands for “contrac-tion”. The connection-contraction algorithm itself consists in recurrent application ofthis step to every graph obtained. That means ifG1 or G2 is not a complete graph, thenwe find another pair of not adjacent vertices and proceed as wedid it with x and y. IfP(G1,λ) = P(G3,λ) + P(G4,λ) andP(G2,λ) = P(G5,λ) + P(G6,λ), then we obtain thatP(G,λ) = P(G3,λ)+P(G4,λ) +P(G5,λ)+P(G6,λ) and so on. When the procedure stops?It stops when we are not able to apply the connection, i.e. allobtained graphs are completegraphs. Therefore,

Graph Coloring 85

b b b b bx y x y xy

G G1 G2

= +

Figure 5.4. Connection-contraction.

P(G,λ) = P(Ki1,λ)+P(Ki2,λ)+ · · ·+P(Kis,λ)

for some integers≥ 1.SinceP(Ki j ,λ) is a polynomial inλ, and the sum of polynomials is a polynomial, we

conclude immediately thatP(G,λ) is a polynomial in λ. Moreover, among all completegraphsKi1, Ki2, . . . , Kis only one is isomorphic toKn; namely that which is obtained byconnections only. All the other complete graphs have< n vertices because they are obtainedfrom G by at least one contraction. Therefore, keeping in mind thatP(Kn,λ) = λ(λ−1)(λ−2) · · · (λ− n+ 1), our next conclusion is that thedegree ofP(G,λ) is n and the majorcoefficient is 1.

However, deeper analysis leads to deeper conclusions. Recall that all vertices ofGhave their names, i.e. labels. When contracting the edges weconcatenated the names andproduced new names for new vertices. As the result, the vertices in every complete graphKi have composite names. Only the names of vertices in the unique graphKn are exactlythe same as they are inG. Notice the following important fact. If the name of a vertexinsomeKi is xyz, for example, then there is a stricti-coloring ofG where verticesx,y, andzare colored with the same color. In other words, every complete graphKi corresponds tosome stricti-coloring ofG, each vertex ofKi corresponds to some color, and the compositename corresponds to all vertices ofG colored with that color.

There are no identical complete graphs with the same number of vertices if we comparethe composite names of vertices. It is so because eachKi was obtained by a unique way fromG. Therefore, thenumber of all complete graphs havingi vertices equals the numberof all feasible partitions of G into i cells,r i(G).

Collecting now all complete graphs on the same number of vertices and doing that forall possible values we obtain the following equality:

P(G,λ) = P(Ki1,λ)+P(Ki2,λ)+ · · ·+P(Kis,λ) =

r1(G)P(K1,λ)+ r2(G)P(K2,λ)+ · · ·+ rn(G)P(Kn,λ) =


n

∑i=1

r i(G)P(Ki,λ). (5.2)

Sincer1 = r2 = · · · = rχ−1 = 0 andP(Ki,λ) = λ(i), we arrive at last to the finalfunda-mental equality:

P(G,λ) =n

∑i=χ

r i(G)λ(i). (5.3)

Equality (5.3) explicitly shows the structure of proper colorings and their relation to fea-sible partitions: fix any feasible partition intoi cells, then count all the colorings that can beobtained from this partition by permutation of the colors (λ(i)), then do that for all feasiblepartitions (r i(G)λ(i)), and at last count that for alli (obtain ∑n

i=χ r i(G)λ(i)). Connection-contraction algorithm not only shows this structure, it also shows the way how to obtainP(G,λ) for any graphG.

bb

b

x z

y

=

bb

b

b

b

x z

y y

xz

+

G G1 = K3 G2 = K2

Figure 5.5. Connection-contraction forG.

Consider an example, see Figure 5.5. LetG = (X,E) be the same graph as in Figure5.1. Apply connection-contraction and immediately obtain

P(G,λ) = P(K3,λ)+P(K2,λ) = λ(3) + λ(2) =

λ(λ−1)(λ−2)+ λ(λ−1) = λ(λ−1)2.

Now without any exhaustive search for the colorings, we obtain, for example, thatP(G,3) = 3(3−1)2 = 12. Moreover, sinceP(G,λ) = P(K3,λ)+P(K2,λ) = 0·P(K1,λ)+1 ·P(K2,λ)+ 1 ·P(K3,λ), we conclude that the chromatic spectrumR(G) = (0,1,1). Thestructure of the colorings can also be seen in Figure 5.5: when i = 3, all the colorings areobtained from permutations of the colors; wheni = 2, all the colorings are obtained by per-mutations of two colors when vertexy is colored with one color and verticesx andz withthe other color.

The connection-contraction algorithm is good for small graphs but it is not efficient forlargen. Since every graph produces two new graphs, the number of graphs is doubling ateach step and is power of 2. The whole procedure can be depicted itself as a graph, whichis a directed tree, see Figure 5.6. Each arc shows which graphis obtained from which; leftdirected arcs denote connection, right directed arcs denote contraction. GraphG is locatedon the zero level, graphsG1 andG2 on the first level, graphsG3,G4,G5, andG6 on the

Graph Coloring 87

b

b b

b b b b

b b b b b b b b

Ki1 Ki2 Kis

G

G1 G2

G3 G4 G5 G6

Figure 5.6. Connection-contraction tree.

second level and so on down. Not all of the graphsKi j are on the last level. Some of themmay appear on higher levels.

Since the chromatic polynomial is unique for any graphG, there is a unique expansionof G into combination of complete graphs as described in equality (5.2). This observationleads us to the conclusion that theorder in which the connection-contraction is imple-mented is not important.

An implementation of the connection-contraction algorithm for cycleC4 is shown inFigure 5.7 in full. Edges to be added at the next level and thencontracted are shown bydashed lines. Multiple edges which appear after contraction are not shown. All four feasiblepartitions ofC4 can be found from the names of vertices of complete graphs. One canconclude that

P(C4,λ) = 0·P(K1,λ)+1·P(K2,λ)+2·P(K3,λ)+1·P(K4,λ) =

λ(2) +2λ(3) + λ(4) = λ(λ−1)+2λ(λ−1)(λ−2)+ λ(λ−1)(λ−2)(λ−3) =

λ(λ−1)[1+2(λ−2)+ (λ−2)(λ−3)] = λ(λ−1)(λ2−3λ+3) =

λ4−4λ3 +6λ2−3λ. (5.4)

andR(C4) = (0,1,2,1).

If for example, 10 colors are available, then the number of all proper 10-colorings isP(C4,10) = 10·9· (102−3·10+3) = 6570.


b b

bb

b b

bb

b

b b

b

b b

b b

b b

b

b

b b

b

a b

cd

a b

cd

a b

cd

a

cbd

ac

d

b a

bd c

ac

bd

C4

Figure 5.7. Connection-contraction forC4.

Exercises 5.3.

1. Apply the connection-contractionalgorithm to find the chromatic polynomial of the followinggraphs:E4, E5, C5, C6, W4, W5, P4, P5, P6 andPn for everyn≥ 6.

2. Apply the connection-contraction algorithm to find all feasible partitions forP4, P5, P6, C5

andW5.

3. Apply the connection-contraction algorithm to find all feasible partitions and the chromaticspectrum for the graphs in Figure 5.8.

4. A manager needs to arrange five people A, B, C, D and E in the offices. Person A is inconflict with B and E; person B is in conflict with A and C; personC is in conflict with B andD; person D is in conflict with A, B, C, E; person E is in conflict with A and D. What is theminimum number of offices and in how many ways the people can bearranged in such a waythat there is no conflict inside any office? Find all the assignments to the offices.

5. There are five cell phone towers A, B, C, D, and E. When transmitting a signal using thesame frequency, tower A interferes with B and E; tower B interferes with A and C; tower C

Graph Coloring 89

b

b

b b

b

b

b

b

b

b

b b

b b

b b

b b

b b

b

b

Figure 5.8.

interferes with B and D; tower D interferes with A, B, C, E; tower E interferes with A and D.What is the minimum number of frequencies necessary to avoidinterference at any momentof time? Find all possible optimal assignments of frequencies to the towers.

6. Which of the graphs in Figure 5.8 is the model for the problems 4 and 5 above?


1. Given a graphG, using a generator of random numbers, find several feasible partitions.

2. Given a graphG, using a generator of random numbers, find the lower bound on the chromaticspectrum.

3. Apply the connection-contraction algorithm to find all feasible partitions, chromatic polyno-mial and the chromatic spectrum for: a) cube; b) Petersen graph.

5.4. Chromatic Polynomial

Proposition 5.4.1 Suppose G is not a connected graph, and let G1,G2, . . . , Gk be the con-nected components, k≥ 2. Then

P(G,λ) = P(G1,λ)P(G2,λ) · · ·P(Gk,λ). (5.5)

Proof. Indeed, each component can be colored independently; sinceG1 hasP(G1,λ) proper colorings,G2 hasP(G2,λ) proper colorings, and so on, the total number ofproper colorings ofG is the product of these numbers. �

The equality (5.5) can be immediately applied to graphEn: since En = nK1, andP(K1,λ) = λ, we obtain

P(En,λ) = λ ·λ · ... ·λ = λn.

On the other hand, if we apply connection-contraction algorithm to En, we obtain thefollowing equality:

λn = S(n,1)λ(1) +S(n,2)λ(2) + . . .+S(n,n)λ(n) (5.6)

whereS(n, i) = r i(En) are some numbers; in other words,S(n, i) equals the number ofpartitions of a set ofn elements intoi subsets. These numbers are known as the so called


Stirling numbers of the second kind. Therefore, the chromatic spectrum ofEn is nothingelse than

R(En) = (S(n,1),S(n,2), . . . ,S(n,n)).

One can check, for example, thatR(E1) = (1),R(E2) = (1,1), R(E3) = (1,3,1),R(E4) = (1,7,6,1) and so on.

In turn, if we expand the expression forλ(n), then we obtain some polynomial withcoefficients:

λ(n) = s(n,1)λ+s(n,2)λ2 + . . .+s(n,n)λn

which are called theStirling numbers of the first kind . For example,λ(4) =−6λ+11λ2−6λ3 + λ4, and therefores(4,1) = −6, s(4,2) = 11, s(4,3) = −6, ands(4,4) = 1.

Generally, Stirling numbers serve as the coefficients to expressλn (or, equivalently,En)in terms ofλ(i) (or, equivalently,Ki) andλ(n) in terms ofλi , i = 1,2, . . . ,n. The latter in amore general setting can be expressed as thedisconnection-contraction algorithm.

The idea of it consists in the following. Recall that in the connection-contraction wehad the equality (5.1):

P(G,λ) = P(G1,λ)+P(G2,λ)

whereG1 is obtained fromG by connection, andG2 by contraction. We re-write this equal-ity as

P(G1,λ) = P(G,λ)−P(G2,λ). (5.7)

We now look as ifG1 is an original graph, andG is obtained fromG1 by deletion of anedge, i.e. disconnection, andG2 is obtained fromG1 by the contraction of that edge. Ap-plying this operation recurrently to each of the graphs obtained as many times as possible,we stop when all graphs on the right side of the equation aboveare the empty graphs.If the connection-contraction algorithm leads to a combination of complete graphs, thedisconnection-contraction algorithm leads to a combination of empty graphs.

An example of disconnection-contraction algorithm applied to the very same graphG,see Figure 5.5, is shown in Figure 5.9. One can see that the same chromatic polynomialmay be obtained in two different ways.

The next theorem describes the behavior of the coefficients of the chromatic polynomial.

Theorem 5.4.1 (Whitney, 1933)The chromatic polynomial P(G,λ) is of degree n(G), withinteger coefficients alternating in sign and beginning with1,−m(G), . . ..

Proof. We prove by induction on the number of edgesm(G). The theorem holds form= 0because in this caseG = En andP(G,λ) = λn. Assume now that the theorem is true for allgraphs with ¡medges and letG be ann-vertex graph withm> 1 edges. Consider arbitraryedgee of G. GraphsG−e andG ·e have fewer edges thanG each. In addition,G ·e hasn−1 vertices. By the induction hypothesis, there exist nonnegative integer numbers{ai}and{bi} such that

P(G−e,λ) = Σni=0(−1)iaiλn−i

Graph Coloring 91

bb

b

=

G

bb

b

b

b

− =

bb

b

b

b b

b

b− − + =

E3−2E2+E1 ⇒ P(G,λ) = λ3−2λ2+ λ = λ(λ−1)2

Figure 5.9. Disconnection-contraction forG.

andP(G ·e,λ) = Σn−1

i=0 (−1)ibiλn−1−i.

Applying one step of disconnection-contraction toG we obtain the equality whichproves the theorem:

P(G,λ) = P(G−e,λ)−P(G ·e,λ)

= λn− (m−1)λn−1+a2λn−2−·· ·+(−1)iaiλn−i · · ·

−(λn−1−b1λn−2 +b2λn−2−·· ·+(−1)ibi−1λn−i · · ·)

= λn−m(G)λn−1+(a2 +b1)λn−2 · · ·(−1)i(ai +bi−1)λn−i · · · .

�

Lemma 5.4.1 If a graph G contains a clique of size k, then the set of all proper coloringsof G can be partitioned intoλ(k) classes having P(G,λ)/λ(k) colorings each.

Proof. Let G = (X,E) be a connected graph with someS⊆ X inducing a clique of sizek. Let us haveλ available colors. Fix a proper coloring ofSwith λ colors and consider allproperλ-colorings ofG that can be obtained by the extension of the coloring ofS. Supposethat the number of such colorings isN1. Fix now another proper coloring ofSwith λ colorsand consider all properλ-colorings ofG that can be obtained by the extension of this newcoloring ofS. Denote the number of such colorings byN2. We claim thatN1 = N2. Indeed,every coloring from the second set of colorings can be obtained from a coloring from thefirst set (and vice versa) by a permutation of colors. Permutation of colorsi and j meansthat all vertices coloredi get color j and all vertices coloredj get colori. SinceS induces a


clique, all colors are different, and any two colorings ofScan be obtained from each otherby a permutation of colors.

Let t = P(GS,λ) = P(Kk,λ) = λ(k). Repeating the reasoning above for each oft color-ings ofS, we arrive to the conclusion thatN1 = N2 = N3 = · · · = Nt. Hence the set of allP(G,λ) colorings ofG is partitioned intot equal classes. This implies that the number ofcolorings in each class isP(G,λ)/λ(k). �

Observe that the statement of the lemma does not hold ifSdoes not induce a clique. IfShask vertices and is not a clique, then it has at least two nonadjacent vertices. The subgraphinduced byShas two different colorings, one withk colors and another withk−1 colors.These colorings cannot be obtained from each other by a permutation of colors. WhenSinduces a clique, it has the same unique feasible partition for every coloring ofG.

Theorem 5.4.2 Let G= (X,E) be a connected graph having a separator S which is a cliqueof size k. Suppose G1 = GX1∪S and G2 = GX2∪S are the two derived subgraphs with respectto S. Then

P(G,λ) =P(G1,λ)P(G2,λ)

λ(k). (5.8)

Proof. SinceGS is a clique of sizek in GX1∪S, by Lemma 5.4.1 the set of allP(GX1∪S,λ)colorings can be partitioned intoP(GS,λ) = λ(k) equal classes. Each class containsP(GX1∪S,λ)/λ(k) colorings. Similarly, theP(GX2∪S,λ) colorings ofGX2∪S can be partitionedinto λ(k) equal classes, and each such class contains exactlyP(GX2∪S,λ)/λ(k) colorings.

X1 S X2

G1 = GX1∪S

G2 = GX2∪S

Figure 5.10. Separation and the chromatic polynomial.

Combining every coloring from each class ofGX1∪S with every coloring from the corre-sponding class ofGX2∪S gives a coloring ofG, see Figure 5.10. Therefore, the total numberof colorings ofG is

P(G,λ) =P(GX1∪S,λ)

λ(k)

P(GX2∪S,λ)

λ(k)λ(k) =

P(GX1∪S,λ)P(GX2∪S,λ)

λ(k). �

Graph Coloring 93

Corollary 5.4.1 Let G= (X,E) be a connected graph having a separator S which is aclique of size k. Suppose G1,G2, . . .Gl are the derived subgraphs with respect to the sepa-rator S. Then

P(G,λ) =P(G1,λ)P(G2,λ) · · ·P(Gl ,λ)

[λ(k)]l−1. (5.9)

Proof. Indeed, sinceSbelongs to every graph, we can repeat the same reasoning for eachGi, i = 1,2, . . . , l . The colorings of graphs all combine in every class of colorings generatedby any single coloring ofS. Since the number of classes isλ(k), the formula follows. �

The formula (5.5) may be regarded as a special case of the formula above if for amoment we accept the point of view that disconnected graph islike “connected” graphhaving a separator which is an empty set, and defineP( /0,λ) = 1.

b

b

b

b

bS

b b

bb

S

b

b

bS

G

G1 G2

x

Figure 5.11. Separation.

An example to the theorem above is shown in Figure 5.11. SinceG1 =C4, G2 = K3 and|S| = 2, we use the formula (5.4) forP(C4,λ) andP(K3,λ) = λ(λ−1)(λ−2) to computethe chromatic polynomial of the graphG:

P(G,λ) =P(G1,λ)P(G2,λ)

λ(2)=

[λ(λ−1)(λ2−3λ+3)][λ(λ−1)(λ−2)]

λ(λ−1)=

λ(λ−1)(λ−2)(λ2−3λ+3).

Corollary 5.4.2 If a connected graph G has a simplicial vertex x of degree k, then

P(G,λ) = (λ−k)P(G−x,λ). (5.10)

Proof. SupposeG is not a complete graph. Then the neighborhoodN(x) is a completeseparator, i.e. a separator induced by a clique onk vertices. In this caseX2 = {x}, seeFigure 5.10. Applying formula (5.8) withG1 = G−x, G2 = Kk+1, andGS = Kk we obtain:

P(G,λ) =P(G1,λ)P(Kk+1,λ)

λ(k)=


P(G−x,λ)λ(k+1)

λ(k)= (λ−k)P(G−x,λ).

If G is a complete graph, thenG = Kk+1, G−x= Kk and the formula follows directly.�

We can apply formula (5.10) to graphG in Figure 5.11 and, sincex is a simplicial vertexandG1 = G−x= C4, compute the chromatic polynomial ofG directly:

P(G,λ) = (λ−2)P(G−x,λ) = λ(λ−1)(λ−2)(λ2−3λ+3).

Theorem 5.4.3 If a graph G has a simplicial vertex x of degree k, then

r i(G) = (i −k)r i(G−x)+ r i−1(G−x).

Proof. Recall thatr i(G) is the number of feasible partitions of graphG into i cells; itcoincides with the number of stricti-colorings ofG if we do not count the permutations ofcolors.

If vertex x is colored with one of the colors already used inG− x, then, because allcolors in the neighborhood ofx are different, we have(i −k)r i(G−x) such possibilities. Ifvertexx is colored with the color not used inG−x, then there arer i−1(G−x) possibilities.Hence the formula follows. �

Corollary 5.4.3 If S(n, i) is the Stirling number of the second kind, then

S(n, i) = iS(n−1, i)+S(n−1, i −1).

Proof. Apply the theorem above to graphEn: r i(En) = S(n, i) and each vertex may beregarded as a simplicial vertex of degree 0. �

Since trivially r1(E1) = S(1,1) = 1, the first five rows and columns of Stirling numbersare:

1 2 3 4 51 1 0 0 0 02 1 1 0 0 03 1 3 1 0 04 1 7 6 1 05 1 15 25 10 1

One can see, for example, that

S(5,3) = 25= 3·S(4,3)+S(4,2) = 3·6+7 = 25.

Exercises 5.4.

1. Apply the disconnection-contraction algorithm to compute the chromatic polynomial ofP3,P4,P5,Pn for n≥ 6, forC4,C5,C6,C7, and forW4,W5,W6.

Graph Coloring 95

b

b b b

bb

b

b b

b

b b b

b b

b

b b

bb

b

b

b

b

G1 G2 G3

Figure 5.12.

2. Compute the Stirling numbers of the second kind corresponding toE6,E7, andE8, and of thefirst kind corresponding toK3, K4,K5, andK6.

3. For each graph in Figure 5.12, by finding a simplicial vertex or a complete separator computethe chromatic polynomial.

Computer Projects 5.4.Write a program for the following algorithmic problems.

1. For any positive integersn andi ≤ n, computeS(n, i).

2. Find a complete separator in a graph.

3. Program the disconnection-contraction algorithm for: a) the cube; b) Petersen graph.

5.5. Coloring Chordal Graphs

Chordal graphs have a very special place in graph coloring because the roots of their chro-matic polynomials have nice properties. Recall that ifG is a chordal graph, then it hasa simplicial elimination ordering, i.e. it can be decomposed by sequential elimination ofsimplicial vertices.

Let G1 = (X,E), |X| = n, be a chordal graph, andσ = (x1,x2, . . . , xn) be a simplicialelimination ordering. LetG2 = G1−x1, G3 = G2−x2, . . . ,Gn = Gn−1−xn−1 = {xn}, suchthatGn+1 = Gn−xn = /0. It means that vertexxi is a simplicial vertex of degree, say,ki , inthe graphGi , i = 1, . . . ,n. Apply sequentially formula (5.10) to each of the graphs in thisorder:

P(G1,λ) = (λ−k1)P(G2,λ) = (λ−k1)(λ−k2)P(G3,λ) · · · =

(λ−k1)(λ−k2)(λ−k3) · · · (λ−kn).

We see that all the roots ofP(G1,λ) are integer numbers equal to the degrees of therespective simplicial vertices. SinceGn consists only of one vertexxn, conclude thatkn = 0and we fix the first rootλ = 0. If G1 is disconnected, then in each component there willbe a last vertex of degree 0; thus the multiplicity of the rootλ = 0 equals the number ofconnected components of the graphG1.


By the definition of clique number,G1 contains a complete subgraph onω(G1) vertices,so maximum among allki ’s is ω(G1)− 1. We now fix the second root of the chromaticpolynomial, namely,λ = ω(G1)−1. Recall thatω(G1) ≤ χ(G1) what implies that there isno coloring using less thanω colors. Consequently, every integer on the closed interval[0,ω− 1] is a root ofP(G1,λ). Therefore, all rootski are the integer numbers from theinterval [0,ω−1], some of them may coincide (have multiplicity) but there areno gaps inthese integers.

Finally, let us show thatχ(G1) = ω(G1) by coloring the vertices in the order inverseto σ, i.e. xn,xn−1,xn−2, . . . ,x2,x1. In general case this procedure is calledonline coloringbecause the vertices of a graph can imaginably be put on the real line and colored from leftto right (or from right to left). The main point is to color thevertices sequentially to obtaina proper coloring of a graph.

To do that, color vertexxn with color 1. Then, ifxn−1 is not possible to color with 1 (i.e.,xn andxn−1 are adjacent), color it with color 2. Forxn−2 try to use color 1, if not possible,try to use color 2, if not possible use the color 3. Continue this procedure each time tryingto use the smallest possible color. If none of the used colorsfits, use the new color. This iswhy sometimes the online coloring is called agreedy coloring. Since the maximum degreeamongki ’s is equal toω−1, we eventually obtain a coloring ofG1 with ω colors. Hence,the conclusion is thatω(G) = χ(G). All these facts can be summarized in the next theorem.

Theorem 5.5.1 If G is a chordal graph, then the chromatic polynomial has thefollowingform:

P(G,λ) = λs0(λ−1)s1(λ−2)s2 · · · (λ−χ(G)+1)sχ−1 (5.11)

where si ≥ 1(i = 0,1, . . . ,χ−1) is the number of simplicial vertices of degree i in the sim-plicial elimination ordering of G.

b

b b b

b

bbb

1 2

34

5

6

7 8

G1

Figure 5.13.

Consider now how this theorem works on the same example of a chordal graph shownin Figure 3.3, see Figure 5.13. As we have seen,G1 has a simplicial elimination orderingσ = (1,2,3,4,5,6,7,8). The degrees of simplicial vertices in eliminationσ are: 2, 2, 3, 2,2, 2, 1, 0. These are the roots of the chromatic polynomial, they all are from the interval[0,3]. Therefore, the chromatic polynomial

P(G1,λ) = (λ−2)(λ−2)(λ−3)(λ−2)(λ−2)(λ−2)(λ−1)(λ−0) =

λ(λ−1)(λ−2)5(λ−3). (5.12)

Graph Coloring 97

Vertices 3, 6, 7 and 8 form the unique maximum clique, soω(G1) = χ(G1) = 4. Onlinecoloring in the order 8, 7, 6, 5, 4, 3, 2, 1 (inverse toσ) produces a proper coloring usingfour colors, see Figure 5.14 (now the numbers are colors). There are other colorings withfour colors; the total number of 4-colorings, for example, is:

P(G1,4) = 4(4−1)(4−2)5(4−3) = 384.

b

b b b

b

bbb

3 2

42

1

3

2 1

G1

Figure 5.14. Online coloring of chordal graph.

It would be difficult to find this number of colorings manuallyby exhaustive search, oreven using connection-contraction algorithm developed for all graphs; structure of chordalgraphs provides an efficient algorithm for computing the number of proper colorings for anynumber of available colors. But nice coloring properties ofchordal graph go much beyondthis. The rest of the section is devoted to such facts that areunimaginable for general graphs.We begin with

Corollary 5.5.1 If Tn is a tree on n vertices, then

P(Tn,λ) = λ(λ−1)n−1. (5.13)

Proof. Trees are special case of chordal graphs, and pendant vertices are the special caseof simplicial vertices. DecomposeTn by sequential elimination of pendant vertices andapply formula (5.11). We obtainn−1 vertices of degree 1 and the last vertex of degree 0,i.e. k1 = k2 = · · · = kn−1 = 1, kn = 0. �

Lemma 5.5.1 For every cycle Cn,n≥ 1

P(Cn,λ) = (λ−1)n +(−1)n(λ−1). (5.14)

Proof. Induction on the number of verticesn. Forn= 1,2,3 verify directly (thoughC1andC2 are not simple graphs, the formula holds). Assume the formula holds for any cycle on< n vertices. Choose any edge ofCn and apply disconnection-contraction:

P(Cn,λ) = P(Tn,λ)−P(Cn−1,λ) =

{apply formula (5.13) and the induction hypothesis} =

λ(λ−1)n−1− [(λ−1)n−1+(−1)n−1(λ−1)] =

(λ−1)n+(−1)n(λ−1). �

We now are able to state the following criterion for chordal graphs:


Theorem 5.5.2 A graph G is chordal if and only if for any induced subgraph G′ (includingG itself) the chromatic polynomial has the following form:

P(G′,λ) = λs′0(λ−1)s′1 . . . (λ−χ′+1)s′χ′−1,

whereχ′ = χ(G′), and s′i ≥ 1 (i = 0,1, . . . ,χ′ − 1) is the number of simplicial vertices ofdegree i in the simplicial elimination ordering of G′.

Proof. ⇒ Every induced subgraph of a chordal graph is chordal as well.Apply Theorem5.5.1.

⇐ If G is not a chordal graph, then it contains a cycleCk,k≥ 4 as an induced subgraph.It is easy to see thatχ(Ck) = 2 if k is even andχ(Ck) = 3 if k is odd. Hence all the rootsof P(Ck,λ) should be from the set{0,1,2}. But by Lemma 5.5.1P(Ck,λ) = (λ− 1)k +(−1)k(λ−1). One can prove thatP(Ck,λ) being a polynomial of degreek≥ 4 has at leastone complex root; a contradiction. �

Let G = (X,E) be a graph,A ⊂ X be an arbitrary separator,G∗1 = (X1,E1), G∗

2 =(X2,E2), . . . ,G∗

k = (Xk,Ek),k ≥ 2, be the connected components obtained after removingvertex setA together with all incident edges fromG. If G is a disconnected graph havinglconnected components, then we assume thatk > l . So, we haveX1∪X2∪ . . .∪Xk∪A = X,Xi ∩Xj = /0, i 6= j. As usually, denote the derived induced subgraphs in the following way:

GX1∪A = G1, GX2∪A = G2, . . . , GXk∪A = Gk, GA = G0.

Theorem 5.5.3 A graph G= (X,E) is chordal if and only if, for any induced subgraphG′ = (X′,E′) (including G itself) and any separator A′ ⊂ X′ of G′,

P(G′,λ) =P(G′

1,λ)P(G′2,λ) . . .P(G′

k,λ)

P(G′0,λ)k−1 . (5.15)

Proof. ⇒ Since every induced subgraph of a chordal graph is chordal too, we prove thestatement forG′ = G. We proceed by induction on|X| = n. The casesn = 2,3,4, can beverified directly. Let the statement be true for all chordal graphs with fewer thann verticeswheren > 4. Let n(G) = n, andx0 ∈ X be a simplicial vertex of degreep in G. There aretwo possible cases.

Case 1.x0 6∈A. Suppose thatx0 ∈X1. SinceA is a separator andx0 is a simplicial vertex,the neighborhoodN(x0) ⊆ X1 ∪A. Therefore,x0 is a simplicial vertex inG1. Applyingformula (5.10) toG1, we obtain:

P(G1,λ) = (λ− p)P(G1−x0,λ).

Applying the same formula toG, we have:

P(G,λ) = (λ− p)P(G−x0,λ).

Notice thatA is a separator inG− x0 with the same number of derived subgraphs. Sincen(G−x0) < n by the induction hypothesis,

P(G−x0,λ) =P(G1−x0,λ)P(G2,λ) . . .P(Gk,λ)

P(G0,λ)k−1 . (5.16)

Graph Coloring 99

Multiplying both sides of (5.16) with the factor(λ− p), we obtain

P(G,λ) =P(G1,λ)P(G2,λ) . . .P(Gk,λ)

P(G0,λ)k−1 .

Case 2.x0 ∈ A, see Figure 5.15. Let|N(x0)∩A| = p1. SinceN(x0) induces a completegraph, the remainingp− p1 = p2 vertices fromN(x0) belong to at most one, sayX1, of thesetsXi,1≤ i ≤ n. Then

{x0}∪N(x0) ⊆ X1∪A, N(x0)∩A⊆ Xi ∪A, 1≤ i ≤ k.

One can see that the vertexx0 is simplicial inG1 with degreep and in all subgraphsGi, i =0,2,3, . . . ,k with degreep1. Apply formula (5.10) toG:

P(G,λ) = (λ− p)P(G−x0,λ).

b

A

X1

X2

Xk

x0 p1

p2

Figure 5.15.

Notice that setA−x0 is a separator inG−x0 with the same number of derived subgraphs.Sincen(G−x0) < n, by the induction hypothesis we have

P(G−x0,λ) =P(G′

1,λ)P(G′2,λ) . . .P(G′

k,λ)

P(G′0,λ)k−1 , (5.17)

where the graphsG′i are obtained fromGi by deletion of vertexx0, i = 0,1, . . . ,k.

On the other hand,

P(G1,λ) = (λ− p)P(G′1,λ), andP(Gi,λ) = (λ− p1)P(G′

i ,λ)

for i = 0,2, . . . ,k.


It immediately follows again that

P(G,λ) =P(G1,λ)P(G2,λ) . . .P(Gk,λ)

P(G0,λ)k−1 .

⇐ For a contradiction, supposeG is not chordal. Then it contains a cycleCk,k ≥ 4, asan induced subgraph. Letx be a vertex ofCk. SinceN(x) is a separator, by formula (5.15),we obtain

P(Ck,λ) = λ(λ−1)2λ(λ−1)k−2λ−2 = (λ−1)k 6= P(Ck,λ),

a contradiction. �

As we have seen, formula (5.15) is known to be true for arbitrary graphs provided aseparator induces a complete graph, see formula (5.9). The main feature of Theorem 5.5.3is that in chordal graphs it holds for any separator. One can see that (5.15) holds for dis-connectedG andGA (provided the number of connected components increases) oreven forthe empty separator. If we acceptP( /0,λ) = 1, then (5.15) turns into the formula for thechromatic polynomial of a disconnected graphG with connected componentsG1, . . . ,Gk

(caseA = /0).From this point of view, chordal graphs have so nice structure of colorings that they

mysteriously look like “complete graphs having separators”. However, (5.15) has an addi-tional final important consequence. Namely, it implies the universal formula for computingthe chromatic polynomial of a chordal graph using anarbitrary elimination ordering whatis impossible for general graphs.

For a graphG, define the function

W(G,λ) =λP(G,λ−1)

P(G,λ).

Theorem 5.5.4 (universal formula) A graph G= (X,E) is chordal if and only if in everyconnected induced subgraph G′ = (X′,E′) for any vertex x∈ X′ the following equalityholds:

P(G′,λ) = P(G′−x,λ)W(G′N(x),λ). (5.18)

Proof. ⇒ SupposeG is a chordal graph. Since every induced subgraph of a chordalgraphis chordal, we prove the necessity forG′ = G. Hence, assume thatG= (X,E) is a connectedchordal graph andx∈ X is an arbitrary vertex.

Case 1.X = {x}∪N(x). Since in any coloring ofG with λ colors vertexx requires aseparate color,

P(G,λ) = λP(G−x,λ−1) =P(G−x,λ)

P(G−x,λ)λP(G−x,λ−1) =

P(G−x,λ)W(GN(x),λ).

Case 2.X 6= {x}∪N(x). SinceG is connected, the subgraphGN(x) is a separator. Applyequality (5.15):

P(G,λ) =P(G−x,λ)P(G{x}∪N(x),λ)

P(GN(x),λ)=

Graph Coloring 101

P(G−x,λ)λP(GN(x),λ−1)

P(GN(x),λ)= P(G−x,λ)W(GN(x),λ).

⇐ SupposeG is not chordal and the formula (5.18) holds for any connectedinducedsubgraph. Then it contains an induced cycleCk,k ≥ 4. Let x be a vertex ofCk which isdenoted by justC. Observe thatC− x is a tree onk− 1 vertices, andCN(x) = E2. We seethatP(C−x,λ) = λ(λ−1)k−2, W(CN(x),λ) = λ−1(λ−1)2, and by the formula

P(C,λ) = P(C−x,λ)W(CN(x),λ) = (λ−1)k 6= P(C,λ),

a contradiction. �

Corollary 5.5.2 If x is a simplicial vertex of degree k≥ 0 in a chordal graph G, then

P(G,λ) = (λ−k)P(G−x,λ).

Proof. Indeed,N(x) = Kk, so

W(GN(x),λ) = W(Kk,λ) =λ(λ−1)(k)

λ(k)= (λ−k).

Applying universal formula (5.18) we obtain

P(G,λ) = P(G−x,λ)W(GN(x),λ) = (λ−k)P(G−x,λ). �

So, decomposition of chordal graphs using a simplicial elimination ordering is a specialcase of a general procedure of decomposition by eliminatingvertices in an arbitrary orderand applying the universal formula (5.18).

b

b b b

b

bbb

1 2

34

5

6

7 8

G

Figure 5.16. Application of the universal formula.

Let us consider how the last theorems work on the example of graphG, see Figure 5.16.Suppose we delete non-simplicial vertex 4 and obtain graphG1. To compute the chromaticpolynomial P(G,λ) from G1, by the universal formula we need to multiplyP(G1,λ) bythe functionW(GN(4),λ). SinceN(4) = {1,5,6}, the induced subgraph is a tree on threevertices (encircled by a dotted curve in the figure). Hence

W(GN(4),λ) =λP(T3,λ−1)

P(T3,λ)=

λ(λ−1)(λ−2)2

λ(λ−1)2 =(λ−2)2

λ−1.

If we decomposeG1 by simplicial elimination in ordering 1, 5, 6, 7, 8, 3, 2, the degreesof simplicial vertices respectively are: 1, 2, 3, 2, 2, 1, 0. We compute

P(G1,λ) = λ(λ−1)2(λ−2)3(λ−3).


Now

P(G1,λ)W(GN(4),λ) = λ(λ−1)2(λ−2)3(λ−3)(λ−2)2

λ−1=

λ(λ−1)(λ−2)5(λ−3) = P(G,λ),

as it was found in formula (5.12).

Exercises 5.5.

1. For graphG in Figure 5.16, compute the chromatic polynomial using separator{5,6,7,8}and respective derived subgraphs.

2. For graphG in Figure 5.16, apply the universal formula for vertex 7.

3. Construct an example of a chordal graph and an ordering of vertices such that online coloringin that order does not give the minimum number of colors.


1. Given a graphG, recognize ifG is chordal.

2. Given a graphG, recognize ifG is chordal, and if yes, compute the chromatic polynomial.

3. Given a graphG, recognize ifG is chordal, and if yes, construct an optimal coloring.

5.6. Coloring Planar Graphs

Recall that for a graphG = (X,E), the Szekeres-Wilf number is

M(G) = maxX′⊆Xminx∈G′d(x)

whered(x) is the degree of vertexx in a subgraphG′ induced by the subset of verticesX′. As we mentioned at the beginning of Section 3.3., this number can easily be found bysequential elimination of vertices of minimum degrees; themaximum degree obtained inthis procedure isM(G).

Theorem 5.6.1 For any graph G,χ(G) ≤ M(G)+1.

Proof. DecomposeG by a sequential elimination of vertices of minimum degree.We obtain a sequence of graphsG1 = G, G2, G3, . . . ,Gn and corresponding se-quence of verticesx1,x2, . . . ,xn such that each vertexxi is a vertex of minimum de-gree in a graphGi, i = 1, . . . ,n. We now apply online coloring toG by recon-structing it in orderingxn,xn−1, . . . ,x2,x1. That means we color the vertices in or-der xn,xn−1, . . . ,x2,x1 each time using the smallest suitable color. In this sequence,the maximum number of colored neighbors for vertexxi is M(G), see Figure 5.17.The worst case occurs when allM(G) neighbors ofxi have distinct colors (other-wise, we just assign toxi the first missing color). We then assign(M(G) + 1)thcolor to xi and continue online coloring. Even if the worst case occurs severaltimes, the total number of colors used in the obtained propercoloring does not exceedM(G)+1. �

Graph Coloring 103

b xiGi+1 N(xi)

Gi

Figure 5.17.Gi+1 is colored,xi is not.

Corollary 5.6.1 If G is a planar graph, thenχ(G) ≤ 6.

Proof. Recall that by Corollary 4.2.3,G contains a vertex of degree at most 5. Anysubgraph ofG is a planar graph. Therefore, in any decomposition ofG by vertices ofminimum degrees, the degrees do not exceed 5. HenceM(G)≤ 5. Applying Theorem 5.6.1results inχ(G) ≤ 6. �

Theorem 5.6.2 (Five Color Theorem, Heawood, 1890)If G = (X,E) is a planar graph,thenχ(G) ≤ 5.

Proof. We prove the theorem by induction onn(G) = n. Evidently, all planar graphs hav-ing ≤ 5 vertices are 5-colorable (color all vertices differently).

Assume now that all planar graphs having< n vertices are 5-colorable. We will provethat under this assumption,χ(G) ≤ 5. By Corollary 4.2.3,G contains a vertexx of degreed(x)≤ 5. By the induction hypothesis,χ(G−x)≤ 5. Consider a proper 5-coloring ofG−x.By this, all the vertices ofN(x) get some colors. We will show that there is always a way toassign a color to vertexx such thatG is 5-colorable.

Case 1: the number of colors used inN(x) is ≤4. We assign the 5th color to x andobtain a 5-coloring ofG.

Case 2: the number of colors used inN(x) is 5. It means thatd(x) = 5. The main ideaof the rest of the theorem is to show that we can construct another proper 5-coloring ofG−x which uses only 4 colors in the neighborhoodN(x).

The situation is equivalent to that shown in Figure 5.17 forGi = G, but now it canbe depicted more specifically in Figure 5.18. Without loss ofgenerality, denoteN(x) ={a,b,c,d,e} and respective colors by 1, 2, 3, 4 and 5.

Subcase 2.1:in the 5-coloring ofG− x, there exists an(a,c)-path P with verticescolored by alternating colors 1 and 3 as shown in Figure 5.18 (numbers are the colors).PathP along with edgesax andxc form a cycleC in the plane with all vertices colored 1and 3, and vertexx uncolored. Switch the colors 2 and 4 on vertices located in the plane


b

b

b b

b b

x

a

b

cd

e

bb

b

b

bb

b

b

b

b

1

2

34

5

31

3

1

31

4

2

4

2

Figure 5.18.

inside cycleC; such re-coloring does not affect vertices colored 2 and 4 which are outsideC. Therefore, vertexd preserves color 4. We obtain another proper 5-coloring ofG− xwhere verticesd andb are colored with color 4. In this new coloring, color 2 is missing inN(x); assign color 2 to vertexx and obtain a proper 5-coloring ofG.

Subcase 2.2:in the 5-coloring ofG−x, there is no(a,c)-path with vertices colored byalternating colors 1 and 3. LetG13 be a subgraph ofG−x induced by the vertices coloredwith colors 1 and 3. G13 is a disconnected graph and verticesa and c are in differentcomponents. Switch the colors 1 and 3 in the component containing vertexa. We obtain aproper 5-coloring ofG−x where color 1 is missing inN(x). Assign color 1 to vertexx andobtain a proper 5-coloring ofG, what proves the theorem. �

Theorem 5.6.3 (Four Color Theorem, Appel, Haken, Koch, 1977) If G is a planargraph, thenχ(G) ≤ 4.

Kempe’s proof, 1879, the most famous error in the history of Graph Theory. We“prove” the theorem by induction onn(G) = n. SinceK5 is not a planar graph, all planargraphs having≤ 5 vertices are 4-colorable.

Assume now that all planar graphs having< n vertices are 4-colorable. We will “prove”that under this assumption,χ(G) ≤ 4. Observe that any plane graph can be complete to atriangulation, i.e. a plane graph with all faces being triangles by just adding some edges.If we prove that any triangulation is 4-colorable, it will imply that any plane graph is 4-colorable, too. Therefore, without loss of generality we assume thatG is a triangulation.

By Corollary 4.2.3,G contains a vertexx of degreed(x) ≤ 5. By the induction hypoth-esis,χ(G−x) ≤ 4. Consider a proper 4-coloring ofG−x. By this, all the vertices ofN(x)get some colors. We “will show” that there is always a way to assign a color to vertexxsuch thatG is 4-colorable.

Graph Coloring 105

Case 1: the number of colors used inN(x) is ≤3. We assign the 4th color to x andobtain a 4-coloring ofG.

Case 2: the number of colors used inN(x) is 4. It means thatd(x) = 4 ord(x) = 5.

Subcase 2.1: d(x) = 4. Apply the reasoning from Theorem 5.6.2 forG′ = G− e (infact, vertexeand color 5 was never used in that proof).

b

b

b b

b b

x

a

b

cd

e

1

2

34

2

bb

b

b

3

1

3

1

bb

b

b

4

1

4

1

P14 P13

G′G′′

2↔ 42↔ 3

Figure 5.19. Kempe’s “proof”.

Subcase 2.2: d(x) = 5. Since the number of colors used inN(x) is 4, two vertices getthe same color. DenoteN(x) = {a,b,c,d,e} and respective colors by 1, 2, 3, and 4. Sincewe assumed thatG is a triangulation, the vertices of the same color cannot be consecutivein the ordera,b,c,d,e. Assume that the colors are distributed as shown in Figure 5.19. LetGi j denote the subgraph induced by vertices coloredi and j, andPi j denote any path havingall vertices colored with alternating (along the path) colors i and j.

Subcase 2.2.1:in the 4-coloring ofG− x, there is no(a,c)-path P13 or there is no(a,d)-pathP14. For example, assume there is no(a,c)-pathP13. Then the subgraphG13

is disconnected, and verticesa andc are in different components. Switching the colors 1and 3 in the component containing vertexa eliminates color 1 fromN(x); we then assigncolor 1 tox and obtain a proper 4-coloring ofG. Similar reasoning applies if there is no(a,d)-pathP14.

Subcase 2.2.2:in the 4-coloring ofG−x, both(a,c)-pathP13 and(a,d)-pathP14 exist,see Figure 5.19. Observe that the componentG′ of G24 containing vertexb is separated fromverticesd ande by the cycle completed by(a,c)-pathP13 and edgesax andxc. Similarly,the componentG′′ of G23 containing vertexe is separated from verticesb andc by the cycle


completed by(a,d)-pathP14 and edgesax andxd. We now permute color 2 with 4 inG′

and color 2 with 3 inG′′ . This recoloring eliminates color 2 fromN(x). Assign color 2 tovertexx and obtain a proper 4-coloring ofG, what “proves” the theorem. �

Where is the trap in this reasoning by Kempe? The answer to this question is depictedin Figure 5.20. The problem is that in Subcase 2.2.2,(a,c)-pathP13 and(a,d)-pathP14 mayintersect at a vertex of color 1. Then the permutation of color 2 with 4 in G′ and of color 2with 3 in G′′ leads to a pair of adjacent vertices colored with color 2 (marked by “?”), i.e.to a non proper coloring ofG.

b

b

b b

b b

x

a

b

cd

e

1

2

34

2

b

b

b

b

4

1

3

1

b

b

b

1

4

3

b

bb b

3 2

4

2 G′

G′′ ?

Figure 5.20. The most famous error in the history of Graph Theory.

The proof above was published by Kempe in 1879, and the error is very instructive.On one hand, it shows an excellent example how drawings may beused in Graph Theoryproofs. On the other hand, it explicitly exhibits the limitsof drawings. Since the result wasconsidered proven until 1890, it also demonstrated the factthat mathematicians of thosetimes liked writing papers more than reading them. But on thetop of all that, nobody couldeven imagine what a dramatic history was ahead.

Formulated first in 1852 by Francis Guthrie (a student (!) of de Morgan), “proved” in1879 by Kempe, refuted in 1890 by Heawood, the Four Color Problem became one of themost famous problems in Discrete Mathematics in 20th century before in 1977 it becamethe Four Color Theorem by Appel, Haken, and Koch. Besides many erroneous proofs, it

Graph Coloring 107

generated many new directions in Graph Theory. For example,only one sub-direction ofchromatic polynomials introduced by Birkhoff in 1912 with the aim to solve the problemby algebraic methods counts more than five hundred research papers.

The main idea of the final proof is quite simple - by induction on the number of vertices;but the number of cases is huge. Though the Kempe’s proof was erroneous, his idea of alter-nating paths and further re-coloring of the respective subgraphs was used in the final proof.A path on which two colors alternate is called aKempe chain. In a plane triangulation,a configuration is a derived subgraph with respect to a separating cycle; it is a subgraphinduced by the cycle and all the vertices which are inside thecycle in the plane. For exam-ple, in Figure 5.20, the subgraph induced by verticesa,b,c,d,e andx is a configuration. Itis basically the same idea that was used in chordal graphs, ormore generally, any graphshaving separators, see Figures 1.34, 5.8, 5.17. In chordal graphs, separators were cliques,in triangulations, separators are cycles.

In a plane triangulation, a configuration isreducible if any 4-coloring of the cycle canbe extended to a 4-coloring of the entire triangulation. There were the following two basicsteps in the proof:

1. Proof that any plane triangulation contains a configuration from a list ofunavoidableconfigurations.

2. Proof that each unavoidable configuration is reducible.

In 1976, Appel, Haken, and Koch, using 1,200 hours of computer time, found 1936 (!)unavoidable configurations and proved that all they are reducible. Historically, it was thefirst time when a famous mathematical problem was solved by extensive use of computers.The final accord in this one-century drama was when the resultwas widely announced, thepaper was published in 1977, and the University of Illinois even announced it by postagemeter stamp of “four colors suffice”, a few errors were found in the original proof. Fortu-nately for the authors, they have been fixed. For the first time, regardless the fact that thereis no human being who would check the entire proof (because itcontains steps that mostlikely can never be verified by humans), the problem is considered completely solved.

In 1996, Robertson, Sanders, Seymour and Thomas improved the proof by finding theset of only (!) 633 reducible configurations. Computers, Kempe chains, and some othertechniques were used in both proofs.

The three consecutive theorems - Corollary 5.6.1 (six colors), Theorem 5.6.2 (five col-ors) and Theorem 5.6.3 (four colors) show the main feature ofgraph coloring: there is avery simple proof for six colors, a relatively simple proof for five colors and an incredibledifficult and complex proof for four colors. The Four Color Problem, formulated by a stu-dent, was first “solved” by a lawyer, and really solved with many contributions of the veryprominent mathematicians of the century.

So we can conclude that every geographic map can be colored with at most four colors.Sometimes one can use less colors. What about the map of USA? Four Color Theoremimplies that it can be colored with four colors. But can it be colored with 3 colors? Theanswer is no. Indeed, take for example the state of Nevada andfive its neighbors: Idaho,Utah, Arizona, California and Oregon. One can easily check,see Figure 5.21, that theiradjacency forms the wheelW6 for whichχ(W6) = 4. West Virginia and Kentucky are also in


b

b

b

bb

b

ID

UT

AZCA

OR NV

Figure 5.21.

the center ofW6 andW8 respectively. Hence a planar graph corresponding to the entire mapof USA contains subgraphs which need four colors each anyway. It means that preciselyfour colors is the minimum for the entire map of USA.

Exercises 5.6.

1. ComputeM(G) whereG is: Pn, Kn, Km,n, Cn, Wn, m,n ≥ 4, tree, Petersen graph, cube andprism, find the bound on the chromatic number and the respective coloring.

2. Arbitrarily draw any number of straight lines in the plane. Use mathematical induction toprove that the obtained regions can be colored with two colors in such a way that no two ofthem with a common segment of line have the same color.

3. Color the map of USA with four colors.

4. Construct a planar graphG with no subgraphs isomorphic toWn,n≥ 4, such thatχ(G) = 4.


1. Given a graphG, find M(G), bound on the chromatic number, and the respective coloring.

2. Given a planar graphG, find an optimal coloring.

3. Given a plane graphG and a proper coloring. Find the longest Kempe chain.

5.7. Perfect Graphs

Any graphG contains a maximum clique which by definition hasω(G) vertices. Since itsvertices are pairwise adjacent, any proper coloring ofG requires at leastω colors, i.e.

χ(G) ≥ ω(G). (5.19)

There are graphs withχ(G) = ω(G), and there are graphs for whichχ(G) > ω(G). Forexample,χ(C4) = ω(C4) = 2 butχ(C5) = 3 > ω(C5) = 2. How large can be the differenceχ(G)−ω(G)? The answer is: it can be arbitrary large.

Graph Coloring 109

Consider the so calledMycielski’s construction. Two vertices are calledcopiesof eachother if they have the same neighborhoods and are not adjacent. Respectively, copying avertex means adding a new vertex adjacent to all neighbors ofa given vertex. A vertex iscalleduniversal if it is adjacent to all other vertices of a graph. In a complete graph allvertices are universal. Generally, adding a new vertex to a graph and making it universalincreases the size of maximum clique by 1. It also increases the chromatic number by 1because the universal vertex requires a new color.

b

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b⇒

Figure 5.22.

Start with graphK2. Copy the vertices, see Figure 5.22, encircled. Then add a newvertex adjacent to all the copies. The last vertex is “universal” to the copies but its additiondoes not create triangles and therefore does not increase the clique number. However, itincreases the chromatic number. This is the main trick of theconstruction. Notice that thegraph obtained isC5.

b

b

b

bb

b

b

b

b

bb

Figure 5.23.

Now repeat the procedure forC5: copy each vertex and add a new vertex adjacent to allof the copies. It is convenient to draw the copies and the “universal” vertex inside the cycle,see Figure 5.23. This graph is called theGrotzsch graph. For it χ = 4, and evidently,ω = 2. One can prove that if we repeat the procedurek times, we obtain a graph withχ−ω = k. It means that from theoretical point of view, there exist graphs with arbitrarylarge differenceχ−ω which are triangle-free. Ifω(G) is the lower bound forχ(G), thenwhat are the graphs for whichχ(G) = ω(G)?


In Graph Theory, when we have some property for a graph, it is convenient to requirethis property to hold for each induced subgraph. The main motivation is that in such casewe can apply mathematical induction by the number of vertices and prove many results.

Definition 5.7.1 A graph G isperfect if χ(G′) = ω(G′) for any induced subgraph G′ in-cluding G itself.

It appears that perfection of graphs is related to the complements of graphs. Recall thatthe clique cover numberθ(G) of a graphG is the minimum number of cliques in graphG such that each vertex belongs to precisely one clique. We saythat the cliques “cover”the vertex set ofG. Since every clique inG induces an independent set in complementG, we conclude that every coloring ofG is equivalent to a clique covering ofG, and thusχ(G) = θ(G). Therefore, ifG is a perfect graph, thenθ(G′) = α(G′) holds for any inducedsubgraphG′ of G includingG itself.

b

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G

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3 2

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G

Figure 5.24. Perfect graphs.

For example, a graphG in Figure 5.24 hasχ(G) = 3, ω(G) = 3, and α(G) = 2 andθ(G) = 2. A proper 3-coloring is shown by numbers, and clique covering is shown bydotted ellipses. Respectively, for its complementG, we haveχ(G) = 2,ω(G) = 2, α(G) = 3andθ(G) = 3. One can easily see that clique covering ofG by two cliques correspondsto a proper 2-coloring ofG and so on. It is also easy to check thatG and G both areperfect graphs. An example of an imperfect graph isC5: χ(C5) = 3,ω(C5) = 2, α(C5) = 2and θ(C5) = 3 (recall thatC5 is isomorphic toC5). In these examples graphs and theircomplements behave similarly with respect to graph perfection. Is this the rule?

May we have the equalityχ(G′) = ω(G′) for all induced subgraphsG′ and inequalityθ(G′) 6= α(G′) for at least one induced subgraph? In other words, may we havea perfectgraphG such thatG is not perfect? The answer is no. This fact was first stated as theWeakPerfect Graph Conjectureby Berge in 1960 and proved by Lovasz in 1972 (the first proofused hypergraph approach).

Theorem 5.7.1 (Weak Perfect Graph Theorem, Lovasz, 1972)A graph G is perfect ifand only if its complementG is perfect.

Graph Coloring 111

Chordal graphs were the first proved to be perfect:

Theorem 5.7.2 If G is a chordal graph, then it is perfect.

Proof. Let G be a chordal graph. Since any induced subgraphG′ is chordal, it is sufficientto prove thatχ(G) = ω(G). SinceG is chordal, it has a simplicial elimination ordering.Apply online (greedy) coloring toG by coloring vertices in inverse ordering as we did it inSection 5.5. Recall that the maximum vertex degree isω(G)−1. Hence,χ(G) = ω(G) andG is perfect. �

Theorem 5.7.3 If G is a quasi-triangulated graph, then it is perfect.

Proof. Let G be a quasi-triangulated graph. Then by definition, see Section 3.5., it hasa decomposition by sequential elimination of vertices thatat each step, are simplicial orco-simplicial (simplicial in the complement). For any induced subgraphG′, this orderingof vertices induces an ordering of vertices ofG′ with the same properties. ThereforeG′

has a decomposition by sequential elimination of vertices that at each step, are simplicialor co-simplicial. It means thatG′ is also a quasi-triangulated graph. Hence, as for chordalgraphs, without loss of generality it is sufficient to prove thatχ(G) = ω(G).

Prove by induction on the number of verticesn(G). The equalityχ(G) = ω(G) is truefor n = 3,4,5. Let it be true for all quasi-triangulated graphs on< n vertices andn(G) = n.By definition,G has a vertexx which is simplicial inG or in G.

Case 1:x is a simplicial vertex inG. Consider a proper coloring ofG−x by χ(G−x) =ω(G− x) colors. If ω(G) = ω(G− x), then|N(x)| ≤ ω(G− x)−1 and we can use a colormissing inN(x) to color vertexx. If ω(G) = ω(G−x)+1, then|N(x)| = ω(G−x) and wemust use a new color forx. In both cases we obtain a proper coloring ofGwith χ(G)= ω(G)colors. Therefore,G is perfect.

Case 2: x is a simplicial vertex inG. Again, graphG− x is quasi-triangulated, has< n vertices, and by the induction hypothesis is perfect. By Theorem 5.7.1, graphG−x isperfect quasi-triangulated graph.G is obtained fromG−x by adding simplicial vertexx.By Case 1,G is perfect. By Theorem 5.7.1,G is perfect. �

There are many classes of perfect graphs. Among them are bipartite graphs,weaklychordal graphs (containing no induced cycles of length≥ 5 inGand inG), strongly perfect(every induced subgraph has an independent set intersecting all maximal cliques),Meynielgraphs (each odd cycle of length≥ 5 has at least two chords),Berge graphs(no inducedodd cycles of length≥ 5 in graph and its complement). Examples and theorem above giverise to the idea that odd cycles of length≥ 5 in graph and its complement prevent a graphfrom being perfect. Stated as theStrong Perfect Graph Conjectureby Berge in 1960, itbecame one of the most important results in graph theory of our time:

Theorem 5.7.4 (Strong Perfect Graph Theorem, Chudnovsky, Robertson, Seymour,Thomas, 2003)A graph G is perfect if and only if it contains no induced C2k+1 andC2k+1,k≥ 2.

One of the fundamental properties of perfect graphs is that they, in contrast to generalgraphs, allow polynomial time algorithm (see Section 12.2.) for such optimization problem


as finding the chromatic number, and consequently, the minimum clique covering, maxi-mum independent set, and the maximum clique.

Exercises 5.7.

1. Determine if Petersen graph and its complement are perfect.

2. Determine which wheels are perfect and which are not.

3. Apply one step of Mycielski’s construction starting withP5, C4, C6, K4, prism.

4. Prove thatθ(C7) > α(C7).

5. Are cube and prism perfect graphs?


1. Given a graphG, determine if it contains inducedC5.

2. Given a quasi-triangulated graphG, find an optimal coloring.

3. Given a quasi-triangulated graphG, find the clique cover numberθ(G) and the respectiveclique covering.

5.8. Edge Coloring and Vizing’s Theorem

In Graph Coloring, we can also color the edges of a graph. In such colorings, parallel(multiple) edges will make a difference, but loops create inconveniences. Therefore theloops are ignored. LetG = (X,E) be a graph with possible parallel edges but withoutloops, andλ ≥ 0 be the number of available colors.

Definition 5.8.1 A proper edge λ-coloring of G is an assignment of a color from set{1,2, . . . ,λ} to every edge of G is such a way that all edges incident to everyvertex havedistinct colors. A graph isedgek-colorable if it has a proper edge coloring with at most kcolors.

Since the maximum number of colors that can be used equals|E|, we are interested inthe minimum number of colors. In order not to confuse with thechromatic number, theminimum number of colors over all proper edge colorings is called thechromatic index ofG and denoted byχ′(G).

In any proper edge coloring ofG = (X,E), edges of the same color represent somematchings. Thus the coloring itself is the partition of the edge setE into a number ofmatchings. The chromatic indexχ′(G) is the minimum number of distinct matchings overall such partitions.

Edge colorings ofG can be expressed as vertex colorings of an auxiliary graph withvertices representing the edges ofG. For graphG = (X,E), construct a simple graphG′ =(X′,E′) such thatX′ = E andE′ is formed in the following way: two vertices inG′ areadjacent if and only if the respective edges have a common vertex in G. GraphG′ is calledthe line graph of G and is denoted byL(G). Not every graph can be a line graph.

Graph Coloring 113

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⇒

L(G)

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edge coloring

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⇒

vertex coloring

Figure 5.25.

An example of a graphG, its line graphL(G), proper edge 4-coloring ofG and propervertex 4-coloring ofG′ = L(G) is shown in Figure 5.25. One can see that every properedge coloring ofG corresponds to a proper vertex coloring ofG′ and vice versa. For thisexampleL(G) is a chordal graph, so for edge colorings ofG we can even compute the chro-matic polynomialP(L(G),λ) = λ(λ− 1)(λ− 2)2(λ− 3) as it follows from the respectivesimplicial decomposition. Notice that every non pendant vertex ofG forms a clique inL(G)but not vice versa: clique{b,c,d,e} in L(G) does not have a corresponding vertex inG. Atlast, observe thatχ′(G) = 4 = χ(L(G)), and moreover, the equalityχ′(G) = χ(L(G)) holdsfor any graphG without loops.

As usually, let∆(G) denote the maximum vertex degree ofG. Evidently,χ′(G)≥ ∆(G).Maximum degree∆ plays a similar role for the chromatic indexχ′ as the clique numberωplays for the chromatic numberχ in vertex coloring whereχ ≥ ω. But how far may be thevalueχ′(G) from ∆(G)? Surprisingly, it must be very close.

Theorem 5.8.1 (Vizing, 1964)If G is a simple graph, then either

χ′(G) = ∆(G), or χ′(G) = ∆(G)+1.

Proof. Simple examples (likeC2k andC2k+1) show that there are graphs with both valuesof χ′. We will show that any graph can be edge colored with∆ +1 colors.

Let G be a simple graph. Take∆ +1 colors and properly color as many edges as possi-ble. If all edges are colored, we are done. Otherwise, we obtain some partial edge coloring.Suppose an edge connecting verticesx andy0 and denoted by justxy0, is uncolored. We willrecolor edges in such a way thatxy0 becomes colored and the number of colors remains thesame. After repeating the procedure as many times as necessary, a complete proper edgecoloring ofG will be obtained. Since we use∆ + 1 colors and∆ is the maximum degree,


at each vertex at least one color is missing. A feature of thisproof is that the reasoning isconstructed in terms of such missing colors.

The best case:a colorc is missing at both verticesx andy0. Color edgexy0 with c.The good case:there is no color missing at bothx andy0 simultaneously. Let a color

c0 be missing atx and colorc1 be missing aty0. If there is an edgexyk coloredc1 with bothends missing a colorci , then re-colorxyk with ci and colorxy0 with c1.

The bad case:there is no edgexyk coloredc1 with both ends missing a color. Edgexy1

is coloredc1; a colorc2 is missing aty1. Edgexy2 is coloredc2, a colorc3 is missing aty2. Edgexy3 is coloredc3, a colorc4 is missing aty3, and so on, see Figure 5.26 (missingcolors are denoted byci). Suppose this sequence continues until a colorcl+1 is missing atboth vertexx andyl . Then we re-color each edgexyi from ci to ci+1, i = 1,2, . . . , l . Call thisre-coloringdownshifting from yl . We arrive to a situation when colorc1 is missing atx.Since it is missing aty0, too, color edgexy0 with c1.

c0b b

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x y0c0 ?

c1c0

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cl+1

c1

c2

c3

ck−1

ck

c0

cl

c0

Figure 5.26.

The worst case:we never reach the situation when some color is missing at both endsof an edge. Then in the sequence of colorsc1,c2,c3, . . ., sooner or later, some colorcl+1

repeats (i.e., we have a loop) because the number of colors isfinite. Let cl+1 = ck be thefirst such coincidence of the colors, 1≤ k≤ l . Observe that colorc0 is present at all coloredneighbors ofx and it is missing atx. Consider now the Kempe pathP consisting of edges of

Graph Coloring 115

two alternating colorsc0 andck and beginning at vertexyl . Since we color the edges, eachvertex may have at most two edges colored withc0 andck; therefore, there is the only suchpath.

Subcase 1:P reaches vertexyk and therefore ends atx. Switch colorsc0 andck alongP. We obtain edgexyk with color ck missing at both ends. Downshifting fromyk results inrelease of colorc1 to be used forxy0.

Subcase 2:P reaches vertexyk−1 and therefore ends atyk−1. Again, switch colorsc0

andck alongP. As a result, colorc0 is missing atyk−1. Since it is also missing atx, weobtain edgexyk−1 with color c0 missing at both ends. Perform downshifting fromyk−1 andagain, release colorc1 to color edgexy0.

Subcase 3:P never reachesyk or yk−1. Sinceck is present atx, P never reaches vertexx.Switch the colorsc0 andck alongP; color c0 becomes missing atyl . Perform downshiftingfrom yl and release colorc1 to color edgexy0. �

The theorem above partitions all simple graphs into two classes: Class 1if χ′(G) =∆(G) andClass 2if χ′(G) = ∆(G)+ 1. Determining whether a simple graph is Class 1 orClass 2 is a hard problem.

If a graphG has multiple (parallel) edges (but not loops), themultiplicity of G, denotedby µ(G), is the maximum number of parallel edges connecting a pair ofvertices. Theorem5.8.1 allows then the following generalization:

Theorem 5.8.2 (Vizing, 1964)If G is a loop-less multigraph with multiplicity µ(G), then∆(G) ≤ χ′(G) ≤ ∆(G)+µ(G).

b

b b

∆ +µ= 6+3 = 9

Figure 5.27. “Fat triangle” requires∆ +µ colors.

The “worst” example when the upper bound is achieved (i.e.χ′(G) = ∆(G) + µ(G))is represented by the so called “fat triangle”, see Figure 5.27. Theorem 5.8.2 is a directgeneralization of Theorem 5.8.1 because for simple graphsµ= 1.

We conclude with

Theorem 5.8.3 (Konig, 1916) If G is a bipartite graph, thenχ′(G) = ∆(G).

Exercises 5.8.

1. Find the chromatic index and an optimal edge coloring of:Kn,Cn,Wn,Km,n, wherem,n≥ 3.


2. Find the chromatic index, chromatic class and an optimal edge coloring of cube, prism andtheir complements.

3. Prove that for the Petersen graphχ′ = 4.

4. In the Petersen graph, replace each edge by three paralleledges and find the chromatic indexand respective edge coloring.


1. Given a tree, find the chromatic index and an optimal edge coloring.

2. Given a graphG, at random generate an edge coloring and test if it is proper.

3. Given a multigraphG, generate a proper edge coloring using a vertex ordering.

5.9. Upper Chromatic Index

In this section we show how replacing the definition of properedge coloring with theop-positeversion (the requirement “all of different colors” is replaced by “at least two of thesame color”) leads to the concept of the upper chromatic index and a formula for it.

Let G = (X,E) be an arbitrary multigraph without loops and, as usually,{1,2, . . . ,λ}be the set of available colors.

Definition 5.9.1 A proper edgeλ-coloring of multigraph G is an assignment of a colorfrom set{1,2, . . . ,λ} to every edge of G in such a way that every non-pendant vertex of Gis incident to at least two edges of the same color.

Following this new definition, we can color all edges with onecolor and it will be aproper coloring; however, we cannot use|E| colors unlessG is just a matching. Let usagree that ifk colors are really used in a proper edge coloring,k ≤ λ, then the coloring iscalledstrict edgek-coloring, or justk-coloring.

Definition 5.9.2 The maximum number k for which there exists a proper edge k-coloringof multigraph G is called theupper chromatic index and denoted byχ′(G).

An example of a multigraphG and its edge 3-coloring is shown in Figure 5.28. We willshow that it uses the maximum number of colors, i.e.,χ′(G) = 3.

Theorem 5.9.1 For a connected multigraph G, the upper chromatic indexχ′(G) = 1 if andonly if the maximum degree∆(G) ≤ 2.

Proof. ⇒ Let χ′(G) = 1. For a contradiction, suppose∆(G) ≥ 3. If G contains a cycle,then color the edges of the cycle with color 1, and all the other edges with color 2. IfG isa tree, color any maximal path with color 1 and all other edgeswith color 2. In both caseswe obtain a proper edge 2-coloring, a contradiction.

⇐ The converse is evident since∆(G) ≤ 2 implies thatG is either a cycle or a path.�

Graph Coloring 117

b b

b

b

12

2

2 3

G

Figure 5.28.

For a connected multigraphG, theorem 5.9.1 immediately implies thatχ′(G)≥ 2 if andonly if ∆(G) ≥ 3.

Consider a proper edgek-coloring of G = (X,E); it partitionsE into k color classes{C1,C2, . . . ,Ck} where eachCi is a set of edges colored with colori. We denote such parti-tion and the respective edge coloring byf .

Let Gi = (Xi,Ci) be the subgraph ofG with a set of verticesXi determined by endpointsof the edges ofCi . Notice that subgraphsGi are not necessarily connected multigraphs andmay have common vertices with each other; however, they do not have common edges.

A vertexx of a subgraphGi is said to besatisfiedby subgraphGi if it is not a pendantvertex inGi.

Let us discuss some properties of edge colorings that useχ′(G) colors.

Theorem 5.9.2 If f is a coloring of G usingχ′(G) colors, then

χ′(Gi) = 1, i = 1, . . . , χ′(G).

Proof. In coloring f , every pendant vertex ofGi is satisfied by some other subgraph.Therefore any proper coloring ofGi with at least two colors leads to a proper(χ′(G)+1)-coloring ofG, a contradiction. �

Corollary 5.9.1 Let f be a coloring of G usingχ′(G) colors. Then each Gi , where1≤ i ≤χ′(G), is either a cycle or a path.

Proof. Indeed, inf , everyGi must be connected because otherwise we could increase thenumber of colors. Any connected multigraph withχ′(G) = 1 is either a cycle or a path(Theorem 5.9.1). �

Theorem 5.9.3 Let f be a coloring of G usingχ′(G) colors. Then for every vertex x of Gthe following implications hold:

1) if x is satisfied by a cycle, then x is satisfied by no path;2) if x is satisfied by two cycles, say Gi and Gj , then x is the only common vertex for Gi

and Gj .

Proof. In 1), otherwise, we could break the path in two paths and increase the number ofcolors.


b

b

x

Gi G j

color i color jnew color

Figure 5.29.

In 2), otherwise, we could split two cyclesGi andG j into one cycle and two paths, seeFigure 5.29, assign colorsi and j to the paths and a new color to the new cycle, and again,increase the number of colors. �

Corollary 5.9.2 Let f be a coloring of G usingχ′(G) colors. If a vertex x of G is satisfiedby more than one Gk, then all Gl s which satisfy x, are all cycles with one vertex in common.

Given aχ′(G)-coloring of an arbitrary multigraphG, we now partition the subgraphsGi of G into the following three classes.

Class A:contains allGi forming the maximum number of (vertex) disjoint cycles inf .Clearly, the numberc′ of subgraphs in A satisfiesc′ ≤ c wherec is the maximum numberof disjoint cycles inG.

Class B:contains the remaining cycles and paths with length at leasttwo each.Class C:containsGi which represent a separate edge each.

Observe that all non-pendant vertices ofG are satisfied by subgraphs from classes A andB, and none is satisfied by any edge from part C; in addition, nopendant vertex is satisfied.

If class B contains cycles, then given the maximum number of disjoint cycles in A, andby Theorem 5.9.2, each of these cycles has a single vertex in common with only one of thecycles in A. The cycles in B may be considered as paths whose endpoints coincide. Noticethat the number of edges in each such cycle or path equals the number of internal verticesplus 1.

The next theorem determines the formula for the upper chromatic index. It may beseen as theopposite to Vizing’s theoremin the sense that finding the minimum number ofcolors is replaced with finding the maximum number of colors.At this point, proper vertexcoloring defined in Section 5.2., proper edge coloring defined in Section 5.8. and properedge coloring defined in this section look unrelated; we willsee however in Part II, that allthey represent special cases of one unifying concept called“mixed hypergraph coloring”.

Theorem 5.9.4 (M.Gionfriddo, Milazzo, Voloshin, 2001)If G = (X,E) is an arbitrarymultigraph with|X| = n, |E| = m, the number of pendant vertices p, and the maximumnumber of disjoint cycles c, then

χ′(G) = c+m−n+ p.

Graph Coloring 119

Proof. We prove first thatχ′(G) ≤ c+m−n+ p. By Corollary 5.9.1, if a coloringf usesχ′(G) colors, then subgraphsGi, 1≤ i ≤ χ′(G), are either cycles or paths, while Corollary5.9.2 states that if a vertexx is satisfied by more than one graphGi , then all these graphshave precisely one vertex in common.

Let c′ be the number of subgraphs, andv′ be the number of satisfied vertices in classA. Clearly, v′ coincides with the number of edges in A because all cycles in Aare vertexdisjoint. Letx be a satisfied vertex inG. If x is satisfied by someGl from class A, then it ispossible to look at all the otherGks which satisfyx, like paths with endpointsx.

If x is not satisfied by any subgraph from class A, then it is satisfied by a subgraph fromclass B; therefore it is possible to consider one such subgraph Gk as a cycle and all othersas paths with endpointsx.

Let r be the number ofGis in class B. Since pendant vertices are not satisfied, thenumber of vertices that are satisfied by subgraphs from B equals n − v′ − p. They are theinternal vertices for the paths or the internal vertices forthe cycles which are consideredas paths with coinciding endpoints. Since the number of edges in each of them equals thenumber of internal vertices plus 1, and the number of all suchsubgraphs equalsr, the totalnumber of edges in B equals

n − v′ − p + r.

Let m′ be the number of edges in classes A and B combined. Then we obtain:

m′ = v′ +(n−v′− p+ r) = n− p+ r.

Since class C containsm−m′ edges, the number of colors inf is:

χ′(G) = c′ + r + (m − m′) = c′ +m−n+ p,

and therefore we obtain

χ′(G) = c′ + m − n + p≤ c+m−n+ p.

Next we prove thatχ′(G) ≥ c+ m− n + p. Let us choosec vertex disjoint cyclesC1, . . . ,Cc and denote the subgraph with vertex setX and edges of these cycles byA0. Colorthe edges of the cycles properly with the colors 1, . . . ,c. Since the number of edgesm(A0) =n(A0), and the number of pendant verticesp(A0) = 0, we obtain a proper coloring ofA0 us-ing c+m(A0)−n(A0)+ p(A0) colors. Therefore,χ′(A0)≥ c+m(A0)−n(A0)+ p(A0), andthe inequality holds.

Now we implement the following coloring augmenting procedure. Choose any satisfiedvertex, sayx, incident to an uncolored edge and construct a path along uncolored edgesuntil the first satisfied vertexy is reached or we get stuck at a pendant vertexy. SinceA0

constitutes the maximum number of disjoint cycles, the new vertices and edges representeither a cycle or a path (if they form a cycle, thenx = y).

Add these new vertices and edges toA0 and denote the obtained subgraph byA1. Colorthe added edges with a new color and declare respective vertices satisfied.

If y is pendant, then the numbers of newly added vertices and edges coincide. Ify is notpendant, then the number of added edges equals the number of added vertices plus 1. So,in either case,

χ′(A1) ≥ χ′(A0)+1≥ c+m(A1)−n(A1)+ p(A1),


and therefore the inequality holds. We repeat this coloringprocedure by constructing sub-graphsA2,A3, . . . until all the edges ofG are colored and all non-pendant vertices satisfied.Since at each coloring step the inequality holds, we obtain that χ′(G) ≥ c+ m− n+ p.Hence the theorem follows. �

For our example, see Figure 5.28, it is easy to see thatc = 1, m= 5, n = 4 andp = 1,and the formula gives:

χ′(G) = c+m−n+ p= 1+5−4+1= 3,

so the coloring in the Figure ismaximal, i.e., using the maximum number of colors. Asthere are several largest sets of vertex disjoint cycles, there are several maximal colorings.Notice that any two parallel edges form cycleC2 and may contribute to the value ofc.

Theorem 5.9.4 allows finding the upper chromatic index for particular classes of multi-graphs.

Theorem 5.9.5 If G is a tree, then

χ′(G) = p − 1.

Proof. Indeed, for any treec = 0 andm = n − 1. �

Exercises 5.9.

b

b

bb

b

G

Figure 5.30.

1. For graphG, see Figure 5.30, find a maximal edge coloring by hand.

2. For graphG, see Figure 5.30, find the largest set of vertex disjoint cycles and determineχ′(G) = c+m−n+ pand respective edge coloring.

3. For graphG, see Figure 5.30, find another largest set of vertex disjointcycles, calculateχ′(G)and show respective edge coloring.

4. For graphG, see Figure 5.30, show that coloring the edges ofC5 with one color does notproduce maximal coloring.

Graph Coloring 121

5. Find the upper chromatic index and a maximal edge coloringof: Cn, Wn, Kn,Km,n, wherem,n≥ 2.

6. Find the upper chromatic index and a maximal edge coloringof Petersen graph, cube andprism and their complements.

7. In Petersen graph, replace each edge by three parallel edges and find the upper chromaticindex and respective edge coloring.

8. InCn,Wn,Kn,Km,n, wherem,n≥ 2, replace each edge by two parallel edges and find the upperchromatic index and respective edge coloring.

Computer Projects 5.9. Write a program for the following algorithmic problems.

1. Given a graphG, find a number of vertex disjoint cycles and an estimate onχ′(G).

2. Given a graphG and a number of properly colored vertex disjoint cycles, implement the edgecoloring procedure as described in the proof of Theorem 5.9.4 to augment the coloring.

3. Generate an edge coloring at random and verify if it is proper.

4. Generate a maximal edge coloring of complete graphKn, n≥ 5.

5. Given a treeTn, construct a maximal edge coloring.

Chapter 6

Traversals and Flows

6.1. Eulerian Graphs

In a graphG, a walk is an alternating sequence of vertices and edges where everyedgeconnects preceding and succeeding vertices in the sequence. A walk starts at a vertex, endat a vertex and has the following form:x0,e1,x1, e2, . . . , ek,xk. There are no restrictions onrepetitions of vertices and edges. The number of edges in a walk is its length. So, a walkbeginning atx0 and ending atxk has the lengthk; it is a (x0,xk)-walk. A walk with x0 = xk

is calledclosed.If no edge is repeating, then a walk is called a(x0,xk)-trail . A trail with norepeating vertices clearly is a path; a path withx0 = xk clearly is a cycle. Thus we recognizethe well known concepts as the special cases of a walk and trail.

A connected graphG is calledEulerian if it has a closed trail containing all edges ofG; such a trail is then called anEulerian trail . Since trails do not repeat edges, Euleriantrail passes through each edge once.

Lemma 6.1.1 If in a graph G the degree of every vertex is at least 2, then G contains acycle.

Proof. If G contains loops or multiple edges, then the statement is evident. Therefore,assumeG is a simple graph. Consider a pathP of maximal length with an endpointx.Observe that all neighbors ofx line onP because otherwise we could extendP. SinceG issimple and the degree ofx is at leats 2, vertexx has at least two distinct neighbors onP.Thus vertexx, the two its neighbors onP and the segment ofP between them form a cycleof length at least 3. �

Theorem 6.1.1 (Euler, 1736)A connected graph G is Eulerian if and only if the degree ofevery vertex is even.

Proof. ⇒ Let T be an Eulerian trail inG.Each timeT passes through a vertex, it uses twoedges. SinceT is closed and uses all edges, every vertex has an even degree.

⇐ Induction on the number of edges ofG: assume the statement is true for all graphswith less thanm(G) edges. Since the degree of every vertex inG is even andG is connected,the degree each vertex is at least 2. Therefore, by Lemma 6.1.1, G contains a cycleC, see


b

b

b b

b

b

b

b

b

b

b

Figure 6.1.

Figure 6.1 (take any cycle asC). Delete the edges ofC from G and obtain a graphG′

in which all vertices have even degrees. GraphG′ may be disconnected but sinceG isconnected, cycleC passes through each component ofG′. Sincem(G′) < m(G), by theinduction hypothesis, every connected component ofG′ has an Eulerian trail. Now wecombine cycleC with an Eulerian trail of each connected component ofG′ in the followingway: traverseC and make a detour at the very first vertex of each component. Since everydetour ends at the same vertex where it started, an Eulerian trail of G is constructed. �

The theorem above implies that every graphG = (X,E) having all vertices of evendegree may be decomposed into cycles, i.e., edge setE may be partitioned into subsetswhere each subset forms a cycle.

The proof of the theorem above explicitly suggests the idea how to construct an Euleriantrail in a given Eulerian graph: choose a closed trail and then recurrently extend it until theinitial graph is obtained.

Exercises 6.1.

1. Which of the graphsKn, Km,n, Wn, cube, prism, the Petersen graph, a tree are Eulerian?

2. The degree equalityΣni=1d(xi) = 2m (See Proposition 1.1.1) implies that in any graphG, the

number of vertices of odd degree is even. How this fact can be used to make any graphEulerian?

3. Show that if a connected graphG containsk vertices of odd degree, then the minimum numberof trails that partition the edges isk/2.

4. Show that if a connected graphG containsk vertices of odd degree, thenk/2 continuouspen-strokes are sufficient to drawG in the plane.

5. What is the minimum number of trails that partition the edge set of the Petersen graph?


1. Given a graphG, recognize if it is Eulerian, and if yes, then construct an Eulerian trail.

Traversals and Flows 125

6.2. Hamiltonian Graphs

In contrast to Eulerian graphs, one may ask if in a graphG, there exists a closed trail passingexactly one time through every vertex ofG. Clearly, such a trail is a spanning cycle whichis then called aHamiltonian cycle. If G contains a Hamiltonian cycle, then it is called aHamiltonian graph .

In contrast (one more) to Eulerian graphs, the problem of recognizing Hamiltoniangraphs is very difficult. Until now there are no criteria for characterization of Hamiltoniangraphs. Typical results about Hamiltonian graphs require significant number of edges ina graph. However, there are Hamiltonian graphs (like cyclesetc.) with small number ofedges. Recall thatd(x) is the degree of a vertexx.

Theorem 6.2.1 (Ore, 1960)If in a simple graph G= (X,E), with |X| = n ≥ 3, for everypair x and y of disjoint vertices

d(x)+d(y) ≥ n,

then G is a Hamiltonian graph.

Proof. By contradiction, assume thatG is not Hamiltonian. Hence it is not a completegraph and we can sequentially add edges toG as long as possible before it becomes Hamil-tonian. The addition of edges preserves the inequality on vertex degrees. Therefore, with-out loss of generality, we can assume thatG is maximal “non-Hamiltonian” graph. Hence itcontains a pathP passing (in order from left to right) through all the vertices x1,x2, . . . ,xn,

b

b

b b

b

b

x1 xn

x2 xn−1

xi xi+1

Figure 6.2.

see Figure 6.2. SinceG is not Hamiltonian, verticesx1 andxn are disjoint what impliesd(x1)+d(xn)≥ n. On the other hand, sincex1 andxn are disjoint,d(x1)≤ n−2 andd(xn)≤n− 2. Consider alln− 3 internal edges of pathP. If for each of them the left end is notadjacent toxn or the right end is not adjacent tox1, then in total sum of degrees forx1 andxn at leastn−3 edges are missing. Therefore

d(x1)+d(xn) ≤ (n−2)+ (n−2)− (n−3) = n−1


what contradicts the condition of theorem. The last impliesthat amongn−3 internal edgesof P there exists one, say,(xi ,xi+1) with xi adjacent toxn andxi+1 adjacent tox1.

Thus the cyclex1,xi+1,xi+2, . . . ,xn,xi ,xi−1, . . . ,x1

is Hamiltonian what completes the proof.�

Corollary 6.2.1 (Dirac, 1952) If in a simple graph G= (X,E), with |X|= n≥ 3, the degreeof each vertex is at least n/2, then G is Hamiltonian.

Proof. The statement follows directly from the theorem above sinced(x)+d(y)≥ n holdsfor every pair of vertices inG.

�

In both the theorem and corollary we include the conditionn ≥ 3 to avoid graphK2

which is not Hamiltonian but satisfies the degree inequality.

Exercises 6.2.

1. For which values ofn andm graphsKn,Km,n, Wn are Hamiltonian?

2. Are the prism, cube, and the Petersen graph Hamiltonian?

3. Is the Grozsch graph Hamiltonian (see Figure 5.23)?

4. Which of the complement of prism, cube, Petersen and Grozsch graphs are Hamiltonian?

5. Which of the graphsC7 andC8 are Hamiltonian?

6. Construct a list of all Hamiltonian cycles ofK4 andK5.


1. Given a graphG with the degree condition as in Theorem 6.2.1, and a path withdisjoint endspassing through all the vertices. For the path, find an edge extension to construct a cycle asin Theorem 6.2.1.

2. Given a graphG and a set of edges, verify if the set of edges forms a cycle.

3. Given a graphG and a cycle, find the procedure of extending the cycle as much as possible.

4. Given a complete graphKn, generate all Hamiltonian cycles.

5. Given a graphG, generaten edges at random and check if they form a cycle.

6. Given a 3-regular graph on 10 vertices, check if it is Hamiltonian.


6.3. Network Flows

Networks are anywhere: in traffic, communications, internet, even in our body. It appearsthat graph theory provides an important mathematica model that allows to find optimalflows in networks.

Recall that a graph in which all edges are ordered pairs (and therefore are called arcs)is called a digraph. A digraphN = (X,A) is called anetwork, if X is a set of vertices(sometimes callednodes), A is a set of arcs, and to each arca ∈ A a non-negative realnumberc(a) is assigned which is called thecapacity of arc a. For any vertexy ∈ X, anyarc of type(x,y) is calledincoming, and every arc of type(y,z) is calledoutcoming. In N,there are two special vertices:u∈X, which is not incident to any incoming arc and is calledthesource, andv∈ X, which is not incident to any outcoming arc and is called thesink.

A network flow F is an assignment toeacharca∈ A a non-negative real numberf (a)(called theflow in a) such that:

1. f (a) ≤ c(a);

2. for any vertexx∈ X, exceptu andv, the followingflow conservation lawholds: thesum of flows of all incoming arcs equals the sum of flows of all outcoming arcs.

An arc a ∈ A for which f (a) = c(a) is calledsaturated; if f (a) < c(a), then arca iscalledunsaturated. Thevalue of the network flow is the sum of all flows in arcs of type(u,x); the flow conservation law implies that it equals the sum of all flows in arcs of type(x,v).

Given a networkN = (X,A), how can we find a maximum flow? Observe that thereis no such concept as the conservation law for the capacities: otherwise we could run themaximum flow right from the source. The answer to this question is closely related to theconcept of acut which is a subset of arcsS⊆ A that separates the source from the sink. Itmeans that every path from the sourceu to the sinkv contains at least one arc fromS. Thecapacity of the cut S is the sum of the capacities of all arcs fromS. Different cuts havedifferent capacities, and evidently, no flow can exceed the smallest capacity over all cuts.Any cut having the smallest capacity is called theminimum cut .

Theorem 6.3.1 (Max-flow min-cut theorem, Ford and Fulkerson, 1956)In each network, the value of maximum flow equals the capacityof minimum cut.

Proof. Since the value of maximum flow does not exceed the capacity ofminimum cut,we prove that for any given maximum flow there exists a minimumcut having the capacityequal to the value of the flow.

Let N = (X,A) be a network with a maximum flow. Consider a sequence of verticesu = x0 → x1 → x2 → ··· → xk such that either(xi ,xi+1) is an unsaturated arc, or(xi+1,xi) isan arc with a non-zero flow,i = 0,1, . . . ,k−1. Denote the set of all such vertices in all suchsequences byY. Sourceu does not have incoming arcs; if all outcoming arcs are saturated,then we are done. If at least one outcoming arc from the sourceis unsaturated, then we haveat least one such sequence andu∈Y. Let Z = X \Y. We claim that the sinkv∈ Z.

By contradiction, assumev∈Y. It means that there exists a sequenceu = x0 → x1 →x2 → ··· → xk = v with the property above. Choose a numberδ > 0 which does not exceed


the value needed to saturate any unsaturated arc of type(xi ,xi+1) and does not exceed theflow in any arc of the type(xi+1,xi). We now increase the flow in all arcs of the first typeand decrease it in all arcs of the second type by the same valueδ. It is clear that the flowconservation law holds, and no capacity is exceeded. Therefore we obtain a flow which isgreater than the maximum flow, a contradiction.

Now, let Sbe the set of all arcs with initial vertex inY and terminal vertex inZ. Evi-dently,S is a cut. Every arc fromS is saturated because otherwise we could move a respec-tive terminal vertex fromZ toY. Each arc fromZ to Y has the zero flow because otherwisewe could move a respective initial vertex fromZ to Y. This implies that the capacity of cutSequals the value of the given maximum flow what completes the proof. �

The idea of the proof in the theorem above is used in the Ford-Fulkerson algorithm forfinding maximum flow in a network. It is illustrated by the example of a network and flowshown in Figure 6.3 with continuation in Figure 6.4.

The pair of numbers (0,3) attached to arc(u,x1) in the initial flow F0 means that theflow in arc (u,x1) is 0, and the capacity is 3. The same “(flow, capacity)” rule holds forevery arc in each flow. As one can see, the initial flowF0 is obtained by running one unit offlow along the path

u→ x4 → x3 → x1 → x2 → v

and assigning flow 0 in all remaining arcs. It immediately saturates the arc(x3,x1) as thepair (1,1) shows.

Flow F1 is obtained from the flowF0 by adding one unit in arc(u,x1), subtracting it inbackward arc(x1,x3), and adding it in arc the(x3,v).

Flow F2 is obtained from the flowF1 by adding two units along the path

u→ x1 → x2 → v.

Flow F3 (see the next figure) is obtained from the flowF2 by adding one unit along thepath

u→ x4 → x3 → v.

At last maximum flowF4 is obtained from the flowF3 by adding one unit along the path

u→ x4 → x3 → x1 → x2 → v.

We conclude that the flow value equals 6, it is maximum becausethe arcs(u,x1), (x3,x1)and (x3,v) form a cut with the capacity 6, so it is the minimum cut. At eachstep weaugmented the flow. One can easily see that there is no furtherflow-augmenting path at thispoint.

Generally, there are many ways to search for the flow augmenting paths.

Exercises 6.3.

1. In the example in Figure 6.3, add arc(x4,x2) with capacity 2 and find the maximumflow.

2. In the example in Figure 6.3, change the capacity of some arcs and find the maximumflow.


b

b

b

b

b

bsourceu sink v

x1 x2

x3x4

(1,5)

(0,3)

(1,7)

(1,1)

(1,4)

(1,6)

(0,2)

b

b

b

b

b

bsourceu sink v

x1 x2

x3x4

(1,5)

(1,3)

(1,7)

(0,1)

(1,4)

(1,6)

(1,2)

b

b

b

b

b

bsourceu sink v

x1 x2

x3x4

(1,5)

(3,3)

(3,7)

(0,1)

(1,4)

(3,6)

(1,2)

initial flow F0

flow F1

flow F2

Figure 6.3.

3. In the example in Figure 6.3, change the direction of an arcand find the maximumflow.

4. Given a network with several sources and sinks, suggest a way to reduce the problem


b

b

b

b

b

bsourceu sink v

x1 x2

x3x4

(2,5)

(3,3)

(3,7)

(0,1)

(2,4)

(3,6)

(2,2)

b

b

b

b

b

bsourceu sink v

x1 x2

x3x4

(3,5)

(3,3)

(4,7)

(1,1)

(3,4)

(4,6)

(2,2)

flow F3

maximum flowF4

Figure 6.4.

to the one with one source and one sink.

5. Given a cube, prism, wheelW7, assign direction to every edge (making them arcs) ina way to obtain a network with one source and one sink, assign the capacity to eacharc, and find the maximum flow.


1. Given a network, find a flow augmenting path.

2. Transform Petersen graph into a network with one source and one sink and find amaximum flow.

3. Given a network with a flow, determine if the flow is maximum.

Part II

HYPERGRAPHS

131

“There is no Mathematics without generalizations...

Mathematics itself is a pure generalization of the world...”

Chapter 7

Basic Hypergraph Concepts

“First was the idea...”

7.1. Preliminary Definitions

b

b b

b b

b

Figure 7.1. This is a hypergraph.

An example of a hypergraph is shown in Figure 7.1. Thebasic ideaof the hypergraphconcept is to consider such ageneralizationof a graph in whichany subsetof a givenset may be an edge rather than two-element subsets. In drawing hypergraphs, vertices arepoints in the plane, edges of size 2 are curves connecting respective vertices (as in graphdrawing), and edges of size different from 2 are closed curves separating a respective subsetfrom the rest of vertices, see Figure 7.1.

In what follows we provide basic hypergraph definitions which generalize the respectivegraph concepts ([6]). LetX = {x1,x2, . . . ,xn} be a finite set, and letD = {D1,D2, . . . ,Dm}be a family of subsets ofX. The pairH = (X,D) is called ahypergraph with vertex setX also denoted byV(H ), and withedge set D also denoted byD(H ). Sometimes, thehypergraphH = (X,D) is called aset-system.


|X| = n is called theorder of the hypergraph, written also asn, or n(H ). The elementsx1,x2, . . . ,xn are called thevertices and the setsD1,D2, . . . ,Dm are called theedges (hy-peredges). The number of edges is usually denoted bym or m(H ). Sometimes we willomit the indices when denoting the vertices and edges if thisevidently does not lead tomisunderstanding. To include the most general case (it may happen in some algorithms),we assume that the set of verticesX and/or the familyD may be empty. A hypergraphwhich contains no vertices and no edges is called theempty set. Some edges may also beempty sets. Some edges may be the subsets of some other edges;in this case they are calledincluded. In some cases some edges may coincide; they are then calledmultiple . A hyper-graph is calledsimple if it contains no included edges. Hence simple hypergraphs do nothave empty and multiple edges. Simple hypergraphs are also known asSperner families.

In a hypergraph, twovertices are said to beadjacent if there is an edgeD ∈ D thatcontains both vertices. The adjacent vertices are sometimes calledneighbor to each other,and all the neighbors for a given vertexx are called theneighborhood of x in a graph orhypergraph. The neighborhood ofx is denoted byN(x). Two edgesare said to beadjacentif their intersection is not empty. If a vertexxi ∈ X belongs to an edgeD j ∈ D, then we saythat they areincident to each other. As one can see, as in graph theory, the adjacency isreferred to the elements of the same kind (vertices vs vertices, or edges vs edges), while theincidence is referred to the elements of different kind (vertices vs edges).

D(x), x∈ X, will denoteall the edges containing the vertexx. The number|D(x)| iscalled thedegree of the vertexx, the number|Di| is called thedegree (size, cardinality)of the edgeDi. Themaximum degreeof the hypergraphH is denoted by

∆(H ) = maxx∈X

|D(x)|.

A hypergraph in which all vertices have the same degreek≥ 0 is calledk-regular. Ahypergraph in which all edges have the same degreer ≥ 0 is calledr-uniform . Therankof a hypergraphH is

r(H ) = maxD∈D

|D|.

An edge of a hypergraph which contains no vertices is called an empty edge. Thedegree of an empty edge is trivially 0. A vertex of a hypergraph which is incident to noedges is called anisolated vertex. The degree of an isolated vertex is trivially 0. An edgeof cardinality 1 is called asingleton (loop), a vertex of degree 1 is called apendant vertex.

A simple hypergraphH with |Di | = 2 for eachDi ∈ D is thus asimple graph, maybewith isolated vertices.

Two simple hypergraphsH1 andH2 are calledisomorphic if there exists a one-to-onecorrespondence between their vertex sets such that any subset of vertices form an edge inH1 if and only if the corresponding subset of vertices forms an edge inH2.

Hypergraph modeling examples. Hypergraphs can model concepts in different sci-ences in a much more general setting than graphs do. In addition, they help to find optimalsolutions for many new optimization problems. While vertices represent elements of a set,the hyperedges represent properties of different subsets,or, even more generally,arbitrarystatements about arbitrary subsets. Let us mention just a few examples.

Basic Hypergraph Concepts 137

Mathematics:

• the vertices are natural numbers from 1 to 100; the hyperedges are the subsets ofnumbers having a common divisor greater than 1, one subset for each common divi-sor;

• the vertices are a finite set of points on the real line; the edges are some subsets of thepoints which form intervals in the ordering of the points, one hyperedge for one suchsubset;

• the vertices are vertices of a graph; the edges are the subsets forming closed neigh-borhoods, i. e., one neighborhood plus the vertex itself foreach vertex;

• the vertices are vertices of a 3-dimensional polyhedron; each face of the polyhedronforms a hyperedge;

• the vertices are points of a finite geometry; each line of the geometry forms a hyper-edge;

• the vertices are points of a block design; each block forms a hyperedge.

Computer science:

• the vertices are computers in a network; the edges are the subsets of computers withdevices from different manufacturers, one subset for everymanufacturer;

• the vertices are all possible inputs for a chip; the edges arethe subsets of inputs whichdetermine some internal defects, one subset for each defect;

• the vertices are files in a data base; the edges are files neededto open for a query, onesubset for every query;

• the vertices are records in a relational data base; the edgesare the subsets of recordsfor which the values of some attributes are “true”, one subset for each attribute.

Genetics:

• the vertices are the elements (A, T, G and C) of a specific DNA sequence; the edgesare the subsets of elements representing genes, one edge fora gene;

• the vertices are species; the edges are the subsets of species having common heredi-tary properties, one edge for a property.

Physics/Chemistry:

• the vertices are the atoms in a molecule; hyperedges of degree 2 correspond to simplecovalent bonds, and hyperedges of degree greater than 2 correspond to polycentricbonds;


• the vertices are chemical compounds produced by a chemical factory; every subset ofcompounds that might explode if combined forms a hyperedge (compare this modelwith its special case discussed in detail at the beginning ofSection 1.2.).

Sociology:

• the vertices are employees in a company; the edges are the subsets of people whohave some common interest, one subset for each interest;

• the vertices are all the interests of employees in a company;the edges are the subsetsof interests which specific employees have, one edge for every employee.

Healthcare:

• the vertices are illnesses; the edges are the subsets of illnesses which can be treatedby some medicines, one hyperedge for each medicine;

• the vertices are medicines; the edges are the subsets of medicines which treat someillnesses, one hyperedge for every illness;

• the vertices are illnesses; the edges are the subsets of illnesses which have somespecific symptoms, one hyperedge for each symptom;

• the vertices are symptoms; the edges are the subsets of symptoms characteristic tosome illnesses, one hyperedge for each illness.

Broadcasting:

• the vertices are the radio transmitters in a region; the edges are the subsets of transmit-ters which transmit on the same frequency right now, one subset for each frequency.

Geographical maps:

• the vertices are cities; the edges are the cities which are onthe same highway;

• the vertices are street crossings in the city map; the edges are the subsets correspond-ing to the bus routes, one edge for each bus route.

Exercises 7.1.

For each of the hypergraphs in Figure 7.2:

1. Find the order and the number of edges;

2. Find included edges (if any).

3. Find multiple edges (if any).

4. Is the hypergraph simple?


b

b b

b b

b

b b

bb

b

b

H1 H2

Figure 7.2.

5. For each pair of vertices, determine if they are adjacent.

6. For every vertex, find the degree and neighborhood.

7. For each pair of edges, determine if they are adjacent.

8. For each edge, find the size.

9. Find the maximum degree.

10. Is the hypergraph regular?

11. Is the hypergraph uniform?

12. Find the rank.

13. Find isolated and pendant vertices (if any).

14. Find a singleton or an empty edge (if any).

15. Name the vertices and edges, re-draw the hypergraph differently and show an iso-morphism.

7.2. Incidence and Duality

“Duality of hypergraphs: a look from the inside out...”

The incidence matrix of a hypergraphH = (X,D) is a matrixI(H ) with n rows thatrepresent the vertices andmcolumns that represent the edges ofH such that

(i, j)-entry=

{

1 if xi ∈ D j ,0 if xi /∈ D j .

An example of the incidence matrix is shown in Figure 7.3. As one can see, in contrast tographs, any (0,1)-matrix is the incidence matrix of a hypergraph. From this point of view,


the hypergraph theory is the theory of (0,1)-matrices. It also follows that empty edges meanzero columns and isolated vertices mean zero rows in the incidence matrix. Let us agreethat if the vertex set of a hypergraph is empty, then the incidence matrix consists only of therow containing the names of the edges; similarly, if the edgeset of a hypergraph is empty,then the incidence matrix consists only of the column containing the names of the vertices.If all the vertices are isolated and all the edges are empty sets, then the incidence matrix istotally zeros.

Let H = (X,D) be a hypergraph withX = {x1,x2, . . . ,xn} and D = {D1,D2, . . . ,Dm}. The dual of the hypergraphH is a hypergraphH ∗ = (Y,Z) whose vertex set isY = {d1,d2, . . . ,dm}, and the edge set is defined as follows:

Z = {X1,X2, . . . ,Xn},

Xi = {d j : xi ∈ D j in H }.

An example of the dual hypergraph is shown in Figure 7.3. It follows that isolated verticesbecome empty edges in the dual hypergraph and vice versa. Since we allow isolated verticesand empty edges, the dual hypergraph may be constructed for any hypergraph. Observethat the incidence matrixI(H ∗) is the transposeI∗ of the incidence matrixI(H ). So,the alternative definition of the dual is the following: a hypergraphH ∗ is thedual of thehypergraphH if I(H ∗) = I∗(H ). Evidently, if we transpose a matrix twice, we receive thesame matrix, therefore(H ∗)∗ = H . Following the agreement above this holds even in thecase whenX or/andD are empty: the column becomes a row and vice versa. Notice that∆(H ) = r(H ∗) and the dual of ak-regular hypergraph isk-uniform. The definition aboveimplies thatdrawing dual hypergraph is a simple procedure: begin with a drawing ofH

⇒ construct incidence matrixI(H ) ⇒ transpose it to obtainI∗(H ) ⇒ draw dualH ∗.For a hypergraphH = (X,D), we define thebipartite representation of H to be the

bipartite graphB(H ) = (X,D;E) with the vertex setX∪D, whereX is the left part,D isthe right part, andE is the edge set; vertexx∈ X is adjacent to vertexD ∈ D in B(H ) ifand only if vertexx∈ X is incident to edgeD ∈ D in H . In this way, any bipartite graph isa bipartite representation of a hypergraph.

It is seen thatB(H ∗) is obtained fromB(H ) simply by interchanging the right part andthe left part while preserving all edges.

The very same hypergraphH , see Figure 7.1, its incidence matrixI(H ), dual hyper-graphH ∗ and bipartite representationB(H ) are shown in Figure 7.3:

H = (X,D), X = {1,2,3,4,5,6}, D = {D1,D2,D3,D4,D5},

D1 = {1}, D2 = {1,2}, D3 = {1,2,4},

D4 = {2,3,5}, D5 = {3,4,5}.

Notice that edgeD1 with |D1| = 1 (singleton) is drawn as the circle containing only vertexx1. So, loops from graph theory become singletons and are drawn differently. An emptyedge is drawn as a circle containing no vertices. These rulesare used throughout the bookunless stated otherwise. InH , edgeD1 is a singleton, vertex 6 is isolated, vertices 3 and5 are incident to the same edges, edgeD2 is included in edgeD3; in I(H ), column D1

contains one 1, row 6 is a zero row, rows 3 and 5 are identical; in H ∗, vertexd1 is of degree


b

b b

b b

b

b b

b b

b

D1 1

D5

2

3

4

5

6

D4

D2

D3

H = (X,D)

D1 D2 D3 D4 D5

1123456

00000

110000

110100

011010

001110

I(H )

d1

d2d3

d4

d5

12

35

4

6

H ∗

b

b

b

b

b

b

1

2

3

4

5

6

b

b

b

b

b

D1

D2

D3

D4

D5

B(H )

Figure 7.3. Example of a hypergraphH , its incidence matrixI(H ), dualH ∗ and bipartiterepresentationB(H )([6]).

1, edge 6 is empty, edges 3 and 5 are multiple; inB(H ), vertexD1 is pendant, vertex 6 isisolated, vertices 3 and 5 have the same neighbors and so on. One can see how the names,degrees, adjacency and incidence, etc. inH are looking in incidence matrixI(H ), dualhypergraphH ∗ and bipartite representationB(H ).

Every picture of a hypergraph is already an isomorphism between the incidence matrixand the points and curves in the plane. In this sense, Figure 7.3 contains four differentpictures of the same structure defined as the hypergraphH . One can easily check thatincidence matrix ofH ∗ is the transpose of the incidence matrix ofH .


b

b b

b

b

1 2

3

45

a

b

c

d

e

G

b b

bb

b c3

a 1b

2

d4

e

G∗

5

Figure 7.4.

Proposition 7.2.1 (Degree equality)For a hypergraphH = (X,D), the sum of all vertexdegrees equals the sum of all edge cardinalities, i.e.,

Σni=1|D(xi)| = Σm

j=1|D j |. (7.1)

Proof. Consider the bipartite representation ofH , i.e. the bipartite graphB(H ) =(X,D;E). If we sum the degrees of vertices in the first part, we obtain the left side ofthe equality; if we sum the degrees of vertices in the second part, we obtain the right sideof the equality. Evidently, they coincide because both are equal to the number of edges inB(H ). �

As one can see, formula (7.1) is a direct generalization of formula (1.1) which wasobtained for graphs. For example in Figure 7.3 it gives:

3+3+2+2+2+0= 12= 1+2+3+3+3.

It is important to mention that any graph as a special case of hypergraphs, has its dual,which is not necessarily a graph. As we shall see, the dualityof hypergraphs is a powerfultool; we are not able to use it if we are restricted by graphs only.

Figure 7.4 shows an example of a graphG such that the dualG∗ is a hypergraph. Noticethat pendant vertex 3 inG becomes loop 3 inG∗, vertex 2 of degree 3 becomes edge 2 ofcardinality 3 and vice versa. In fact, these two different drawings represent two differentpictures of the same structure taken from different points of view. Points, lines and curvesin the plane are just simplest tools to visualize it.

Thus, drawings, (0-1)-matrices and bipartite graphs may beused to identify hyper-graphs. However, it would be wrong to study only the matricesand bipartite graphs insteadof hypergraphs. To see this, the reader should try to formulate the notion of the chromaticnumber of a graph (a simple special case of a hypergraph) in the language of matrices orbipartite graphs. It depends on the problem however, and forsimplicity it is sometimesconvenient to use matrices and/or bipartite graphs.

Edge lists. Besides incidence matrices and any structure suitable for bipartite graphs,there is one more simple and convenient way to store hypergraphs in computer memory. Itrepresents the list of all edges and therefore is called theedge list. For hypergraphH in


Figure 7.3, the edge list is:

L = {{1},{1,2},{1,2,4},{2,3,5},{3,4,5}}.

The edge list should be accompanied by the indication if there are isolated vertices;otherwise they might be lost. In the example above, vertex 6 is isolated.

In computer memory, the edge list is usually represented by aone-dimensional array(the list itself) and one additional two-dimensional arrayindicating the beginning and theend for each edge.

Adjacency matrix. An n×n matrix A = (ai j ) is an adjacency matrix of a hypergraphH , denoted byA(H ), if

ai j =

{

1 if D(xi)∩D(x j) 6= /0,0 otherwise.

This definition is the same as for graphs, see Section 1.3. However, in contrast to graphs,it is not possible to draw a hypergraph from its adjacency matrix in a unique way. In otherwords, there is no one-to-one correspondence between hypergraphs and their adjacencymatrices. That is why the adjacency matrices have a limited use in hypergraph theory.

Exercises 7.2.

1. For hypergraphsH andH ∗ in Figure 7.3, find the neighborhood of each vertex.

2. For hypergraphsH and H ∗ in Figure 7.3, find the degree of each vertex and thecardinality of each edge.

3. Find the rank ofH andH ∗ in Figure 7.3.

4. Write down an arbitrary (0,1)-matrix and draw the respective hypergraph and its dual.

5. Write down an arbitrary edge list and draw the respective hypergraph and its dual.

6. Apply degree equality (7.1) to graphG and its dualG∗ in Figure 7.4.

7. Construct the edge list for graphG and its dualG∗ in Figure 7.4.

8. Construct bipartite representationsB(B(H )) (Figure 7.3), andB(G) (Figure 7.4).

9. Construct the dual of:En,Kn,Pn,Cn, andWn, n = 3, 4,5, 6,7.

10. Construct the dual of cube, prism and Petersen graph.

Computer Projects 7.2.Using an appropriate hypergraph representation, write a pro-gram for the following algorithmic problems.

1. Given the incidence matrix of a hypergraph, construct theedge list.

2. Given the incidence matrix of a hypergraph, draw the hypergraph on the screen.


3. Given the incidence matrix of a hypergraph, construct theadjacency matrix.

4. Given the edge list of a hypergraph, construct the adjacency matrix.

5. Given the edge list of a hypergraph, construct the edge list of the dual.

6. Given a hypergraphH , find its rank.

7. Given a hypergraphH , construct bipartite representationB(H ).

7.3. Basic Hypergraph Classes

Complete hypergraphs.For 0≤ r ≤ n, we define thecompleter-uniform hypergraph tobe the simple hypergraphKr

n = (X,D) such that|X| = n andD(Krn) coincides with all the

r-subsets ofX. Thus a complete graph onn vertices is a complete 2-uniform hypergraphK2n

also denoted byKn. The completer-uniform hypergraph and the family of its edges bothare denoted byKr

n, i.e. Krn = (X,Kr

n), where|X| = n.All five complete hypergraphs on four vertices are shown in Figure 7.5. ForK0

4, wedraw an empty edge to emphasize the empty set. As one can see, the total number of alledges is 1+ 4+ 6+ 4+ 1 = 16= 24 = 2n what is the number of all subsets on the set offour vertices. This holds for anyn≥ 0 and allr such that 0≤ r ≤ n. It is important to noticethat the number of edges inKr

n is:

|D(Krn)| =

(

nr

)

=n!

r!(n− r)!=

n(n−1)(n−2) · · · (n− r +1)

r!.

For example, inK34 we have

|D(K34)| =

(

43

)

=4·3·21·2·3

= 4.

Among all complete hypergraphs on four vertices, onlyK24 is a simple graph. At last,

observe that all complete hypergraphs arer-uniform and(n−1

r−1

)

-regular hypergraphs.Paths and Cycles.In a hypergraphH = (X,D), an alternating sequence

µ= x0D0x1D1x2 . . .xt−1Dt−1xt

of distinct verticesx0,x1,x2, . . . ,xt−1 and distinct edgesD0,D1,D2, . . . , Dt−1 satisfyingxi ,xi+1 ∈ Di, i = 0,1, . . . , t −1, is called apath connecting the verticesx0 andxt , or, equiv-alently,(x0,xt)-path; it is called acycle if xt = x0. The value oft is called thelength of thepath/cycle respectively.

For example, in Figure 7.3, the sequenceµ1 = 1D22D43D55 is a (1,5)-path of length 3,and the sequenceµ2 = 1D22D43D54D31 is a cycle of length 4.

Connected hypergraphs.The hypergraphH = (X,D) is calledconnectedif for anypair of its vertices there is a path connecting them. IfH is not connected, then it consists oftwo or moreconnected componentseach of which is a connected hypergraph. An isolatedvertex, a vertex incident to loops only and an empty edge are also considered connectedcomponents. Therefore,H , H ∗ andB(H ) have the same number of connected components.


b b

bb

K04

b b

bb

K14

b b

bb

K24

b b

bb

K34

b b

bb

K44

Figure 7.5. All complete hypergraphs on four vertices.

In Figure 7.5, the hypergraphK04 has 5 connected components,K1

4 has 4 connectedcomponents,K2

4,K34 andK4

4 are connected. If we look at Figure 7.5 as one hypergraph, thenit is a disconnected hypergraph having 12 connected components.

Bipartite hypergraphs. A hypergraphH = (X,D) is calledbipartite if its vertex setX can be partitioned into two disjoint setsX1 andX2 (called parts) in such a way that eachhyperedge of cardinality≥ 2 contains vertices from both parts. It means that there is nosuchhyperedge insideX1 and there is no such hyperedge insideX2. In other words, the verticesof a bipartite hypergraph can be colored with two colors in such a way that no hyperedgeof size≥ 2 is monochromatic. Bipartite hypergraphs are also calledbi-chromatic or 2-colorable hypergraphs. As it follows from the definition, singletons and empty edges playno role in bipartition; they are usually ignored when discussing the properties of 2-colorablehypergraphs.

A completer-partite hypergraph is anr-uniform hypergraphH = (X,D) such that setX can be partitioned intor non-empty parts, each edge contains precisely one vertex fromeach part, and all such subsets formD. The completer-partite hypergraphs are usuallydenoted byKr

n1,n2,...,nrwhereni is the number of vertices in partXi. Bipartite hypergraphs

generalize bipartite graphs.Isomorphic hypergraphs. Two (not necessarily simple) hypergraphsH = (X,D) and

H ′ = (X′,D ′) are calledisomorphic, written H ∼= H ′, if there is a one-to-one correspon-dence between the setsX andX′ and a one-to-one correspondence between the setsD andD ′ such that for every vertexx∈X and for every edgeD ∈ D we have thatx∈D if and only


if for the corresponding vertexx′ ∈ X′ and the corresponding edgeD′ ∈ D ′ the inclusionx′ ∈ D′ holds. Using graph terminology, one could say that two hypergraphs are isomorphicif and only if their bipartite representations are isomorphic as graphs (preserving respectivebipartition).

Exercises 7.3.

1. Draw all complete hypergraphs on 0, 1, 2, 3, 5, 6 and 7 vertices.

2. Draw the duals to all complete hypergraphs on four vertices, Figure 7.5.

3. Write down the incidence matrix of:K34, K3

5, K15,K5

6, andK47.

4. Write down the edge list ofK34, K3

5, K15, K5

6, andK47.

5. DrawK31,2,3 and write down the incidence matrix.

6. Explain when isomorphic hypergraphs have identical incidence matrix.

Computer Projects 7.3.Using an appropriate hypergraph representation, write a pro-gram for the following algorithmic problems.

1. Given a hypergraphH , determine if it is a completer-uniform hypergraph.

2. Forn1,n2,n3 ≥ 1, generate an edge list ofK3n1,n2,n3

.

3. Given a hypergraphH , determine if it is connected.

4. Given a bipartite graphB(H ), drawH and construct the incidence matrix.

5. ∗ Given a 3-uniform hypergraph, recognize if it is bipartite.

7.4. Basic Hypergraph Operations

As in graph theory, there are a few basic operations which allow to obtain one hypergraphfrom another. They are helpful in proofs of many theorems anduseful in many algorithmsfor solving optimization problems on hypergraphs.

Strong deletion of a vertex.Let us have a hypergraphH = (X,D) and a vertexx∈ X,see Figure 7.6. Astrong deletionof x from H is the removing of all the edges containingx from D and removing ofx from X.

Recall thatD(x) denotes the set of edges containing vertexx in hypergraphH . IfX1 = X−{x}, andD1 = D −D(x), then strong deletion ofx from D results in obtainingthe hypergraphH1 = (X1,D1). We write this operation asH1 = H − x. In H1, we canchoose and strongly delete another vertex to obtain a hypergraphH2 and so on; sequentialstrong deletion of vertices results in a sequence of hypergraphs. As in graphs, this approachis common and very helpful in developing of many algorithms and proving of a series oftheorems by using mathematical induction.


b

b b

b

H

x

b b

b

H1 = H −x

deletex strongly

Figure 7.6. Strong deletion ofx from H .

Sometimes there is a need to strongly delete an entire subsetof vertices; it is equivalentto a sequential strong deletion of the respective vertices in any order. We will see thathypergraphs obtained by strong deletions of vertices play acrucial role in the theory. Suchdeletions of vertices are called “strong” because the vertices are removed from a hypergraphalong with all incident edges.

b

b b

b

H

x

b b

b

H1 = H −x

deletex weakly

Figure 7.7. Weak deletion ofx from H .

Weak deletion of a vertex. As in the previous case, let us have a hypergraphH =(X,D) and a vertexx∈ X. A weak deletionof x from H is the removing ofx from setXand from each hyperedge ofD(x). If the very same vertexx is weakly deleted fromH , seeFigure 7.7, we obtain a different hypergraphH1. We also write this operation asH1 = H −xwith understanding that the meaning of deletion “weak”.

In our example, see Figure 7.7, the loop atx becomes an empty edge, the edge of size 2incident tox becomes a loop, and the edge of size 3 becomes an edge of size 2 connectingthe remaining two vertices. Recall that in hypergraph theory, it is common to draw edgesof size two as line segments and not as closed curves.

In graph theory, loops were curves connecting vertex to itself; in hypergraph theory,the loops become singletons drawn as circles; this reflects the basic idea of hypergraphs ascollections of sets rather than lines and curves.


b

b

b

b

b

D

b b

deleteD strongly

H H1 = H −D

Figure 7.8. Strong deletion of hyperedgeD from H .

Strong deletion of a hyperedge.It is removing of a hyperedge from the list of edgesand then weak deletion of all of its vertices from the vertex set. Strong deletion of the edgeD is shown in Figure 7.8. In such case we writeH1 = H −D and indicate that the deletionis strong.

Weak deletion of a hyperedge.It is the simplest operation of deletion in a hypergraph:we just remove a hyperedge from the list of edges. All the restremains unchanged. Weakdeletion of an empty edge is calledclearing. In many algorithms clearings usually accom-pany weak deletions of the vertices. Figure 7.9 shows an example of weak deletion of thesame edgeD from the same hypergraphH .

b

b

b

b

b

D

b

b

b

b

bdeleteD weakly

H H1 = H −D

Figure 7.9. Weak deletion of hyperedgeD from H .

Important observation. Let us compare the deletions. In the incidence matrix, strongdeletion of a vertex corresponds to removing of the respective row and further removingof all columns that have 1 at intersection with the row. In turn, the weak deletion of avertex corresponds to just removing of the respective row. As we know, transposition ofthe incidence matrix results in the incidence matrix of the dual hypergraph. Rows becomethe columns and columns become the rows. Respectively, vertices become the edges, andedges become the vertices. So, removing of rows correspondsto removing of the columnsin the transposed matrix, or, equivalently, deletion of vertices corresponds to the deletions ofedges in the dual. In other words, the strong (weak) deletionof any vertex in a hypergraph isnothing else than the strong (weak) deletion of the respective edge in the dual. Comparingweak and strong deletions, notice that only for isolated vertices, as for empty edges, strong


and weak deletions are equivalent.The hypergraphH , see Figures 7.6 and 7.7, and hypergraphH , see Figures 7.8 and

7.9, aredual to each other. Observe now that strong deletion of vertexx in Figure 7.6 isthe same as the strong deletion of the edgeD in Figure 7.8. Similarly, weak deletion of thevertexx in Figure 7.7 is nothing else than the weak deletion of the hyperedgeD in Figure7.9. Consequently, the hypergraphH1 in Figure 7.6 and hypergraphH1 in Figure 7.8 aredual to each other. In the same way,H1 in Figure 7.7 andH1 in Figure 7.9 are dual to eachother, too.

In graph theory, such fundamental concept as duality is missing because dual graphsare not properly graphs, they are hypergraphs. This is the first (but not last) case when theduality helps in understanding, simplifying and unifying many combinatorial relations.

Contraction of a hyperedge.Let D be an edge in a hypergraphH = (X,D). A con-traction of the edgeD consists in the following two steps, see an example in Figure7.10:

1. weakly deleteD from H ;

2. replace all vertices ofD by one vertex belonging to eachD′ ∈ D such thatD∩D′ 6= /0.

b

b

b

b

b

D

b b

b

contractD

H H1 = H D

Figure 7.10. Contraction of a hyperedge.

Contraction of an edge may significantly change the structure of a hypergraph. Forexample, notice that in Figure 7.10,H1 is isomorphic to its dualH ∗

1 , while H is not iso-morphic toH ∗. As in graph theory, sequential application of deletions todecompose ahypergraph and then to reconstruct it in inverse order is widely used in many algorithms.

Exercises 7.4.For hypergraphH in Figure 7.11, do the following:

1. Find all distinct (1,13)-paths.

2. Find the shortest (1,6)-path.

3. Find the shortest and the longest cycle.

4. Determine ifH is bipartite.

5. Strongly delete vertices 1, 2, 3, and 7.

6. Weakly delete vertices 1, 2, 3, and 7.


b b

b

b

b

b

b

bb

b

b

b

bb

b 12

34

5

6

7

89

10

11

12

13

14

15

H

Figure 7.11.

7. Strongly delete edges{7,11,12} and{3,5,9}.

8. Weakly delete edges{7,11,12} and{3,5,9}.

9. Contract edges{7,11,12} and{3,5,9}.

10. Construct the dual hypergraphH ∗ and compare it with Petersen graph.

11. Construct the dual hypergraph to every hypergraph obtained in 5, 6, 7, 8, and 9.

Computer Projects 7.4. For hypergraphH in Figure 7.11, using an appropriate hy-pergraph representation, write a program for the followingalgorithmic problems.

1. Sequentially strongly delete vertices in an order determined by a user; draw eachintermediate step on the screen.

2. Sequentially weakly delete vertices in an order determined by a user; draw each in-termediate step on the screen.


3. Sequentially strongly delete edges in an order determined by a user; draw each inter-mediate step on the screen.

4. Sequentially weakly delete edges in an order determined by a user; draw each inter-mediate step on the screen.

5. Sequentially contract hyperedges in an order determinedby a user; draw each inter-mediate step on the screen.

7.5. Subhypergraphs

By strong and weak deletions of vertices and edges from a hypergraph one can obtaindifferent types of subhypergraphs. As we shall see, many concepts in this section (as in theothers, too) have their original ideas coming from Graph Theory.

Subhypergraphs. Let us have a hypergraphH = (X,D). Any hypergraphH ′ =(X′,D ′) such thatX′ ⊆ X, andD ′ ⊆ D is called asubhypergraph of H . In such case,we writeH ′ ⊆ H . Evidently,H ′ can be obtained fromH by strong deletion of the verticesfrom setX−X′ (sequentially in any order, or at once) and further weak deletion of the re-maining edges fromD −D ′ (sequentially in any order, or at once). In Figure 7.12, bothH1

andH2 are subhypergraphs ofH : H1 is obtained by strong deletion of vertex 5 and 1, andH2 is obtained by strong deletion of vertex 4 and weak deletion of edge{1,2,5}. Noticethat the order in which the vertices and edges are deleted is not important.

Induced subhypergraphs. A hypergraphH ′ = (X′,D ′) is called aninduced subhy-pergraph of a hypergraphH = (X,D) if X′ ⊆ X and all edges ofH completely containedin X′ form the familyD ′. Sometimes we say thatH ′ is a subhypergraphinduced by X′.Induced subhypergraphH ′ may be obtained fromH by strong deletion of verticesX−X′

(sequentially in any order, or at once). Induced subhypergraph is a special case of subhy-pergraph. A subhypergraph is not induced if at least one hyperedge ofH being a subset ofX′, is missing. In a hypergraphH , it is convenient to denote asubhypergraph induced bya setY ⊆ X by HY. In Figure 7.12, ifY = {2,3,4}, thenH1 = HY.

Partial subhypergraph. For a hypergraphH = (X,D), any subhypergraphH ′ ⊆ H

such thatH ′ = (X,D ′) is called apartial subhypergraph. Thus partial subhypergraphshave the same vertex set as the hypergraph itself and may be obtained only by weak dele-tions of edges. Any spanning subgraph of a graph is a partial subhypergraph.

Stable (independent) sets.Let H = (X,D) be a hypergraph. A subset of verticeswhich contains no edge ofH is called thestable set,or theindependent set.Independentset of vertices induces an empty subhypergraph. There are maximal by inclusion and notmaximal by inclusion stable sets. The largest size of a stable set over all maximal byinclusion stable sets is called thestability (independence) number, denoted byα(H ).For any hypergraphH with not all vertices singletons, 1≤ α(H ) ≤ |X|. For a hypergraphH without singletons, if we weakly delete all vertices of a stable set, then in the obtainedsubhypergraph no edge is empty.

For hypergraphH , see Figure 7.12, vertices 2 and 5 form a maximal by inclusionstableset, but it is not a maximum stable set. Vertices 1,2 and 3 forma maximum independentset, soα(H ) = 3.


b

b b

b b b

b

b

b

bb

b1

2

34

52

34

1

2

3

5

H H1 H2

Figure 7.12. HypergraphH , induced subhypergraphH1 and subhypergraphH2.

Strongly independent (stable) sets.For a hypergraphH = (X,D), a subset of verticesS⊆ X is called astrongly independent (stable) setif |S∩D| ≤ 1 for every hyperedgeD ∈D. The cardinality of a maximum strongly independent set is denoted byα(H ). Evidently,for a graphG, α(G) = α(G).

Transversals. A set T ⊆ X is called atransversal of a hypergraphH = (X,D) if|T ∩D| ≥ 1 for every edgeD ∈ D. The cardinality of a minimum transversal is denoted byτ(H ). Following the definition, every setS= X −T is an independent set. Therefore wehave the following equality:

α(H )+ τ(H ) = |X|.

In Figure 7.12, vertices 4 and 5 form a minimum transversal, so τ = 2. The comple-mentary set of vertices 1, 2, and 3 as we have seen, is an independent set andα(H ) = 3.Evidently,α(H )+ τ(H ) = 3+2 = 5 = |X|.

If we strongly delete from a hypergraph the vertices of a transversal, then we obtain ahypergraph without edges, i.e. an empty hypergraph. Transversals are sometimes calledblocking sets, or node-covers.

Matchings. In a hypergraphH , a set of edges which pairwise have no vertices incommon is called amatching. A perfect matching is a matching which contains everyvertex of a hypergraph. The maximum size of a matching (over all matchings) is denotedby ν(H ). Any matching ofH is a strongly independent set of vertices in the dualH ∗.Therefore,ν(H ) = α(H ∗). Since any matching is a set of pairwise non-intersecting edges,any transversal must have at least one vertex from each edge of the matching. This factimplies that for any hypergraphH ,

τ(H ) ≥ ν(H ). (7.2)

We say thatH satisfies theKonig property if

τ(H ) = ν(H ). (7.3)

In hypergraphH , see Figure 7.12, edge{2,3,5} forms a maximal by inclusion match-ing; however, edges{1,2,5} and {3,4} form a perfect matching, soν(H ) = 2. Anytransversal ofH must have at least one vertex from each of the edges{1,2,5} and{3,4},soτ(H )≥ 2= ν(H ). Sinceτ(H ) = ν(H ), H satisfies the Konig property. Notice that if a


transversal containsν(H ) vertices, then it is a minimum transversal; similarly, if a match-ing containsτ(H ) hyperedges, then it is a maximum matching. Inequality (7.2)is veryimportant in combinatorial optimization because almost all problems in this area may bere-formulated in terms ofτ or ν for some hypergraph. The Konig property for hypergraphshas its roots in Theorem 2.4.3 for bipartite graphs.

Coverings. For a hypergraphH = (X,D), a subset of edgesD ′ is called acovering ifthe union of all edges fromD ′ coincides withX. We say thatD ′ coversH . The minimumnumber of edges in a covering is denoted byρ(H ). One can see that each covering ofH isa transversal inH ∗ and vice versa, therefore

ρ(H ) = τ(H ∗).

A hypergraphH has thedual Konig property if ρ(H ) = α(H ).For hypergraphH in Figure 7.12, edges{1,2,5} and{3,4} form a minimum covering,

so ρ(H ) = 2. Vertices 1 and 3 form a maximum strongly independent set, soα(H ) = 2.Sinceρ(H ) = α(H ), hypergraphH satisfies the dual Konig property.

GraphC5 is an example of a hypergraph withτ = 3 > 2 = ν andρ = 3 > α = α = 2.Therefore,C5 does not satisfy Konig property, neither it satisfies the dual Konig property.

Hypergraph minors. A hypergraphH ′ is a minor of a hypergraphH if it can beobtained fromH by a sequence of weak or strong vertex or edge deletions, identification oftwo vertices of a hyperedge and the replacement of a hyperedge by any subset of vertices.

Exercises 7.5.For hypergraphH in Figure 7.11, do the following:

1. Draw subhypergraphH ′ = (X′,D ′), where X′ = {1,2,3, 4,5,7,8} and D ′ ={{1,3,7},{2,4,8}}.

2. Draw subhypergraphH ′ induced by vertex subsetX′ = {1,2,3, 4,5, 7,8}.

3. Draw any partial subhypergraph.

4. Find a maximal independent set.

5. Findα(H ).

6. Find a maximum strongly independent set.

7. Find a minimal transversal.

8. Findτ(H ).

9. Find a maximal matching.

10. Findν(H ).

11. Find aρ(H ) and respective minimum covering.

12. Draw a hypergraph minor ofH on five vertices which is not a subhypergraph ofH .


Computer Projects 7.5. For hypergraphH in Figure 7.11, write a program for thefollowing algorithmic problems.

1. Draw a subhypergraph induced by a set of vertices determined by the user.

2. Generate all maximal independent sets.

3. Generate all minimal transversals.

4. Findα(H ).

5. Findτ(H ).

6. Findν(H ).

7. Findρ(H ).

8. Find a minor ofH .

7.6. Conformality and Helly Property

Many properties of hypergraphs can be modeled and explainedusing the “language ofgraphs”.

The line graph L(H ) of a hypergraphH = (X,D) (sometimes called theintersectiongraph of a family D) is the graph with setD as the vertex set and two vertices are adjacentif and only if the respective edges intersect:

L(H ) = (D,E), where(Di,D j) ∈ E ⇔ Di ∩D j 6= /0.

The2-section(H )2 of a hypergraphH = (X,D) is the graph with the same vertex setX, and two vertices are adjacent if and only if they both belongto an edge:

(H )2 = (X,E) where{xi ,x j} ∈ E ⇔ D(xi)∩D(x j) 6= /0.

Theorem 7.6.1 For any hypergraphH ,

(H )2 = L(H ∗).

Proof. Indeed, graph(H )2 has vertex setX which becomes the edge set inH ∗, what inturn, becomes the vertex set forL(H ∗); so they have the same vertex set. Two vertices in(H )2 are adjacent if and only if they have an edge in common inH , what occurs if and onlyif the respective edges intersect in the dual hypergraphH ∗; the last happens if and only ifthe respective vertices inL(H ∗) are adjacent. �

Corollary 7.6.1 For any hypergraphH ,

L(H ) = (H ∗)2.

Proof. Apply equality(H ∗)∗ = H and Theorem 7.6.1 for hypergraphH ∗. �


Theorem 7.6.2 For any hypergraphH ,

ν(H ) = α(L(H )) andν(H ∗) = α((H )2).

Proof. By definition, ν(H ) is the maximum number of edges ofH which are pairwisedisjoint; by definition ofL(H ), this is the maximum number of vertices which represent anindependent set. Similarly, maximum number of pairwise disjoint edges inH ∗ equals themaximum number of pairwise non-adjacent vertices inH what coincides withα((H )2). �

b

b b

b b b

b

b

b

bb

b

b

b

1

2

34

5

a

b

c

d

a

bc

d

1

2

3

4

51

2

34

5

H H ∗ (H )2 = L(H ∗)

Figure 7.13.

Figure 7.13 shows a hypergraphH , its dual H ∗ and the graph which is(H )2 andL(H ∗). Edges 1 and 4 inH ∗ for example, form a maximum matching withν(H ∗) = 2;respectively, vertices 1 and 4 in graphL(H ∗) represent a maximum independent set withα(L(H ∗)) = 2.

One can also check thatL(H ) and(H ∗)2 are isomorphic to the graph obtained fromC4 on verticesa,b,c,d by adding diagonalbd. Edgesa andc in H form a matching withν(H ) = 2; respectively, verticesa andc in L(H ) represent a maximum independent setwith α(L(H )) = 2, and so on. Observe that generally, for a hypergraphH , graphsL(H )and(H )2 are different graphs.

A hypergraphH has theHelly property (is Helly, for short) if foreverysubfamily of itsedges the following implication holds:if every two edges of the subfamily have a nonemptyintersection, then the whole subfamily has a nonempty intersection.

A hypergraphH is called anintersecting family if all of its edges pairwise intersect. Itmeans thatL(H ) is a complete graph. A hypergraphH is called astar if there is a vertexwhich belongs to all hyperedges. Clearly, stars are the special case of intersecting fami-lies. In a hypergraphH , the number of edges in a maximum intersecting subfamily equalsω(L(H )). The simplest example of an intersecting family which is nota star is triangleK3: all three edges pairwise intersect but there is no common vertex. In this terminology, ahypergraph is Helly if every partial subhypergraph representing an intersecting family, is astar.

In multigraphs, the intersecting families are the stars or the triangles with possibly mul-tiple edges. The Helly multigraphs do not contain triangles.

The Helly property plays a crucial role in optimization problems. Let us consider theproblem of finding a minimum transversal of a hypergraphH , or, equivalently, a maximumindependent set ofH . Every clique in graphL(H ) represents an intersecting family inH


b

b b

b b b

b

b

b

bbb

b

a

b

c

d

a b

cd

a b

cd

H L(H ) L(H )

Figure 7.14.

and vice versa. IfH is Helly, then every such family is a star. Therefore, every transversalof H represents a set of cliques inL(H ) which cover all vertices, i.e, clique covering ofL(H ). Immediate conclusion is thatτ(H ) = θ(L(H )), and finding a minimum transversalof H is now reduced to finding a minimum clique covering ofL(H ). In its turn as weknow, the last is equivalent to finding the chromatic numberχ(L(H )) We can see that thereis a relation between transversals ofH and feasible partitions of the complementary graphL(H ). If H is not Helly, we do not have this relation and we are not able touseL(H ) forfinding τ(H ).

HypergraphH in Figure 7.13 is not Helly: edgesb,c, andd form an intersecting family(“triangle”) which is not a star. Let us extend edged and consider a hypergraphH inFigure 7.14. It becomes Helly withL(H ) andL(H ) as in Figure 7.14. One can see thatevery clique (not necessarily maximal) inL(H ) has a corresponding vertex inH and oneminimum covering by cliques ofL(H ) is: {a,b,d} and{c}. It means that one minimaltransversal ofH is obtained by taking one vertex from intersecting family{a,b,d} andone vertex from edgec in H , so τ(H ) = θ(L(H )) = 2. Such covering corresponds to aproper coloring of(L(H )) with χ(L(H )) = 2 colors: vertices{a,b,d} with the first colorand vertexc with the second. There are several such optimal solutions.

A hypergraphH is calledconformal if all the maximal cliques of the graph(H )2 are allthe maximal by inclusion edges ofH . If H does not contain included edges, it is conformalif and only if the edges ofH are precisely the maximal cliques of 2-section(H )2. In otherwords, any conformal hypergraph may be obtained from a simple graph by taking the set ofmaximal cliques as hyperedges and possibly adding a number of included hyperedges. Forthe simplicity reason, let us agree that in the next lemma andtheorems hypergraphsdo notcontain isolated vertices.

Lemma 7.6.1 A hypergraphH = (X,D) is conformal if and only if for any Y⊆X inducinga clique in(H )2 there is an edge D∈ D such that Y⊆ D.

Proof. ⇒ Let H = (X,D) be a conformal hypergraph,G = (H )2, andY ⊆ X be any setof vertices such thatGY is a clique. Since any clique is a subgraph of at least one maximalclique, there exists a setZ ⊆ X such thatY ⊆ Z andGZ is a clique. ThenZ = D for someD ∈ D becauseH is conformal. HenceY ⊆ D.


⇐ Let the condition of Lemma hold, i.e., for anyY ⊆X inducing a clique in(H )2 thereis an edgeD ∈ D such thatY ⊆ D. To prove thatH is conformal, we need to prove that thefamily D ′ of maximal edges ofH and the familyC of maximal cliques ofG coincide, i.e.D ′ = C .

Case 1: Prove thatD ′ ⊆ C . Let D ∈ D ′. By definition of G, all vertices ofD arepairwise adjacent, i.e.GD is a clique. It is contained in a maximal cliqueC∈ C , i.e. D ⊆C.By the condition of Lemma, forC there exists an edgeD′ such thatC ⊆ D′. So we haveD ⊆ C ⊆ D′. But bothD andD′ are maximal what implies thatD = C = D′ andD ∈ C .Consequently,D ′ ⊆ C .

Case 2:Prove thatC ⊆ D ′. LetC∈ C be an arbitrary maximal clique inG. The verticesof C are pairwise adjacent. By the condition of Lemma, there exist an edgeD ∈ D withC ⊆ D. In turn, there exists an edgeD′ ∈ D ′ such thatD ⊆ D′. Since vertices ofD′ arepairwise adjacent inG, there exists a maximal cliqueC′ such thatD′ ⊆C′. Since bothC andC′ are maximal, andC ⊆ D′ ⊆C′, it follows thatC = D = C′ andC ∈ D ′. Consequently,C ⊆ D ′. �

Theorem 7.6.3 (Gilmore, 1961)A hypergraphH = (X,D) is conformal if and only if forany three edges D1,D2,D3, there is an edge D such that

(D1∩D2)∪ (D1∩D3)∪ (D2∩D3) ⊆ D. (7.4)

Proof. ⇒ Let H = (X,D) be a conformal hypergraph andD1,D2,D3 be arbitrary edges.The vertices of(D1∩D2)∪(D1∩D3)∪(D2∩D3) are pairwise adjacent and therefore inducea clique. By Lemma 7.6.1, there exists an edgeD such that

(D1∩D2)∪ (D1∩D3)∪ (D2∩D3) ⊆ D.

⇐ Assume that the inclusion 7.4 holds and show that then the condition of Lemma 7.6.1fulfils. Induction on|Y| whereY is the set of pairwise adjacent vertices of graph(H )2. If|Y| = 1, a vertex is contained in some edge ofH (otherwise it is isolated). If|Y| = 2, thentwo vertices ofY are adjacent and by definition of(H )2 there is an edgeD such thatY ⊆ D.Let the condition of Lemma 7.6.1 hold for allY such that|Y| < k, k ≥ 3, and letY be asubset of vertices inducing a clique of sizek in (H )2.

Since|Y| ≥ 3, choose verticesx1,x2,x3 ∈Y and putYi =Y−{xi}, i = 1,2,3. Notice that

Y = (Y1∩Y2)∪ (Y1∩Y3)∪ (Y2∩Y3).

By the induction hypothesis, there exist edgesD1,D2,D3 such thatYi ⊆ Di, i = 1,2,3.This implies

Y ⊆ (D1∩D2)∪ (D1∩D3)∪ (D2∩D3).

Applying relation (7.4), we obtainY ⊆D for someD∈ D. Thus, any subset of pairwiseadjacent vertices of graph(H )2 is contained in some edge ofH . By Lemma 7.6.1,H isconformal. �

Theorem 7.6.4 A hypergraphH is conformal if and only if its dualH ∗ is Helly.


b

b b

H = K3

not conformal, not Helly

b

b b

H1 - conformal, not Helly

b

b b

b

H2∼= H ∗

1 - Helly,not conformal

b

b b

b

H3∼= H ∗

3 - Helly, conformal

Figure 7.15. Conformality and Helly property.

Proof. ⇒ Let H = (X,D) be a conformal hypergraph. Choose any intersecting sub-family of edges inH ∗, sayX1,X2, . . . ,Xk. In H , they correspond to pairwise adjacent ver-ticesx1,x2, . . . ,xk. SinceH is conformal, by Lemma 7.6.1 there is an edgeD such thatx1,x2, . . . ,xk ∈ D. It means that inH ∗, there is a vertexd such thatd ∈ X1, d ∈ X2, . . . ,d ∈ Xk, or, equivalently,d ∈ X1∩X2∩ ·· ·∩Xk. Consequently, every intersecting subfamilyof H ∗ is a star, i.e.H ∗ is Helly.

⇐ Assume now thatH ∗ is a Helly hypergraph. Then any intersecting family ofH ∗ isa star. Equivalently, inH , any set of pairwise adjacent vertices is contained in an edge. ByLemma 7.6.1,H is a conformal hypergraph. �

The simplest example to show the relation between Helly property and conformalityis the triangleK3, see Figure 7.15. Apparently,K3 itself is neither Helly, nor a conformalhypergraph. Indeed, it is an intersecting family but not a star; it is a clique but not contained


in any hyperedge. If we add an edge containing all vertices, then it becomes a conformalhypergraphH1. Otherwise, if we add a vertex incident to all three edges, weobtain a HellyhypergraphH2. One can easily check thatH2

∼= H ∗1 . At last, if we add toH2 an edge

containing all vertices, we obtain a hypergraphH3 which is both Helly and conformal.A hypergraph is calledself-dual if H ∼= H ∗; evidently, the incidence matrix of a self-

dual hypergraph is square and symmetric about main diagonal. A hypergraphH is calledbi-conformal if both H and H ∗ are conformal. Consequently, ifH is a bi-conformalhypergraph, then bothH andH ∗ are Helly hypergraphs.

Notice that hypergraphH3 in Figure 7.15 is bi-conformal and self-dual, i.e.H3∼= H ∗

3 .

Exercises 7.6.

1. Construct line graphL(H ) for hypergraphH , see Figure 7.11 and show that it isisomorphic to Petersen graph.

2. Construct 2-section(H )2 for hypergraphH , see Figure 7.11.

3. Prove that any bipartite graph is Helly.

4. Prove that Petersen graph is a Helly hypergraph.

5. For whichn, p,q, r ≥ 1, the graphsPn,Cn,Wn,En,Kn, K3p,q,r are or are not Helly ?

6. For whichn, r ≥ 1 complete hypergraphKrn is Helly?

7. Prove that cube is and prism is not a Helly hypergraph.


1. Given a hypergraphH , constructL(H ).

2. Given a hypergraphH , construct(H )2.

3. Given a hypergraph, recognize if it is Helly.

4. Given a hypergraph, recognize if it is conformal.

5. Given a hypergraph, recognize if it is self-dual.

Chapter 8

Hypertrees and ChordalHypergraphs

“–I see many trees, but where is the forest?...”

8.1. Hypertrees and Chordal Conformal Hypergraphs

Trees do not have cycles and represent the simplest class of graphs. We know the definitionof a cycle in a hypergraph and it might seem that a direct generalization would be the bestone for introducing and studying hypertrees. However, basic properties of trees hold formuch more general structures than just hypergraphs withoutcycles.

A host graph for a hypergraph is a connected graph on the same vertex set, such thatevery hyperedge induces a connected subgraph of the host graph.

Definition 8.1.1 A hypergraphH = (X,D) is called ahypertree if there exists a host treeT = (X,E) such that each edge D∈ D induces a subtree in T .

In other words, any hypertree is isomorphic to some family ofsubtrees of a tree. Re-spectively,H is not a hypertree if for any tree on the same vertex set, at least one hyperedgeof H induces a disconnected subgraph, i.e. a forest. Any tree is ahypertree because it isa host tree for itself. In contrast, a cycle cannot be a hypertree because there is always anedge inducing a forest. This is shown in Figure 8.1: a treeT can be re-drawn in such a waythat the edges become ellipses. We will always draw the host trees by dashed lines. In ourexample, a host tree coincides withT itself. If we do the same for the cycleC4, on the hosttree at least one edge (lower edge in the figure) ofC4 induces a disconnected subgraph. Onecan try different trees on four vertices ofC4 to be the host tree: however, there will alwaysbe at least one disconnected subgraph of the host tree induced by an edge ofC4. Since thesame reasoning applies to any cycle, no graph with cycles is ahypertree.

An example how the definition of a hypertree works in general hypergraphs is shownin Figure 8.2. One can see that hypergraphH1 is a hypertree, but in drawing ofH2 thereis an edge inducing a forest isomorphic to the empty graphE3 in the host tree. May beanother tree on the same vertex set can serve as a host tree? The first problem that arises inhypertrees is how to recognize them: given a hypergraphH , how can we find a host tree if


b

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b

b

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b

b

T T

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b b

b b

b

C4 C4

Figure 8.1.

it exists? Alternatively, ifH is not a hypertree, how that can be proven? We need a few newdefinitions and theorems to develop an algorithm for recognizing of hypertrees and findingsolutions to some other optimization problems as well.

Lemma 8.1.1 (Buneman, 1974, Gavril, 1974, Walter, 1972)The line graph of a hyper-tree is chordal.

Proof. Let H = (X,D) be a hypertree and letT = (X,E) be a host tree. We use inductionon the number of vertices ofT; when there is only one, the line graph is a clique and ischordal. WhenT is larger, letx be a pendant vertex ofT. Let L′ be the line graph of thehypergraphH ′ obtained fromH by weak deletion ofx; if {x} is a subtree itself, we removeit from the family of edges. Since the host tree forH ′ is T − x, the hypergraphH ′ is ahypertree too, and the induction hypothesis implies thatL′ is chordal.

If no hyperedge ofH coincides with{x}, thenL′ = L, andL is chordal. If{x} occursby itself as a subtree, then the neighbors of this vertex inL form a clique, since as subtreesthey all containx. ThusL is obtained fromL′ by adding a simplicial vertex which cannotintroduce a chordless cycle. HenceL is chordal. �

Lemma 8.1.2 (Berge, 1973)Every hypertree is a Helly hypergraph.

Proof. The usual proof is by induction; we present a short argument by Lehel (1972). LetH = (X,D) be a hypertree andT = (X,E) be a host tree. To show that every intersectingfamily of H is a star, we prove the contrapositive. LetF ⊆ D be an intersecting familywhich is not a star. Then eachx∈X misses someD∈ F . Mark the edge ofT that is the firston the path fromx to D. Since we placen(T) = |X| marks onn(T)−1 edges, some edgeis marked twice. This edge belongs to a path inT between some membersDi andD j of F ,which therefore do not intersect. Hence,F is not an intersecting family, a contradiction.�

Hypertrees and Chordal Hypergraphs 163

b

b b

b

H1

b b

bb

H2

Figure 8.2.

Definition 8.1.2 A hypergraphH is calledchordal if every cycle of length≥ 4 has twonon-consecutive vertices which are adjacent.

This definition directly extends the definition of chordal graphs. ThusH is a chordal hy-pergraph if and only if its 2-section(H )2 is a chordal graph. Ifx is a simplicial vertex inthe 2-section(H )2, then the neighbors ofx are pairwise adjacent inH .

In a hypergraphH = (X,D), a vertexx is pendant to a vertex y if D(x) ⊆ D(y).Sometimes, pendant vertex is calledhyper-pendant. Thus, a pendant vertex in a graphis either isolated or pendant to its unique neighbor. Pendant vertices in a hypergraphHcorrespond to included edges in the dual hypergraphH ∗.

Theorem 8.1.1 A hypergraphH is chordal and conformal if and only ifH ∗ is a hypertree.

Proof. [6] ⇒ Let H = (Y,D) be a chordal conformal hypergraph. We prove thatH ∗ isa hypertree by induction onn = n(H ). For n ≤ 2, every hypergraph is a hypertree. Nowconsidern > 2, and suppose that the claim holds for smaller hypergraphs.The idea of theproof consists in weak deleting of a simplicial vertex of(H )2 from H , finding a host treein the obtained dual hypergraph, and then reconstructingH by sequential expansions of theedges (operation inverse to weak deleting of the vertex) andconstructing a host tree forH ∗.

Step 1: constructing hypergraphH0 and its 2-section(H0)2. SinceH is a chordalhypergraph,(H )2 is a chordal graph. Therefore,(H )2 has a simplicial vertexx. The closedneighborhood ofx in (H )2 (the neighborhood plusx itself) induces a maximal clique in(H )2. By the conformality ofH , this set forms an edgeD0 in H . Sincex has no additionalneighbors, every edge inH that containsx is a subset ofD0.

Let H0 be the hypergraph obtained fromH by weakly deletingx (see an example inFigure 8.3); thus,H0 has vertex setY−{x} and edge set{D−{x}: D ∈ D}. Sincex issimplicial in (H )2 and the edgeD0−{x} in H0 contains all neighbors ofx in H , we have(H0)2 = (H )2− x (strong deletion ofx), and thusH0 is chordal. The maximal cliques in(H0)2 are the same as those in(H )2, except that the clique with vertex setD0 is lost, and the


clique with vertex setD0−{x} is gained (unless it is not maximal). SinceH0 hasD0−{x}as an edge,H0 is conformal. In the example (Figure 8.3), edgeD becomes a singleton, andedgeD0 becomes an edge of size 2 inH0.

b

b

b

bx

a

b

cD

D0

Eb

b

b

a

b

c

E

D0

D

H H0

deletex

Figure 8.3.

Step 2: induction hypothesis.SinceH0 is chordal and conformal and contains< nvertices, we may apply the induction hypothesis toH0 to conclude thatH0

∗ is a hypertree.Thus there is a host treeT with vertex setD ′ = V(H0

∗), such that for everyx′ ∈V(H0) =

D(H0∗), the elements ofD ′ that containx′ comprise a set of vertices ofT that induces a

subtree.

Step 3: induction step. It remains to show thatH ∗ is also a hypertree. We do this inseveral steps. First, define a hypergraphH1 by expanding the edgeD0−{x} to includexand becomeD0. This adds a vertex of degree 1 toH0 and thus a singleton toH0

∗. SinceD0−{x} is one vertex inH0

∗, it is also one vertex inT. Adding this vertexd0 as a singletonX as a subtree shows thatH1

∗ is a hypertree.

Good case:d is adjacent tod0 in host tree T. Let D ′x be the set of edges inH0, other

thanD0−{x}, obtained by deletingx from the edges ofD(x) in H . To obtainH from H1,we must expand these edges to includex. For each vertexd of T that is a neighbor ofd0

and corresponds to someD ∈ D ′x, we expand the edgeX of H ∗

1 to included. This yields ahypertreeH ∗

2 with host treeT, in which the edgeX corresponding tox is a star with centerd0. The pendant vertices of this star are vertices inH ∗

2 that correspond to edges inH2 whichwe have expanded to includex.

Figure 8.4 shows the same example in dual form. EdgesD, D0 andE become verticesd, d0 ande; similarly, verticesx,a,b andc become edgesX,A,B, andC in dualH0. NoticethatX andC are the singletons. Host treeT is drawn by dashed lines.

Bad case:d is not adjacent tod0 in host tree T. Let H3 be the hypergraph obtainedfrom H2 by includingx in an edgeD∈D ′

x that we have not yet expanded. The edgeD of H2

corresponds to a vertexd in H ∗2 that is pendant tod0, because the edges ofH ∗

2 containingd or d0 are the vertices ofH2 contained inD or D0, respectively, and we haveD ⊂ D0.The vertexd in H ∗

2 occurs as a vertex ofT not adjacent tod0. Adding the edge(d,d0) toT creates a unique cycle. Deleting the edge(d,d′) on that cycle (withd′ 6= d0) yields adifferent treeT3.


b

b

b

d

d0 X

B e

C

Ab

b

b

d

d0

X

B e

C

A

expandX

H ∗0 H ∗

Figure 8.4.

SinceX consists only ofd0 and neighbors ofd0, we can now extendX to included;this corresponds to expandingD to includex, and the vertices of the extendedX still inducea star inT3. In order to complete the proof thatH ∗

3 is a hypertree, we must show that theother edges ofH ∗

3 induce a subtree even when(d,d′) is deleted. LetX′ be an edge ofH ∗2

that containsd andd′. In H2, the vertexx′ belongs toD andD′. Sinced is pendant tod0 inH ∗

2 , all the edges ofH ∗2 that containedd also containd0. Thus the subtree inT consisting

of the vertices ofX′ contains not onlyd andd′ but also the path fromd′ to d0. This impliesthat the vertices ofX′ still form a subtree inT3.

The operation of changing the host treeT (drawn by dashed lines) and further expansionof edgeX is shown in Figure 8.5: vertexeplays the role ofd′.

b

b

b

d

d0 X

B eC

Ab

b

b

d

d0

X

B e

C

A

expandXchangeT

H ∗0 H ∗

Figure 8.5.

Repeating this procedure of enlarging the remaining edges of D ′x yields a sequence of

hypertreesH ∗3 ,H ∗

4 , . . . with host treesT3,T4, . . .. At each step, the conditions are preservedso that the argument made forH ∗

3 holds also for the next step. After all these edges areexpanded, the resulting hypertree isH ∗.

⇐ Let H ∗ be a hypertree. By Lemma 8.1.1,L(H ∗) is chordal. Since(H )2 = L(H ∗),we conclude that(H )2 is a chordal graph, and thusH is a chordal hypergraph.

Let R be a set of pairwise adjacent vertices inH . These become pairwise intersectingedges inH ∗. By Lemma 8.1.2,H ∗ is Helly, and hence these edges inH ∗ have a commonvertex. Thus inH the elements ofR lie in a common edge. This shows that the vertices ofevery clique in(H )2 lie in an edge ofH , and thusH is conformal. �


b

b b

b

ba b c

deb

b b

b

ba b c

de

X Y

Z b

b

b

x

z

y

AE

D

C

B

G H H ∗

Figure 8.6.

Corollary 8.1.1 (Flament, 1978) A hypergraphH is a hypertree if and only ifH is Hellyand L(H ) is chordal.

Proof. Theorem 8.1.1 states thatH is a hypertree if and only ifH ∗ is conformal and hasa chordal 2-section. The corollary follows becauseH has the Helly property if and only ifH ∗ is conformal (Theorem 7.6.4), and because(H ∗)2 = L(H ) (Corollary 7.6.1). �

The clique hypergraph of a graphG is the hypergraph with vertex setV(G) whoseedge set is the family of vertex sets of maximal cliques in graph G. Note that the 2-sectionof the clique hypergraph of a graphG is G. Further, by construction every clique hypergraphis conformal. With Theorem 8.1.1, these observations yield:

Corollary 8.1.2 The dual of the clique hypergraph of a chordal graph is a hypertree.

Corollary 8.1.3 The dual of a chordal graph G is a hypertree if and only if G is a tree.

Proof. If G is a tree, then the clique hypergraph ofG is G, and Corollary 8.1.2 yields theresult. IfG∗ is a hypertree, thenG as a hypergraph is conformal. Since all edges ofG havesize 2, conformality requires thatG has no clique of size greater than 2, and thusG is a tree.

�

Corollary 8.1.4 (Gavril, 1974) A graph G is chordal if and only if it is the line graph of ahypertree.

Proof. If G is chordal, then Corollary 8.1.2 says that the dual of its clique hypergraph isa hypertree. The vertices of the hypertree are the maximal cliques ofG, and the cliquescontaining a particular vertex ofG form a subtree of the host tree. Furthermore, verticesof G are adjacent if and only if they appear together in a maximal clique, meaning that thesubtrees containing them have a common vertex in the host tree. The converse is Lemma8.1.1. �

A simple example to illustrate these statements is the triangle K3 which is chordal butnot a conformal hypergraph, nor it is a hypertree, see Figure7.15. Adding an edge con-taining all the vertices makes it conformal, while adding a vertex contained in all the edges


makes it a hypertree; they are dual to each other. Another example, see Figure 8.6, showshow to find a tree representation, i.e., a hypertree, for a chordal graphG: construct a cliquehypergraphH of G, take the dualH ∗ and construct a respective tree. The host treeT onverticesx,y andz is shown by dashed lines. Verticesa,b,c,d, ande in G andH becomehyperedgesA,B,C,D, andE in H ∗ and each induce a subtree inT. One can easily see thatG = L(H ∗). Drawings ofH andH ∗ are simply two different drawings of a (0,1)-matrix(namely, the incidence matrix ofH ) and its transpose.

Chordal conformal hypergraphs are also known asα-acyclic hypergraphs. In [2], hy-pertrees are called “arboreal hypergraphs” and chordal conformal hypergraphs are called“co-arboreal hypergraphs”.

Exercises 8.1.

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Figure 8.7.

1. For each of the hypergraphs in Figure 8.7, construct the line graph.

2. Which of the hypergraphs in Figure 8.7 are Helly hypergraphs?

3. Which of the hypergraphs in Figure 8.7 are hypertrees?

4. Which of the hypergraphs in Figure 8.7 are chordal?

5. Which of the hypergraphs in Figure 8.7 are chordal and conformal?

6. For each of the hypergraphs in Figure 8.7, construct the dual.

7. For each of the hypergraphs in Figure 8.7, verify thatL(H ) = (H ∗)2.

8. Which of the dual hypergraphs to every hypergraph in Figure 8.7 is a hypertree?


1. Given a graphG, construct the clique hypergraph ofG.

2. Given a hypertree, construct the line graph.

3. Given a chordal conformal hypergraph, construct the 2-section.

4. Given a chordal graphG, construct a hypertreeH such thatG = L(H ).

5. Given a hypergraph, determine if it is chordal.


8.2. Algorithms on Hypertrees

In a hypergraphH = (X,D), a vertexx ∈ X is called atransversal vertex if there existan edgeD ∈ D such thatD = {x} (i.e., D is a singleton). It means thatx belongs to everytransversal ofH , including minimum ones. Transversal vertex inH becomes an edgeX inH ∗ which contains a vertexd belonging only toX.

Recall that in a hypergraphH = (X,D), a vertexx ∈ X is calledpendant (hyper-pendant) if there exist a vertexy ∈ X,y 6= x, such thatD(x) ⊆ D(y). Such a vertexy isthen called atwin vertex ofx. In dual hypergraphH ∗, hyper-pendant vertexx becomesan edgeX which is contained as a subset in an edgeY corresponding to the twiny, i.e. itis an included edge. IfH contains at least two vertices, then any isolated vertex is hyper-pendant becauseD(x) = /0 for any isolated vertexx. Any pendant vertex in a graph ishyper-pendant. Hyper-pendant vertex cannot be transversal, and transversal vertex cannotbe hyper-pendant.

Theorem 8.2.1 If x ∈ X is a hyper-pendant vertex in a hypergraphH = (X,D), and hy-pergraphH1 is obtained fromH by weak deletion of x, then:

1. there exists a maximum independent set S⊆ X such that x∈ S;

2. τ(H ) = τ(H1);

3. α(H ) = α(H1)+1;

4. ν(H ) = ν(H1);

5. if H is a hypertree, thenH1 is a hypertree.

Proof. 1. By definition of hyper-pendant vertex, there exists a twinvertex y such thatD(x) ⊆ D(y). Consider two cases.

Case a):y belongs to a minimum transversalT ⊆ X. Thenx 6∈ T because otherwise thesetT1 = T−{x} is a transversal with|T1|= |T|−1= τ(H )−1, a contradiction. Therefore,x∈ X−T = SwhereS is a maximum independent set.

Case b):ybelongs to no minimum transversal ofH . If there exists a minimum transver-salT1 such thatx∈ T1, then setT2 = (T1−{x})∪{y} is a minimum transversal what contra-dicts the condition of the case. Hence again, there exists a maximum independent setS⊆ Xsuch thatx∈ S.

2. Weak deletion of a vertex from a hypergraph cannot decrease the size of a minimumtransversal, soτ(H ) ≤ τ(H1). On the other hand, 1. implies that inH there is a mini-mum transversal not containingx; weak deletion ofx cannot increase it, soτ(H ) ≥ τ(H1).Therefore,τ(H ) = τ(H1).

3. Sinceα(H )+ τ(H ) = |X|, α(H1)+ τ(H1) = |X|−1, andτ(H ) = τ(H1), concludethatα(H ) = α(H1)+1.

4. Any maximum matching ofH is a matching inH1, soν(H ) ≤ ν(H1). On the otherhand, inH1 no matching contains two edges fromD(x) becausex is hyper-pendant. Thisimplies that every matching inH1 is a matching inH , i.e.,ν(H ) ≥ ν(H1).


5. By Corollary 8.1.1,H is a hypertree if and only if it is Helly andL(H ) is a chordalgraph. Sincex is hyper-pendant,H1 is a Helly hypergraph, too, and, moreover,L(H ) =L(H1). Consequently,H1 is a hypertree. �

Theorem 8.2.2 If x ∈ X is a transversal vertex in a hypergraphH = (X,D), and hyper-graphH1 is obtained fromH by strong deletion of x, then:

1. x belongs to any transversal ofH ;

2. τ(H ) = τ(H1)+1;

3. α(H ) = α(H1);

4. ν(H ) = ν(H1)+1;

5. if H is a hypertree, thenH1 is a hypertree.

Proof. 1. It follows from the definition of transversal and the fact that{x} ∈ D.2. LetT1 ⊂ X1 = X−{x} be a minimum transversal ofH1. Then setT2 = T1∪{x} ⊂ X

is a transversal ofH , what impliesτ(H )≤ τ(H1)+1. On the other hand, ifT is a minimumtransversal ofH , thenT0 = T −{x} is a transversal ofH1, what impliesτ(H )≥ τ(H1)+1.

3. Sinceα(H )+ τ(H ) = |X|, α(H1)+ τ(H1) = |X|−1, andτ(H ) = τ(H1)+ 1, con-clude thatα(H ) = α(H1).

4. Sincex is a transversal vertex, for any maximum matchingF of H1 the familyF ∪{x} is a matching ofH , what impliesν(H ) ≥ ν(H1)+1.

Strong deletion of a vertex from a hypergraph cannot decrease the size of a maximummatching by more than 1; henceν(H )−1≤ ν(H1) what implies the required equality.

5. Any induced subhypergraph of a hypertree is also a hypertree; even ifH1 is a dis-connected hypergraph, each connected component is a hypertree, and the host tree for theentireH1 can be obtained by sequential connecting of the respective host trees by edges.�

Proposition 8.2.1 If H = (X,D), |X| ≥ 2, is a hypertree, then it contains at least two ver-tices such that each of them is either transversal or hyper-pendant.

Proof. Indeed, sinceH is a hypertree, there exists a treeT which is a host graph. TreeThas at least two pendant vertices. Evidently, each of these vertices is either hyper-pendant,or transversal. �

Theorem 8.2.3 If H is a hypertree, then it satisfies the Konig property, i.e.

τ(H ) = ν(H ). (8.1)

Proof. SinceH is a hypertree, by Proposition 8.2.1 it contains either a hyper-pendantvertex, or a transversal vertex. We can decomposeH sequentially by weak deleting of ahyper-pendant vertex or strong deleting of a transversal vertex. Theorems 8.2.1 and 8.2.2assure that in such decomposition we obtain a sequence of hypertrees, and, moreover, thecardinality of the minimum transversalτ changes or does not change simultaneously withthe cardinality of maximum matchingν. At the very end of decomposition we obtain a


one-vertex hypergraph; either it is a hyper-pendant (isolated) vertex, thenτ = ν = 0, or atransversal vertex, thenτ = ν = 1. In all steps we have the equality, soτ(H ) = ν(H ). �

The statement above could be seen in a different way. Recall that Theorem 5.7.1 statesthat graphG is perfect if and only if its complementG is perfect. Theorem 5.7.2 states thatany chordal graph is perfect. Hence for chordal graphsχ = ω andθ = α. We know thatif H is a hypertree, thenL(H ) is chordal. Thereforeθ(L(H )) = α(L(H )). BecauseHis Helly, the cardinality of minimum transversalτ(H ) = θ(L(H )). By definition ofL(H ),the size of maximum matchingν(H ) = α(L(H )). So in total, we have the same equalityτ(H ) = ν(H ). Moreover, it holds for every partial subhypergraphH ′ ⊆ H and for everyinduced subgraph ofL(H ) as well.

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1-2

3-2

2-4

H1 = H H2

H3

H4

T4

T3

T2 T1 = T

Figure 8.8. Recognition of a hypertree.

Next we propose an algorithm for recognizing hypertrees. Itis based on Proposition8.2.1 and weak deletions of vertices and edges.

Algorithm 8.2.1 Recognition of a hypertree.

INPUT: An arbitrary hypergraphH = (X,D).OUTPUT: A host tree T ifH is a hypertree, or answer “No” otherwise.


1. Find a hyper-pendant vertex or such a transversal vertex that becomes hyper-pendantafter weak deleting of all incident singletons. If there areno such vertices, output“No” and end.

2. Delete weakly the hyper-pendant vertex and fix the twin vertex. Continue steps 1 and2 until one vertex remains.

3. Starting with one vertex construct the host tree T by adding vertices in inverse order;each time add the vertex and connect it with its twin.

4. Output tree T .

An example how the algorithm works is shown in Figure 8.8. Thearrow with 1-2 on itmeans that hyper-pendant vertex 1 is weakly deleted fromH , its twin vertex 2 is fixed andH2 is the hypergraph obtained. HypergraphH2 does not contain hyper-pendant vertices,however, vertex 3, for example, becomes hyper-pendant withthe twin 2 after weak deletionof the singleton{3}. Similarly, vertex 2 inH3 becomes hyper-pendant with the twin 4 afterweak deletion of the singleton{2}. At last, hypergraphH4 has the only vertex. So, theordering of vertices is: 1, 3, 2, 4. At this point we start constructing the host tree beginningwith T4. Sequentially adding vertices in inverse order 4, 2, 3, 1 andconnecting them to theirtwins, finally, we arrive to treeT which is the host tree forH .

b b

bb

b b

bxy y

Figure 8.9. Hyper-pendant verticesx andy.

Applying Algorithm 8.2.1 to hypergraphH2, shown in Figure 8.2, see Figure 8.9, wefind that vertexx is hyper-pendant with twiny; after weak deletion ofx we obtain thetriangle which contains neither transversal nor hyper-pendant vertices. Algorithm stopswith the output “No”.

It is important to notice that the output of Algorithm 8.2.1 does not depend on theorder in which vertices and edges are deleted. Another observation is that in the incidencematrix of a hypergraph, a hyper-pendant vertex can be recognized by the following: its rowcontains 1 in the same columns as the row of respective twin.


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G

Figure 8.10.

Algorithm 8.2.2 Finding minimum transversal and maximum matching in a hypertree.INPUT: HypertreeH = (X,D).OUTPUT: minimum transversal T⊆ X and maximum matchingF ⊆ D.

1. Put T= /0, F = /0.

2. Find a hyper-pendant or transversal vertex.

3. If the vertex is hyper-pendant, delete it weakly.

4. If the vertex is transversal, include it in T , the incidentsingleton inF , and delete thevertex strongly.

5. Repeat steps 2-4 as many times as possible.

6. Output T ,F .

Theoretical base for Algorithm 8.2.2 is provided by Theorems 8.2.1 and 8.2.2. It onlyremains to notice that in practical realization of the algorithm one needs to keep track aboutall changes in the edges; including a singleton in the maximum matching really meansincluding an original edge from which the singleton was obtained by all previous deletions.

A remarkable instructive feature of Algorithm 8.2.2, however, is that it works for muchmore general hypergraphs than just hypertrees. A clue to this fact is that strong deletionsof the vertices may change the structural properties of a hypergraph in such a way that itbecomes “like hypertree” not being hypertree at all in the beginning. An example of such


a case is demonstrated in Figure 8.10. GraphG is not a hypertree; it has the only hyper-pendant (=pendant) vertex 6. After weak deletion of it, vertex 1 becomes transversal. Strongdeletion of vertex 1 produces a graph with two pendant vertices, 2 and 5. Weak deletion ofvertex 2 makes vertex 3 transversal in the next graph. Strongdeletion of vertex 3 and furtherweak deletion of vertex 4 results in one transversal vertex 5. The output of algorithm is theminimum transversalT = {1,3,5} and maximum matchingF = {{1,6},{2,3},{4,5}}, soτ(G) = ν(G). Maximum matchingF is shown by double lines in the second drawing ofG.One can compare graphG with cycleC5 whereτ(C5) = 3 > 2 = ν(C5).

Exercises 8.2.

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Figure 8.11.

1. Find all pendant and transversal vertices in Figure 8.11.

2. Apply Algorithm 8.2.1 to each hypergraph in Figure 8.11.

3. Computeτ, α andν and find a minimum transversal, maximum independent set andmaximum matching for each hypergraph in Figure 8.11.

4. For each of the hypergraphs in Figure 8.11, construct the dual hypergraph and solvethe problems 1-3 above.


1. Given a hypergraphH , recognize if it is a hypertree.

2. Given a hypertreeH , find minimum transversal, maximum independent set and max-imum matching.

3. Given a hypergraphH that can be decomposed by weak deletions of hyper-pendantand strong deletions of transversal vertices. Findτ(H ), α(H ) andν(H ).

4. Given a chordal conformal hypergraphH , find a minimum coveringρ(H ).


8.3. Cyclomatic Number of a Hypergraph

Cycles in graphs and hypergraphs represent the main cause ofcomplexity for optimizationproblems. Therefore it is important to know how many cycles are in a given graph orhypergraph, or how they are structured, i.e. what additional properties they have. Recall(see Section 2.1.) that for a simple connected graphG, the cyclomatic numberΛ(G) =m− n+ 1. The same formula generalizes to multigraphs if we add toG any number ofloops or multiple edges; each such addition forms a unique cycle with respect to a chosenspanning tree and increases the number of edgesm= m(G). Any additional edge, even loop,is considered as a “chord”, or “diagonal” of the spanning tree. In this way the cyclomaticnumber is a measure of “how far” the multigraph is from a tree.We now show that thecyclomatic number can be generalized to hypergraphs, and hypertrees play a role similar tothat what trees play for multigraphs.

Let H = (X,D) be a hypergraph. A multigraphG = (X,E) is called thegeneralized2-sectionof H , denoted by[H ]2, if it is obtained fromH in the following way:

1. all loops and edges of size 2 ofH are included inE;

2. every hyperedge of size≥ 3 of H is replaced by a complete graph on the samevertices and all edges of the complete graph are included inE.

Hence,(H )2 = [H ]2 if and only if H itself is a simple graph. Further, let us call anymultigraph without cycles of length≥ 3 a multi-forest. The weight of a multi-forestT,denoted byw(T), is the number of edges of size 2 inT; the loops have weight 0. IfT isa multi-forest, thenT denotes a forest obtained fromT by weak deletion of all loops andreplacing each set of parallel edges by an edge connecting the same pair of vertices. ThusT = (T)2. A vertexx is calledpendant in a multi-forestT if it is pendant inT.

Let H = (X,D) be a hypergraph andT = (X,E) be a spanning multi-forest of themultigraph [H ]2. For brevity, throughout this section call every such a spanning multi-forest by just “forest”.

Proposition 8.3.1 For any forest T of a multigraph[H ]2

w(T) ≤ ∑D∈D

(|D|−1). (8.2)

Proof. By definition of T, any single edgeD ∈ D may provide toT maximum|D| − 1edges of size 2. �

Any singleton in a hypergraph, or, equivalently, any loop ina multigraph is consideredas a cycle of length 1 (by default we assume that hypergraphs do not contain empty edges).Let l(H ,T) denote the number of singletons (loops) of a hypergraphH which are notincluded in a forestT. It is convenient to consider them as “chords” ofT.

Definition 8.3.1 Generalized cyclomatic numberof a hypergraphH with respect to for-est T is the value

Λ(H ,T) = ∑D∈D

(|D|−1)−w(T)+ l(H ,T). (8.3)


b

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H [H ]2

b

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forestT

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forestT

Figure 8.12. ForestsT andT.

Let H be a connected multigraph (with loops and multiple edges); then[H ]2 = H . Ifwe take any spanning tree (clearly without loops and multiple edges) as a forestT, thenw(T) = |X|−1= n−1. Since all loops ofH will contribute tol(H ,T), we obtain:

Λ(H ,T) = ∑|D|=2

(|D|−1)−w(T)+ l(H ,T)

= |D|− (|X|−1) = m−n+1.

This equality shows thatΛ(H ,T) generalizes usual cyclomatic number of a multigraph.An example of a hypergraphH , its generalized 2-section[H ]2, a forestT and a forest

T is shown in Figure 8.12. ForestT contains three edges of size 2, an isolated vertex andonly one of the two loops ofH , sow(T) = 3. One can compute:

Λ(H ,T) = ∑D∈D

(|D|−1)−w(T)+ l(H ,T)

= 2+2+0+0−3+1= 2.

From the definition of[H ]2 it follows that for a fixed forestT = (X,E), edges of acomplete graph generated by an edgeD are split into two subsets: those which belong toT(“lie on T”) and those which do not (the remaining edges). Evidently, if |D|−1 edges of size2 belong toT (possible maximum), then the subgraph that they induce onT is connected.If less than|D| − 1 edges belong toT, then those edges induce a disconnected subgraph,


i.e. a forest. LetED be the set of edges of size 2 which are generated by the 2-section of Dand lie onT, andc(T,D) be the number of connected componentsin such a subgraph.

Theorem 8.3.1Λ(H ,T) = ∑

D∈D

(c(T,D)−1)+ l(H ,T). (8.4)

Proof. First, notice that for every edgeD such that|D| ≥ 2, the following equality holds :

c(T,D) = |D|− |ED|. (8.5)

Second, recall that by definition,

w(T) = ∑|D|≥2

|ED|.

Further, since for any singletonD, evidentlyc(T,D) = 1, we obtain:

Λ(H ,T) = ∑D∈D

(|D|−1)−w(T)+ l(H ,T)

= {ignore the singletons} = ∑|D|≥2

(|D|−1)−w(T)+ l(H ,T)

= {insert the weight of the forest}

= ∑|D|≥2

(|D|−1)− ∑|D|≥2

|ED|+ l(H ,T)

= {unite the sum} = ∑|D|≥2

(|D|− |ED|−1)+ l(H ,T)

= {use (8.5)} = ∑|D|≥2

(c(T,D)−1)+ l(H ,T)

= {bring in singletons back} = ∑D∈D

(c(T,D)−1)+ l(H ,T). �

For the example in Figure 8.12, formula (8.4) gives:

Λ(H ,T) = (2−1)+ (1−1)+ (1−1)+1= 2.

Proposition 8.3.2 For any hypergraphH and forest T ,Λ(H ,T) ≥ 0.

Proof. Apply inequalities (8.2) andl(H ,T) ≥ 0. �

Theorem 8.3.2 (Acharya, Las Vergnas, 1982)For a connected hypergraph H ,Λ(H ,T) = 0 if and only if H is a hypertree and T is a multi-tree of maximumweight containing all loops ofH .


Proof. ⇒ AssumeΛ(H ,T) = 0. Formula (8.4) impliesc(T,D) = 1 for any edgeD ∈D, and, in addition,l(H ,T) = 0. It means that 2-section of any edgeD ∈ D induces aconnected subgraph inT, and all possible loops ofH are inT. In other words, sinceH isconnected there exists a treeT (obtained fromT by replacing multiple edges with singleedges and removing singletons) such that every edge ofH induces a subtree ofT. HenceT is a host tree andH is a hypertree.

Further, sincec(T,D) = 1, inequality (8.2) becomes equality

w(T) = ∑D∈D

(|D|−1).

ThereforeT is the multi-tree of maximum weight. Since it contains all loops ofH , thetheorem follows.

⇐ Immediately we havel(H ,T) = 0. SinceT is a multi-tree of maximum weight,w(T) = ∑D∈D(|D|−1), and by (8.3) equalityΛ(H ,T) = 0 holds. �

Theorem 8.3.3 (Voloshin, 1987)LetH = (X,D) be a hypergraph, x∈X a hyper-pendantvertex,D(x) ⊆ D(y) for some y∈ X, and T be a forest of[H ]2. Then in[H ]2 there exists aforest T1 such that

1. x is adjacent to y and pendant inT1;

2. w(T1) = w(T).

Theorem 8.3.4 (Voloshin, 1987)LetH = (X,D) be a hypergraph, x∈ X, T be a forest in[H ]2, DT(x) denote the set of edges of T incident to x, and m2 be the number of edges ofsize 2 fromH not included in T . If hypergraphH1 is obtained fromH by weak deletion ofx, and forest T1 is obtained from T by strong deletion of x, then:

Λ(H ,T) = Λ(H1,T1)+ |D(x)|− |DT(x)|−m2. (8.6)

Corollary 8.3.1 Let the conditions of Theorem 8.3.4 hold. Then

Λ(H ,T) = Λ(H1,T1) if and only if |D(x)| = |DT(x)|+m2.

An example of a hypergraphH with a vertexx, and a forestT is shown in Figure 8.13.According to the definition,

Λ(H ,T) = 2+2+1+0+0−3+1= 3.

If we weakly deletex from H and strongly deletex from T, we obtain respectively a hyper-graphH −x and a forestT −x. Again, according to the definition,

Λ(H −x,T −x) = 2+1+0+0−2+1= 2.

Now notice that|D(x)| = 3, |DT(x)| = 1, andm2 = 1. Hence,

Λ(H −x,T −x)+ |D(x)|− |DT(x)|−m2 = 2+3−1−1= 3 = Λ(H ,T).


b

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b

H T ⊆ [H ]2

xx

b

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H −x T −x

Figure 8.13.

Theorem 8.3.5 Let H = (X,D) be a hypergraph, T be a forest of maximum weight in[H ]2, x∈ X be a vertex pendant and adjacent to y inT . If hypergraphH1 is obtained fromH and forest T1 is obtained from T by weak deletion of x, then T1 is a forest of maximumweight in[H1]2, and moreover,

w(T1) = w(T)−|DT(x)|.

Proof. Assume there is another forest, say,T ′1 which is of maximum weight in[H1]2.

Evidently,w(T ′1)≥w(T1). If T1 is not a forest of maximum weight, thenw(T ′

1)≥ w(T1)+1.Consider forestT ′

1 in [H ]2. To make it spanning add vertexx and connect it withy by|DT(x)| edges. Denote the obtained forest byT ′. Then

w(T ′) = w(T1)+ |DT(x)| ≥ w(T1)+1+ |DT(x)| = w(T)+1

what contradicts the maximality ofT. Therefore,w(T ′1) = w(T1) and the theorem follows.

�

Generalized cyclomatic numberΛ(H ,T) depends on the weight of the forestT whichcan be chosen in many ways from multigraph[H ]2. How can we get rid of that dependence?The answer is suggested by Theorem 8.3.2: one need to consider forests of the maximumweight and include all the loops. For a hypergraphH , let

w(H ) = maxT

w(T).


Definition 8.3.2 Thecyclomatic number of a hypergraphH is called the value

Λ(H ) = minT

Λ(H ,T) = ∑D∈D

(|D|−1)−w(H ).

Theorem 8.3.6 Let H = (X,D) be a hypergraph with hyper-pendant vertex x∈ X, andhypergraphH1 be obtained fromH by weak deletion of x. Then

Λ(H ) = Λ(H1).

Proof. Let y ∈ X be a twin vertex forx. By Theorem 8.3.3, there exists a forestT ofmaximum weight in[H ]2 such thatx is pendant and adjacent toy in T. If x is incidentto edges of size 2 inH , then all such edges connectx with y becausex is the hyper-pendant vertex. SinceT is the forest of maximum weight, all such edges belong toT.We are able now to apply Theorem 8.3.4 (or, equivalently, Corollary 8.3.1): m2 = 0 and|D(x)| = |DT(x)|, and therefore

Λ(H ) = Λ(H ,T) = Λ(H1,T1).

By Theorem 8.3.5, forestT1 is the forest of maximum weight in[H1]2 what impliesΛ(H1,T1) = Λ(H1). Hence,Λ(H ) = Λ(H1). �

Theorem 8.3.7 For a hypergraphH , the following statements are equivalent:

1. H can be decomposed by sequential weak deletions of hyper-pendant vertices andsingletons;

2. Λ(H ) = 0;

3. H is a hypertree.

Proof. 1⇒ 2: Apply Theorem 8.3.6 and evident fact that singletons do not contribute toΛ(H ).

2⇒ 3: Apply Theorem 8.3.2.3⇒ 1: Apply Theorem 8.2.1. �

We now conclude by presenting an algorithm for computingΛ(H ).

Algorithm 8.3.1 Computation ofΛ(H )

INPUT: A hypergraphH .OUTPUT:Λ(H ).

1. Construct 2-section[H ]2.

2. In [H ]2, delete all loops and replace multiple edges with single edges having weightequal to the multiplicity. Obtain a weighted graph G.


3. In G, find a spanning tree T of maximum weight w(H ).

4. ComputeΛ(H ) = ∑D∈D(|D|−1)−w(H ).

5. OutputΛ(H ).

Step 3 of the algorithm may use Kruskal’s Algorithm 2.3.1 formaximum spanning tree,see Section 2.3.

Exercises 8.3.

b

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H1 H2

Figure 8.14.

1. For each of the hypergraphs in Figure 8.14, construct the generalized 2-section,choose a (multi)-forest and compute its weight. Verify the inequality (8.2).

2. Compute the generalized cyclomatic number with respect to forest chosen in 1.

3. In each case of 2., verify the equality (8.4).

4. In each of the hypergraphs in Figure 8.14, weakly delete a vertex and verify theequality (8.6).

5. For each of the hypergraphs in Figure 8.14, compute the cyclomatic number.


1. Given a hypergraphH , apply Algorithm 8.3.1 to compute cyclomatic numberΛ(H ).

2. Given a hypergraphH , find a set of edges which after weak deletion leaves a hyper-tree.

Chapter 9

Some Other RemarkableHypergraph Classes

“Beauty is in the eyes of the beholder...”

There are several classes of hypergraphs which have nice structural properties. Most ofthem have been introduced as generalizations of bipartite graphs and therefore are relatedto colorings. In this section, we consider a few of them; proper hypergraph colorings willbe discussed in the next chapter.

9.1. Balanced Hypergraphs

Let us recall that in a hypergraphH = (X,D), an alternating sequence

µ= x0D0x1D1x2 . . .xt−1Dt−1xt

of distinct verticesx0,x1,x2, . . . ,xt−1 and distinct edgesD0,D1,D2, . . . Dt−1 satisfyingxi ,xi+1 ∈ Di , i = 0,1, . . . , t − 1, is called a cycle ifxt = x0. The value oft is called thelength of the cycle. A cycle is called odd or even if its lengthis odd or even respectively. Asone can see, in cycles vertices and edges play a similar role:every edge of a cycle containstwo consecutive vertices, and every vertex is contained in two consecutive edges. This factis widely used in duality of hypergraphs.

A hypergraphH is calledbalancedif every oddcycle of length≥ 3 has an edge con-taining three vertices of the cycle; it is calledtotally balanced if every cycle of length≥ 3has an edge containing three vertices of the cycle. A cycle itself is calledbalancedif it con-tains an edge having three vertices of the cycle. Evidently,the totally balanced hypergraphis balanced.

An example of a balanced cycle is shown in Figure 9.1. Notice that hypergraph itself isnot balanced.

Since cycles represent partial subhypergraphs (with possible isolated vertices), everypartial subhypergraph of a (totally) balanced hypergraph is (totally) balanced.

Proposition 9.1.1 The dual of a (totally) balanced hypergraph is (totally) balanced.


b

b

bb

b

bb

bb

b

Figure 9.1. Balanced cycle.

Proof. Indeed, if an edge contains three vertices of the cycle, it means that one vertex iscontained in three edges of the cycle. When taking dual hypergraph, cycles become cyclesof the same length, so the assertion follows. �

Theorem 9.1.1 (Berge)A hypergraphH is balanced if and only if every subhypergraphH ′ obtained fromH by weak deletions of vertices is 2-colorable.

The hypergraph shown in Figure 9.1 is not balanced because weak deletion of all pen-dant vertices and one lower vertex results in a graph having atriangle.

Theorem 9.1.2 (Berge, Las Vergnas, 1970)A hypergraph is balanced if and only if everypartial subhypergraph has the Konig property.

Since balanced hypergraphs are invariant with respect to duality, they also posses thedual Konig property.

Corollary 9.1.1 (Berge) Every balanced hypergraph has the Helly property and is confor-mal.

Proof. Let H be a balanced hypergraph and letH ′ ⊆ H be an intersecting family. ByTheorem 9.1.2,τ(H ′) = ν(H ′). But ν(H ′) = 1 because any matching contains maximumone edge from every intersecting family. Thus,τ(H ′) = 1 what means there exists a vertexcommon to all the edges ofH ′, i.e. H ′ is a star. ThereforeH is a Helly hypergraph. Sincethe dual of a balanced hypergraph is balanced, we conclude that H is conformal. �

Corollary 9.1.2 Every totally balanced hypergraph is a hypertree.

Proof. Let H be a totally balanced hypergraph. Since every cycle of length > 3 has anedge incident to three vertices of the cycle, the line graphL(H ) is a chordal graph. SinceH is balanced, it is a Helly hypergraph. Corollary 8.1.1 implies thatH is a hypertree. �

Some Other Remarkable Hypergraph Classes 183

Corollary 9.1.3 Every totally balanced hypergraph is a chordal conformal hypergraph.

Proof. Let H be a totally balanced hypergraph. Since every cycle of length > 3 has anedge incident to three vertices of the cycle, the 2-section(H )2 is a chordal graph andHis a chordal hypergraph. SinceH is balanced, it is a conformal hypergraph. HenceH is achordal conformal hypergraph. �

Exercises 9.1.

1. Which of the graphsKn,, Km,n, Wn, prism, cube and Petersen graph are balancedhypergraphs?

2. Which graphs are totally balanced hypergraphs?

3. Give an example of a hypertree which is not a totally balanced hypergraph.

4. Give an example of a chordal conformal hypergraph which isnot a totally balancedhypergraph.

9.2. Interval Hypergraphs

A hypergraphH = (X,D) is called aninterval hypergraph if there exists a linear orderingof the verticesx1,x2, ...,xn such that everyD ∈ D induces an interval in this ordering. Inother words, the vertices ofX can be placed on the real line such that every hyperedge is aninterval, see Figure 9.2.

b b b b b b b

b b b b

Figure 9.2. Interval hypergraph.

Theorem 9.2.1 If H is an interval hypergraph, then it is Helly and graph L(H ) is chordal.

Proof. Any interval hypergraphH is a hypertree because in this case a host graph is asimple path. Apply Corollary 8.1.1. �

Theorem 9.2.2 If H is an interval hypergraph without included edges, then the dual H ∗

is an interval hypergraph, too.


Proof. Indeed, suppose the vertices ofH are placed on the real line in such an orderingthat each edge is an interval. Order the edges ofH by the very left vertex in each edge.Since there are no included edges, every vertex is the very left vertex for at most one edge.So, we have uniquely determined linear ordering of the edges. Now observe that for eachvertex all incident edges appear in this ordering. That means the dual hypergraphH ∗ is aninterval hypergraph. �

Figure 9.2 shows two interval hypergraphs which are dual to each other.

Theorem 9.2.3 Every interval hypergraph is totally balanced.

Proof. SupposeH is an interval hypergraph and the vertices are points on the real linein order from left to right. Consider any cycle of length≥ 3. The very left edge of thecycle intersects with the second edge, the second with the third and so on around the cycle.Sooner or later the last edge must intersect the very first edge; it means it contains at leastthree vertices of the cycle. HenceH is totally balanced.

�

b

b

b

b

b

H1

b

b

b b

H2

Figure 9.3.

There are totally balanced hypergraphs which are not the interval hypergraphs, and thereare hypertrees which are not balanced, see Figure 9.3,H1 andH2 respectively.

Exercises 9.2.

1. Which graphs are interval hypergraphs?

2. Explain why cycleCn,n≥ 3 is not an interval hypergraph.

3. Construct an example of a hypertree on≥ 4 vertices which is not an interval hyper-graph.

4. Construct an example of a totally balanced hypergraph on≥ 4 vertices which is notan interval hypergraph.

5. A graphG is called aninterval graph if there exists an interval hypergraphH suchthatG = L(H ). Evidently, any interval graph is chordal. Construct an example of achordal graph which is not an interval graph.


9.3. Normal Hypergraphs

Let H = (X,D) be a hypergraph. Thechromatic index of H is the minimum number ofcolors needed to color the edges ofH such that no two intersecting edges have the samecolor. As for graphs, we denote it byχ′(H ). Clearly,χ′(H ) = χ(L(H )). Now we have thefollowing relation:

χ′(H ) ≥ ω(L(H )) ≥ ∆(H ).

We state that hypergraphH has theedge-coloring property if χ′(H ) = ∆(H ). Theedge-coloring property means that if we take dual hypergraph H ∗, then χ((H ∗)2) =ω((H ∗)2).

At last, a hypergraphH is callednormal if every partial hypergraphH ′ ⊆ H has theedge-coloring property, i.e.

χ′(H ′) = ∆(H ′).

One can prove that any balanced hypergraph is normal and therefore has edge-coloringproperty. The example of hypergraphH2 in Figure 9.3 shows that the converse is not true:H2 is a normal hypergraph but it is not balanced.

Theorem 9.3.1 Any hypertree is a normal hypergraph.

Proof. Indeed, every partial hypergraph of a hypertree is a hypertree; in addition, it isHelly and the line graph is a chordal graph which is perfect. Therefore for a hypertreeHitself and for any partial hypergraph we have the required equality:

χ′(H ) = χ(L(H )) = ω(L(H )) = ∆(H ). �

Theorem 9.3.2 (Fournier, Las Vergnas, 1972)Every normal hypergraph is 2-colorable.

Theorem 9.3.3 (Lovasz, 1972)For any hypergraphH , the following conditions are equiv-alent:

1. H is normal;

2. every partial hypergraphH ′ ⊆ H has the Konig property.

Corollary 9.3.1 A hypergraphH is normal if and only ifH satisfies the Helly propertyand L(H ) is a perfect graph.

Proof. ⇒ Let H be a normal hypergraph. By Theorem 9.3.3 every partial hypergraphH ′ ⊆ H has the Konig property. In particular, ifH ′ is an intersection family, thenτ(H ′) = ν(H ′) = 1 what implies thatH is a Helly hypergraph. The last means thatH ∗ isconformal. Further,χ′(H ) = ∆(H ′) meansχ((H ∗)2) = ω((H ∗)2), and this equality holdsfor every induced subgraph of(H ∗)2, i.e. graph(H ∗)2 is perfect. SinceL(H ) = (H ∗)2,the implication follows.

⇐ Assume thatH satisfies the Helly property andL(H ) is a perfect graph. Then indual hypergraphH ∗ the maximal edges are the maximal cliques of graph(H ∗)2 = L(H ).Observe thatχ((H ∗)2) = ω((H ∗)2) becauseL(H ) is perfect; this means thatχ′(H ) =


∆(H ). Since by the same reason the last equality holds for everyH ′ ⊆ H , the hypergraphH is normal. �

In fact, normal hypergraphs have been introduced by Lovaszin 1972 as Helly hyper-graphs having a perfect line graph. They represent a remarkable example when an out-standing graph-theoretic problem was first solved using hypergraph approach. The Berge’sweak perfect graph conjecture that graph is perfect if and only if its complement is perfect(see Theorem 5.7.1), was first proved via normal hypergraphs. Namely, it followed fromTheorem 9.3.3 by establishing such a fundamental fact that in normal hypergraphs everypartial hypergraph has the Konig property. After that, a pure graph-theoretic proof wasfound. This is not a unique case when a problem was first solvedusing hypergraphs andthen the solution was re-phrased using a different terminology.

Exercises 9.3.

b

b b

b b

b

b

b

bb

b

b b

b

b

b

H1 H2

Figure 9.4.

1. Which of the graphsKn, Km,n, Cn, Wn, cube, prism and Petersen graph are normalhypergraphs?

2. Which of the hypergraphs in Figure 9.4 is normal or not and why?

3. Which of the hypergraphs in Figure 9.4 satisfies the Konigproperty?

4. For each of the hypergraphs in Figure 9.4 find the chromaticindex and respectiveedge coloring.

5. Which of the hypergraphs in Figure 9.4 has the edge-coloring property?


9.4. Planar Hypergraphs

Let H = (X,D) be a hypergraph. Recall that a bipartite representation ofH is the bipartitegraphB(H ) with vertex setX ∪D. The vertexx ∈ X is adjacent to the vertexd ∈ D inB(H ) if and only if x∈ D in H . A hypergraphH is calledplanar if and only if B(H ) is aplanar graph.

b b b b=

b

b

b

b

b

b

b

b

1

2

3

4

1

2

3

4

f1

f2

D1 D2 D1 D2

H

b

b

bb b b

1

2

3

4

D1 D2

H

b

b

b

b

b

b

1

2

3

4

D1

D2

B(H ) B(H )

f1

f2

b bd1 d2

1 32

f1

4 f2

H ∗

Figure 9.5. Planar hypergraphH .


Thus, planar graphs are the special case of planar hypergraphs in which all edges havesize 2. As one may see, a planar hypergraph admits anembedding in the plane in sucha way that each vertex corresponds to a point in the plane, andevery edge corresponds toa closed region homeomorphic to a disk. The region contains (in its boundary) the pointscorresponding to the vertices of the edge, and does not contain the points correspondingto the other vertices. Furthermore, two such regions intersect exactly in the points thatcorrespond to the vertices in the intersection of the corresponding edges. In this way, theconnected regions of the plane which do not correspond to theedges form thefacesof theembedding of the planar hypergraph. As for graphs, thesizeof a face (region) is the numberof vertices on the boundary.

An example of a planar hypergraphH is shown in Figure 9.5. First, we may draw edgesof size two as ellipses containing the respective vertices and vice versa, see the top of thefigure. Each time we draw edges of size two, we keep in mind thisunderstanding. Further,if a hypergraph is planar, we show how the usual drawing of edges ofH (namely, edgesD1

andD2) can be re-drawn in such a way that hyperedges intersect onlyat the neighborhoodsof common vertices. In this way, the regions corresponding to the two facesf1 and f2appear. One can see that in the plane embedding ofH face f1 is an interior face of size 2while f2 is an unbounded face of size 4. If for every edge we put a vertex“at the center” andconnect it with the original vertices of the very same edge, we obtain a plane embeddingof the bipartite representationB(H ). The figure also shows two drawings ofB(H ), withoriginal vertices 1, 2, 3, and 4, and the vertices corresponding to edgesD1 andD2.

Using properties of bipartite representationB(H ), one can derive many properties of aplane embedding of the hypergraphH .

Proposition 9.4.1 A hypergraphH is planar if and only if dual hypergraphH ∗ is planar.

Proof. If H is planar hypergraph, then bipartite representationB(H ) is a planar graph.Bipartite representation ofB(H ∗) is obtained fromB(H ) by simply interchanging the rolesof parts. HenceB(H ∗) is planar, and thereforeH ∗ is planar. �

In contrast to graphs, in plane embeddings of hypergraphs the singletons are drawn ascircles and not as loops. Therefore the singletons, if addedto a hypergraph do not form newfaces. With this agreement the plane embedding of the dual hypergraphH ∗ is also shownin Figure 9.5.

Recalling that the degree of a vertexx ∈ X in H is |D(x)| we obtain the followinggeneralization of Euler’s formula for hypergraphs:

Theorem 9.4.1 (Euler’s formula) Let H = (X,D), |X| = n, |D| = m, be a planar hyper-graph embedded in the plane with f faces. Then

n−m

∑i=1

(|Ei|−1)+ f = m−n

∑j=1

(|D(x j)|−1)+ f = 2.

Proof. [6] Construct the planar embedding of the bipartite graphB(H ). It containsn′ =n+mvertices,

m′ =m

∑j=1

|D j | =n

∑i=1

|D(xi)| (9.1)


edges andf ′ = f faces. SinceB(H ) is a planar graph, by Theorem 4.2.1, Euler’s formulagives:

n′−m′+ f ′ = 2.

Therefore, using the first equality of (9.1), we have

n−m

∑i=1

(|Di |−1)+ f = 2, (9.2)

and, using the second equality of (9.1), we obtain

m−n

∑j=1

(|D(x j)|−1)+ f = 2. (9.3)

�

For example, for the plane embedding ofH in Figure 9.5 we have:

n−m

∑i=1

(|Ei |−1)+ f = 4− (2+2)+2= 2

and

m−n

∑j=1

(|D(x j)|−1)+ f = 2− (0+1+0+1)+2) = 2.

A planar hypergraph is calledmaximal if it is simple (i.e., does not contain includededges) and adding any new edge of size≥ 2 makes it non planar. Consequently, an em-bedding of a planar hypergraph is calledmaximal if and only if every face has size 2, orequivalently, if and only if in the corresponding embeddingof B(H ) every face has size 4.

This maximality is relative in the sense that in every such face one can insert an addi-tional edge of size 2. However, if a planar hypergraphH is not maximal, then there is atleast one face of size at least 3, and therefore one can insertan additional edge of size atleast 3 in that face.

If we draw the faces of a maximal planar hypergraph as curves connecting the respectivetwo vertices, then we obtain a plane graph whose faces correspond to the edges of the initialhypergraph. In this way, a plane graph corresponds to a planar embedding of a maximalhypergraph such that the faces of the graph correspond to theedges of the hypergraph.

Notice that this “face-hyperedge” duality is different from both the hypergraph duality(“vertices - hyperedges”) and the classic planar graph duality (“vertices - faces”, see Section4.5.).

An example of maximal embedding of a planar hypergraph is shown in Figure 9.6.It is complete 3-uniform hypergraphK3

4. Regions representing faces are denoted byf1, f2, f3, f4, f5 and f6. Notice that all the faces have size 2, and facef6 is unbounded.Facesf1, . . . , f6 can be seen as the edges of size 2 connecting the respective pairs of ver-tices: facef1 “connects” vertices 1 and 2, facef2 “connects” vertices 1 and 4 and so on.Unbounded facef6 “connects” vertices 1 and 3. In this way the figure turns into aplaneembedding of the simple complete graphK4 which is drawn below.

There are six edges in the plane embedding ofK4, and each corresponds to a face fromthe plane embedding ofK3

4 and vice versa. There are four faces in the plane embedding of


b b

b b

f1

f2 f3 f4

f5f6

1 2

34

b b

bb

1 2

34

Figure 9.6. Maximal planar hypergraphK34 and its “face-hyperedge” dualK4.

K4, and each corresponds to a hyperedge from the plane embedding of K34 and vice versa.

In this way the duality “face-hyperedge” is observed directly and it is different from thehypergraph duality and the classic planar graph duality ofK3

4 andK4 respectively.


Exercises 9.4.

1. For each hypergraphH in Figure 9.7, construct its dualH ∗, bipartite representationB(H ) andB(H ∗).

b

b b

b b

b

b

b

bb

b

b b

b

b

b

H1 H2

Figure 9.7.

2. Determine which of the hypergraphs in Figure 9.7 is planarand draw both a planeembedding ofH andH ∗. In both cases, verify the Euler’s formula.

3. For planar hypergraph in Figure 9.7, add a hyperedge of size 3 to make its embeddingmaximal.

4. For a non-planar hypergraph in Figure 9.7, weakly (strongly) delete an edge/a vertexand find a plane embedding (if it exists) of an obtained hypergraph and its dual.


1. Given a hypergraphH , construct bipartite representationB(H ).

2. Given a planar hypergraphH , construct its plane embedding.

3. Given a plane embedding of a hypergraphH , determine if it is maximal.

4. Given a hypergraphH , determine if it is planar, and if yes, construct a plane embed-ding.

5. Given a plane embedding of a hypergraphH , and a facef , construct such a planeembedding ofH that f is an unbounded face.

Chapter 10

Hypergraph Coloring

“Graph coloring unfolding: unforeseen features of unforeseen generalizations...”

10.1. Basic Kinds of Classic Hypergraph Coloring

In this chapter we use some parts adapted and updated from research monograph [6]. Asin graph coloring, let{1,2, . . . ,λ} be the set of available colors. Aproper λ-coloring of ahypergraphH = (X,D) is a labeling of its verticesX with the colors from set{1,2, . . . ,λ}in such a way that every edgeD ∈ D such that|D| ≥ 2 hasat least two vertices coloreddifferently . In other words, in any proper coloringno edge of size≥ 2 is monochromatic.We do not necessarily have to use allλ colors. When considering the colorings we ignorethe edges of size≤ 1, which is equivalent to the preliminary weak deletion of all suchelements from the familyD. A properλ-coloring sometimes is called aweak coloringof a hypergraph. The minimumλ for which there exists a properλ-coloring is called thechromatic number of H and is denoted byχ(H ). Since every vertex gets one color,the maximum number of different colors that may actually be used in anyλ-coloring is atmostn(H ). If λ > n, then in everyλ-coloring at least one color remains unused. Properλ-colorings exist for every finiteλ ≥ n.

If H is a simple hypergraph with all edges of size two, then it is a simple graph and weobtain a usual classic graph coloring as studied in Chapter 5. Since in a proper coloring ofa graph no edge is monochromatic, this requirement directlygeneralizes to any hyperedgeof size≥ 2. As the loops are ignored in graph coloring, the singletons are ignored in hyper-graph coloring. An important point, however, is that graph coloring, as we shall see, mayhave many different generalizations.

A λ-coloring ofH = (X,D) which uses preciselyk≤ λ colors defines afeasible parti-tion of X into k stable setsS1,S2, . . . ,Sk calledcolor classes. Each color classSi representsa set of vertices colored with colori. Therefore there is no edge inside any ofSi . Thus wehave the following:

X = ∪ki=1Si , Si 6= /0, Si ∩Sj = /0, i 6= j. (10.1)


b

b

bb

b 1

2

3

3

3

Figure 10.1.

Proposition 10.1.1 Let H be a hypergraph of order n with stability numberα(H ),transversal numberτ(H ) and chromatic numberχ(H ). Then the following inequalitieshold:

α(H )χ(H ) ≥ n;

χ(H ) ≤ τ(H )+1 = n−α(H )+1.

Proof. The first inequality follows from equality (10.1) because inan optimal proper col-oring of H with χ(H ) colors,|Si | ≤ α(H ).

The second inequality is obtained by the following reasoning: choose a minimumtransversal ofH and color it withτ(H ) colors; then color the remaining vertices ofH

with a new color. Since we obtain a proper coloring ofH and useτ + 1 colors, the chro-matic numberχ cannot be greater thanτ+1= n−α+1. �

An example of a hypergraphH and a proper 3-coloring is shown in Figure 10.1. As onecan easily see,τ(H ) = 2, respectivelyα(H ) = 3, andχ(H ) = 2. The coloring produces afeasible partition of the vertex set into three color classes. Vertices of color 1 and 2 representa minimum transversal, and vertices of color 3 represent a maximum independent set. Theinequalityα(H )χ(H ) ≥ n turns into 3·2≥ 5, andχ(H ) ≤ τ(H )+1 into 2≤ 2+1.

There are some restrictive types of hypergraph coloring that are regularly encounteredin the literature, see for example [1, 2].

Strong colorings. A strongλ-coloring ofH is a coloring of the vertices using at mostλ colors in such a way that every edgeD ∈ D is polychromatic, i.e. has all vertices coloreddifferently. Thestrong chromatic number γ(H ) is the smallestλ for which there exists astrongλ-coloring ofH . It follows thatγ(H ) ≥ χ(H ) because every strong coloring is alsoa weak coloring. Evidently the strong and weak colorings coincide whenH is a graph. Inaddition, strong coloring is nothing else than a properλ-coloring of the graph(H )2, the2-section ofH . Since any graph, as a special case of a hypergraph, is a 2-section of itself,strong colorings do not add anything new to coloring theory.

Hypergraph Coloring 195

Equitable colorings. An equitableλ-coloring of H = (X,D) is a partition ofX intoλ stable setsSi , i = 1,2, . . . ,λ, such that for everyD ∈ D, and for everyi the followinginequalities hold:

⌊

|D|

λ

⌋

≤ |D∩Si| ≤

⌈

|D|

λ

⌉

,

wherebrc is the largest integer not greater thanr, anddre is the smallest integer not smallerthanr.

Good colorings. A good λ-coloring of H is a partition ofX into λ stable setsSi ,i = 1,2, . . . ,λ, such that eachD ∈ D has

min{|D|,λ}

colors. Ifλ ≤ min

D∈D|D|,

then everySi forms a transversal ofH . If λ ≥ maxD∈D |D|, then a goodλ-coloring is astrongλ-coloring. Finally, for everyλ, any equitableλ-coloring is also a goodλ-coloring.

Uniform colorings. For a hypergraphH of ordern, a properλ-coloring(S1,S2, . . . , Sλ)is uniform if the number of vertices of the same color is always the same (to within one),i.e.

⌊nλ

⌋

≤ |Si | ≤⌈n

λ

⌉

.

The problem of the existence of uniform colorings arises in numerous scheduling problems.I-regular colorings. For every edgeD j ∈ D of size at least two inH , let there be two

corresponding integersa j ,b j such that

0≤ a j ≤ b j ≤ |D j |.

An I-regularλ-coloring ofH is a partition ofX into λ stable setsSi , i = 1,2, . . . ,λ, insuch a way that for everyD j ∈ D, and everyi = 1,2, . . . ,λ we have

a j ≤ |D j ∩Si| ≤ b j .

Notice that each weak coloring is an I-regular coloring ifa j = 0 andb j = max{1, |D j | −1};every strong coloring is an I-regular coloring witha j = 0 andb j = 1; an arbitrary equitable

coloring is an I-regular coloring witha j =⌊

|D j |λ

⌋

andb j =⌈

|D j |λ

⌉

.

Remark. On one hand, I-regular colorings seem to be the most general.On the otherhand, it is not so. Suppose we want to express that for such colorings someD j must have atleast two vertices of the same color. HenceD j may be monochromatic. If it is monochro-matic and we use at least two colors, then some color is missing and some color uses allthe vertices ofD j . Therefore, in the language of I-regular colorings, we mustput a j = 0,b j = |D j |. The last is equivalent to having no constraint on the coloring of D j , i.e. it isequivalent to considering all I-regular colorings ofH without D j . To express the condi-tion above it is necessary to require the following: in everycoloring there exists ani suchthat |Si ∩D j | ≥ 2. The point of this remark is that all graph generalizations ofcoloringsdescribed above miss a case.

As in graphs, cycles play an important role in hypergraph colorings. Recall thatΛ(H )denotes the cyclomatic number of a hypergraph, see Section 8.3.


Theorem 10.1.1For any hypergraphH = (X,D),

χ(H ) ≤ Λ(H )+2.

Proof. We use Algorithm 8.3.1 that computes the cyclomatic number of H . Let T =(X,E) be the spanning tree in the weighted graphG constructed in Step 2. Starting atany vertex, color the vertices ofT with two colors by alternating the colors along the tree.Those edges ofH which have two adjacent vertices inT are colored properly. The numberof remaining edges is at mostΛ(H ). Using a new color for each of them, we obtain aproper coloring ofH with Λ(H )+2 colors. �

Next, we cite some significant results in this direction. Fordetails we refer the readerto [2, 8].

Theorem 10.1.2 (Erdos, Hajnal, 1966) For any natural numbers h, k, l , all ≥ 2, there ex-ists an h-uniform hypergraphH = (X,D) such that χ(H ) = k andH contains no cyclesof length< l .

A hypergraphH is edge-critical if it contains no isolated vertices,χ(H ) = k, k ≥ 3,and weak deletion of any edge results in a hypergraphH ′ with χ(H ′) = k−1.

Lemma 10.1.1 (Zykov, 1974)Any hypergraphH with χ(H ) ≥ 3 can be transformed intoan edge-critical hypergraph by weak deletion of edges and vertices of degree≤ 1.

The intersections of edges in cycles appear to be also important:

Theorem 10.1.3 (Zykov, 1974)In each edge-critical hypergraphH with χ(H ) = 3 thereexists an odd cycle such that no three of its edges share a common vertex.

Corollary 10.1.1 (Fournier, Las Vergnas, 1972)If in a hypergraphH every odd cyclehas three edges that share a common vertex, thenχ(H ) = 2.

Numerous papers study the smallest number of edges (or the largest number of edges)which anr-uniform hypergraph onn vertices can have ifχ(H ) > k (χ(H ) ≤ k); these areoften referred to as “extremal problems related to the chromatic number of a hypergraph”.In most papers the results are obtained by probabilistic methods.

Nevertheless, one of the well developed direction in classic hypergraph coloring is theinvestigation of bi-chromatic hypergraphs, i.e. hypergraphs withχ(H ) = 2, as generaliza-tions of bipartite graphs. A detailed exposition of this topics can be found in the last chapterof [2].

Exercises 10.1.

1. For hypergraphH in Figure 10.2, find the chromatic numberχ(H ), a respectiveproper coloring and a feasible partition.

2. For hypergraphH in Figure 10.2, determineα(H ), τ(H ) and verify the inequalitiesof Proposition 10.1.1


b b

b

b

b

b

b b

H

Figure 10.2.

3. For hypergraphH in Figure 10.2, find the strong chromatic numberγ(H ), a respec-tive strong proper coloring and a feasible partition.

4. For hypergraphH in Figure 10.2, construct examples of equitable, good, uniform,andI -regular colorings, if they exist.

5. For hypergraphH in Figure 10.2, compute the cyclomatic numberΛ(H ) and verifythe inequality of Theorem 10.1.1.

6. Is hypergraphH in Figure 10.2 edge-critical?


1. Given a hypergraphH , a random coloring of the vertices and a numberλ ≥ 1. De-termine if the coloring is a properλ-coloring.

2. Given a hypergraphH , find an upper bound on the chromatic number by generatingcolorings at random.

10.2. Greedy Algorithm for the Lower Chromatic Number

Let H = (X,D) be a hypergraph, and, as usually, letD(x) be the set of edges containingx ∈ X. A star in H which has a unique center is called amonostar. A vertexx can be acenter for many stars or even monostars; we are interested inthe largest number of edges ina monostar withx as the center.

Definition 10.2.1 Themono-degreem(x,H ) of a vertex x∈X in a hypergraphH = (X,D)is the maximum cardinality of a subfamilyD1(x) ⊆ D(x) such that:

Di ,D j ∈ D1(x) ⇒ Di ∩D j = {x}.


In other words, the mono-degree of a vertexx is the maximum number of edges of amonostar with vertexx as the center. IfH is a graph without loops, then mono-degreecoincides with the usual degree of a vertex.

Consider the valueM(H ) = max

Y⊆Xminx∈Y

m(x,HY).

It can be computed by considering all induced subhypergraphs of H , choosing a vertex ofminimum mono-degree in each of them and then taking the maximum mono-degree overall induced subhypergraphs.

If applied to graphs,M(H ) equals the Szekeres-Wilf number of a graph, see Section3.3. Recall thatω denotes the maximum cardinality of a clique, and the Szekeres-Wilfnumber is at leastω−1, see Proposition 3.3.1. We have seen in Theorem 3.3.1 that whenconsidered for graphs the valueM(G) is closely related to chordal graphs.

Next we consider a greedy hypergraph coloring algorithm which is related to the valueM(H ). The idea is to find a good ordering of the vertices by first decomposingH usingthe mono-degrees of the vertices. Then greedily colorH successively, by adding verticesin reverse ordering. At each step we use the first suitable color in the set of colors. Inthe worst case we can’t use any color from the set of colors. Then we assign a new colorto the next vertex and add the color to the set of colors. Sincewe are looking for theminimum number of colors, at each worst case we lose one color. In detail, the algorithmlooks as follows. We use the notationc(x) for the color of a vertexx ∈ X and the vectorc = (c(x1),c(x2), . . . ,c(xn)) for a coloring ofH ; c(x) = 0 means thatx is not colored.

Algorithm 10.2.1 (greedy hypergraph coloring)

INPUT: An arbitrary hypergraphH = (X,D), X = {1,2, . . . ,n}.OUTPUT: A proper coloring c= (c(1),c(2), . . . ,c(n)) of H .

1. Set C= (0,0, . . . ,0), i = n, Hn = H . Find a vertex of minimum mono-degree inHn

and label it xn.

2. Put i := i −1; if i = 0, then go to step 5.

3. Strongly delete the vertex xi+1 and form an induced subhypergraphHi = Hi+1−xi+1.

4. Find a vertex of minimum mono-degree inHi and label it xi ; go to step 2.

5. Color x1 with the first color: c(x1) = 1, i = 1.

6. Put i := i +1; if i = n+1, then go to step 8.

7. Color xi in Hi with the smallest suitable color from{1,2, . . . , n}; go to step 6.

8. Output c= (c(1),c(2), . . . ,c(n)). End.

Remark. It is important to observe that the algorithm is greedy in thesense that itnever requires re-coloring (backtracking) of vertices that have been colored. The worstcase results in a new color and at any step a proper coloring isobtained. Though simple,the algorithm has one complex point, namely that of determining a vertex of minimum


mono-degree (step 4). Let us suppose thatx is an arbitrary vertex in an arbitrary hyper-graph. Determining its mono-degree is equivalent to findingthe maximum monostar withx as the central vertex. The latter is equivalent to finding themaximum matching in thehypergraph obtained by weak deletion ofx from the subhypergraph induced by the neigh-borhood ofx. This leads to the problem of finding a maximum stable set in the 2-sectionof the dual hypergraph, which is generally difficult to do. However, if the degrees of thevertices are bounded by a constant, for example if∆(H ) is bounded, then some polynomialbounds may be derived for the complexity. In practice some modifications might be madeto avoid exhaustive searching at the cost of accuracy depending on the structural propertiesof hypergraphs.

Theorem 10.2.1The maximum value of the minimum mono-degree generated by steps 1-4of the greedy hypergraph coloring algorithm equals M(H ).

Proof. Let t be the maximum value of the minimum mono-degree over all the vertices inthe ordering generated by steps 1-4. It is clear thatt ≤M(H ). We must show thatt ≥M(H )also holds. There is an induced subhypergraphHY and a vertexy∈Y such that

m(y,HY) = minz

m(z,HY) = M(H ).

Let k be the step when the first vertex from the setY was deleted. HenceHY is an inducedsubhypergraph ofHk. Therefore

M(H ) = m(y,HY) ≤ m(xk,Hk) ≤ t.

Consequently,t = M(H ). �

Corollary 10.2.1 For any hypergraphH = (X,D)

χ(H ) ≤ M(H )+1.

Proof. From the algorithm and Theorem 10.2.1 it follows that the maximum value ofthe mono-degree obtained by the algorithm coincides withM(H ). The maximum numberof colors which we are not allowed to use to color the vertexxi at step 7 is not greaterthanM(H ). Indeed, if we cannot use the colors 1,2, . . . , t it means that we have the edges,sayD1,D2, . . . ,Dt in Hi , which are mono-colored (except the vertexxi) with the respectivecolors 1,2, . . . , t. These colors are different, therefore the edgesD1,D2, . . . ,Dt have a uniquecommon vertexxi and represent a monostar. Since we use the next color forxi , the theoremfollows. �

Notice that Theorem 10.2.1 is a direct generalization of theprocedure for computingM(G) described at the beginning of Section 3.3., and Corollary 10.2.1 is a direct general-ization of Theorem 5.6.1

Figure 10.3 shows the application of Algorithm 10.2.1 to thehypergraph in Figure 10.1.At the very beginning,H5 = H andX = {1,2,3,4,5}. In H , the mono-degrees of verticesare:

m(1,H ) = 2,m(2,H ) = 2,m(3,H ) = 2,m(4,H ) = 1,m(5,H ) = 1.


b

b

bb

b1 2

34

5

H5 = H

b

b

bb

1 2

34

H4 = H5−5

b

b

b

1 2

3

H3 = H4−4

b

b

2

3

H2

b2

H1

bcolor=1

H1

b

b

1

2

H2

b

b

b

2 1

2

H3

b

b

bb

2 1

21

H4

b

b

bb

b2 1

21

1

H5 = H

Figure 10.3.

Since min{2,2,2,1,1} = 1, either vertex 4 or vertex 5 can be deleted to obtain hypergraphH4. Choose vertex 5:H4 = H5−5. In H4, vertex 4 has the minimum mono-degree 1. Con-tinuing decomposition of the original hypergraphH we obtain the sequence of hypergraphsH5,H4,H3,H2 andH1 = ({2}, /0) which corresponds to the ordering of vertices

5,4,1,3,2.

At this point the algorithm starts coloring the hypergraphH1 by assigning color 1 to vertex 2at Step 5. Next it reconstructs the original hypergraph by adding vertices in inverse ordering

2,3,1,4,5,

and coloring each of them by the smallest suitable color. At this segment of work thenumbers in the Figure mean the colors. Algorithm ends with the output of the followingproper coloring of the original hypergraphH :

c = (2,1,2,1,1).


Maximum among all minimal mono-degrees found by Algorithm is 2. Hence by Theo-rem 10.2.1,M(H ) = 2. At last, one can easily see thatχ(H )=2≤ M(H )+1=3.

Exercises 10.2.

b

b b

b

b

b bH

Figure 10.4.

1. In hypergraphH in Figure 10.4, find the mono-degree of each vertex.

2. Apply Algorithm 10.2.1 to hypergraphH in Figure 10.4 to find a proper coloring,M(H ) andan upper bound on the chromatic numberχ(H ).

3. For hypergraphH in Figure 10.4, find the exact value ofχ(H ).


1. Given a hypergraphH and a vertex, find the mono-degree of the vertex and a respectivemono-star.

2. Given a hypergraphH , find M(H ).

3. Given a hypergraphH , apply Algorithm 10.2.1 to find an upper bound on the chromaticnumberχ(H ) and a respective proper coloring.

10.3. Basic Definitions of Mixed Hypergraph Coloring

Until now, we discussed the classic graph and hypergraph coloring. The basic problem wasto find the chromatic number, i.e. the minimum number of colors, of a graph or a hypergraphand the corresponding optimal coloring. Since for proper coloring the maximum numberof colors isn if we consider strict colorings, or isλ if we consider all proper colorings, theproblem of finding the largest number of colors over all proper colorings never occurred.In this section we introduce the basic concepts of mixed hypergraph coloring where, incontrast, problems on both the minimum and maximum number ofcolors occur [6].


Let X = {x1,x2, . . . ,xn},n ≥ 1, be a finite set, and letC = {C1,C2, . . . ,Cl} and D ={D1,D2, . . . ,Dm} be two families of subsets ofX such that thesize of every member ofC ∪ D is at least 2. Each ofC , D may be empty, and any subset of vertices may becontained in bothC and D. If C 6= /0, then denoteI = {1,2, . . . , l}, and if D 6= /0, thendenoteJ = {1,2, . . . ,m}.

Definition 10.3.1 A mixed hypergraph is a triple H = (X,C ,D) where X is called thevertex set, also denoted by V(H ), C is the family of subsets calledC -edges, also denotedby C (H ), andD is the family of subsets calledD-edges, also denoted byD(H ).

As in graphs and hypergraphs, let us callλ ≥ 1 natural numbers{1,2, . . . ,λ} theset ofcolors. Any labeling of elements ofX by colors is called acoloring. More formally, wewill call any coloring amapping c from the vertex setX into the set of colors and denote itby

c : X →{1,2, . . . ,λ}.

In such a mapping, the numberc(x),x∈ X, is called thecolor of vertex x.

Definition 10.3.2 A proper λ-coloring of a mixed hypergraphH = (X,C ,D) is a mappingc : X →{1,2, . . . ,λ} such that the following two conditions hold:

1) every C∈ C has at least two vertices of aCommon color;2) every D∈ D has at least two vertices ofD ifferent colors.

Again, as in graphs, we also use the terms “proper coloring”,or just “coloring”, if thevalue ofλ is not important and this does not lead to misunderstanding.Note that Defini-tion 10.3.2 of a proper coloring generalizes all those contained in Chapter 5 and Sections10.1. and 10.2. that correspond to the caseC = /0. In contrast to graphs and hypergraphs,however, not every mixed hypergraph is colorable:

Definition 10.3.3 A mixed hypergraphH is calledcolorable if it admits at least one propercoloring; otherwiseH is calleduncolorable.

b b1 2

C

D

Figure 10.5. The smallest uncolorable mixed hypergraph.

The example of the smallest uncolorable mixed hypergraph isshown in Figure 10.5:H = (X,C ,D) whereX = {1,2}, C = {{1,2}}, andD = {{1,2}}. Indeed, according toDefinition 10.3.2,C -edgeC = {1,2} requires the vertices 1 and 2 to be of a common color;however,D-edgeD = {1,2} requires the vertices 1 and 2 to be of different colors. Thiscontradiction cannot be reconciled with any set of available colors{1,2, . . . ,λ}.

In what follows we assume that mixed hypergraphs are colorable unless stated other-wise.


Definition 10.3.4 The minimumλ for which there exists a properλ-coloring of a mixedhypergraphH is called thelower chromatic number of H , denoted byχ(H ).

The chromatic number of a graph or hypergraph as defined in Chapter 5 and Sections10.1. and 10.2. simply becomes the lower chromatic number ofa mixed hypergraph withC = /0.

Two proper coloringsc1,c2 of a mixed hypergraphH are said to bedifferent , if thereexists at least one vertexx∈ X that changes the color, i.e.c1(x) 6= c2(x). Hence any permu-tation of colors results in a new proper coloring since it represents a different mapping. LetP(H ,λ) be thenumber of different proper λ-colorings of a mixed hypergraphH . As ingraphs, we need to identify proper colorings which use all the colors; in this case we useifor the number of colors.

Definition 10.3.5 A proper i-coloring of a mixed hypergraphH is called a strict i-coloring, if each of the i colors is used.

Strict i-colorings exist only for 1≤ i ≤ n. Note that a properχ(H )-coloring is necessar-ily a strict coloring sinceχ(H ) is the minimum number of colors over all proper colorings.The maximum number of colors may now be specified as follows:

Definition 10.3.6 The maximum i for which there exists a strict i-coloring of a mixed hy-pergraphH is called theupper chromatic number of H , denoted byχ(H ).

In a coloring of a mixed hypergraphH = (X,C ,D), a subset of verticesY ⊆ X iscalledmonochromatic if all the colorsc(y),y ∈ Y, have the same value, and it is calledpolychromatic if all the colorsc(y),y ∈ Y, are pairwise different. In a proper coloringof a mixed hypergraph,D-edges are non-monochromatic subsets, while theC -edges arenon-polychromatic subsetsof vertices.

Any strict i-coloring of H induces a partition of the vertex setX into i nonemptymonochromatic subsets calledcolor classes. Therefore we can interpret strict coloringsusing the language of partitions:

Definition 10.3.7 In a mixed hypergraphH , a partition of X into i nonempty sets X1, X2,. . . ,Xi is called afeasible partition of H if every C -edge has at least two vertices in aCommon set and everyD-edge has at least two vertices inD ifferent sets of the partition.

Sometimes we will write the feasible partition or stricti-coloringc asc= X1∪X2∪ . . .∪Xi.Let r i(H ) = r i ,1≤ i ≤ n, denote thenumber of feasible partitions of a mixed hypergraphH into i sets.

Definition 10.3.8 The integer vector

R(H ) = (r1, r2, . . . , rn)

is called thechromatic spectrum of the mixed hypergraphH .


In fact, since we have lower and upper chromatic numbers for every colorable mixedhypergraphH , there are no feasible partitions into less thanχ color classes, and there nofeasible partitions into more thanχ color classes. Therefore, chromatic spectrum generallyhas the following form:

R(H ) = (0, . . . ,0, rχ, . . . , rχ,0, . . . ,0).

Further, each feasible partition intoi color classes determinesi! strict i-colorings ob-tained from each other by a permutation of colors. Hence the number of stricti-coloringsequalsr i i!. In general, if we haveλ ≥ i colors, then to count properλ-colorings we have(λ

i

)

ways to choose the subset ofi colors. Consequently the number of properλ-colorings

generated by all feasible partitions intoi subsets is(λ

i

)

r i i! = r iλ(λ− 1) . . . (λ− i + 1) =

r iλ(i). To obtain the number of all properλ-coloring it is sufficient then to add all thesenumbers for everyχ(H ) ≤ i ≤ χ(H ). Thus we arrive to the following conclusion:

Proposition 10.3.1 The number of properλ-colorings P(H ,λ) of a colorable mixed hy-pergraphH is a polynomial inλ and has the following formula:

P(H ,λ) =χ(H )

∑i=χ(H )

r i(H )λ(i). (10.2)

Therefore we callP(H ,λ) the chromatic polynomial of the mixed hypergraphH .Chromatic polynomials of graphs, see Section 5.3., represent an important special case, andequality (10.2) directly generalizes the fundamental equality (5.3). Therefore we will call itthe fundamental equality in mixed hypergraph coloring.

Corollary 10.3.1 For any colorable mixed hypergraphH , the degree of the chromaticpolynomial equalsχ, and the leading coefficient equals rχ.

Proof. Indeed, the major term in equality (10.2) equals

rχλ(λ−1)(λ−2) · · · (λ− χ+1)

what proves the statement. �

Definition 10.3.9 For an uncolorable mixed hypergraphH we set

χ(H ) = χ(H ) = 0, R(H ) = (0,0, . . . ,0), P(H ,λ) = 0.

For a mixed hypergraphH = (X,C ,D), the partial mixed subhypergraphH = (X,C , /0),also denoted byHC = (X,C ), is called aC -hypergraph, and the partial mixed subhyper-graphH = (X, /0,D), also denoted byHD = (X,D), is called aD-hypergraph.

Remark. We will use the prefixesC - and D- when discussing different coloringand structural properties ofHC and HD respectively. For example, aD-graph is a D-hypergraph with all theD-edges of size 2, i.e. it is a classic graph. It is also convenientto use the term “edge” for any element fromC ∪D. Thus theprefixes express the type ofcoloring of a subset of vertices.


b

b

b b

b

1

2

34

5

C1

C2

D1

D2

Figure 10.6. Mixed hypergraphH .

In this framework, the colorings of classic hypergraphs arethe colorings ofD-hyper-graphs; the chromatic number of a hypergraph becomes the lower chromatic number ofHD .Their opposites are the colorings ofC -hypergraphs; the main problem here will be to findχ(HC ). Notice that, for any mixed hypergraphH (colorable or not), the partial hypergraphsHC andHD are both colorable, trivially with one color andn(H ) colors respectively. Obvi-ously,χ(HC ) = 1, r1(HC ) = 1, andχ(HD) = n(H ), rn(HD) = 1. Generally, for a colorablemixed hypergraphH = (X,C ,D), the following evident inequalities hold:

1≤ χ(HD) ≤ χ(H ) ≤ χ(H ) ≤ χ(HC ) ≤ n. (10.3)

Figure 10.6 shows an example of a mixed hypergraphH = (X,C ,D) withX = {1,2,3,4,5}, C = {C1,C2} = {{1,4,5},{1,3,4}}, and D = {D1,D2} ={{1,3,4}{1,2,3}}. The C -edges are drawn by dashed (closed) curves; we will fol-low this rule unless stated otherwise. One can easily find that χ(H ) = 2, while χ(H ) = 4.Notice thatC2 = D1 = {1,3,4} what means that in any proper coloring the subset ofvertices {1,3,4} can be neither monochromatic nor polychromatic; since it has threeelements, precisely two colors must be used. A coloring which uses the minimum numberof colors isc1 = (1,2,2,1,1); a strict coloring which uses the maximum number of colorsis c2 = (1,2,3,1,4). There are also other proper colorings. TheC -hypergraphHC isobtained fromH be weak deletion of allD-edges; theD-hypergraphHD is obtained fromH be weak deletion of allC -edges. Thus the inequalities (10.3) turn into the following:

1≤ χ(HD) = 2≤ χ(H ) = 2≤ χ(H ) = 4≤ χ(HC ) = 4≤ n = 5.

It is easy to construct uncolorable mixed hypergraphs. BothH1 andH2, see Figure 10.7,are uncolorable mixed hypergraphs.H1 containsD-graphK3 which requires three differentcolors; however, the uniqueC -edge containing the same three vertices requires two verticesto be of the same color what cannot be satisfied. Generally, any C -edge with all the verticesinducing a completeD-graph will be called anevidently uncolorableC -edge.

Mixed hypergraphH2 contains a tree consisting ofC -edges. This implies that all fourvertices must have the same color. However, the uniqueD-edge being the set of all verticesrequires at least two different colors. Again, we have a contradiction that leads to the


b

b

b

D1 D2

D3

C1b

b

b b

C1

C2 C3

D1

H1 H2

Figure 10.7. Evidently uncolorableC - andD-edges.

uncolorability. Generally, anyD-edge with each pair of the vertices connected by aC -pathconsisting ofC -edges of size 2, will be called anevidently uncolorable D-edge. Noticethat suchC -paths may not be inside theD-edge.

Evidently, any mixed hypergraph containing an evidently uncolorable edge of any typeis uncolorable. Recall thatin mixed hypergraph coloring, by default, we do not consideredges of size≤ 1 of any kind.

One more direct observation is that any mixed hypergraph with χ(HC ) < χ(HD) isuncolorable. We will discuss uncolorable mixed hypergraphs in Section 10.6.

In a mixed hypergraph, if a subset of vertices is aC -edge and aD-edge at the same time,then it is abi-edge. For example, in Figure 10.6, the subset of vertices{1,3,4} is aC -edgeand aD-edge at the same time, i.e. it is a bi-edge. A mixed hypergraph H = (X,C ,D) is abihypergraph if C = D.

Let us denote the underlying families ofH = (X,C ,D) by E = C ∪D. We say thatH ′ = (X,E) is theunderlying hypergraph ofH . Using underlying hypergraphs, manygeneral structural concepts and agreements (such as drawing rules, basic operations, etc.)introduced for hypergraphs will be applied to mixed hypergraphs. For example, any pathin the underlying hypergraphH ′ is amixed path in H , and it is aC -path (D-path) in H

if it is a path inHC (HD), respectively.H is anr-uniform mixed hypergraph ifH ′ is anr-uniform hypergraph.H is connectedif H ′ is connected; otherwiseH is disconnected.For any subsetY ⊆X, a mixed hypergraphHY = (Y,C ′,D ′) is theinduced subhypergraphof H if the underlying hypergraph ofHY is the induced subhypergraph inH ′; this meansthatC ′ andD ′ consist of all those members ofC and ofD, respectively, which are entirelycontained inY. Clearly, any induced subhypergraph of a mixed hypergraph can be obtainedby strong deletions of the respective vertices.

In a mixed hypergraphH = (X,C ,D), a set of verticesS⊆ X is C -stable or C -independent (D-stable,D-independent)if it contains noC -edge (noD-edge) as a subset.

Exercises 10.3.

1. For mixed hypergraphH in Figure 10.8, find a few proper colorings and respectivefeasible partitions; estimate the chromatic spectrumR(H ).


b

b b

bb

H

Figure 10.8.

2. For mixed hypergraphH in Figure 10.8, find the lower and upper chromatic numbers,respective strict colorings and respective feasible partitions.

3. What is the degree of the chromatic polynomialP(H ,λ)?

4. For mixed hypergraphH in Figure 10.8, drawHC and HD . Find their lower andupper chromatic numbers.

5. For mixed hypergraphH in Figure 10.8, verify inequalities (10.3).

6. For mixed hypergraphH in Figure 10.8, draw a few induced and partial subhyper-graphs.


1. Given a mixed hypergraphH , randomly generate a series of proper colorings andfind an estimate on the chromatic spectrumR(H ).

10.4. Greedy Algorithm for the Upper Chromatic Number

In this section we discuss a greedy algorithm for the upper chromatic number as the oppositeto the greedy algorithm for the lower chromatic number described in Section 10.2.

Let H = (X,C , /0) be aC -hypergraph, abbreviated simply byH = (X,C ) and letC (x)denote the set ofC -edges containing vertexx∈ X. Two verticesx andy are adjacent if andonly if C (x)

⋂

C (y) 6= /0. Call the setC (x)∩C (y) a bistar of the vertex x∈ X with respectto the vertexy. So, every vertexy that is adjacent tox defines a bistar. Some bistars of agiven vertex may coincide. Furthermore, call the value

b(x,H ) = maxy

{|C (x)∩C (y)| : y∈ X, y 6= x}


the bidegreeof vertexx. In this way, the bidegree of a vertex is formed by a maximumbistar. We will see that the bidegree inC -hypergraphs in some sense plays the role of themonodegree inD-hypergraphs as defined in Definition 10.2.1.

Call the valueo(x,H ) = |C (x)|−b(x,H ) ≥ 0

theoriginality of a vertex x in theC -hypergraphH .Thuso(x,H ) = 0 implies that there is another vertexy ∈ X which is contained in all

C -edges containingx. The term “originality” is justified as follows. If the vertices of ahypergraph represent different objects in real life, and the C -edges correspond to sets ofobjects that have a common property (each property is oneC -edge), then all the propertiesof the objectx are C (x). The objectx with o(x,H ) = 0 is “not original” because thereexists at least one other object with the same properties. So, the originality of a vertex isa measure of “similarity with its neighbors”. The higher theoriginality, the less similar avertex is to its neighbors. The upper limit here is|C (x)|−1 if x is the center of a monostar.

Definition 10.4.1 The value

O(H ) = maxY⊆X

minx∈Y

o(x,HY)

is called theresistanceof a hypergraphH .

We will see thatO(H ) plays the role which is similar to the role ofM(H ).

Definition 10.4.2 In a coloring of a mixed hypergraphH = (X,C ,D), a set M⊆ X of ver-tices is called amonochromatic componentof a vertex x∈ M, and is denoted by MC(x),if all the vertices of M have the same color as x and are reachable from x byC -paths.

Now, in order to find a lower bound for the upper chromatic number, we propose a quasi-greedy algorithm for an arbitraryH = (X,C ). As for the lower chromatic number, the ideais to find a good ordering of the vertices and greedily colorH successively, maximallyusing the local information. But this time, at each step we use a new color for the nextvertex and verify if the coloring obtained is proper. If the coloring is wrong (i.e. thereare polychromaticC -edges), then we re-color a monochromatic component starting at theneighborhood of the given vertex in order to guarantee the correctness of the new coloringand minimize the losses of used colors. The unavoidable backtracking in this approachjustifies the more precise term “quasi-greedy”.

Algorithm 10.4.1 (quasi-greedyC -hypergraph coloring)

INPUT: An arbitraryC -hypergraphH = (X,C ), |X| = n.OUTPUT: A strict coloring ofH in a number of colors.

1. Set i= n, Hn = H ; find a vertex of minimum originality and label it xn.

2. Put i := i −1; if i = 0, then go to step 5.

3. Strongly delete the vertex xi+1 and form an inducedC -hypergraphHi = Hi+1−xi+1.


4. Find a vertex of minimum originality inHi and label it xi ; go to step 2.

5. Set the list of used colors U= {1}, color the vertex x1 with the color 1; set i=1,new= 2.

6. Put i := i + 1; if i = n, then go to step 10; color the vertex xi of Hi with color new;put U := U ∪{new}, new:= new+1.

7. Verify the correctness of the coloring ofHi ; if there are no polychromaticC -edges,then go to step 6.

8. In Hi , choose a neighbor y of the vertex xi , which generates a largest bistar of xi withall theC -edges polychromatic. If xi was not yet re-colored inHi , then re-color xi withthe color of y, put U:= U −{new}, new:= new−1, and go to step 7.

9. Re-color all the vertices from the monochromatic component MC(y) with the colorof xi and go to step 7.

10. Renumber the colors of U in increasing order; end.

Remark. The monochromatic component re-coloring used in the algorithm is theop-posite to the bi-chromatic chain re-coloring by Kempe,see Section 5.6.

Complexity. Let us suppose that the hypergraphH = (X,C ), with |X|= n, |C | = k, isrepresented by its incidence matrixI(H ) of the sizen×k. Since finding the originality of avertex requiresO(nk) steps, finding the minimum originality requiresO(n2k) steps. Hence,steps 1-4 may be implemented, in the worst case, inO(n3k) steps.

To check if the coloring is proper, it is necessary to look through every column, sothe complexity of step 7 isO(nk). Simultaneously, one can label the wrongly coloredC -edges. Finding the neighborhood ofxi and simultaneously identifying the vertexy takesO(nk) steps. Finding and re-coloring the monochromatic component MC(y) can also beimplemented inO(nk) steps. Since, in the worst case, all the vertices from the neighborhoodmay require re-coloring, the complexity of steps 7-9 isO(n2k). Steps 7-9 are repeatedntimes; consequently, the complexity of the second part and of the whole algorithm does notexceedO(n3k). One can improve this bound by using special data structures and techniques.

Example 10.4.1Consider theC -hypergraphH = (X,C ) (see Figure 10.9) such thatX =

{1, 2, 3, 4, 5}, C = {C1, C2, C3, C4, C5}, C1 = {1,2,3}, C2 = {2, 3, 4}, C3 = {3,4,5},C4 = {4,5,1}, andC5 = {5,1,2}. SetH5 = H . All the vertices have the same originality1, therefore, let us start with the first vertex:x5 = 1.

Form the inducedC -hypergraphH4 = (X4, C4) with X4 = {2,3,4,5}, C4 = {C2,C3}.The first vertex with minimum originality isx4 = 2.

Form the inducedC -hypergraphH3 = (X3, C3) with X3 = {3,4,5}, C3 = {C3}. Thefirst vertex with minimum originality isx3 = 3.

Form the inducedC -hypergraphH2 = (X2, C2) with X2 = {4,5}, C2 = { /0}. The firstvertex with minimum originality isx2 = 4.


b

b

bb

b

1

2

34

5

C1

C2

C3C4

C5

Figure 10.9.C -hypergraphH = (X,C ) of Example 10.4.1.

Form the inducedC -hypergraphH1 = (X1, C1) with X1 = {5}, C1 = { /0}. The lastvertex isx1 = 5.

These are the results of steps 1-4. As usual, letc(i) be the color of vertexi, c(i) = 0means the vertex is not colored,i = 1, . . . ,5.

Set c = (c(1), c(2),c(3), c(4), c(5)) = (0,0,0,0,0). Start coloring. Step 5:c =

(0,0,0,0,1). Step 6: c = (0,0,0,2,1). Step 7: there are no polychromaticC -edges inC -hypergraphH2.

Step 6:c = (0,0,3,2,1). Step 7:C -edgeC3 is polychromatic.Step 8: re-coloring:c = (0,0,2,2,1).Step 7: there are no polychromaticC -edges in theC -hypergraphH3.

Step 6:c = (0,3,2,2,1).Step 7: there are no polychromaticC -edges in theC -hypergraphH4.

Step 6:c = (4,3,2,2,1).Step 7: theC -edgesC1,C4,C5 are each polychromatic in theC -hypergraphH5.Step 8: vertices 2 and 5 in theC -hypergraphH5 generate the largest bi-stars of vertex

1 with all the C -edges polychromatic; choose vertex 5; vertex 1 was not yet re-colored;re-color it with the colorc(5) = 1: c = (1,3,2,2,1).

Step 7: theC -edgeC1 is still polychromatic.Step 8: vertices 2 and 3 in theC -hypergraphH5 are contained in one polychromatic

C -edge; choose vertex 3.Step 9: re-color the monochromatic componentMC(3)={3,4} : c=(1,3,1,1,1).Step 7:H1 = H is colored properly.


Step 10: renumber the colors in increasing order:c = (1,2,1,1,1). End.

Remark. Note that in the process of coloring, whenH5 = H is reconstructed andthe coloringC = (4,3,2,2,1) is obtained, for vertexx5 = 1 none from the available colors1,2,3,4,5. . . is appropriate. At this point we reach adeadlock, i.e. the situation when noneof the available colors can be used to color the next vertex ina C -hypergraph. SinceH isevidently colorable, the re-coloring becomes unavoidable.

Theorem 10.4.1The maximum value of the minimum originality generated by steps 1-4 ofthe Algorithm 10.4.1 equals O(H ).

Proof. Let t be the maximum value of minimum originality over all the vertices in theorder generated by steps 1-4. It is clear thatt ≤ O(H ).

Suppose thatt ≤ O(H )− 1. Hence in an induced subhypergraphH ′ ⊆ H there is avertexy such that

o(y,H ′) = minz

o(z,H ′) = O(H ) ≥ t +1.

It is easy to see that the originality of any vertex is a monotone function with respect to theinduced subhypergraph inclusion. This implies that the first vertex ofH ′ that was deletedby the algorithm had originality≥ t +1, a contradiction. Consequently,t = O(H ). �

In some sense, Theorem 10.4.1 is similar to Theorem 10.2.1.

Theorem 10.4.2The number of colors that may be lost at steps 6-10 of Algorithm 10.4.1does not exceed the value O(H )+1.

Proof. Let us suppose that we have the worst case at step 8; i.e., all theC -edges containingxi in Hi have all their vertices polychromatic. Recall that inHi , for the vertexxi , there is aneighbory forming the largest bistar ofxi . If we re-color vertexxi with c(y), then we arelosing one color (new) and at the same time are properly coloringb(xi ,Hi) C -edges. Hence,in the worst case, there remain at mosto(xi ,Hi) = |C (xi)|−b(xi ,Hi) C -edges that are stillcolored improperly. Every suchC -edge may form a separate bistar ofxi . Therefore, whenre-coloring one monochromatic component for each of theseC -edges, we are again losingat mosto(xi ,Hi) colors. Consequently, the total number of colors lost is notgreater thano(xi ,Hi)+1. Since for anyi, 1≤ i ≤ n,

o(xi ,Hi ,) ≤ O(H ) = maxY⊆X

minx∈Y

o(x,H /Y,)

the theorem follows. �

Corollary 10.4.1 If O(H ) = 0, then there exists an ordering of the vertex set X such thatAlgorithm 10.4.1 may be implemented without re-colorings of monochromatic components.

Proof. Indeed, the vertices of originality 0 are pendant to some neighbors, and the algo-rithm uses the color of the neighbor. �

Corollary 10.4.2 In Algorithm 10.4.1, if|U | = p, thenχ(H ) ≥ p.


We described Algorithm 10.4.1 which is a consecutive quasi-greedy coloring algorithmsearching forχ(H ) and a corresponding coloring for an arbitraryC -hypergraphH . Letus compare it to the classical greedy coloring Algorithm 10.2.1 for the lower chromaticnumber of aD-hypergraph. Both use the same information and are consecutive; i.e., wedecompose theD-hypergraph orC -hypergraph first by deleting the vertices following somegreedy rules. Then we reconstruct the initialD-hypergraph orC -hypergraph by addingvertices in reverse order and coloring them using the neighborhoods. Each time, we assignthe first free color if we have aD-hypergraph and a new color if we have aC -hypergraph.However, the results are different. The greedy algorithm needs no re-coloring; in the worstcase, we simply assign a new color to the next vertex. The quasi-greedy algorithm, in theworst case, encounters the deadlock and requires re-coloring of vertices which are alreadycolored. The greedy algorithm gives a possibility to obtainan upper bound for the lowerchromatic number directly. It is not the case with Algorithm10.4.1. We must implementit, obtain a number of colors, and only then write a trivial inequality. One can considerAlgorithm 10.2.1 as opposite to Algorithm 10.4.1.

The comparison above explicitly shows that if for finding themaximum number ofcolors we apply the same approach as for the minimum, we obtain much less.

The valueO(H ) was called the “resistance” of aC -hypergraph because it shows howthe structure of aC -hypergraph may resist the quasi-greedy coloring algorithm. Smallerresistance indicates that fewer colors are lost in the worstcase when re-coloring. So fromthis view point, hypergraphs withO(H ) = 0 deserve special attention.

Theorem 10.4.3 If H = (X,C ) is a hypertree, then O(H ) = 0.

Proof. Induction on|X| = n. For n = 2,3 the assertion is trivial. Assume it holds for allhypertrees on< n vertices. Consider a vertexx that is pendant in the host tree. Since anyC -edge ofH has cardinality at least 2,o(x,H ) = 0. From this andO(HY) = 0 for anyY ⊂X(by the induction hypothesis sinceHY is also a hypertree), it follows thatO(H ) = 0. �

Corollary 10.4.3 If H = (X,C ) is a hypertree, then Algorithm 10.4.1 requires no re-coloring of monochromatic components.

Proof. Apply Theorem 10.4.3 and Corollary 10.4.1. �

Thus, hypertrees are the first class of hypergraphs that playa special role in sequentialC -hypergraph coloring.

Exercises 10.4.

1. ForC -hypergraphH in Figure 10.10, find bidegree and originality of each vertex.

2. ForC -hypergraphH in Figure 10.10, find the resistanceO(H ).

3. ForC -hypergraphH in Figure 10.10, apply Algorithm 10.4.1 to find a proper color-ing and a lower bound on the upper chromatic number.

4. For C -hypergraphH in Figure 10.10, find the upper chromatic numberχ(H ), therespective proper coloring and feasible partition.


bb b

b

b

b

b

H

Figure 10.10.

5. ForC -hypergraphH in Figure 10.10, explain why Algorithm 10.4.1 requires no re-coloring of monochromatic components.


1. Given aC -hypergraphH and a random coloring, determine if the coloring is proper.

2. Given aC -hypergraphH , implement Algorithm 10.4.1.

10.5. Splitting-Contraction Algorithm

Recall that we use the following notation. For a mixed hypergraph H = (X,C ,D), thevertex setX = {x1,x2, . . . ,xn},n ≥ 1, the edge familiesC = {C1,C2, . . . ,Cl} and D ={D1,D2, . . . ,Dm}; if C 6= /0, thenI = {1,2, . . . , l}, and if D 6= /0, thenJ = {1,2, . . . ,m}. Anedgecontaininganother edge as a subset is calledincluding; an edge which iscontainedas a subset in another edge is calledincluded.

Now, in order to computeP(H ,λ) andR(H ) for an arbitrary mixed hypergraphH =(X,C ,D), we provide the following 5 rules.

1. If H contains an evidently uncolorable edge of any type, then it is uncolorable andcan be removed from further considerations (elimination).

2. If Ci ⊆Cj , then P(H ,λ) = P(H −Cj ,λ), R(H ) = R(H −Cj), i, j ∈ I ; it means thatincluding edgeCj can be weakly deleted because the respective coloring constraintsare satisfied in the included edgeCi (C -clearing).

3. If Di ⊆ D j , thenP(H ,λ) = P(H −D j ,λ), R(H ) = R(H −D j), i, j ∈ J; it means thatincluding edgeD j can be weakly deleted because the respective coloring constraintsare satisfied in the included edgeDi (D-clearing).


4. If {xk,xl} 6∈ D and{xk,xl} 6∈ C , then

P(H ,λ) = P(H1,λ)+P(H2,λ), R(H ) = R(H1)+R(H2),

whereH1 = (X,C ,D1), D1 = D ∪{xk,xl},

H2 = (X,C1,D), C1 = C ∪{xk,xl} (splitting ).

In other words, we can split the colorings ofH into the colorings ofH1 and thecolorings ofH2; in H1 verticesxk andxl have different colors, and inH2 verticesxk

andxl have the same color.

5. If Ct = {xk,xl}, for somet ∈ I andxk,xl ∈ X, such thatCt 6= Ds for anys∈ J, then

P(H ,λ) = P(H1,λ), R(H ) = R(H1), where

H1 = (X1,C1,D1), X1 = (X\{xk,xl})∪{y}, y is a new vertex;

if xk ∈ D j , or xl ∈ D j , j ∈ J, thenD1j = (D j\{xk,xl})∪{y}, otherwiseD1

j = D j ;

if xk ∈Ci , or xl ∈Ci, i ∈ I , i 6= t, thenC1i = (Ci\{xk,xl})∪{y}, otherwiseC1

i = Ci ;

C1 = C −Ct (contraction).

Any C -edge of size 2 can be contracted unless it coincides with aD-edge of size 2.

Remark. In 5., n(H ) = n(H1) + 1 and the equalityR(H ) = R(H1) means thatr i(H ) = r i(H1) for i = 1,2, . . . ,n−1 andrn(H ) = 0 sinceH contains oneC -edgeCt of size2. We will use the equalities of chromatic spectra in this sense when discussing differentoperations on mixed hypergraphs.

The algorithm that allows us to computeP(H ,λ) andR(H ) for any mixed hypergraphH = (X,C ,D) is a generalization of the connection-contraction algorithm for graphs (seeSection 5.3.) which was first explicitly suggested by Zykov in 1949. The “embryos” ofconnection-contraction may be seen in the papers by Birkhoff and Lewis (1946), by Whit-ney (1932), and even by Birkhoff (1912) where the chromatic polynomials were introduced(in the language of maps!).

Our idea is to find a pair of vertices that is neither aC -edge nor aD-edge, and to splitall the colorings ofH into two classes with respect to this pair of vertices. Further, byimplementing elimination,C -clearing,D-clearing and contraction (the order is important)the initial problem is reduced to the same problem for the newpair of “simpler” mixedhypergraphs (in the sense that one of them has fewer verticesand the other has moreD-edges of cardinality 2). Next we obtain a list of complete graphs with labeled vertices andfinally form a list of all strict colorings. We call this algorithm the “splitting-contractionalgorithm” and present it in the following form:

Algorithm 10.5.1 (splitting-contraction)

INPUT: An arbitrary mixed hypergraphH = (X,C ,D) with X labeled1,2, . . . ,n.OUTPUT: A list L of all strict colorings, the chromatic spectrum R(H ), the chromatic

polynomial P(H ,λ), the chromatic numbersχ(H ) and χ(H ).


1. Set lists L= Z = Y = /0, R(H ) = (0,0, . . . ,0), P(H ,λ) = 0, χ(H ) = χ(H ) = 0. AddH to Y.

2. Verify the condition of elimination for each element fromY; delete evidently uncol-orable mixed hypergraphs from Y.

3. PerformC -clearing andD-clearing where possible in Y .

4. Perform contraction where possible in Y; when contracting, amalgamate the labelsof the respective vertices.

5. Perform one splitting in each element of Y where possible;move completeD-graphsfrom Y to Z; if splitting is performed at least once then go to step 2.

6. Form a list L of all strict colorings using the labels of vertices of completeD-graphsfrom Z.

7. Compute the chromatic spectrum R(H ) by counting the numbers of completeD-graphs in Z having exactly i vertices, i= 1,2, . . . ,n.

8. Compute the chromatic polynomial P(H ,λ) using fundamental equality (10.2).

9. Determineχ(H ), χ(H ) using R(H ).

10. OUTPUT: list L, vector R(H ), polynomial P(H ,λ), numbersχ(H ), χ(H ). End.

Proposition 10.5.1 For any mixed hypergraphH = (X,C ,D), Algorithm 10.5.1 is cor-rect.

Sketch of Proof. Correctness of the algorithm follows from the evident rulesof elimina-tion, clearing, splitting and contraction, and from the fact that every labeled vertex of anycomplete graph in the listZ corresponds to the monochromatic subset of the respective ver-tices in precisely one feasible partition ofH . The last means that any feasible partition isreachable fromH by the splitting-contraction algorithm. �

Example 10.5.1The example in Figure 10.11 illustrates the splitting-contraction algorithmand some of the new properties of colorings as well. TheC -edges are shown by dashedellipse and lines. We have here a mixed hypergraphH = (X,C ,D), whereX = {1,2,

3,4}, C = {C} = {{1,2,3,4}}, D = {D1,D2,D3, D4} = {{1,2}, {2,3}, {3,4},{4,1}}.As the result of splitting-contraction, we obtain

Z = {K3,K3,K2}, R(H ) = (0,1,2,0),

P(H ,λ) = 2λ(3) + λ(2) = 2λ3−5λ2 +3λ,

χ = 2, χ = 3, and the corresponding listL of three strict colorings in the form(c(1),c(2),c(3),c(4)) is the following (now the numbers are colors):

L = {(1,2,3,2),(1,2,1,3),(1,2,1,2)}.


b

b

b

b

D1

D2D3

D4

1

2

3

4

Cb

b

b

b

1

2

3

4

b

b

b

b

1

2

3

4

b

b

b

b

1

2

3

4

b

b

b

b

1

2

3

4 b

b

b

13

24

b

b

b

1

3

24 b b

b

4 2

13

b b

b

4 2

13

b

b

b

1

3

24 b b

b

4 2

13

b

b

24

13

= + =

+ +=

+ + =

+ +

splitting 13

spl. 24

C -clearing+ contraction 13

elimination C -clearing+ contraction 24

splitting 24

contraction 24

Figure 10.11. Splitting-contraction algorithm.

The running time of Algorithm 10.5.1 is exponential (see Section 12.2.). It shows that thereexists a common unified procedure different from the exhaustive search which enables usto compute all strict colorings and to solve the problems on the minimum and maximum


number of colors simultaneously. Using special propertiesof mixed hypergraphs, moreefficient procedures for the same goals may be found in some cases. One can see thatC -edges were present implicitly in the classic connection-contraction algorithm (namely atthe point of contraction) and in this way were implicitly used since 1912. Finally, for anuncolorable mixed hypergraph, the output contains the empty list and zeros.

We can see that the class of polynomials that may be chromaticfor mixed hypergraphsis much larger than the class for usual graphs and hypergraphs because of the interactionsbetweenC -edges andD-edges.

Example 10.5.2Let H = (X,C ,D), whereX = {1,2,3,4,5}, C = {{1,2,3}, {1,3,4},{1,4,5},{1,5,2}}, D = {{3,5}}; we have thatχ(H ) = 3 and, after adding theD-edge{2,4}, we obtain the new mixed hypergraphH1 for which χ(H1) = 2. It is easy to seethat, in general, adding oneC -edge toH can increaseχ(H ) and adding oneD-edge candecreaseχ(H ).

There is one more unusual property of mixed hypergraph colorings, which is impossiblein D-hypergraphs. As we have seen, clearing operations eliminate the edges which containother edges of the same type, while the chromatic spectrum and the chromatic polynomialremain unchanged. In mixed hypergraphs however, some edgesnot containing other edgesof the same type may also be eliminated with the same effect.

Definition 10.5.1 In a mixed hypergraphH = (X,C ,D), a C -edge C∈ C (D −edge D∈D) is calledredundant if it does not contain any otherC -edge (D-edge) as a subset, and

R(H ) = R(H −C) (R(H ) = R(H −D)).

Example 10.5.3Let H = (X,C , /0), where X = {1,2,3,4} and C = {{1,2,3},{1,3,4},{1,2,4}, {2,3,4}}, see Figure 10.12. One can see that anyC -edge inH is re-dundant because any threeC -edges provide that at most two colors are used for the missingtriple. Therefore,

R(H ) = R(H −Cj) = (1,7,0,0), j = 1,2,3,4.

Moreover, if we add toH the family

D = {{1,2},{2,3},{3,4},{4,1}}

forming a simple cycle, then in the mixed hypergraphH ′ = (X,C , D), anyD-edge is alsoredundant because three otherD-edges guarantee different colors at the ends of a missingD-edge.

Proposition 10.5.2 If H is a disconnected mixed hypergraph andH1,H2, . . . , Hk k ≥ 2,are the connected components, then

χ(H ) = max{χ(H1),χ(H2), . . . ,χ(Hk)},

χ(H ) = χ(H1)+ χ(H2)+ . . .+ χ(Hk),

P(H ,λ) = P(H1,λ)P(H2,λ) . . .P(Hk,λ).


b b

bb

b b

bb

=

b b

bb

b b

bb

=

H

H ′

Figure 10.12.

Proof. The formulas follow from the possibility to color the components independently.�

Definition 10.5.2 A mixed hypergraphH = (X,C ,D) is called reduced if no edge is asubset of any other edge of the same type and the size of eachC -edge is at least 3.

Remark 10.5.1 As follows from the splitting-contraction algorithm,C -clearing, D-clearing and contractions ofC -edges of size two do not change anything; the coloring prop-erties of a mixed hypergraph can be derived from the respective reduced mixed hypergraph.Hence, without loss of generality,C -edges of size 2 and edges containing other edges of thesame type may be ignored, i.e. we only may consider reduced mixed hypergraphs.

Exercises 10.5.

1. For mixed hypergraphsH1 andH2 in Figure 10.13, apply splitting-contraction Algo-rithm 10.5.1 to compute both the chromatic polynomials and the chromatic spectra.Find all feasible partitions.


2. Apply the splitting-contraction algorithm toH1 = ({1,2,3,4},K34 , /0) and H2 =

({1,2,3,4}, /0,K34 ) to find the feasible partitions, chromatic polynomials, andchro-

matic spectra.


1. Given a mixed hypergraphH and a random coloring, determine if the coloring isproper.

b

b b

b b

b

b

b

H1 H2

Figure 10.13.

2. Given a mixed hypergraphH , determine if it is reduced.

3. For mixed hypergraphsH1 andH2 in Figure 10.13, by generating colorings at ran-dom, find an estimate on the chromatic spectrum, the chromatic polynomial, andcompare with the exact values.

4. Given a mixed hypergraphH , by generating colorings at random, find an estimate onthe chromatic spectrumR(H ) and the chromatic polynomialP(H ,λ).

10.6. Uncolorability

Let H = (X,C ,D) be a mixed hypergraph andC = {C1,C2, . . . , Cl} and D = {D1, D2,. . . , Dm} be two nonempty families of edges. As usual, denoteI = {1,2, . . . , l} andJ ={1,2, . . . ,m}. In this section we assume that mixed hypergraphH is simple, i.e. no edge iscontained in any other edge of the same type.

Recall that a mixed hypergraph is called uncolorable if it admits no proper coloring;otherwise, it is called colorable. Next we formulate the colorability problem:

Definition 10.6.1 Given a mixed hypergraphH = (X,C ,D), the problem of decidingwhether there exists at least one proper coloring ofH is called thecolorability problem .

The colorability problem represents a new type of problem incoloring theory. It con-tains, as a special case, the problem of deciding whether a classic hypergraph admits aproper coloring with a given number of colors. Namely, to anyD-hypergraph, we can


addC -edges of the complete hypergraphKk+1n . Then aD-hypergraphH = (X,D), is k-

colorable if and only if the mixed hypergraphH ′ = (X,( X

k+1

)

,D) is colorable.Minimal uncolorable mixed hypergraphs. In this subsection we show that quite dif-

ferent methods are required to determine the conditions forcolorability in different classesof mixed hypergraphs. Evidently, if a mixed hypergraph contains an uncolorable subhyper-graph of any kind (induced or partial), it is uncolorable as well. Therefore, one of the basicgoals is to find the list of all minimal uncolorable mixed hypergraphs from a given classand describe the colorability in terms of forbidden subhypergraphs.

In particular, we prove that there are uncolorable mixed hypergraphsH with arbitrarydifference between the upper chromatic number of the partial C -hypergraphHC and thelower chromatic number of the partialD-hypergraphHD .

Definition 10.6.2 An uncolorable mixed hypergraph is called (inclusion-wise) minimaluncolorable if it is connected and becomes colorable after the weak removal of anyC -edgeor anyD-edge.

Notice that minimal uncolorable mixed hypergraphs do not contain isolated vertices. Iftwo verticesx1,x2 form a bi-edge, thenH is uncolorable, by the evident conflict of con-straints on{x1,x2}. Thus, minimal uncolorability in this situation immediately implies thatin H there are no further vertices,C -edges, orD-edges. Evidently, a minimal uncolorablemixed hypergraph becomes colorable if we strongly delete any vertex.

Theorem 10.6.1Every uncolorable mixed hypergraphH = (X,C ,D) contains aninclusion-wise minimal uncolorable (induced, partial) subhypergraph.

Proof. One can sequentially delete vertices (strongly),C -edges (weakly) andD-edges(weakly) until we get a colorable subhypergraph; then restore the last element removed anddelete other elements, and so on. SinceH is finite, the assertion follows. �

Definition 10.6.3 Given an uncolorable mixed hypergraphH = (X,C , D), a partial sub-hypergraphH ′ (induced subhypergraphH ′) is called amaximal partial colorable sub-hypergraph (maximal induced colorable subhypergraph) ofH if adding anyC -edge orD-edge (or any vertex with all the incident edges of both types) of H to H ′makesH ′ un-colorable.

So, minimal uncolorable subhypergraphs and maximal colorable subhypergraphs of un-colorable mixed hypergraphs determine a critical border ofcolorability. The interaction ofC andD on the same vertex setX is not simple:HC andHD may be very easy to colorseparately, while the wholeH is uncolorable.

To show that, let us consider the following problem:For every integer k≥ 0, let v(k)denote the smallest natural number n such that there exists an inclusion-wise minimal un-colorable mixed hypergraphH = (X,C ,D), |X|= n, for whichχ(HC )−χ(HD) = k. Whatare the values of v(k) for k = 0,1,2, . . . .?

Observe first that if minimality is not required, then one caneasily construct an uncol-orable mixed hypergraphH with largeχ(HC )−χ(HD) by taking just one bi-edge of size2 together withn− 2 = k+ 1 isolated vertices ; then the lower chromatic number ofHD


b b

bb

Figure 10.14. The “most uncolorable” mixed hypergraph on four verticesU4.

is 2 and the upper chromatic number ofHC is n− 1. Second, ifk is negative, then everymixed hypergraph is evidently uncolorable. For everyn = |X| ≥ 2, the “most uncolorable”example (if minimality is not required) is the mixed hypergraphUn = (X,

(X2

)

,(X

2

)

) whichis uncolorable together with each of its induced subhypergraphs of order≥ 2. In Un, eachpair of vertices is aC -edge and aD-edge at the same time. Figure 10.14 shows the mostuncolorable mixed hypergraph on four verticesU4.

For nonnegativek, the theorem below gives the characterization of the numbers v(k).

Theorem 10.6.2 (Tuza, Voloshin 2000)For every k≥ 0,

v(k) = k+4.

Proof. Let H = (X,C ,D) be an inclusion-wise minimal uncolorable mixed hypergraphsuch that|X|= n = v(k), andχ(HC )−χ(HD) = k. We have to prove thatn = v(k) = k+4.

We show first thatn ≥ k+ 4. For a contradiction, suppose thatn < k+ 4. SinceH

is uncolorable,χ(HD) ≥ 2. If χ(HD) ≥ 3, then χ(HC ) ≥ k+ 3, which impliesn = k+ 3,χ(HC ) = n; therefore,H contains noC -edges and thus it is colorable, a contradiction.

Hence,χ(HD) = 2. Then we have only two possibilities for the number of vertices:n = k+2 or n = k+3.

Similarly to the previous case, forn= k+2 andχ(HC ) = k+2 it follows that the mixedhypergraphH contains noC -edges, and therefore is not uncolorable. Hence, considerthe last casen = k + 3. Sinceχ(HC ) = k+ 2 = n− 1, the partialC -hypergraphHC =(X,C ) is a bistar, i.e., aC -hypergraph having two vertices, sayx1 andx2, that belong toall C -edges. If this pair{x1,x2} were not aD-edge inHD , then we could colorx1,x2

with the first color and the remaining vertices all differently, which again contradicts theuncolorability of H . Consequently, the pair{x1,x2} is a D-edge inH . SinceH is anuncolorable hypergraph minimal under inclusion, noC -edge may coincide with{x1,x2},and therefore the cardinality of eachC -edge is at least 3.

Consider an arbitrary proper 2-coloring ofHD = (X,D). It is, at the same time, acoloring of the initial mixed hypergraphH , because eachC -edge contains at least three


b

b b

b b

b b

Figure 10.15. Minimal by inclusion uncolorable mixed hypergraph,k = 3.

vertices. Thus, again, we obtain thatH is colorable. This contradiction implies thatv(k) =n≥ k+4.

Now, in order to prove the converse inequalityv(k) ≤ k+ 4, we construct a series ofexamples of minimal uncolorable mixed hypergraphs withχ(HC )− χ(HD) = k andn =k+4, k = 0,1,2, . . . . The construction will depend on the parity ofk; we first describe theparticular casesk = 0,1 that can be verified directly.

k = 0.Let H = (X,C ,D), whereX = {1,2,3,4}, C = {{1,2,3}, {1,2, 4}}, D = {{1,2},

{2,3},{2,4},{3,4}}.

k = 1.ConsiderH = (X,C ,D), whereX = {1,2,3,4,5}, C = {{1,2,3}, {1,2,4}, {1,2,5}},

D = {{1,2},{3,4},{4,5},{3,5}}.

k = 2l , l ≥ 1.Construct the mixed hypergraphH = (X,C ,D), whereX = {1,2,3, . . . , k+ 4}, C =

{{1,2, i} : 3≤ i ≤ k+4}, andD = {{i, i +1} : 1≤ i ≤ k+3}∪{k+4,2}.In other words,HC = (X,C ) represents a 3-uniform bistar in which vertices 1,2 belong

to all C -edges, and thereforeχ(HC ) = n− 1 = k+ 3. Moreover,HD = (X,D) is the oddcycle(2,3,4, . . . ,k+4,2) with the pendantD-edge{1,2}, so thatχ(HD) = 3.


Let us try to construct a proper coloringc of H . As usual,c(i) means the color ofvertex i, i = 1,2, . . . ,n. In any possible coloring ofH , vertices 1 and 2 have to be coloreddifferently, sayc(1) = 1, c(2) = 2. Since{2,3} is a D-edge, we havec(3) 6= c(2) and,because of theC -edge{1,2,3}, the unique possibility for vertex 3 to get colored isc(3) =c(1) = 1. In the same way,c(4) 6= c(3) and, because of theC -edge{1,2,4}, the uniquepossibility for vertex 4 to be colored isc(4) = c(2) = 2.

It is now clear that the colors have to alternate on the cycle(2,3,4, . . . ,k+ 4). Sincec(k+ 3) = 1 andc(2) = 2, we can color vertexk+ 4 neither with color 1 nor with color2. However, any other colorc(k+ 4) is infeasible because of theC -edge{1,2,k + 4}.Consequently,H is uncolorable. One can easily check that it is minimal underinclusion.k = 2l +1, l ≥ 1.

Construct the mixed hypergraphH = (X,C ,D), whereX = {1,2,3, . . . ,k+ 4}, C ={{1,2, i} : 3≤ i ≤ k+4}, andD = {1,2}∪{{i, i +1} : 3≤ i ≤ k+3}∪{k+4,3}.

Again, HC = (X,C ) represents a 3-uniform bistar with vertices 1,2 shared by allC -edges, so thatχ(HC ) = n−1 = k+ 3. In the present case,HD = (X,D) is a disconnectedgraph havingD-edge{1,2} as the first component and the odd cycle(3,4, . . . ,k+ 4,3) asthe second component, yielding againχ(HD) = 3. Casek = 3 is shown in Figure 10.15.

Let c(1) = 1, c(2) = 2. Forc(3) there are only two possibilities:c(3) = 1, or c(3) = 2.By symmetry, we may assumec(3) = 1. Then, similarly to the argument above, we obtainc(4) = 2, c(5) = 1, c(6) = 2, and so on; i.e., the colors have to alternate along the odd cycle.Since vertexk+4 cannot be colored with any color (because of theC -edge{1,2,k+4} andtheD-edges{k+3,k+4} and{k+4,3}), we conclude thatH is uncolorable. Minimalityis also easily seen. Hence, the theorem follows. �

Complete uncolorable mixed hypergraphs.For 2≤ l ,m≤ n = |X|, let

K (n, l ,m) = (X,C ,D) = (X,

(

Xl

)

,

(

Xm

)

).

Hence,|C | =(n

l

)

and|D| =(n

m

)

. Call K (n, l ,m) thecomplete(l ,m)-uniform mixed hy-pergraph of ordern. In other words, inK (n, l ,m) every l vertices form aC -edge, andeverym vertices form aD-edge. Evidently, for givenn, l ,m there exists exactly one (up toisomorphism)K (n, l ,m). Complete (4,2)-uniform mixed hypergraphK (4,4,2) is shown inFigure 10.16.

b

b b

b

Figure 10.16.K (4,4,2).


Theorem 10.6.3 (Tuza, Voloshin, 2000)K (n, l ,m) is uncolorable if and only if

n≥ (l −1)(m−1)+1.

Proof. ⇒ Let n≤ (l −1)(m−1). We colorm−1 vertices with the first color, the nextm−1 vertices with the second color, etc. Sincen≤ (l −1)(m−1), this procedure requiresat mostl −1 colors, and a proper coloring ofK (n, l ,m) is obtained; i.e., the hypergraph iscolorable.⇐ Let n≥ (l −1)(m−1)+1. Suppose there exists a proper coloring ofK (n, l ,m). Sinceeachm-tuple is aD-edge and eachl -tuple is aC -edge, the number of vertices in any onecolor class does not exceedm− 1 and the total number of colors does not exceedl − 1.Sincen≥ (l −1)(m−1)+1, there is at least oneC -edge orD-edge colored improperly, acontradiction. �

If we fix l andm, and letn approach the infinity, then we have the fixed number (namely,(l −1)(m−1)) of colorable mixed hypergraphsK (n, l ,m) and any number of uncolorableK (n, l ,m). In other words, whenn is growing, the share of colorableK (n, l ,m) becomessmaller and smaller. This can be stated as the following corollary.

Corollary 10.6.1 (Tuza, Voloshin 2000) For fixed(l ,m), almost allK (n, l ,m) are uncol-orable.

A completely different conclusion is obtained, however, ifwe do not fix the valuesl andm. In the analysis below it will turn out that the proportion ofuncolorable complete mixedhypergraphs of ordern tends to zero asn gets large.

Theorem 10.6.4 (Tuza, Voloshin, 2000)For unrestricted(l ,m), almost allK (n, l ,m) arecolorable.

Proof. In order to simplify the formulas, let us calculate for mixedhypergraphs of ordern+1 instead ofn. Sincel = 1 andm= 1 are excluded by definition, we haven2 possibilitiesto choose the pair(l ,m) in the range 2≤ l ≤ n+1, 2≤m≤ n+1. Applying Theorem 10.6.3,we obtain thatK (n, l ,m+1) is uncolorable if and only if

(l −1)(m−1) ≤ n.

Here the smallest possible value ofm−1 is 1. Thus, for eachl ≥ 2, there are preciselyb nl−1c

uncolorable complete mixed hypergraphs of ordern+1. Consequently, the total numberNn

of complete uncolorable mixed hypergraphs onn+1 vertices equals

Nn =n

∑k=1

⌊nk

⌋

' nlogn

where the asymptotic equation is meant asn tends to infinity (here and next we use sim-ple calculus formulas). Thus, the proportion of uncolorable complete mixed hypergraphsequals

limn→∞

Nn

n2 = limn→∞

lognn

= 0,

implying that almost all large complete mixed hypergraphs are colorable. �


l \m 2 3 4 5 6 7 82 – – – – – – –3 – – – + + + +4 – – + + + + +5 – + + + + + +6 – + + + + + +7 – + + + + + +8 – + + + + + +

Figure 10.17. The uncolorability ofK (8, l ,m).

Example 10.6.1The behavior of the uncolorability ofK (8, l ,m) is shown in Figure 10.17.The signs “+” and “–” mean that for givenl ,m, the mixed hypergraphK (8, l ,m) is colorableor uncolorable, respectively. The statements above may be commented on in the followingway. Whenn tends to infinity, and ifl ,m are fixed, then the uncolorability zone sooner orlater reaches anyK (n, l ,m); however, ifl ,mare not fixed, then the ratio of the uncolorabilityzone in the entire “big square” tends to zero. The table in Figure 10.17 contains 16 cases ofuncolorability and 33 cases of colorability; therefore theratio of uncolorability is 16/49.

Colorability of mixed hypertrees. A mixed hypergraphH = (X,C ,D) is called amixed hypertree if there exists a host treeT = (X,E) such that everyC ∈ C and everyD ∈ D induces a subtree inT. Clearly, it is a direct generalization of the hypertree conceptstudied in Chapter 8 which now corresponds to the caseC = /0.

Recall that in a mixed hypergraphH = (X,C ,D), a D-edgeD ∈ D, is called evidentlyuncolorable if each pair of verticesx,y ∈ D is connected by aC -path ofH consisting ofC -edges of size 2.

Theorem 10.6.5 (Tuza, Voloshin 2000)A mixed hypertreeH = (X,C ,D) is uncolorableif and only if it contains an evidently uncolorableD-edge.

Proof. ⇒ Let H = (X,C ,D) be an uncolorable mixed hypertree. By contradiction,suppose it does not contain evidently uncolorableD-edges. Observe that if it contains noC -edges of size 2, then it is colorable. Indeed, consider the corresponding host treeT andcolor it as usual, starting at any vertex and alternating colors 1 and 2 along the tree. Thecoloring obtained is, at the same time, a proper coloring ofH . If H = (X,C ,D) containsC -edges of size 2, then each of them coincides with an edge ofT. Now we repeat theprevious procedure with the following exception: if we encounter aC -edge of size 2, thenwe do not change color along this edge ofT (i.e., an edge ofT becomes properly coloredif and only if it is not aC -edge inH ). Since there are no evidently uncolorableD-edges inH , we again obtain a proper coloring ofH . Therefore,H is colorable, a contradiction

⇐ Obvious. �

We end the section with the following evident

Corollary 10.6.2 Every reduced mixed hypertree is a colorable mixed hypergraph.


b b

b

b

b

b

b

Figure 10.18. Uncolorable mixed hypertree.

Corollary 10.6.3 If H is a colorable mixed hypertree, thenχ(H ) ≤ 2.

An example of uncolorable mixed hypertree is shown in Figure10.18. As usual,C -edges are drawn by dashed curves, andD-edges are drawn by solid curves; the host tree isnot shown.

Exercises 10.6.

b

b

bb

b

H

Figure 10.19.

1. For mixed hypergraphH in Figure 10.19, find the lower chromatic number ofHD

and the upper chromatic number ofHC .

2. Prove that mixed hypergraphH in Figure 10.19 is uncolorable.

3. For mixed hypergraphH in Figure 10.19, find a minimal uncolorable subhypergraph.

4. For mixed hypergraphH in Figure 10.19, find a maximal colorable subhypergraph.



1. Given a mixed hypergraphH , by generating colorings at random, determine if it iscolorable.

2. Given an uncolorable mixed hypergraphH , by generating colorings at random, finda maximal colorable (induced / partial) subhypergraph.

10.7. Unique Colorability

Let H = (X,C ,D) be an arbitrary mixed hypergraph and no edge is included, i.e. containedin any other edge of the same type.

Definition 10.7.1 A mixed hypergraphH is calleduniquely colorable (uc hypergraph oruc for short) if it has precisely one strict coloring apart frompermutation of colors. Theclass of uniquely colorable mixed hypergraphs is denoted byUC .

Equivalently,H is uc if it allows exactly one feasible partition of the vertex setX intocolor classes. Let us agree that the expression “unique coloring” means “unique partition”into the corresponding number of color classes. This means that we ignore the permutationof colors unless stated otherwise. The term “uniquely colorable” is inherited from graphcoloring; the meaning of it is “uniquely partitionable”.

Evidently, if H is a uc hypergraph, then

χ(H ) = χ(H ) = χ, rχ(H ) = 1,

andR(H ) = (0, . . . ,0,1,0, . . . ,0).

Therefore,P(H ,λ) = λ(λ−1)(λ−2) · · · (λ−χ+1) = λ(χ).

A classic graphG as aD-graph, is uniquely colorable if and only ifG is a complete graphKn (this is true for allD-hypergraphs). In other words, the ucmixed hypergraphs representmerely generalizations of cliques. As we know, the completer-uniform hypergraphKr

n is ageneralization ofKn; however, the coloring properties ofKr

n are far to be as nice as those ofuc mixed hypergraphs. This reflects the fact thatKr

n is uc if and only ifr = 2.

Proposition 10.7.1 Given a uc mixed hypergraphH = (X,C ,D) with a unique strict col-oring c, then the union of any two color classes contains aD-edge; any partition of anycolor class encounters aC -edge having exactly two vertices in common with this color classand at most one vertex in common with any other color class.

Proof. This follows directly from the definition of a proper coloring of a mixed hyper-graph. �


Definition 10.7.2 In a mixed hypergraphH = (X,C ,D), a sequence of vertices x=x0,x1, . . . ,xk = y,k≥ 1, is called an(x,y)-invertor if and only if xi 6= xi+1 and(xi ,xi+1)∈ D

for every i= 0,1, . . . ,k−1, and, moreover, the following implication holds:

x j 6= x j+1 6= x j+2 6= x j ⇒{x j ,x j+1,x j+2} ∈ C , j = 0,1, . . . ,k−2.

The(x,y)-invertor is called odd or even if k is odd or even, respectively; it is called cyclicif x = y; and if k≥ 2, then x1, . . . ,xk−1 are termed internal vertices.

If a mixed hypergraph has an even(x,y)-invertor, then in every proper coloring, theverticesx andy have the same color; in this case, the(x,y)-invertor “transports” the colorof vertexx to vertexy. In contrast, if there is an odd(x,y)-invertor, then in every coloring,the verticesx and y have different colors; in this case, the(x,y)-invertor “excludes” thecolor of vertexy to be the color of vertexx. Notice that invertors may be induced or partialsubhypergraphs, and induce either a uc subhypergraph or an uncolorable one.

b b b b b b

b

b

x0 x1 x2 x3

H1

1 2 1 2

x0 x1 = x3 x4

x2

1 2

1

1

H2

Figure 10.20. Invertors.

Two examples of invertors are shown in Figure 10.20.H1 is an odd invertorx0,x1,x2,x3, while H2 is an even invertorx0,x1,x2, x3,x4. The colors are shown by numbers 1and 2. While inH1, theD-edges represent a simple path connectingx0 andx3, in H2, the(x0,x4)-path uses edge{x1,x2} twice, i.e. it is not a simple path. The important featurehowever, is that every three different consecutive vertices in such path form aC -edge. BothH1 andH2 are uc mixed hypergraphs with the unique strict coloring shown in the figure. Inevery proper coloring ofH1, the verticesx0 andx3 have different colors. In contrast, in anyproper coloring ofH2, the verticesx0 andx4 have the same color. We say thatH1 excludesvertexx3 from having the color of vertexx0; and,H2 “transports” the color of vertexx0 tovertexx4. One can easily see that for both invertorsχ = χ = 2, and, moreover

R(H1) = R(H2) = (0,1,0,0),P(H1,λ) = P(H2,λ) = λ(λ−1).

The latter means that from coloring point of view, the invertors simply extend the propertiesof complete graphK2.

Embeddings into uc mixed hypergraphs.In D-hypergraphs, for eachn we have theonly uc graphKn; the fundamental property is that any induced subgraph ofKn is a ucgraph as well. There is a very different situation in mixed hypergraphs. In this subsection,


we show that every mixed hypergraph having at least one coloring may be an inducedsubhypergraph of a uniquely colorable mixed hypergraph. Namely, we embed a mixedhypergraphH into a mixed hypergraphH ′ such thatH ′ has a unique strict coloring. Moreprecisely:

Theorem 10.7.1 (Tuza, Voloshin, Zhou 2002)Let H = (X,C ,D) be a colorable mixedhypergraph and c= X1∪ ·· ·∪Xt be a strict t-coloring. Then there exists a uc mixed hyper-graphH ′ = (X′,C ′,D ′) with the following properties:

1. H is an induced subhypergraph ofH ′;

2. χ(H ′) = χ(H ′) = t and an extension of c is the unique strict coloring ofH ′.

Proof. To extendH into someH ′ in the required way, we first choose a dummy span-ning treeTi inside each partition classXi of c. For every edgexy of Ti we put an even(x,y)-invertor with internal vertices not inX. Different even invertors should be internallydisjoint. These invertors ensure that vertices fromXi will always have the same color.

At this point, the internal vertices of the invertors may getnew colors; hence, the numberof possible colors forH ′ may increase. In order to avoid this, we insert oneC -edge of size 3,{x= x1,x2,z}, for each even(x,y)-invertor, wherez is any vertex such thatx andzbelong todistinct color classesXi underc. TheseC -edges imply that no new color will occur outsideX.

Next, for each pair(i, j) such thatXi andXj are two distinct classes ofc, we choosetwo verticesxi ∈ Xi andx j ∈ Xj , and build an odd(xi ,x j)-invertor on them. Then any twovertices belonging to distinct classes ofc are assigned distinct colors. Again, the newlyadded intermediate vertices in odd invertors should be distinct for the distinct vertex pairsof H , and should not be inX. Since colors on invertors alternate, the colors on an oddinvertor are all the same as the two colors at the endpoints. Hence, no new colors outsideXcan occur on internal vertices of newly created invertors.

It is clear that such extension ofc is the unique strict coloring ofH ′ with preciselytcolor classes because each class ofc is monochromatic (due to the presence of even inver-tors) and no pair of vertices belonging to distinct color classes ofc can get the same color(by the odd invertors). �

Uniquely colorable mixed hypertrees.In a mixed hypergraphH , for two verticesx,ythere may be many(x,y)-invertors (a subhypergraph induced by an(x,y)-invertor is alsotermed an(x,y)-invertor). Two(x,y)-invertors are different if they represent two differentsequences of vertices. The shortest(x,y)-invertor contains the minimum number of vertices.Recall that an(x,y)-invertor with x = y is called a cyclic invertor. We show here thatinvertors play an important role in uniquely colorable mixed hypertrees. Also recall thatreduced mixed hypertrees are colorable (Corollary 10.6.2).

Definition 10.7.3 In a mixed hypertree, a cyclic invertor is calledsimple if all C -edges aredifferent, anyD-edge is used along the invertor precisely two times, and therepetition issequential.


Let µ = (z0,z1, . . . ,zk = z0), k ≥ 6 be a simple cyclic invertor in a mixed hypertree.Without loss of generality, assume thatz0 6= z1 6= z2 6= z0. From the definition of a simplecyclic invertor, it follows thatz0 6= z2 66= . . . 6= zk−2 andz1 = z3 = . . . = zk−1 = y, wherey isthe center of a star in the host treeT.

Theorem 10.7.2 If H = (X,C ,D) is a reduced mixed hypertree with|D| ≤ n− 2, thenr2(H ) ≥ 2.

Proof. Let T = (X,E) be a host tree of the mixed hypertreeH . Since |D| ≤ n− 2,in T there exists an edgee = {x,y} 6∈ D. Starting with verticesx,y, we can constructtwo different colorings with colors 1 and 2 generating two different partitions. First, putc(x) = c(y) = 1 and color all the other vertices alternatively along the treeT with the colors2, 1, 2, . . . . Second, apply the same procedure starting withc(x) = 1 andc(y) = 2. �

Theorem 10.7.3 (Niculitsa, Voloshin, 2000)A reduced mixed hypertreeH = (X,C ,D) isuniquely colorable if and only if for every two vertices x,y∈X there exists an(x,y)-invertor.

Proof. ⇒ Let c be the unique feasible partition of the mixed hypertreeH . Recall thatc(x) denotes the color class of vertexx in the partition, or, equivalently, the color of vertexx. We show that for any two verticesx,y∈ X there exists an(x,y)-invertor.

SupposeH has two verticesu,v∈X such that there is no(u,v)-invertor inH . Considerthe unique(u,v)-path in the host tree T ofH . The assumption implies that either, inH ,there is noD-path connectingu andv or, in the sequenceu = x1,x2, . . . ,xp = v, there existsa triple of pairwise different verticesx j ,x j+1,x j+2 not belonging toC . If there is noD-pathconnectingu and v, then, by Theorem 10.7.2,H has two different feasible partitions, acontradiction.

Assume now that in the sequenceu = x1,x2, . . . ,xp = v, each pair of consecutive ver-tices is aD-edge, and there is a triple of pairwise different verticesx j ,x j+1,x j+2 such that{x j ,x j+1,x j+2} 6∈ C . Evidently,x j+1 is not pendant inT. Let T1 andT2 be two connectedcomponents obtained after deletion of vertexx j+1 from the host treeT. There are two cases.

1) c(x j ) = c(x j+2). Evidently, the number of color classes in the unique partition c of H

is 2. Re-color the vertexx j+2 and all vertices of even distance fromx j+2 in the componentT2 with the new color. The obtained coloring is a proper coloring of H inducing a feasiblepartition different fromc, a contradiction.

2) c(x j) 6= c(x j+2). Since{x j ,x j+1},{x j+1,x j+2} ∈ D, c(x j ) 6= c(x j+1) 6= c(x j+2). Con-sequently,H is colored with at least three colors. But every mixed hypertree can also becolored with two colors, and we again have two different feasible partitions, a contradiction.

⇐ Assume that any two verticesx,y ∈ X are joined by an(x,y)-invertor. SupposeHhas at least two distinct feasible partitionsc1 andc2. Then there are two vertices, sayx′,y′,such thatc1(x′) = c1(y′) butc2(x′) 6= c2(y′). Without loss of generality, consider an(x′,y′)-invertorx′ = x0,x1, . . .xk = y′. From the definition of an invertor, it follows that, ifk is even,then in all possible colorings the verticesx′ andy′ have the same color; ifk is odd, then inall possible colorings the verticesx′ andy′ have distinct colors. Therefore, in all colorings,eitherc(x′) = c(y′) or c(x′) 6= c(y′), a contradiction. �

Corollary 10.7.1 If H = (X,C ,D) is a reduced uniquely colorable mixed hypertree withthe host tree T= (X,E), thenD = E .


Recall that in a mixed hypergraphH , theC -edgeC∈ C is called redundant ifR(H ) =R(H −C).

Corollary 10.7.2 If H = (X,C ,D) is a reduced uniquely colorable mixed hypertree, thenno D-edge is redundant.

Corollary 10.7.3 In a reduced uniquely colorable mixed hypertreeH = (X,C ,D), anyC -edge of size≥ 4 is redundant.

Proof. No invertor contains such aC -edge. �

Theorem 10.7.4 In a reduced uniquely colorable mixed hypertreeH = (X,C ,D), a C -edge C of size 3 is redundant if and only if there exists a simple cyclic invertor containingC.

Proof. ⇒ Let C = {x1,x2,x3} be the redundantC -edge. By definitionH ′ = (X,C ′,D)whereC ′ = C \{C} is a uniquely colorable mixed hypertree. Then for the verticesx1 andx3

in H ′ there exists an(x1,x3)-invertor: x1 = z0,z1, . . . ,zk = x3. Construct the(x1,x1)-invertorin the following way:x1 = z0,z1, . . . ,zk = x3,x2,x1. This invertor is a simple cyclic invertorof H containingC.

⇐ Conversely, suppose that theC -edgeC = {x1,x2,x3} is contained in a simple cyclicinvertorx1 = z0,z1, . . . ,zk = x3,x2,x1. Then the verticesx1 andx3 are joined by two different(x1,x3)-invertors: {x1,x2,x3} = C andx1 = z0,z1, . . . ,zk = x3 = (x1,x3)

′-invertor. In each(x,y)-invertor containingC, replace thisC -edge by the(x1,x3)

′-invertor. Thus,H ′ = (X,C \{C},D) is uniquely colorable; i.e., theC -edgeC is redundant. �

Consider a mixed hypergraphH = (X,C ,D). Let X = X1∪X2∪ . . .∪Xi be a properi-coloring ofH , χ(H ) ≤ i ≤ χ(H ) and choose anyXj .

Definition 10.7.4 Thetouching graph of a color class Xj is the graph Lj = (Xj ,E j) wherethe edge set Ej is defined in the following way:{x,y} ∈E j if and only if some C∈ C satisfiesC∩Xj = {x,y}, and |C∩Xk| ≤ 1 for any k 6= j.

Theorem 10.7.5Given a mixed hypergraphH and an arbitraryχ-coloring ofH , then allχ touching graphs are connected.

Proof. If at least one touching graph is disconnected, then we can construct a new strictcoloring ofH with χ+1 colors by assigning a new color to the vertices of one component.

�

Notice that the connectedness of a touching graph for a colorclass means that we can’t splitthe color class into two smaller ones.

Corollary 10.7.4 Given a uniquely colorable mixed hypergraphH and the unique parti-tion c, then allχ(H ) touching graphs are connected.

Corollary 10.7.5 If a reduced mixed hypertreeH = (X,C ,D) is uniquely colorable, thenin its 2-coloring, the touching graphs L1 and L2 are connected.


Theorem 10.7.6Given any uniquely colorable mixed hypergraphH = (X,C ,D), then

|C | ≥ n−χ.

Proof. Let H be a uniquely colorable mixed hypergraph,χ(H ) = χ. Consider a uniqueχ-coloringX = X1∪X2∪ . . .∪Xχ and construct the touching graphs

L1 = (X1,E1), L2 = (X2,E2), . . . ,Lχ = (Xχ,Eχ).

The minimum number of edges inLi required forLi to be connected is|Xi|−1, and, in thiscase, eachLi is a tree,i = 1,2, . . . ,χ. Since every edge inLi corresponds to someC -edge ofH , we obtain that the minimum number ofC -edges is:

|X1|−1+ |X2|−1+ . . .+ |Xχ|−1= |X|−χ. �

Corollary 10.7.6 A mixed hypergraphH with |C | < n−χ(H ) is not uc.

Corollary 10.7.7 The minimum number ofC -edges in any reduced uniquely colorablemixed hypertreeH = (X,C ,D) is n−2.

Corollary 10.7.8 In a reduced uniquely colorable mixed hypertreeH = (X,C ,D), thenumber of redundantC -edges is|C |−n+2.

Proof. Indeed, for each touching graphLi, construct a spanning treeTi , i = 1,2. Eachelementary cycle inLi generates some simple cyclic invertor inH . Therefore, everyC -edge ofH is redundant if it has size≥ 4, or corresponds to some edge ofLi which is achord with respect toTi. �

Remark. A redundantC -edge may become irredundant after deleting some other re-dundantC -edges fromC .

Definition 10.7.5 A reduced mixed hypertreeH = (X,C ,D) is calledcomplete if everyedge of the host tree T forms aD-edge, and every path on three vertices of T forms aC -edge inH .

Therefore, having the host treeT = (X,E) for the complete mixed hypertreeH =(X,C ,D), we see thatD = E . As in graphs and hypergraphs, here completeness meansthat we can’t add any additionalC - or D-edges and preserveH being reduced.

Denote byM the number ofC -edges of a complete mixed hypertreeH = (X,C ,D). Then

M = ∑x∈T

d(x)≥2

(

d(x)2

)

,

whered(x) is the degree of vertexx in the host treeT. Examples show that for anyk > 1,one can construct a mixed hypertreeH = (X,C ,D) with |D|= n−1, n−2≤ |C | ≤M andχ(H ) = k. Therefore, these bounds on|D| and|C | are not sufficient for a mixed hypertreeto be uniquely colorable.


Theorem 10.7.7 (Niculitsa, Voloshin, 2000)Let H = (X,C ,D) be a reduced uniquelycolorable mixed hypertree with the minimum number ofC -edges, T= (X,E) be a hosttree, T1 and T2 be touching graphs which are trees. Then there exists a vertex x∈ X, simul-taneously pendant in T and in T1 or T2, such thatH − x, obtained by strong deletion of xfrom H , is a uc mixed hypertree.

Proof. Let us first suppose that every pendant vertex of the host treeT is not pendant ineitherT1 or T2. This means that every such vertex belongs to at least twoC -edges. Then,it is possible to start at any such vertex and construct a cyclic invertor (sinceT is finite).This implies that oneTi contains a cycle; i.e.,H is not a uc mixed hypertree with theminimum number ofC -edges, a contradiction. Hence, there exists a vertexx ∈ X whichis pendant inT and in, sayT1. Then x belongs to precisely oneC -edgeC = {x,y,z}.Evidently, {x,y},{y,z} ∈ D. Consequently,c(x) = c(z) in the unique partitionc of H .Strongly deletex from H , thus obtainingH − x. Clearly, H − x is a mixed hypertree,|C (H − x)| = |C (H )| − 1. If it is not uc, then it admits at least two strict colorings. Inany extension of a proper coloring ofH − x to a proper coloring ofH , we always havec(x) = c(z). Therefore both colorings ofH − x induce different strict colorings of theinitial mixed hypertreeH , a contradiction. �

Definition 10.7.6 A mixed hypergraphH = (X, C ,D) is calleduc-orderable if there ex-ists an orderingσ = (x1,x2, . . . ,xn) of the vertex set X such that every mixed subhypergraphHi , induced by vertices

{x1,x2, . . . ,xi}, i = 1,2, . . . ,n

is uniquely colorable.

It is easy to see that if we add vertices in the orderσ and color them successively, theneach time, there is precisely one possibility to color the next vertex. Equivalently, we candecomposeH by elimination of vertices in order inverse toσ, i.e. (xn,xn−1, . . . ,x1), and ateach step we obtain a uc mixed hypergraph. Notice that not every uc mixed hypergraph isuc-orderable, but any uc-orderable mixed hypergraph is uc.

From Theorem 10.7.7, we conclude that a uc-orderable mixed hypertreeH can be rec-ognized by consecutive elimination of pendant vertices of the D-graphHD in a specialordering, by applying the following:

Algorithm 10.7.1 (uc-ordering)

• INPUT: A reduced mixed hypertreeH = (X,C ,D), σ (n-dimensional empty vector).

• OUTPUT: A uc-orderingσ of H or an indication thatH is not uc.

• Idea: Simultaneous decomposition ofHD , spanning trees T1 and T2 of touchinggraphs L1,L2, respectively, by pendant vertices.

Iterations:

1. If there is a vertex x∈ X not belonging to aC -edge of size 3 or aD-edge of size 2,then returnNON uc. Otherwise, remove fromC all elements of size≥ 4.


b b b

b

b

b b

b

bb

x1 x2 x3

x4

x5

1 2 1

2

2

x1 x3

x2

x4

x5

L1

L2

H

Figure 10.21. Uniquely colorable mixed hypertree.

2. ColorD-graphHD with two colors.

3. Construct touching graphs L1 and L2.

4. If Li, i = 1,2, is not connected, then returnNON uc.

5. For Li, construct spanning tree Ti , i = 1,2.

6. i := 1.

7. While in Ti there exists a vertex x pendant in both Ti and HD , delete it from Ti andHD , and include x inσ.

8. If at least one of T1 and T2 is not empty, assign i:= 3− i and go to 7; otherwise,return uc, σ=uc-ordering.

Remark. All chords of the graphLi, with respect to the spanning treeTi , i = 1,2,correspond to redundantC -edges inH . The treesT1 and T2 provide the existence of aunique(x,y)-invertor for anyx,y∈X. Theorem 10.7.7 ensures, at any step of the algorithm,the existence of a vertex, sayx, pendant in bothHD and one ofT1 or T2. Notice that notevery elimination of pendant vertices inHD generates a uc-ordering inH .

An example of a uc mixed hypertreeH and respective touching graphsL1 andL2 isshown in Figure 10.21. GraphL1 is a tree, soT1 = L1. GraphL2 is not a tree; it meansthat H has a redundantC -edge. Any of threeC -edges{x1,x3,x4},{x4,x3,x5},{x2,x3,x5}is redundant. In graphL2, choose edge{x2,x5} to be a chord of spanning treeT2. At thebeginning, vertexx1 is pendant in bothHD andL1. After deleting ofx1, graphL1 containsthe only vertexx3. However,x3 is not pendant inHD . Therefore, Algorithm 10.7.1 switchesspanning treeT1 to T2. Now vertexx2 is pendant in bothHD andL2, so it is deleted next,and so on. Thus the algorithm decomposesH in the orderingx1,x2,x4,x3, andx5. Hencethe uc-ordering is:

σ = (x5,x3,x4,x2,x1).


From Theorem 10.7.7 and Algorithm 10.7.1 we have

Corollary 10.7.9 A reduced mixed hypertree is uniquely colorable if and only if it is uc-orderable.

Therefore, combining Theorems 10.7.3, 10.7.9 and the relation between the chromaticpolynomial and chromatic spectrum, we obtain the following:

Corollary 10.7.10 Let H = (X,C ,D) be a reduced mixed hypertree. Then the followingfive statements are equivalent:

(1) R(H ) = (0,1,0, . . . ,0);(2) P(H ,λ) = λ(λ−1);(3) H is uniquely colorable;(4) Every two vertices x,y∈ X are joined by an(x,y)-invertor;(5) H is uc-orderable.

Exercises 10.7.

b b

bb

H1

b b b b

b

bb

bH2

Figure 10.22.

1. Show that mixed hypergraphH1 in Figure 10.22 is not uc.

2. Embed mixed hypergraphH1 (Figure 10.22) into a uc-mixed hypergraph.

3. For mixed hypergraphH2 in Figure 10.22, find a proper coloring and construct touch-ing graphs for every color.

4. Determine if mixed hypergraphs in Figure 10.22 are mixed hypertrees.

5. Apply Algorithm 10.7.1 to mixed hypergraphH2 in Figure 10.22, to determine ifH2

is uniquely colorable. If yes, find a uc-ordering.

6. For mixed hypergraphH2 in Figure 10.22, verify the conditions of Corollary 10.7.10.



1. Given a mixed hypertreeH , apply Algorithm 10.7.1 to determine ifH is uniquelycolorable.

10.8. Perfection

In graph coloring theory, perfect graphs (see Section 5.7.)provide an important theoreticand algorithmic topic of research. In the language of mixed hypergraphs, aD-graphG iscalled perfect if, for every one of its induced subgraphsG′ (including G itself), the lowerchromatic number equals the size of its largest clique, i.e.χ(G′) = ω(G′).

The notion of graph perfection is difficult to extend to general D-hypergraphs because,from the point of view of colorings, there is no natural and simple analogue of completegraphs (cliques). For example, theD-hypergraphK r

n , r ≥ 3, being a hypergraph general-ization of the cliqueKn and considered as a mixed hypergraphH = (X, /0,

(Xr

)

), has manyproper colorings and its properties are not so nice as those of the cliqueH = (X, /0,

(X2

)

).In contrast, we find a natural notion of perfection of an arbitrary mixed hypergraph with

respect to the upper chromatic number. Hence, when talking about graph perfection wemean perfection with respect to the lower chromatic number,while talking abouthyper-graph perfection we mean perfection with respect to the upper chromatic number.

Recall that by definition in a mixed hypergraphH = (X,C ,D) a setS⊆ X is said to beC -stable (C -independent) if it contains noC -edgeC∈ C .

Definition 10.8.1 The cardinality of a maximumC -stable set inH is called theC -stability(C -independence) numberα

C(H ).

It follows from this definition that a maximumC -stable set is the largest set that couldpossibly be polychromatic over all proper colorings. One can compare it with the notion of amaximumD-stable set which in turn is the largest set that could possibly be monochromaticover all proper colorings.

Proposition 10.8.1 For every mixed hypergraphH = (X,C ,D),

χ(H ) ≤ αC(H ). (10.4)

Proof. Consider a strictχ-coloring ofH and choose one vertex from each color class toform a setS. SinceS is a polychromatic set, it does not contain anyC -edge. It means thatS is aC -stable set. The cardinality of a maximumC -stable set cannot be smaller; hence theinequality follows. �

Mixed hypergraphs withχ(H ) = αC(H ) may be constructed easily, and it is now seen

thatαC(H ) plays a role forχ(H ) analogous to that played by the maximum clique number

ω for the chromatic numberχ in a classic graphG. Namely, it is opposite, in some sense,to the inequalityχ(G) ≥ ω(G), see Section 5.7. The significant difference, however, is thatthe latter holds only for graphs, i.e. 2-uniformD-hypergraphs, while the inequality (10.4)holds for any mixed hypergraph.


b

b b

b

b

b

b 1

2 3

4

5

2k+4

2k+5

Figure 10.23.

We know from Mycielski’s construction (Section 5.7.) that there are graphs (even with-out triangles) such that the differenceχ(G)−ω(G) is arbitrarily large. Something similarcan be stated with respect to the upper chromatic number and theC -stability number, evenfor C -hypergraphs.

Theorem 10.8.1For any k≥ 0 and χ ≥ k+ 1, there exists a 3-uniformC -hypergraphH

such thatα

C(H )− χ(H ) > k.

Proof. Let H = (X,C ) be a 3-uniformC -hypergraph withX = {1,2, . . . ,2k+ 5}, andC1 = {1,2,3},C2 = {1,4,5}, . . . ,Ck+2 = {1,2k+4,2k+5} such that

Ci ∩Cj = {1}, i, j ∈ I = {1,2, . . .k+2}, i 6= j.

In other words,H is a 3-uniform monostar having vertex 1 as the unique center,seeFigure 10.23. Vertex 1 represents the minimum transversal,therefore all other verticesform a stable set. Henceα

C(H ) = 2k+4. Further, since no two edges have two vertices in

common, in every proper coloring we must repeat one color in each edge. Since the numberof edges isk+2, the number of colors in any proper coloring is at most 2k+5− (k+2) =k+3, i.e. χ(H ) = k+3. Hence,α

C(H )− χ(H ) = 2k+4− (k+3) = k+1 > k. �

Compare Figure 10.23 with Figure 5.23: they both show that the lower and upper chro-matic numbers can be arbitrarily far from their respective bounds.

Definition 10.8.2 A colorable mixed hypergraphH = (X,C ,D) is called perfect if forevery induced subhypergraphH ′ (includingH itself) the following equality holds:

χ(H ′) = αC(H ′). (10.5)

Example 10.8.1Every colorable mixed hypergraphH = (X,(X

r

)

,D), r ≥ 2, is perfect. In-deed,χ(HY) = r−1= α

C(HY) for each|Y| ≥ r, andχ(HY) = |Y|= α

C(HY) for each|Y|< r.


A mixed hypergraphH = (X,C ,D) with C 6= /0 is called aC -bistar if there are twovertices (called the center) common to allC -edges.

Theorem 10.8.2EveryC -bistar H = (X,C ,D) is a perfect mixed hypergraph.

Proof. Obviously, we can color the center with one color and the remaining vertices withall different colors, soχ(H ) = n−1= αC (H ). Consider anyY ⊆ X. If HY contains at leastoneC -edge, thenχ(HY) = |Y|−1 = αC (HY). Otherwise,χ(HY) = |Y| = αC (HY). �

Minimal non-perfect mixed hypergraphs. Recall that we denote byτ(HC ) thetransversal number ofHC , i.e., the cardinality of the smallest subset of vertices that containsat least one vertex from everyC -edge of a mixed hypergraphH = (X,C ,D). Let τ2(HC )denote thebitransversal number of HC which is the cardinality of the smallest subsetof vertices that contains at least two vertices from everyC -edge of a mixed hypergraphH = (X,C ,D).

A C -monostar is a mixed hypergraph which has exactly one vertex in common withall of its C -edges. That vertex is called thecenter. Therefore, ifH = (X,C ,D) is a C -monostar, thenτ(HC ) = 1, andτ2(HC )≥ 3. However, ifH = (X,C ,D) is aC -bistar, then ithas two vertices in common with all theC -edges, what impliesτ(HC ) = 1, andτ2(HC ) = 2.The classes ofC -monostars andC -bistars have an empty intersection.

Theorem 10.8.3A C -monostarH = (X,C ,D) is not a perfect mixed hypergraph.

Proof. It follows from the definition ofC -monostar thatαC(H ) = |X|−1. Since also by

definition H contains at least twoC -edges, in any proper coloring either we have threevertices including the center colored with the same color, or we have one color repeatedin oneC -edge and another color repeated in anotherC -edge. In both cases the maximumnumber of colors is not greater than|X|−2. This implies thatχ(H ) ≤ |X|−2, so χ(H ) 6=α

C(H ). �

Every subhypergraph of aC -monostarH which itself is aC -monostar (i.e., it containsat least twoC -edges ofH ) is not a perfect mixed hypergraph. It means that the minimalnot perfect subhypergraph that becomes perfect after strong deletion of a vertex representsjust twoC -edges sharing precisely one vertex.

Definition 10.8.3 An r-uniformC -hypergraphH = (X,C ), |X| = n≥ 3, r ≥ 2, is calleda cycloid and denoted by Crn if X = {0,1, . . . ,n−1} andC = {{i, i +1(mod n), . . . , i + r −1(mod n)} : i = 0,1, . . . ,n−1}.

In other words, one can say that for a cycloid there exists a host graphCn = (X,E)representing a simple cycle without chords, such thatC coincides with the family of allpaths of lengthr − 1 onCn. Thus, the usual cycleCn (considered as aC -hypergraph) isC2

n for anyn≥ 3. Note that the example given for Algorithm 10.4.1, see Figure10.9, is thecycloidC3

5 considered as the mixed hypergraph(X,C35, /0).

Theorem 10.8.4A cycloid Crn = (X,C ), 3≤ r ≤ n, is perfect if and only if2r ≥ n+2.


Proof.Case 1: 2r ≥ n+ 2. Since for r = n the theorem is evident, letr ≤ n− 1. Hence,

|Ci ∩Cj | ≥ 2, i, j ∈ I . Let x1 be a neighbor ofx2, x3 be a neighbor ofx4, and the pair{x1,x2}be opposite to the pair{x3,x4} on the host cycle ofCr

n. The inequality 2r ≥ n+ 2 impliesthat for anyCi ∈ C either |Ci ∩{x1,x2}| ≥ 2 or |Ci ∩{x3,x4}| ≥ 2. Thus,Cr

n is the unionof two C -bistars with disjoint centers. Fromr ≤ n−1, we conclude that there are no twovertices belonging to allC -edges, soχ(Cr

n) < n−1. On the other hand, we can colorx1,x2

with the first color,x3,x4 with the second color and all the remainingn−4 vertices with alldifferent colors. Therefore,χ(Cr

n) = n− 2. SinceCrn is not aC -monostar, and verticesx1

andx3 form a minimum transversal,τ(Crn) = 2. It implies α(Cr

n) = n−2 = χ(Crn). At last,

every induced subhypergraph ofCrn is aC -bistar and, by Theorem 10.8.2, is perfect. Hence,

Crn is perfect.

Case 2: 2r = n+ 1. Observe thatCrn does not contain anyC -monostar as an induced

subhypergraph. Forr = 3,4, it can be verified directly thatCrn is notC -perfect. Hence, let

r ≥ 5. Sinceτ(Crn) = 2, it follows thatα(Cr

n) = n−2. We show thatχ(Crn) < n−2. For a

contradiction, assume thatχ(Crn) = n−2. Consider a strict coloring usingχ colors. There

are two possibilities: either one pair of vertices is colored with one color and another pairof vertices is colored with another color and all othern−4 vertices colored differently, or,there are three vertices colored with one color and all the remainingn−3 vertices coloreddifferently. We consider this as the following two subcases.

Subcase 1:pair x1,x2 is colored with one color and pairx3,x4 is colored with anothercolor. ThenCr

n is the union of twoC -bistars with centers that do not intersect. Assumethat x1,x2 ∈ X represent a center of the firstC -bistar, andx3,x4 a center of the secondC -bistar, and, moreover,x1,x2,x3,x4 are placed onCr

n clockwise in this order. Letni j be thenumber of vertices betweenxi andx j , i, j = 1,2,3,4. If n12 = 0 andn34 = 0, then aC -edgeC ∈ C may be found easily such that|C∩{x1,x2}| ≤ 1 and|C∩{x3,x4}| ≤ 1; so, assumethatn12+n34 ≥ 1. We haven12+n23+n34+n41+4= n = 2r −1. Since anyC -edge mustcontain either{x1,x2} or {x3,x4}, it follows that

n12+n23+n34+2 < r,

n12+n14+n34+2 < r.

By summing the above two inequalities, we have 2r −1+ n12 + n34 < 2r which gives thecontradictionr < r.

If the verticesx1,x2,x3,x4 are placed onCrn in any other order, then aC -edgeC ∈ C

may easily be found such that|C∩{x1,x2}| ≤ 1 and|C∩{x3,x4}| ≤ 1. Consequently,Cr2r−1

cannot be the union of twoC -bistars with disjoint centers for anyr ≥ 3.Subcase 2:verticesx1,x2, andx3 are colored with the same color and placed onCr

nclockwise in this order. Thenx1,x2,x3 is a minimumC -bitransversal. Letni j be the numberof vertices betweenxi andx j , i, j = 1,2,3. We haven12+n23+n31+3= n = 2r −1. SinceanyC -edge must contain two vertices amongx1,x2 andx3, it follows that

n12+1+n23 < r,

n23+1+n31 < r,


n31+1+n12 < r.

Summing these inequalities impliesr < 5, a contradiction. Consequently,Cr2r−1 cannot be

colored withn−2 colors in such a way that three vertices have the same color and all theother vertices have different colors. This holds for anyr ≥ 3 and Subcase 2 is proved.

Case 3: 2r ≤ n. In this case, allC -edges containing a fixed vertex form an inducedsubhypergraph which is aC -monostar, and hence,Cr

n is not perfect. �

Next we provide a description of one more example of minimal non perfect C -hypergraph found by Kral’. Letr ≥ 3 be a fixed integer. We define aC -hypergraphH r = (X,C ) in the following way. Let

X = {1,2,3, . . . ,2r},

and the vertices be drawn in the plane in cyclic clockwise order. The edge family has 2r +2edges,

C = {C1,C2, . . . ,C2r ,Co,Ce}

such thatC1 = {1,2,4,6,8, . . . ,2r −2}.

ThusC1 contains consecutive vertices 1 and 2, skips vertex 3 and contains all (except 2r)even vertices thereafter,r vertices in total. Further, edgeC2 is obtained when shifting(=rotating)C1 by 1 clockwise around the cycle, edgeC3 is obtained when shiftingC2 by 1in the same direction around the cycle, and so on,. . . , edgeC2r is obtained when shiftingC2r−1 by 1 around the cycle. In other words, edgesC2,C3, . . . ,C2r are all copies ofC1

obtained by shifting around the cycle. At last, edgeCo contains all odd vertices, and edgeCe contains all even vertices. Clearly,H r is anr-uniform (r +1)-regularC -hypergraph on2r vertices.

For r = 3 the example is shown in Figure 10.24. To avoid confusion, edgesC2, . . . ,C6

are not shown, and edgesCo andCe are shown separately.

Theorem 10.8.5 (Kral’, 2003) TheC -hypergraphH r contains no cycloid on2r − 1 ver-tices and no monostar as induced subhypergraphs; moreover

χ(H r) = 2r −4 < 2r −3 = αC (H r).

Perfection of mixed hypertrees. A hypergraphH is calledbi-Helly if for any sub-family of edges the following implication holds: if every two edges of the subfamily haveintersection of cardinality at least two, then the whole subfamily has intersection of cardi-nality at least two.

A mixed hypergraphH = (X,C ,D) is called aneclipse if the intersection of allC -edges induces a completeD-graph. Evidently, any eclipse has at least oneC -edge; if it hasexactly oneC -edge, then it is uncolorable. AnyC -monostar is a special case of an eclipse.Every C -bistar where each center is aD-edge of size 2 is an eclipse. An example of aC -bistar and an eclipse which is obtained from thatC -bistar is shown in Figure 10.25.


b b

bb

b b

1 2

36

45

shift

b b

bb

b b

C1

Co Ce

Figure 10.24. Kral’s construction.

Theorem 10.8.6Every eclipseH is not a perfect mixed hypergraph.

Proof. Indeed, since the intersection of allC -edges is not empty,τ(HC ) = 1, henceα(HC ) = n− 1. If H contains only oneC -edge, then it is uncolorable andχ(H ) =0. If H has more than oneC -edges, then, in any proper coloring, the intersectionof all C -edges must be polychromatic. This implies that no twoC -edges may repeatthe same color, therefore no coloring withn− 1 colors exist. In all casesχ(H ) 6=n−1. �

Clearly, if an eclipse is a hypertree, or any subhypergraph of a hypertree, then it is eithera C -monostar or aC -bistar havingD-graphK2 as the unique center.

Theorem 10.8.7 If a mixed hypertreeH = (X,C ,D) contains no eclipse then it is perfect.

Proof. Let H = (X,C ,D) be a colorable mixed hypertree without eclipses. Since everyinduced subhypergraph of a mixed hypertree is also a mixed hypertree, it is sufficient toshow that we can colorH with αC (H ) colors.

By Theorem 8.1.2 the hypertreeHC = (X,C , /0) satisfies the Helly property; i.e., for anysubfamilyC1 ⊆ C with C∩C′ 6= /0 for each pairC,C′ ∈ C1, it follows that

⋂

C∈C1

C 6= /0.

In fact, for every such subfamily|⋂

C∈C1| ≥ 2 because otherwiseC1 forms an eclipse. Thus

HC is a bi-Helly hypergraph, and moreover,C∩C′ 6= /0 implies|C∩C′| ≥ 2 for anyC,C′ ∈C .Consider the problem of finding a minimum transversal ofHC . For this, construct the

line graph ofHC , that is the graphL(HC ) = (C ,E), where(C,C′) ∈ E ⇔C∩C′ 6= /0 in H .Because of the Helly property, minimum covering ofL by cliques corresponds to a mini-mum transversal ofHC . Let K1,K2, . . . ,Kt be the cliques of such a minimum covering ofL.Thusτ(HC ) = t andαC (H ) = |X|− t.


b b

b

b

b b

b b

b

b

b b

C -bistar eclipse

Figure 10.25.C -bistar and eclipse.

Because of the bi-Helly property, every clique of the graphL forms aC -bistar as a partialsubhypergraph inH with at least one center (C -bitransversal). Since by the condition ofthe theoremH contains no eclipses, every suchC -bistar has at least one center that is not aD-edge.

Let the pair{xi ,yi} be theC -bitransversal corresponding to the cliqueKi, i = 1,2, . . . , tsuch that{xi ,yi} 6∈D. They all are different becauseKi 6= K j implies that{xi ,yi}∩{x j ,y j}=/0. Color the verticesx1,y1 with the first color,x2,y2 with the second color,. . . ,xt ,yt withthe tth color; after that, color all the remaining vertices each with a different color fromt + 1, t + 2, . . . , |X| − t. Thus, we obtain a proper coloring ofH with |X| − t = αC (H )colors, and the theorem follows. �

Corollary 10.8.1 If H = (X,C ,D) is a perfect mixed hypertree,ν(HC ) andτ(HC ) are themaximum cardinality of a matching and the minimum cardinality of a transversal ofHC

respectively, then

χ(H ) = αC (H ) = |X|− τ(HC ) = |X|−ν(HC ).

Proof. SinceHC is also a hypertree, it fulfills the well known Konig property, see Theo-rem 8.2.3, saying that

ν(HC ) = τ(HC ).

Hence, the assertion follows. �

Remark. The presence or absence of monostars inC -hypergraphs does not deter-mine perfection directly. Consider, for example, theC -hypergraphH = (X,C ) whereX = {1,2,3,4,5}, C = {C1,C2,C3}, C1 = {1,2,3}, C2 = {2,3,4,5}, C3 = {1,4,5}. Wesee thatC1∩C3 = {1} and we have a monostar as a partial subhypergraph. However,H isperfect withχ(H ) = αC (H ) = 3. On the other hand, the cycloidC3

5 contains monostars as


partial subhypergraphs, does not contain them as induced subhypergraphs and, at the sametime, is not perfect (Theorem 10.8.4).

We end the section with explicit description of six known examples of minimal nonperfect 3-uniformC -hypergraphs:

V1 = ({1,2,3,4},{{1,2,3},{1,3,4},{1,2,4}}) (monostar);V2 = ({1,2,3,4,5},{{1,2,3},{1,4,5}})(monostar);V3 = ({1,2,3,4,5},{{1,2,3},{1,3,4},{1,4,5}}) (monostar);V4 = ({1,2,3,4,5},{{1,2,3},{1,3,4},{1,4,5}, {1,2,5}}) (monostar);V5 = ({1,2,3,4,5},{{1,2,3},{2,3,4},{3,4,5}, {4,5,1},{5, 1, 2}}) (cycloid C 3

5 );K1 = ({1,2,3,4,5,6},{{1,2,4},{2,3,5},{3,4,6}, {4,5,1}, {5,6, 2}, {6,1,3},

{1,3,5},{2,4,6}})(Kral’s example).As one can see, the perfection of hypergraphs is much more complex than the perfection

of graphs discussed in Section 5.7.

Exercises 10.8.

b

b b

b

H1

b

b b

b

b

H2

Figure 10.26.

1. For each of mixed hypergraphsH1 andH2 in Figure 10.26, findC -stability numberαC(H ) and upper chromatic numberχ(H ). Verify inequality (10.4).

2. Which of the mixed hypergraphsH1 andH2 in Figure 10.26 is perfect and which isnot, and why?

3. For each of mixed hypergraphsH1 andH2 in Figure 10.26, find bi-transversal numberτ2(H1) andτ2(H2).


1. Given aC -hypergraphH , recognize ifH is a monostar, a cycloid, or the Kral’shypergraph.

2. Given a mixed hypergraphH , recognize ifH is aC -bistar, or an eclipse.

3. Given a mixed hypergraphH , find all monostars.


10.9. Chromatic Spectrum

Every stricti-coloring of a mixed hypergraphH induces a feasible partition of the vertexset intoi nonempty color classes. By definition, the numbers of all feasible partitions intoi color classes, 1≤ i ≤ n, form the chromatic spectrum. The positive components of thechromatic spectrum begin with the number of feasible partitions into the lower chromaticnumber of classes and end with the number of feasible partitions into the upper chromaticnumber of classes. It is easy to see that for anyC -hypergraph or for anyD-hypergraphthis segment of the chromatic spectrum contains no gaps, i.e. no zeroes between positivenumbers. However, the chromatic spectra of mixed hypergraphs may have gaps. Thismeans that if we know the lower and upper chromatic numbers, we don’t know if thereare colorings using any intermediate number of colors. In a more formal language theseproperties are explained as follows.

The set of valuesk such thatH has a strictk-coloring is called thefeasible setof H ,denoted byS(H ). In other words,S(H ) is the set of indicesk such thatrk(H ) > 0.

Definition 10.9.1 A mixed hypergraphH = (X,C ,D) has agap at k if S(H ) containselements larger and smaller than k, but omits k. The chromatic spectrum R(H ) is calledcontinuous (gap-free)if S(H ) has no gaps. Otherwise, it is calledbroken.

Proposition 10.9.1 For any mixed hypergraphH = (X,C ,D), both HC = (X,C ) andHD = (X,D) have continuous chromatic spectra.

Proof. Indeed, forHC , start with any strictχ-coloring and sequentially as long as possibleunite any two color classes; we end when there is only one color class. Since at each step aproper coloring is obtained, the chromatic spectrum ofHC is gap-free.

For HD , start with any strictχ-coloring and sequentially as long as possible split anycolor class having at least two vertices; we end when there are |X| color classes of cardi-nality 1 each. At every step a proper coloring is obtained, hence the chromatic spectrum ofHD is continuous. �

Notice that the last part of the statement above is a direct generalization of Theorem5.2.1 about the chromatic spectrum of graphs.

The simplest example. One of the simplest ways to construct a mixed hypergraphwith a gap in the chromatic spectrum is the following. First,define an operation called the“inflation” of a D-edge. Let us have aD-edge{a,b}, see Figure 10.27. We double thevertices, i.e. replace verticesa,b with the pairsa,a′ andb,b′ respectively. Thus,D-edge{a,b} becomes a quadruple, i.e.D-edge{a,a′,b,b′}.

Next, we add all triples on verticesa,a′,b,b′ as C -edges. IfH denotes the mixedhypergraph obtained, then simplyH = (X,K3

4,K44) whereX = {a,a′,b,b′}. The inflation

consists in the replacing ofD-edge{a,b} by H . One can easily see that

R(H ) = (0,7,0,0)

and all seven feasible partitions are induced by the following seven strict 2-colorings (inordera,a′,b,b′):

1122,1112,1121,1222,2122,1212,1221.


b

b

b

b

b

ba

b

a a′

b b′

inflation

Figure 10.27. Inflation of edge{a,b}.

Observe that among all seven colorings there is only one coloring, namely, the first one,when vertexa has the same color asa′ (color 1), and vertexb has the same color asb′ (color2). This fact is the key to the gap in the chromatic spectrum.

The final step in constructing the example is the following: take a completeD-graphK4 on verticesa,b,c,d and by doubling the vertices inflate all of its sixD-edges, see Figure10.28. To avoid any confusion, theC -edges are not shown. Denote the mixed hypergraphobtained byH .

Consider now the proper colorings ofH . It has four pairs of vertices, namely,{a,a′},{b,b′}, {c,c′}, and{d,d′}. In any proper coloring ofH , if there is a pair, say{a,a′},colored with different colors, say, 1 and 2, then each of the remaining vertices is coloredwith one of the colors 1 or 2. This occurs because any new color, say 3, at any of the vertices,say c, immediately results in a polychromatic triple{a,a′,c} which is aC -edge. So, itremains to consider the proper colorings when each pair is monochromatic. Surprisingly,there is only one coloring of such type (if we do not count the permutations of colors):

11223344.

Indeed, no two pairs of vertices may have the same color because each “pair of pairs” isa D-edge, see Figure 10.28. The fundamental conclusion at thispoint is thatH has severalstrict 2-colorings (r2 ≥ 2), a strict 4-coloring (r4 = 1) and no strict 3-coloring (r3 = 0), i.e.there is a gap in the chromatic spectrum. Since by the reasoning above there are no strict5-,6-,7- and 8-colorings, the chromatic spectrum has then the following form:

R(H ) = (0, r2,0,1,0,0,0,0).


b

b b

b

b

b b

b

a

a′ b′

b

c

c′d

d′

H

Figure 10.28. Inflation ofK4.

The mixed hypergraphH is not the minimal one. One can observe that the gap maybe obtained by starting with the sameD-graphK4 and doubling only verticesc andd (i.e.inflating D-edge{c,d}) and adding triples{a,c,c′}, {a,d,d′}, {b,c,c′}, {b,d,d′} asC -edges. Thus these triples become bi-edges. In addition, recall that inC -hypergraphK3

4 anyedge is redundant, see Figure 10.12; it means that oneC -edge can be removed from theinflation of edge{c,d}, and we still have the gap. Later on this smallest example will bedenoted byH2,4 where index meansS(H ) = {2,4}.

Developing this idea further, the construction of minimal examples of mixed hyper-graphs with the gaps in their chromatic spectrum in a more general setting is considerednext.

The smallest mixed hypergraphs with gaps.Let s, t be two integer numbers suchthat 2≤ s≤ t − 2. In this subsection we construct a mixed hypergraphHs,t with feasiblesetS(Hs,t) = {s, t} and prove thatHs,t has the fewest vertices among alls-colorable mixedhypergraphs with a gap att −1; this minimum number of vertices is 2t −s.

We begin with an explicit construction of a mixed hypergraphwith 2t −2 vertices andfeasible set{2, t}. As usual,Kn is viewed as the mixed hypergraph(X, /0,

(X2

)

); trivially,S(Kn) = {n}. First, we describe the construction informally. Beginning with Kt , expandt−2 of the vertices into pairs, leaving two special vertices unexpanded. TheD-edge consistingof the two special vertices remains, and the otherD-edges expand intoD-edges of size 3 or4 (special vertex plus pair, or union of two pairs). Add, asC -edges, all triples consisting ofthree vertices arising from two original vertices (specialvertex plus pair, or three verticesfrom two pairs).


This describes the construction completely, but we presentit more formally to facilitatethe proofs. The smallest instance is fort = 4; from K4, a 6-vertex mixed hypergraph withspectrum{2,4} is produced. Let[m] = {1, . . . ,m}.

Construction 1. Define a hypergraphH2,t with vertex set

{x1,x2,a1, . . . ,at−2,b1, . . . ,bt−2}.

Let T be the set of triples of the formxraibi , for r ∈ {1,2} andi ∈ [t −2]. LetU be the setof quadruples of the formaibia jb j for i, j ∈ [t −2]. Let W be the union, overi, j ∈ [t −2],of the sets of four triples contained in{ai ,bi ,a j ,b j}. TheC -edges inH2,t areT ∪W. TheD-edges areT ∪U ∪{x1x2}.

Lemma 10.9.1 S(H2,t) = {2, t}.

Proof. Let c be an arbitrary coloring ofH2,t . If c(ai) 6= c(bi) for somei ∈ [t −2], then theC -edges inT andW that containai andbi force all other vertices to have the same color asai or bi . Thus, in this case there are at most two colors. The existence of D-edges preventsa proper 1-coloring, and setting allc(ai) = c(x1) = 1 andc(bi) = c(x2) = 2 completes aproper 2-coloring.

Hence, we may assume thatc(ai) = c(bi) for all i ∈ [t − 2]. Now theD-edges inUforce these colors to be distinct for alli, and theD-edges inT along with{x1x2} requireadditional colors forx1 andx2. This completely forces the coloring, which usest colors andis proper. �

In order to extend this construction to the lower chromatic numbers, we use a simplelemma about combining feasible sets. Thejoin of two mixed hypergraphs(X1,C1,D1)and (X2,C2,D2) with disjoint vertex sets is the mixed hypergraph(X,C ,D) defined byX = X1∪X2, C = C1∪C2, andD = D1∪D2∪R, whereR is the set of all pairs consistingof one vertex fromX1 and one fromX2.

Lemma 10.9.2 If H1 andH2 are mixed hypergraphs, then the feasible set of the join ofH1

andH2 is {i + j: i ∈ S(H1), j ∈ S(H2)}.

Proof. The D-edges added between the vertex sets ofH1 and H2 prohibit colors fromappearing in both sets. Thus, every proper coloring of the join consists of proper coloringsof H1 andH2 using disjoint sets of colors. �

Call the joinH ′ of an arbitrary mixed hypergraphH = (X,C ,D) with one new vertexz 6∈ X an elementary shifting of the chromatic spectrum ofH . Evidently, elementaryshifting results in the chromatic spectrum being shifted tothe right by one position for allpositive components and adding one zero in the very left position. That is, if R(H ) =(r1, r2, . . . , rn), thenR(H ′) = (0, r1, r2, . . . , rn). Joining with the complete graphKi resultsin i elementary shiftings.

Theorem 10.9.1 If H is an s-colorable mixed hypergraph with a gap at t− 1, then n≥2t −s, and this is sharp.


Proof. Consider a proper coloring ofH usingk colors, wherek is the smallest element ofthe feasible set larger thant −1. If n < 2t − s, then using at leastt colors requires havingat leasts+ 1 color classes of size 1. Two such color classes can be combined to obtain aproper coloring usingk−1 colors unless they form aD-edge of size 2. Sincek−1 is notin the feasible set,H containsKs+1. Now H is not s-colorable; the contradiction yieldsn≥ 2t −s.

For s= 2, Lemma 10.9.1 shows that Construction 10.9. achieves the bound. Fors> 2,defineHs,t to be the join ofKs−2 andH2,t−s+2. By Lemma 10.9.2, the feasible set ofHs,t is{s−2}+{2, t −s+2} = {s, t}. The number of vertices inHs,t is s−2+2(t −s+2)−2 =2t −s. �

Corollary 10.9.1 The minimum number of vertices in a mixed hypergraph with a gap in itsfeasible set is 6, achieved byH2,4.

Proof. Every mixed hypergraph with a gap in its feasible set iss-colorable with a gap att −1, for somes, t with t −1 > s≥ 2 (notice thats≥ 2, otherwise there are no gaps). Thust ≥ 4 andt −s≥ 2. By Theorem 10.9.1,n≥ t +(t −s) ≥ 6. �

As we mentioned earlier, closer analysis allows oneC -edge in the 6-vertex example tobe dropped without changing the spectrum. Thus, 7C -edges and 6D-edges suffice. Themixed hypergraphH2,4 has the following structure:X = {1,2,3,4,5,6}, C = {{1,2,3},{1,4,5},{6,2,3}, {6,4,5},{2,3,4}, {3,4,5},{2,4,5}}, D = {{1,6}, {1,2,3}, {1,4,5},{6,2,3}, {6,4,5}, {2,3, 4,5}}. Subsets{1,2,3}, {1,4,5},{6,2,3}, {6,4,5} are the bi-edges. In Figure 10.29, the bi-edges are drawn by doubled curves. One can easily checkthat the four strict 2-colorings in order of vertices 1,2,3,4,5,6 are 112122, 112212, 212121,212211, the unique strict 4-coloring is 122334, hence the chromatic spectrum

R(H2,4) = (0,4,0,1,0,0),

and the chromatic polynomial

P(H ,λ) = λ(λ−1)(λ2−5λ+10).

H2,4 is the minimal mixed hypergraph with a gap in the following sense: any deletionof a vertex or an edge of any type eliminates the gap. As Corollary 10.9.1 states, there areno gaps if a mixed hypergraph has less than six vertices. However, we will see that evenH2,4 is not the smallest one if the total number of edges is considered.

Feasible sets and doubling-shifting algorithm.The gaps in the chromatic spectrumraise the question regarding which sets of positive integers are feasible sets of mixed hy-pergraphs. In this subsection we characterize all feasiblesets and suggest a conceptualalgorithm for constructing a mixed hypergraph realizing a given feasible set.

The n-vertex trivial mixed hypergraph(X, /0, /0) has, evidently, feasible set{1, . . . ,n}. We construct mixed hypergraphs realizing all other feasible sets using the trivialmixed hypergraphs, the join operation of Lemma 10.9.2, and one additional operation. Thisoperation is similar to the construction ofH2,t from Kt. In Construction 10.9., expandingtwo of the vertices was avoided in order to create few vertices. Here the construction ishuge, so we prefer the simplicity gained by expanding all vertices into pairs.


b b

b

bb

b

6 1

2

34

5

Figure 10.29. The minimal mixed hypergraph with a gapH2,4.

Construction 2. Let H = (X,C ,D) be a mixed hypergraph. We construct a mixedhypergraphH ′ = (X′,C ′,D ′) with X′ =

⋃

v∈X{v−,v+}. For each edgeD ∈ D, we addD′ =

⋃

v∈D{v−,v+} to D ′. For each edgeC ∈ C , we addC′ = {v− : v∈C} to C ′. Finally,for each ordered pairu,v∈ X, we add the triples{v−,v+,u−} and{v−,v+,u+} to C ′.

The application of Construction 2 may be calleddoubling. It has the effect of appendingelement 2 to the feasible set. This is what Construction 1 didto Kt, and the analysis heregeneralizes Lemma 10.9.1.

Lemma 10.9.3 Let H be a mixed hypergraph with feasible set S. Ifχ(H ) ≥ 2, then themixed hypergraphH ′ obtained fromH via Construction 2 has feasible set S∪{2}.

Proof. Let c be a proper coloring ofH ′. If c(v−) 6= c(v+) for a vertexv of H , then theC -edges that are triples containingv−, v+ force all other vertices to have colorc(v−) orc(v+). Thus, such a coloring uses exactly two colors. We obtain a strict 2-coloring bysettingc(u−) = c(v−) andc(u+) = c(v+) for all u. Since each member ofD ′ consists offull pairs, the constraints onD-edges are satisfied. Also, each member ofC ′ contains twovertices with superscripts of the same type.

It remains to consider proper colorings withc(v−) = c(v+) for each vertexv of H . Letc be the coloring ofH defined by ˜c(v) = c(v−). For each member ofD ′, the coloringconstraint is satisfied byc if and only if c satisfies the constraint for the correspondingmember ofD. The same statement holds for members ofC ′ that arise from members ofC . By construction, the new triples inC ′ are automatically satisfied. Thus,c is a propercoloring ofH ′ if and only if c is a proper coloring ofH . Note that ˜c uses the same numberof colors asc.


Similarly, we can extend each proper coloring ofH to a proper coloring ofH ′ usingthe same number of colors, by copying the color of each vertexonto both of its copies. Thisimplies that every integer greater than 2 is feasible forH if and only if it is feasible forH ′.

�

Corollary 10.9.2 LetH andH ′ be as above, and

R(H ) = (0, . . . ,0, rχ, . . . , rχ,0, . . . ,0).

ThenR(H ′) = (0, r2(H

′),0. . . . ,0, rχ, . . . , rχ,0, . . . ,0).

Proof. Indeed, the proof of Lemma 10.9.3 establishes a bijection between strict coloringsof H with at least three colors and strict colorings ofH ′ with at least three colors. Thenumber of 0s on the right part of the chromatic spectrum ofR(H ′) is greater byn(H ) sincen(H ′) = 2n(H ). �

Using shiftings (joins with cliques) and doublings, one canproduce all feasible sets.

Theorem 10.9.2 (Jiang et al., 2002)A finite set of positive integers is the feasible set for amixed hypergraph if and only if it omits the number 1 or is an interval containing 1.

Proof. It remains only to consider the sets not containing 1. We produce a mixed hyper-graphH (T) with feasible setT. We use induction on the size of setT, and within eachsize, we use induction on the smallest elementt of T. ForT = {t}, we setH (T) = Kt .

For |T| > 1 andt = 2, we letH (T) be the mixed hypergraph obtained by applyingConstruction 2 toH (T −{2}). Lemma 10.9.3 implies that this works.

For |T| > 1 andt > 2, we letH (T) be the join ofKt−2 with the mixed hypergraphH (T ′), whereT ′ is obtained fromT by subtractingt−2 from each element. Lemma 10.9.2implies that this works. �

Next we present a conceptual algorithm which, for an arbitrary increasing sequence ofinteger numbers, constructs a mixed hypergraph having the respective feasible set.

Algorithm 10.9.1 (doubling-shifting algorithm)

INPUT: A set S= {n1,n2, . . . ,np} of increasing integer numbers, n1 ≥ 2.OUTPUT: A mixed hypergraphH with the feasible set S.Initialization: i = np−1; H = K3.Iteration: while i 6= 1, do:

1. If i ∈ S, do doubling.

2. If i 6∈ S, do elementary shifting.

3. i = i −1.

End.


b b b b

b b b

1 2 3 4

5 6 7

Figure 10.30. The smallest 3-uniform bihypergraph with a gap.

Uniform bihypergraphs. All examples discussed so far represented mixed hyper-graphs withC -edges andD-edges of different sizes, i.e. not uniform. As next theoremshows the uniform bihypergraphs may also have gaps in their chromatic spectra.

Theorem 10.9.3 (L. Gionfriddo, Voloshin, 2002)The minimum number of vertices overall 3-uniform bihypergraphs with a gap in the chromatic spectrum is 7; the minimum num-ber of edges over all 3-uniform bihypergraphs on 7 vertices with a gap in the chromaticspectrum is 9.

Proof. We first prove that there is no 3-uniform bihypergraph on 6 vertices with a brokenchromatic spectrum.

Let H = (X,E) be a colorable 3-uniform bihypergraph withX = {1,2, . . . , 6}, withall the elements ofE being triples (bi-edges of size 3), andR(H ) = (r1, r2, . . . , r6). For acontradiction, suppose thatR(H ) is broken.

Evidently,r1 = 0, r6 = 0. If r5 6= 0, then in a strict 5-coloring ofH , all bi-edges contain afixed monochromatic pair of vertices, say 1,2, and all the other color classes are singletons.These color classes can easily be combined, sor4 6= 0, r3 6= 0, r2 6= 0, a contradiction.Hencer5 = 0. If r4 6= 0, then in a strict 4-coloring ofH , there are at least two singletoncolor classes which impliesr3 6= 0, a contradiction. Hence, there are no gaps inR(H ).

We now show that there exists a 3-uniform bihypergraph on 7 vertices with a brokenchromatic spectrum. LetH = (X,E) be a 3-uniform bihypergraph defined as follows (seeFigure 10.30, bi-edges are drawn as classic hyperedges):X = {1,2, . . . ,7}, E = {{1,2,3},{1,2,5}, {1,2,6}, {1,2,7}, {3,4,2}, {3,4,5}, {3,4,6}, {3,4,7}, {5,6,7}}. In everyproper coloring, there are two possibilities for vertices 1and 2: eitherc(1) 6= c(2) or c(1) =c(2). Let {A,B,C,D, . . .} be the set of colors. Ifc(1) = A, c(2) = B, then all the remaining


vertices are colored with the colors from{A,B}; i.e., all the colorings are 2-colorings. Ifc(1) = c(2) = A, thenc(3) = c(4) = B, and consequently,c(5),c(6),c(7) ∈ {C,D}; i.e., wehave a strict 4-coloring. Therefore, no strict 3-coloring exists; thus,S(H ) = {2,4} (closeranalysis shows thatR(H ) = (0,12,0,3,0,0,0),P(H ,λ) = 3λ(λ−1)(λ2−5λ+10)).

Finally, we prove that 9 is the minimum number of edges for a 3-uniform bihypergraphwith a gap in the chromatic spectrum over all 3-uniform bihypergraphs on 7 vertices. LetH = (X,E) be a 3-uniform bihypergraph withX = {1,2, . . . ,7} and |E | ≤ 8. For a con-tradiction, suppose thatR(H ) = (r1, r2, . . . , r7) has at least one gap. We immediately haver1 = r7 = 0. If r6 6= 0 (respectivelyr6 = 0, r5 6= 0), then at least 3 color classes are singletonswhich can easily be combined to obtain a strict coloring with3, 4 and 5 (respectively 3 and4) colors, soR(H ) contains no gaps, a contradiction.

Therefore, supposer7 = r6 = r5 = 0, r4 6= 0. In a strict 4-coloring, if the number ofsingleton color classes≥ 2, then r3 6= 0, a contradiction. Hence, the only type of a 4-coloring to examine is

(A,A,B,B,C,C,D).

If {1,2,7} 6∈ E (respectively{3,4,7} 6∈ E , {5,6,7} 6∈ E), then there exists a 3-coloring(A,A,B,B,C,C,A) (respectively(A,A,B,B, C,C,B), (A,A,B,B, C, C,C)) and thusr3 6= 0,a contradiction. So, we have that in any case

{1,2,7},{3,4,7},{5,6,7} ∈ E .

Observe further that the remaining, at most, 5 triples ofH belong to the following family:

123 341 561124 342 562125 345 563126 346 564.

Since it consists of 3 columns, there is one column which contains, at most, one tripleof H . Choose the first column and let{1,2,3} 6∈ E , {1,2,4} 6∈ E ,{1,2,5} 6∈ E . If{3,4,1},{3,4,2} 6∈ E , then the following 3-coloring exists: (A,A,A,A,B,B,C), a contra-diction. Therefore{3,4,1} ∈ E , or {3,4,2} ∈ E . If {3,4,5}, {3,4,6} 6∈ E then thefollowing 3-coloring exists: (A,A,A,B,C,C,B), a contradiction. Hence{3,4,5} ∈ E or{3,4,6} ∈E . If {5,6,1} 6∈E (respectively,{5,6,2} 6∈E), then the following 3-coloring ex-ists: (B,A,C,C,B,B,A) (respectively (A,B,C,C,B,B,A)), acontradiction. Therefore{5,6,1},{5,6,2} ∈ E . If {1,2,6} ∈ D then all 8 bi-edges ofH are exhausted and the 3-coloring(A,A,B,B,A,C,C) completes the proof.

Suppose{1,2,6} 6∈ E . If {3,4,1} 6∈ E (respectively{3,4,2} 6∈ E), then the follow-ing 3-coloring exists: (A,B,A,A,C,C,B) (respectively (B,A,A,A,C,C,B)), a contradiction.Therefore both{3,4,1}, {3,4,2} ∈ E . Hence all 8 bi-edges ofH are exhausted, and the3-coloring (A,B,C,C,A,B,B) completes the proof. �

We conclude the section with the following important results:

Theorem 10.9.4 (Kral’ et al., 2000) If H is a mixed hypertree, then the chromatic spec-trum is continuous.


Theorem 10.9.5 (Kral’ et al., 2003) If H is a mixed hypergraph with each vertex of degree2, then the chromatic spectrum is continuous.

If we take dual to the mixed hypergraph from the theorem above, then the dual is amultigraph with two types of vertices:C -vertices andD-vertices. A proper coloring ofHbecomes a coloring of the edges of the respective multigraphin such a way that eachC -vertex has two edges of the same color and eachD-vertex has two edges of different colors.Let us call such a multigraph amixed multigraph . Thus the theorem above leads to thefollowing important conclusion:

Corollary 10.9.3 (Kral’ et al., 2003) The chromatic spectrum in edge coloring of anymixed multigraph is continuous.

Notice that in such colorings there are lower and upper chromatic indexes; the specialcase gives the formula for the upper chromatic indexχ′(G) = c+ m− n+ p, see Part I,Section 5.9.

Exercises 10.9.

b

b b

b

b

bb

b

H

Figure 10.31.

1. For mixed hypergraphH in Figure 10.31, find feasible setsS(HC ) andS(HD).

2. For mixed hypergraphH in Figure 10.31, find feasible setS(H ) and prove that thechromatic spectrum is gap-free.

3. Given set of integersS= {2,3,5}, apply Algorithm 10.9.1 to construct a mixed hy-pergraphH with S(H ) = S.


1. Given a mixed hypergraphH , by generating colorings at random, find an estimate onfeasible setS(H ).

2. Implement Algorithm 10.9.1.


10.10. Coloring Planar Hypergraphs

Let H = (X,C ,D) be a mixed hypergraph. Denote the underlying family of edgesof H byE = C ∪D. Let us agree that if a subset of vertices is a bi-edge, then itappears inE onlyonce. Recall that hypergraphH ′ = (X,E) is the underlying hypergraph ofH . As we knowfrom Section 9.4., a hypergraphH ′ is planar if and only if bipartite representationB(H ′) isa planar graph.

Definition 10.10.1 A mixed hypergraphH = (X,C ,D) is calledplanar if the underlyinghypergraphH ′ = (X,E) is planar.

This can be viewed as follows: we can embedH ′ in the plane and label all hyperedgeswith B, C or D appropriately, according to whether they are bi-edges,C -edges orD-edges.Note that in the plane,C -edges of size 2 can be contracted as described in the splitting-contraction Algorithm 10.5.1, and bi-edges of size 2 lead touncolorability; so, in general,it suffices to only consider reduced mixed hypergraphs.

A first discussion on the coloring of planar hypergraphs can be found in a paper byZykov [8]. The main results discussed there may be reformulated in the language of mixedhypergraphs as follows.

Theorem 10.10.1 (Bulitco [8]) The four color theorems for planarD-graphs and for pla-nar D-hypergraphs are equivalent.

Theorem 10.10.2 (Burshtein, Kostochka [8])If a planar D-hypergraph contains at mostoneD-edge of size 2, thenχ(H ) ≤ 2.

There are uncolorable planar mixed hypergraphs. A planar embedding of the smallestnon-trivial (reduced) uncolorable planar mixed hypergraph H = (X,C ,D)= (X,

(X3

)

,(X

2

)

)is shown in Figure 10.32:X = {1,2,3},C = {C} = {{1,2,3}}, D = {D1,D2,D3} ={{1,2},{2,3}, {1,3}}, the four faces aref1, f2, f3, f4. It is not difficult to extend thisexample to an infinite family of uncolorable planar mixed hypergraphs.

The general structure of uncolorable planar mixed hypergraphs is unknown. However,if a planar mixed hypergraph is colorable, then naturally the problem of determining thelower and upper chromatic number arises. Finding the lower chromatic number is difficultsince if we allowD-edges of cardinality 2, it contains the four color problem as a specialcase, see Section 5.6. As to the upper chromatic number, the simplest interesting case isthat of 3-uniformC -hypergraphs. This case could be viewed as an analogue to thefourcolor problem in the sense that we consider the maximum number of colors rather than theminimum. The situation, however, is different than the fourcolor problem since the upperchromatic number of a 3-uniformC -hypergraph must depend on the number of verticesn.

Therefore, we next consider an important simple case of planar mixed hypergraphs,namely, maximal 3-uniform planar bihypergraphs.

Since every face of a maximal planar hypergraph is of size 2, we can associate a graphG(H ) on the same vertex set withH : replace every face inH by an edge inG, so that everyedge inH becomes a face ofG. H is maximal 3-uniform, soG must be a triangulation inthe usual sense, see Section 4.5. We useH andG interchangeably, and since every edge ofH is a bi-edge, we will refer to them asbitriangulations .


b

b b

C

D1

D2

D3

f1

f2

f3

f4

1

23

Figure 10.32. The smallest reduced uncolorable planar mixed hypergraph.

We now study the colorings of bitriangulations; we want to color the vertices of trian-gulationG so that every face has exactly two vertices of the same color.

Let us define that a coloringc1 is arefinementof a coloringc2 if every color class ofc1

is contained in a color class ofc2. If for every color class of the coloringc2 we construct atouching graph (see Section 10.7., Definition 10.7.4), thenthe connected components of thetouching graph represent the possibility of all further refinements ofc2. We say that a colorclass of a coloringc is non-partitionable if it is a color class in every refinement ofc; i.e.,its touching graph is connected. In other words, there is no way to split the vertices of thecolor class and obtain another proper coloring. If every color class ofc is non-partitionable,thenc is said to be amaximal coloring. Notice that the number of colors used in a maximalcoloring is not necessarilyχ.

Lemma 10.10.1A color class in a coloring of a bitriangulation is non-partitionable if andonly if it induces a connected subgraph.

Proof. In a triangulation two vertices are together in a face if and only if they are adja-cent. Therefore, a non-partitionable color class must induce a connected subgraph, sinceotherwise we can simply re-color one of the components. Conversely, a connected colorclass can not be refined further, since re-coloring some of its vertices results in two adjacentvertices from the old color class receiving distinct new colors. Since all faces are of size 3,this leads to a polychromatic face. �

The duality of planar graphs in the following theorems refers to the classic planar duality“vertices - faces” (see Section 4.5.). In fact, starting with a maximal planar 3-uniform mixedbihypergraphH (a bitriangulation), we consider its corresponding “hyperedges - faces”dual which is the graphG (also called a bitriangulation), and then we proceed to the dual“vertices - faces” graphG∗. In this way, a proper coloring of the original maximal planar3-uniform bihypergraphH becomes a coloring of the vertices of the graphG in such a waythat each face has two vertices of the same color, which, in turn, becomes a coloring of thefaces of the graphG∗ in such a way that each vertex belongs to two faces of the same color.


Theorem 10.10.3 (Kundgen et al., 2002)There is a 1-1 correspondence between the k-colorings of a bitriangulation G and the k-face-colorings of the 2-factors in the dual G∗.In this correspondence, a coloring c1 of G is a refinement of a coloring c2 if and onlyif the corresponding 2-factors are identical and the face-coloring associated with c1 is arefinement of the face-coloring associated with c2.

Proof. The main idea of the proof is accredited to Penaud (1975), whoessentially showedthat there is a 1-1 correspondence between 2-colorings ofG and 2-factors ofG∗ (see Corol-lary 10.10.2).

A 2-factor ofG∗ is simply a collection of closed Jordan curves; it partitions the planeinto regions, inducing a partition ofV(G) into non-empty sets. Thus, every proper face-coloring of this 2-factor withk colors corresponds to ak-coloring ofV(G). Such a coloringis in fact ak-coloring of the bitriangulationG, since it follows from the face-coloring beingproper that every face ofG is colored with precisely two colors.

Conversely, given ak-coloring, we can recover the 2-factor and its face-coloring. Sincein every face ofG there are exactly two vertices of the same color, we get a 2-regular span-ning subgraph, i.e. a 2-factor ofG∗, by taking the dual edge of every edge inG that isincident to vertices of different colors. Now, if two vertices are in the same region (gen-erated by the 2-factor), then there is a curve connecting them, that passes only throughvertices in this region. But then consecutive vertices on this curve must be on the sameface and therefore adjacent. The edge joining these vertices can not be the dual of an edgein the 2-factor, since otherwise it would follow from the Jordan Curve Theorem 4.1.1 thatthey are in different regions. By the definition of the 2-factor, it thus follows that consec-utive vertices on this curve must be of the same color, and that therefore every vertex in agiven region has the same color. Since every region of the 2-factor must contain at least onevertex, we can therefore uniquely define the coloring of the regions, and thisk-coloring isa proper coloring, since faces are separated by dual edges and thus adjacent faces containadjacent vertices of different colors.

For the second part of the proof, observe that a refinement of the face-coloring of thedual graph clearly leads to a refinement of the coloring of thebitriangulation. For theconverse, suppose thatc1 is a refinement ofc2. Following the construction of the dual2-factor, it follows that the 2-factor forc1 must contain the 2-factor forc2, from which itfollows that they are identical. Finally, the face-coloring corresponding toc1 must be arefinement of the coloring corresponding toc2. �

As in Section 5.4., letS(n,k) denote the Stirling numbers of the second kind, i.e. thenumber of ways of partitioning a set ofn elements into exactlyk sets. Also definefk(G∗) tobe thenumber of 2-factorsof G∗ that consist of exactlyk components (i.e,k vertex disjointcycles), and letf (G∗) = ∑i≥1 fi(G∗) be the total number of 2-factors ofG∗.

Corollary 10.10.1 Every coloring of a bitriangulation G can be refined to a unique maxi-mal coloring and there are exactly fk−1(G∗) maximal k-colorings of G.

Proof. By the Jordan Curve Theorem 4.1.1, a given 2-factor consisting of k−1 disjointcycles divides the plane intok regions and, by Lemma 10.10.1, the coloring that assigns adifferent color to each face must be the unique maximal coloring for this 2-factor, since (as


shown in the proof above) the vertices in every region inducea connected subgraph. Thesecond statement follows immediately. All refinements of a given coloring correspond tothe same 2-factor, so that the first statement also follows. �

Corollary 10.10.2 Every bitriangulation G has exactly f(G∗) strict 2-colorings. In gen-eral, the components of the chromatic spectrum R(G) are defined by

rk(G) = ∑i≥1

S(i,k−1) fi(G∗), 1≤ k≤ n(G),

and the chromatic polynomial is given by

P(H ,λ) = ∑i≥1

fi(G∗)λ(λ−1)i .

Proof. The first statement follows from both summation formulas, bysettingk = 2 orλ = 2 respectively. For the first formula, it suffices, by Theorem10.10.3, to show that every2-factor consisting ofi cycles can bek-face-colored in exactlyS(i,k−1) ways. To see this,create a graph whose vertices are the faces in the dual of the 2-factor, and two verticesare adjacent if and only if the corresponding faces are separated by a 2-factor. This graphis connected and hasi edges. By the Jordan curve theorem, it has exactlyi + 1 verticesand must therefore form a treeT. Let rk(T) be the number of properk-colorings ofT.To see thatrk(T) = S(e(T),k−1), observe thatr1(K1) = 1 andrk(K1) = 0 for k ≥ 2. Byremoving a pendant vertexx, we can see thatrk(T) = (k−1)rk(T − x)+ rk−1(T − x), theusual recursion for the Stirling numbers, as shown by Theorem 5.4.3 and Corollary 5.4.3.For the second formula, recall that the chromatic polynomial for a tree oni + 1 vertices isλ(λ−1)i (Theorem 5.5.1). �

Corollary 10.10.3 The chromatic spectrum of every bitriangulation G is continuous,χ(G) = 2 and χ(G) = 1+max{k : fk(G∗) ≥ 1}.

Proof. SinceG∗ is a 3-regular bridgeless graph it follows from Petersen’s theorem (seee.g. [7, p.124]) that it has a 2-factor. So, by Corollary 10.10.2, every bitriangulation is2-colorable, and therefore must have lower chromatic number 2. A coloring achievingthe upper chromatic number must be maximal, so that the valueof χ(G) follows fromCorollary 10.10.1. Ifk = χ(G), then fk−1(G∗) ≥ 1; so, sinceS(k−1, i −1) ≥ 1 for every2≤ i ≤ k, we get thatr i(G)≥ 1 in this range and that the chromatic spectrum is continuous.Furthermore, ani-coloring can be obtained from ani-coloring of the tree. �

Corollary 10.10.4 Every planar mixed hypergraph without edges of size 2 is 2-colorable.

Proof. We may assume that the mixed hypergraph is a maximal bihypergraph, sinceaddingC - or D-edges only decreases the number of 2-colorings. Similarly, if G containsany faces of size larger than 3, then they can be divided into faces of size 3 by adding graphedges to obtain a bitriangulation. The result now follows from Corollary 10.10.3. �

Corollary 10.10.5 Every uniquely colorable planar mixed hypergraph must havean edgeof size 2.


Proof. Suppose thatG is uniquely colorable and free of edges of size 2. Again, wemay assume thatG is a bitriangulation. By Corollary 10.10.3,χ(G) = 2; then, by Corol-lary 10.10.2,G∗ has a unique 2-factor that must be a Hamiltonian cycle. But this contradictsTheorem 10.10.4 below. �

Theorem 10.10.4 (Thomason, 1978, Tutte, 1946, “Smith’s Theorem”)The number of Hamiltonian cycles containing a given edge of acubic graph is even.

Proof. We sketch the elegant proof of Thomason. Letuv be the given edge. Considerthe graph whose vertices are the Hamiltonian paths startingat u with edgeuv. Two suchpaths are adjacent if one can be obtained from the other by adding an edge at the end of thepath and deleting a different edge. Now, vertices of degree 1in this graph correspond toHamiltonian cycles containinguv, and all other vertices have degree 2. Thus, the numberof Hamiltonian cycles containinguv is even. �

As with the class of uncolorable planar mixed hypergraphs, there are infinitely manyuniquely colorable planar mixed hypergraphs. One can easily see this, for example, byconstructing a planar embedding of an(x,y)-invertor of arbitrary length (see Section 10.7.)which represents a uniquely colorable mixed hypergraph.

Theorem 10.10.3 and the corollaries above can be illustrated by the following example.

Example 10.10.1Let H be a maximal planar 3-uniform bihypergraph, i.e. a bi-triangulation, such that

H = (X,

(

X3

)

,

(

X3

)

)

whereX = {1,2,3,4},C = D = {{1,2,3},{2,3,4},{3,4,1},{4,1,2}};

the drawing ofH and its “hyperedges - faces” dual which isK4 is shown in Section 9.4.,Figure 9.6.

The drawing ofK4 in Figure 10.33 is the continuation of Figure 9.6. In the planeembedding ofK4 there are four faces denoted bya,b,c, andd. The dual “vertices - faces”is drawn by dotted curves and then redrawn asK∗

4. GraphK∗4 has three 2-factors which all

have one component (i.e., are hamiltonian) and shown next. Edges “not participating” in2-factors are drawn by dashed curves. In the first 2-factor, faces 1 and 4 are outside thecycle and colored with colorA; faces 2 and 3 are inside the cycle and colored with colorB.In the second 2-factor, faces 1 and 3 are inside the cycle and colored with colorA; faces 2and 4 are outside the cycle and colored with colorB. At last, in the third 2-factor, faces 1and 2 are inside the cycle and colored with colorA; faces 3 and 4 are outside the cycle andcolored with colorB.

This leads to the conclusion that there are only three strictcolorings of the verticesof K4 andH , and all they are the following (in order of vertices 1,2,3,4): ABBA, ABAB,andAABB. Thus,χ(H ) = χ(H ) = 2, the chromatic spectrumR(H ) = (0,3,0,0), and thechromatic polynomialP(H ,λ) = 3λ(λ−1).


b b

bb

1 2

34

a

b

c

d

b b

bb

a b

cdK4 K∗

4

b b

bb

12

34

b b

bb

b b

bb

12

34

12

34

12

34

2-factors

A A

B

B

A B

B

A

A B

A

B

1st coloring:

ABBA

2nd coloring:

ABAB

3rd coloring:

AABB

⇒ R(H ) = (0,3,0,0), P(H ,λ) = 3λ(λ−1)

χ(H ) = χ(H ) = 2

Figure 10.33. Continuation of Figure 9.6.

Gap in the chromatic spectrum. Planar mixed hypergraphs with a gap in their chro-matic spectrum were first constructed by Kobler and Kundgen:

Lemma 10.10.2 (Kobler, Kundgen, 2001)Let H ′2,4 = (X,C ,D) be the mixed hypergraph

with X = {1,2, . . . ,6}, C = {C1, . . . ,C4} = {{1,2,3}, {2,3,4}, {2,4,5}, {4,5,6}}, andD = {D1, . . . ,D6} = {{1,2},{1,5},{1,6},{2,4},{3,6},{4,6}}. H ′

2,4 is planar and has


feasible set S(H ′2,4) = {2,4}.

Proof. Figure 10.34 shows an embedding ofH ′2,4 in the plane. TheD-edges are drawn

as line segments and a curve, theC -edges as regions having size 3, and the faces are notindicated. Letc be a strict coloring ofH ′

2,4. If c(2) 6= c(3), then the remaining colors areforced and we have the 2-coloring{1,3,4}∪{2,5,6}. If c(2) = c(3), thenc(4) = c(5) andc(2) 6= c(4), which results in the strict 4-coloring{1}∪{2,3}∪{4,5}∪{6}. One can seethat these feasible partitions are unique. �

Consequently, the chromatic spectrum

R(H ′2,4) = (0,1,0,1,0,0),

and the chromatic polynomial

P(H ′2,4,λ) = λ(λ−1)(λ2−5λ+7)

(compare toH2,4 in Figure 10.29 and to the bihypergraph in Figure 10.30).

b b b b

b

b

1 2

3

4

5

6D1

D2

D3

D4 D6

D5

C1 C2

C3 C4

Figure 10.34. Planar mixed hypergraph with a gap,H ′2,4.

Theorem 10.10.5 (Kobler, Kundgen, 2001)A non-empty set of positive integers S is thefeasible set of some planar mixed hypergraph if and only if S is an interval{s,s+1, . . . , t}with 1≤ s≤ 4 or of the form{2,4,5, . . . , t}.

Proof. Let G = (X,D) be at-vertex planarD-graph. The feasible setS(G) = {χ(G),χ(G) + 1, . . . , t}, whereχ(G) ≤ 4. This shows the sufficiency of the condition whenS is


an interval. WhenShas a gap at 3, consider the mixed hypergraph obtained fromH ′2,4 by

taking the vertices{7, . . . , t +2} and placing them in the region containing vertices 2 and 3.Then include theC -edges{{2,3,7},{2,3,8}, . . . ,{2,3, t +2}}. If we have a 2-coloring onH ′

2,4, then this only extends to a 2-coloring of the larger graph, whereas from the 4-coloring,we obtain all other values in the feasible set.

It remains to prove the necessity of the condition. So, consider a planar mixed hyper-graphH , with feasible setS 6= /0. If 1 ∈ S, thenS trivially forms an interval. Letc be astrict t-coloring of H , wheret is the largest value inS. We will construct a planar mixedhypergraphH ′ with {4,5, . . . , t} ⊂ S(H ′) ⊂ S. So, by the choice oft, it follows thatS is ofthe required form.

H ′ will have the same vertex set asH . We will keep every edge of size 2, and sinceH

is colorable this must be either aC -edge or aD-edge. Now consider all edges containing≥ 3 vertices. If the corresponding region is aC -edge, then it contains verticesu,v withc(u) = c(v). We replace the region by aC -edge{u,v}. If the region is aD-edge, then itcontains verticesu,v with c(u) 6= c(v). We replace the region by aD-edge{u,v}. If theregion is a bi-edge, then it contains verticesu,v,w with c(u) = c(v) 6= c(w). We replacethe region by aC -edge{u,v} and aD-edge{v,w}. The mixed hypergraphH ′ we obtainis planar and still hasc as a strictt-coloring. Furthermore, in obtainingH ′ from H nocoloring constraints are lost, so that every coloring ofH ′ is a coloring ofH .

We now have a planar mixed hypergraph with all edges (C - or D-) of size 2. Contractevery C -edge and obtain a loopless planarD-graphG. There is a 1-1 correspondencebetween strict colorings ofG and strict colorings ofH ′. Since the feasible setS(G) ={χ(G), χ(G)+1, . . . , t}, we see that{4,5, . . . , t} ⊂ S(H ′). �

The importance of the theorem above is that the gaps in the chromatic spectrum ofplanar mixed hypergraphs may occur only at 3.

Exercises 10.10.b

b

bb

b

G

Figure 10.35.

1. Let H = (X,C ,D) be a bi-triangulation such that its “hyperedges - faces” dual isgraphG in Figure 10.35. Construct vertices-faces dual graphG∗, find all 2-factors ofG∗, and the chromatic spectrumR(H ) and the chromatic polynomialP(H ,λ).

2. Determine if mixed hypergraphH ′2,4 in Figure 10.34 is perfect.



1. Given a mixed hypergraphH , determine ifH is planar.

2. Given a planar mixed hypergraphH , determine if its chromatic spectrum is broken.

Chapter 11

Modeling with Hypergraphs

11.1. List Colorings without Lists

For cellular telephones, the frequencies are assigned by zones. Every zone is assigned a listof frequencies that can be used in the zone. If two zones interfere, they cannot use the samefrequency at any time. Suppose we have three zones in a region, sayZ1,Z2 andZ3. ZoneZ1 must use frequenciesa or c, zoneZ2 must use frequenciesa or b, and zoneZ3 must usefrequenciesb or c. ZonesZ1 andZ2 interfere, zonesZ2 andZ3 interfere, and zonesZ1 andZ3 do not interfere.

In how many ways and in which ways can we assign frequencies tothe zones? Constructa graphG with three verticesZ1, Z2 andZ3, see Figure 11.1; two vertices are adjacent ifthe respective zones interfere. The list of admissible colors {a,c} is assigned to vertexZ1,the list of admissible colors{a,b} is assigned to vertexZ2, and the list of admissible colors{b,c} is assigned to vertexZ3. Now the original problem can be formulated as follows: inhow many ways and in which ways can we color the vertices of graph G such that adjacentvertices have different colors and each vertex is colored with the color from its list? Suchgraph colorings are called thelist colorings. We show that in turn, the list colorings can bemodeled by the mixed hypergraph colorings without any lists.

Construct a mixed hypergraphH = (X,C ,D) in the following way, see Figure 11.1.Since we have three colorsa,b,c in total, add three new verticesa,b, andc and putX ={Z1,Z2,Z3,a,b,c}. SetD to include all edges ofG plus all edges forming complete graphon verticesa,b andc. At last, setC = {{Z1,a,c},{Z2,a,b}, {Z3,b,c}}.

One can easily see that every list coloring ofG corresponds to a proper coloring of themixed hypergraphH , and vice versa. Indeed, since the firstC -edge{Z1,a,c} must have atleast two vertices of the same color, and vertices corresponding to colors form a completegraph, in any proper coloring ofH , vertexZ1 will be colored either with a color of vertexaor with a color of vertexc. The similar is true for allC -edges.

As a consequence, graphG admits at least one list coloring if and only if mixed hy-pergraphH is colorable. Clearly, the names of colors (=the values of frequencies) do notmatter; the things that matter are the structure of graphG (=interferencies between thezones in the region), the lists of colors with assignments, and the total number of colors(frequencies).

For our simple example in Figure 11.1, one can manually find that χ(H ) = χ(H ) =


b

b

b

Z1

Z2

Z3

{a,c}

{a,b}

{b,c}G

b

b

bb b

b

Z1

Z2

Z3

a b

c

H = (X,C ,D)

Figure 11.1. List colorings without lists.

3, chromatic spectrumR(H ) = (0,0,3,0, 0,0), chromatic polynomialP(H ,λ) = 3λ(λ−1)(λ− 2), and moreover, the three list colorings ofG in order(Z1,Z2,Z3) are as follows:(a,b,c),(c,b,c),(c,a,c).

11.2. Resource Allocation

Consider the following example. Let us haven = 4 elementary jobsX = {x1,x2,x3,x4} which are to be executed by allocatingm = 5 available resourcesY = {y1,y2,y3,y4,y5}. Suppose that time is discrete and each of these elementary jobs can be executedduring one unit of time. A set of resourcesS(xi) ⊆ Y must be available for jobxi ∈ Xto be performed. For execution, jobx1 requires resourceS1 = {y1}, job x2 requires re-sourcesS2 = {y1,y2,y3}, job x3 requires resourcesS3 = {y3,y4}, job x4 requires resourcesS4 = {y2,y4,y5}. For any resource used by at least two jobs simultaneously, a penalty mustbe paid.

Management requirements are: jobsx1,x2,x3 must be performed for at most two unitsof time, and jobsx1,x2,x4 must also be performed for a maximum two units of time without

Modeling with Hypergraphs 265

b

b

b

b

b

b

b

b

b

jobs resourcesx1

x2

x3

x4

y1

y2

y3

y4

y5

b

b

b b

x1

x2

x3 x4

G

H = (X,C ,D)

Figure 11.2. Bipartite graphG and mixed hypergraphH for resource allocation.

penalty. Is there a resource allocation for all the jobs without any penalty?Construct a bipartite graph “jobs – resources”G, see Figure 11.2, and then construct

a mixed hypergraphH as follows. SinceS1 ∩S2 6= /0, construct the firstD-edgeD1 ={x1,x2}. SinceS2∩S3 6= /0, construct the secondD-edgeD2 = {x2,x3}. SinceS2∩S4 6= /0,construct the thirdD-edgeD3 = {x2,x4}. SinceS3∩S4 6= /0, construct the fourthD-edgeD4 = {x3,x4}. SinceS1∩S3 = /0 andS1∩S4 = /0, no otherD-edge can be added toH .

Management requirements give us the following twoC -edges:C1 = {x1,x2,x3}, C2 ={x1,x2,x4}. Put C = {C1,C2}, D = {D1,D2,D3,D4}. Thus, the mixed hypergraphH =(X,C ,D) is obtained.

One can easily see thatH is uncolorable. This means that no allocation of resources,without any penalty is possible. However, if we violate, forexample, the constraint ex-pressed by theC -edgeC1, then we can color the verticesx1,x2,x3,x4 respectively with thecolors 1,2,3,1. Therefore, in this case, the minimum numberof time units is 3, and themaximum number of time units is also 3, for all the jobs to be performed. In any case, onepenalty must be paid. If we are interested in the performancewithout any penalty, then onemight ignore, for example, jobx3. This gives the minimum and maximum time of 2 units


for the remaining jobs to be executed, etc.In general case, every proper coloring ofH represents an allocation of resources with-

out penalty and vice versa; i.e., the resource allocation problem is an instance of a col-orability problem. Respectively, the minimum and maximum time for executing all thejobs without penalty correspond to the lower and upper chromatic numbers. The maxi-mization of the number of jobs or the minimization of penalties represent the respectiveoptimization problems on coloring the mixed hypergraphH . Also, the information aboutthe chromatic spectrumR(H ) and all the strict colorings will answer the question if thereexists an allocation such that each resource is used continuously. Notice that constraints ex-pressed byC -edges may generally increase the minimum duration in resource allocation. Inaddition,the possibility of gaps in the chromatic spectrum gives no guarantee that resourceallocation is possible for each of the intermediate values betweenχ(H ) and χ(H ).

In computer science applications, the jobs could be, for example, the computationalproblems, and the resources could be the processors in parallel computations. Or, the jobscould be the queries, and the resources could be the files in a data base, and so on.

Chapter 12

Appendix

“What did you like most during the study at the university?- Great school, great professors...What did you hate most?- Mathematical induction...”

12.1. What Is Mathematical Induction

One of the most popular methods of proof in graph theory is theproof by mathematicalinduction. The idea is simple. Suppose we want to prove a statementPn depending onnwheren = 1,2,3, . . . . This is equivalent to prove an infinite list of statementsP1,P2, . . . tobe true. Instead of proving the infinite list of theorems, we prove just the following two:

1. ProveP1.

2. For anyk≥ 1, prove that ifPk is true, thenPk+1 is true.

It is easy to see that this is sufficient to conclude thatPn is true for alln = 1,2, . . ., or,equivalently, all statementsP1,P2, . . . are true.

Step 1. is called thebasis of induction, see Fig. 12.1. Step 2. is called theinductionstep. It can be abbreviated asPk →Pk+1 and called animplication. In step 2., the statementPk is called theinduction hypothesis.In notationPn, n is called theinduction parameter.

Mathematical induction is a bright example how generalizations work in mathematics:instead of provingP1 → P2, thenP2 → P3, and so on (up to infinity!), one just prove thegeneral casePk → Pk+1. In this way, the induction step replaces infinitely many proofs.To have a complete proof, we only need an initial condition which is the induction basis.Evidently, any positive integer greater than 1 may serve as the induction basis.

Summarizing these arguments, one can describe the following sequence of steps inwriting the proofs by mathematical induction in Graph Theory:

1. Write down the statementPn, i. e. the theorem to be proved.

2. Induction basis: write down and prove the statementPn0, where propertyPn0 isdirectly observed; usually,n is the number of vertices, andn0 = 3, orn0 = 4.


0 1 2 3 k k+1 n

P1 P2 P3 Pk Pk+1 Pn

Induction basis Induction hypothesis Induction step

→

1. P1

2. Pk → Pk+1

Pn is true for alln = 1,2, . . .

...

...

...

...

Figure 12.1. Mathematical induction.

3. Write down the statementPk.

4. Write down the statementPk+1.

5. Induction step: prove that ifPk is true, thenPk+1 is true. In the proof, the assumptionthatPk is true,must be used. To provePk+1, one consider a graph onk+1 vertices.Then apply to it an operation such as removing a vertex or contraction an edge to ob-tain a graph havingk vertices. At this point it is very important to assure that propertyPk (for a graph onk vertices) holds. Finally, considering the inverse operation oneprove that propertyPk+1 (for a graph onk+1 vertices) fulfills.

6. Conclude thatPn is true for alln≥ n0.

0 1 2 3 k k+1 n

P1 P2 P3 Pk Pk+1 Pn

Induction basis Induction hypothesis Induction step

1. P1

2. P1,P2, ...,Pk → Pk+1

Pn is true for alln = 1,2, ...

...

...

...

...

Figure 12.2. Strong mathematical induction.

There is another form of this method which is called thestrong mathematical induc-tion. The difference is in the induction step. Namely, instead ofproving Pk → Pk+1, the

Appendix 269

strong mathematical induction requires the following: prove that ifP1 is true, and ifP2 istrue, and ifP3 is true, and so on, and ifPk is true, thenPk+1 is true. This can be abbreviatedasP1,P2, . . . ,Pk → Pk+1. Implicitly it is clear that if we started atP1 and arrived toPk, thenall intermediate statementsP2,P3, . . . ,Pk−1 must be true. In some complicated cases weneed all these values to provePk+1, and that is exactly the motivation of applying the strongmathematical induction. The idea is depicted in Figure 12.2. As in simple mathematicalinduction, evidently, any positive integer greater than 1 may serve as the induction basis.

In Graph Theory proofs, sometimes indexk is omitted or hidden; the assumption is “letP hold for all graphs on< n vertices” and then one proveP for a graph onn vertices.

12.2. Graph Theory Algorithms and Their Complexity

Generally, analgorithm is a finite set of precise instructions for solving a problem.Ingraph theory, algorithms consist of sequences of numbered steps (with possible repetitionsand checking logical conditions) describing what to do to solve a problem for a graphor hypergraph. In all cases, a graph or a hypergraph is in the input, and a number or aspecial subgraph in many cases is in the output. An algorithmcan be programmed and theprogram can be run on the computer. If we try the same program for different graphs, thenwe find that time for computations depends on the number of vertices n, i.e., it is somefunction f (n). If we have another algorithm for the same problem, then the running timeis another function ofn, say,g(n). How to compare the algorithms? The first algorithm isbetter if f (n) ≤ g(n) beginning with some fixed number of vertices, sayn≥ N1. If the firstalgorithm is run on a computer which isc1 times faster, and the second algorithm is run ona computer which isc2 times faster, then the first is better whenf (n)/c1 ≤ g(n)/c2 holdsfor all n≥ N2 whereN2 is some other natural number. This is equivalent tof (n) ≤Cg(n)for some constantC = c1/c2. Evidently, it holds for any other constantC′ ≥ C and anyother N ≥ N2. In other words, no matter how fast the new computers would be, if theinequality f (n) ≤Cg(n) holds for some sufficiently large constantC and alln≥ N, then thefirst algorithm is better. This reasoning explains the meaning of the following notation incomparison of the complexity of algorithms.

We say thatf (n) is O(g(n)) (read “f (n) is big-oh ofg(n)”, sometimes denoted byf (n)= O(g(n))) if there are constantC and a numberN such that

f (n) ≤Cg(n)

for all n≥N. The basic idea of this definition is thattime is the measure of the complexityof algorithms, and that time should not depend on the speed ofa computer. The comparisonof algorithms isasymptotical, i.e., what occurs beginning with someN on, or, as we say,whenn, the number of vertices, approaches infinity.

In practice, when estimating algorithms, one compute the number of elementary oper-ations (addition, multiplication, comparison, etc.) as a function ofn in the worst caseatevery step of the algorithm. That is an upper bound for the complexity. One then say thatthe complexity of the algorithm isO(g(n)) whereg(n) is that very same upper bound.

If g(n) = n, then the algorithm is calledlinear-time and its complexity is denoted byO(n). Generally, ifg(n) is a polynomial of degreek on n, then the algorithm is called


polynomial-time and complexity is denoted byO(nk). If g(n) is an exponential function,then the algorithm is calledexponential-time and complexity is denoted byO(an) wherea > 1. There is also aconstant complexity denoted byO(1), logarithmic complexitydenoted byO(logn) and evenfactorial complexity denoted byO(n!).

For example, let a graphG be given by its adjacency matrixA(G) of sizen×n and theproblem is to find the vertex degrees. Since the matrix is symmetric, it is sufficient to checkevery entry of the lower triangle if it is 0 or 1. The number of such checks is(n2 − n)/2which is the polynomial ofn. Therefore the complexity of such procedure isO(n2).

However, many problems in graph theory are much more complex. The best knownalgorithms for finding the chromatic number of a general graph, or determine if two graphsare isomorphic, for example, are exponential-time.

For hypergraphs, the complexity usually is expressed as a function of the sum of all edgecardinalities, not justn. For example, if we need to find vertex degrees of a hypergraphonn vertices andm edges given by its incidence matrix, then we need to check every of mnentries if it is 0 or 1. So the total time isO(mn). However, if the hypergraph is given by itsedge list, then we may scan the lists and determine the degreeof every vertex. In this case,the total time isO(the sum of all edge cardinalities). As one can see, the complexity ofmany algorithms depends on the graph or hypergraph representation in computer memory.

In general, a problem that can be solved by polynomial-time algorithm is calledtractable, otherwise it is calledintractable. Tractable problems form the so calledclass Pof problems. There are many problems thatno polynomial-time algorithmcan solve them,but a solution (if known) can be checked in polynomial time. Such problems form the socalledclass NP. At last, there is a class of problems with the property that if any of theseproblems can be solved in polynomial time, then all of them can be solved in polynomialtime because there is a polynomial-time transformation from each other. They form the socalled class ofNP-complete problems.

12.3. Answers and Hints to Selected Exercises

CHAPTER 1

Section 1.11. n(G1) = 4,m(G1) = 3,n(G2) = 5,m(G2) = 7,n(G3) = 8,m(G3) = 12. 5. Degree se-quences:G1 : (1,1,1,3),G2 : (2,2,2, 4,4), G3 : (2,2,3,3,3,3,4,4).

Section 1.21. L(G1) = {{2},{1,3,4},{2,4},{2,3}}.2.

A(G1) =

0 1 0 01 0 1 10 1 0 10 1 1 0

3.

Appendix 271

I(G1) =

1 0 0 01 1 0 10 1 1 00 0 1 1

4. J(G1) = {{1,2},{2,3},{3,4},{2,4}}.

Section 1.45. G5,G6,G7,G8,G9. 7. BecauseKs,r is obtained fromKr,s by just interchanging the parts.8. G1

∼= G2, G3∼= G4, G7

∼= G8∼= G9. 9. When the names and order of vertices correspond

to the given isomorphism.10. When the names and order of vertices correspond to anyisomorphism.

Section 1.56. η(T) = 2,η(C5) = 3,η(W5) = 4,η(K2,3 = 3. 10. If we reverse the order in degreesequence ofG and add it with the degree sequence ofG as two vectors, then we obtain avector with all components equal ton−1.

Section 1.611. In K4,4: C4,C8, in cube:C4,C8, in Petersen graph:C5,C9. 13. For graphs in Figure 1.21:ω(G1) = ω(G2) = ω(G3) = ω(G4) = 3, ω(G5) = ω(G6) = ω(G7) = ω(G8) = ω(G9) = 2.14. For graphs in Figure 1.21:α(G1) = α(G2) = α(G3) = α(G4) = 2, α(G5) = α(G6) = 4,α(G7) = α(G8) = α(G9) = 3. 15. For graphs in Figure 1.21:τ(G1) = τ(G2) = 2, τ(G3) =τ(G4) = 3, τ(G5) = τ(G6) = 4, τ(G7) = τ(G8) = τ(G9) = 2. 16. For graphs in Figure 1.21:ν(G1) = ν(G2) = ν(G3) = ν(G4) = 2, ν(G5) = ν(G6) = 4, ν(G7) = ν(G8) = ν(G9) = 2.17. If n is even:α(Cn) = τ(Cn) = ν(Cn) = n/2, if n is odd: α(Cn) = ν(Cn) = (n− 1)/2,τ(Cn) = (n+1)/2, for all n: ω(Cn) = 2.

Section 1.73. κ(G) = 2. 4. k≥ 2. 12. 2.

CHAPTER 2

Section 2.11. Λ(En) = 0, Λ(Cn) = 1, Λ(Kn) = (n−1)(n−2)/2, Λ(Wn) = n−1.

Section 2.21. diam(T) = 9, radius= 5.

Section 2.32. Minimum weight = 28.3. Maximum weight = 49.4. nn−2.

Section 2.42. m= n. 4. τ(Km,n) = ν(Km,n) = min{m,n}. 6. τ = ν = 13.


CHAPTER 3

Section 3.22. G1 andG3 are chordal;G2 is not chordal. 3. G2. 7. θ(G1) = α(G1) = 3, θ(G2) =α(G2) = 5, θ(G3) = α(G3) = 3. 8. For cube: 6.9. From cube: 5.Section 3.31. M(G1) = 3,ω(G1)−1 = 2; M(G2) = 4,ω(G2)−1 = 2; M(G3) = 2,ω(G3)−1 = 2. 2.G1,G2 are not chordal;G3 is chordal.3. M(Cn) = 2, M(Kn) = n−1, M(Wn) = 3; for cube,prism and Petersen graph:M(G) = 3. 4. Take empty graphEn with sufficient largen; adda vertex and make it adjacent to all the vertices ofEn; repeat the procedure. Observe thatω = 2 for all obtained graphs, andM(G) is increasing by 1 every step until it reachesn.Evidently, the graph obtained isKn,n. 5. Delete vertices by minimum degree; since graph isk-regular, at any step except the first, there is a vertex of degree≤ k−1.

Section 3.42. diam(G) = 6, radius = 3.

Section 3.51. G1 andG2 are quasi-triangulated,G3 is not. 2. C5 in G3. 5. Quasi-triangulated graphs:Cn for n = 3,4, Kn for n≥ 3, Wn for n = 4,5.

CHAPTER 4

Section 4.11. G1: 5; G2: 7. 3. Yes.

Section 4.21. See Figure 1.30 for prism and Figure 1.17 for cube.

Section 4.31. For K3,3: 0,1,2,3; forK5: 0,1,2,3,4.6. 1.

Section 4.41. Kn: n = 3,4; Km,n: m= 1,n≥ 3, or m= 2,n≥ 3; Wn: n≥ 4. 3. G1 is not planar;G2 isplanar.

CHAPTER 5

Section 5.21. χ(En) = 1, χ(Kn) = n, χ(Km,n) = 2, χ(Tn) = 2, χ(C2n) = 2, χ(C2n+1) = 3, χ(W2n) = 4,χ(W2n+1) = 3. 4. λ ≥ 136. 6. χ(G) ≤ 6. 7. χ(G) = 5.

Section 5.31. P(E4,λ) = λ4; P(C5,λ) = (λ−1)5− (λ−1); P(W4,λ) = λ(4); P(Pn,λ) = λ(λ−1)n−1.6. The last one.

Section 5.42. S(7,1) = 1, S(7,2) = 63, S(7,3) = 301, S(7,4) = 350, S(7,5) = 140, S(7,6) = 21,S(7,7) = 1; s(3,1) = 2, s(3,2) = −3, s(3,3) = 1. 3. P(G1,λ) = λ(λ− 1)3(λ− 2)2(λ2−3λ+3)2; P(G2,λ) = λ(λ−1)(λ−2)4; P(G3,λ) = λ(λ−1)(λ−2)4(λ2−3λ+3).

Appendix 273

Section 5.53. Graph G = (X,E) with X = {1,2,3,4,5} and adjacency listL(G) = {{2,4},{1,3,4,5},{2,5}, {1,2,5},{2,3,4}}; the order of online coloring: 1,2,3,4,5.

Section 5.61. For Petersen graph:M = 3, thereforeχ ≤ 4. 2. Idea: switch the colors in the regions onone side of a newly added line.

Section 5.71. None of them is perfect because each contains inducedC5. 2. W2k−1 are perfect,W2k arenot,k≥ 3.

Section 5.81. χ′(Kn) = n if n≥ 3 is odd,χ′(Kn) = n−1 if n≥ 2 is even.χ′(Cn) = 3 if n≥ 3 is odd,χ′(Cn) = 2 if n≥ 2 is even.χ′(Wn) = n−1 for all n≥ 4. χ′(Km,n) = max{m,n}.

Section 5.92. χ′(G) = 8. 5. χ′(Cn) = 1, χ′(Wn) = n−1. 6. For Petersen graphχ′ = 7, for cubeχ′ = 6,and for prismχ′ = 5.

CHAPTER 6

Section 6.12. Connect each pair of such vertices by an edge.

Section 6.21. Kn : n≥ 3; Km,n : m= n; Wn : n≥ 4. 2. Prism and cube - yes; Petersen graph - not.3.Yes.

Section 6.31. 8. 4. Add a new source and connect it with outcoming arcs to all sources; add a new sinkand connect it with incoming arcs from all sinks.

CHAPTER 7

Section 7.11. n(H1) = 6, m(H1) = 5; n(H2) = 6, m(H2) = 11. 2. Two in H1, seven inH2. 3. None.4.None.9. ∆(H1) = 3, ∆(H2) = 5. 10. None.11. None.12. r(H1) = 3; r(H2) = 4. 13. H1:one isolated vertex, no pendant vertices;H2: no isolated and no pendant vertices.14. H1:one singleton, no empty edges;H2: no singletons and no empty edges.

Section 7.21. In H : N(1) = {2,4}, N(2) = {1,3,4,5}, N(3) = {2,4,5}, N(4) = {1,2,3,5},N(5) = {2,3,4}, N(6) = /0; in H ∗: N(d1) = {d2,d3}, N(d2) = {d1,d3,d4}, N(d3) ={d1,d2,d4,d5}, N(d4) = {d2,d3,d5}, N(d5) = {d3,d4}. 3. r(H ) = r(H ∗) = 3. 7. L(G) ={{1,5},{1,2},{2,3},{2,4},{4,5}}; L(G∗) = {{a,b}, {b,c,d},{c}, {d,e},{a,e}.} 9. E∗

nconsists ofn empty edges.


Section 7.34. L(K3

4) = {{1,2,3},{1,3,4}, {2,3,4}, {1,2,4}}. 6. When order of the vertices is thesame under the isomorphism.

Section 7.42. 1→ 7→ 11→ 6. 10. H ∗ is the Petersen graph.

Section 7.54. For example, S = {3,4,5,6,7,8,10,11,13,14}. 5. α(H ) = 10. 7. For ex-ample, T = {1,2,9,12,15}. 8. τ(H ) = 5. 9. For example, D ′ = {{6,11,15},{9,13,14},{1,3,7},{2,8,4}}. 10. ν(H ) = 4. 11. ρ(H ) = 6.

Section 7.63. Since bipartite graphs do not contain triangles, every intersecting family is a star.4.Petersen graph does not have triangles.7. Cube does not have triangles, while prism does.

CHAPTER 8

Section 8.12. 3rd, 4th and 5th.3. 3rd, 4th and 5th.4. 2nd, 3rd, 4th and 5th.5. 3rd, 4th and 5th.8. 3rd,4th and 5th.

Section 8.23. τ(H1) = 2, α(H1) = 4, ν(H1) = 2; τ(H2) = 3, α(H2) = 5, ν(H2) = 3.

Section 8.35. Λ(H1) = 1, Λ(H2) = 0.

CHAPTER 9

Section 9.11. Kn : n = 1,2; Km,n : m,n ≥ 1; Wn : none; prism: no; cube: yes; Petersen graph:no. 2. Graphs without cycles of length≥ 3. 3. H = (X,D) with X = {1,2,3,4} andD = {{1,2,3}, {1,2,4},{2,3,4}}. 4. H = (X,D) with X = {1,2,3} andD = {{1,2,3},{1,2},{2,3},{1,3}}.

Section 9.21. PathsPn, n≥ 1. 2. In any linear ordering of the vertices ofCn, n ≥ 3, there is alwaysan edge which is not an interval in the ordering.3. H = (X,D) with X = {1,2,3,4}, D ={{1,4},{2,4},{3,4}}. 4. H = (X,D) with X = {1,2,3,4}, D = {{1,4},{2,4},{3,4}}.5. H = (X,D) with X = {1,2,3,4}, D = {{1,4},{2,4},{3,4}}.

Section 9.32. H1: not because it is not Helly;H2: not becauseL(H2) = C5 which is not perfect.3.None.4. χ′(H1) = 3; χ′(H2) = 3. 5. None.

Appendix 275

Section 9.42. H1.

CHAPTER 10

Section 10.11. χ(H ) = 2. 2. α(H ) = 6; τ(H ) = 2. 3. γ(H ) = 4. 5. Λ(H ) = 0. 6. No.

Section 10.22. M(H ) = 2, χ(H ) ≥ 3. 2. χ(H ) = 3.

Section 10.32. χ(H ) = 2, χ(H ) = 3. 3. 3. 4. χ(HC ) = 1, χ(HC ) = 3; χ(HD ) = 2, χ(HD) = 5.

Section 10.42. 0. 3. χ(H ) ≥ 3. 4. χ(H ) = 3. 5. Because the resistanceO(H ) = 0.

Section 10.51. P(H1,λ) = 3λ(3) +6λ(2), R(H1) = (0,6,3,0); P(H2,λ) = λ(2), R(H2) = (0,1,0,0).

Section 10.61. χ(HD) = 3, χ(HC ) = 3.

Section 10.74. Yes.5. Yes.

Section 10.81. αC (H1) = 3, χ(H1) = 3; αC (H2) = 3, χ(H2) = 3. 2. H1 is perfect,H2 is not. 3.τ2(H1) = 2, τ2(H2) = 3.

Section 10.91. S(HC ) = {1,2,3,4,5,6}, S(HD) = {2,3,4,5,6,7,8}. 3. S(H ) = {2,3,4,5}.

Section 10.102. No.

12.4. Glossary of Additional Concepts

This glossary contains informal definitions of additional concepts that are most often usedin the literature.

• Acyclic graph: graph without cycles

• Acyclic orientation: replacing edges by arcs which produces no directed cycles

• Almost always true: a property which has asymptotic (asn→ ∞) probability 1


• Antihole: a subgraph induced byCk

• Automorphism: a permutation of the vertices that keeps the adjacency

• Binary tree: a tree with a root in which every non-pendant vertex has at most twoneighbors further from the root

• Bipartite Ramsey number: given a bipartite graphG, the minimumn such that any2-coloring of the edges ofKn,n produces a monochromatic copy ofG

• Block: maximal 2-connected subgraph

• Block designs: also known as BIBDs (Balanced Incomplete Block Designs), are thek-uniform hypergraphs with vertices called points, edges called blocks, such thatevery point is contained in exactly r blocks, any subset of points of a given size iscontained in exactlyλ blocks

• Cactus: graph in which no two cycles share an edge

• Chinese Postman Problem: to find a shortest closed walk passing through each edgeof a weighted graph

• k-choosable graph: when for any lists of colors of lengthk assigned to vertices, thereexists a proper list coloring

• Claw: graphK1,3

• Cograph: graph with no induced pathP4

• Color-critical: a graph for which every proper subgraph hasthe smaller chromaticnumber (index)

• (n,k,λ)-configuration: ak-uniform hypergraph of ordern such that every pair ofvertices is contained in preciselyλ edges

• Density: ratiom(G)/n(G)

• Dominating set: a subsetSof vertices in a graph such that every vertex not inShas aneighbor fromS

• Domination number: the size of a smallest dominating set

• Eigenvalue of a graph: eigenvalue of the adjacency matrix

• Extremal combinatorics: study of the smallest (largest) number of edges which anr-uniform hypergraph of ordern can have provided that some given property holds

• Fano plane: Steiner Triple SystemSTS(7), or Steiner SystemS(2,3,7)

• Finite projective plane of rankr: a hypergraphH = (X,D) with |X|= |D|= r2− r +1, vertices called points, edges called planes, such that every point belongs to exactlyr lines, every line contains exactlyr points, any two distinct points are on preciselyone line, and any two distinct lines have precisely one pointin common

Appendix 277

• Genus of a graph: the minimum genus of a surface on which the graph can be em-bedded without crossings of edges

• Genus of a surface: the number of handles added to the sphere

• Girth: the length of a shortest cycle

• Greedy algorithm: an algorithm for finding an optimal solution of a problem thattakes the best possible choice at each step; it does not guarantee the optimal solutionfor the whole problem

• Hereditary class: a class of graphs such that any induced subgraph of any graph fromthe class is also in the class

• Hole: induced subgraph isomorphic toCk,k≥ 4

• Homomorphism: a mapf : V(G1) →V(G2) that keeps adjacency

• Hypergraph removal lemma: for a givenr-uniform hypergraphH , in any largerr-uniform hypergraph containing a bounded number of copies ofH , one can removebounded number of edges to construct a hypergraph with no copy of H

• List coloring: proper vertex coloring of a graph in which every vertex has a list ofadmissible colors

• Matroid: discrete hereditary structure generalizing linear independence in vectorspaces; it has many equivalent formulations; for example, graphic matroid for a graphG consists of edges as “elements” and subsets of edges not forming any cycle as“bases”

• Non-orientable surface: a surface which does not have two different sides, for exam-ple like Mobius strip

• Orientable surface: a surface with two different sides, forexample like sphere ortorus

• Outerplanar graph: a planar graph that can be embedded in theplane with all thevertices on the unbounded face

• Partition of a set: dividing a set into a number of nonempty subsets

• Pigeonhole principle: ifn items are put intom pigeonholes andn > m, then at leastone pigeonhole contains more than one item; or, ifn vertices are colored withmcolors andn > m, then there are at least two vertices of the same color

• Probabilistic graph theory: the study of existence of graphs with some propertiesusing probability theory; usually, the graphs are not constructed, but one prove thatwith positive probability they exist

• Ramsey numberR(p,q): the minimumn such that any 2-coloring of the edges ofKn

produces either a monochromatic copy ofKp or a monochromatic copy ofKq; forexample,R(3,3) = 6


• Random graph: a graph in which every pair of vertices forms anedge with probabilityp

• Regularity Lemma (also known as Szemeredi regularity lemma): every graph havingsufficiently many vertices and edges can be approximated by some composition ofalmost regular parts; in other words, the vertex set can be partitioned into a numberof almost equal in size subsets having special properties with sufficiently small set ofleftover vertices; it has a generalization tor-uniform hypergraphs

• Satisfiability problem: the problem of finding truth values for logical variables suchthat some logical formula becomes true

• Spectrum of a graph: the set of eigenvalues with their multiplicity

• Steiner Quadruple SystemSQS(v): Steiner SystemS(3,4,v)

• Steiner SystemS(t,k,v): a k-uniform hypergraphH = (X,B) with |X| = v, hyper-edges called blocks, such that anyt distinct vertices appear together in precisely oneblock

• Steiner Triple SystemSTS(v): Steiner SystemS(2,3,v)

• System of distinct representatives: for a collection of sets, a choice of one elementfrom each set such that all chosen elements are distinct

• Thickness of a graphG: minimum number of planar graphs into whichG can be split

• Topological graph theory: the study of graph drawings on different surfaces

• Torus: the orientable surface of genus 1; equivalently, a sphere with added handle

• Total coloring: coloring of both vertices and edges so that no adjacent and no incidentelements have the same color

• Tournament: digraph obtained fromKn by replacing edges with arcs (orientation)

• Transitive digraph: if there are arcs(x,y) and(y,z), then there is arc(x,z)

• Transversal hypergraphTrH : for a given hypergraphH = (X,E), TrH = (X,D)whereD is the family of all minimal transversals ofH

• Traveling Salesman Problem: to find a shortest spanning cycle in a weighted graph

• Traversal: visiting all the vertices or edges of a graph in a special way

• Turan graph: the complete multipartite graph with all parts of almost the same size(different by at most 1)

• Turan numberT(n, p, r): the smallest number of edges in anr-uniform hypergraphonn vertices such that every set of vertices of cardinalityp contains at least one edge

• Turan’s theorem: for a givenn, the Turanr-partite graph contains the maximumnumber of edges and does not containKr+1

References

[1] C. Berge.Graphs and Hypergraphs.North-Holland, 1973.

[2] C. Berge.Hypergraphs: combinatorics of finite sets.North-Holland, 1989.

[3] J.A. Bondy and U.S.R Murty.Graph Theory.Springer, 2008.

[4] G. Chartrand and P. Zhang.Introduction to Graph Theory.Walter Rudin StudentSeries in Advanced Mathematics, 2004.

[5] R. Diestel.Graph Theory.Springer, 2006.

[6] V. Voloshin. Coloring Mixed Hypergraphs: theory, allgorithms and applications.AMS, Providence, 2002.

[7] D. B. West.Introduction to Graph Theory.Prentice Hall, 2001.

[8] A. A. Zykov. Hypergraphs.Uspekhi Mat. Nauk29 (1974), 89–154 (in Russian).

Index

(x,y)-invertor, 228(x,y)-path, 18, 144(x,y)-separator, 34(H )2, 2-section, 154B(H ), bipartite representation, 140E(G), edge set, 7E(x), set of edges containing vertexx, 7G, graph, 6L(G), line graph, 112L(H ), line graph, 154N∞(x), farthest set of vertices, 41O(g(n)), big-oh, 269P(G,λ), chromatic polynomial, 80P(H ,λ), chromatic polynomial of a hy-

pergraph, 204R(G), chromatic spectrum, 81R(H ), chromatic spectrum, 203V(G), vertex set, 7V(H ), vertex set, 135[H ]2, generalized 2-section, 174∆(G), maximum degree, 6, 113∆(H ), maximum degree of hypergraph,

136Λ(G), cyclomatic number, 40Λ(H ,T), generalized cyclomatic num-

ber, 174Λ(H ), cyclomatic number of a hyper-

graph, 179α(G), stability number, 31α(H ), stability (independence) number,

151α

C(H ), C -stability number, 236

χ(H ), upper chromatic number, 203χ′(G), upper chromatic index, 116χ′(G), chromatic index, 112χ′(H ), chromatic index, 185

χ(G), chromatic number, 80χ(H ), lower chromatic number, 203χ(H ), chromatic number, 193κ(G), connectivity, 34λ(n), falling factorial, 84µ(G), multiplicity of a graph, 115ν(G), maximum size of a matching, 32ω(G), clique number, 31τ(G), transversal number, 31τ2(HC ), bitransversal number, 238θ(G), clique cover number, 56k-connected graph, 34k-factor, 32k-factorization, 32k-regular graph, 21k-regular hypergraph, 136l(H ,T), number of loops ofH not in for-

estT, 174m= m(H ), the number of edges, 136n= n(H ), order ofH , the number of ver-

tices, 136r-uniform hypergraph, 136, 206r i(G), number of feasible partitions, 81r i(H ), number of feasible partitions, 203w(T), weight of a tree, 43C -bistar, 238C -clearing, 213C -hypergraph, 204C -monostar, 238C -perfection, perfection, 237C -stability number,α

C(H ), 236

C -stable set, 206D-clearing, 213D-graph, 204D-hypergraph, 204D(H ), edge set, 135

281

282 Index

D(x), set of edges containing vertexx,136

H , hypergraph, 135H ∗, dual hypergraph, 140HY, subhypergraph induced byY, 1512-colorable hypergraph, 1452-section,(H )2, 154

acyclic graph, 275acyclic orientation, 275adjacency list, 12adjacency matrix, 13, 143adjacent edges, 7adjacent edges in hypergraph, 136adjacent vertices, 6adjacent vertices in hypergraph, 136algorithm, 269almost always, 275antihole, 276applications, 8, 17, 137, 263, 264arc, 16automorphism, 276

back edge, 46backtracking, 46balanced cycle, 181balanced hypergraph, 181Berge graph, 111bi-chromatic hypergraph, 145bi-conformal hypergraph, 159bi-edge, 206bi-Helly hypergraph, 240bidegree, 208bihypergraph, 206binary Ramsey number, 276binary tree, 276bipartite graph, 21bipartite hypergraph, 145bipartite representation,B(H ), 140bistar, 207, 221bitransversal number,τ2(HC ), 238bitriangulation, 254block, 276block design, 276blocking set, 152breadth-first search algorithm, 45

bridge, 36broken chromatic spectrum, 82, 244Brooks’ theorem, 82

cactus, 276capacity of a cut, 127capacity of an arc, 127cardinality of edge, 136cell, 81center, 41Chinese Postman Problem, 276choosable graph, 276chord of a cycle, 51chord with respect to the spanning tree,

40chordal graph, 51chordal hypergraph, 163chromatic index,χ′(H ), 185chromatic index,χ′(G), 112chromatic number,χ(G), 80chromatic number,χ(H ), 193chromatic polynomial of a hypergraph,

P(H ,λ), 204chromatic polynomial,P(G,λ), 80chromatic spectrum,R(H ), 203chromatic spectrum,R(G), 81class 1 graphs, 115class 2 graphs, 115class NP, 270class P, 270clearing, 148clique, 30clique cover number,θ(G), 56clique covering, 56clique hypergraph, 166clique number,ω(G), 30closed Jordan curve, 67closed walk, 123co-simplicial vertex, 62cograph, 276color class, 81, 193, 203color-critical graph, 276colorability problem, 219colorable mixed hypergraph, 202coloring, 80, 202

Index 283

complement of a graph, 28complete (l ,m)-uniform mixed hyper-

graph, 223completer-partite hypergraph, 145completer-uniform hypergraph, 144complete bipartite graph, 21complete graph, 20complete hypergraph, 144complete mixed hypertree, 232complexity, 269computation of cyclomatic number, algo-

rithm, 179configuration, 107, 276conformal hypergraph, 156connected component, 19, 144connected graph, 18connected hypergraph, 144connected mixed hypergraph, 206connected region, 67connection-contraction algorithm, 84connectivity,κ(G), 34constant complexity, 270continuous chromatic spectrum, 82, 244contractible graph, 28contraction, 214contraction of a hyperedge, 149contraction of an edge, 27copy of a vertex, 109covering, 153crossing, 67crossing number, 68cube, 12, 22cubic graph, 21cut, 127cycle, 19, 30cycle in hypergraph, 144cycloid, 238cyclomatic number of a hypergraph,

Λ(H ), 179cyclomatic number,Λ(G), 40

deadlock, 211degree of a vertex, 6degree of edge, 136degree sequence, 7

deletion of a vertex, 25deletion of an edge, 26density, 276depth-first algorithm, 46derived subgraph, 35diagonal of a cycle, 51diameter, 41diametral path, 41different proper colorings, 203digraph, 16directed graph, 16disconnected graph, 18disconnection-contraction algorithm, 90disjoint edges, 7disjoint vertices, 6distance, 41dominating set, 276domination number, 276doubling, 249drawing dual hypergraph, 140dual graph to a plane graph, 77dual hypergraph,H ∗, 140dual Konig property, 153

eccentricity , 41eclipse, 240edge, 5edgek-colorable graph, 112edge coloring, 112, 116edge list, 14, 142edge-coloring property, 185edge-critical hypergraph, 196edge-cut, 36edge-intersection in the plane, 67edge-separator, 36eigenvalue of a graph, 276elementary cycle, 40elementary shifting, 247elimination, 213embedding of hypergraph, 188empty edge, 136empty graph, 18equitable coloring, 195Euler’s formula for hypergraphs, 188Eulerian graph, 123

284 Index

Eulerian trail, 123even cycle, 19evidently uncolorableC -edge, 205evidently uncolorableD-edge, 206exponential-time, 270extremal combinatorics, 276

face, 188face of a plane graph, 67factor, 32factorial complexity, 270falling factorial,λ(n), 84family, 6Fano plane, 276feasible partition, 81, 193, 203feasible set, 244finding maximum stable set in chordal

graph, algorithm, 55finding minimum transversal and maxi-

mum matching in a hypertree,algorithm, 172

finite projective plane, 276flow conservation law, 127flow in an arc, 127forest, 21four color problem, 79fragment, 75friendship graph, 12fundamental equality, 204fundamental relation, 86

gap in chromatic spectrum, 244gap-free chromatic spectrum, 82generalized 2-section,[H ]2, 174generalized cyclomatic number,

Λ(H ,T), 174genus of a graph, 277genus of a surface, 277girth, 277good coloring, 195Grotzsch graph, 109graph, 6graph applications, 8, 17graph minor, 32graph perfection vs hypergraph perfec-

tion, 236

greedy C -hypergraph coloring, algo-rithm, 208

greedy algorithm, 277greedy coloring, 96greedy hypergraph coloring, 198

Hadwiger number, 28Hall’s theorem, 47Hamiltonian cycle, 125Hamiltonian graph, 125Helly property, 155hereditary class, 277hole, 277homeomorphic graphs, 73homomorphism, 277host graph, 161hyper-pendant vertex, 163, 168hyperedge, 136hypergraph, 135hypergraph applications, 136, 263, 264hypergraph minor, 153hypergraph removal lemma, 277hypertree, 161hypertree recognition algorithm, 170

I-regular coloring, 195incidence matrix, 14, 139incident vertices and edges, 7, 136included hyperedge, 136, 213including hyperedge, 213incoming arc, 127independent (stable) set, 31induced subgraph, 30induced subhypergraph, 151, 206induction, 267inflation, 244initial vertex, 16intersecting family, 155intersection graph, 154interval graph, 11, 184interval hypergraph, 183intractable problem, 270isolated vertex, 15, 136isomorphic graphs, 23isomorphic hypergraphs, 136, 145isomorphism, 23

Index 285

join, 247Jordan curve, 67

Konig property, 152Konig’s theorem, 45, 48Kempe chain, 107, 209Kempe’s proof, 104Kruskal’s algorithm for minimum span-

ning tree, 43Kuratowski’s theorem, 74

latticed graph, 62leaf, 41length of a path, 18length of the cycle, 19length of the path/cycle, 144line graph,L(H ), 154line graph,L(G), 112linear-time, 270list coloring, 263, 277logarithmic complexity, 270loop, 15loop in hypergraph, 136lower chromatic number,χ(H ), 203

matching, 32, 152matching covering, 47matroid, 277maximal (minimal) versus maximum

(minimum), 31maximal by inclusion complete subgraph,

30maximal clique, 30maximal colorable subhypergraph, 220maximal coloring, 120maximal embedding, 189maximal planar graph, 76maximum degree of hypergraph,∆(H ),

136maximum degree,∆, 6Menger’s theorem, 37Meyniel graph, 111minimal separator, 34minimal uncolorable mixed hypergraph,

220

minimum (maximum) spanning treeproblem, 43

minimum cut, 127mixed hypergraph, 202mixed hypertree, 225mixed multigraph, 253monochromatic component, 208monochromatic subset, 203monodegree, 197monostar, 197multi-forest, 174multigraph, 15multiple edges, 15multiple hyperedge, 136multiplicity, 15multiplicity of a graph,µ(G), 115Mycielski’s construction, 109

neighbor, 7neighborhood, 7neighborhood in hypergraph, 136neighborhood of a subset, 46network, 127network flow, 127node, 127non-orientable surface, 277normal hypergraph, 185NP-complete problems, 270

odd cycle, 19online coloring, 96orientable surface, 277originality of vertex, 208outcoming arc, 127outerplanar graph, 277

parallel edges, 15partial subhypergraph, 151partition, 277path, 18pendant vertex, 41, 136, 163, 174perfect elimination ordering, 54perfect graph, 110perfect matching, 32, 152Petersen graph, 22pigeonhole principle, 277

286 Index

planar graph, 67planar hypergraph, 187planar mixed hypergraph, 254planarity testing algorithm, 74plane embedding, 67plane graph, 67plane triangulation, 76polychromatic subset, 203polynomial-time, 270prism, 22probabilistic graph theory, 277properλ-coloring, 80, 193proper coloring, 80, 202proper edgeλ-coloring, 112

quasi-triangulated graph, 62

radius, 41Ramsey number, 277random graph, 278rank of hypergraph,r(H ), 136reduced mixed hypergraph, 218reducible configuration, 107redundant edge, 217refinement, 255regular graph, 21regular hypergraph, 136regularity lemma, 278resistance ofC -hypergraph, 208root in a tree, 46

satisfiability problem, 278satisfied vertex, 117saturated arc, 127self-dual hypergraph, 159separator, 34set-system, 135simple cycle, 30simple graph, 16simple hypergraph, 136simple invertor, 229simplicial decomposition, 54simplicial elimination ordering, 54simplicial vertex, 51singleton, 136sink, 127

size of a face, 67size of edge, 136size of face, 188source, 127spanning subgraph, 31spanning tree, 32spectrum of a graph, 278Sperner family, 136splitting, 214splitting-contraction algorithm, 214stability (independence) number,α(H ),

151stability (independence) number,α(G),

31stable (independent) set, 31, 151star, 155Steiner Quadruple System, 278Steiner System, 278Steiner Triple System, 278stereographic projection, 70Stirling numbers of the first kind, 90Stirling numbers of the second kind, 90strict i-coloring, 81, 203strict edge coloring, 116strong chromatic number, 194strong coloring, 194strong deletion of a hyperedge, 148strong deletion of a vertex, 26strong deletion of an edge, 26strong deletion of vertex, 146strong perfect graph conjecture, 111strongly independent (stable) set, 152strongly perfect graph, 111subgraph, 29subhypergraph, 151symmetric matrix, 13System of distinct representatives, 278Szekeres-Wilf number, 57Szemeredi regularity lemma, 278

terminal vertex, 16thickness of a graph, 278topological graph theory, 278torus, 278total coloring, 278

Index 287

totally balanced hypergraph, 181touching graph, 231tournament, 278tractable problem, 270trail, 123transitive graph, 278transposition of a matrix, 13transversal, 31, 152transversal hypergraph, 278transversal number,τ(G), 31transversal vertex, 168Traveling Salesman Problem, 278traversal, 123, 125, 278tree, 20triangle, 19trivial graph, 41Turan graph, 278Turan number, 278Turan’s theorem, 278twin vertex, 168

uc, 227uc-orderable mixed hypergraph, 233uc-ordering algorithm, 233unavoidable configuration, 107unbounded face, 67uncolorable mixed hypergraph, 202underlying hypergraph, 206undirected graph, 16uniform coloring, 195uniform hypergraph, 136uniquely colorable mixed hypergraph,

227universal vertex, 109unsaturated arc, 127upper chromatic index,χ′(G), 116upper chromatic number,χ(H ), 203

value of the network flow, 127vertex, 5, 136vertex cover, 31vertex cut, 34Vizing’s theorem, 113

walk, 123weak coloring, 193

weak deletion of a hyperedge, 148weak deletion of a vertex, 26weak deletion of vertex, 147weak perfect graph conjecture, 110weakly chordal graph, 111weakly cyclic vertex, 62weight of a multi-forest,w(T), 174weight of a tree,w(T), 43weight of an edge, 43weighted graph, 43wheel, 20Whitney’s theorem, 90

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