UNIVERSITY of CALGARY Graph Colouring and Forbidden

UNIVERSITY OF CALGARY

Graph Colouring and Forbidden Subgraphs

Jia Shen

A THESIS

SUBMITTED TO THE FACULTY OF GRADUATE STUDIES

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE

DEGREE OF DOCTOR OF PHILOSOPHY

DEPARTMENT OF MATHEMATICS AND STATISTICS

CALGARY, ALBERTA

APRIL, 2008

°c Jia Shen 2008

ISBN: 978-0-494-38239-4

Abstract

Concepts and questions of graph colouring arise naturally from practical problems and have found applications in many areas, including Information Theory and most notably Theoretical Computer Science.

This thesis mainly concerns graph colouring problems. Given a graph F , a graph

H is said to be F -free if H does not contain F as an induced subgraph. We will be concerned with what we call the F -avoiding colouring problem: for a given graph

F , colour the vertices of a graph G such that each maximal F -free subgraph of G receives more than one colour. This colouring is motivated by questions raised by

Erd˝os-Gallai-Tuza, and Duﬀus-Kierstead-Trotter.

The following question will be resolved in full: Given a graph F , does there exist a constant c, depending only on F , such that every graph G has a vertex-colouring with at most c colours such that each nontrivial maximal F -free subgraph of G receives more than one colour? Using a similar technique, the related transversal problem is also completely resolved.

We also investigate the precise number of colours needed for various classes of graphs F and G. The corresponding problems in the setting of partially ordered sets are studied as well.

iii Acknowledgements

First I thank Dr. Bill Sands for accumulating discussions towards the results in this thesis. Dr. Sands read through the draft of this thesis many times and corrected many of the inaccuracies. I also would like to thank Dr. Sands for lending me some precious books in his collection, which are very helpful.

Thanks also go to the faculty and the staﬀ of the Department of Mathematics and Statistics at the University of Calgary for all the help I got during the past years.

Finally, I would like to thank my dear parents and my beloved wife, Jingjing Zhou.

It is your encouragement, patience, support and love that made my studies possible.

iv Table of Contents

Abstract iii

Acknowledgements iv

Table of Contents v

1 Introduction 1

1.1 Subject of Research ...... 1

1.2 Motivation and Historical Notes ...... 2

1.3 Relations and Diﬀerences ...... 5

1.4 Structure of the Thesis ...... 7

2 Deﬁnitions, Notation and Basic Facts 8

2.1 The Basic Notation ...... 8

2.2 Graphs ...... 8

2.3 Graph Colouring ...... 12

2.4 Hypergraphs ...... 13

2.5 Basic Properties of F -avoiding Colouring ...... 15

3 Clique Colouring 21

3.1 Constant Bounds ...... 21

3.2 General Bounds ...... 24

3.3 Complexity ...... 25

3.4 Tree-decompositions and Chordal Graphs ...... 26

v vi

4 A Complete Characterization 31

4.1 Bounded Avoiding Chromatic Numbers ...... 31

4.2 Unbounded Avoiding Chromatic Numbers ...... 33

4.3 General Bounds ...... 38

4.4 Transversals ...... 44

5 Graphs F without Isolated Vertices 51

5.1 General Results ...... 51

5.2 Various Classes of Graphs ...... 60

6 Small F with Isolated Vertices 70

6.1 Characterization of Maximal Subgraphs ...... 70

6.2 Union of Cliques ...... 71

6.3 Sparse Graphs ...... 74

6.4 Induced Long Path ...... 77

6.5 Some Other Classes of Graphs ...... 80

7 Partially Ordered Sets 93

7.1 Preliminaries for Partially Ordered Sets ...... 93

7.2 Colouring of Posets ...... 98

7.3 Bounds ...... 100

7.4 Decomposition of Partially Ordered Sets ...... 102

7.5 Relations between Various Bounds ...... 109

7.6 An Approach towards Two Colourable Posets ...... 117

8 Remarks and Open Problems 126 vii

8.1 Open Problems ...... 126

8.2 Classes of Graphs F ...... 128

Bibliography 130 List of Figures

2.1 A disjoint union of Kk’s...... 18

3.1 a tree-decomposition (T, ) = (T, (V ) ∈ ) ...... 27 V t t T 3.2 a tree-decomposition of G ...... 29

4.1 F F and an edge ...... 35 1 ∨ 2

4.2 3-cycle v1e1ve2v2ev1 ...... 37 4.3 colouring scheme ...... 39

5.1 diam(G) 3...... 52 ≥

5.2 the graph K3 + e ...... 54

5.3 K2 and a vertex ...... 55 5.4 A list of graphs: K ,P ,K ,P ,C ,K + e, K e ...... 56 2 3 3 4 4 3 4 − 5.5 N[v] and a vertex x ...... 57

5.6 F is a star and G is not a star ...... 62

5.7 The colouring of wheel Wn ...... 66

6.1 acP3 (G) for union of cliques ...... 71 6.2 ac (G) where g(G) 5 ...... 75 P3 ≥

6.3 acP3 (P5)...... 80

6.4 acP3 (Pk)...... 81

6.5 acP3 (Cn)...... 82

6.6 acP3 (G) for bigraphs G with δ(G)=1...... 87

7.1 A partition of a linearly indecomposable poset ...... 103

viii ix

7.2 the 17-element poset Q ...... 112

7.3 A poset P with ack(P ) = 3 ...... 113

7.4 Example 7.1 ...... 116

7.5 the relation m n ...... 119 ≺Q

7.6 a1 Q a2 Q Q ak Q a1 ...... 120 2k+1≺ ≺ · · · ≺ ≺ 7.7 ( 2)-free ...... 121 X1 k 7.8 ( n)-free ...... 123 X1 k 7.9 ( (2 1))-free ...... 124 ⊕ X1 k 7.10 ( (1 2))-free ...... 125 ⊕ X1 Chapter 1

Introduction

In this thesis, all graphs are ﬁnite, undirected and simple (i.e., contain no multiedges and loops) unless otherwise speciﬁed.

1.1 Subject of Research

When considering relational structures such as graphs and partially ordered sets, subsets which do not contain a prescribed, induced “forbidden” substructure are often of interest. For example, if the forbidden subgraph is K2 (i.e. two independent vertices), then subgraphs that do not contain K2 are all complete subgraphs. Thus a maximal K2-free subgraph is a maximal complete subgraph, or a maximal clique. In general, for a given graph F , a graph is said to be F -free if it does not contain

F as an induced subgraph. The main subject of this thesis is to investigate, for a given “forbidden” graph F , maximal F -free subgraphs of a graph G. Our main interest will be what we call the F -avoiding colouring problem1: for a given graph

F , colour the vertices of a graph G such that each maximal F -free subgraph of G receives more than one colour.

We will be concerned with the following question:

Question 1:

Given a graph F , does there exist a constant c, depending only on F ,

1The term “F -free colouring” has been deﬁned by other researchers for a diﬀerent, yet related, colouring problem. We will review this concept in Section 1.4.

1 2

such that every graph G has a vertex-colouring with at most c colours

so that each nontrivial (i.e. with more than one vertex) maximal F -free

subgraph of G receives more than one colour?

A related question is the F -free subgraph transversal: a subset T V (G) of ⊆ vertices of a graph G is said to be an F -free subgraph transversal if T meets every maximal F -free subgraph of G. For a given “forbidden” graph F , the F -free subgraph transversal number of a graph G, denoted by τF (G), is the minimum cardinality of an F -free subgraph transversal of G. We will also be considering the following question:

Question 2:

For which graphs F is the F -free subgraph transversal number τF (G) bounded by a constant fraction of the order of G, namely, τ (G) c G F ≤ | | for some constant c < 1 independent of G?

1.2 Motivation and Historical Notes

The subject of this thesis is motivated by a problem raised by Erd˝os,Gallai and

Tuza.

In [30] Erd˝os, Gallai and Tuza proposed to investigate the so-called clique- transversal problem, namely, the problem of estimating the minimal cardinality of a subset of V (G) which meets every maximal clique of G. In particular, according to

Erd˝os(cf. [2, 24]), Gallai asked whether a triangulated graph (or chordal graph) on n vertices has a subset of size n/2 which intersects every nontrivial maximal clique.

In fact, the following stronger result is true: The vertices of a chordal graph can be 3

2-coloured so that all nontrivial maximal cliques receive two colours. We will give a proof of it in Chapter 3.

Aigner and Andreae [2] noted that, as can be easily seen, a comparability graph

G has a subset of size at most G /2 which intersects every nontrivial maximal clique. | | They also raised the question: is this true for a cocomparability graph?

Lonc and Rival [38] conjectured something stronger: a cocomparability graph is

2-colourable so that each nontrivial maximal clique receives both colours. If this statement were true, then the smaller of the two colour classes would have size at most n/2 and would intersect every maximal clique.

Duﬀus, Sands, Sauer, and Woodrow gave a counterexample in [25] to both questions. However, Duﬀus, Kierstead and Trotter [24] proved that the next best thing to the Lonc-Rival conjecture is true, i.e., three colours instead of two are enough.

The vertices of a cocomparability graph can be 3-coloured so that all

nontrivial maximal cliques receive at least two colours.

In the language of partially ordered sets, what Duﬀus et al. have proven is that the elements of any partially ordered set can be 3-coloured so that all nontrivial maximal antichains receive more than one colour. We shall give further generalizations of this result in Chapter 7.

Duﬀus et al. [25] also raised a much more general question.

Does there exist a constant c such that every perfect graph can be c-

coloured so that each nontrivial maximal clique receives at least two

colours?

More generally, the following question arises: 4

For which classes of graphs does there exist a constant c, depending G only on , such that every graph G can be c-coloured so that each G ∈ G nontrivial maximal clique receives at least two colours?

This question soon became interesting in its own right and received much attention. Bacs´oet al. [5] and D´efossez[20] have investigated this problem for various classes of graphs, and we shall survey their results in Chapter 3. For related results, also see e.g. [33, 36, 37, 41, 42].

This kind of question motivates the subject of this thesis. The problems above can be asked in a more general setting.

We thus make the following deﬁnition. Given the “forbidden graph” F , the F - avoiding chromatic number acF (G) of a graph G is the smallest integer k so that G has a vertex colouring of k colours such that each nontrivial maximal F -free subgraph of G receives more than one colour.

Remark. We note that if F > 2, then any maximal F -free subgraph of a graph G | | with G 2 has at least two vertices. Thus if F has more than two vertices we can | | ≥ ignore the requirement of maximal F -free subgraphs being nontrivial in the above deﬁnition.

In addition, we denote

ac(F ) := sup ac (G): G is any ﬁnite graph . { F }

We note that the above supremum could be + . If this is the case, then we say ∞ that ac(F ) does not exist.

Using the above notation, the ﬁrst question in Section 1.1 says that 5

Question 1: For a given graph F , does ac(F ) exist?

In this thesis, we shall resolve questions 1 and 2 of Section 1.1 in full. Further- more, we investigate the values of acF (G) for various graphs F and various kinds of graphs G. We also investigate the analogue of the question in the context of partially ordered sets in Chapter 7, and partially resolve the following question:

Question 3:

For a given partially ordered set F , is acF (P ) bounded for all partially ordered sets P ?

1.3 Relations and Diﬀerences

By deﬁnition, an F -avoiding colouring is a colouring of the vertex set of a graph G so that each colour class contains no maximal F -free subgraphs of G, that is, each monochromatic F -free subgraph of G can be extended to a larger F -free subgraph by adding in a certain vertex with another colour. From this point of view, for a given graph F , the parameter acF (G) concerns the extendability of F -free subgraphs of G. It can be regarded as one of the measures for such extendability—the transversal of maximal F -free subgraphs of G, which we will deal with in Chapter 4, is another such measure.

The F -avoiding chromatic number, in general, is quite diﬀerent from the ordinary chromatic number χ(G) of a graph G in many aspects.

First, the F -avoiding chromatic number acF (G) is not a global parameter of graphs, but rather, it is determined by very local structure of a graph. This is evidenced by many theorems and examples scattered throughout the thesis (see e.g. 6

Example 2.1, Theorem 6.4). By comparison, the ordinary chromatic number χ(G) of a graph G is determined by global structures of graphs—we know that χ(G) can be arbitrarily large even if G is locally tree-like, or locally 3-colourable, see e.g. [26, 28].

Secondly, acF (G) is not monotonic relative to either subgraphs or induced sub- graphs2; while the chromatic number χ(G) is monotonic relative to subgraphs.

Furthermore, for any graph G with acF (G) arbitrarily large (see Chapter 4), a

0 0 new graph G with acF (G ) = 2 can be constructed from G by adding one vertex to G. Actually, G0 is obtained by adding an isolated vertex to G, if F has a dominating vertex (see Lemma 2.9); if F has an isolated vertex, we consider the complement F which has a dominating vertex.

This shows that the parameter acF (G) changes dramatically; while the ordinary chromatic number χ(G) is “continuous” in that

χ(G) χ(G v) + 1 for all v V (G). ≤ − ∈

In summary, the parameter acF (G) has properties diﬀerent than that of χ(G).

The simplest case, acK2 (G), received much attention in the setting of perfect graphs.

Now we review a related, yet diﬀerent, concept for F -avoiding colouring, namely

F -free colouring (c.f. [11, 13, 14, 15, 16]). Given a graph F , a graph G is F -free k-colourable if the vertices of G can be k-coloured so that each induced copy of F in

G receives more than one colour. The F -free chromatic number fcF (G) of G is the

2A graph invariant f( ) is monotone increasing (respectively, monotone decreasing) relative to · subgraphs if f(G1) f(G2) (respectively, f(G1) f(G2)) whenever G1 is a subgraph of G2; f( ) is monotone increasing≤ (respectively, monotone≥ decreasing) relative to induced subgraphs if · f(G1) f(G2) (respectively, f(G1) f(G2)) whenever G1 is an induced subgraph of G2. For example,≤ the ordinary chromatic number≥ χ(G) is monotone increasing relative to subgraphs; the cochromatic number ([3]), the F -free chromatic number (see below), etc., are monotone increasing relative to induced subgraphs, but not relative to subgraphs. 7 least integer k so that G is F -free k-colourable.

F -free colouring is essentially diﬀerent from F -avoiding colouring in the following sense. In an F -free colouring, each colour class is F -free; while in an F -avoiding colouring, each colour class is maximal “F -free subgraph” free. The F -free chromatic number is monotone increasing relative to induced subgraphs. Furthermore, the

F -free chromatic number fc (G) is “continuous” (it is easy to see that fc (G) F F ≤ fc (G v) + 1 for any v V (G)), while ac (G) and ac (G v) can be arbitrarily F − ∈ F F − far apart.

1.4 Structure of the Thesis

The thesis is organized as follows. In Chapter 2 we collect deﬁnitions and notation that will be used throughout the thesis, and give basic facts about the F - avoiding chromatic number. In Chapter 3, we summarize the known results on clique-colouring of graphs. In Chapter 4 we prove that if F has neither a dominating vertex nor an isolated vertex, then acF (G) is bounded; otherwise, acF (G) is not bounded. In this same chapter we also study the corresponding F -free subgraph transversal problem. In Chapters 5 and 6, we investigate acF (G) for graphs F having no isolated vertices and for small F having an isolated vertex. In Chapter 7 we consider the F -avoiding colouring of partially ordered sets. We conclude the thesis with some open questions in Chapter 8. Chapter 2

Deﬁnitions, Notation and Basic Facts

We provide some background material in this chapter. We start by giving some basic deﬁnitions and notation which will appear throughout the thesis. For general notation and terminology, we follow the books [7, 19, 21].

In this thesis, we concentrate on finite structures. The graphs, hypergraphs, and posets are all finite unless specified otherwise.

2.1 The Basic Notation

By N we denote the set of natural numbers, including zero. Z denotes the set of integers.

A set = A ,...,A of pairwise disjoint subsets of a set A is a partition of A A { 1 k} if A = k A and A = for every i. By [A]k we denote the set of all k-element i=1 i i 6 ∅ S subsets of A. We write [n] for the set 1, 2, . . . , n . { } For a real number x we denote by x the greatest integer less than or equal to b c x, and by x the least integer greater than or equal to x. d e

2.2 Graphs

A graph is a pair G = (V,E) of sets satisfying E [V ]2; thus, the elements of E are ⊆ 2-element subsets of V . The elements of V are called the vertices of the graph G; the elements of E are the edges of G. Sometimes for clarity, V and E will be written

8 9

V (G) and E(G), respectively. The complement G of a graph G is the graph on V with edge set [V ]2 E. \ The number of vertices of a graph G is its order, written as G ; its number of | | edges is denoted by e(G) := E(G) . A graph of order 0 or 1 is called trivial. | | Two vertices u, v of G are adjacent, or neighbours, if u, v E(G), and we say { } ∈ that uv is an edge of G; u, v are called the endvertices of the edge uv. A vertex v is incident with an edge e if v is an endvertex of e. The set of neighbours of a vertex v in G is denoted by N (v), or brieﬂy by N(v). More generally for U V (G), the G ⊆ neighbours in V (G) U of vertices in U are called neighbours of U, and their set is \ denoted by N (U). We write N[v] := N(v) v , and N[U] := N(U) U. G ∪ { } ∪ For two subsets X,Y V (G), the set of all edges xy of G, where x X and ⊆ ∈ y Y , is denoted by E(X,Y ), and e(X,Y ) := E(X,Y ) . The set of all edges ∈ | | of G containing a vertex v is denoted by EG(v), or brieﬂy by E(v). The degree d (v) = d(v) of a vertex v is the number E(v) of edges containing v. G | | A vertex of degree 0 is isolated. The number

δ(G) := min d(v) v V { | ∈ } is the minimum degree of G, the number

∆(G) := max d(v) v V { | ∈ } is its maximum degree. The number

1 d(G) := d(v) V | | Xv∈V is the average degree of G. 10

Clearly, we have the relation

δ(G) d(G) ∆(G). ≤ ≤

If all the vertices of G have the same degree k, then G is k-regular, or simply regular. A 3-regular graph is also called cubic.

A set U of vertices in G is dominated by a set W V (G) (and W dominates U) ⊆ if every u U is adjacent to some w W . A set D is a dominating set in G if D ∈ ∈ dominates V D in G. \ A graph G is complete if all vertices of G are pairwise adjacent. A complete graph on n vertices is denoted by Kn. A set of vertices is independent if no two of its elements are adjacent.

Let G = (V,E) and G0 = (V 0,E0) be two graphs. If G0 G, that is V 0 V ⊆ ⊆ and E0 E, then G0 is a subgraph of G. If G0 G and furthermore G0 contains all ⊆ ⊆ the edges xy E(G) with x, y V 0, then G0 is an induced subgraph of G. We say ∈ ∈ that V 0 induces or spans G0 in G and write G0 = G[V ]. Thus if U V is any set of ⊆ vertices, then G[U] denotes the graph on U whose edges are precisely the edges of G with both ends in U.

If U is a set of vertices of G, we write G U for G[V U]. If U = v is a singleton, − \ { } we write G v rather than G v . Instead of G V (G0) we simply write G G0. − − { } − − For a subset F of [V ]2 we write G F := (V,E F ) and G+F := (V,E F ); G e − \ ∪ −{ } and G + e are abbreviated to G e and G + e. { } − We set G G0 := (V V 0,E E0) and G G0 := (V V 0,E E0), where G = (V,E) ∪ ∪ ∪ ∩ ∩ ∩ and G0 = (V 0,E0). If V V 0 = , then G and G0 are disjoint. If graphs G and H ∩ ∅ are disjoint, we denote by G + H the disjoint union of G and H. The join G H of ∨ 11 disjoint graphs G and H is the graph obtained from G + H by joining each vertex of G to each vertex of H.

A path is a non-empty graph P = (V,E) of the form

V = x , x , . . . , x , { 1 2 k}

E = x x , x x , . . . , x − x , { 1 2 2 3 k 1 k} where the xi are all distinct. The number of edges of a path is its length. A path on

1 k vertices is denoted by Pk . We often refer to a path by the sequence of its vertices, writing, say, P = x1x2 . . . xk, and calling P a path from x1 to xk (or between x1 and xk). If P = x x . . . x is a path and k 3, then the graph C := P + x x is called 1 2 k ≥ 1 k a cycle. As with paths, we often denote a cycle by its cyclic sequence of vertices, writing, say, C = x1x2 . . . xkx1. The length of a cycle is its number of edges (or vertices); the cycle of length k is called a k-cycle and denoted by Ck. The minimum length of a cycle in a graph G is the girth g(G) of G; the maximum length of a cycle in G is its circumference. An edge which joins two vertices of a cycle but is not itself an edge of the cycle is a chord of that cycle. Thus, an induced cycle in G is one that has no chords.

The distance d(x, y) of two vertices x, y is the length of a shortest path from x to y in G; if no such path exists, we set d(x, y) := . The greatest distance between ∞ any two vertices in G is the diameter of G, denoted by diam(G).

A non-empty graph is called connected if any two of its vertices are linked by a path in G. A maximal connected subgraph of G is called a component of G.

1 Graph theorists often denote by Pk the path of length k. In this thesis we use Pk to denote the path on k vertices, that is, of length k 1. − 12

G is called k-connected if G > k and G X is connected for every set X V | | − ⊆ with X < k. The greatest integer k such that G is k-connected is the connectivity | | κ(G) of G.

Given a graph F , a graph G is F -free if G does not contain an induced subgraph isomorphic to F .

2.3 Graph Colouring

A vertex colouring of a graph G = (V,E) is a map c : V S. The elements of S are → called the available colours. If, in addition, the colouring satisﬁes that c(v) = c(w) 6 whenever v and w are adjacent, then the colouring is said to be proper.

A vertex colouring c : V 1, . . . , k is called a k-colouring. Note that a k- → { } colouring is nothing but a vertex partition into at most k sets, called colour classes.

The smallest integer k such that G has a proper k-colouring is the chromatic number of G; it is denoted by χ(G). A graph G with χ(G) = k is called k-chromatic; if χ(G) k, we call G k-colourable. ≤ Since a colouring of G is deﬁned to be a mapping (or function) c : V (G) S, → then for a subset X V (G), c(X) naturally denotes the range (co-domain) of X ⊆ under the mapping c; if the range c(X) consists of only one element a, we often simply write c(X) = a instead of c(X) = a . { } Using greedy colouring (or sequential colouring), we can easily obtain that χ(G) ≤ ∆(G) + 1. In this inequality, equality holds for complete graphs and odd cycles. In all other cases, χ(G) is actually bounded above by ∆(G), which is a classic result of

Brooks [10]. 13

Theorem 2.1 ([10]) Let G be a graph which is neither complete nor an odd cycle.

Then

χ(G) ∆(G). ≤

Somewhat interestingly, large chromatic number can occur as a purely global phenomenon (see [26, 28]). The following classic theorem, due to Erd˝os[26], gives evidence that the chromatic number χ(G) cannot be deduced from local considera- tions: locally (around each vertex) a graph of large girth looks just like a tree, and is, in particular, 2-colourable there.

Theorem 2.2 ([26]) For any integers k and `, there exists a graph G with girth g(G) > ` and chromatic number χ(G) > k.

List colouring is a more general version of vertex colouring: the set of colours available at each vertex may be restricted. This model was introduced independently by Vizing and Erd˝os-Rubin-Taylor.

For each vertex v in a graph G, let L(v) denote a list of colours available at v.A list colouring or choice function is a proper colouring f such that f(v) L(v) for all ∈ v. A graph is k-choosable or list k-colourable if every assignment of k-element lists to the vertices permits a proper list colouring. The list chromatic number, choice number, or choosability χ`(G) is the minimum k such that G is list k-colourable.

2.4 Hypergraphs

A hypergraph := (V, ) consists of a vertex set V and a family of subsets of H E E V . The elements e are called hyperedges. A hypergraph is k-uniform if every ∈ E 14 hyperedge has size k.A k-uniform hypergraph is called complete or a clique, if H every k vertices of form a hyperedge. H A cycle of length ` (` 2) in is an alternating sequence v , e , . . . , v , e , v of ≥ H { 1 1 ` ` 1} distinct vertices and hyperedges such that v e e for i = 1, . . . , ` (subscripts i+1 ∈ i ∩ i+1 are taken mod `). The girth of a hypergraph , denoted by g( ), is the length of H H its smallest cycle. If contains no cycle, then g( ) := . H H ∞ A transversal of is a subset T V that intersects each hyperedge. The mini- H ⊆ mum size of a transversal of is the transversal number τ( ) of . An independent H H H set of is a subset S V that contains no hyperedges. The independence number H ⊆ α( ) of is the minimum size of an independent set of . H H H Let = (V, ) be a hypergraph and let k 2 be an integer. A k-colouring2 H E ≥ of the vertices of is a mapping c : V ( ) 1, . . . , k such that every nontrivial H H → { } hyperedge is not monochromatic; that is, c(e) 2 for every e with e > 1. | | ≥ ∈ E | | (The only monochromatic hyperedges are therefore the trivial hyperedges, namely those hyperedges of cardinality one.)

The chromatic number χ( ) of a hypergraph is the smallest integer k for which H H admits a k-colouring. H The chromatic number of a complete k-uniform hypergraph is easy to ﬁnd, see, e.g. [8, p. 117]:

Proposition 2.3 Let be a complete k-uniform hypergraph of order n. Then H n χ( ) = . H k 1 − 2Here, in the context of hypergraphs, by k-colouring of , what we really mean is a “proper” k-colouring. H 15

2.5 Basic Properties of F -avoiding Colouring

In this section we will give the basic properties of acF (G) for various graphs F and G. All of the results in this section are straightforward. We will be using some of them to obtain other propositions in later chapters.

2.5.1 Hypergraph standpoint

It will be convenient to adopt a viewpoint of hypergraphs to deﬁne the F -avoiding chromatic number. Given a graph F , for any graph G we can form a hypergraph

= (G) as follows. The vertex set of is the vertex set V (G) of the graph H HF H G; the hyperedges of are all maximal F -free subgraphs of G. We note that if H F 2, there may be trivial hyperedges in (G), that is, hyperedges consisting of | | ≤ HF only one vertex. The F -avoiding chromatic number acF (G) of the graph G is thus, as can be easily seen, the chromatic number of the so-defined hypergraph (G). HF The definition of hypergraph chromatic number rules out trivial hyperedges. This is consistent with our definition of F -avoiding chromatic number3.

In Chapter 8, we will study the F -avoiding colouring for partially ordered sets.

We don’t explicitly make use of hypergraphs to deﬁne F -avoiding chromatic number for posets. But it is still convenient to keep in mind that it is also the chromatic number of a special hypergraph.

2.5.2 Basic properties

First, let us start with the following very basic observation.

3Indeed this is one of the reasons that we require that each nontrivial F -free subgraph gets more than one colour. 16

Lemma 2.4 Given any graphs F and G, acF (G) = acF (G).

Proof. We ﬁrst show that ac (G) ac (G) for any graph G. For this, we set F ≤ F k := acF (G). Then the vertices of G can be k-coloured such that every maximal F -free subgraph receives at least two colours. It is easy to see that, with the same colouring, each maximal F -free subgraph of G receives at least two colours. For, if H is a maximal F -free subgraph of G, then the vertices of H induce a maximal F -free subgraph in G, thus V (H) is not monochromatic. This shows that ac (G) k, F ≤ namely ac (G) ac (G). F ≤ F Similarly, we have ac (G) ac (G). Thus ac (G) = ac (G). F ≤ F F F Using the notation that we have previously deﬁned, we can rephrase the above proposition as follows.

Corollary 2.5 For any graph F , ac(F ) = ac(F ).

Another immediate observation is that for a given graph F , if a graph G does not contain F as an induced subgraph, then G is itself maximal F -free. Thus we have the following observation.

Lemma 2.6 For a given graph F , if a nontrivial graph G does not contain F as an induced subgraph then acF (G) = 2.

Proof. Since G does not contain F as an induced subgraph, then G contains only one maximal F -free subgraph of G, namely G itself. Thus acF (G) = 2 (any 2-colouring of V (G) will do). 17

Lemma 2.7 Let F be any graph. Then

1 if F = 2,  | |   acF (F ) = 3 if F = 3,  | | 2 otherwise.    Proof. If F = 2, then F = K or K . Thus each maximal F -free subgraph of F | | 2 2 is a single vertex, i.e., all maximal F -free subgraphs of F are trivial. Consequently, acF (F ) = 1. It is easy to see that ac (F ) = 3 if F = 3, because every two vertices in F form F | | a maximal F -free subgraph. Thus acF (F ) = χ(K3) = 3. Now consider the case when F 4. In this case, we colour two vertices with | | ≥ colour 1, and the remaining vertices with colour 2. It is easy to see that neither colour class contains a maximal F -free subgraph of F . Thus acF (F ) = χ(K3) = 2.

The next result concerns a relationship between F -avoiding chromatic number for diﬀerent graphs F .

Lemma 2.8 Let F1 and F2 be two graphs. For any graph G, if each maximal F1-free subgraph of G contains a maximal F -free subgraph of G, then ac (G) ac (G). 2 F1 ≤ F2

Proof. Set k := acF2 (G). Then G can be coloured with k colours such that each maximal F2-free subgraph receives at least two colours. In this same colouring of

G, each maximal F1-free subgraph also receives at least two colours as it contains a maximal F -free subgraph of G. Thus ac (G) k = ac (G) as desired. 2 F1 ≤ F2 18

Remark. In Chapter 8, using the idea behind this lemma, we will generalize the results of Lonc-Rival and Duﬀus-Kierstead-Trotter to the k-element antichain and chain, for arbitrary k 2. ≥

Lemma 2.9 Let F be a connected graph. If the graph G is disconnected, then acF (G) = 2.

Proof. Let G ,...,G (k 2) denote the connected components of G. We colour G 1 k ≥ 1 with colour 1, and G2 through Gk with colour 2. Then each maximal F -free subgraph of G will receive at least two colours. To see this, we let H be a maximal F -free subgraph of G. If H is coloured 1, then H is contained in G . For any x V (G ), 1 ∈ 2 we claim that H x is still F -free. For, if there were any copy of F contained in ∪ { } H x , it must involve x, contradicting the fact that F is connected. Similarly H ∪ { } cannot be coloured 2. Thus acF (G) = 2.

Example 2.1 Let F = K3, and let G be a disjoint union of at least two Kk’s. Then acK3 (G) = 2 while acK3 (Kk) = k.

Figure 2.1: A disjoint union of Kk’s.

Proof. Since in a Kk, each edge is a maximal K3-free subgraph of the Kk, acK3 (Kk) =

χ(Kk) = k. 19

On the other hand, making use of Lemma 2.9, we know that acK3 (G) = 2.

Remark. Using the same idea, for any graph F with a dominating vertex, we can construct a disconnected graph G with acF (G) = 2 but acF (Gi) arbitrarily large for

0 0 all connected components Gi of G. To begin, we take a graph G with acF (G ) = k

(F has a dominating vertex, so such a graph exists, see Chapter 4), and take Gi’s to

0 be copies of G . Let G be a disjoint union of Gi’s, then acF (G)=2, and acF (Gi) = k for each component Gi. Thus the parameter acF (G) can be 2 although G contains very complex substructure (for example, a substructure which has acF (Gi) arbitrarily large). There will be more examples in the later chapters showing this phenomenon.

Proposition 2.10 ac(K ) = (r 2), and thus equivalently, ac(K ) = . r ∞ ≥ r ∞

Proof. We will ﬁrst show that ac(K ) = . 2 ∞ This can be justiﬁed by considering triangle-free graphs. A maximal clique in a triangle-free graph must be an edge, thus acK2 (G) = χ(G). Moreover triangle-free graphs can have arbitrarily large chromatic numbers [27]. Hence acK2 (G) is large, G being any triangle-free graph with large chromatic number.

n Now let us assume that r 3. We will show that ac r (K ) = . In the graph ≥ K n d r−2 e K , any r 1 vertices of K form a maximal K -free subgraph. If G is coloured with n − n r less than n colours, then there must be a colour class containing at least r 1 d r−2 e − vertices which constitute a maximal Kr-free subgraph of Kn, a contradiction. Thus

n ac r (K ) . K n ≥ d r−2 e On the other hand, K can be coloured with n colours such that each colour n d r−2 e class has at most r 2 vertices. With this colouring, each maximal K -free subgraph − r of Kn receives more than one colour. 20

n Thus ac r (K ) = which can be arbitrarily large as n tends to inﬁnity. K n d r−2 e Consequently ac(K ) = . r ∞ Chapter 3

Clique Colouring

In this chapter, we shall focus on clique colouring. A k-clique-colouring of a graph

G is a vertex-colouring of G with k colours such that each nontrivial maximal clique

1 of G receives more than one colour. The clique chromatic number acK2 (G) of G is the least integer k for which G admits a k-clique-colouring.

As was pointed out in [41], in general, clique colouring is harder than ordinary vertex colouring: it is coNP-complete even when restricted to checking whether a vertex 2-colouring is a 2-clique-colouring [5]; it is NP-hard to decide whether a perfect graph is 2-clique-colourable [41].

In Sections 3.1–3.3, we survey known results about clique colouring. In Section

3.4, as was mentioned in Chapter 1, we prove a conjecture raised by Gallai.

3.1 Constant Bounds

In this section we will survey the known bounds of the clique chromatic number for various graphs G. It is worth pointing out that the following classes of graphs are all hereditary in that any induced subgraph of a graph G is still in the class if G is.

1To date, there is no standard notation for the clique chromatic number. The authors of [5] denote the clique chromatic number by κ(G), which is often used to denote the connectivity of a graph G in the literature. The clique chromatic number, being a special case of the more general

F -avoiding chromatic number, is naturally denoted by acK2 (G) throughout this thesis.

21 22

Planar graphs

B. Mohar and R. Skrekovskiˇ [42] proved an upper bound of 3 for the clique chromatic number for planar graphs. Their method was based on an extension of Gr¨otzsch’s celebrated theorem [34] that any triangle-free planar graph is 3-colourable.

Theorem 3.1 ([42]) ac (G) 3 for any planar graph G. Furthermore, there is a K2 ≤ polynomial time algorithm for ﬁnding such a 3-colouring.

The authors of [42] also considered the list-colouring version of the clique colouring problem. They proved that any planar graph G is 4-list-clique-colourable; moreover, the same result holds for the class of locally planar graphs on arbitrary surfaces.

P4-free graphs

In [36], Ho´angand McDiarmid considered a slightly diﬀerent colouring—what is referred to by them as “strong division”. Among other results, they proved:

Theorem 3.2 ([36]) Every P4-free graph G is such that acK2 (G) = 2.

(P5,C5)-free graphs

A graph is said to be (P5,C5)-free if it does not contain P5 or C5 as an induced subgraph. In [33], Graver and Skrekovskiˇ studied the chromatic number for (P5,C5)- free graphs, and proved the following result. For an easier proof, the reader is referred to [20].

Theorem 3.3 ([33]) Every (P5,C5)-free graph G is such that acK2 (G) = 2. 23

Co-paw-free graphs

2 The graph P3 + K1 is called a co-paw . Co-paw-free graphs have been studied in [33] (see also [20]).

Theorem 3.4 ([33]) For any co-paw-free graph G = C , ac (G) = 2. 6 5 K2

Diamond-free graphs

D´efossez[20] investigated diamond-free graphs. A graph is called a diamond if it is

K4 minus an edge. An edge is called ﬂat if it does not lie in a triangle. A graph of odd order which is at least 5, containing a Hamiltonian cycle none of whose edges is

ﬂat, is said to be a bad cycle. The following bounds were proven.

Theorem 3.5 ([20]) For all (diamond, bad-cycle)-free graphs G, acK2 (G) = 2.

The following result was proved by Bacs´oet al in [5].

Theorem 3.6 ([5]) For all (diamond, ﬂat-edge)-free perfect graphs G, acK2 (G) = 2.

An odd hole in a graph G is an induced odd cycle of length at least 5. It is known

[46, 51] that a diamond-free graph is perfect if and only if it does not contain odd holes. D´efossezproved the following.

Theorem 3.7 ([20]) For all (diamond, odd hole)-free graphs G, ac (G) 4. K2 ≤ 2Its complement is called a paw. 24

(Bull, odd hole)-free graphs

A bull is the graph with vertices a, b, c, d, e and edges ab, bc, be, cd and ce. D´efossez

[20] proved the following result.

Theorem 3.8 ([20]) For all (bull, odd hole)-free graphs G, acK2 (G) = 2.

Furthermore, in [20], the author also gave a polynomial time algorithm for ﬁnding such a 2-colouring for (bull, odd hole)-free graphs.

Claw-free graphs

A claw is the graph K1,3. Chv´ataland Sbihi [18] gave a structural decomposition result on claw-free perfect graphs. Making use of their result, Bacs´oet al. [5] found the exact value of acK2 (G) for claw-free perfect graphs G.

Theorem 3.9 ([5]) For all claw-free perfect graphs G, acK2 (G) = 2.

Generalized split graphs

3 In [5], Bacs´oet al. determined acK2 (G) for all generalized split graphs .

Theorem 3.10 ([5]) For all generalized split graphs G, acK2 (G) = 2.

3.2 General Bounds

In the last section, constant bounds were given for acK2 (G) for various graphs G. On the other hand, bounding the clique chromatic number in terms of various graph invariants has been investigated by Bacs´oet al. [5].

3For the deﬁnition of a generalized split graph, cf. Chapter 6. 25

Theorem 3.11 ([5]) ac (G) γ(G)+1, where γ(G) denotes the domination num- K2 ≤ ber of G.

Bounding the clique chromatic number of a graph G in terms of the order n of G has also been studied by various authors. Kotlov proved in an unpublished manuscript (see [5]) that ac (G) √2n . In [5], a weaker bound was proved. K2 ≤ d e

Theorem 3.12 ([5]) ac (G) 2 √n . K2 ≤ d e

3.3 Complexity

In general, the clique colouring problem is NP-hard [5].

Instance: Graph G = (V,E) and integer k.

Question: Is ac (G) k? K2 ≤

In fact, much stronger results are true for 2-colourability. Kratochv´ıland Tuza

[37] showed that the following problem is NP-complete.

Instance: A perfect graph G with clique number ω(G) 3. ≤

Question: Is acK2 (G) = 2?

Bacs´oel al. [5] also showed that the following problem is NP-complete.

Instance: Graph G with ∆(G) 3. ≤

Question: acK2 (G) = 2? 26

3.4 Tree-decompositions and Chordal Graphs

It is natural to ask about the clique chromatic number for chordal graphs as maximal cliques in a chordal graph are well understood. It seems that, to date, the clique chromatic number for chordal graphs has not appeared in the literature. In this section we shall, using a tree-decomposition technique, show that acK2 (G) = 2 for any chordal graph G.

3.4.1 Tree-decompositions

In the proof of the monumental Graph Minor Theorem, due to N. Robertson and

P. Seymour in a series of papers starting from [48], which “dwarfs any other result in graph theory and may doubtless be counted among the deepest theorems that mathematics has to oﬀer,”4 the technique of tree-decomposition plays a crucial role. Roughly speaking, tree-decomposition (and the concept of tree-width) permits generalizations of tree properties to general tree-like graphs, see [49].

Let G be a graph, T a tree, and let = (V ) ∈ be a family of vertex sets V t t T V V (G) indexed by the vertices t of T . We follow the interpretation in [21]. t ⊆ The pair (T, ) is called a tree-decomposition of G if it satisﬁes the following three V conditions:

(T1) V (G) = t∈T Vt; S (T2) for every edge e E(G) there exists a t T such that both ends of e lie in V ; ∈ ∈ t

(T3) V V V whenever t , t , t T satisﬁes t t T t (the path from t to t1 ∩ t3 ⊆ t2 1 2 3 ∈ 2 ∈ 1 3 1

t3 in T ). 4Quoted from [21, Chap. 12] 27

Figure 3.1: a tree-decomposition (T, ) = (T, (V ) ∈ ) V t t T

We call these subgraphs G[V ] and the sets V themselves the parts of (T, ) and t t V say that (T, ) is a tree-decomposition of G into these parts. For an example, see V Figure 3.1.

One of the most important features of a tree-decomposition is that it transfers the separation properties of its tree to the graph decomposed 5:

Proposition 3.13 ([21]) Let t1t2 be any edge of T and let T1, T2 be the components of T t t , with t T and t T . Then V V separates U := V from − 1 2 1 ∈ 1 2 ∈ 2 t1 ∩ t2 1 t∈T1 t S U2 := t∈T2 Vt in G. 5 S For A, B V (G), a subset X V (G) of vertices separates A and B in G if every path from A to B contains⊆ a vertex from X.⊆ More generally we say that X is a separating set in G if X separates two vertices of G X in G. − 28

Another easy fact is that tree-decompositions are passed on to subgraphs:

Proposition 3.14 ([21]) If (T, (Vt)t∈T ) is a tree-decomposition of G, and H is a subgraph of G, then the pair (T, (V V (H)) ∈ ) is a tree-decomposition of H. t ∩ t T

3.4.2 Chordal graphs

The following theorem deals with the K2-avoiding chromatic number acK2 (G) of chordal graphs G. A graph is chordal (or triangulated) if each of its cycles of length at least 4 has a chord, i.e., it contains no induced cycles other than triangles.

Theorem 3.15 Let G be a chordal graph. Then acK2 (G) = 2.

We will use mathematical induction to prove this result. To make our argument go through, we will need a structural result for chordal graphs, as speciﬁed below.

Chordal graphs have well understood structure. Their maximal cliques are nicely characterized in the setting of tree-decomposition. Tree-decomposition thus leads to a very simple structural characterization of chordal graphs.

Proposition 3.16 ([21]) G is chordal if and only if G has a tree-decomposition into complete parts.

Proof of Theorem 3.15. By Proposition 3.16, we may ﬁrst assume that G has a tree-decomposition (T, ) such that G[V ] is complete for every t T . We apply V t ∈ induction on T . If T 1, then G is complete and hence ac (G) = 2. So let | | | | ≤ K2 T 2, and let t be a leaf of T . Denote by t the only neighbour of t in T , | | ≥ 1 2 1 and U := V V . Deﬁne G := G[V U], and G := G G , as depicted in t1 ∩ t2 1 t1 \ 2 − 1 0 0 Figure 3.2. Let T := T t . Since (T , (V ) ∈ 0 ) is a tree-decomposition of G − 1 t t T 2 29

PSfrag replacements U G1

Figure 3.2: a tree-decomposition of G

into complete parts, Proposition 3.16 yields that G2 is chordal. By the induction

0 0 hypothesis acK2 (G2) = 2 as (T , (Vt)t∈T ) is a tree-decomposition of G2 into fewer parts. Hence there is a colouring c : V (G ) [2] which assigns two colours to each 1 2 →

maximal clique of G2. Now we colour the vertices of G1 with a colour so that G[Vt1 ] is two coloured. That is, if G[U] is coloured i (1 i 2), then we colour G with ≤ ≤ 1

i + 1 (addition is modulo 2); and if G[U] is two coloured, then colour G1 arbitrarily. Now let us show that, with this colouring, any maximal clique receives two

colours. Let H be a maximal clique of G. If H G , then H must be a maxi- ⊆ 2

mal clique of G2, and hence is two coloured. Suppose H contains vertices of G1. By

Proposition 3.13, V V separates V = V (G U) from 0 V = V (G G ) in t1 ∩ t2 t1 1 ∪ t∈T t − 1 S G. Since H is a clique, H cannot contain vertices of V U, otherwise H would have t2 \ a cutset V V smaller than itself, a contradiction. Thus H G[V ] which is a t1 ∩ t2 ⊆ t1 clique. It follows that H = G[Vt1 ] and hence is two coloured.

On page 2 we mentioned that Gallai asked whether a chordal graph G on n

vertices has a clique-transversal of size at most n/2. The above theorem answers 30 this question in the aﬃrmative: the smaller of the two colour classes has size at most n/2 and meets every maximal clique of G. Chapter 4

A Complete Characterization

In this chapter, we give necessary and suﬃcient conditions to ensure that the F - avoiding chromatic number, ac(F ), is bounded. In Sections 4.1 and 4.2 we charac- terize the graphs F for which ac(F ) is bounded. Precisely, we will prove the following theorem.

Theorem 4.1 Let F be a graph.

(i) If F has neither an isolated vertex nor a dominating vertex, then ac(F ) 3. ≤ (ii) If F has either an isolated vertex or a dominating vertex, then ac(F ) = + . ∞

In Section 4.3 we also give upper bounds for acF (G), F being a graph with either an isolated vertex or a dominating vertex, in terms of various graph invariants of G.

4.1 Bounded Avoiding Chromatic Numbers

It is relatively easy to prove that when F has neither an isolated vertex nor a dominating vertex, ac(F ) 3. In doing so, we will show that for any graph G there ≤ is always a certain vertex v V (G), and a partition of V (G) into partition classes ∈ (colour classes) such that we can add v to any monochromatic F -free subgraph of

G without it resulting in any induced copy of F . This property yields the desired colouring.

31 32

Proof of Theorem 4.1 (i). We show that ac (G) 3 for every graph G. Let G be a F ≤ graph. We choose an arbitrary vertex v V (G), and colour the vertex v with colour ∈ 1, the neighbours N(v) of v with colour 2, and the set M(v) := V (G) N[v] of all \ other vertices with colour 3. Formally, the colouring c is deﬁned by

1 if x = v,   c(x) := 2 if x N(v),  ∈ 3 if x M(v).  ∈   Note that the condition on F in statement (i) of the theorem means that F = K 6 2 or K , thus F > 2. Then every maximal F -free subgraph of G has at least two 2 | | vertices1. Now we claim that any maximal F -free subgraph receives at least two colours. Suppose H is a maximal F -free subgraph of G. Colour class 1 consists of one vertex, thus H is not contained in it. If H is coloured 2, then H N(v). ⊆ H v is still F -free, as if there were any copy of F in H v , it must involve ∪ { } ∪ { } v, and thus would contain a dominating vertex v, contradicting the fact that F contains no dominating vertex. Consequently colour class 2 contains no maximal

F -free subgraph of G. Similarly, if H is coloured 3, then H M(x). H v ⊆ ∪ { } is still F -free, as if there were any copy of F in H v , it must involve v, and ∪ { } thus would contain an isolated vertex v, contradicting the fact that F contains no isolated vertex. Hence colour class 3 contains no maximal F -free subgraph of G.

This completes the proof of part (i) of the theorem.

Remark. The bound 3 is tight, as acP4 (C5) = 3 by Example 5.3.

1We are, of course, assuming that G is not trivial, i.e. G > 1. | | 33

4.2 Unbounded Avoiding Chromatic Numbers

In the last section we have bounded ac(F ) whenever F is a graph with no isolated vertices or dominating vertices. In this section, we will prove that for the remaining cases of graphs F , ac(F ) is unbounded.

To prove the theorem, we need some preliminary results. For one thing, our argument relies on the celebrated theorem of Erd˝osand Hajnal [29]. The theorem states that, roughly speaking, there are sparse hypergraphs with high chromatic number.

Erd˝os-Hajnal Theorem For any integers k 2, ` 2, and r 1, there exists a ≥ ≥ ≥ k-uniform hypergraph = (X, ) with the following properties: H E

1. g( ) `; H ≥

2. χ( ) > r. H

Secondly, we will need a technique introduced by Neˇsetˇriland Rödl[43]. Basically it says that we can derive an ordinary graph from a given n-uniform hypergraph H by replacing hyperedges of with graphs on n vertices. H Neˇsetˇril-RödlConstruction Let H be a graph of order n. Given an n-uniform hypergraph = (V, ), for each hyperedge e , let f : V (H) e be a fixed H E ∈ E e → bijection. Define the ordinary graph G = (V,E(G)) such that xy E(G) iff there ∈ exist e and zt E(H) such that f (z), f (t) = x, y . ∈ E ∈ { e e } { } This graph G is called the amalgamation of and H, and will be denoted by H G = (V, ) H or G = H. E ∗ H ∗ 34

Let F be a graph with either an isolated vertex or a dominating vertex. We will take advantage of the Erd˝os-Hajnal Theorem to construct graphs G with large

F -avoiding chromatic number. In doing so, we will apply the Neˇsetˇril-R¨odlCon- struction to replace hyperedges of a given hypergraph , which has large girth and H large chromatic number, with copies of an appropriately chosen graph H.

Since for any given graph F , the identity ac(F ) = ac(F ) holds, to show that ac(F ) is unbounded when F has either an isolated vertex or a dominating vertex, we may restrict our attention to the case when F has an isolated vertex.

Henceforth we will assume that F has an isolated vertex. And thus we can rewrite F as the disjoint union of an (arbitrary) graph F 0 and an isolated vertex,

0 i.e. F = F + K1. The following lemma develops the properties of the graph H that we will use to replace hyperedges of . H

Lemma 4.2 Let F = F 0 + K be a graph and H = F 0 F 0 the join of two copies of 1 ∨ F 0. Then

(i) H is F -free;

(ii) if we add in a new vertex v to H such that e(v, V (H)) 1, then H v ≤ ∪ { } contains F as an induced subgraph.

Proof. For clarity, let F and F be two copies of the graph F 0, and let H = F F , as 1 2 1 ∨ 2 depicted in Figure 4.1. It is easy to see that H is F -free. For, if H contains an induced copy of F , say F˜, then F˜ cannot lie in either V (F1) or V (F2), i.e. V (F˜) * V (F1) and V (F˜) * V (F2). Note that the graph induced by the edges between F1 and F2 35 is a complete bipartite graph, so any induced subgraph of H that contains vertices from both F1 and F2 is connected, and thus is not isomorphic to F . Now let us show that (ii) holds. If v is nonadjacent to any vertex of H, then obviously H v contains a copy of F = F 0 + K — for example, F v . If V ∪ { } 1 1 ∪ { } is adjacent to precisely one vertex y V (H), without loss of generality we assume ∈ y F , then v is nonadjacent to any vertex of F , and hence F v is a copy of ∈ 1 2 2 ∪ { } 0 F = F + K1.

PSfrag replacements y F1 F2

Figure 4.1: F F and an edge 1 ∨ 2

We now utilize the aforementioned Erd˝os-Hajnal Theorem and the Neˇsetˇril-R¨odl

Construction to construct graphs with large avoiding chromatic numbers.

Note that there have been constructive arguments found for the Erd˝os-Hajnal

Theorem, see e.g. [39, 44]. Hence our argument is actually constructive, that is, given a graph F with an isolated vertex, for any integer r, the proof can give a construction of G so acF (G) > r.

Proof of Theorem 4.1 (ii). Suppose k := F 0 is the order of F 0. As earlier mentioned, | | using Lemma 2.4 that acF (G) = acF (G), without loss of generality we may assume

0 that F has an isolated vertex. Thus we write F = F + K1. By the Erd˝os-Hajnal Theorem, there is a 2k-uniform hypergraph = (V, ) with H E 36 g( ) > 3 and χ( ) > r. We note that in , no pair of vertices lies in more than one H H H hyperedge as we have g( ) > 3, and thus every two hyperedges of can intersect H H in at most one vertex.

We will place on each hyperedge e an appropriate graph H(e) (deﬁned below) ∈ E as guaranteed by the Neˇsetˇril-R¨odlConstruction, and will show that in the resulting graph G, each such graph H(e) resulting from the hyperedge e is a maximal ∈ E F -free subgraph of G. This graph G will have large F -avoiding chromatic number.

Let H := F 0 F 0. Using the Neˇsetˇril-R¨odlConstruction, we replace each hyper- ∨ edge e with a copy H(e) of H, and thus get a graph G in which two vertices x, ∈ E y are adjacent if and only if they are adjacent in some copy H(e), e . ∈ E Since each pair of vertices of lies in no more than one hyperedge, then in the H vertex set V (H(e)) of each copy H(e), e , there won’t be any additional edges ∈ E resulting from copies of H on other hyperedges, and hence V (H(e)) indeed induces an induced copy of H in G.

Now we show that each such copy H(e) of H = F 0 F 0, e , is a maximal ∨ ∈ E F -free subgraph of G. For this, we need the following result.

Fact. For each hyperedge e ( ), and for any v V e, we have ∈ E H ∈ \

e (v, H(e)) 1. G ≤

Recall that in each pair of vertices lies in no more than one hyperedge. It H follows that each edge of the resulting graph G lies within precisely one hyperedge of . If there were two edges from v to H(e), then there would be two hyperedges H e (i = 1, 2) from v to v e respectively (i = 1, 2). This would yield a 3-cycle i i ∈ 37 v e ve v ev in the hypergraph , as depicted in Figure 4.2, contradicting the fact 1 1 2 2 1 H that g( ) > 3. H

PSfrag replacements e1 v1 e v

v2 e2

Figure 4.2: 3-cycle v1e1ve2v2ev1

Thus, using Lemma 4.2, for each e , since e (v, H(e)) 1 for any v V ( ) e, ∈ E H ≤ ∈ H \ H(e) = F 0 F 0 is a maximal F -free subgraph of G. ∨ Now we show that ac (G) χ( ). F ≥ H Let c be a vertex-colouring of G which assigns at least two colours to each maximal

F -free subgraph of G. Since each hyperedge e ( ) of results in a maximal ∈ E H H F -free subgraph of G, then c assigns at least two colours to the vertices of the hyperedge. This implies that the colouring c is indeed a proper colouring of the hypergraph . Therefore ac (G) χ( ). H F ≥ H Since χ( ) > r by our construction, we have H

ac(F ) ac (G) χ( ) > r, ≥ F ≥ H as desired. 38

4.3 General Bounds

In this section we will bound acF (G) in terms of invariants of G for certain graphs

F . First, using a variant of sequential colouring, we can bound acF (G) by roughly the independent domination number of G, when F has an isolated vertex. The

2 independent domination number of a graph G, denoted by γI(G), is the size of a smallest independent dominating set of G.

Theorem 4.3 Let F be a graph with an isolated vertex. For every graph G,

(i) ac (G) γ (G) + 2, if F is not an independent set; F ≤ I (ii) ac (G) γ (G) + 1, if F is an independent set. F ≤ I

Proof. We ﬁrst partition V (G) into subsets, and then deﬁne the colouring on the subsets. Suppose γ (G) =: k, and D := x , x , . . . , x is a minimum independent I { 1 2 k} dominating set of k vertices in G. We recursively construct subgraph Gi as follows:

1. Deﬁne G0 := G;

2. When G − is defined, set G := G − N i−1 [x ]; i 1 i i 1 − G i With subgraphs G defined as above, we define the colouring c : V (G) i → 1, . . . , k + 2 as follows. { } Case 1. F is not an independent set.

For vertices not in D, deﬁne3

c(NGi−1 (xi)) = i, for i = 1, . . . , k.

2There is no standard notation for independent domination number. We follow [22] to denote by γI (G) the independent domination number of a graph G. 3As mentioned in Chapter 2, for a subset X V (G), if the range c(X) consists of only one element a, we often simply write c(X) = a instead⊆ of c(X) = a . { } 39

For the vertices belonging to D, deﬁne, for j = 1, . . . , k,

k + 1 if j = k;  6 c(xj) =   k + 2 if j = k.  

PSfrag replacements 1

x1 x2 xk−1 xk k + 1 k + 1 k + 1 k + 2

Figure 4.3: colouring scheme

The colouring is indicated in Figure 4.3. Now we proceed to show that each maximal F -free subgraph of G receives at least two colours.

Let H be a maximal F -free subgraph of G. If H is coloured ` where 1 ` k, ≤ ≤ then H x is still F -free. For, if H x contains an induced copy of F , it ∪ { `} ∪ { `} must involve x`, since H itself is F -free. This contradicts the fact that F contains an isolated vertex, since x is adjacent to all vertices in V (H) N − (x ). ` ⊆ Gi 1 i

If H is coloured k + 1, then H x , . . . , x − and thus must in fact be an inde- ⊆ { 1 k 1} pendent set. Since F is not an independent set, H x is still F -free, contradicting ∪{ k} the maximality of H.

Since the colour class k + 2 consists of only one vertex, it does not contain H

( H > 1). | | 40

This shows that any maximal F -free subgraph of G receives at least two colours in the colouring c. Thus ac (G) k + 2 = γ (G) + 2. F ≤ I Case 2. F is an independent set.

We first assume that the graph G does not have isolated vertices, that is, δ(G) ≥ 1. With subgraphs G defined as above, the colouring c : V (G) [k+1] is as follows. i → For vertices not in D, define

c(NGi−1 (xi)) = i, for i = 1, . . . , k, and colour all the vertices belonging to D with colour k + 1, i.e., for i = 1, . . . , k,

c(xi) = k + 1.

Now we proceed to show that each maximal F -free subgraph of G receives at least two colours.

Let H be a maximal F -free subgraph of G with H > 1. If H is monochromatic, | | and c(H) = k+1, then V (H) D, and thus H consists of pairwise non-adjacent ver- ⊆ tices. Let x be a vertex of H, and y N (x ), where N (x ) = by assumption. h h ∈ G h G h 6 ∅ Thus H y induces a subgraph of G which is the disjoint union of single vertices ∪ { h} and an edge x y . Hence H y induces an F -free subgraph of G, contrary to h h ∪ { h} the maximality of H. If H is coloured `, where 1 ` k, then H x is still ≤ ≤ ∪ { `} F -free. For, if H x contains an induced copy of F , it must involve x , but x ∪ { `} ` ` is adjacent to all vertices of H, contradicting the fact that F is an independent set.

Thus ac (G) k + 1 = γ (G) + 1. F ≤ I 0 Now let us assume that G has isolated vertices x1, x2, . . . , xr. Let G denote the subgraph of G induced by V (G) x , x , . . . , x . Then G0 has no isolated vertices. \{ 1 2 r} 41

0 0 Using the above result, we see that G can be coloured with γI(G ) + 1 colours

0 0 1, 2, . . . , γI(G ) + 1 such that each maximal F -free subgraph of G receives at least

0 two colours. Now colour the vertices x1, x2, . . . , xr with r new colours γI (G ) +

0 2, . . . , γI(G ) + r + 1. Then each maximal F -free subgraph of G receives at least two colours under this colouring. Since γ (G) = γ (G0) + r, we see that ac (G) I I F ≤

γI(G) + 1.

Remarks. (a) If the graph F contains no isolated vertices, then the bound in

Theorem 4.3 need not hold. This can be seen by setting F = K3 and G = Kn. Since any two vertices in Kn form a maximal K3-free subgraph of G, they must be two coloured. Thus acK3 (Kn) = χ(Kn) = n (while γI (Kn) = 1).

(b) In Theorem 4.3 (i), the bound γI (G)+2 is tight, as can be seen by the example acP3 (K2 + K2) = 4 (see Chapter 6) while γI (K2 + K2) = 2; in Theorem 4.3 (ii), the bound γI(G) + 1 is nearly tight, as can be seen by acK3 (Kn) = γI (Kn) = n.

Making use of Theorem 4.3, we can bound acF (G) in terms of the order of G as below.

Theorem 4.4 Let F be a graph with either an isolated vertex or a dominating vertex, and F itself not an independent set or a complete graph. For every graph G of order n, ac (G) 2 √n + 1. F ≤ d e

Proof. Once again, using Lemma 2.4 that acF (G) = acF (G), we may assume that F has an isolated vertex and F is not itself an independent set.

We construct an independent dominating set D = v , . . . , v recursively as { 1 k} follows: 42

(i) choose v V (G) with N(v ) √n; 1 ∈ | 1 | ≥

Note that D can be empty. By construction, we know that D is an independent set of G. Let us ﬁrst assume that D is not the empty set. We consider the subgraph G[N[D]] of G induced by the vertex set N[D]. Note that condition (ii) implies D < √n, as otherwise the graph G would have more than n vertices. | | Since D is an independent dominating set in G[N[D]], and D < √n, Theorem 4.3 | | implies that there is a colouring c of G[N[D]] with D + 2 √n + 1 colours, 1 | | ≤ d e say 1, 2,..., √n + 1 , which assigns at least two colours to each maximal F -free { d e } subgraph of G[N[D]].

On the other hand, in the subgraph G G[N[D]] induced by the vertex set − V N[D], the degree of every vertex is strictly smaller than √n by the construction \ of D. Hence we can properly colour this subgraph with ∆(G G[N[D]])+1 √n − ≤ d e colours, say √n + 2,..., 2 √n + 1 . We refer to this colouring as c . {d e d e } 2 Now combine c and c to get the colouring c : V (G) 1,..., 2 √n + 1 : 1 2 → { d e }

c (x) if x N[D];  1 ∈ c(x) :=   c (x) if x V N[D]. 2 ∈ \   Let H be a maximal F -free subgraph of G. If H G[N[D]], then H must be ⊆ a maximal F -free subgraph of the graph G[N[D]] and thus receives at least two 43 colours in the colouring c . If H meets both G[N[D]] and G G[N[D]], then H is 1 − obviously at least two coloured as G[N[D]] and G G[N[D]] receive diﬀerent sets − of colours. Now suppose that H G G[N[D]]. If H is not an independent set, ⊆ − then H receives at least two colours as it contains at least one edge, which is not monochromatic under the proper colouring c2. If H is an independent set, then H D is still an independent set as D is independent and H, D are independent ∪ of each other—there are no edges between D and V N[D]. Consequently H D is \ ∪ still F -free as F is not an independent set, contradicting the maximality of H as D is nonempty.

Now let us suppose that D = . By the deﬁnition of D, we see that ∆(G) < √n. ∅

Choose any vertex v1, and colour v1 with colour 1, N(v1) with colour 2. In the subgraph G N[v ] induced by V N[v ], the degree of every vertex is strictly smaller − 1 \ 1 than √n, so we can properly colour this subgraph with ∆ + 1 √n colours, say ≤ d e 3,..., √n + 2 . { d e } In this colouring each maximal F -free subgraph receives at least two colours. To see this, let H be a maximal F -free subgraph, and suppose that H is monochromatic.

H cannot be coloured 1 since F 3 implies that H 2. If H is coloured 2, then | | ≥ | | ≥ V (H) N(v ). Then H v is still F -free as if there were a copy of F contained ⊆ 1 ∪ { 1} in H v , it must involve v , since H itself is F -free. This contradicts the fact ∪ { 1} 1 that F contains an isolated vertex as v is adjacent to all vertices of H G[N(v )]. 1 ⊆ 1 Now let us assume that H is not coloured 1 or 2, hence V (H) V N[v ]. If H ⊆ \ 1 is not an independent set, then H is two coloured as H contains at least one edge which is not monochromatic by the deﬁnition of the colouring. If H is an independent set, then H v is still an independent set as there are no edges between v and ∪ { 1} 1 44

V N[v ], and thus H v is still F -free, contradicting the maximality of H. \ 1 ∪ { 1} Remark. If F is an independent set or a clique, then the above bound need not hold. This can be seen by the aforementioned equalities acK3 (Kn) = χ(Kn) = n and acK3 (Kn) = χ(Kn) = n. Let us add a few comments at the end of this section. We have proven upper bounds for acF (G), for a given graph F with an isolated vertex, in terms of the minimal cardinality of an independent dominating set and the order of G. It is easily seen that the bounds γ (G) + 2 and 2 √n + 1 do not imply one another. I d e

Indeed, these parameters can be arbitrarily far apart: for example, for the path Pn,

γI(Pn) is roughly n/3; while for a clique Kn, γI(Kn) = 1. A natural question here is whether the bound O(√n) in Theorem 4.4 can be improved to the asymptotically better bound o(√n). The answer to this question is still unknown even for the simplest graph F = P3.

4.4 Transversals

The F -avoiding colouring which we are mainly concerned with is motivated by the clique-colouring problem, which was, in turn, motivated by the clique-transversal problem, as earlier mentioned in Chapter 1.

It will be convenient for us to utilize the language of hypergraphs. Given graphs

F and G, denote by (G) the hypergraph (V (G), ) whose vertex set is V (G) and HF E whose hyperedges are the maximal F -free subgraphs of G. We will write

τ (G) := τ( (G)) F HF 45 for the transversal number of (G). Thus, in our notation, the clique-transversal HF number of a graph G is τK2 (G).

Remark. Notice that the hypergraphs (G)) and (G) have the same vertex HF HF set and the same hyperedges. Thus (G) = (G), and hence τ( (G)) = HF HF HF τ( (G)), i.e. τ (G) = τ (G). We will be using this identity frequently to simplify HF F F our arguments.

In [30], Erd˝oset al. proved that, among other results, there are graphs G on n vertices such that τ (G) n o(n), as n . On the other hand, the following K2 ≥ − → ∞ upper bound was also proved.

Theorem 4.5 Let G be a graph on n vertices. Then τ (G) n √2n + 3 . K2 ≤ − 2 Moreover, they raised the following question (cf. problems4 2 and 4 in [30]):

Problem 1. For which classes of graphs , is the clique-transversal number τ (G) G K2 ≤ λ G , for some absolute constant λ < 1, and for all G ? | | ∈ G This problem has received considerable attention ([2, 4, 5, 24, 38, 52]) and has been studied for various classes of graphs, including chordal graphs, line graphs and their complements, as well as comparability graphs and cocomparability graphs.

Nevertheless, until the present there was no result concerning τF (G) for graphs F other than K2.

In this section, we will consider the behaviour of τF (G) for general graphs F . The general problem that we are interested in is whether τF (G) is bounded by a constant fraction of G . | | 4The precise form of their problem 2 is this: Suppose that each clique of G has at least k = k(n) vertices. Which value of k(n) insures that τK (G) is less than (1 c)n (for some absolute constant 2 − c), or is o(n) or O(nα) for a given α, 0 < α < 1? 46

Problem 2. For which graphs F does there exist a constant λ < 1, depending only on F , such that τ (G) λ G for any graph G? F ≤ | |

For example, the aforementioned result from [30] shows that τK2 (G) is not bounded above by a constant fraction of G . | | In this section, we shall give a complete characterization of graphs F for which

τ (G) is bounded by a fraction of G . To be precise, we shall prove the following F | | theorem.

Theorem 4.6 Let F be a graph.

(i) If F contains neither an isolated vertex nor a dominating vertex, then τ (G) F ≤ 1 ( G + 1) for any graph G; 2 | | (ii) if F contains either an isolated vertex or a dominating vertex, then there exist

graphs G of order n such that τ (G) = n o(n), n . F − → ∞

For F with either an isolated vertex or a dominating vertex, we shall also prove the following general bound.

Theorem 4.7 Let F be a graph with either an isolated vertex or a dominating vertex, and F itself not an independent set or a complete graph. For every graph G of order n, τ (G) n 1 √n + 1 . F ≤ − 2 2

Our approach for both Theorem 4.6 and 4.7 is quite diﬀerent from that used by

Erd˝oset al. It should be pointed out that Theorem 4.7 does not imply the bound in

Theorem 4.5. 47

The simplest form of Theorem 4.6 (ii), namely the unboundedness of clique- transversal number, admits a direct construction not involving hypergraphs. More- over, the construction guarantees the stronger property that each maximal clique has at least nc/ log log n vertices. The interested reader is referred to [30].

4.4.1 Proof of Theorem 4.6

Proof of (i). Let F be a graph with neither an isolated vertex nor a dominating vertex, and let G be a graph. We choose an arbitrary vertex v V (G), and consider ∈ the partition V = v N(v) M(v), where M(v) := V (G) N[v]. { } ∪ ∪ − From the proof of Theorem 4.1 (i), we know that any maximal F -free subgraph of G meets at least two partition classes. Thus both N(v) v and M(v) v are ∪ { } ∪ { } transversals of (G). HF Since N(v) + M(v) = G 1, we see that either | | | | | | − G 1 N(v) | | − | | ≤ 2 or G 1 M(v) | | − . | | ≤ 2 Hence

τ (G) min N(v) + 1, M(v) + 1 F ≤ {| | | | } G + 1 | | , ≤ 2 as desired.

Remark. We note that the above bound ( G + 1)/2 is tight. It is easily seen that | | τ (C ) = 3 = ( C + 1)/2 (Example 5.3), where P denotes the path on 4 vertices. P4 5 | 5| 4 48

To prove Theorem 4.6 (ii), we will need the following theorem, due to Erd˝osand

Hajnal [29], that there are sparse hypergraphs with small independence number.

Erd˝os-Hajnal Theorem II For any integers k 3, ` 2, there exists a real ≥ ≥ number > 0 and an integer N, such that for every n > N, there exists a k-uniform hypergraph = (V, ) on n vertices with the following properties: H E

1. g( ) `; H ≥

2. α( ) < n1−. H

Proof of (ii). As in the proof of Theorem 4.1 (ii), using the fact that τF (G) = τF (G),

0 we may assume that F contains an isolated vertex, and hence write F as F + K1. Suppose k := F 0 is the order of F 0. By the Erd˝os-Hajnal Theorem, there is a | | 2k-uniform hypergraph = (V, ) on n vertices with g( ) > 3 and α( ) < n1−. H E H H We will place on each hyperedge e a graph H(e) := F 0 F 0. Using the ∈ E ∨ argument essentially identical to that of Theorem 4.1 (ii), we see that H(e) = F 0 F 0 ∨ is a maximal F -free subgraph of G.

Now we show that τ (G) n o(n). F ≥ − Let T be a subset of V (G) which meets every maximal F -free subgraph of G.

Since each hyperedge e ( ) of results in a maximal F -free subgraph of G, it ∈ E H H follows that T is a transversal of the hypergraph . Therefore τ (G) τ( ). H F ≥ H By the Gallai Identity τ( ) = V ( ) α( ) (cf. [8], [31, p. 405]), we have H | H | − H

τ (G) τ( ) n n1− F ≥ H ≥ − = n o(n), − as desired. 49

4.4.2 Proof of Theorem 4.7

Using the identity τF (G) = τF (G), we may suppose that F has an isolated vertex and is not itself an independent set.

As in the proof of Theorem 4.4, construct an independent dominating set D =

v , . . . , v recursively as follows: { 1 k} (i) choose v V (G) with N(v ) √n; 1 ∈ | 1 | ≥

(ii) when v has been chosen, i 1, choose v V (G) N[v ] such that i ≥ i+1 ∈ \ j≤i j S N(v ) N[v ] √n; | i+1 \ j≤i j | ≥ S (iii) if for all v V (G) N[v ], N(v) N[v ] < √n, then set k := i and ∈ \ j≤i j | \ j≤i j | S S D := v , . . . , v . { 1 k} Using an argument almost identical to that of Theorem 4.4, V (G) can be partitioned into at most 2 √n + 1 parts (colour classes there) such that V (G) V meets d e \ j every maximal F -free subgraph of G, where Vj is the largest partition part, and furthermore, from the proofs of Theorem 4.3 (i) and Theorem 4.4, at least one partition class consists of a singleton by our construction. Hence

2 √n V n 1, d e · | j| ≥ − and thus n 1 V (G) V n − | \ j| ≤ − 2 √n d e n 1 n − ≤ − 2( √n + 1) b c 1 n ( √n 1) ≤ − 2 b c − √n 1 = n b c + . − 2 2 50

Since the set V (G) V is a transversal of maximal F -free subgraphs of G, we see \ j that √n 1 τ (G) n b c + . F ≤ − 2 2

Remark. If F is an independent set or a complete graph, then the above bound does not hold. This can be seen by setting F = K and G = K . Since any t 1 t n − vertices in K form a maximal K -free subgraph of G, τ (K ) = n t+2. Similarly, n t Kt n − we also have τ t (K ) = n t + 2. K n − Chapter 5

Graphs F without Isolated Vertices

In this chapter we will investigate acF (G), F being a graph with no isolated vertices, for various classes of graphs G. We shall ﬁrst give bounds of acF (G) for general graphs G. Then we shall look at special classes of graphs G, including bipartite graphs, cycles, wheels, etc.

5.1 General Results

In Chapter 4 we have proven that if a graph F has neither an isolated vertex nor a dominating vertex then ac (G) 3 for any graph G; otherwise, ac (G) can be F ≤ F arbitrarily large. The following theorem implies that, if F has a dominating vertex

1 and acF (G) is large then G must have diameter at most two .

Theorem 5.1 If the forbidden graph F has no isolated vertices and diam(G) 3, ≥ then acF (G) = 2.

Proof. Since diam(G) 3, there exist two vertices u, v V (G) at distance at least ≥ ∈ 3 in G, as depicted in Figure 5.1. We colour N[u](u and all its neighbours) with colour 1 and everything else with colour 2. Formally, the colouring c is deﬁned by

1 if x N[u],  ∈ c(x) :=   2 otherwise.  1Most graphs have diameter two. Indeed, it can be proven that almost all graphs have diameter precisely two, see e.g. [9].

51 52

PSfrag replacements v u

N(v) N(u)

Figure 5.1: diam(G) 3 ≥

Assume that H is a maximal F -free subgraph of G. We show that H is not monochromatic under the colouring c. If H were coloured 1, then H N[u], and ⊆ hence H v would still be F -free since F contains no isolated vertex. If H were ∪ { } coloured 2, then H V (G) N[u], and thus H u would also be F -free. In either ⊆ − ∪{ } case, it contradicts the maximality of H, thus H is 2-coloured.

This shows that acF (G) = 2.

Remark. By Proposition 6.9 we know that acP3 (C6) = 3. Thus if F has an isolated vertex then the conclusion need not hold.

A very special case of Theorem 5.1 is when G is not connected. In this case, diam(G) = , and thus we obtain the following result. This will be useful in the ∞ sequel, so we list it as a corollary.

Corollary 5.2 Let F be a graph with no isolated vertices. If a graph G is disconnected, then acF (G) = 2.

The following result allows us to calculate an upper bound for acF (G) recursively.

Given a graph G, we will bound acF (G) in terms of acF (G[N(v)]), where G[N(v)] is 53 the subgraph induced by the neighbourhood N(v), F being a graph with no isolated vertices. We note here that this bound is a rough estimate as it is monotone relative to induced subgraphs while acF (G) is not.

Lemma 5.3 If F is a graph with no isolated vertices, then

acF (G) min acF (G[N(v)]) + 2. ≤ v∈V (G){ }

Proof. We show that for any v V (G), ac (G) ac (G[N(v)]) + 2. Let k := ∈ F ≤ F ac (G[N(v)]). We colour v with colour 1, V (G) N[v] with colour 2, and N(v) with F \ k colours 3, . . . , k + 2 so that each maximal F -free subgraph of G[N(v)] gets at { } least two colours. Let H be a maximal F -free subgraph of G. If V (H) N(v), ⊆ then H must be itself a maximal F -free subgraph of G[N(v)], and thus is 2-coloured.

If H V (G) N[v], then H v is still F -free as if there were any copy of F in ⊆ \ ∪ { } H v , it must involve v, contrary to the fact that F has no isolated vertices. It ∪ { } follows that H is not maximal F -free, a contradiction. Thus each maximal F -free subgraph of G receives at least two colours, and ac (G) ac (G[N(v)]) + 2. F ≤ F We now turn our attention to graphs G with small minimum degrees. We ﬁrst prove the following result which shows that for most graphs F with no isolated vertices, acF (G) = 2 provided the minimum degree of G is small. We write K3 + e for the graph indicated in Figure 5.2. We also let K e denote K with an edge 4 − 4 deleted.

Theorem 5.4 Let F be a graph with no isolated vertices.

(i) If F is not K ,P ,P or K + e, then ac (G) = 2 for any graph G ( G > 1) 2 3 4 3 F | | with minimum degree δ(G) 2, except ac (K ) = 3. ≤ K3 3 54

Figure 5.2: the graph K3 + e

(ii) If F is K ,P ,or P , then there are graphs G with minimum degree δ(G) 2 2 3 4 ≤ such that ac (G) 3. F ≥

Proof. Our approach is this. We ﬁrst prove that ac (G) = 2 for any G ( G > 1) F | | with δ(G) 2 and all graphs F except the ones in a ﬁnite list. After this is done, ≤ we will deal with the graphs F in the list one by one. To prove (ii), we will present examples for F = K , P and P , to show ac (G) 3. 2 3 4 F ≥ We prove (i) through the following series of facts.

Fact 1. If F has no isolated vertices and G contains an isolated vertex, and G > 1, | |

then acF (G) = 2.

This is easily seen by Corollary 5.2.

Fact 2. If F = P has no isolated vertices and G is a graph with minimum degree 6 3

δ(G) = 1, then acF (G) = 2.

Since δ(G)=1, there is a vertex v V (G) such that d(v) = 1. We denote by u ∈ the unique neighbour of v in G. The colouring c : V (G) [2] is deﬁned by →

1 if y u, v ,  ∈ { } c(y) :=   2 otherwise.   55

For any x V (G) u, v , there are two possible graphs that u, v, x can induce, ∈ \{ } { } as depicted in Figure 5.3.

PSfrag replacements PSfrag replacements

v u x v u x

Figure 5.3: K2 and a vertex

Since F is not either of these two graphs, the subgraph G[N[v] x ] is F -free for ∪{ } all x V (G). It follows that the subgraph G[N[v]] does not contain a maximal F -free ∈ subgraph of G, and hence no maximal F -free subgraph is coloured 1. If a maximal

F -free subgraph H is coloured 2, then V (H) V (G) N[v]. The graph induced by ⊆ \ V (H) v is still F -free, for if there were a copy of F in V (H) v , it must ∪ { } ∪ { } involve v, contrary to the supposition that F has no isolated vertex. Consequently

acF (G) = 2.

Fact 3. If a graph F has no isolated vertices and is not one of the graphs listed in

Figure 5.4 and G is a graph with δ(G) = 2, then acF (G) = 2.

Let us assume that v is a vertex with d(v) = 2. We ﬁrst note that if G = K3 then ac (G) = 2, as F = P ,K , and ac (K ) = 2 for any F with F > 3, by Lemma F 6 2 3 F 3 | | 2.6. Now suppose G = K , then V (G) N[v] = . We colour N[v] with colour 1 and 6 3 \ 6 ∅ V (G) N[v] with colour 2. This colouring will assign two colours to each maximal \ F -free subgraph of G. To see this, let H be a maximal F -free subgraph of G.

If H is coloured 2, then H v is still F -free as if H v were to contain a copy ∪{ } ∪{ } of F , then the copy must involve v, contradicting the supposition that F contains

no isolated vertices. 56

Figure 5.4: A list of graphs: K ,P ,K ,P ,C ,K + e, K e 2 3 3 4 4 3 4 − 57

v x v x PSfrag replacements PSfrag replacements

Figure 5.5: N[v] and a vertex x

Let us now suppose that H is coloured 1, and thus H N[v]. For any vertex ⊆ x V (G) N[v], noting that d(v) = 2, there are six possible graphs that N[v] x ∈ \ ∪ { } can induce, as depicted in Figure 5.5.

By assumption, F has no isolated vertices and is not a graph listed in Figure 5.4.

It is easily veriﬁed that F is not an induced subgraph of any graph listed in Figure 5.5.

Thus, for any vertex x V (G) N[v], the subgraph G[N[v] x ] is F -free, and hence ∈ \ ∪{ } 58

H x is still F -free as H N[v], contrary to the maximality of H. Consequently, ∪ { } ⊆ acF (G) = 2.

Fact 4. If F = K , C , or K e and G is a graph with δ(G) = 2, then ac (G) = 2, 3 4 4 − F

except acK3 (K3) = 3.

Let v V (G) be a vertex of G with d(v) = 2. We write N[v] =: w , w for ∈ { 1 2} the neighbours of v in G. The colouring c : V (G) [2] is deﬁned as follows. For a → vertex x V (G), ∈ 1 if x = v or w1, c(x) :=   2 otherwise.  Suppose G = K . Let H be a maximal F -free subgraph of G. If H is coloured 2, 6 3 then H v is still F -free as v is a vertex with d(v) 1 in H v while F , being one ∪{ } ≤ ∪{ } of K , C and K e, has no vertex of degree less than 2. If H is coloured 1, then H 3 4 4 − must equal vw as the graph F has at least three vertices. For any x V (G) N[v], 1 ∈ \ H x is K -free, as the only possible graphs induced by V (H) x are listed in ∪ { } 3 ∪ { } Figure 5.3 which do not include K ; H x is also C -free and (K e)-free as it 3 ∪ { } 4 4 − has three vertices. Thus, with this colouring c, each maximal F -free subgraph of G receives two colours, F being K , C or K e. Thus ac (G) = 2. 3 4 4 − K3

If G = K3, then by Lemma 2.6, acC4 (K3) = 2, acK4−e(K3) = 2. Since every two vertices in K3 form a maximal K3-free subgraph, we see that acK3 (K3) = 3.

(ii) Now we show that, if F = K2,P3, or P4, then the conclusion of (i) does not hold. This can be justiﬁed by the following examples.

Example 5.1 acK2 (C5) = 3. 59

Proof. Note that each maximal K2-free subgraph is a maximal independent set.

We enumerate the cycle as C5 =: v1v2v3v4v5 and suppose that acK2 (C5) = 2. It is easy to see that, for any 1 i 5, the pair v , v (where addition is modulo 5) ≤ ≤ i i+3 forms a maximal independent set in C5, and thus must get more than one colour. If acK2 (C5) = 2, without loss of generality, we assume that v1 is coloured 1, then v3 and v4 must be coloured 2, and hence v2 and v5 must be coloured 1. This yields a monochromatic maximal independent set v , v . Thus ac (C ) > 2. { 2 5} K2 5

Now we colour the vertices v1, v2, v3, v4, v5 with colours 1, 1, 2, 2, 3 respectively. Each colour class is an edge or a singleton, and thus does not contain a maximal independent set. So acK2 (C5) = 3.

Example 5.2

(i) acP3 (C4) = 4;

(ii) acP3 (P3) = 3.

Proof. (i) We enumerate the cycle as C =: v v v v . Since each pair v , v forms a 4 1 2 3 4 { i j} maximal P3-free subgraph of C4, it must get more than one colour. Thus we conclude that acP3 (C4) = χ(K4) = 4.

(ii) We note that in the graph P3, every two vertices form a maximal P3-free subgraph. Thus acP3 (P3) = 3.

Example 5.3 acP4 (C5) = 3.

Proof. Let us enumerate the cycle as C5 =: v1v2v3v4v5. Since every three vertices of

C5 form a maximal P4-free subgraph, and every 2-colouring of C5 must colour three vertices the same, we see that ac (C ) 3. P4 5 ≥ 60

Now let us show that 3 colours are enough. We colour v1, v2, v3, v4, v5 with 1, 1, 2, 2, 3 respectively. Then each colour class consists of two vertices and hence contains no maximal P4-free subgraph of C5—each maximal P4-free subgraph has at least three vertices.

Thus acP4 (C5) = 3.

Remarks. (a) From Theorem 4.1 we know that for graphs F with a dominating vertex, ac (G) can be arbitrarily large. Theorem 5.4 implies that, if F = K ,P , or F 6 2 3

K3 + e, then such graphs G must have minimum degree at least 3. (b) Theorem 4.1 yields ac(P ) 3, and thus we know from Example 5.3 that 4 ≤ ac(P4) = 3. This is the only graph F we know that has neither isolated vertices nor dominating vertices and attains the upper bound 3.

(c) The case of F = K3 +e is, at this point, unknown. However, from Facts 1 and 2 in the above proof we know that ac (G) = 2 for any graph G with δ(G) 1. K3+e ≤

Question 5.5 Is acK3+e(G) = 2 for every graph G with minimum degree 2?

5.2 Various Classes of Graphs

5.2.1 Bipartite graphs

In this subsection we will investigate acF (G), F being a graph with no isolated vertices, for bipartite graphs G. We shall prove that as long as F = P , ac (G) = 2 6 3 F for any bipartite graph G; and for F = P3, this result does not hold.

Theorem 5.6 Let F be a graph without isolated vertices. If G is a bipartite graph, 61

then acF (G) = 2, except in the following cases:

ac (K ) = 3, k 2 P3 1,k ≥ ac (K ) = 4, k, ` 2. P3 k,` ≥

Proof. Let us ﬁrst assume that G = (A, B) is not a complete bipartite graph, where

(A, B) is a bipartition of G. Then there exist x A and y B such that xy / E(G), ∈ ∈ ∈ and thus d(x, y) 3. By Theorem 5.1, ac (G) = 2 as diam(G) 3. ≥ F ≥ Now suppose that G = (A, B) is a complete bipartite graph.

Case 1. F = K2. If G is a star K1,k, we colour the centre x and a leaf y of G with colour 1, and everything else with colour 2. Since a nontrivial maximal

K2-free subgraph of G is a maximal independent set, it receives two colours. Thus acK2 (K1,k) = 2. If G is not a star, we 2-colour G in such a way that A and B both get both colours. Since each K2-free subgraph of G is a maximal independent set, it must be either A or B as G is a complete bipartite graph. Consequently the colouring assigns two colours to each maximal K2-free subgraph of G. Thus acK2 (G) = 2 in this case.

Case 2. F is not a star. We colour A with colour 1, and B with colour 2. Let H be a maximal F -free subgraph of G. If H is coloured 1, then V (H) A, and hence ⊆ for any vertex y B, H y is still F -free as F is not a star. Similarly H is not ∈ ∪ { } coloured 2. Thus acF (G) = 2 in this case. Case 3. F is a star K (` 3). If G is a star K (k 2), we colour the centre x 1,` ≥ 1,k ≥ and a leaf y of G with colour 1, and everything else with colour 2. A maximal F -free subgraph H cannot be coloured 1 as F 4. If H is coloured 2, it can be extended | | ≥ to the set of all leaves which is an independent set, contrary to the maximality of H. 62

If G is not a star, we choose two vertices x A and y B. Colour x with colour 1, ∈ ∈ y with colour 2, A x with colour 2, and B y with colour 1. Now we show that, \{ } \{ } with this colouring, any maximal F -free subgraph of G gets two colours. Suppose that a maximal F -free subgraph H of G is coloured 1. If V (H) B, then H y ⊆ ∪ { } is an independent set and thus F -free. If x V (H), as indicated in Figure 5.6, then ∈ H is itself a star in which x is the centre of H. It follows that V (H) B < ` as | ∩ | otherwise H contains F . Hence for any vertex u A x , H u is still F -free, as ∈ \{ } ∪ { }

F is not K2 or K1,2. Similarly, H is not coloured 2.

This shows that acF (G) = 2.

x y PSfrag replacements

A B

Figure 5.6: F is a star and G is not a star

Case 4. F = P3. First of all, we show ac (K ) = 3 for stars K (k 2). Denote by x the centre P3 1,k 1,k ≥ of K1,k, and write B for the set of leaves. If K1,k were coloured with two colours and each maximal P3-free subgraph of K1,k receives two colours, then B would be 2-coloured as it is a maximal P -free subgraph. Then there would be a vertex y B 3 ∈ coloured with the same colour as x. Hence the edge xy would be a monochromatic maximal P -free subgraph of K . This contradiction shows that ac (K ) 3. 3 1,k P3 1,k ≥ 63

To see acP3 (K1,k) = 3, we colour the vertex x with colour 1, and colour B with colours 2, 3 arbitrarily. If a maximal P -free subgraph is coloured i (2 i 3) then 3 ≤ ≤

V (H) ( B, and thus H can be extended to a larger P3-free subgraph B, contrary to the maximality of H. Hence acP3 (K1,k) = 3.

Secondly, we show that acP3 (G) > 3 for any complete bipartite graph G = (A, B), where A and B are the bipartition classes of G, A , B > 1. Since A and B are | | | | maximal P3-free subgraphs of G, they must be both at least 2-coloured. If G were coloured with three colours and each maximal P3-free subgraph of G receives more than one colour, then there must be two vertices x A and y B which receive ∈ ∈ the same colour, say 1. Thus the edge xy forms a monochromatic maximal P3-free subgraph of G as G is complete bipartite, a contradiction. To see ac (G) 4, we P3 ≤ colour A with colours 1 and 2, and B with colours 3 and 4, in any way as long as each colour appears. If a maximal P -free subgraph H is coloured i (1 i 4), 3 ≤ ≤ then H can be extended to the bipartition class A or B, contrary to the maximality of H. Thus acP3 (G) = 4. This completes the proof.

Remarks. (a) In view of Theorem 4.1, we see that for a given graph F with a dominating vertex, if acF (G) is large then G has to contain odd cycles.

(b) In Chapter 6 we shall show that if F = P3 and G is a bipartite graph, then acP3 (G) is still bounded above by 4.

(c) A further question is to determine the upper bound of acF (G) for graphs F with isolated vertices and bipartite graphs G. 64

5.2.2 Cycles

We know from the previous theorem that for any F with no isolated vertices, acF (C) = 2 for any even cycle C except C4. The following theorem determines acF (C) for graphs F with no isolated vertices and for any cycle C.

Proposition 5.7 Let F be a graph without isolated vertices and C be a cycle. Then acF (C) = 2, except that

acK2 (K3) = 1,

acK2 (C5) = 3,

acP4 (C5) = 3,

acK3 (K3) = 3,

acP3 (C4) = 4.

Proof. If the cycle has at least six vertices, then we make use of Theorem 5.1 and get acF (C) = 2. Now we consider the cases when C = K3, or C4 or C5.

If C = K3, then the nontrivial induced subgraphs of C are K2 and K3. Since K3 has no nontrivial K2-free subgraphs, acK2 (K3) = 1. If F = K3 then any maximal

K3-free subgraph of K3 is an edge. Hence acK3 (K3) = 3.

If C = C4, then the nontrivial induced subgraphs of C with no isolated vertices are K2, P3 and C4. By Lemma 2.7 or Theorem 5.6, we know that acC4 (C4) = 2.

Example 5.2 yields that acP3 (C4) = 4; Theorem 5.6 yields that acK2 (C4) = 2.

If C = C5, then the nontrivial induced subgraphs of C with no isolated vertices are K2, P3, P4 and C5. By Lemma 2.7, we know that acC5 (C5) = 2. Example 5.1 yields that acK2 (C5) = 3; Example 5.3 yields that acP4 (C5) = 3. Now consider the 65

case that F = P3. We colour two adjacent vertices of C with colour 1, everything else with colour 2. Since each maximal P3-free subgraph of C consists of an edge and a vertex independent of the edge, it receives two colours, so acP3 (C5) = 2.

Remark. In Chapter 6 we will also determine the values of acP3 (C) for all cycles C.

5.2.3 Wheels

A wheel W on n vertices is deﬁned by W := K C − , n 4. In this section, we n n 1 ∨ n 1 ≥ shall determine the precise value of ac (W ) for all graphs F with F 3 and for F n | | ≥ all wheels Wn.

Theorem 5.8 Let F ( F 3) be a graph. Then ac (W ) = 2 except for the cases | | ≥ F n

ac (W ) = 3 for n 5, P3 n ≥

acK3 (W4) = 4.

Proof. Suppose W = v C − , where v is the centre. We enumerate the vertices n { } ∨ n 1 of C − as x , . . . , x − . If F has no dominating vertices, then we colour v with n 1 1 n 1 { } colour 1, and everything else with colour 2. If a maximal F -free subgraph H is coloured 2, then H v is still F -free as if there were any copy of F , it would have ∪ { } to involve v, contrary to the condition that F has no dominating vertices. This yields that H could not be a maximal F -free subgraph, a contradiction. Thus acF (Wn) = 2 in this case.

Now let us consider the case when F has a dominating vertex.

Case 1. F = P3.

If n = 4, then P3 is not an induced subgraph of W4 and thus acP3 (W4) = 2. PSfrag replacements PSfrag replacements 66

x 3 1 x 9 x1 8 1 3 x8 x1 2 x2 x7 3 x7 2 3 2 2 1 x2 x9 v v x x6 3 x6 1 1 2 x3 2

x4 x5 x5 2 1 2 x4 1

Figure 5.7: The colouring of wheel Wn

Now suppose n 5. We ﬁrst show that ac (W ) 3. Assume that the wheel ≥ P3 n ≥

Wn is coloured with two colours, and each maximal P3-free subgraph receives two colours. Without loss of generality, assume that the centre v is coloured 1. Then in

the cycle Cn−1, no consecutive vertices are coloured 1, as otherwise there will be a

K3 coloured 1 which is maximal P3-free in Wn. Thus the vertices in the cycle Cn−1

that are coloured 1 are independent and divide C − into segments P ,...,P , k 1, n 1 1 k ≥ each of which is a monochromatic path (might be a trivial path) with colour 2.

0 Fact 1. Each segment Pi has a maximal P3-free subgraph Pi of Pi that contains

both ends of Pi. (In the case that Pi is a trivial path, the ends of Pi coincide.)

This is because one may pick both ends of Pi and extend them if necessary to a

0 maximal P3-free subgraph Pi of Pi.

k 0 Fact 2. The union i=1 Pi induces a maximal P3-free subgraph of the wheel Wn. S To see this fact, we show that, for any x / k P 0, the subgraph of the wheel ∈ i=1 i S W induced by k P 0 x contains a P . We consider the following cases. n i=1 i ∪ { } 3 S 67

Case a). The vertex x is coloured 2. Then x P for some 1 j i, and the ∈ j ≤ ≤ segment Pj has at least three vertices (if Pj is a K1 or K2 then, by construction, it is in k P 0). Thus P 0 x has a P as P 0 is a maximal P -free subgraph of P i=1 i j ∪ { } 3 j 3 j S and P contains P ( P 3). It thus follows that k P 0 x contains a P . j 3 | j| ≥ i=1 i ∪ { } 3 S Case b). The vertex x is coloured 1. Let us ﬁrst assume that x is in the cycle Cn−1.

Hence, as noted before, the neighbours of x in Cn−1 must be both coloured 2. Since these neighbours are contained in k P 0, the subgraph induced by k P 0 x i=1 i i=1 i ∪ { } S S contains a P3, i.e., the vertex x and its two neighbours in Cn−1. Now assume that x = v, the centre of W . If k 2, then the ends of the segment P are independent n ≥ 1 k 0 of those of P2, all of which are contained in i=1 Pi , by construction. Thus the S subgraph induced by k P 0 x contains a P , i.e., the vertex v and an end from i=1 i ∪ { } 3 S each of P1 and P2. If k = 1, there is only one segment P1, which has at least three vertices as n 5. Thus the ends of P are nonadjacent and contained in P 0. It ≥ 1 1 follows that P 0 v has a P , i.e., the centre v and both ends of P . 1 ∪ { } 3 1

k 0 Fact 2 yields that the subgraph induced by i=1 Pi is a monochromatic maximal S P -free subgraph of W in the 2-colouring. Thus ac (W ) 3 for n 5. 3 n P3 n ≥ ≥ To see that ac (W ) = 3 for n 5, we colour the vertices of W as indicated in P3 n ≥ n Figure 5.7. Formally, if n 1 is even, then the colouring c : V (W ) [3] is deﬁned − n → by

3 if x = v or xn−1,   c(x) := 1 if x = xi, i < n 3 is odd,  − 2 otherwise.    68

If n 1 is odd, the colouring is −

3 if x = v or xn−1,   c(x) := 1 if x = xi, i < n 1 is odd,  − 2 otherwise.    Let H be a maximal P -free subgraph of W . If H is coloured 1, then H x − 3 n ∪ { n 1} is still P -free; if H is coloured 2, then H x is still P -free; if H is coloured 3, 3 ∪ { 1} 3 then H x is still P -free. ∪ { 1} 3 Thus ac (W ) = 3, n 5. P3 n ≥

Case 2. F = K3.

Since W4 = K4, then each edge is a maximal K3-free subgraph of W4. Thus acK3 (W4) = χ(K4) = 4. Now assume n 5. We colour the vertices v and x with colour 1, and the ≥ 1 vertices x , . . . , x − with colour 2. { 2 n 1} Let H be a maximal K -free subgraph of W . If H were coloured 1, then H 3 n ⊆ v, x , and hence H x is still K -free as n 5. If H were coloured 2 then { 1} ∪ { 3} 3 ≥ H would be a subgraph of the cycle and hence could be extended to the cycle

x . . . x − x which is still K -free as n 5. Thus ac 3 (W ) = 2 for n 5. 1 n 1 1 3 ≥ K n ≥ Case 3. F 4. | | ≥ We ﬁrst note that if n = 4, then the only induced subgraph of order 4 of the graph W4 is itself, and by Lemma 2.7, we know that acW4 (W4) = 2. Hereinafter we assume that n 5. Note that F has a dominating vertex. We colour v, x with ≥ { 1} colour 1, and everything else with colour 2. Let H be a maximal F -free subgraph of

W (n 5). Since H 3, H is not coloured 1. If H is coloured 2, then H can be n ≥ | | ≥ 69

extended to the cycle C − which does not contain a dominating vertex as n 5. n 1 ≥ Thus ac (W ) = 2 for all F with a dominating vertex and F 4. F n | | ≥ Chapter 6

Small F with Isolated Vertices

In Chapter 4 we proved that if F has an isolated vertex then acF (G) is unbounded.

In Chapter 3 we surveyed results about acK2 (G), that is, clique colouring. In this chapter, we will investigate acF (G), for certain classes of graphs G and for the next simplest graph F with an isolated vertex—the disjoint union of a singleton and an edge, i.e. K1 + K2. It is easy to see that this graph can also be regarded as P3.

To make the notation more readable when we put it in subscript, we will write P3 instead of K1 + K2 throughout this chapter.

6.1 Characterization of Maximal Subgraphs

Lemma 6.1 Let G be a graph. Then every maximal P3-free subgraph of G is a maximal subgraph which is a disjoint union of cliques.

Proof. Let H be a maximal P3-free subgraph of G. Since H is P3-free, the connected components of H must be all cliques (singletons are regarded as degenerate cliques).

It follows that each maximal P3-free subgraph of G is an inclusionwise maximal subgraph of G which is a disjoint union of cliques.

Corollary 6.2 Let G be a graph. Then every maximal P3-free subgraph of G is a maximal complete multipartite subgraph of G.

Proof. Let H be a maximal P3-free subgraph of G. Then H is a maximal P3-free subgraph of G, and hence by Lemma 6.1 is a maximal subgraph of G which is

70 71 a disjoint union of cliques. It follows that H is a maximal complete multipartite subgraph of G.

6.2 Union of Cliques

In this section, we will completely determine the value for acP3 (G) for all disjoint unions of cliques G. For convenience, we will write

r G = `K + K , t 2, r 0, ` 0 1 ti i ≥ ≥ ≥ Xi=1 to denote that G is the disjoint union of ` isolated vertices and r nontrivial cliques

Kti .

Proposition 6.3 If G = `K + r K , t 2, is a disjoint union of cliques, then 1 i=1 ti i ≥ P acP3 (G) = 2 except that

4 if G = K + K , t , t 2,  t1 t2 1 2 ≥  ac (G) =  r P3 3 if G = K1 + Kti , r 1, ti 2,  i=1 ≥ ≥ P r 3 if G = K i , r 3, t 2.  i=1 t ≥ i ≥   P

......

Figure 6.1: acP3 (G) for union of cliques

Proof. The graph G and a general colouring scheme are depicted in Figure 6.1. 72

We prove the formula through a series of facts.

Fact 1. If a graph G = `K (` 2) is an independent set, then ac (G) = 2. 1 ≥ P3

To see this, we note that G is itself P3-free, and then apply Lemma 2.6.

Fact 2. If a graph G = K , t 2, then ac (G) = 2. t ≥ P3

Kt is itself P3-free, so Lemma 2.6 applies.

Fact 3. If a graph G = K + K , where t, s 2, then ac (G) = 4. t s ≥ P3 We ﬁrst show that ac (G) 4. Suppose to the contrary that G is coloured P3 ≥ with 1, 2, 3 and that each maximal P3-free subgraph of G receives more than one colour. Since each of the connected components Kt and Ks is a maximal P3-free subgraph of G, it must be at least two coloured. Because in total there are three colours available, there must be a colour appearing in both components, say 1; and suppose that u K and v K are two vertices coloured 1 in the two components. ∈ t ∈ s Then the pair u, v forms a maximal P -free subgraph of G and is monochromatic, { } 3 a contradiction to the supposition. Thus ac (G) 4. P3 ≥

If we colour the Kt and the Ks with colours 1, 2 and 3, 4, respectively, so that each component is 2-coloured, then each maximal P3-free subgraph H of G receives two colours. For, if H is coloured i, 1 i 4, then H is contained in the clique K ≤ ≤ t

(if i = 1, 2) or Ks (if i = 3, 4), and thus can be extended to a larger P3-free subgraph

Kt or Ks, contrary to the maximality of H. Consequently acP3 (G) = 4.

Fact 4. If a graph G = K + r K , where r 1, t 2, then ac (G) = 3. 1 i=1 ti ≥ i ≥ P3 P To see ac (G) 3, we suppose to the contrary that G is coloured with 1 and P3 ≥

2, and that each maximal P3-free subgraph of G receives two colours. We write v for the isolated vertex (i.e. the connected component K1) of G. Since the nontrivial 73

connected component Kti is a maximal P3-free subgraph of G, it must be 2-coloured. Hence there is a vertex u K (1 i r) coloured with the same colour as v, and i ∈ ti ≤ ≤ thus the set v, u , . . . , u forms a monochromatic maximal P -free subgraph of G, { 1 r} 3 contrary to the supposition.

If we colour the nontrivial connected components Kti with colours 1 and 2 so that each gets two colours, and colour the isolated vertex v with 3, then each maximal

P3-free subgraph of G receives more than one colour. Let H be a maximal P3-free subgraph of G. If H contains two vertices from the same Kti , then the maximality of

H guarantees that Kti is a subgraph of H, so H is at least 2-coloured. If H contains at most one vertex from each Kti , then H is a maximal independent set and hence v V (H), implying that H is at least 2-coloured. Thus H gets at least two colours, ∈ and acP3 (G) = 3.

Fact 5. If a graph G = r K , where t 2, r 3, then ac (G) = 3. i=1 ti i ≥ ≥ P3 P To see ac (G) 3, we suppose to the contrary that G is coloured with 1 and P3 ≥

2, and that each maximal P3-free subgraph of G receives two colours. Since each nontrivial connected component Kti is a maximal P3-free subgraph of G, it must be 2-coloured. Hence there is a vertex u K coloured with colour 1. Thus the graph i ∈ ti induced by the set u .u , . . . , u is a monochromatic maximal P -free subgraph of { 1 2 r} 3 G, a contradiction.

If we colour the graph G with 3 colours such that each connected component Kti receives precisely two colours, and the ﬁrst three components receive colours 1 and

2, 2 and 3, 1 and 3, respectively, then each maximal P3-free subgraph of G receives more than one colour. To see this, let H be a maximal P3-free subgraph of G. If H 74

contains two vertices from the same Kti , then the maximality of H guarantees that

Kti is a subgraph of H, so H is at least 2-coloured. If H contains at most one vertex from each K , then H is a maximal independent set and hence v V (H), implying ti ∈ that H is at least 2-coloured. Thus H receives at least two colours, and acP3 (G) = 3.

Fact 6. If a graph G = `K + r K , where `, t 2, r 1, then ac (G) = 2. 1 i=1 ti i ≥ ≥ P3 P We colour the nontrivial connected components Kti with colours 1 and 2 so that each gets both colours, and colour the independent set `K1 with colours 1 and 2 so it is 2-coloured, as indicated in Figure 6.1. Let H be a maximal P3-free subgraph of G. If H contains two vertices from the same Kti , then the maximality of H guarantees that Kti is a subgraph of H, so H is 2-coloured. If H contains at most one vertex from each Kti , then H is a maximal independent set and hence all isolated vertices are in V (H), implying that H is 2-coloured. Thus H receives two colours and acP3 (G) = 2. In summary, we see that the formula holds.

6.3 Sparse Graphs

As we mentioned before, the parameter acF (G) is quite diﬀerent from the ordinary chromatic number χ(G). In general acF (G) is not a global parameter of a graph G, but rather, it is determined by very local structures of G. This opinion will be evidenced once again by Theorem 6.4, which states that acP3 (G) is bounded by 4 if g(G) 5. In contrast, the ordinary chromatic number χ(G) is unbounded regardless ≥ of how big g(G) is.

Theorem 6.4 If G is a connected graph with girth at least 5, then ac (G) 4. P3 ≤ PSfrag replacements

1 2 3 0 1 2

V1 V2 V3 V4 V5 V6

Figure 6.2: ac (G) where g(G) 5 P3 ≥

Proof. Let G be a connected graph. We choose a vertex v V (G), then partition ∈ V (G) into subsets in which all vertices have the same distance to v, as indicated in

Figure 6.2. Formally, deﬁne

V := x V (G): d(x, v) = i , i = 0, 1, 2,... i { ∈ }

Then we alternate colours on the subsets Vi using colours 0, 1, 2 and 3. Precisely, deﬁne the colouring c : V (G) [4] as →

0 if i 0 (mod 4),  ≡   1 if i 1 (mod 4),  ≡ c(Vi) :=   2 if i 2 (mod 4), ≡   3 if i 3 (mod 4).  ≡   Now we show that, with this colouring, every maximal P3-free subgraph gets at

least two colours. Let H be a maximal P3-free subgraph of G. By Corollary 6.2, H is a maximal complete multipartite subgraph of G. We note that if H is a complete

multipartite subgraph of G with at least one edge, then H must be a maximal star

(of the form N[x] for some vertex x V (G)), as otherwise H has a 3-cycle or a ∈ 76

4-cycle, contrary to the condition g(G) 5. Thus H is either a maximal star, or ≥ a maximal independent set of G (if this independent set cannot be extended to a larger complete multipartite subgraph of G). If H is a maximal star, then H must contain vertices in V and V for some i 0, and hence gets at least two colours, i i+1 ≥ by the deﬁnition of the colouring. Now suppose that H is a maximal independent set. Let

k := max d(x, v): x V (G) . { ∈ } If k 3, then H cannot be coloured 1 since each vertex in H is adjacent to v, and ≤ thus H v is still a complete multipartite graph, contradicting the maximality of ∪ { } H. If H is coloured `, 2 ` 3, then H v is still an independent set, contrary ≤ ≤ ∪ { } to the maximality of H. Thus H must get at least two colours.

If k 4, then H cannot be coloured `, 0 ` 3; any vertex x V , where ≥ ≤ ≤ ∈ i i ` 2 (mod 4), 1 i k, is nonadjacent to any vertex of H, by deﬁnition of the ≡ − ≤ ≤ subsets V . Thus H x is still an independent set, contradicting the maximality i ∪ { } of H.

This proves that each maximal P3-free subgraph of G receives at least two colours, and thus ac (G) 4. P3 ≤ Remark. From Theorem 4.1 we know that if the condition of g(G) 5 is dropped, ≥ then acP3 (G) can be arbitrarily large. Propositions 6.8 and 6.9 show that acP3 (P5) =

3 and acP3 (C6) = 3. Thus acP3 (G) is equal to either 3 or 4. If G is disconnected then the value 4 can be attained: in view of Proposition 6.3, we see that acP3 (K2+K2) = 4. In the above theorem we restricted our attention to connected graphs G. If the graph G is disconnected, we can get the following weaker bound. 77

Corollary 6.5 If G is a disconnected graph with girth at least 5, then ac (G) 8. P3 ≤

Proof. Since G is disconnected, it has at least two connected components. From

Theorem 6.4 we know that the component G1 can be coloured with colours 1, 2, 3, and 4 such that each maximal P3-free subgraph of G1 receives at least two colours. Using Theorem 6.4, we can colour each of the components G (2 i k) with colours i ≤ ≤

5, 6, 7, and 8 in such a way that each maximal P3-free subgraph of Gi receives at least two colours as well. If any of the Gi is trivial we just colour it with one of the four allowed colours.

Let H be a maximal P3-free subgraph of G. As earlier mentioned, H is either a maximal star or a maximal independent set of G. If H is contained in a single component G (1 i k) of G, it must be a maximal P -free subgraph of G as i ≤ ≤ 3 i well, and hence receives at least two colours. If H intersects diﬀerent components, it must be a maximal independent set, and thus meets all components. Thus H is also at least 2-coloured in the colouring.

Hence ac (G) 8. P3 ≤

6.4 Induced Long Path

Theorem 6.6 For any connected graph G, if there is a vertex v V (G) such that ∈ v is not an endvertex of an induced P , then ac (G) 3. 4 P3 ≤

Proof. Let G be a connected graph and let v V (G) be a vertex that is not ∈ an endvertex of an induced P . Deﬁne the colouring c : V (G) [3] as follows: 4 → 78

u V (G), ∀ ∈ 1 if u = v,   c(u) = 2 if u N(v),  ∈ 3 otherwise.    Notice that N(v) does not contain any maximal P3-free induced subgraph H of G, since if V (H) N(v) is P -free, then H v is also P -free. Thus if H were a ⊆ 3 ∪ { } 3 monochromatic maximal P -free induced subgraph of G, then V (H) V (G) N[v]. 3 ⊆ \ If H is an independent set, then H v is still independent, contradicting the ∪ { } maximality of H. If H is a complete multipartite induced subgraph with at least one edge, and V (H) V (G) N[v], then H is contained in some certain connected ⊆ \ component C of the subgraph G N[v] induced by V (G) N[v]. − \ Since G is connected, there exists w N(v) adjacent to some vertex of C, but not ∈ all the vertices of C, otherwise H will not be a maximal P3-free induced subgraph of G. Thus neither of N(w) C and V (C) (N(w) C) is the empty set. As C is a ∩ \ ∩ connected component, there are x N (w), y V (C) N (w) so that xy E(G). ∈ C ∈ \ C ∈

Therefore vwxy is an induced P4, contrary to the assumption that v is not an endvertex of a P4.

Remarks. (a) The converse of Theorem 6.6 is not true, as can be seen by acP3 (Pk) = 2 for k 7, which will be proved later in Proposition 6.8. ≥ (b) If the graph G is not connected, then the bound of 3 need not hold, as can be seen by acP3 (K2 + K2) = 4 by Proposition 6.3.

Using mathematical induction, we can prove the following generalization of The- orem 6.6. 79

Theorem 6.7 For any connected graph G, if there is a vertex v V (G) such that ∈ v is not an endvertex of an induced P (k 4), then ac (G) 2k 5. k ≥ P3 ≤ −

Proof. Let G be a connected graph.

The case k = 4 has been proved in Theorem 6.6. Now let k > 4, and assume the theorem is true for all connected graphs with a vertex that is not an endvertex of any induced Pk−1. Suppose v is not the endvertex of an induced Pk. For each component C of G N[v], pick u N(v) that is adjacent to some vertices of C. − C ∈ Denote C0 := C u . Note that the vertex u is not an endvertex of an induced ∪ { C } C 0 Pk−1 in G[C ], as otherwise adding v to the path will yield an induced Pk starting with v. Thus by induction hypothesis, ac (G[C0]) 2k 7 for all C0. P3 ≤ − Now suppose for every component C of G N[v], ac (G[C0]) 2k 7. For each − P3 ≤ − C, choose a (2k 7)-colouring r of G[C0], using colours 1, 2,..., 2k 7, such that − C − 0 each maximal P3-free subgraph of G[C ] is at least two coloured. Then we colour N(v) with colour 2k 6 (u ’s are recoloured with 2k 6 too), and colour v with − C − colour 2k 5. −

Now let us prove that, with this colouring, each maximal P3-free subgraph H of G receives more than one colour. If H is an independent set, then H is not contained in

N(v) or V (G) N[v], as otherwise H v would yield a larger complete multipartite \ ∪{ } subgraph of G. Consequently H is at least two coloured.

If H is a maximal complete multipartite induced subgraph with at least one edge, then H is not contained in N(v) as otherwise H v would induce a larger ∪ { } complete multipartite subgraph of G. Now let us assume that H V (G) N[v]. ⊆ \ Since H is connected, it must be contained in some connected component C of 80

V (G) N(v). Thus H C0, and hence it must be a maximal complete multipartite \ ⊆ induced subgraph of G[C0] as well. Thus H is not monochromatic by the choice of the colouring rC .

6.5 Some Other Classes of Graphs

6.5.1 Paths

From Theorem 6.4, we know that ac (P ) 4 for all paths P . In this section, we P3 ≤ completely determine the values acP3 (P ) for all paths P .

Proposition 6.8 If P is the path on k vertices (k 2), then k ≥

2 if k = 5,  6 acP3 (Pk) =   3 if k = 5. PSfrag replacements   (b)

x x x x x PSfrag replacements 1 2 3 4 5 (a) (a)

x1 x2 x3 x4 x5 (b)

Figure 6.3: acP3 (P5)

Proof. In a path Pk, each maximal P3-free subgraph of Pk is either P3 or a maximal independent set. 81

We ﬁrst consider P5 =: x1x2x3x4x5, as indicated in Figure 6.3. Suppose acP3 (P5) =

2; then x1 and x4 must receive diﬀerent colours as they form a maximal P3-free sub-

graph of P5. Without loss of generality, we suppose x1 is assigned 1 and x4 is

assigned 2. Now x2 and x5 must receive diﬀerent colours. If x2 is coloured 1 and

x5 is coloured 2, as indicated in Figure 6.3 (a), then x3 cannot be coloured without

forming a monochromatic P3. If x2 is coloured 2 and x5 is coloured 1, as depicted in

Figure 6.3 (b), then again x3 cannot be coloured without forming either a monochro-

matic P3 or a monochromatic maximal independent set. Thus acP3 (P5) > 2. To see

PSfrag replacementsacP3 (P5) = 3, we colour P5 with 3 colours: x1, x2, x3, x4, x5 are, respectively, coloured

with 1, 1, 2, 2, and 3. Clearly no P3 or maximal independent set is monochromatic.

Thus acP3 (P5) = 3.

x1 x2 x3 xk−2 xk−1 xk 3 1(2) 11 2 1 2 11 22

Figure 6.4: acP3 (Pk)

Now let us show that ac (P ) = 2 for k = 5. Denote P =: x x , as indicated P3 k 6 1 ··· k in Figure 6.4.

For k = 2, 3, Lemma 2.6 yields that acP3 (Pk) = 2. If k = 4, we colour x1, x2, x3, x4

with colours 1, 1, 2, 2, respectively. It is easy to see that each P3 and maximal independent set gets two colours.

Now we suppose k 6. We colour x , x and x with colours 1, 1, 2, respectively; ≥ 1 2 3

and colour xk−2, xk−1 and xk with colours 1, 2, 2, respectively. Then alternate colours 82

1 and 2 on the remaining vertices with x4 coloured 1. Let H be a maximal P3-free subgraph of Pk. If H contains edges, then H is a P3 in Pk, and it is easy to see that each P3 receives two colours. Now suppose that H is an independent set. If H is coloured 2, then H x is still independent; if H is coloured 1, then H x ∪ { 1} ∪ { k} is still independent. Thus each maximal independent set receives two colours. This proves that each maximal P3-free subgraph of Pk is 2-coloured.

6.5.2 Cycles

Proposition 6.9 Let C be a cycle of order n 3. Then n ≥

3 if n = 6, 9, acP3 (Cn) =   2 otherwise.   3 3 2 x1 1 x2 xn 2 x 2 3 1 PSfrag replacements 1 2 1 2 2 1 2 1 1 2

Figure 6.5: acP3 (Cn)

Proof. We enumerate the vertices of C as x , , x , as indicated in Figure 6.5, and n 1 ··· n prove the proposition through the following series of facts. We note that, in a cycle 83

C , if n 6, each maximal P -free subgraph of C is a P or a maximal independent n ≥ 3 n 3 set; the maximal P3-free subgraphs of C5 are P3’s.

Fact 1. acP3 (C6) > 2.

Suppose the cycle C6 is coloured with two colours such that each maximal P3-free subgraph receives both colours. Since x , x , x is a maximal P -free subgraph of { 1 3 5} 3

C, it cannot be monochromatic. Without loss of generality, we assume that x1, x3, x5 are coloured 1, 1, 2, respectively. This yields that one of x4 and x6 must be coloured 1 as otherwise x , x , x forms a monochromatic P . By symmetry we may suppose { 4 5 6} 3 that x is coloured 1, then x , x forms a monochromatic maximal independent 4 { 1 4} set. This contradiction shows that acP3 (C6) > 2.

Fact 2. acP3 (C9) > 2.

Suppose to the contrary that C9 is coloured with two colours, and each maximal

P3-free subgraph receives both colours. Without loss of generality, assume that x1, x2 are coloured 1. Then x3, x9 must be coloured 2 as every P3 must be 2-coloured. Since x , x , x is a maximal independent set, x must be coloured 1, hence x and x { 3 6 9} 6 5 7 cannot be both coloured 1. Since x , x , x , x is a maximal independent set, x { 3 5 7 9} 5 and x7 cannot be both coloured 2 either. By symmetry we assume that x5 and x7 are coloured 1 and 2 respectively. Then x must be coloured 1 as x , x , x form a 8 { 7 8 9} P . This yields a maximal independent set x , x , x coloured 1. 3 { 2 5 8}

Fact 3. acP3 (Cn) = 3, n = 6, 9.

We colour x1, x2, x3 and xn with colours 3, 3, 1, 2, respectively, and then alternate colours 2 and 1 on the remaining vertices, starting with x4, as indicated in Figure

6.5. It is easy to see that each P3 gets more than one colour. Now we show that each 84 maximal independent set receives at least two colours. If a maximal independent set contains both x3 and xn, then obviously it gets at least two colours; if it contains at most one of x , x , then it must contain at least one of x , x , and hence receives { 3 n} { 1 2} at least two colours as well.

Fact 4. ac (C ) = 2, k 1. P3 4k ≥

We colour the vertices x1, . . . , x4k with colours 1, 1, 2, 2 alternately. If k = 1, then P3 is not an induced subgraph of C4, and hence the only maximal P3-free subgraph of C is C itself, which is 2-coloured in this colouring. Now suppose k 2. 4 4 ≥

It is easy to see that each P3 receives two colours. Now we show that each maximal independent set also receives two colours. Assume that S is a monochromatic maximal independent set of the cycle C4k. By symmetry, suppose that S is coloured 1 and x S. Then x / S and hence x is not adjacent to any vertex of S. Thus 1 ∈ 2 ∈ 3 S x is a larger independent set, contrary to the maximality of S. ∪ { 3} Fact 5. ac (C ) = 2, k = 2. P3 4k+1 6

We ﬁrst show acP3 (C5) = 2. Colour x1, x2, x3, x4, x5 with 1, 1, 2, 1, 2, respectively.

Then each P3 receives both colours; in C5 each maximal independent set can be extended to a P3, and hence is not maximal P3-free. Thus each maximal P3-free subgraph of C5 receives both colours in this colouring.

If k 3, we colour the vertices x , . . . , x − with colours 1, 1, 2, 2 alternately, ≥ 1 4(k 1) and the vertices x4k−3, . . . , x4k+1 with colours 1, 1, 2, 1, 2 respectively. It is easy to see that each P3 receives two colours. Assume that S is a monochromatic maximal independent set of the cycle C4k+1. If S is coloured 2, we consider the vertices x3, x4. If x / S then S x is a larger independent set, contrary to the maximality of 3 ∈ ∪ { 2} 85

S; if x S then x / S, and hence S x is a larger independent set, contrary 3 ∈ 4 ∈ ∪ { 5} to the maximality of S. If S is coloured 1, we consider the vertices x , x . If x / S 5 6 5 ∈ then S x is a larger independent set, contrary to the maximality of S; if x S ∪ { 4} 5 ∈ then x / S, and x / S since k 3 implies that x is coloured 2. Hence S x 6 ∈ 8 ∈ ≥ 8 ∪ { 7} is a larger independent set, contrary to the maximality of S.

Fact 6. ac (C ) = 2, k 2. P3 4k+2 ≥

We colour the vertices x1, . . . , x4k with colours 1, 1, 2, 2 alternately, and the vertices x4k+1, x4k+2 with colours 1 and 2 respectively. It is easy to see that each

P3 receives two colours. Assume that S is a monochromatic maximal independent set of the cycle C4k+2. By symmetry, we may assume that S is coloured 2, and we consider the vertices x , x . If x / S then S x is a larger independent set, 3 4 3 ∈ ∪ { 2} contrary to the maximality of S. If x S then x / S, and hence S x is a 3 ∈ 4 ∈ ∪ { 5} larger independent set, contrary to the maximality of S.

Fact 7. ac (C ) = 2, k 1. P3 4k+3 ≥

We colour the vertices x1, . . . , x4k with colours 1, 1, 2, 2 alternately, and the vertices x4k+1, x4k+2, x4k+3 with colours 1, 1 and 2 respectively. It is easy to see that each P3 receives two colours. Assume that S is a monochromatic maximal independent set of the cycle C4k+3. Exactly as in the previous case, we see that

S cannot be coloured 2. If S is coloured 1, we consider the vertices x1 and x4k+2. If x S then S x is a larger independent set. If x S then S x 1 ∈ ∪ { 3} 4k+2 ∈ ∪ { 4k} is a larger independent set, both contrary to the maximality of S. If x / S and 1 ∈ x / S, then S x is a larger independent set, contrary to the maximality 4k+2 ∈ ∪ { 4k+3} of S. 86

Thus we see that ac (C ) = 2 for n = 6, 9. P3 n 6

6.5.3 Bipartite graphs

Proposition 6.10 ac (G) 4, for any bipartite graph G. P3 ≤

Proof. Let the bipartition be (A, B). If the graph G is a complete bipartite graph then P G, and by Lemma 2.6 ac (G) = 2. If G is not complete bipartite, we 3 6⊆ P3 arbitrarily colour the set A with two colours 1 and 2, and the set B with colours 3 and 4 such that A and B are both 2-coloured (as long as it is feasible, i.e. A 2, | | ≥ B 2; if A (or B) contains only one vertex then colour it with 1 (or 3)). Let | | ≥

H be a maximal P3-free subgraph of G. If H meets both A and B, then obviously H receives at least two colours. If H A (or H B), then we have A 2 (or ⊆ ⊆ | | ≥ B 2) as H has at least two vertices, and H must be a maximal independent set | | ≥ of G. Thus H is in fact A or B, and hence receives two colours.

Remark. Noting that acP3 (K2 + K2) = 4 by Proposition 6.3, the bound 4 is sharp. If we restrict our attention to connected bipartite graphs with a leaf, the following result holds.

Proposition 6.11 Let G be a bipartite graph that contains a connected component

C with C 3 and δ(C) = 1. Then ac (G) 3. | | ≥ P3 ≤

Proof. We suppose that u C and d(u) = 1. Let v be the vertex adjacent to u. ∈ We colour u and v with colour 3, and colour the remaining vertices properly with colours 1 and 2, as indicated in Figure 6.6.

Let H be a maximal P3-free subgraph of G. If H contains an edge, then H is 2-coloured. For, if H is monochromatic, then it must be coloured 3, thus V (H) = 87

PSfrag replacements u v

3 3 1 2 1 2 1

Figure 6.6: acP3 (G) for bigraphs G with δ(G) = 1

u, v . Since C is connected and C 3, then v must be adjacent to a vertex other { } | | ≥ than u, say w. Hence H can be extended to u, v, w which, being P , is a larger { } 3

P3-free subgraph of G, a contradiction. If H is a maximal independent set, then H must contain either u or v, and hence is also 2-coloured. Therefore ac (G) 3. P3 ≤

Remark. The upper bound 3 is sharp, as can be seen by acP3 (P5) = 3 (Proposition 6.8). Using the almost identical argument, we can generalize the above result as

follows: Let G be a graph that contains a connected component C with C 3 and | | ≥ δ(C) = 1. Then ac (G) χ(G) + 1. P3 ≤

6.5.4 Perfect graphs

As earlier mentioned in Chapter 3, whether acK2 (G) is bounded by a constant for all perfect graphs G is still open. Constant bounds for various classes of perfect

graphs have been found (see Chapter 3).

In this section, we consider the corresponding problem for F = P3. We will show that for almost all (in a sense which will be made precise below) perfect graphs G,

acP3 (G) is bounded. We ﬁrst consider a class of special perfect graphs—split graphs. A graph is a 88 split graph if its vertices can be partitioned into a clique and an independent set. It is well-known that if G is a split graph then both G and G are chordal [53, p. 345].

Proposition 6.12 If G is a split graph, then ac (G) 3. P3 ≤

Proof. By Lemma 2.6 we suppose that G 3. If G is an independent set, then | | ≥

P3 is not a subgraph of G, and thus acP3 (G) = 2 by Lemma 2.6. Now suppose that G is not an independent set and that V (G) can be partitioned into subsets X and

S, where X induces a clique and S induces an independent set. If X consists of one vertex u then, since G is not an independent set, u is adjacent to some vertex v S. We colour u and v with colours 1 and 2 respectively, and colour everything ∈ else with colour 3. If a maximal P3-free subgraph H of G is monochromatic, it must be coloured 3, and hence can be extended to a larger independent set S. Hence we may assume X 2. If S = 1 then we colour S with colour 1, and colour X with | | ≥ | | colours 2 and 3. If a maximal P3-free subgraph H of G is monochromatic, it must be coloured 2 or 3, and hence can be extended to a larger clique X. Now we suppose that X , S 2. | | | | ≥ Case 1. e(X,S) = 0. Thus G is the disjoint union of the clique G[X] and the independent set S. Note that X 2 and the only maximal P -free subgraphs are | | ≥ 3 G[X] and S x for any x X. We colour G with two colours 1 and 2 such that X ∪ { } ∈ and S are both 2-coloured. It is easy to see that with this colouring each maximal

P3-free subgraph receives both colours. Case 2. e(X,S) 1. Thus there is an edge from X to S. Let e = uv, where ≥ u X and v S, be such an edge. We colour u and v with colour 1, the remaining ∈ ∈ vertices of X with colour 2, and the remaining vertices of S with colour 3. If H 89

is coloured with 1, then H = e, and thus it can be extended to a larger P3-free subgraph of G by adding a vertex w from X, regardless of the adjacency of w and v. If H is coloured 2, then H u is still P -free as H u X; If H is coloured ∪ { } 3 ∪ { } ⊆ 3, then H v is still P -free as H v S, contradicting the maximality of H. ∪ { } 3 ∪ { } ⊆ Thus ac (G) 3 for all split graphs. P3 ≤

Remark. The bound 3 is sharp, as it can be seen by Proposition 6.3 that acP3 (Kn + K ) = 3 for n 2. 1 ≥

We now proceed to show that acP3 (G) is bounded above by 6 for almost all perfect graphs G. For this, let us ﬁrst introduce some more terminology and background information.

Given a graph property Q, we say that almost all graphs satisfy the property Q in the probability space Ωn consisting of a certain class of graphs on n vertices, each graph being equally likely in Ωn, if

Pr(G Ω : G satisﬁes Q) 1 as n . ∈ n → → ∞

An odd hole in a graph G is an induced odd cycle of length at least 5; an odd antihole in G is the complement of an odd hole. A graph is a Berge graph if it has no odd hole or odd antihole. Chudnovsky, Seymour, Robertson and Thomas [17] proved the following theorem in 2002.

Theorem 6.13 (The Strong Perfect Graph Theorem)

A graph is perfect if and only it is a Berge graph.

In this section we will actually show that ac (G) 6 for almost all C -free P3 ≤ 5 graphs. To this end, we need the concept of generalized split graphs. 90

A graph G is a generalized split graph if either G or G has a vertex partition into subsets A, B (1 i k) such that G[A] and all G[B ]’s are cliques, and furthermore, i ≤ ≤ i e(B ,B ) = 0 if i = j. It is easy to see that a generalized split graph is obtained i j 6 by expanding the vertices of the independent set of a split graph to diﬀerent cliques and then possibly taking the complement. Generalized split graphs were introduced by Pr¨omeland Steger in [47], and were shown to be perfect graphs. We shall take advantage of the following result in [47].

Theorem 6.14 Almost all C5-free graphs are generalized split graphs.

An immediate consequence of the above theorem is that almost all C5-free graphs are perfect graphs.

We make use of the above theorem to prove that ac (G) 6 for almost all P3 ≤ perfect graphs G. Our approach is as follows. We show that ac (G) 6 for all P3 ≤ generalized split graphs G. Thus, using Theorem 6.14, ac (G) 6 for almost all P3 ≤

C5-free graphs G. Let Ωn be the probability space consisting of all perfect graphs

0 of order n, each graph being equally likely in Ωn, and let Ωn be the probability space consisting of all C5-free graphs of order n, each graph being equally likely in

0 0 Ωn. Thus Ωn is a subspace of Ωn, and we write Pr(Ωn) for the probability of this

0 subspace in the probability space Ωn. 91

Note that Theorem 6.14 implies that Pr (Ω ) 1, as n . We have n → → ∞

Pr G Ω : ac (G) > 6 = G Ω : ac (G) > 6 / Ω ∈ n P3 ∈ n P3 | n| G Ω0 : ac (G) > 6 / Ω ≤ ∈ n P3 | n| G Ω0 : ac (G) > 6 / Ω0 = ∈ n P3 | n| Ω / Ω0 | n| | n| Pr G Ω0 : ac (G) > 6 = ∈ n P3 0. Pr (Ωn) →

The last asymptotic behaviour is because

Pr G Ω0 : ac (G) > 6 0, n , ∈ n P3 → → ∞ and

Pr (Ω ) 1 = 0, n . n → 6 → ∞ Hence we have

Pr G Ω : ac (G) 6 1, n , ∈ n P3 ≤ → → ∞ which, by deﬁnition, means that acP3 (G) is bounded for almost all perfect graphs G1.

Now let us prove the main result of this section.

Theorem 6.15 If G is a generalized split graph, then ac (G) 6. P3 ≤

Proof. Let G be a generalized split graph. By definition, we know that V (G) can be partitioned into subsets A, B (1 i k) such that G[A] and the G[B ]’s are all i ≤ ≤ i 1The above is a direct argument. In the theory of random graphs, the following is a funda- 0 mental result. Let Q be a graph property, and let Ωn be a subspace of Ωn. Then Pr(Q) = 0 Pr(Q Ωn) Pr(Ωn) + Pr(Q Ωn) Pr(Ωn) Pr(Q Ωn) Pr(Ωn), and thus Pr(Q) 0 in Ωn implies that | | ≥ | → Pr(Q Ωn) 0 in Ωn, as long as Pr(Ωn) c = 0, n . This result will immediately yield that, | → → 6 → ∞ setting the property Q to be acP (G) > 6, if acP (G) 6 for almost all C5-free graphs G, then 3 3 ≤ ac (G) 6 for almost all perfect graphs G. P3 ≤ 92 cliques (or independent sets), and e(B ,B ) = 0 (e(B ,B ) = B B , respectively) i j i j | i|| i| if i = j. We colour A arbitrarily with colours 1 and 2, B with colours 3 and 4, each 6 1 B (i 2) with colours 5 and 6 such that both colours are used for A and B ’s if i ≥ i possible. Let H be a maximal P3-free subgraph of G. If H is coloured with some `, 1 ` 4, then H must be a clique (an independent set). Note that H 2 and ≤ ≤ | | ≥ hence the subset (A or B1) containing H has at least two vertices. Thus H can be extended to a larger clique (independent set) by adding in the remaining vertices in the same subset. If H is coloured 5, then either H is contained in a single B (i 2) i ≥ and hence is a clique (independent set), or H meets different Bi’s and hence is an independent set (a complete multipartite subgraph) of G. In the first case, H can be extended to Bi which is a larger clique (independent set). In the second case, H v for any v B will yield a larger independent set (complete multipartite ∪ { } ∈ 1 subgraph) of G. Thus each maximal P3-free subgraph of G gets at least two colours, and ac (G) 6. P3 ≤ Chapter 7

Partially Ordered Sets

In this chapter, we investigate avoiding colouring of partially ordered sets. Results in Sections 7.5 and 7.6 are joint work with Dr. Bill Sands.

7.1 Preliminaries for Partially Ordered Sets

Our terminology and notation are from [19].

Deﬁnitions

A partially ordered set (or poset for short) is a pair P = (X, ), where is a binary ≤ ≤ relation on the set X which is:

(i) reﬂexive: x x for all x X; ≤ ∈

(ii) antisymmetric: x y and y x imply x = y; ≤ ≤

(iii) transitive: x y and y z imply x z. ≤ ≤ ≤

An ordered relation on P gives rise to a relation < of strict inequality: x < y in

P if and only if x y and x = y. It is possible to re-state conditions (i)–(iii) above ≤ 6 in terms of <, and so to regard < rather than as the fundamental relation; see ≤ [19].

Other notation associated with is predictable. We use x y and y x ≤ ≤ ≥ interchangeably, and write x y to mean “x y is false,” and so on. The symbol ≤

93 94

is used to denote non-comparability in the context of partially ordered sets: we k write x y if x y and y x. If x y for all y = x in P , then x is an isolated element k k 6 of P .

Let P = (X, ) be a partially ordered set and let x, y X. We say that x is ≤ ∈ covered by y (or y covers x), and write x <: y or y :> x, if x < y and x z < y ≤ implies z = x. The latter condition demands that there be no element z of P with x < z < y. If x is not smaller than any other element in P , then x is a maximal element of P ; the term minimal element is deﬁned dually. The set of maximal and minimal elements of P are denoted max(P ) and min(P ), respectively. If x is the unique maximal (minimal) element in P , then x is called a maximum (minimum).

Let P = (X, ) be a partially ordered set and let Y X be a subset of X. Then ≤ ⊆ Y inherits an order relation from P ; given x, y Y , x y in Y if and only if ≤|Y ∈ ≤ x y in X. We say in these circumstances that Q = (Y, ) is an (induced) ordered ≤ ≤|Y subset or subposet of P = (X, ), and write Q P . We also write instead of . ≤ ⊆ ≤ ≤|Y

Chains and antichains

Let P = (X, ) be a partially ordered set. Then P is a chain if, for all x, y X, x ≤ ∈ and y are comparable (that is, either x y or y x). Alternative names for a chain ≤ ≤ are linearly ordered set and totally ordered set.

At the opposite extreme from a chain is an antichain. A partially ordered set

P is an antichain if x y in P only if x = y, that is, any two elements are non- ≤ comparable. Clearly, any subposet of a chain (an antichain) is a chain (an antichain).

We write k for the k-element chain and k for the k-element antichain. 95

Hasse diagram

Let P be a partially ordered set. We can represent P by a conﬁguration of circles

(representing the elements of P ) and interconnecting lines (indicating the covering relation). The construction goes as follows.

1. To each point x P , associate a point p(x) of the Euclidean plane R2, depicted ∈ by a small circle with centre at p(x).

2. For each covering pair x <: y in P , take a line segment `(x, y) joining the circle

at p(x) to the circle at p(y).

3. Carry out (1) and (2) in such a way that

(a) if x <: y, then p(x) is “lower” than p(y) (that is, in standard Cartesian

coordinates, has a strictly smaller second coordinate);

(b) the circle at p(z) does not intersect the line segment `(x, y) if z = x and 6 z = y. 6

A conﬁguration satisfying (1)–(3) is called a Hasse diagram (or diagram) of P .

Order-isomorphisms

To be able to recognize that two partially ordered sets P = (X, ) and Q = (X 0, 0) ≤ ≤ are “essentially the same”, we need the concept of order-isomorphism. We say that

P and Q are order-isomorphic, and write P ∼= Q, if there exists a map φ from X onto Y such that x y in P if and only if φ(x) φ(y) in Q. The map φ is called ≤ ≤ an order-isomorphism. 96

Such a map faithfully mirrors the order structure. It is necessarily bijective (that is, one-to-one and onto).

Maps between partially ordered sets

Order-isomorphisms are a very special type of map between partially ordered sets.

In this subsection, order-preserving maps are deﬁned more generally.

Let P and Q be partially ordered sets. A map φ : P Q is said to be →

(i) order preserving (or, alternatively, monotone) if x y in P implies φ(x) φ(y) ≤ ≤ in Q.

(ii) an order embedding if x y in P if and only if φ(x) φ(y) in Q. ≤ ≤

We note that a map φ : P Q is an order-isomorphism if it is an order embedding → which maps P onto Q.

The dual of a partially ordered set

Given any partially ordered set P we can form a new partially ordered set P d, the dual of P , by deﬁning x y to hold in P d if and only if y x holds in P . For P ≤ ≤ ﬁnite, we obtain a Hasse diagram for P d simply by “turning upside down” a Hasse diagram for P .

Down-sets and up-sets

Let P = (X, ) be a partially ordered set. For x X, we write ≤ ∈

D(x) := D( x ) = a X : a < x , { } { ∈ } 97 and

U(x) := U( x ) = a X : a > x , { } { ∈ } to denote the down-set and up-set of x, respectively. Also

D[x] := D(x) x , ∪ { } and

U[x] := U(x) x . ∪ { } Given x < y in P , we say

(x, y) := U(x) D(y), ∩ is an interval in P , and correspondingly,

[x, y] := (x, y) x y . ∪ { } ∪ { }

Sums

There are several diﬀerent ways to join two ordered sets together.

Suppose that P and Q are disjoint partially ordered sets. The disjoint union

P + Q of P and Q is the partially ordered set formed from P Q by deﬁning x y ∪ ≤ in P + Q if and only if either x, y P and x y in P or x, y Q and x y in Q. ∈ ≤ ∈ ≤ A diagram for P + Q is formed by placing side by side diagrams for P and Q.

Again let P and Q be disjoint partially ordered sets. The linear sum P Q is ⊕ deﬁned by taking the following order relation on P Q: x y if and only if ∪ ≤ x, y P and x y in P, ∈ ≤ x, y Q and x y in Q, ∈ ≤ or x P and y Q. ∈ ∈ 98

A diagram for P Q (when P and Q are ﬁnite) is obtained by placing a diagram ⊕ for P directly below a diagram for Q and then adding a line segment from each maximal element of P to each minimal element of Q. For example, 1 P represents ⊕ P with a new bottom element added; P 1 represents P with a new top element ⊕ added.

Height and width

Let P be a partially ordered set. The height of P is the number of elements in a maximum chain. The width of P is the size of a maximum antichain. We have the following well-known max-min result.

Observation 7.1 (Folklore) A poset P of height h can be partitioned into h antichains.

As can be easily seen, this result is trivial. While its dual version for chains, due to Dilworth, is deﬁnitely non-trivial and admits an elegant combinatorial proof.

Theorem 7.2 (Dilworth, 1950) A poset P of width w can be partitioned into w chains.

Dilworth’s theorem is typically grouped with other combinatorial max-min theorems with a common LP ﬂavour, such as Hall’s matching theorem, Tutte’s 1-factor theorem, the K¨onig-Egervary theorem and Menger’s theorem.

7.2 Colouring of Posets

Let P be a partially ordered set. As in the context of graphs, given a forbidden partially ordered set F , we can deﬁne acF (P ). 99

Deﬁnition. For a “forbidden partially ordered set” F , acF (P ) of the partially ordered set P is the smallest integer k so that the elements of P can be k-coloured such that each nontrivial maximal F -free subposet of P receives more than one colour.

Moreover, we denote

ac(F ) := sup ac (P ): P is a ﬁnite partially ordered set . { F }

In Chapter 4, we gave a complete characterization of graphs F for which ac(F ) is ﬁnite. But for posets, we haven’t succeeded in obtaining such a characterization.

Let us begin with the following easy fact.

d Lemma 7.3 Let F be a partially ordered set. Then we have acF (P ) = acF d (P ) for any partially ordered set P . Thus, ac(F ) = ac(F d).

d Proof. Given a partially ordered set P , set acF d (P ) =: k. Let Q be a maximal F -free subposet of P . Since Q is F -free in P , then Qd is F d-free in P d. For any x P Q, Q x contains a copy of F in P , and thus Qd x contains a copy ∈ − ∪ { } ∪ { } of F d in P d. Thus Qd is a maximal F d-free subposet of P d. This yields that each maximal F -free subposet of P induces a maximal F d-free subposet of P d.

d d Since k = ac d (P ), there is a colouring c : P [k] such that each maximal F → F d-free subposet of P d receives at least two colours. It follows that, with the same colouring c, each maximal F -free subposet of P receives at least two colours. Thus

d d ac (P ) ac d (P ). Similarly we have ac d (P ) ac (P ). Therefore ac (P ) = F ≤ F F ≤ F F d d acF d (P ). And hence, by deﬁnition of ac(F ), we have ac(F ) = ac(F ).

We ﬁrst consider the simplest F ’s. If we take F to be the 2-element antichain

2, then for any partially ordered set P , a maximal 2-free subposet of P will be a 100

maximal chain. Thus ac2(P ) is the smallest integer k so the elements of P can be k-coloured such that each nontrivial maximal chain of P receives more than one colour.

If we take F to be the 2-element chain 2, then for any partially ordered set P , a maximal 2-free subposet of P will be a maximal antichain. Thus ac2(P ) is the smallest integer k so the elements of P can be k-coloured such that each nontrivial maximal antichain of P receives more than one colour.

The following bounds for F = 2 and F = 2 are known.

Proposition 7.4 ([2]) ac(2) = 2.

This result was probably ﬁrst noted by Aigner and Andreae [2]. Indeed, the above value can be easily proven: we colour the maximal elements of P with colour

1, everything else with colour 2. Then clearly each nontrivial maximal chain of P , if any, receives both colours. The argument of the theorem below is more diﬃcult.

Theorem 7.5 ([24, 25]) ac(2) = 3.

Note that these two results establish ac(F ) for all posets F with F = 2. When | | F 3, for a given poset P with P 2, all maximal F -free subposets of P are | | ≥ | | ≥ nontrivial, as maximal F -free subposets of P have at least F 1 2 elements. | | − ≥

7.3 Bounds

In this section we give bounds for ac(F ) for posets F with F > 2. So in what | | follows all maximal F -free subposets of a poset P are nontrivial. 101

A poset is said to be bounded if it has both a maximum and a minimum; otherwise it is non-bounded. If P has a linear decomposition P = Q R, then P is said to be ⊕ linearly decomposable; P is linearly indecomposable if it has no linear decompositions.

First of all, we present the following theorem whose proof is easy to ﬁnd.

Theorem 7.6 If F is a non-bounded poset with no isolated elements then ac(F ) 3. ≤

Proof. By Lemma 7.3, we may assume that F does not contain a unique minimum.

We choose a minimal element x P , and colour x with colour 1, the set C(x) ∈ of all elements comparable to x with colour 2, and the set I(x) of all elements incomparable to x with colour 3. Formally, the colouring c is deﬁned by

1 if y = x,   c(y) := 2 if y C(x),  ∈ 3 if y I(x).  ∈   Now we claim that any maximal F -free subposet receives at least two colours.

Suppose Q is a maximal F -free subposet of P . Since colour class 1 consists of only one element, Q is not contained in colour class 1. If Q is coloured 2, then Q C(x). ⊆ Q x is still F -free, as if there were any copy of F in Q x , it must involve ∪ { } ∪ { } x, and thus would contain a unique minimum element x, contradicting the fact that

F contains no unique minimum. Consequently colour class 2 contains no maximal

F -free subposet of P . Similarly, if Q is coloured 3, then Q I(x). Q x is still ⊆ ∪ { } F -free, as if there were any copy of F in Q x , it must involve x, and thus would ∪ { } contain an isolated element x, contradicting the fact that F has no isolated elements.

Hence colour class 3 contains no maximal F -free subposet of P . 102

Remarks. (a) If F is a non-bounded poset with isolated elements, then ac(F ) 3 ≤ need not hold. For example, in the poset 2 + 2, every two elements form a maximal

(2 + 1)-free subposet. Thus ac2+1(2 + 2) = 4. Another example is F = 3 and P = n (n 3). Since in the poset n, any two elements form a maximal 3-free subposet, ≥ they must receive diﬀerent colours. Then ac3(n) = n.

(b) We do not know of any example of F showing that ac(F ) = 3.

7.4 Decomposition of Partially Ordered Sets

In the last section, we found a bound for ac(F ) for any poset F that is not bounded and contains no isolated elements. In this section, we investigate the bounds for bounded posets F . To do so, we introduce a decomposition technique that enables us to give a partial answer to the question. We shall see that, if F is bounded, but contains no splitting elements other than the minimum and the maximum, then 10 colours are enough.

We ﬁrst develop the following procedure that gives a decomposition of any linearly indecomposable poset. The idea of this decomposition is from [24].

For M P , we deﬁne ⊆

I(M) := x P M : x is incomparable to some element of M . { ∈ − }

For any poset P that is linearly indecomposable, we deﬁne a sequence

x > x > > x 1 2 ··· k of elements of P with certain properties that we make use of to colour the elements of P . Precisely, we have the following proposition. 103

Proposition 7.7 Let P be a poset that is linearly indecomposable. Then there exists a sequence x > x > > x in P such that 1 2 ··· k

(i) I(x ) max(P ) = and I(x ) min(P ) = ; 1 ∩ 6 ∅ k ∩ 6 ∅ (ii) for all 1 i k + 1, there exists z P incomparable to every element in ≤ ≤ i ∈

[xi, xi−1], where [x1, x0] := U[x1], [xk+1, xk] := D[xk];

(iii) for any a, b P , each of which is incomparable to some element in the se- ∈ quence x > x > > x , let j and i be the least indices so that a x , b x , 1 2 ··· k k j k i respectively. If i j 3, then we have a < x < b in P . ≤ − i+2

max(P )

¡ ¡ x1

£ ¢£ x2 ¢

PSfrag replacements M2

¤¥

¦§

min(P ) Mk Figure 7.1: A partition of a linearly indecomposable poset

Proof. We ﬁrst note that I(max(P )) = , as if not, then P = (P max(P )) 6 ∅ − ⊕ max(P ) is a linear decomposition. Choose x P to be minimal in I(max(P )), thus 1 ∈ 104

I(x ) max(P ) = . Note that x / max(P ) by deﬁnition of I(max(P )). Denote 1 ∩ 6 ∅ 1 ∈ M := max(P ) I(x ) = , 0 ∩ 1 6 ∅ M := min(I(x )) = . 1 1 6 ∅ If M min(P ) = , then continue to deﬁne x as below; otherwise, stop and set 1 ∩ ∅ 2 k := 1.

Assume that x − has been chosen and M − min(P ) = , where M − := i 1 i 1 ∩ ∅ i 1 min(I(x − )) (note that I(x − ) = ). It follows that D(x − ) = , because x − i 1 i 1 6 ∅ i 1 6 ∅ i 1 ∈ min(P ) implies M − min(P ) = , a contradiction. Also, we have Y := P i 1 ∩ 6 ∅ −

D(x − ) = U[x − ] I(x − ) = . Choose x D(x − ) to be minimal such that i 1 i 1 ∪ i 1 6 ∅ i ∈ i 1

I(x ) M − = . Such an x exists, as if not, then any element of D(x − ) is i ∩ i 1 6 ∅ i i 1 comparable to all elements of Mi−1 and hence is less than all elements of Mi−1; consequently any element of D(xi−1) is less than all elements of Y , and thus P =

D(x − ) Y is a linear decomposition, a contradiction. Denote i 1 ⊕

Mi := min(I(xi)).

If M min(P ) = , then continue to deﬁne x ; otherwise, set k := i and stop. The i ∩ ∅ i+1 procedure is depicted in Figure 7.1.

Let us now show that the sequence x1 > x2 > . . . satisﬁes (i)–(iii). Since x > x > > x > . . . , and P is a ﬁnite poset, we will, at some step, get 1 2 ··· i a certain k such that M min(P ) = , where M := min(I(x )), thus (i) holds. k ∩ 6 ∅ k k If i = 1, then [x , x ] = U[x ], and we choose z M . For any a U(x ), if z 1 0 1 ∈ 0 ∈ 1 is comparable to a then it must be z > a x , a contradiction. Similarly, if i = k, ≥ 1 then [x , x ] = D[x ], and we choose z M min(P ). For any a D(x ), if z is k+1 k k ∈ k ∩ ∈ k comparable to a then it must be z < a x , a contradiction. ≤ k 105

Now we assume that 2 i k. Choose any element z M − I(x ) = . For ≤ ≤ ∈ i 1 ∩ i 6 ∅ any a [x , x − ], if z > a, then z > a x , contradicting the fact that z I(x ); ∈ i i 1 ≥ i ∈ i if z < a, then z < a x − , contradicting the fact that z M − := min(I(x − )). ≤ i 1 ∈ i 1 i 1 Thus (ii) also holds.

Now let us prove (iii). Let a I(x ), b I(x ), i j 3, where i and j are the ∈ j ∈ i ≤ − least indices so that a x and b x respectively. We know that a < x − x and, k j k i j 1 ≤ i+2 by the choice of xi+1 among the elements of D(xi), every element below xi+1 is less than all elements in I(xi), thus xi+2 is less than all elements of Mi = min(I(xi)). As b I(x ) is greater than or equal to at least one of M = min(I(x )), x < b as ∈ i i i i+2 desired.

Having found this decomposition, we can now deal with posets P which are not linearly decomposable. We ﬁrst prove the following lemma which deals with F -free subposets in linear sums.

An element of a poset P is a splitting element if it is comparable to every element of P . A splitting element of a poset P is called an interior splitting element if it is not the maximum or the minimum of P .

Lemma 7.8 Let F be a poset with no interior splitting elements. If Q = Q Q 1 ⊕ 2 is an F -free poset, then Q0 := Q x Q is F -free as well. 1 ⊕ { } ⊕ 2

Proof. Let us suppose, to the contrary, that Q0 contains a copy F 0 of F . Thus F 0 must involve x as Q is F -free. If the element x is the minimum of F 0, then for any y Q ,(F 0 x ) y is a copy of F in Q, contrary to the fact that Q is F -free. ∈ 1 \{ } ∪ { } Similarly, x is not the maximum of F 0. 106

Now let us assume that x is neither the minimum nor the maximum of F 0. Hence

0 0 F meets both Q1 and Q2, and thus x is a splitting element of F , contrary to the fact that F has no interior splitting elements.

With these tools in hand, we can prove the following bound for linearly indecomposable posets P .

Theorem 7.9 Let F be a bounded poset with no interior splitting elements. If P is a partially ordered set that is linearly indecomposable, then ac (P ) 5. F ≤

Proof. Theorem 7.5 yields that ac2(P ) 3 for any poset P . Thus we may assume ≤ that F = 2. Proposition 7.7 yields a sequence x > x > > x in P satisfying 6 1 2 ··· k the properties (i)–(iii) there.

We colour the poset P as follows. Colour all xi’s with colour 1. If an element y = x (1 i k) is comparable to all the x ’s, colour y with colour 2. Otherwise y 6 i ≤ ≤ i is incomparable to some element x . Choose s to be the least i so that y x . Colour i || i y with colour 3 if s 1 (mod 3), colour it with colour 4 if s 2 (mod 3), and colour ≡ ≡ it with colour 5 if 3 s. | Formally, we deﬁne the colouring c as follows:

1 if y = x , for some 1 i k,  i ≤ ≤   2 if y is comparable to all the xi’s,    c(y) = 3 if s 1 (mod 3),  ≡ 4 if s 2 (mod 3),  ≡   5 if 3 s.   |  107

We argue that this colouring assigns at least two colours to each maximal F -free subposet. Suppose Q is a monochromatic maximal F -free subposet.

If Q is coloured 1, then Q x , . . . , x and thus Q is a chain. The proof of ⊆ { 1 k} property (ii) of Proposition 7.7 yields an element z M incomparable to x . If ∈ 0 1 z > max(Q), then Q z is a chain, and hence is F -free as F = 2 and F has no ∪ { } 6 interior splitting elements. If z max(Q), then Q z has no maximum, and any || ∪ { } subposet of Q z involving z is either a chain or has no maximum. Thus Q z ∪ { } ∪ { } is still F -free as F has a maximum and contains no interior splitting elements.

k+1 Now suppose Q is coloured 2, then Q (x , x − ), where (x , x ) := U(x ), ⊆ i=1 i i 1 1 0 1 S (xk+1, xk) := D(xk). If Q contains elements of (xi, xi−1) and (xj, xj−1) for i < j, then Q x is still F -free by Lemma 7.8. Hence we suppose Q (x , x − ), for ∪ { i} ⊆ j j 1 some 1 j k + 1. Proposition 7.7 (ii) yields an element z P incomparable to ≤ ≤ ∈ all q Q. It follows that Q z is still F -free, since if there were any copy of F in ∈ ∪ { } Q z , it must involve z which is incomparable to all elements of Q, contradicting ∪ { } the fact that F (being bounded) has no isolated elements. Thus Q is not a maximal

F -free subposet.

Suppose Q is coloured 3. If Q is contained in I(xi) for a single index i, then Q x is still F -free, since if there were any copy of F in Q x , it must involve ∪ { i} ∪ { i} xi, contradicting the fact that F has no isolated elements. It follows that Q is not a maximal F -free subposet of P . Thus there must be at least two indices i and j so that Q I(x ) = and Q I(x ) = . Without loss of generality, we assume that ∩ i 6 ∅ ∩ j 6 ∅ i < j. Hence i j 3 ( i j (mod 3) by deﬁnition of the colouring). For any ≤ − ≡ a Q I(x ), b Q I(x ), we know from the deﬁnition of the colouring c that i ∈ ∩ j ∈ ∩ i and j are the least indices such that a I(x ) and b I(x ). Thus a < x < b by ∈ j ∈ i i+2 108

Proposition 7.7 (iii).

From Lemma 7.8, Q x is still an F -free subposet of P , contradicting the ∪ { i+2} supposition that Q is a maximal F -free subposet of P .

An argument similar to the above shows that Q cannot be coloured 4 or 5.

Lemma 7.10 Let F be a bounded partially ordered set with no interior splitting elements. If P = P P P is a linearly decomposable poset, then P is 1 ⊕ 2 ⊕ · · · ⊕ k

2-colourable so that every maximal F -free subposet of P which is not in a single Pi (1 i k) receives both colours. ≤ ≤

Proof. We colour Pi’s alternatingly with colours 1 and 2. Formally, we deﬁne the colouring c : P [2] as follows. →

1 if y P , i is odd,  ∈ i c(y) :=   2 if y P , i is even. ∈ i   Suppose Q is a maximal F -free subposet of P which is not in a single Pi. If Q is monochromatic, then Q intersects P and P where i j 2. Thus for any x P , i j ≤ − ∈ i+1 P < x < P . Consequently Q0 = Q x is still F -free by Lemma 7.8, contradicting i j ∪ { } the maximality of Q.

Now we are ready to prove the bound 10 for bounded posets F which contain no interior splitting elements.

Theorem 7.11 Let F be a bounded poset with no interior splitting elements. Then ac(F ) 10. ≤

Proof. Let P be a partially ordered set. Suppose P = P P P where each P 1 ⊕ 2 ⊕· · ·⊕ k i (1 i k) is linearly indecomposable. Using Theorem 7.9, there exists a colouring ≤ ≤ 109 c : P [5] such that c assigns at least two colours to every maximal F -free 1 → 1|Pi subposet in P for each 1 i k. Lemma 7.10 guarantees a colouring c : P [2] i ≤ ≤ 2 → such that every maximal F -free subposet receives at least two colours if it is not contained in a single Pi. Consider the colouring

c = c c : P [5] [2]. 1 × 2 → ×

Suppose Q is a maximal F -free subposet of P . If Q is in a single P (1 i k), i ≤ ≤ where Pi is linearly indecomposable, then c1 assigns at least two colours to Q. If Q is not contained in a single Pi, then c2 assigns at least two colours to Q. Thus all maximal F -free subposets of P get at least two colours under the colouring c.

Remark. If F is a bounded poset with interior splitting elements, then the bound

10 need not hold. For example, if F = 3 and P = n (n > 10), then in the poset n any two elements form a maximal 3-free subposet. It follows that any two elements must receive diﬀerent colours. Thus ac3(n) = n > 10.

7.5 Relations between Various Bounds

The particular maximal F -free subposet depends on F . However, we have the following relation.

Theorem 7.12 Let F be a poset. Then every maximal (2 F )-free subposet of a ⊕ poset P contains a maximal (1 F )-free subposet. ⊕

Proof. Let F 0 = 1 F and F 00 = 1 F 0 = 2 F . Let Q be a maximal F 00-free ⊕ ⊕ ∼ ⊕ 0 subposet of P . We note that the poset Q may contain copies of F . Denote by MQ 110 the set of all elements of Q which are minimums of copies of F 0 in Q. In notation, we have

M := q Q : q is the minimum of at least one copy of F 0 in Q . Q { ∈ }

We ﬁrst note that M is an antichain in P . Given any a, b M , if a < b, Q ∈ Q 0 then we have that, by deﬁnition of MQ, b is the minimum of a copy L of F , hence L a Q is a copy of F 00, contrary to our supposition that Q is F 00-free. ∪ { } ⊆ Now we denote Q0 := Q M and proceed to show that Q0 is a maximal F 0-free − Q subposet of P . Clearly Q0 is F 0-free, since if Q0 has a copy L of F 0, then the minimum

0 element of L would be in Q (and hence not in MQ), contradicting the deﬁnition of

MQ. For any element u P Q0, we show that Q0 u contains a copy of F 0. If ∈ − ∪ { } u Q, then we have u M since u / Q0. Thus u is the minimum of some copy ∈ ∈ Q ∈ 0 K of F in Q, by deﬁnition of MQ. Since MQ is an antichain (as was shown earlier), K u Q0. Hence Q0 u contains K, which is a copy of F 0. − { } ⊆ ∪ { } If u P Q, then Q u contains a copy τ(F 00) of F 00, since Q is maximal ∈ − ∪ { } F 00-free, where τ : F 00 Q u is a ﬁxed order embedding of posets. Denote the → ∪ { } minimum of F 00 by x , and the upper cover of x in F 00 by y. We distinguish the following two cases.

Case 1. u = τ(x). Then τ(F 0) τ(y) Q0 since τ(y) M and M is an − { } ⊆ ∈ Q Q antichain. But F 0 = (τ(F 0) τ(y) ) u Q0 u ; that is, Q0 u contains a ∼ − { } ∪ { } ⊆ ∪ { } ∪ { } copy of F 0.

Case 2. u = τ(x). Then either τ(x) Q0 or τ(y) Q0, since τ(x) < τ(y) 6 ∈ ∈ and M is an antichain. In either case, F 0 = τ(F 00) τ(y) Q0 u or F 0 = Q ∼ − { } ⊆ ∪ { } ∼ 111

τ(F 00) τ(x) Q0 u is a copy of F 0 contained in Q. − { } ⊆ ∪ { } Remark. The stronger conclusion that each maximal (1 F )-free subposet of a ⊕ poset P contains a maximal F -free subposet is not true. For example we take F to be 2, and P to be 1 2. It is easily seen that the maximal elements of P form ⊕ a maximal (1 2)-free subposet that does not contain a maximal chain; that is, a ⊕ maximal 2-free subposet of P .

The preceding theorem implies the following relation.

Corollary 7.13 Let F be a poset. Then

ac(2 F ) ac(1 F ). ⊕ ≤ ⊕

Proof. If ac(1 F ) = , the inequality is obviously true. If ac(1 F ) is ﬁnite, set ⊕ ∞ ⊕ k := ac(1 F ). Thus we can colour the elements of any poset P with k colours so that ⊕ every maximal 1 F -free subposet receives at least two colours. Since every maximal ⊕ (2 F )-free subposet contains a maximal 1 F -free subposet, then each maximal ⊕ ⊕ (2 F )-free subposet receives at least two colours as well. Thus ac(2 F ) k. ⊕ ⊕ ≤

Corollary 7.14 For any k 2, each maximal k-free subposet of any poset P con- ≥ tains a maximal antichain.

Proof. If k = 2, a maximal 2-free subposet of P is itself a maximal antichain.

When k 3, we use induction. Suppose that each maximal (k 1)-free subposet of ≥ − P contains a maximal antichain. By Theorem 7.12, each maximal k-free subposet of P contains a maximal (k 1)-free subposet which, by the inductive hypothesis, − contains a maximal antichain. Thus, we conclude that each maximal k-free subposet of P contains a maximal antichain of P . PSfrag replacements

112

Figure 7.2: the 17-element poset Q

Let P and Q be posets. The lexicographic product P Q of P and Q is deﬁned × on the Cartesian product P Q by: (x , x ) (y , y ) if x < y or (x = y × 1 2 ≤ 1 2 1 1 1 1 and x y ). Thus the lexicographic product P Q is obtained by replacing each 2 ≤ 2 × element p of P by a copy Qp of Q, and putting all points of Qp below all points of

0 Qp0 if p < p in P .

Theorem 7.15 ac(k) = 3 for any k 2. ≥

Proof. By Theorem 7.5, any poset P can be coloured with 3 colours such that each nontrivial maximal antichain of P receives at least two colours. Using Corollary 7.14, we know that with the same colouring, each maximal k-free subposet of P receives at least two colours as well. Thus ac(k) 3. ≤ Now we show that, for any k-element chain k (k 2), we can construct a poset ≥ P such that ack(P ) = 3. Our objective is to construct a poset P such that any

2-colouring of it will yield a monochromatic maximal k-free subposet. In [25], a

17-element poset Q is constructed, as depicted in Figure 7.2, that has ac2(Q) = 3. 113

Figure 7.3: A poset P with ack(P ) = 3

We consider the product P = Q C in the lexicographic order, where C = 2k 3 × − is a (2k 3)-element chain, as indicated in Figure 7.3. For q Q, denote C := − ∈ q (q, x): x C . { ∈ } Suppose to the contrary that there is a colouring c : P [2] such that each → maximal k-free subposet of P receives at least two colours. This colouring c of P induces a colouring c0 : Q [2] as follows. For each q Q, we colour q with the → ∈ colour that appears on at least k 1 elements of C in P . We shall show that, − q with this colouring c0, any maximal antichain A of Q receives two colours, and thus ac2(Q) = 2, contradicting the fact that ac2(Q) = 3. 114

Let A be a maximal antichain of Q. For any q Q, let S C be a (k 1)- ∈ q ⊂ q − element chain in Cq. First note the following fact.

0 Fact. A := q∈A Sq yields a maximal k-free subposet of P . S It is clear that A0 is a k-free subposet of P . To see that A0 is a maximal subposet of P with this property, we show that for any element (q, x) P A0, A0 (q, x) ∈ − ∪ { } has a k-element chain. By construction of P ,(q, x) C where q Q. If q A ∈ q ∈ ∈ then (q, x) C S since (q, x) / A0. It follows that A0 (q, x) has a k-element ∈ q\ q ∈ ∪ { } chain since S is a (k 1)-element chain and C is a (2k 3)-element chain. Now q − q − suppose q / A. Because A is a maximal antichain of Q, there exists an element ∈ q0 A such that either q < q0 or q0 < q. Without loss of generality we assume ∈ 0 0 q < q . By definition of P , each element of S 0 A is greater than (q, x) C . Thus q ⊆ ∈ q 0 A (q, x) has a k-element chain, i.e., the elements of S 0 and (q, x). ∪ { } q Now in the colouring c, for each element q A, there is a set D C of k 1 ∈ q ⊆ q − 0 elements that is monochromatic. Using the preceding fact, we see that A := q∈A Dq S is a maximal k-free subposet of P and thus is 2-coloured. Since Dq is monochromatic for each q A, there must be q , q A such that D and D get different colours. ∈ 1 2 ∈ q1 q2 Thus, by definition of the colouring c0 : Q [2], the two elements q and q in A get → 1 2 different colours and hence A is 2-coloured. This shows that the colouring c0 assigns two colours to each maximal antichain of Q, and hence ac2(Q) = 2, a contradiction.

Thus ack(P ) 3, and so ack(P ) = 3. Therefore ac(k) = 3 for any k 2. ≥ ≥

To investigate ack(P ) for the k-element antichain k, we ﬁrst prove the following result which has the same ﬂavour as Theorem 7.12. 115

Theorem 7.16 Let P be a partially ordered set. Then every maximal k-free subposet

(k 3) of P contains a maximal (k 1)-free subposet of P . ≥ −

Proof. Let A be a maximal k-free subposet of P . Then A has width at most k 1. − We know from Dilworth’s Theorem that A can be decomposed into at most k 1 − disjoint chains.

Of all such Dilworth decompositions of A, we choose one for which the length of the shortest chain C is minimum.

We claim that Q := A C is a maximal subposet of P with width at most k 2. − − Note that Q has width at most k 2. For any element x P , if x / Q, then either − ∈ ∈ x A Q or x P A. If x A Q = C, then the width of Q x must be at ∈ − ∈ − ∈ − ∪ { } least k 1 as otherwise it contradicts the choice of C. − If x P A, then the width of Q x is at least k 1 as well, as otherwise the ∈ − ∪ { } − width of A x would be at most k 2 + 1 = k 1, contradicting the maximality ∪ { } − − of A. Thus the subposet Q is a maximal (k 1)-free subposet of P . −

Corollary 7.17 Let P be a partially ordered set. Then every maximal k-free subposet (k 2) of P contains a maximal chain. ≥

Proof. If k = 2, then a maximal 2-free subposet is itself a maximal chain of P .

When k 3, we use induction. Suppose that each maximal (k 1)-free subposet of ≥ − P contains a maximal chain. By Theorem 7.16, each maximal k-free subposet contains a maximal (k 1)-free subposet which, by the inductive hypothesis, contains a − maximal chain of P . Thus, we conclude that each maximal k-free subposet contains a maximal chain of P . 116

At this point, one might naturally wonder if Theorem 7.16 could be strengthened

to a stronger form similar to what is in Theorem 7.12. The following example shows

that, in general, the counterpart of Theorem 7.12 does not hold.

Example 7.1 Let F = 2 + 1 and F 0 = F + 1. Then in the poset P indicated in

Figure 7.4, the circled elements constitute a maximal F 0-free subposet of P that does

not contain a maximal F -free subposet of P . PSfrag replacements

a1 For example, in Figure 7.4, the F -free subposet a2, a7 can be extended by a4 a2 { } or a , and thea F -free subposet a , a , a can be extended by a . 8 3 { 3 6 7} 8 a4 a5 a6 a7 a PSfrag8 replacements F F 0

a5 a6 a7 a8

F a1 a2 a3 a4 F 0

Figure 7.4: Example 7.1

Theorem 7.18 ack(P ) = 2, k 3, for each poset P = n. ≥ 6

Proof. Let P be any partially ordered set other than an antichain. We colour all

maximal elements of P with colour 1, everything else with colour 2. Then every 117 nontrivial maximal chain receives at least two colours. Hence every maximal k-free subposet of P is two coloured, since it contains a nontrivial maximal chain. Thus ack(P ) = 2.

Remark. If P is an antichain, then the above bound does not hold. For example, if P = n (n k), then every k 1 elements of P are a maximal k-free subposet of ≥ − P . Thus the minimum number of colours needed is n/(k 1) , and ac(k) = + . d − e ∞

7.6 An Approach towards Two Colourable Posets

In Theorem 7.6, we have shown that if a non-bounded poset F has no isolated element, then ac(F ) 3. In this section, we shall utilize an idea from the last ≤ section to further investigate which posets F are such that ac(F ) = 2.

Corollary 7.17 shows that if F is an antichain, then for any poset P , each maximal

F -free subposet of P contains a maximal element. Consequently, ack(P ) = 2 for any

P other than an antichain (Theorem 7.18). Along the same lines, the question that we are concerned with in this section is

Question:

Which posets F are such that, for all posets P , all maximal F -free subposets of P contain a maximal element of P ?

We say that a poset F satisfying the above property has the “maximal element property.” We ﬁrst give a construction showing that for a large number of posets F , the above property does not hold.

Proposition 7.19 Let F be a poset with a connected component A so that F A − 118 does not contain a subposet isomorphic to A. Then there is a poset P so that P minus its maximal elements is a maximal F -free subposet of P .

Proof. The poset P is constructed as follows. Let a be a maximal element of A. Let

F1 and F2 be two copies of F , and let a1 and a2 be the copies of a in F1 and F2, respectively. Then P is the disjoint union of F1 and F2, with the additional relations that a is greater than all elements of F a , and a is greater than all elements 1 2 − { 2} 2 of F a . 1 − { 1} Note that a and a are the only maximal elements of P , and P a , a is 1 2 − { 1 2} the disjoint union of two copies of A a and two copies of F A, and therefore − { } − P a , a is F -free (because by supposition A is connected and F A does not − { 1 2} − contain a copy of A). However P a and P a each contains a copy of F . − { 1} − { 2} Thus P a , a is maximal F -free. − { 1 2} In particular, this construction shows that no connected poset F has the maximal element property (here A = F and F A is empty). Also, no disconnected poset − F whose largest component appears only once has this property. The major posets left to look at are, for example, posets consisting of at least two copies of the same poset. The simplest such poset is 2 + 2. The following theorem shows that the poset

2 + 2 does satisfy the maximal element property.

Theorem 7.20 Let P be any poset. Then each maximal (2 + 2)-free subposet of P contains a maximal element of P .

Proof. Let Q be a maximal (2 + 2)-free subposet of P . We assume to the contrary that Q contains no maximal elements of P . Thus for any maximal element m P , ∈ 119

m m0 PSfrag replacements

Figure 7.5: the relation m n ≺Q

Q m contains a copy of 2 + 2 in which the element m is a maximal element. We ∪ { } write m0 for the other maximal element of this copy of 2 + 2, as indicated in Figure

7.5. By assumption, m0 is not a maximal element of P , hence there exists a maximal element n P such that m0 < n, as indicated in Figure 7.5. ∈ For a given subposet Q, the above description gives rise to a relation in the set of maximal elements of P : for two maximal elements m, n P , we say m precedes ∈ n with respect to Q, and write m n, if n > m0 where m0 is the maximal element, ≺Q other than m, in some copy of 2 + 2 contained in Q m . ∪ { } Since, by supposition, Q contains no maximal elements of P , then for any maximal element a P , there exists a maximal element b P such that a b. Because ∈ ∈ ≺Q the poset P is ﬁnite, there must be a cycle

a a a a 1 ≺Q 2 ≺Q · · · ≺Q k ≺Q 1 in P , k 2, where the elements a are maximal elements of P , as depicted in Figure ≥ i 7.6. Note that a , b , a0 , c = 2 + 2, for all 1 i k. { i i i i} ∼ ≤ ≤

0 Fact 1. ci ai−1, i = 1, . . . , k, where subscripts are taken modulo k. PSfrag replacements

120

a1 a2 a3 ak

0 0 0 0 ak a1 a2 ak−1 ...... b1 b2 b3 bk

ck c1 c2 ck−1

Figure 7.6: a a a a 1 ≺Q 2 ≺Q · · · ≺Q k ≺Q 1

0 0 Since a − a , we have a > a ; since a , b , a , c is a copy of 2 + 2 in i 1 ≺Q i i i−1 { i i i i} Q a , we have a c . Thus, we see that c a0 . ∪ { i} ik i i i−1 0 Fact 2. ci < ai+1, i = 1, . . . , k, where subscripts are taken modulo k.

Since a0 , c , a0 , c is not 2 + 2, at least one of the inequalities c < a0 and { i i i+1 i+1} i+1 i 0 0 0 ci < ai+1 holds. Using Fact 1, ci+1 ai, and thus there must be ci < ai+1.

0 Fact 3. c1 < ai, i = 2, . . . , k.

0 0 We use induction on i. Fact 2 yields that c1 < a2. Suppose that c1 < ai for some 2 i < k; we show that c < a0 . Since a0 , c , a0 , c is not 2 + 2, at least one ≤ 1 i+1 { i 1 i+1 i+1} 0 0 0 of the inequalities ci+1 < ai and c1 < ai+1 holds. Fact 1 yields that ci+1 ai, and

0 thus there must be c1 < ai+1.

0 0 Fact 3 yields that c1 < ak in P , while we know from Fact 1 that c1 ak, a contradiction. This contradiction shows that the maximal (2 + 2)-free subposet Q

must contain a maximal element of P .

Corollary 7.21 ac(2 + 2) = 2.

Proof. Let P be a partially ordered set. If P is an antichain, ac2+2(P ) = 2 since P does not contain 2 + 2. If P is not an antichain, then a maximal 2 + 2-free subposet 121

of P cannot be an antichain. Note that the set of maximal elements of P is not the

entire poset P . Thus we colour the maximal elements of P with colour 1, and other

elements (which must exist) with colour 2. Theorem 7.20 yields that each maximal

(2 + 2)-free subposet of P receives both colours.

Now we know that the poset 2 + 2 has the maximal element property. What

about other posets F that are disjoint unions of chains? We ﬁrst note that if F is 2k+1 the disjoint union of an odd number of 2’s, i.e. F = 2,(k 1), then it does not ≥ X1 have the maximal element property.

2k+1 Example 7.2 For any k 1, there is a poset P such that some maximal ( 2)- ≥ X1 free subposet of P contains no maximal element of P .

PSfrag replacements ......

C1 C2 C3k+1

2k+1 Figure 7.7: ( 2)-free X1

Proof. We consider the poset shown in Figure 7.7, where the width of the poset is 2k+1 3k + 1. Then it is easily seen that the circled elements of P constitute a ( 2)-free X1 2k+1 subposet Q of P . For, if there were a copy F of 2 contained in Q, then between X1 122 any two 2’s of F there must be an element of Q not contained in F , otherwise there are comparabilities between the two copies of the 2’s, a contradiction. It follows that the subposet Q would have at least 2(2k+1)+2k+1 = 6k+3 elements, contradicting 2k+1 the fact that Q = 6k +2. Thus Q is indeed ( 2)-free. If a maximal element of P , | | X1 which by symmetry we may suppose to be the solid point at the upper right corner, 2k+1 is added to Q, then the new subposet contains 2, as is shown by the thick lines X1 in the diagram.

Remark. The construction in Example 7.2 does not work for a disjoint union of an 2k even number of 2’s, i.e. F = 2, k 1. This is because in order for the circled ≥ X1 2k part to be ( 2)-free, its width must be at most 3k 1 (if the circled part has width − X1 2k 3k, it will contain 2). In this case, adding in any maximal element of P will not X1 result in a new 2 disjoint from others. The case of a disjoint union of an even number of 2’s is left open. We do not even know the answer to the next simplest such F , i.e., F = 2 + 2 + 2 + 2.

If F is the disjoint union of at least two n’s, (n 3), then the following example ≥ shows that F does not have the maximal element property.

Example 7.3 For any k 2 and n 3, there is a poset P such that some maximal k ≥ ≥ ( n)-free subposet of P contains no maximal element of P . X1 Proof. Consider the poset shown in Figure 7.8, where the width of the poset is 2k 1, − and the chains Ci (1 i 2k + 1) all have n + 1 elements. Then it is easily seen ≤ ≤ k that the circled elements of P constitute a ( n)-free subposet Q of P . For, if there X1 123

......

PSfrag replacements

......

C1 C2 C2k−3 C2k−2 C2k−1

k Figure 7.8: ( n)-free X1

k were a copy F of n contained in Q, then between any two n’s of F there must X1 be another n of Q not contained in F , otherwise there are comparabilities between

these two copies of the n’s, a contradiction. It follows that the subposet Q would

have at least k + k = 2k n’s, contradicting the deﬁnition of Q. Thus Q is indeed k ( n)-free. If a maximal element of P , which by symmetry we may suppose is the X1 solid point at the upper right corner, is added to Q, then the new subposet contains k n, as is shown by the thick lines in the diagram. X1 k The following example resolves the case F = (2 1), k 2. ⊕ ≥ X1 124

Example 7.4 For any k 2, there is a poset P such that some maximal k ≥ ( (2 1))-free subposet of P contains no maximal element of P . ⊕ X1

PSfrag replacements ......

C1 C2 C2k−3 C2k−2 C2k−1

k Figure 7.9: ( (2 1))-free ⊕ X1

Proof. We consider the poset shown in Figure 7.9, where the number of 4-element

chains is 2k 1. Then, analogous to the arguments in Examples 7.2 and 7.3, it is − k easily seen that the circled elements of P constitute a ( (2 1))-free subposet Q of ⊕ X1 P . If a maximal element of P , which by symmetry we may suppose is the solid point k at the upper right corner, is added to Q, then the new subposet contains (2 1), ⊕ X1 as is shown by the thick lines in the diagram. k Having seen the above example for F = (2 1), we would naturally think of ⊕ X1 k k the question: what about the dual of (2 1), i.e., F = (1 2)? The following ⊕ ⊕ X1 X1 k example shows that the poset F = (1 2) does not satisfy the maximal element ⊕ X1 property either. 125

PSfrag replacements

......

C1 C2 C3 C2k−2 C2k−1

k Figure 7.10: ( (1 2))-free ⊕ X1

Example 7.5 For any k 2, there is a poset P such that some maximal k ≥ ( (1 2))-free subposet of P contains no maximal element of P . ⊕ X1 Proof. Consider the poset shown in Figure 7.10, where the total number of disjoint

3-element chains is 2k 1. Then it is easily seen that the circled elements of P k − constitute a ( (1 2))-free subposet Q of P . If a maximal element of P , which ⊕ X1 by symmetry we may suppose is the solid point at the upper right corner, is added k to Q, then the new subposet contains (1 2), as is shown by the thick lines in ⊕ X1 the diagram.

In summary, the only posets that we know with the maximal element property

are antichains and 2+2. In the near future, we hope to be able to solve the question

of determining which posets have the maximal element property. Chapter 8

Remarks and Open Problems

8.1 Open Problems

Exact Values

As mentioned in Chapter 1, the F -avoiding chromatic number is diﬀerent from the ordinary chromatic number. For the ordinary chromatic number, it is trivially true that for any integer k there exist graphs G such that the chromatic number of G equals k. However, for the F -avoiding chromatic number, F being a graph with either an isolated vertex or a dominating vertex, the corresponding question is nontrivial.

We have shown in this thesis that for a graph F with either an isolated vertex or a dominating vertex, and for any k 2, there exists a graph G such that ac (G) > k. ≥ F

Problem 1 Given an integer k 3, for a graph F with either an isolated vertex or ≥ a dominating vertex, does there exist a graph G such that acF (G) = k?

For example, for F = K (r 3), we know from the proof of Proposition 2.10 that r ≥ n ac r (K ) = . Thus for any integer k 2, there is an n such that ac r (K ) = k. K n d r−2 e ≥ K n

Analogously, for F = Kr, the answer is positive. For all other graphs F , we do not know the answer to Problem 1.

It is worth pointing out that the hypergraph construction that is used in Theorem

4.1 would be problematic if we require ac (G) = k rather than ac (G) k. This is F F ≥ because when constructing the graph G, hyperedges of a certain hypergraph are H 126 127 replaced by certain graphs, and new maximal F -free subgraphs might be produced in the resulting graph G. It thus follows that ac (G) χ( ) instead of ac (G) = χ( ). F ≥ H F H

Computational Complexity

An important aspect is the complexity of computing the F -avoiding chromatic number. The simplest case is whether the following question is NP-hard.

Problem 2 Let F be a graph with either an isolated vertex or a dominating vertex.

Instance: A graph G

Question: Is acF (G) = 2?

Asymptotically Better Bounds

We have proven in Theorem 4.4 that 2 √n +1 is an upper bound of ac (G)(n = G ) d e F | | when F has either an isolated vertex or a dominating vertex, but is not itself an independent set or a complete graph. A natural question is

Problem 3 Let F be a graph with either an isolated vertex or a dominating vertex, but which is not an independent set or a complete graph. Does acF (G) = o(√n) hold for every graph G?

The requirement that F is not an independent set or a complete graph is neces-

n sary; we know from the proof of Proposition 2.10 that ac r (K ) = ac (K ) = . K n Kr n d r−2 e

2-colourability for Bounded F

In Chapter 4 we have proven that if F has neither an isolated vertex nor a dominating vertex, then ac(F ) is bounded by 3. Determining which graphs F have ac(F ) = 3 is a natural question. 128

Problem 4 For a graph F with neither an isolated vertex nor a dominating vertex, is ac(F ) = 2 or 3?

We know from Theorem 5.3 that acP4 (C5) = 3, and thus ac(P4) = 3. This is the only such graph F that we know of that satisﬁes ac(F ) = 3, and I do not know of any graph F with ac(F ) = 2.

In the setting of partially ordered sets, the corresponding question of determining which posets F have ac(F ) = 2 is interesting as well.

Problem 5 Is it true that for all posets F that are not a chain or an antichain, ac(F ) is bounded?

Due to lack of counterexamples, we suspect that the answer for this problem is yes.

8.2 Classes of Graphs F

The concept of F -avoiding colouring can be generalized to a family of graphs. F Given a family of graphs, a subgraph of G is said to be -free if it does not F F contain any member of as an induced subgraph. For example, on page 22 we F deﬁned (P5,C5)-free graphs.

The notation acF (G) allows the concept of avoiding colouring to include other concepts of colourings. For example, if we take to be K ,K , then acF (G) is F { 2 3} the ordinary chromatic number of a graph G, namely the least number of colours needed so that all edges get diﬀerent colours. In this thesis, we study the avoiding 129 chromatic number only for a single graph F ( = 1). Many more questions arise |F| if we consider > 1. |F| Bibliography

[1] D. Achlioptas, J.I. Brown, D.G. Corneil and M.S.O. Molloy, The existence of

uniquely -G colourable graphs, Discrete Mathematics 179 (1998), 1-11.

[2] M. Aigner and T. Andreae, Vertex sets that meet all maximal cliques of a graph,

1986, manuscript.

[3] N. Alon, M. Krivelevich and B. Sudakov, Subgraphs with large cochromatic

number, J. Graph Theory 25 (1997), 295-297.

[4] T. Andreae, M. Schughart and Z. Tuza, Clique-transversal sets of line graphs

and complements of line graphs, Discrete Math. 88 (1991), 11-20.

[5] G. Bacsó,S. Graver, A. Gyárfás,M. Preissmann, and A. Seb˝o,Coloring the

maximal cliques of graphs, SIAM J. Discrete Math. 17 (2004), 361-376.

[6] G. Bacs´oand Z. Tuza, Dominating cliques in P5-free graphs, Periodica Mathe- matica Hungarica 21 (1990), 303-308.

[7] C. Berge, Graphs and Hypergraphs, North Holland, New York, 1979.

[8] C. Berge, Hypergraphs, North Holland, New York, 1989.

[9] B. Bollob´as, Random Graphs, Cambridge University Press, 2nd ed., New York,

2001.

[10] R.L. Brooks, On colouring the nodes of a network, Proc. Cambridge Phil. Soc.

37 (1941), 194-197.

130 131

[11] J.I. Brown and D.G. Corneil, On generalized graph colorings, J. Graph Theory

11 (1987), 87-99.

[12] A. Bondy and U.S.R Murty, Graph Theory with Applications, Elsevier North-

Holland, New York, 1976.

[13] J.I. Brown and V. R¨odl,A Ramsey type problem concerning vertex colourings,

J. Combin. Theory Ser. B 52 (1991), 45-52.

[14] J.I. Brown and D.G. Corneil, On uniquely -G k-Colourable graphs, Quaestiones

Math. 15 (1992), 477-487.

[15] J.I. Brown and V. R¨odl,Folkman numbers for graphs of small order, Ars Com-

binatoria 35 (1993), 11-27.

[16] J.I. Brown, The complexity of generalized graph colourings, Discrete Appl.

Math. 69 (1996), 245-253.

[17] M. Chudnovsky, N. Robertson, P. Seymour and R. Thomas, The strong perfect

graph theorem, Ann. of Math. 164 (2006), 51-229.

[18] V. Chv´ataland N. Sbihi, Recognizing claw-free Berge graphs, J. Combin. Theory

Ser. B, 44 (1988), 154-176.

[19] B.A. Davey and H.A. Priestley, Introduction to Lattices and Order, 2nd ed.,

Cambridge University Press, New York, 2002.

[20] D. D´efossez,Clique-coloring some classes of odd-hole-free graphs, J. Graph The-

ory 53 (2006), 233-249. 132

[21] R. Diestel, Graph Theory, 3rd ed., Springer-Verlag, New York, 2005.

[22] W. Duckworth and N.C. Wormald, On the independent domination number of

random regular graphs, Combinatorics, Probability and Computing 15 (2006),

513-522.

[23] D. Duﬀus, B. Sands, and P. Winkler, Maximal chains and antichains in the

Boolean lattice, SIAM J. Discrete Math. 3 (1990), 197-205.

[24] D. Duﬀus, H. A. Kierstead, and W. T. Trotter, Fibers and ordered set coloring,

J. Combin. Theory Ser. A 58 (1991), 158-164.

[25] D. Duﬀus, B. Sands, N. Sauer, and R. Woodrow, Two-colouring all two-element

maximal antichains, J. Combin. Theory Ser. A 57 (1991), 109-116.

[26] P. Erd˝os,Graph theory and probability, Canad. J. Math. 11 (1959), 34-38.

[27] P. Erd˝os,Graph theory and probability II, Canad. J. Math., 13 (1961), 346-352.

[28] P. Erd˝os,On circuits and subgraphs of chromatic graphs, Mathematika 9 (1962),

170-175.

[29] P. Erd˝osand A. Hajnal, On chromatic numbers of graphs and set systems, Acta

Math. Acad. Sci. Hungar. 17 (1966), 61-99 .

[30] P. Erd˝os,T. Gallai and Z. Tuza, Covering the cliques of a graph with vertices,

Discrete Math. 108 (1992), 279-289.

[31] R.L. Graham, M. Gr¨otschel, and L. Lov´asz, Handbook of Combinatorics, Vol I,

II, North-Holland, New York, 1995. 133

[32] R.L. Graham, B.L. Rothschild, and J.H. Spencer, Ramsey Theory, 2nd ed.,

Wiley-Interscience, New York, 1990.

[33] S. Graver and R. Skrekovski,ˇ Coloring the clique hypergraph of graphs without

forbidden structure, manuscript (2003).

[34] H. Gr¨otzsch, Ein Dreifarbensatz f¨urdreikreisfreie Netze auf der Kugel, Wiss.

Z. Martin Luther Univ, Halle-Wittenberg, Math. Naturwiss. Reihe 8 (1959)

109-120.

[35] P.L. Hammer and B. Simeone, The splittance of a graph, Combinatorica 1

(1981), 275-284.

[36] C. Ho´angand C. McDiarmid, On the divisibility of graphs, Discrete Math. 242

(2002), 145-156.

[37] J. Kratochv´ıland Z. Tuza, On the complexity of bicoloring clique hypergraphs

of graphs, J. Algorithms 45 (2002), 40-54.

[38] Z. Lonc and I. Rival, Chains, antichains, and ﬁbres, J. Combin. Theory Ser. A

44 (1987), 207-228.

[39] L. Lov´asz,On chromatic number of ﬁnite set systems, Acta Math. Acad. Sci.

Hungar. 19 (1968), 59-67.

[40] T.Luczakand A. Ruci´nski,Ramsey properties of random graphs, J. Combin.

Theory Ser. B 56 (1992) 55-68.

[41] D. Marx, Complexity of clique-coloring and related problems, manuscript

(2004). 134

[42] B. Mohar and R. Skrekovski,ˇ The Gr¨otzsch theorem for the hypergraph of max-

imal cliques, Electron. J. Combin. 6 (1999), R26.

[43] J. Neˇsetˇriland V. R¨odl,Partitions of vertices, Comm. Math. Univ. Carolinae

17 (1976), 85-95.

[44] J. Neˇsetˇriland V. R¨odl,A short proof of the existence of highly chromatic

hypergraphs without short cycles, J. Combin. Theory Ser. B 27 (1992) 225-

227.

[45] J. Neˇsetˇril, Ramsey theory, in Handbook of Combinatorics (R.L. Graham,

M. Grotschel, and L. Lovasz, eds.), Elsevier Science, New York, 1995.

[46] K.R. Parthasaraty and G. Ravindra, The validity of the strong perfect graph

conjecture for (K e)-free graphs, J. Combin. Theory Ser. B 26 (1979), 98-100. 4 −

[47] H.J. Pr¨omeland A. Steger, Almost all Berge graphs are perfect, Combin. Probab.

Comput. 1 (1992), 53-79.

[48] N. Robertson and P. Seymour, Graph minors. I. Excluding a forest, J. Combi-

natorial Theory, Ser. B 35 (1983), 39-61.

[49] N. Robertson and P. Seymour, Graph minors. II. Algorithmic aspects of tree-

width, J. Algorithms 7 (1986), 309-322.

[50] V. R¨odland A. Ruci´nski,Threshold functions for Ramsey properties, J. Amer.

Math. Soc. 8 (1995) 917-942.

[51] A. Tucker, Coloring perfect (K e)-free graphs, J. Combin. Theory Ser. B 43 4 − (1987), 313-318. 135

[52] Z. Tuza, Covering all cliques of a graph, Discrete Math. 86 (1990), 117-126.

[53] D. West, Introduction to Graph Theory, 2nd ed., Prentice Hall, New Jersey,

2001.