DEGREE SEQUENCES, FORCIBLY CHORDAL GRAPHS, AND COMBINATORIAL PROOF SYSTEMS

Transcription

1 DEGREE SEQUENCES, FORCIBLY CHORDAL GRAPHS, AND COMBINATORIAL PROOF SYSTEMS DISSERTATION Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of the Ohio State University By Christian Altomare, B.S. Graduate Program in Mathematics The Ohio State University 2009 Dissertation Committee: Dr. G. Neil Robertson, Advisor Dr. John Maharry Dr. Akos Seress

2

3 ABSTRACT We study the structure theory of two combinatorial objects closely related to graphs. First, we consider degree sequences, and we prove several results originally motivated by attempts to prove what was, until recently, S.B. Rao s Conjecture, and what is now a theorem of Paul Seymour and Maria Chudnovsky, namely, that graphic degree sequences are well quasi ordered. We give a new, surprisingly non-graph theoretic proof of the bounded case of this theorem. Next, we obtain an exact structure theorem of degree sequences excluding a square and a pentagon. Using this result, we then prove a structure theorem for degree sequences excluding a square and, more generally, excluding an arbitrary cycle. It should be noted that taking complements, this yields a structure theorem for excluding a matching. The structure theorems above, however, are stated in terms of forcibly chordal graphs, hence we next begin their characterization. While an exact characterization seems difficult, certain partial results are obtained. Specifically, we first characterize the degree sequences of forcibly chordal trees. Next, we use this result to extend the characterization to forcibly chordal forests. Finally, we characterize forcibly chordal graphs having a certain path structure. Next, we define a class of combinatorial objects that generalizes digraphs and partial orders, which is motivated by proof systems arising in mathematical logic. We ii

4 give what we believe will be the basic theory of these objects, including definitions, theorems, and proofs. We define the minors of a proof system, and we give two forbidden minors theorems, one of them characterizing partial orders as proof systems by forbidden minors. iii

5 To Moomar. iv

6 ACKNOWLEDGMENTS First and foremost, I wish to thank Neil Robertson, my advisor. It is every student s wish to have an advisor with such depth of understanding, breadth of knowledge, and raw intuition for his field of expertise. I have gained from him not only knowledge, but an understanding of how research mathematics is carried out. His ability to find the right generalization to prove, the right special case to consider, the right approach to try, and the right question to ask at all, has continually amazed me. Second, I would like to thank S.B. Rao for a beautiful conjecture. Third, I would like to thank Christopher McClain for his generous and patient help related to typesetting and document preparation, which are not my strong suits. Fourth, I wish to thank Akos Seress and John Maharry for their time and effort participating in my thesis committee. Fifth, I would like to thank everyone in the Ohio State University Mathematics Department who has helped me in my time since I started taking mathematics courses here as a high school student. In particular, in the order in which I met them, I am thankful to John Maharry, Alexander Dynin, Judie Monson, Yung-Chen Lu, Vitaly Bergelson, Randall Dougherty, Tim Carlson, Cindy Bernlohr, Boris Pittel, and once again my advisor for the amount of time, effort, and patience they were willing to spend toward my career and development. I thank the countless others in the v

7 department who have helped me as well. Without their help, this would not be possible. Sixth, I thank my parents, Richard and Karen Altomare. vi

8 VITA April 7, Born - Columbus, OH Undergraduate, The Ohio State University B.S. in Mathematics, The Ohio State University 2001-Present Graduate Teaching Associate, The Ohio State University FIELDS OF STUDY Major Field: Mathematics Specialization: Graph Theory vii

9 TABLE OF CONTENTS Abstract Dedication Acknowledgments Vita ii iv v vii CHAPTER PAGE 1 Introduction Introduction to Degree Sequences Degree Sequence Basics, Notation, and Conventions Introduction to Combinatorial Proof Systems The Bounded Case of Rao s Conjecture Excluding Matchings and Cycles Forcibly Chordal Trees Forcibly Chordal Forests Forcibly Chordal Graphs Combinatorial Proof Systems Introduction Proof Closure The Merge Preceding Set Proof Systems Examples Motivation for Definition of Proof Proof Definition and Basics viii

10 7.4.4 Autonomous Sets Axioms The Information in the Set of Proofs Autonomous Systems Introduction Definition of Proof Revisited The Canonical Orders Canonical Order Definition and Basics Descendability Canonical Order Basic Theorems And Examples Partial Orders As Ausyses Well Founded Autonomous Systems Blocking The Blocking Order Blocking In Posets Ausys Lexicographic Sum Subausys, Dot, Homomorphisms, and Minors Relations to Matching and Connectivity Bibliography ix

11 CHAPTER 1 INTRODUCTION This work studies two classes of objects. The first class we study is the class of degree sequences of finite graphs. The second class we study is a class of combinatorial proof systems we call autonomous systems. 1.1 Introduction to Degree Sequences We assume familiarity with basic graph theory. Definitions and conventions are as in [3] unless otherwise stated. Our graphs are finite, simple, and undirected throughout unless otherwise stated. Definition Let G be a graph with vertices v 1,..., v n, listed such that d(v 1 ) d(v n ). Then the degree sequence of G, denoted by D(G), is the sequence (d(v 1 ),..., d(v n )). We make no use of the fact that, according to our definition, the degree sequence is a decreasing sequence. It is rather simply the easiest way to make the degree sequence of a graph unique, so we can refer to the degree sequence D(G) of G, as opposed to a degree sequence of G. We note that while a degree sequence does not technically have any vertices, it can be 1

12 very suggestive to think of the vertices of a degree sequence, which we sometimes do. The degree sequence (2, 2, 2, 1, 1), for instance, would be said to have three vertices of degree 2 and two of degree 1. Definition Let D be a degree sequence and let G be a graph. We say that G realizes D, or that G is a realization of D, if D(G) = D. We denote by R(D) the set of realizations of D. Myriad theorems in combinatorics, and in particular graph theory, study the graphs not containing a fixed graph, for various notions of containment. It is fruitful to define a notion of containment for degree sequences as well so that similar questions may be asked and theorems proved. Definition Let D 1 and D 2 be degree sequences. We write D 1 D 2 if there is a graph G 1 in R(D 1 ) and a graph G 2 in R(D 2 ) such that G 1 is an induced subgraph of G 2. The reader may check that is a reflexive, transitive relation. One motivation for making this definition is that the induced subgraph relation for graphs can be extremely difficult to work with, even for questions that are tractable if the induced subgraph relation is replaced with another containment relation. The relation for degree sequences is similar to, but more tractable in many cases than, the induced subgraph relation for graphs. 2

13 A discussion of claw free graphs and degree sequences best illustrates this point. A claw is the unique graph up to isomorphism with degree sequence (3, 1, 1, 1). Suppose we wish to find the structure of claw free graphs. What claw free means of course depends on the containment relation used. If we work with the minor relation, we are asking which graphs have no claw as a minor. It is trivial that a graph is claw free in this sense iff it has no vertices of degree three or more. The claw free graphs are trivially then exactly the disjoint unions of paths and cycles. If instead of working with the minor relation, we rather work with the induced subgraph relation, the structure of claw free graphs is then a deep and difficult theorem of Chudnovsky and Seymour, proved in a series of five papers totalling over 200 pages. Now, if instead of working with graphs excluding a claw as an induced subgraph, we instead ask which degree sequences exclude the degree sequence of a claw, then the structure theorem given by Robertson and Song can be proved in under six pages. Thus, in passing from induced subgraphs to the relation on degree sequences, we have a theorem that is motivated by induced subgraphs, yet still more amenable to analysis. With this motivation, degree sequence analogues of questions asked for graphs are often asked for degree sequences. The celebrated Minor Theorem of Robertson and Seymour says that finite graphs are well quasi ordered under the minor relation. A well quasi order is a reflexive, transitive relation T on a set X such that if x 1, x 2,..., x n,... is an infinite sequence in X then there exist i and j with i < j such that x i T x j. Analogous to the Minor Theorem, S.B. Rao s famous conjecture, first stated in 1971 [10] and proved in 2008 by M. Chudnovsky and P. Seymour and to appear in [1], says that 3

14 degree sequences of graphs are well quasi ordered under. We refer to this theorem as Rao s Conjecture throughout. In chapter 2, we give a proof that for each positive integer k, Rao s Conjecture holds for degree sequences of maximum degree at most k. Our proof was obtained independently of Chudnovsky and Seymour s proof of Rao s Conjecture, and our proof makes no use of the structure theory for degree sequences of those authors. In fact, our proof has surprisingly little graph theory at all, which leads us to believe we may be able to obtain results in a far more abstract, general setting in future works. Just as Rao s Conjecture is natural in light of the Minor Theorem, it is also natural, in light of the many graph theorems excluding minors, topological minors, and so on, to attempt to find the structure of degree sequences excluding a given degree sequence. In chapter 3, we characterize degree sequences excluding (the degree sequence of) certain matchings and cycles. These exclusion results we obtain are stated in terms of pentagons, hexagons, the complete bipartite graph K 3,3, the split graphs first defined in [5], a binary operation we call the half join first defined in [16] to characterize degree sequences with at most one realization up to isomorphism, and in terms of forcibly chordal graphs. A graph is chordal if no induced cycle has four or more vertices. A graph is forcibly chordal if every graph with the same degree sequence is chordal. While our exclusion theorems are exact, they are only valuable structure theorems to the extent we understand the structure of the pentagons, hexagons, K 3,3, split graphs, and forcibly chordal graphs they are stated in terms of. Pentagons, hexagons, and K 3,3 may be considered well understood. The structure of split graphs has been 4

15 found by Chudnovsky and Seymour in their proof of Rao s Conjecture to appear in [1]. We may thus take the structure of split graphs as known. That leaves the forcibly chordal graphs. While the forcibly P and potentially P degree sequences have been determined for many properties P (see [11] and [12] for excellent surveys), the forcibly chordal graphs have not, to our knowledge, yet been characterized. Our partial characterization of forcibly chordal graphs occupies us for the next three chapters. In Chapter 4, we characterize the forcibly chordal trees. In Chapter 5, we use these results to extend the characterization to forcibly chordal forests. In Chapter 6, we characterize connected, forcibly chordal graphs having a path structure, in a sense to be defined in that chapter. We believe these results can be extended in upcoming work to fully characterize forcibly chordal graphs. 1.2 Degree Sequence Basics, Notation, and Conventions In order to make our presentation self contained and more efficient, we give the basic notation, theorems, definitions, and conventions here for easy reference. First, we must eliminate any possibility of ambiguity in the containment relation we will use throughout chapters 2 through 6. Definition We say a graph G excludes a graph H if G contains no induced subgraph isomorphic to H. We say a degree sequence D 2 excludes a degree sequence D 1 if D 1 D 2. We say that a degree sequence D excludes a graph G if D excludes D(G). Note that while the subgraph and minor relations are far more commonly used in graph theory than the induced subgraph relation, in light of the above definitions, 5

16 we will work exclusively with the induced subgraph relation. In light of this fact, we make certain conventions to simplify wording throughout. If we say G contains H, we mean as an induced subgraph, if we say G contains a hole, we mean an induced hole, and so on. As such, when it causes no confusion, we will often forget to say induced. Another consequence of the fact that we work strictly with the induced subgraph relation is that we can often simplify presentation by identifying the set X and the induced subgraph G[X] of the graph G, which we often do when no confusion arises. If we say a subset X of a graph G has a certain graph property, we mean that G[X] does. Moreover, if G is a graph and X is a subset of V (G), we permit ourselves to say X is a subset of G. Since we do not distinguish between X and G[X] in this work, the reader should note that in particular, when we write X G, we always mean that X is an induced subgraph of G. We use the notation G 1 G2 to denote the disjoint union of graphs G 1 and G 2. Similarly, n i=1 G i denotes the disjoint union of graphs G 1,..., G n. If k is a nonnegative integer, we use the notation k G or kg to denote the disjoint union of k isomorphic copies of G. Given subsets X and Y of a graph G, we say that X is complete to Y if each x in X is adjacent to each y in Y. We say that X is complete if all pairs of distinct vertices in X are adjacent. We say x in G is a universal vertex, or simply that x is universal, if x is complete to G x. If G is a graph, G c denotes the complement. If the degree sequence D is realized by a graph G, we may speak of the complementary degree sequence D c as the degree 6

17 sequence of G c. Though D may in general have more than one graph realizing it, it is simple to check this definition of D c does not depend on the choice of the realizing graph, and D c is thus well defined. The set X is anti-complete in G if X is complete in G c. The set X is anti-complete to Y in G if X is complete to Y in G c. In general, a graph or set is said to be anti-p if property P holds on taking complements. An anti-hexagon, for instance, is the complement of a hexagon, an anti-forcibly chordal graph is the complement of a forcibly chordal graph, and so on. Chapters 2 through 6 make extensive use of switchings, which we now define. Definition Let G be a graph. A switching is a tuple (a, b, c, d) of distinct vertices in G such that a and b are adjacent, b and c are nonadjacent, c and d are adjacent, and d and a are nonadjacent. The edges of the switching are ab and cd. The nonedges of the switching are bc and da. If (a, b, c, d) is a switching in G then the graph G ab + bc cd + da is said to arise from G by a switching in G. If there is a sequence of graphs G 1,..., G n such that G i+1 arises from G i by a switching in G i for each i with 1 i < n then G n is said to arise from G 1 by a sequence of switchings. Another way to state that (a, b, c, d) is a switching in G is that ab and cd are edges of G while bc and da are nonedges of G. It is very important to note this definition says nothing about whether or not ac is an edge or nonedge of G, and similarly for bd. Moreover, we stress that if we say xy is not an edge of a switching, xy may or may not be an edge of G. Similarly, if we say e is not a nonedge of the switching, 7

18 while it is tempting to see this statement as a double negation equivalent to e being an edge of the switching, this is not the case. The edge e may or may not be an edge of the switching. The reason for this behavior is simple. A switching has exactly two edges and two nonedges. This leaves two pairs of vertices in {a, b, c, d} that are either edges of G yet not edges of the switching, or nonedges or G yet not nonedges of the switching. While care is needed on these points, no confusion arises if such care is taken, and we speak of switchings rather informally by listing the two edges and the two nonedges. We are rarely so formal as to present a switching as a tuple as in the definition. We call two graphs equivalent if they have the same degree sequence. The reader may note that if H arises from G by a switching in G then D(H) = D(G). Moreover, by induction on the number of switchings, one sees that if H arises from G by a sequence of switchings then D(H) = D(G). The following converse is a theorem first proved in [6]. It is used at key points in chapter 2 and extensively throughout chapters 3 through 6 as our primary tool. Theorem Graphs G and H are equivalent iff H arises from G by a sequence of switchings. We now fix notation and conventions regarding the most important types of graphs we use. Definition A graph G is called a split graph if V (G) can be partitioned into 8

19 (possibly empty) cells A and B such that G[A] is complete and G[B] is anti-complete. The partition (A, B) is called a split partition. We note the above definition allows for possibly empty split graphs. In general, we allow empty graphs, but in cases where no problems arise, we casually disregard empty graphs without comment if doing otherwise would unnecessarily complicate a statement with trivialities. We let C k denote a cycle on k vertices, P k denote a path on k vertices (not k edges), M k denote the matching kp 2, and K k the complete graph on k vertices. We often say triangle for C 3, square for C 4, and so on. A hole in a graph is an induced cycle on at least four vertices. A graph is called chordal if it has no holes. 1.3 Introduction to Combinatorial Proof Systems In the second part of this work, we define and study certain combinatorial proof systems that we call autonomous systems. No background in logic is required to understand this part, though a basic understanding of partial orders, linear orders, well foundedness and well orders, and transfinite induction and recursion is needed at some points. The necessary facts may be found in [14] and [9]. The fact that we need assume no previous exposure to logic from the reader arises from our abstract approach, which of necessity starts from scratch, diverges early from that typically studied by logicians, and soon far more closely resembles structural graph and partial order theory than classical proof theory. Proof theory is one of the main branches of mathematical logic. While proof theory as 9

20 understood by mathematical logicians does indeed study proofs, it is just as fair to say that proof theorists study syntax and semantics, for the statements of typical results in proof theory would be impossible to formulate, let alone prove, without syntactic and semantic notions. While proof theory has many deep and difficult results, they are deep and difficult results for proof systems with a great deal of structure beyond the proofs themselves. In 2001, the author had the goal of studying proofs in as general and abstract a setting as possible. A proof is considered a (not necessarily finite or even well founded) partial order such that for all x, the set of elements less than x is enough information to infer x. We take, is enough information to infer, as a primitive notion. More precisely, the proof system is a set together with a set of pairs (S, x), with S a subset of and x a point in the domain, which we take to mean that S implies x. We explicitly note that in this context, we have no syntax or semantics. We have only implication and proofs. While it may seem a priori that this is too general to prove anything, we in fact obtain nontrivial results. It is fair to say we obtain no logical theorems. Our theorems are purely combinatorial. This was, in fact, a great surprise to the author, who intended to prove logical results and found himself instead working in structural combinatorics. Roughly, just as there are rooted and unrooted trees, there are also rooted and unrooted proof systems. While rooted trees have singleton roots, rooted proof systems allow arbitrary root sets, which are in fact the axioms of the proof system. Unrooted proof systems generalize directed graphs. Roughly speaking, if a directed edge from x to y is taken to mean x implies y, we can instead allow directed edges 10

21 from an arbitrary set X to a point y to mean that set X of formulas implies y. An unrooted proof system could, therefore, be thought of as a directed hypergraph. Rooted proof systems, on the other hand, generalize well founded partial orders. This work focuses on a generalization of rooted proof systems, which we will call autonomous systems, and which we will define without reference to unrooted proof systems. The topic of unrooted proof systems will be addressed in future work of the author. We give the basic definitions, theorems, and constructions related to these autonomous systems that have proved useful in their study. We give three distinct axiomatizations of autonomous systems, give numerous characterizations of partial orders as autonomous systems, and define what we call the canonical orders that encode context dependent needing in a proof system and turn out to be an important structural tool in proving even statements making no mention of these orders. We define the notions of weak and strong aut descendability, two finiteness conditions on which many autonomous system theorems and proofs depend. We define homomorphisms and two containment relations that allow us to define the minors of a proof system. We then use the canonical orders to prove two forbidden minors theorems that hold under the assumption of strong aut descendability. (In particular, they hold for finite and even finitary autonomous systems.) We also extend the definition of partial order lexicographic sum to autonomous systems and prove the basic properties of the lexicographic sum 11

22 CHAPTER 2 THE BOUNDED CASE OF RAO S CONJECTURE In this chapter, we answer a question posed by N. Robertson, who asked if graphic degree sequences of bounded degree can be realized as disjoint unions of graphs with bounded size components. Our answer in the affirmative implies the bounded case of S.B. Rao s Conjecture, which we state now. Theorem Graphic degree sequences of bounded degree are well quasi ordered. There is surprisingly little graph theory in our proof. In fact, the graph theory only comes in the initial lemmas constructing graphs with certain prescribed degree sequences. Though there is an existence proof of all these initial lemmas using the Erdös-Gallai inequalities proved in [4], our goal is to give a detailed construction from first principles. We therefore avoid using any outside results in this proof. We now turn to the proof. Definition A graph G is called k-regular if every vertex has degree k. A graph is called regular if it is k-regular for some k. Lemma Let k be an even integer. Then there is an integer L k such that for all L L k there is a k-regular graph G on L vertices. 12

23 Proof. k is even, so let k = 2l, and let L k = k + 1 = 2l + 1. For each L L k, we define a graph G on the integers 0, 1,..., L 1 by letting E(G) = {xy : 1 x y mod L l } Obviously, G has L vertices, and since L is at least 2l + 1, it follows that for all x, the 2l vertices x l, x l + 1,..., x 1, x + 1,..., x + l are parwise distinct. Therefore each x in G has degree 2l = k. Therefore G is a k-regular graph on L vertices as needed, thus completing the proof. The graphs in the above proof are called circulants. Lemma Let k be an odd integer. Then there is an integer L k such that for all even L L k there is a k-regular graph G on L vertices. Proof. Let L k = 2k. It is enough to construct, for each even L L k, a k-regular bipartite graph G on L vertices. So take an even L L k and let L = 2l. Note then that l k. Take disjoint sets A = {v 1,..., v l } and B = {w 1,..., w l }. Define G as the graph with vertex set A B and edge set E(G) = {v i w j : 0 w j v i mod l k 1} G is then a k-regular graph on L vertices, and the proof is complete. Lemma Given a positive integer k and a nonnegative integer j, there is a graph G with exactly 2j vertices of degree k 1 and all other vertices of degree k. 13

24 Proof. Let m = max{k, j}. Take disjoint sets A = {v 1,..., v m } and B = {w 1,..., w m }. Define G as the graph with vertex set A B and edge set E(G) = {v i w j : 0 w j v i mod m k 1} Let G = G v 1 w 1 v 2 w 2 v j w j. Then d G (v i ) = d G (w i ) = k 1 for 1 i j and d G (v i ) = d G (w i ) = k for i > j. The claim follows. Lemma Given a positive integer k and nonnegative integer j such that 2j k, there is a graph G with exactly one vertex of degree 2j and all other vertices of degree k. Proof. By the previous lemma, there is a graph G with exactly 2j vertices of degree k 1 and all other vertices of degree k. (Note the G of this lemma is the G of the previous lemma.) Let v be a point not in G. Let G be the graph on V (G ) {v} such that xy E(G) iff one of the following conditions holds: i xy E(G ). ii x = v and d G (y) = k 1. It follows by definition of G that v is adjacent in G to exactly the vertices of degree k 1 in G. By choice of G, there are exactly 2j of these. So d G (v) = 2j. It is enough to show d G (x) = k for all other vertices in G, so take x v. Then d G (x) is k or k 1. If d G (x) = k then it follows by definition of G that for all y in V (G), the edge xy is in G iff it is in G. Therefore d G (x) = d G (x) = k. If d G (x) = k 1 then it follows from the definition of G that N G (x) = N G (x) {v}. Therefore d G (x) = d G (x) + 1 = k = k. 14

25 Lemma Given nonnegative integers j and k, there is a graph G with exactly 2j + 1 vertices of degree 2k and all others of degree 2k + 1. Proof. Let m = max{2k + 1, k + j}. Let A = {v 1..., v m } and B = {w 1,..., w m } be disjoint sets. Define a graph G on A B by letting A and B be anti-complete and letting v i w j E(G ) iff 0 (w j v i ) mod m 2k. Note G is a 2k + 1-regular graph. Let G = G v 1 w 1 v 2 w 2 v k+j w k+j. Then for 1 i k + j, we have d G (v i ) = d G (v i ) 1 = (2k+1) 1 = 2k. Similarly, d G (w i ) = 2k. For i > k+j, we see from the definition of G that N G (v i ) = N G (v i ) therefore d G (v i ) = d G (v i ) = 2k + 1. Similarly, for i > k + j, we see that d G (w i ) = 2k + 1. Let z be a point not in G. Let G be the graph on V (G ) {z} such that E(G) = E(G ) {zv i 1 i k} {zw i 1 i k}. We show G has the desired properties. First, note that since z is not in G, it follows directly from the definition of G that d G (z) = 2k. Now consider v i with 1 i k. Then N G (v i ) = N G (v i ) {z} therefore d G (v i ) = d G (v i ) + 1 = 2k + 1. Similarly, d G (w i ) = 2k + 1 for 1 i k. By similar reasoning, the reader may check that d G (v i ) = d G (w i ) = 2k+1 if k+j+1 i m and that d G (v i ) = d G (w i ) = 2k if k + 1 i k + j. Therefore G has exactly 2j + 1 vertices of degree 2k and the rest of degree 2k + 1 as claimed. Lemma Given distinct, nonnegative integers integers j and k, there is a graph G with exactly one vertex of degree 2j +1 and all other vertices of degree 2k

26 Proof. By the previous lemma, there is a graph G with exactly 2j + 1 vertices of degree 2k and all other vertices of degree 2k+1. (The G of this lemma is the G of the previous lemma.) Take y not in G. Define G as the graph with vertex set V (G ) {y} and edge set E(G ) {yx x V (G ) and d G (x) = 2k}. We show G has the desired property by showing d G (y) = 2j + 1 and all other vertices have degree 2k + 1 in G. First, by choice of G, we know there are exactly 2j + 1 elements x in G such that d G (x) = 2k. Since N G (y) consists, by definition of G, of exactly these elements, we see that d G (y) = N G (y) = 2j +1. We show all other vertices of G have degree 2k +1 in G. So take x V (G) y. Then d G (x) is 2k or 2k + 1. If d G (x) = 2k then by definition of G, we see that N G (x) = N G (x) {y}. Therefore d G (x) = d G (x) + 1 = 2k + 1. If d G (x) = 2k +1 then it follows by the definition of G that N G (x) = N G (x). Therefore d G (x) = d G (x) = 2k + 1. Therefore all vertices other than y have degree 2k + 1 in G, as was to be shown. We note that in the following lemma, i and j may or may not be distinct. The possibility that i = j must be allowed for use in a later proof. Lemma Let i and j be nonnegative integers. Let k be a positive, even integer. Then there is a graph G with vertices v w such that d G (v) = 2i + 1, d G (w) = 2j + 1 and d G (x) = k for all other vertices x in G. Proof. We know there is a graph G with a vertex y of degree 2i+2j+2 and d G (x) = k for all other x. Let G be obtained from G by splitting the vertex y into nonadjacent 16

27 vertices v and w such that v is adjacent in G to 2i + 1 of the G neighbors of y and w is adjacent in G to the remaining 2j + 1 G neighbors of y. Definition Let U be a class of graphs. U is called productive if the following conditions hold: (i) For every odd, nonnegative integer k, there is an integer L U,k such that for all even L L U,k, there is a k-regular graph G in U of cardinality L. (ii) For every even, nonnegative integer k, there is an integer L U,k such that for all L L U,k, there is a k-regular graph G in U of cardinality L. (iii) Given positive integers j, k, with 2j k, there is a graph in U with exactly one vertex of degree 2j and all other vertices of degree k. (iv) Given distinct, nonnegative integers j and k, there is a graph with exactly one vertex of degree 2j + 1 and all other vertices of degree 2k + 1. (v) Given nonnegative integers i and j, and a positive, even k, there is a graph G in U with vertices v w such that d G (v) = 2i + 1, d G (w) = 2j + 1 and d G (x) = k for all other vertices x in G. Corollary The class of finite graphs is productive. Proof. This is a restatement of the lemmas thus far proved. 17

28 Definition We call a class U of graphs finitely representable if there is a finite subset F of U such that for every graph G in U, there is a graph G such that D(G ) = D(G) and G is the disjoint union of graphs in F. Lemma The finite union of finitely representable classes is also finitely representable. Proof. Let U 1,..., U n be finitely representable and let U = n i=1 Since each U i is finitely representable, there is for each i a finite subset F i of U i such that every graph in U i has the same degree sequence as a disjoint union of graphs in F i. Let F = n i=1 Then given a graph G in U, there is i such that G is in U i. Therefore G has the same degree sequence as the disjoint union of some graphs in F i. Since F contains F i, we see G has the same degree sequence as the disjoint union of some graphs in F. Since G is an arbitrary graph in U and F is finite, we see that U is finitely representable, as was to be shown. U i F i Lemma If U is a finite set of graphs then U is finitely representable. 18

29 Proof. Let F = U. We make use of the following basic fact from number theory. Lemma Let S be a nonempty set of positive integers and let g be its greatest common divisor. Then there is a finite subset F S of S and a positive integer n such that for all n n, we can write n g as a 1 s a p s p for some positive integer p, some nonnegative integers a 1,..., a p, and some elements s 1,..., s p of S. Theorem Let U be a class of graphs and let U k be the class of k-regular graphs in U. Then U k is finitely representable. Proof. If U k is empty then it is vacuously true that U k is finitely representable, so suppose U k is nonempty. Let S be the set of cardinalities of graphs in U k. Then S is nonempty. Let g be its greatest common divisor. By the previous lemma, there exists n such that for all n n, we can write n g as a 1 s a p s p for some p positive integer p, some positive integers a 1,..., a p, and some elements s 1,..., s p of S. Since each s i is in S, it follows by the definition of S that there are graphs G 1,..., G p in U k such that G i = s i for each i. Let F = {G 1,..., G p } {H U k : H < ng}. Note that since there are only finitely many graphs on less than ng vertices, F is a finite set. By definition of finite representability, it is enough to show that given G in U k, there is a graph G with the same degree sequence as G such that G is the disjoint union of graphs in F. So take a graph G in U k. 19

30 If G < ng then G is in F by definition of F, so we see that G itself is a graph with the same degree as G that is the disjoint union of elements of F. So suppose G ng. Then we may write G = a 1 s a p s p as in the previous lemma. Consider the graph G = p a i G i i=1 Then G = p i=1 a i G i = p i=1 a is i = G. Also note that G and G are both k- regular. Since G and G are k-regular graphs of the same cardinality, they have the same degree sequence. Clearly, we have expressed G as the disjoint union of graphs in F. This completes the proof of the lemma. Definition A degree class sequence C is an infinite sequence c 0, c 1, c 2, c 3,... with values in {1, 2, 3,..., } such that c i is eventually 1. Definition Let U be a class of graphs and C = (c i ) i=1 a degree class sequence. Then U C denotes the class of graphs G in U such that for all i, the graph G has less than c i vertices of degree i. Definition Let X = {x i } i 0 be a sequence. We define the support S(X) of X as {i : x i 1}. We define the infinity support S (X) of X as {i : x i = }. Lemma Let U be a productive class. Let C be a degree class sequence such that S(C) is finite. Then U C is finitely representable. 20

31 Proof. The proof is by induction on S (C). If S (C) = 0 then U C is finite, so we know by Lemma that U C is finitely representable. Suppose the result is true for all degree class sequences with infinity support of cardinality at most N. We must prove that U C is finitely representable for each degree class sequence C such that S (C) = N + 1. For every proper subset X of S (C) and every positive integer M, let X M be the degree class sequence such that X M (i) = C(i) for all i except that X M (i) = M for all i in S (C) X. Since the infinity support of X M is a proper subset of the infinity support S (C), we know by the induction hypothesis that U XM is finitely representable for each such X M. Since the finite union of finitely representable classes is finitely representable, we see that for each M, the class X S (C) U XM is finitely representable. To show U C is finitely representable, it is therefore enough to show that W M := U C X S (C) U XM is finitely representable for some M. Note that W M is the class of graphs G in U C such that if C(i) = then G has at least M vertices of degree i. We have only to show this class is finitely representable for some large enough M. This is immediate from the definition of productivity and Theorem If M is large enough, we simply take out vertices in G whose degree d is in S(C) S (C) by using the almost regular graphs whose vertices all have the same degree except possibly one or two. More precisely, we subtract the degree 21

32 sequence of these almost regular graphs from the degree sequence D of G. Call the remaining degree sequence D, which we do not yet know is graphic. The degree sequence D has an even number of vertices of odd degree. We may pair them up. (More formally, we partition the set of vertices of odd degree into doubletons.) For each such pair {2i+1, 2j+1} in turn, by condition (iv) of Definition , we may choose a graph G i,j with all vertices of degree 2i + 1 except one of degree 2j + 1. Let D i,j be the degree sequence of G i,j. Let D be the degree sequence resulting from subtracting each D i,j with i paired to j from the degree sequence D. It is again important to note that, at this point, we have not yet shown that D or D is realizable. However, by choosing M large enough, D and D are indeed realizable. Our degree sequence remaining has an even number of vertices of each odd degree and at least M vertices of each degree in S (C). By Theorem , D may therefore be realized as the disjoint union of finitely many regular graphs. Letting G realize G, it is clear from the definitions of D and D that we may unite G with almost regular graphs to obtain a graph graph H with the same degree sequence as G. Since S(C) is finite, only finitely many such almost regular graphs are used. W M is therefore finitely representable as needed. Theorem For any fixed bound k, degree sequences bounded by k are finitely realizable. 22

33 Proof. The class of degree sequences with all degrees at most k is simply U C where U is the class of finite graphs and C is the degree class sequence satisfying C(i) = if i k and C(i) = 1 for i > k. Since U is productive, we may apply the previous result. We now prove Theorem Proof. We know that degree sequences of degree at most k can be realized as disjoint unions from a finite set F of graphs. Let G 1, G 2,... be a sequence of graphs each of which is a disjoint union of graphs in F = {F 1,..., F t }. Then for each i, we may write G i = c i,1 F 1 ci,t F t, for some nonnegative integers c i,t. We may choose a strictly increasing sequence (i n ) n 0 such that c in,j is an increasing sequence of n for each j. Then G in is an increasing sequence of graphs under the induced subgraph relation. Since the sequence G 1, G 2,... was an arbitrary sequence of disjoint unions in F, this shows finite disjoint unions in F are well quasi ordered under induced subgraph. In particular, their degree sequences are well quasi ordered under. 23

34 CHAPTER 3 EXCLUDING MATCHINGS AND CYCLES In this chapter, we derive structure theorems for some classes of degree sequences excluding the matching M 2 and/or cycles. More precisely, we first recall the characterization of split graphs by forbidden induced subgraphs. Next, we use this to characterize the degree sequences that exclude the matching M 2 and a square. We next use this result to characterize degree sequences excluding only a square and, more generally, degree sequences excluding an arbitrary cycle. For each theorem one proves characterizing the degree sequences having a property X by excluding graphs in the set S, one may also prove the complementary theorem that the degree sequences whose complementary degree sequence has property X are exactly those that exclude graphs whose complement is in S. Taking the complementary theorem to the result on excluding a square, we characterize degree sequences excluding the matching M 2. However, each of these theorems is stated in terms of another class: split graphs, forcibly chordal graphs, and, generalizing forcibly chordal graphs, the class of graphs which forcibly have all chordless cycles of length at most k. This leads naturally to the problem of characterizing forcibly chordal graphs, which we address in the following chapters. 24

35 We make use of the following propositions, the first of which is a folklore theorem that may be taken as an exercise. Proposition The following are equivalent for a graph G: (i) G excludes M 2, C 4, and C 5. (ii) G excludes M 2 and all holes. (iii) G is a split graph. Corollary Split graphs are chordal. Proof. By Proposition , split graphs contain no holes. Therefore, they are chordal. Proposition The following are equivalent for a degree sequence D: (i) D excludes the degree sequences (1, 1, 1, 1), (2, 2, 2, 2), and (2, 2, 2, 2, 2). (ii) D excludes the degree sequence (1, 1, 1, 1) and the degree sequences of all cycles on at least 4 vertices. (iii) D is the degree sequence of a split graph. 25

36 Proof. D satisfies condition (i) of this theorem iff it is realized by a graph that satisfies condition (i) of Proposition Similarly for conditions (ii) and (iii). Since the three conditions of Proposition are equivalent, it thus follows that the three conditions of this theorem are equivalent. The following proposition follows from the well known characterization of split graphs as those graphs for which at least one of the Erdös-Gallai inequalities is equality. Proposition Let D be the degree sequence of a split graph. Then every realization of D is a split graph. In other words, the above proposition states that every split graph is forcibly split. In particular, we know the following. Corollary Every split graph is forcibly chordal. Proof. If a graph is split then, by Proposition it has no holes and is thus chordal. Every split graph is forcibly split, therefore every realization of the degree sequence of a split graph is split, and hence chordal. Since every realization of the degree sequence of a split graph is chordal, it follows that every split graph is forcibly chordal. Our next lemmas will make use of the notion of half join, which we define below. Informally, the half join is obtained by joining an arbitrary graph H completely to 26

37 the complete part of a split graph S and anti-completely to the anti-complete part of S. Definition Let S be a split graph with partition into a complete part A and anti-complete part B. Let H be an arbitrary graph. Then the half join (S, A, B, H) of S and H with respect to the split partition (A, B) is defined as the graph with vertex set V (S) V (H) and edge set E(S) E(H) {xy : x H and y A} The above definition of half join arises naturally and often when working with split graphs, and is used, for instance, in [16] to state a decomposition theorem for split graphs, though we make no use of this theorem. Tyshkevich does not use the word half join, or any other word, simply using notation to denote the operation, but we find it convenient to have a word denoting it, so we choose half join. We also note that A and B are mentioned in addition to S because a split graph may have more than one split partition, but in practice, when talking about half joins, we are usually far less formal, and simply say the half join of S and a pentagon and similar. We will permit this abuse of language when it causes no confusion. We will use the following lemma several times in proving the structure theorems of this chapter. Lemma Let S be a split graph with split partition (A, B) and let H be an arbitrary graph. Let G be a (not necessarily connected) graph on at least three vertices 27

38 with no induced triangles, no isolated vertices, and which is not a star. If G is an induced subgraph of the half join (S, A, B, H) then G is an induced subgraph of S or H. Proof. Since (A, B) is a split partition, A is by definition complete. Therefore A G is complete and thus any three vertices of A G comprise an induced triangle in G. Since G is triangle free by assumption, it follows that A G is empty, has exactly one vertex, or has exactly two vertices. We consider these three cases. First, assume A G is empty. B is anti-complete to H and B itself is anti-complete, therefore every vertex of B G is an isolated vertex of G. Since G has no isolated vertices by assumption, it follows that B G is empty. Therefore G is an induced subgraph of H as the lemma claims. Second, assume A G contains exactly one vertex x. Consider any other two vertices y and z in G. We show that y and z are non-adjacent. First, suppose one of y or z is in B. Without loss of generality, we may assume y is in B. Note that z is in B or H since x is the only element of A G. Since B is anti-complete and anti-complete to H, and since z is in either B or H, it follows that y and z are not adjacent. Now suppose neither y nor z is in B. Then both y and z are in H. Since H is complete to A, it follows that y and z are both adjacent to x. Since G has no induced triangles by assumption, it therefore follows that y and z are not adjacent. This shows that for all choices of y and z in G distinct from x, the vertices y and z are non-adjacent. Now take any element y of G distinct from x. If y is in H then y is adjacent to x since H is complete to A. Otherwise, y is in B. Since G has no isolated vertices, we see that y must be adjacent to some vertex of G. Since vertices of B are 28

39 at most adjacent to vertices of A, we see that y is adjacent to some element of A G. Since x is the only such vertex by assumption, we see that y is adjacent to x. We have thus shown that x is complete to G x. Since we have also shown G x is anti-complete, we see that G is a star, contrary to assumption. This contradiction shows that A G can not have exactly one vertex. Finally, assume A G contains exactly two vertices. Call them x and y. Since A is complete, x and y are adjacent. Suppose H G contains a vertex z. Since A is complete to H, it follows that z is adjacent to x and y and hence x, y, z comprise a triangle, contrary to choice of G as triangle free. Therefore H G is empty, which means G is an induced subgraph of S as claimed. In all three cases, G is an induced subgraph of H or S, thus completing the proof. In the following proposition, we characterize graphs excluding M 2 and C 4. This proposition is notable in two ways. First, we prove the result for graphs, which is stronger than simply proving the analogous result for degree sequences, and it is somewhat surprising a nice characterization exists for graphs at all. Later in the chapter, for instance, when we exclude M 2 alone, it will be quite necessary to use degree sequences rather than graphs. Second, we have pointed out each exclusion theorem has a complementary theorem, but the following proposition is self complementary. The results later in the chapter lose the property of self complementarity as well. Theorem The following are equivalent for a graph G: 29

40 (i) G excludes M 2 and C 4. (ii) G is a split graph or the half join of a split graph and a pentagon. Proof. To see that (ii) implies (i), note that by Proposition , we know that if G is a split graph then G has no induced M 2 and no induced holes, and in particular no induced square. Now suppose G is the half join of a split graph and a pentagon. Note that since M 2 and C 4 have at least three vertices, have no induced triangles, no isolated vertices, and are not stars, it follows from Lemma that if M 2 or C 4 is an induced subgraph of the half join of a split graph and a pentagon then M 2 or C 4 must be an induced subgraph of a split graph or an induced subgraph of a pentagon. It is easy to see a pentagon contains no induced M 2 or C 4, and we have already noted a split graph contains no induced M 2 or C 4, therefore the half join of a split graph and a pentagon has no induced M 2 or C 4. For the other direction, suppose G has no induced M 2 and no induced C 4. If G also has no induced pentagon then by Proposition , we know that G is a split graph as desired. So suppose G has an induced pentagon C. We show that G[V (G) C] is a split graph and that G is the half join of G[V (G) C] and C. Toward this end, we first show that every vertex x of V (G) C is either complete or anti-complete to C. So let x be in V (G) C. Let C = {a, b, c, d, e} with the vertices in that cyclic order. Suppose x is adjacent to at least one vertex in C. Without loss of generality, x is adjacent to a. If x has degree 1 in G[C x] then {x, a} and {c, d} are independent edges and G thus has an induced M 2, contrary to hypothesis. Suppose 30

41 x has degree 2 in G[C x]. Then x is either adjacent to a vertex of C adjacent to a or a vertex of C at distance 2 from a in C. Suppose x is adjacent to a vertex adjacent to a. Without loss of generality, x is adjacent to b. Then {x, a} and {c, d} are again independent edges, contrary to hypothesis that G has no induced M 2. So suppose x is adjacent to a vertex at distance 2 from a in C. Without loss of generality, x is adjacent to c. Then x, a, b, c is an induced 4 cycle in G, contrary to hypothesis. Both possible graphs in which x has degree 2 in G[C x] result in a contradiction, thus x has degree greater than 2 in this graph. Now suppose x has degree 3 or 4 in G[C x]. Consider the complement K of G[C x]. The complement of a pentagon is a pentagon, so K consists a pentagon together with a vertex adjacent to either 1 or 2 vertices of that pentagon. By the previous paragraph, such a graph contains an induced M 2 or C 4, so K has an induced M 2 or C 4. If K has an induced C 4 then by taking complements, G[C x] has an an induced M 2. Similarly, if K has an induced M 2 then by taking complements, G[C x] has an induced C 4. Both these possibilities are contrary to assumption that G has no induced M 2 or C 4. Thus x can not have degree 3 or 4 in G[C x] either. The only remaining possibility is that x has degree 5 in G[C x], or in other words, x is complete to C. We have thus shown that if x is not anti-complete to C then x is complete to C. Since x was arbitrary, we have shown that every vertex outside C is either complete or anti-complete to C. Let A be the set of vertices outside C and complete to C, and let B be the set of vertices outside C and anti-complete to C. We know that A B C = G. We show that A is complete and B is anti-complete. 31