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Forbidden configurations Raggi, Miguel 2011-08-09

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Forbidden Con gurations by Miguel Raggi a thesis submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the faculty of graduate studies (Mathematics) The University Of British Columbia (Vancouver) August 2011 c Miguel Raggi, 2011Abstract In this work we explore the  eld of Forbidden Con gurations, a problem in Extremal Set Theory. We consider a family of subsets of f1; 2; :::;mg as the corresponding f0; 1g-incidence matrix. For f0; 1g-matrices F , A, we write F  A if A has a submatrix which is a row and column permutation of F . We say a f0; 1g-matrix is simple if it has no repeated columns. Let kAk denote the number of columns of A. A f0; 1g-matrix F with row and column order stripped is a con guration. Given m 2 N and a family of con gurations F , our main function of study is forb(m;F) := maxfkAk : A simple and for all F 2 F we have F  Ag: We give a general introduction to the main ideas and previous work done in the topic. We develop a new more computational approach that allows us to tackle larger problems. Then we present an array of new results, many of which were solved in part thanks to the new computational approach. We use both new ideas and new spins on old ideas to tackle the problems. The new results include  nding exact bounds on small con gurations that were previously unknown, and proving some previously conjectured asymptotic bounds for \boundary" con gurations. We also develop a relationship between Forbidden Con gura- tions and Patterns, which we use to prove some results. iiPreface Most of the results in this thesis were done jointly with both Dr. Richard Anstee and Dr. At- tila Sali and we produced four (submitted) papers. There are three papers with Dr. Anstee and Dr. Sali: \Evidence for a Forbidden Con guration Conjecture; one more case solved," [ARS10a], \Forbidden Con gurations: Quadratic Bounds," [ARS11] and \Forbidden Con-  gurations and Product Constructions," [ARS10b]. There is one with Dr. Anstee, \Genetic Algorithms applied to problems of Forbidden Con gurations," [AR11]. The preprints for these papers can be downloaded from http://www.math.ubc.ca/ anstee/. A more detailed description is in order. Chapter 1 and Chapter 2 have an introductory character and the ideas contained in them were thought of before I was involved. Chapter 3 is mostly my doing, under the supervision of Dr. Anstee. I wrote all the code and developed the algorithms, but of course, with good advice and encouragement from my supervisor. From Chapter 4, Section 4.1 (from [AR11]) and Section 4.2 (from [ARS10b]) were done mostly by me as well, again with good advice from Dr. Anstee. In particular I produced a list of conjectures for all unknown exact bounds for 3 4 con gurations (using a Genetic Algorithm as described in Chapter 3), and Dr. Anstee gave me advice on which ones to pursue, as he felt were more likely to produce results. He was right. Section 4.3 was done jointly with Dr. Anstee. All results from Chapter 5 (from [ARS10a] and [ARS11]) were done jointly with Dr. Anstee and Dr. Sali. The results on Chapter 6 and Chapter 7 (from [ARS10b]) were started o by Dr. Anstee and myself, but were completed by Dr. Anstee and Dr. Sali in Hungary (although using some of my computations). iiiContents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Extremal Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Basic De nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3.1 Matrices of 0’s and 1’s . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3.2 Con gurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.3.3 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.4 The Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.5 Table of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 1.6 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.6.1 Extremal Combinatorics and Set Theory . . . . . . . . . . . . . . . . 29 1.6.2 Forbidden Con gurations . . . . . . . . . . . . . . . . . . . . . . . . . 32 2 Basic Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.1 Standard Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.2 What Is Missing? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.3 Implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 iv3 Computer Program Developed . . . . . . . . . . . . . . . . . . . . . . . . . 42 3.1 Representation of a Con guration . . . . . . . . . . . . . . . . . . . . . . . . 43 3.2 Subcon gurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.3 Determining X(F) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.4 Boundary Cases to Classify Con gurations . . . . . . . . . . . . . . . . . . 47 3.5 Finding What Is Missing and Forbidden . . . . . . . . . . . . . . . . . . . . 48 3.6 Guessing Forb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.6.1 Brute Force Greedy Search . . . . . . . . . . . . . . . . . . . . . . . . 55 3.6.2 Hill Climbing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 3.6.3 Genetic Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 4 Exact Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 4.1 Two 3x4 Exact Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 4.1.2 The Bound for W . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 4.1.3 The Bound for V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 4.2 Exact Bound for Ten Products . . . . . . . . . . . . . . . . . . . . . . . . . . 79 4.3 Critical Substructures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 5 Three Asymptotic Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 5.2 Quadratic Bound for a 4-rowed Con guration . . . . . . . . . . . . . . . . . 89 5.2.1 What Is Missing? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.2.2 Case Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 5.2.3 Linear Bound for the Inductive Children . . . . . . . . . . . . . . . . 100 5.3 Quadratic Bound for a 5-rowed Con guration . . . . . . . . . . . . . . . . . 103 5.3.1 Applying Standard Induction . . . . . . . . . . . . . . . . . . . . . . 103 5.3.2 What Is Missing? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 5.3.3 Linear Bound for the Inductive Children . . . . . . . . . . . . . . . . 107 5.4 Classi cation of 6-rowed Quadratic Bounds . . . . . . . . . . . . . . . . . . . 118 6 Patterns and Splits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 6.1 Patterns and Splits in 2-Dimensions . . . . . . . . . . . . . . . . . . . . . . . 123 6.2 Patterns and Splits in d-Dimensions . . . . . . . . . . . . . . . . . . . . . . . 128 7 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 v7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 7.2 Submatrices of TxT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 7.3 Submatrices of IxI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 7.4 Submatrices of IxT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 7.5 Submatrices of IxTxIc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 7.6 Fractional Exponent Bound for a Family of Con gurations . . . . . . . . . . 143 7.7 All Pairs of Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 8 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 8.1 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 8.1.1 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 8.1.2 A Common 4-rowed Subcon guration . . . . . . . . . . . . . . . . . . 154 8.1.3 Other 3x4 Exact Bounds . . . . . . . . . . . . . . . . . . . . . . . . . 154 8.1.4 Critical Substructures . . . . . . . . . . . . . . . . . . . . . . . . . . 156 8.2 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 viList of Tables Table 1.1 Classi cation: 1 row. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Table 1.2 Classi cation: 2 rows. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Table 1.3 Classi cation: 3 rows. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Table 1.4 Classi cation: 4 rows. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Table 1.5 De nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Table 8.1 Conjectured values of forb(m;Vi) . . . . . . . . . . . . . . . . . . . . . . . 155 Table 8.2 Conjectured values for forb(m;F3) and forb(m;K4) . . . . . . . . . . . . . 157 Table 8.3 Conjectured values for forb(m; 2  K24) . . . . . . . . . . . . . . . . . . . . 157 viiAcknowledgements First, I would like to thank my supervisor, Richard Anstee, for his endless patience and the incredible amount of time he spent working with me. It was fun. I would also like to thank my committee members, Stephanie Van Willigenburg, and Joel Friedman, as well the University Examiners and the External Examiner, all for taking the time to go through my rather long (but quite fun) thesis. Also deserving of a thankful mention is my other co-author, Attila Sali. I would also like to thank the Department of Mathematics at UBC. In particular, I would like to thank all professors, specially those who I took (at least) one course from: Richard Anstee, Kai Behrend, Patrick Brosnan, Jingyi Chen, Joel Feldman, Joel Friedman, Richard Froese, Christoph Hauert, Kalle Karu, Richard Kenyon, Kevin Leyton-Brown (from Comp. Sci.), Greg Martin, Wayne Nagata, Ed Perkins, Malabika Pramanik, Laura Scull, Andrew Rechnitzer, Dale Rolfsen, Denis Sjerve, Lior Silberman, Jozsef Solymosi and Vinayak Vatsal. The acknowledgements wouldn’t be complete without thanking all the people who were my friends during my time in Vancouver. By no means a comprehensive list: Yuri, Dennis, Hardeep, Ramon, Itzia, Ignacio, Felipe, and all the others. In my long life I have met many nice people who have done me good. I’ll be eternally grateful to all of them. In my undergraduate thesis acknowledgements (or \agradecimientos") I gave my best shot to mention as many of them as I could. To mention all of them again would undoubtedly make me hit the much dreaded 173 page mark, quite unlucky indeed. Therefore, I would like to say the following: \input{agradecimientos.tex} Some individuals, though, were so instrumental in my (alleged) success that I feel they deserve the mention twice. My family: Mal u, Gerardo, Tanja, Daniel, Grola, O’Kuri, Medo, Gole and, even though she doesn’t live with us anymore, Kima. I’m sure she is running happily in a farm upstate somewhere. I feel like I’m missing someone... someone important... viiiDedication A Teresa. Aunque esta tesis represente el tiempo que no estuve contigo, terminarla signi ca poder regresar a ti. ixChapter 1 Introduction 1.1 Summary We give a brief description of the contents of each chapter of the thesis. For a more in-depth description, each chapter contains some introductory remarks. For this summary we try to use as few concepts as possible, but in order to describe the contents of some chapters it is unavoidable to use some key concepts de ned in Section 1.3. Forbidden Con gurations is at its core an extremal set theory problem: We attempt to maximize the size of a family of subsets of f1; 2; :::;mg subject to some restriction, namely that the family doesn’t contain a particular object which we call a con guration. The function forb(m;F ) represents the maximum size of a family of subsets of f1; 2; :::;mg with no con guration F . For proper de nitions, see Section 1.3. We start in Chapter 1 by giving a general introduction and motivation to the  eld of Forbidden Con gurations (in Section 1.2) and some basic de nitions (in Section 1.3) that are used throughout the whole thesis. In Section 1.4 we describe a conjecture of Anstee and Sali (Conjecture 1.4.1) which has been a driving force in studying Forbidden Con gurations. There is a brief history of the topic in Section 1.6 including related extremal problems. In Chapter 2, we explain three techniques that have been quite fruitful in the  eld of Forbidden Con gurations, namely those of Standard Induction, What Is Missing and Implications. We have greatly increased the reach of Standard Induction (and to a lesser extent, Implications) with new ways to use them in solving problems. We have increased the usability of What Is Missing by developing a computer program to do it automatically. In Chapter 3 we describe the computer program we use to push the boundaries by increasing the size of the problems we are able to deal with. We’ll often refer to this section 1throughout the rest of the thesis. We used the software in three fundamentally di erent ways. First and most importantly, we used it to obtain relevant information about con gurations (or family of con gurations), by  nding What Is Missing as described in Section 2.2. Second, we used some speci c-purpose code to perform case analysis when we found it hard or tedious to do so by hand, helping in various proofs. Lastly, we used some local search strategies (described in Section 3.6) to guess extremal matrices and structures. Section 3.6 could be removed entirely and all proofs would still be correct, but a reader might be left wondering \How did they come up with this answer?". This would be a perfectly valid question for which the only reasonable answer was \the computer told us so." Chapter 4 provides some new exact values for forb(m;F ) (see De nition 1.3.24) for some con gurations F for which the bound was not previously known. In Section 4.1 in particular we prove exact bounds for two of the smallest con gurations for which the answer wasn’t known. These results can be found in [AR11]. In Section 4.2 we prove an exact bound for a family of 10 con gurations, a result that can be found in [ARS10b]. Finally, in Section 4.3 we give all critical substructures (see De nition 4.3.1) of K4, as well as a conjecture for all critical substructures of Kk. In this chapter we make heavy use the program described in Section 3.6 to guess extremal con gurations and then proceed to prove they are indeed extremal, mostly using Standard Induction and What Is Missing followed by case analysis. In Chapter 5 we give proofs of some boundary cases (see De nition 1.4.3), which serve as further evidence for our motivating Conjecture 1.4.1. We give the asymptotic bound for a 4-rowed con guration F8(t) in Section 5.2 (appears in [ARS10a]), a 5-rowed con guration F7 in Section 5.3 and a 6-rowed con guration G6 3 in Section 5.4. This last one in particular is proven to be the unique quadratic 6-rowed boundary case. The last two results appear in [ARS11]. For Chapter 6 we give an entirely di erent extremal combinatorics problem with a more geometric  avour. This chapter is independent of all previous chapters, but is used in Chapter 7 for some proofs, as some problems considered in Chapter 7 can be interpreted as problems of patterns. We dedicate Chapter 7 to the study of product constructions. We de ne a sister function to ‘forb’ and prove various interesting results about it. Then we prove an unexpected bound (for ‘forb’) of a family of con gurations, which on a  rst glance would seem to be a coun- terexample of Conjecture 1.4.1 (but isn’t). Most of the results of Chapter 6 and Chapter 7 can be found in [ARS10b]. Finally, Chapter 8 contains some open problems, as well as some ideas and thoughts on 2Forbidden Con gurations and what the future might yield. The results of Section 4.1.3 and Section 5.3 and its application to Section 5.4 provide a good sample of the  avour of the new results given this thesis. 1.2 Extremal Problems A general extremal problem in combinatorics has the following character: For a given prop- erty P , what is the maximum size of a family that satis es P? For example, if the objects of study are families of subsets of f1; 2; :::;mg and property P is that no two subsets have two elements in common, the question becomes: What is the maximum number of subsets of f1; 2; :::;mg such that no two subsets have two elements in common? For this particular example we will give the answer later in Example 1.3.14. In the  eld of Forbidden Con gurations, we consider a particular extremal problem that arises as a generalization to celebrated extremal results in combinatorics, namely those of Erd}os and Stone [ES46] and Erd}os and Simonovits [ES66]. They consider the following problem: Given m 2 N and a  xed (small) graph F ,  nd the maximum number of edges in a (simple) graph G with m vertices that avoids having a subgraph isomorphic to F . For example, Mantel’s result answers the question \what is the maximum number of edges in a triangle-free graph on m vertices?" and can be viewed as a special case. We will discuss this further in Section 1.6. There are a number of ways to generalize the above problem to hypergraphs, but in this thesis we consider the following generalization: Given m 2 N and a hypergraph F ,  nd the maximum number of edges in a simple hypergraph H on m vertices that avoids having a subhypergraph (or trace) isomorphic to F . We consider the notion of subhypergraph as follows: The vertices of a subhypergraph are a subset of the vertices of the original hypergraph. For the edges, we choose a subset of the edges from the original hypergraph and consider the multiset consisting of intersections of the chosen edges with the vertices of the subhypergraph (more on this later). As stated before, there are alternate generalizations of this problem. For example, instead of considering the edges of the subhypergraph to be intersections of edges of the original hy- pergraph with the vertices of the subhypergraph, one might consider a subhypergraph to have only edges that are fully contained in the chosen subset of the vertices. This alternate de nition makes sense especially when considering k-uniform hypergraphs, in the sense that all edges have size k (a graph is a 2-uniform hypergraph). This alternate view of subhyper- 3graphs is consistent with the way subgraphs of graphs are usually de ned. In contrast, the way we see subhypergraphs when restricted to 2-uniform hypergraphs would suggest that when considering a subset of the vertices, we might have to consider \edges" of size 1 or size 0, even though we started with all edges of size exactly 2, since the intersection of an edge of size 2 with the vertices of the subhypergraph might have size 1 or 0. Moreover, the subhypergraph need not be simple, even if the original hypergraph was. When studying di cult extremal problems, such as Forbidden Con gurations, an exact answer to the extremal question might not come easily. In such cases, it’s common practice to settle for asymptotic bounds, given in Big-O notation: If f; g : N ! N, we de ne the following notation:  We say f is O(g) if there exists a constant c > 0 and a number N for which for all n  N , we have f(n) < c  g(n).  We say f is  (g) if there exists a constant d > 0 and a number N for which for all n  N , we have f(n) > d  g(n).  We say f is  (g) if f is both O(g) and  (g). That is, if there exists constants c > 0, d > 0 and N for which for all n > N , d  g(n) < f(n) < c  g(n). In such case we say f has the same asymptotic growth as g. 1.3 Basic De nitions 1.3.1 Matrices of 0’s and 1’s First some standard notation. Let X be a  nite set. We denote by 2X the power set of X: the set that consists of all subsets of X (note that there are 2jXj of them). Denote all subsets of X of a given size t by  X t  . A set system A on a set X is a family of subsets of X. In other words, A  2X . We might view a set system on X as a simple hypergraph on the vertices X. We  nd it convenient to encode hypergraphs or set systems in the language of matrices. For a given order of the elements of X, and an ordering of the elements of a family A, we can encode a set system as a matrix in the following way: A single subset   X can be thought of as the incidence vector of 0’s and 1’s with jXj entries, where the entry i is 1 if and only if i 2  . The family A gives rise to a collection of f0; 1g-vectors with jXj components. Consider these f0; 1g-vectors as column vectors and concatenate them together to form a 4f0; 1g-matrix. Since the elements of X don’t play any role in our investigations, we might as well take X = [m] := f1; 2; :::;mg with m being the size of X. For example, the matrix A = 2 6 6 6 6 4 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 0 0 1 0 3 7 7 7 7 5 represents the family A = ff1g; f2; 3g; f1; 3g; f2; 3; 4g; ;g. Notice that in this encoding no two columns are equal: if two columns were equal, they would represent the same subset of X. Di erent orderings of the columns of A give rise to di erent ways of ordering the sets of the family A. We usually think of the set [m] as having a canonical order, but we can think of di erent row orders as giving rise to equivalent hypergraphs. We will de ne the notion of equivalence of f0; 1g-matrices formally in the next section, but for now let us talk about matrices a little longer. De nition 1.3.1 We say a f0; 1g-matrix A is simple if no two columns of A are equal. The size of the family A (i.e. the number of edges in the hypergraph) is represented by the number of columns of matrix A. Since we will refer to this number often, we  nd it convenient to create a short notation for it. De nition 1.3.2 For a f0; 1g-matrix A, we de ne kAk to be the number of columns of A. The size of each edge in the hypergraph is represented by the number of 1’s in the corresponding column. De nition 1.3.3 For a f0; 1g-matrix A we denote by  0(A) the number of 0’s in matrix A and by  1(A) the number of 1’s. In the case of a column  , we say  has column sum k (and write  1( ) = k) if it has exactly k ones. For m-rowed A, we have  0(A) +  1(A) = m  kAk: De nition 1.3.4 De ne 0m to be the m-rowed column with all 0’s and 1m to be the m- rowed column of all 1’s. We now de ne some operations of f0; 1g-matrices that will be useful for constructing new matrices from previously constructed ones. 5De nition 1.3.5 Let A be a f0; 1g-matrix. We denote Ac the f0; 1g-complement of A. That is, the matrix that results from replacing every 0 in A by a 1, and every 1 by a 0. For example, A = 2 6 6 6 6 4 0 1 0 1 0 0 0 0 1 0 1 1 0 1 0 1 3 7 7 7 7 5 =) Ac = 2 6 6 6 6 4 1 0 1 0 1 1 1 1 0 1 0 0 1 0 1 0 3 7 7 7 7 5 : De nition 1.3.6 Let A and B be f0; 1g-matrices with the same number of rows. De ne the concatenation [AjB] to be the con guration that results from taking all columns of A together with all columns of B. For t 2 N, we de ne the product t  A := [ A j A j    j A | {z } t times ]: The operation  has precedence over j, so that [t  A j B] = [(t  A) j B]. De nition 1.3.7 Let A and B be f0; 1g-matrices. We construct the product A  B by taking each column of A, and putting it on top of each column of B. So if the columns of A are  1; :::;  a and the columns of B are  1; :::;  b, then A B is the matrix with ab columns: A  B = 2 6 6 6 6 6 6 6 6 6 4 j j j j j j j j j  1  1 :::  1  2  2 :::  2 ::: :::  a  a :::  a j j j j j j j j j j j j j j j j j j  1  2 :::  b  1  2 :::  b ::: :::  1  2 :::  b j j j j j j j j j 3 7 7 7 7 7 7 7 7 7 5 : Here is an example of a product: A = " 0 1 1 0 0 1 # ; B = " 1 0 0 1 # =) A  B = 2 6 6 6 6 4 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 0 1 0 0 1 0 1 0 1 3 7 7 7 7 5 : Often we will consider single columns and rows of f0; 1g-matrices. 6De nition 1.3.8 For a column  and a f0; 1g-matrix A, we de ne the multiplicity of  in A, written as  ( ;A), as the number of columns of A which are equal to  . For example,  0 B @ 2 6 4 1 0 0 3 7 5 ; 2 6 4 0 1 1 1 1 0 1 0 0 0 1 0 3 7 5 1 C A = 2: De nition 1.3.9 Let A and B be m-rowed f0; 1g-matrices. We de ne subtraction A B to be the m-rowed f0; 1g-matrix such that for each m-rowed column  , A B satis es  ( ;A B) = maxf0;  ( ;A)  ( ;B)g: The order in which the columns appear in A  B isn’t important, but just so that it is a well de ned operation of matrices, we might choose the order of the columns of A  B to be the same as for A. This operation corresponds to set di erence of the corresponding set systems. Here is an example of the di erence of two matrices: 2 6 4 0 1 1 1 1 1 0 1 0 0 0 0 1 0 0 3 7 5 2 6 4 1 1 0 1 1 1 0 0 0 1 1 0 3 7 5 = 2 6 4 0 1 1 1 0 0 0 0 0 3 7 5 : Of the columns of A, the column (0; 1; 0)T has multiplicity 1 in A and 0 in B, so it must have multiplicity 1 in A B. The column (1; 0; 0)T has multiplicity 3 in A and 1 in B, so it must have multiplicity 3 1 = 2 in A B. Finally, the column (1; 1; 1) has multiplicity 1 in A and 1 in B so it must have multiplicity 0 in A B. De nition 1.3.10 Let A be a f0; 1g-matrix. Given a subset of the rows S, we de ne the restriction AjS to be a f0; 1g-matrix formed from rows S of A. For example, A = 2 6 6 6 6 4 0 1 0 1 0 1 0 0 1 1 1 1 0 1 0 0 1 1 0 0 1 1 0 1 3 7 7 7 7 5 =) Ajf2;4g = " 0 0 1 1 1 1 0 0 1 1 0 1 # : 7The following notation is convenient. De nition 1.3.11 Let  and  be m  1 columns. Then we say    if for every row r for which the r-th entry of  is a 1, the r-th entry of  is also a 1. For example, 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5  2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 but 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5  2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 : De nition 1.3.12 Let  be an m 1 column and A be an m-rowed matrix. We use  2 A if  is a column of A. The following three special matrices turn out to be remarkably useful in our investigations. We shall see why they are important in Conjecture 1.4.1 De nition 1.3.13 We de ne the following three matrices:  The identity matrix Im is be the m  m matrix with 1’s in the diagonal and 0’s everywhere else. It corresponds to the set system ff1g; f2g; :::fmgg. For example, I4 = 2 6 6 6 6 4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 3 7 7 7 7 5 :  The identity complement Icm is the f0; 1g-complement of Im. That is, every entry is a 1, except for the diagonal entries, which are 0’s. It corresponds to the set system f[m] n f1g; [m] n f2g; :::; [m] n fmgg. For example, Ic4 = 2 6 6 6 6 4 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 3 7 7 7 7 5 : 8 The tower matrix Tm is the m (m+ 1) matrix that corresponds to the set system f;; [1]; [2]; [3]; :::; [m]g. For example, T4 = 2 6 6 6 6 4 0 1 1 1 1 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 3 7 7 7 7 5 : For these three matrices, when the number of rows is implicit we might omit the subindex m and write I; Ic or T . Example 1.3.14 Returning to the problem of  nding the maximum number of subsets of [m] such that no two subsets have two elements in common, we can restate it as the problem of  nding the most number of columns a simple matrix with m rows can have without having the following submatrix: F = " 1 1 1 1 # : The answer to this particular example is straightforward: The maximum number of columns is  m 0  +  m 1  +  m 2  : Indeed, we can take one column with no 1’s (the empty set), all columns with one 1 and all columns with two 1’s. For every pair of rows there is at most one column with two 1’s in that pair. Thus the number of columns we can have with more than one 1 is at most  m 2  . For example, for m = 3, this is the matrix: A = 2 6 4 0 1 0 0 1 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 1 3 7 5 : This concludes Example 1.3.14. Example 1.3.15 Now consider the following question. What is the maximum number of subsets of f1; 2; :::;mg such that no three sets each have an element that the other two sets don’t have? This is a very similar problem. We wish to  nd the most number of columns a 9f0; 1g-matrix can have, having neither of the following six submatrices: 2 6 4 1 0 0 0 1 0 0 0 1 3 7 5 nor 2 6 4 1 0 0 0 0 1 0 1 0 3 7 5 nor 2 6 4 0 1 0 1 0 0 0 0 1 3 7 5 nor 2 6 4 0 1 0 0 0 1 1 0 0 3 7 5 nor 2 6 4 0 0 1 1 0 0 0 1 0 3 7 5 nor 2 6 4 0 0 1 0 1 0 1 0 0 3 7 5 : In other words, we wish to  nd the maximum number of columns in a matrix that doesn’t have any row and column permutation of I3 as a submatrix. We will answer this question later (in Proposition 4.3.3). This demonstrates the need to get rid of row and column ordering. We do so in the next section. 1.3.2 Con gurations De nition 1.3.16 (informal) A con guration is a f0; 1g-matrix with column and row order stripped. More formally, letMm be the set of all m-rowed matrices. De ne an equivalence relation: If A;B 2Mm, we say A  B i A is a permutation of the rows and columns of B. Consider the set of con gurations Cm by taking the quotient: Cm :=Mm=  ; and if F is a f0; 1g-matrix, de ne ~F to be the equivalence class (or con guration) to which F belongs. We will usually abuse notation and not distinguish between a con guration and a f0; 1g- matrix representative, always remembering that we can permute the order of the rows and columns without altering the con guration. For example, the six matrices given in Exam- ple 1.3.15 represent the same con guration. Notice that if F , G are con gurations and t is a number, the operations we de ned for f0; 1g-matrices kFk, F  G, F c, t  F are well de ned in con gurations, but concatenation [F jG] and subtraction F  G are not. We will therefore use the notation [F jG] and F  G carefully, when the order of rows is understood. The notion of being a simple con guration 10is also well de ned: a con guration for which any representative is a simple matrix. We have that F  G = G F as con gurations, and if F and G are simple, F  G is also simple. As with matrices, we might refer to the multiplicity of a column  (and use the notation  ( ; F )). What we mean by this is that for a speci c representative of a con guration F , there are k columns that are equal to  . Of course, if we choose a di erent representative for F (i.e permute the rows and columns), we will have k copies of a row permutation of  . De nition 1.3.17 Denote by Kk the unique k  2k simple \complete" con guration cor- responding to the set system 2[k]. That is, the one representing the power set of [k]. The notation was chosen to mirror the common notation for the \complete" graph Kk. As an example, here is K3: K3 = 2 6 4 0 1 0 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 0 1 0 1 1 1 3 7 5 : Denote by Ksk the k   k s  con guration with k rows and all possible columns of column sum s. Here is K23 : K23 = 2 6 4 1 1 0 1 0 1 0 1 1 3 7 5 : We also use notation such as K sk (resp. K  s k ), meaning the con guration with k rows and all possible columns of column sum  s (resp  s). Observe that for k; ‘ 2 N, we have Kk  K‘ = Kk+‘. De nition 1.3.18 For a con guration F and a f0; 1g-matrix A (or a con guration A), we say that F is a subcon guration of A, and write F  A if there is a representative of F which is a submatrix of A. We say A has no con guration F (or doesn’t contain F as a con guration or avoids F ) if F is not a subcon guration of A. The notion of subhypergraph in our matrix notation becomes the notion of a subcon g- uration. Observe that  de nes a partial ordering in the set Cm of con gurations, since  is re exive (F  F ), anti-symmetric (if F  G and G  F , then F = G as con gurations), and transitive (F  G  H implies F  H). 11De nition 1.3.19 Let Avoid(m;F ) := fA 2 Cm : A is simple and F  Ag: In other words, the set of all m-rowed simple con gurations with no F as a subcon guration. For example, Avoid  2; " 1 0 0 1 #! = (" 0 0 # ; " 1 0 # ; " 1 1 # ; " 0 1 0 0 # ; " 0 1 0 1 # ; " 1 1 0 1 # ; " 0 1 1 0 0 1 #) De nition 1.3.20 Our main object of study is the following extremal function. We de ne forb(m;F ) := maxfkAk : A 2 Avoid(m;F )g: In other words, forb(m;F ) is the maximum number of columns a simple m-rowed f0; 1g- matrix can have with no F as a subcon guration. For example, we may conclude that forb  2; " 1 0 0 1 #! = 3; by looking at all con gurations in Avoid  2; " 1 0 0 1 #! . Remark 1.3.21 Observe that forb is an \increasing" function in the second variable, mean- ing F  G =) forb(m;F )  forb(m;G): Indeed, if F  G, then every matrix containing G as a con guration also contains F . Thus, Avoid(m;F )  Avoid(m;G), which means forb(m;F )  forb(m;G):  Interestingly, it is an important open problem to see if forb(m;F ) is increasing in the  rst variable in the sense that m < n implies forb(m;F )  forb(n; F ). We have strong evidence to believe this is the case, but it hasn’t been proven. More about this in Section 8.1. Remark 1.3.22 Observe that forb(m;F ) = forb(m;F c) by symmetry. We may also consider forbidding a family of con gurations F : 12De nition 1.3.23 Let F be a family of con gurations. De ne Avoid(m;F) := \ F2F Avoid(m;F ): In other words, the set of all m-rowed simple matrices that avoid all con gurations in F . De nition 1.3.24 We now de ne the analogous extremal function forb(m;F) := maxfkAk : A 2 Avoid(m;F)g: Note that according to this de nition, there exists a simple matrix of size m forb(m;F) that doesn’t contain any F 2 F as a subcon guration, but every matrix of size m  (forb(m;F) + 1) contains at least one F 2 F . Also note that (F ; ) can be considered as a partially ordered set, and if H := minimal elements of F , then Avoid(m;F) = Avoid(m;H). De nition 1.3.25 We de ne the extremal matrices ext(m;F) := fA 2 Avoid(m;F) : kAk = forb(m;F)g: Problem 1.3.26 (The Main Problem) We are interested in determining forb(m;F). We would like to  nd forb(m;F) exactly for any family, but we’ll settle for asymptotic bounds when it’s too hard to  nd forb(m;F) exactly. This problem was  rst introduced by Anstee in [Ans85] for jFj = 1. We deal mostly with the case jFj = 1. We now give some de nitions that will be helpful in our investigations. De nition 1.3.27 We say a family A is laminar if for every X; Y 2 A we have that either X  Y , Y  X or X \Y = ;. Analogously, we say a f0; 1g matrix A is laminar if for every two columns  ;  of A we have either    , or    or there are no rows in which both  and  have a 1. Equivalently, we say m-rowed A is laminar if A 2 Avoid(m;F ) for F := 2 6 4 1 1 1 0 0 1 3 7 5 : 13For example, 2 6 6 6 6 4 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 3 7 7 7 7 5 is laminar but 2 6 6 6 6 4 1 0 1 1 0 0 1 1 0 1 1 1 0 1 0 1 3 7 7 7 7 5 isn’t. 1.3.3 Basic Properties While studying forb(m;F) for some family of con gurations F , we usually give lower bounds and upper bounds on forb(m;F). A lower bound usually consists of an example of a f0; 1g- matrix with no F 2 F as a con guration. An upper bound is usually accomplished by a theoretical argument. Proposition 1.3.28 We have that forb(m;0k) = forb(m;1k) =  m k  1  +  m k  2  + :::+  m 0  and forb(m; 2  0k) = forb(m; 2  1k) =  m k  +  m k  1  + :::+  m 0  : Thus, if 0k  F or 1k  F for some con guration F , we have that forb(m;F ) is  (mk 1) and if 2  0k  F or 2  1k  F , then forb(m;F ) is  (mk). Proof: The bound for forb(m;1k) is easy. Consider an extremal matrix A such that 1k  A. Note that in A, we cannot have columns with column sum k or more. This leaves only  m k 1  +  m k 2  + :::+  m 0  columns (the ones with column sum k  1 or less). The bound for forb(m; 2  1k) is also easy. For the columns with column sum k or higher, for each k-subset of the rows there must be at most one column with 1’s in that subset of the rows, so there are at most  m k  columns, plus the columns with column sum < k, which cannot contribute to create 2  1k. The bounds for forb(m;0k) and forb(m; 2  0k) are analogous.  Here is an important result proved independently by V. Vapnik and A. Chervonenkis, N. Sauer and M. Perles and S. Shelah: 14Theorem 1.3.29 [VC71][Sau72][She72] We have that forb(m;Kk) = forb(m;1k) =  m k  1  +  m k  2  + :::+  m 0  ; and so forb(m;Kk) is  (mk 1). Thus for any simple k-rowed con guration F , forb(m;F ) is O(mk 1). An important generalization of this result from Anstee and F uredi is: Theorem 1.3.30 [AF86] For  xed k, t, we have that as m!1, forb(m; t  Kk) = forb(m; t  1k) = t 2 k + 1  m k  (1 o(1)) +  m k  +  m k  1  +    +  m 0  : This in particular says that if F has k rows, we have that forb(m;F ) is O(mk) [F ur83], since F is contained in t  Kk for some t. 1.4 The Conjecture Anstee and Sali conjectured that the \best" asymptotic constructions in terms of avoiding a single con guration F would be formed from products of I; Ic and T . There is ample evidence for this conjecture, but no proof or counterexample has been found yet. The research in forbidden con gurations is often guided by this conjecture. Conjecture 1.4.1 [AS05] Let F be a con guration. Let Pr(a; b; c) := Ir  ::: Ir| {z } a times  Icr  ::: I c r| {z } b times  Tr  ::: Tr| {z } c times ; De ne X(F ) to be the largest number such that there exist numbers a; b; c 2 N with 15a+ b+ c = X(F ) such that for all r 2 N, F  Pr(a; b; c): Then forb(m;F ) is  (mX(F )): Observe that X(F ) is always an integer. Also note that kPr(a; b; c)k = ra+b  (r + 1)c which is  (rX(F )), so by taking r = dm=X(F )e (and perhaps deleting some rows in case X(F ) - m), we have that kPr(a; b; c)k is  (mX(F )), so the fact that forb(m;F ) is  (mX(F )) is built into the conjecture. In order to prove the conjecture, all that would be required would be to prove that forb(m;F ) is O(mX(F )) for every F . A disproof would be easier, as only a counterexample would be required. A valid objection is that  nding X(F ) given F is not a trivial task, but for relatively small con gurations F we have a computer program that yields the answer very quickly. We can compute X(F ) for F having less than  10 rows in just a few seconds. This task takes merely exponential time, not doubly exponential. A simple (but surprising) corollary of the conjecture is that repeating columns more than twice in F has no e ect on the asymptotic behavior of forb(m;F ). In other words, assuming the conjecture were true, the multiplicity of a column in a con guration would not a ect the asymptotic bound, and it for asympotic bounds, it would only matter if a column is not there (has multiplicity 0), appears once (has multiplicity 1), or appears \multiple times" (has multiplicity 2 or more). Lemma 1.4.2 Let Ft = [Gjt  H] with G and H simple f0; 1g-matrices that have no columns in common. Then X(F2) = X(Ft) for all t  2. In particular, if the conjecture were true, then forb(m;Ft) and forb(m;F2) would have the same asymptotic behavior. Proof: It su ces to show that given t, G, H, a, b and c, there exists an R such that for every r  R, we have F2 = [Gj2  H]  Pr(a; b; c) () Ft = [Gjt  H]  Pr(a; b; c): 16Since F2  Ft, we only need to prove that if F2  Pr(a; b; c) for some r, then Ft  PR(a; b; c) for some R. Suppose then F2 is contained in the product Pr(a; b; c) for some r. The idea is to  nd a subcon guration of Pr(a; b; c) in which there are some columns with multiplicity 1, and for the columns with multiplicity 2 or more, the multiplicity depends on r. We need r large enough so that the multiplicity of any one column (with multiplicity of 2 or more) is larger than t. Let x be the number of rows of Ft. Notice the following three facts, which include de nitions for EI , EIc and ET . EI(x; r) := [(r  x)  0x j Ix]  Ir EIc(x; r) := [(r  x)  1x j Icx]  I c r ET (x; r) := j r x k  Tx  Tr: The  rst and second facts are easy to see; just take any subset of x rows from Ir or Icr . The third statement is true by taking the br=xc-th row of Tr, the 2br=xc-th row of Tr, etcetera, up to the xbr=xc-th row. For example, if r = 5 and x = 2, we may take the second and fourth row from T5: T5 = 2 6 6 6 6 6 6 4 0 1 1 1 1 1 0 0 1 1 1 1 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 3 7 7 7 7 7 7 5 =) T5jf2;4g = " 0 0 1 1 1 1 0 0 0 0 1 1 # = ET (2; 5) Note that in the three con gurations EI(x; r); EIc(x; r) and ET (x; r), we have that there are some columns of multiplicity 1 and there are some columns for which their multiplicity can be made as large as we wish by making r large. Formally, let E(x; r) be one of EI(x; r) or EIc(x; r) or ET (x; r). We have that for every x-rowed column  there are three possibilities: either  ( ;E(x; r)) = 0 for all r, or  ( ;E(x; r)) = 1 for all r, or lim r!1  ( ;E(x; r)) =1: If  is a column for which lim r!1  ( ;E(x; r)) = 1, we may conclude that there is an R for which  ( ;E(x; r))  t for every r  R. Since F2 is contained in Pr(a; b; c) for some r, the columns in H will have multiplicity at least 2 in some subset of the rows of Pr(a; b; c). By taking PR(a; b; c), we see that Ft is also a subcon guration of PR(a; b; c).  17Conjecture 1.4.1 has been a driving force behind the  eld of Forbidden Con gurations, speci cally when forbidding a single con guration (see [Ans]). We search for maximal and minimal forbidden con gurations with a speci c asymptotic bound, since we have that if both forb(m;F ) and forb(m;G) are  (mk), and we have a con guration H such that F  H  G, then forb(m;H) is also  (mk). For a given k and s, we search for minimal and maximal s-rowed con gurations F for which forb(m;F ) is  (mk). We hope to use this to classify the asymptotic bounds for all con gurations. De nition 1.4.3 We say a con guration Ft = [Gjt  H] (for t  2) with G and H simple with no columns in common is a predicted boundary case if for any column  not present in H, X([Ftj ]) > X(Ft). We say Ft is a boundary case if forb(m;Ft) is  (mk) but for any column  not present in H, forb(m; [Ftj ]) is  (mk+1). Conjecture 1.4.1 is that predicted boundary cases are the same as boundary cases. Boundary cases for s up to 4 were classi ed by Fleming ([Ans]). The conjecture has been proven for all k ‘ con gurations F with k = 1; 2; 3 and many others in various papers. For k = 2 in [AGS97]. For k = 3 in [AGS97], [AFS01] and was completed in [AS05]. For k = 4, the case when F is simple was completed in [AF86]. For k = 4 and F non-simple, there were only three cases left to do. We completed one of them ([ARS10a]) and the proof is given in Section 5.2. For ‘ = 2, the conjecture was veri ed in [AK06]. For k = 5, there are nine 5 6 predicted boundary con gurations F for which X(F ) = 2 ([ARS11]). We prove one of them in Section 5.3. For k = 6, we give a complete classi cation of the con gurations F for which forb(m;F ) is quadratic ([ARS11]). If we only consider simple con gurations, then we may also consider a restricted version of a boundary case. De nition 1.4.4 We say a simple con guration F is a predicted s-boundary case if for any column  not present in F , we have that X([F j ])  X(F ) + 1. We say it is a s-boundary case if forb(m;F ) is  (mk) but forb(m; [F j ]) is  (mk+1) for any column  not present in F . The classi cation of s-boundary cases for  ve rows was done by Ryan [Ans]. We now give here a classi cation of the minimal and maximal boundary cases. For t  2, 18Minimal Maximal Constant [0]; [1] [0 1] Linear [0 0]; [1 1] t  [0 1] Table 1.1: Classi cation: 1 row. Minimal Maximal Constant " 1 0 # " 1 0 # Linear " 0 0 # , " 1 1 # " 1 0 0 1 # , " 1 1 0 0 # " 0 0 1 0 1 1 t  " 1 0 ## , " 0 0 t  " 1 0 0 1 ## Quadratic " 0 0 0 0 # , " 1 1 1 1 # , " 0 1 1 0 0 1 0 0 0 1 1 1 # , t  K2 Table 1.2: Classi cation: 2 rows. 19Minimal Maximal Linear 2 6 4 0 0 1 3 7 5, 2 6 4 1 1 0 3 7 5 2 6 4 1 0 1 0 0 1 1 1 0 0 0 1 3 7 5 Quadratic 2 6 4 0 0 0 3 7 5, 2 6 4 1 1 1 3 7 5, 2 6 4 1 0 0 0 1 0 0 0 1 3 7 5, 2 6 4 0 1 1 1 0 1 1 1 0 3 7 5, 2 6 4 1 1 0 0 0 0 3 7 5, 2 6 4 1 1 1 1 0 0 3 7 5, 2 6 4 1 0 1 0 0 1 0 1 0 0 1 1 3 7 5 2 6 4 0 0 1 1 0 0 0 1 0 1 1 1 t  2 6 4 1 0 1 0 0 1 1 1 0 0 0 1 3 7 5 3 7 5, 2 6 4 0 0 1 1 0 0 1 1 0 1 0 1 t  2 6 4 1 0 1 0 0 1 0 1 0 0 1 1 3 7 5 3 7 5, 2 6 4 0 0 0 t  2 6 4 1 0 0 1 1 0 1 0 1 0 0 0 1 0 1 3 7 5 3 7 5, 2 6 4 1 1 1 t  2 6 4 0 1 1 0 0 1 0 1 0 1 1 1 0 1 0 3 7 5 3 7 5 Cubic 2 03, [2 K13 jK 2 3 ], [2 K 1 3 jK 3 3 ], [K03 j2  K 2 3 ], [K 1 3 j2  K 2 3 ], 2  13 t  K3 Table 1.3: Classi cation: 3 rows. 20Before showing the table for 4 rows, let us de ne some con gurations: B1 = 2 6 6 6 6 4 0 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 1 1 1 1 3 7 7 7 7 5 ; B2 = 2 6 6 6 6 4 0 0 0 1 1 0 0 0 1 1 0 0 1 1 1 0 1 1 0 1 3 7 7 7 7 5 ; B3 = 2 6 6 6 6 4 0 0 0 1 1 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 3 7 7 7 7 5 ; B4 = 2 6 6 6 6 4 0 0 0 0 1 0 0 1 1 1 0 0 1 1 1 0 1 0 1 1 3 7 7 7 7 5 ; B5 = 2 6 6 6 6 4 0 0 0 1 1 0 0 1 1 1 0 0 1 0 1 0 1 0 1 1 3 7 7 7 7 5 ; B6 = 2 6 6 6 6 4 0 0 0 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 1 3 7 7 7 7 5 : D = 2 6 6 6 6 4 0 0 0 0 0 0 0 1 1 1 1 0 0 0 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 3 7 7 7 7 5 : 21Minimal Maximal Linear 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 0 1 0 0 3 7 7 7 7 5 Quadratic 2 6 6 6 6 4 1 1 1 1 0 0 0 0 3 7 7 7 7 5 , 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 , 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 , 2 6 6 6 6 4 1 0 1 0 0 1 0 1 3 7 7 7 7 5 , 2 6 6 6 6 4 1 0 0 0 1 0 0 0 1 1 1 1 3 7 7 7 7 5 , 2 6 6 6 6 4 0 1 1 1 0 1 1 1 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 0 1 1 0 0 0 1 1 0 0 0 1 t 2 6 6 6 6 4 1 0 1 1 1 1 0 0 0 1 0 1 0 0 1 0 3 7 7 7 7 5 3 7 7 7 7 5 (Conjectured) 2 6 6 6 6 4 1 0 1 1 0 1 1 1 0 0 1 0 0 0 0 1 t 2 6 6 6 6 4 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 0 3 7 7 7 7 5 3 7 7 7 7 5 (Conjectured) 2 6 6 6 6 4 1 0 1 0 0 1 0 1 0 0 1 1 0 0 1 1 t 2 6 6 6 6 4 0 1 1 0 1 1 0 0 3 7 7 7 7 5 3 7 7 7 7 5 Solved in Section 5.2 22Minimal Maximal Cubic 2 6 6 6 6 4 1 0 0 0 1 0 0 0 1 0 0 1 3 7 7 7 7 5 , 2 6 6 6 6 4 0 1 1 1 0 1 1 1 0 1 1 0 3 7 7 7 7 5 , 2 6 6 6 6 4 1 0 0 0 1 0 0 0 1 0 0 0 3 7 7 7 7 5 , 2 6 6 6 6 4 0 1 1 1 0 1 1 1 0 1 1 1 3 7 7 7 7 5 , 2 6 6 6 6 4 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 3 7 7 7 7 5 , 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 , 2 6 6 6 6 4 1 0 0 1 0 1 0 1 0 0 1 1 1 1 1 0 3 7 7 7 7 5 , 2 6 6 6 6 4 0 1 1 0 1 0 1 0 1 1 0 0 0 0 0 1 3 7 7 7 7 5 , 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 , 2 6 6 6 6 4 1 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 3 7 7 7 7 5 , 2 6 6 6 6 4 1 1 1 1 1 1 0 0 3 7 7 7 7 5 , 2 6 6 6 6 4 0 0 0 0 0 0 1 1 3 7 7 7 7 5 All these are conjectured. [K4 j t  [K4  Bi]] for i = 1; 2; :::6, [K04 j t  D] [K04 j t  D] c Quartic All others (see [Ans] for a list) t  K4 Table 1.4: Classi cation: 4 rows. For  ve rows, the minimal and maximal matrices haven’t been classi ed, except for the quadratic case, and these are the nine maximal  ve-rowed con gurations which are conjectured to have a quadratic bound. Of these, only F7 is known to have a quadratic bound, proved in Section 5.3. F3 = 2 6 6 6 6 6 6 4 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F4 = 2 6 6 6 6 6 6 4 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F5 = 2 6 6 6 6 6 6 4 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 1 0 3 7 7 7 7 7 7 5 23F6 = 2 6 6 6 6 6 6 4 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F7 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 0 1 0 3 7 7 7 7 7 7 5 F8 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 3 7 7 7 7 7 7 5 F9 = 2 6 6 6 6 6 6 4 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1 1 0 1 1 1 1 0 0 3 7 7 7 7 7 7 5 F10 = 2 6 6 6 6 6 6 4 1 1 0 1 0 0 1 0 1 1 1 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F11 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 1 1 3 7 7 7 7 7 7 5 241.5 Table of Notation For the following de nitions, A;B are m-rowed f0; 1g matrices, F and G are con gurations, F is a family of con gurations,  is a column, and S is a subset of rows. We only give informal de nitions. For full de nitions, see Section 1.3. Name Notation Informal Defn. Example Simple Con-  guration N/A A con guration with no repeated columns. " 1 1 0 0 1 1 # p " 1 1 1 0 1 0 #  Number of Columns kFk The number of columns of a repre- sentative.           " 1 1 1 0 1 0 #          = 3 Number of 0’s/1’s  0( );  1( )  0(A);  1(A) The number of 0’s/1’s in a column or matrix.  0  " 1 0 0 0 #! = 3  1 0 B @ 2 6 4 1 0 1 3 7 5 1 C A = 2 Column of 0’s/1’s 0m;1m An m-rowed column of 0’s/1’s 03 = 2 6 4 0 0 0 3 7 5 13 = 2 6 4 1 1 1 3 7 5 Complement F c Replace 1’s for 0’s and 0’s for 1’s. " 1 0 1 0 0 1 #c = " 0 1 0 1 1 0 # 25Name Notation Informal Defn. Example Concatenation [AjB] The columns of A to- gether with those of B. "" 1 0 0 1 #      " 0 0 0 1 ## = " 1 0 0 0 0 1 0 1 # Product t  F [ F j F j    j F | {z } t times ] 2  " 1 0 1 1 # = " 1 0 1 0 1 1 1 1 # Product F  G The con guration that consists of all combinations of each column of F on top of each column of G A = " 0 1 1 0 0 1 # ; B = " 1 0 0 1 # =) A  B = 2 6 6 6 6 4 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 0 1 0 0 1 0 1 0 1 3 7 7 7 7 5 Multiplicity  ( ;A) Number of columns of A which are equal to    " 1 0 # ; " 0 1 1 1 1 0 1 0 #! = 2 Di erence A B Columns of A minus columns of B " 1 0 0 1 #  " 1 1 0 1 # = " 0 1 # 26Name Notation Informal Defn. Example Restriction AjS Rows S of A put to- gether A = 2 6 4 0 0 1 1 0 1 0 1 0 0 1 0 3 7 5 =) Ajf1;3g = " 0 0 1 1 0 0 1 0 # Identity Im 0’s everywhere except for diagonal I3 = 2 6 4 1 0 0 0 1 0 0 0 1 3 7 5 Identity Complement Icm 1’s everywhere except for diagonal Ic3 = 2 6 4 0 1 1 1 0 1 1 1 0 3 7 5 Tower Tm 1’s on the upper-right corner and 0’s on the lower left corner. T3 = 2 6 4 0 1 1 1 0 0 1 1 0 0 0 1 3 7 5 Complete Km All possible columns K2 = " 0 1 0 1 0 0 1 1 # Complete of sum k Kkm All columns of column sum k K23 = 2 6 4 1 1 0 1 0 1 0 1 1 3 7 5 27Name Notation Informal Defn. Example Sub- Con guration F  G There is a row and col- umn permutation of F which is a submatrix of G " 1 0 0 1 #  2 6 4 0 0 1 0 1 1 1 1 0 3 7 5 " 1 0 0 1 #  2 6 4 1 1 1 1 0 1 0 0 1 3 7 5 Avoid Avoid(m;F) The set of m-rowed simple matrices which avoid all F 2 F . Avoid  2; " 1 0 #! = (" 0 0 # ; " 1 1 # ; " 0 1 0 1 #) Forb forb(m;F) Maximum number of columns of a matrix in Avoid(m;F). forb  m; " 1 0 #! = 2 Ext ext(m;F) Set of matrices in Avoid(m;F) with exactly forb(m;F) columns ext  2; " 1 0 #! = (" 0 1 0 1 #) Laminar Con-  guration N/A 2 6 4 1 1 1 0 0 1 3 7 5  A 2 6 4 1 1 0 1 0 0 0 0 1 3 7 5 Table 1.5: De nitions 281.6 History 1.6.1 Extremal Combinatorics and Set Theory Extremal Combinatorics deals with problems of  nding the maximum or minimum size of a combinatorial class. As such, the path taken by the study of Extremal Combinatorics is  lled with alleyways, short roads with many stop signs and quite a few landmarks. The reason for the sinuous nature of this  eld is that coming up with extremal combinatorial problems is extremely easy, but coming up with answers to said problems often turns out to be very hard. Very similar problems can have vastly di erent solutions. It is impossible to pinpoint exactly when the  rst time such questions were studied, but some (like [Juk01]) argue it was Euler who  rst studied such problems systematically. The last few decades however have seen an explosion of problems of such ilk, one of the reasons being that the hungry monster of Computer Science has posed and keeps posing many such questions with practical applications. Extremal Combinatorics (at least maximization problems) often involves a race to provide better and better upper bounds (which are generally accomplished by theoretical arguments), and better and better lower bounds (which are generally accomplished by constructions). The  eld of Forbidden Con gurations is no di erent and we often try to \sandwich" forb(m;F) by providing simple matrices with no con guration F 2 F and then using a theoretical argument for the upper bound. Many di erent problems have been studied by a diverse collection of mathematicians. Here we mention a few examples of problems and theorems. Theorem 1.6.1 [Man07] Let G = (V;E) be a graph on n vertices with the property that there is no triangle. Then jEj  n2=4. A generalization came from Tur an. Theorem 1.6.2 (Tur an theorem)[Tur41] Let G = (V;E) be a graph on n vertices such that G does not contain a subgraph isomorphic to Kk. Then, jEj  k  2 k  1  n2 2 =  1 1 k  1   n2 2 : 29And further generalization is due to Erd}os and Stone, asking for the maximum number of edges that avoids a given graph H. This question was answered asymptotically by Erd}os, Stone, and Simonovits. Theorem 1.6.3 [ES46] [ES66] (Erd}os-Stone-Simonovits) Given a number n and a graph H, let forb(n;H) denote the maximum number of edges of a graph in n vertices with no H as a subgraph. Let  (H) denote the chromatic number of H. Then for every  > 0 there exists N such that for all n  N we have  1 1  (H) 1    n2 2  forb(n;H)   1 1  (H) 1 +   n2 2 The related problem of Zarankiewicz asks for the maximum number of edges from a given complete bipartite graph for which a smaller given complete bipartite graph is avoided. A bound was given by K}ov ari, S os and Tur an ([KST54]). A better bound was given by F uredi in [F ur96]. We use this result in our investigations in Chapter 7. There is a rich class of results involving intersecting families (i.e. every two sets in the family intersect). The following are some highlights. Theorem 1.6.4 Let A be an intersecting family on [n]. Then jAj  2n 1. Theorem 1.6.5 [EKR61] (Erd}os, Ko, Rado) Let A be an intersecting k-uniform family on [n]. Then jAj   n 1 k 1  . A generalization of this theorem to t-intersecting families is due to Ahlswede and Kacha- trian ([AK97]) who completely characterized the extremal families. A stability result by Anstee and Keevash ([AK06]) was obtained for certain values of the parameters. This was used to establish the asymptotics of forb(m;F ) for k  2 con gurations F . 30Here is a foundational antichain result. Theorem 1.6.6 [Spe28] (Sperner’s Theorem) Let A be a set system on [n] for which there are no A;B 2 A such that A  B. Then jAj   n dn=2e  : The following problem is one of many in the  eld of Ramsey Theory. Many generalizations of this problem also exist and there are many interesting results. Problem 1.6.7 [Ram30] Given a number k,  nd the maximum number of vertices n for a graph G such that neither G nor Gc contain Kk. By [Ram30], it is known that such number n exists for any k. The full Ramsey Theorem considers colouring all t-sets of [n] using ‘ colours. The Theorem states that for every k there is n such that any such graph on n vertices has a full monochromatic k ‘clique’. We may use this, together with the idea of Section 2.2 to conclude that for m large enough there is a clique (of a  xed wanted size) of rows of the matrix A for which the same possibility of \What Is Missing" occurs. We now give a very small sampling of extremal problems in many other areas of combi- natorics. In graph theory, Problem 1.6.8 Given a graph G, what is the minimum number of colors one needs in order to colour the vertices of G so that two vertices who share an edge do not have the same colour? It is well-known that 4 colours is enough for any planar graph [AHK77]. For general graphs, the number of colours needed is called the chromatic number. To additive combinatorics, 31Problem 1.6.9 [TV06] Given an Abelian group G, what is the maximum size of a sum-free set A  G? (sum-free means there are no a; b; c 2 A with a+ b = c). In computational geometry, Problem 1.6.10 [O’R87] (Art Gallery Problem) Given a polygon P , what is the mini- mum number of lights needed to fully illuminate the interior of the polygon? As the reader might have realized by now, there are probably more problems in extremal combinatorics than there are mathematicians ( nd the minimum number of mathematicians to state all extremal combinatorics problems?) and so we could go on and on. 1.6.2 Forbidden Con gurations The  eld of Forbidden Con gurations began when Vapnik and Chervonenkis, Sauer, Perles and Shelah independently proved Theorem 1.3.29, each for di erent purposes. The theorem was proved  rst by Vapnik and Chervonenkis in 1968 in [VC68]. Sauer and Bollobas attribute the original problem to Erd}os. This theorem and the related concepts of VC-dimension and shattered sets have applications in  elds as diverse as Machine Learning and Pattern Recognition ([BEHW89], [Vap00], [WD81]), Probability ([Ste78]), Combinatorial Geometry ([Ver05], [Mat02]), Extremal Set Theory ([MZ07], [BKS05]), etc. The general question: \For a f0; 1g-matrix F , what is the maximum number of columns a simple f0; 1g-matrix A can have so that A doesn’t contain a row and column permutation of F as a submatrix" was  rst posed by Anstee in [Ans85]. Since then, many people have studied this question. The  rst paper studying this problem in a systematic way is [AGS97]. Both asymptotic and exact bounds were considered for small forbidden con gurations (small in the number of rows). A large number of exact bounds have been proven in [AFS01], [AK07], [ABS11], [AK10]. A particular 4 2 con guration is given in [ABS11] for which  nding an exact bound is highly unlikely. A clearer explanation is in [AK10]. In a private communication, Dukes has obtained tight bounds on the leading coe cient of the associated quadratic bound. 32Conjecture 1.4.1 arose from the various investigations of asymptotic bounds. Product constructions were introduced in [AGS97]. The importance of three building blocks I, Ic and T were also noted in results by Balogh and Bollob as [BB05]. The reader might note that I, Ic and T arise repeatedly in proofs, such as in the results of Section 5.3, and product con- structions arise even in exact bounds, such as in the results of Section 4.1.3. Conjecture 1.4.1 was  rst stated in [AS05], but Anstee already believed that forbidden con guration bounds would be  (mk) for some integer k (that depended on F ) long before that. Since then, much of the progress in asymptotic forbidden con gurations has been guided by this conjecture. 33Chapter 2 Basic Techniques We have a number of powerful techniques that work in many cases and that we will use them repeatedly in this thesis. A careful look at the particular properties of each family of con gurations we wish to study is typically required. We will describe some of these techniques in generality here and we will refer to them when used to solve particular examples. 2.1 Standard Induction Let F be a family of con gurations and let A 2 Avoid(m;F). We wish to do induction on m, but if we delete a row r from A, we might run into trouble: the resulting matrix after the deletion might not be simple. Let Cr = Cr(A) be the matrix that consists of the m  1-rowed columns that have multiplicity 2 in the matrix that results from deleting row r from A. If we permute the rows and columns of A (i.e. we take a representative of the con guration A) so r becomes the  rst row, the columns of A n frow rg can be divided into four blocks: A = r " 0    0 1    1 Br Cr Cr Dr # : (2.1.1) where Br = Br(A) are the columns that appear with a 0 on row r, but don’t appear with a 1, Dr = Dr(A) are the columns that appear with a 1 but not a 0, and Cr are the columns that appear with both. We call this the standard decomposition of A. Note that [BrjCrjDr] is simple and since F  A, we have that F  [BrjCrjDr]. So [BrjCrjDr] 2 Avoid(m 1;F). This means any upper bound on kCrk (as a function of m), 34automatically yields an inductive upper bound on kAk: kAk  forb(m 1;F) + kCrk We now study the structure of Cr. Note that [0 1] Cr is in A. De nition 2.1.1 We say a con guration H is an inductive child of a con guration F if F  " 0 ::: 0 1 ::: 1 H H # : Note that if H is an inductive child of F and F  A, then H  Cr. Therefore if we de ne HF to be the minimal elements of the family of inductive children of F under order  , we know Cr 2 Avoid(m 1;HF). Proposition 2.1.2 Let F be a family of con gurations. Then forb(m;F)  m 1X i=1 forb(i;HF): Proof: Let A 2 ext(m;F). Then we know that for any row r, forb(m;F) = kAk = kCrk+ k[BrjCrjDr]k  kCrk+ forb(m 1;F)  forb(m 1;HF) + forb(m 1;F)  m 1X i=1 forb(i;HF) This concludes the proof.  Since this proposition gives an upper bound on forb(m;F) with respect to forb(m;HF), and HF consists of smaller con gurations (but perhaps more of them), we may repeat this process and  nd a bound instead on forb(m;HHF ), and so on. 35Given F , constructing HF by hand is easy. We have a computer program that does this for us, but it almost never gets used since it’s so easy to do it by hand. The inductive children of a con guration must have eithr the same number of rows, or one less that the original. Then we must look at columns which get \repeated". Perhaps an example would help to clarify this. Example 2.1.3 The inductive children of F = 2 6 4 0 0 1 1 0 0 1 1 1 1 0 1 3 7 5 are " 0 0 1 1 1 1 0 1 # ; " 0 0 1 0 0 1 # ; and 2 6 4 0 1 1 0 1 1 1 0 1 3 7 5 : These are obtained by deleting a row of F and only taking one for each column that appears with both a 0 and a 1 in the deleted row of F , as well as taking the repeated columns of F only once.  Sometimes this bound is not enough because forb(m;HF) is too large. What we actually need in order to proceed as above by induction is that kCr(A)k is small for some choice of r, not necessarily that forb(m;HF) is small. Often it is the case that we must search for the row r with the least amount of repetition, one for which kCr(A)k is as small as possible. If we can prove that for every A 2 Avoid(m;F) there must always be a row r with kCr(A)k being \small enough", we may appeal to induction and proceed as above. There is another method we might use when kCr(A)k is too large. We may delete a limited number of columns (without deleting any row) before proceeding to do induction. For example, if A 2 Avoid(m;F), we might select some columns U from A, so that A = [A0 jU ] which satisfy the property that kUk is small, and Cr(A0) is also small (and therefore we might do induction on A0). In this thesis we also use a new method in order to prove the results of Section 5.3 and Section 7.6 that involves considering only \essential" rows (in some sense). 36De nition 2.1.4 For a matrix A and a row r, let L(r) be a minimal subset of the rows of Cr(A) such that CrjL(r) is a simple con guration. This involves some choice. For each row r we  x this choice. This concept turns out to be quite useful in our investigations, and it arose by considering deleting \non-essential" rows of Cr(A) and looking at the structure of the remaining rows. In the next section we will give an alternate way to study the structure of a matrix A 2 Avoid(m;F). 2.2 What Is Missing? Let F be a ( nite) family of forbidden con gurations and let t be the maximum multiplicity of any column in any F 2 F . In other words, t := maxf ( ; F ) :  is a column, and F 2 Fg: Suppose we have a f0; 1g-matrix A 2 Avoid(m;F). Given s 2 N, consider all s-tuples of rows from A, and for each s-tuple of rows S, consider the matrix AjS formed from rows S of A as in De nition 1.3.10. Without any restriction, AjS could have all 2s possible columns, each with \high" multi- plicity. But with the restriction that F  A for each F 2 F , in particular F  AjS, so some columns have to be missing: we can’t have more than t of all 2s columns, or else we would de nitely have all F 2 F as subcon gurations. That is, some columns have multiplicity 0 and some must have multiplicity less than t. Perhaps an example might clarify this idea. Suppose F consists of a single con guration F := 2 6 4 1 0 0 2  2 6 4 1 1 0 3 7 5 3 7 5 = 2 6 4 1 1 1 0 1 1 0 0 0 3 7 5 ; so in this case, t = 2. Let A = 2 6 6 6 6 4 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 1 3 7 7 7 7 5 : 37For S = f2; 3; 4g we see that AjS = 2 6 4 0 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 1 3 7 5 : In AjS, the columns (1; 0; 0)T and (0; 0; 1)T have multiplicity 0, the columns (0; 0; 0)T , (1; 1; 1)T , (0; 1; 1)T and (1; 1; 0)T have multiplicity 1, and the columns (0; 1; 0)T and (1; 0; 1)T have multiplicity 2. Now imagine what would happen if all columns had multiplicity 2 in AjS instead. Then F would be a subcon guration of AjS, which would be a contradiction. So some columns must have multiplicity less than 2. This motivates the following de nition. De nition 2.2.1 Given a matrix A, a number s 2 N and an s-tuple S of the rows of A, we say an s-rowed column  is absent if  ( ;AjS) = 0. We say it is in short supply if  ( ;AjS) < t. We say  is in long supply if  ( ;AjS)  t. In the example, we can conclude that for any A 2 Avoid(m;F ), in each triple of rows (a; b; c) of A there is an ordering (x; y; z) of (a; b; c), for which the columns marked by \no" must be absent, the columns marked with < 2 must be in short supply and the rest may potentially be in long supply. no no no x y z 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 or < 2 < 2 < 2 x y z 2 6 4 1 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 or no no < 2 < 2 x y z 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 : Of course if there are no columns of sum 1 in AjS (the  rst case), then F  AjS. The same is true if all columns of sum 2 are in short supply (the second case). The third case might be a little harder to see, but if we take a look at the columns potentially in long supply, we see why: absent short supply long supply no no 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 ; < 2 < 2 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 =) l:s: l:s: l:s: l:s: 2 6 4 0 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 1 0 3 7 5 2 6 4 1 1 1 3 7 5 38Clearly F is not a subcon guration: no matter how many times we repeat each column marked in long supply, if we are only allowed to have the columns marked in short supply only once. Notice that in our example, since A does not have F as a subcon guration, in AjS we see that the third case is the one that occurs, with row order (x; y; z) being (2,3,1), so AjS satis es no no < 2 < 2 2 6 4 0 0 1 3 7 5 2 6 4 1 0 0 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 0 3 7 5 : Some e ort is required to determine What Is Missing when we avoid all F 2 F for a given F . In the case above, it is straightforward to check the list of 3 cases is complete. In this particular example s = 3 and the number of rows of F was also 3, but s and the number of rows of F do not need to be equal. When s is less than the number of rows of F , every column is in long supply. We can choose s to be anything, but s being the number of rows of F is often used. Larger s provides more information, but there is a trade o : it’s often harder to analyze such information. 2.3 Implications Sometimes while studying the set of possibilities for What Is Missing from an 3-tuple of rows fx; y; zg, we  nd that one of the possibilities has this restriction: < t < t x y z 2 6 4 0 1 0 3 7 5 2 6 4 0 1 1 3 7 5 Notice that this means in particular that in rows x; y there are at most 2t  2 columns that have a 0 in row x and a 1 in row y: < t < t x y z 2 6 4 0 1 0 3 7 5 2 6 4 0 1 1 3 7 5 =)  2t 2 x y " 0 1 # 39This observation motivates the following de nition. De nition 2.3.1 Given a matrix A and a function f : N N! N, we say for a row x and a row y that we have the implication x! y, if the following is satis ed on the pair of rows x; y:  f(t) x y " 0 1 # (2.3.1) Notice that x! y means that if in some column of A there is a 0 in row x, then there is usually a 0 in row y, except perhaps for f(t) columns. De nition 2.3.2 We say a column violates an implication x ! y if it has a 0 in row x and a 1 in row y. Thus, x ! y means there are at most f(t) columns that violate the implication. We call implications that never get violated pure implications, and implications that get violated at least once impure implications. Proposition 2.3.3 Let A be a f0; 1g-matrix and consider the directed graph G, where the vertices are the rows, and the arrows are the implications. Suppose we had an implication x! y and we also had a path of implications x = x0 ! x1 ! x2 ! :::! xn = y: Then we may conclude that if a column of A violates x ! y, it must also violate an implication of the form xi 1 ! xi for some i with 1  i  n. Proof: Indeed, if a column  has a 0 in row x and a 1 in row y, consider the smallest index i for which  has a 1 in row i. Then the column must have a 0 in row i  1 and therefore violates xi 1 ! xi.  Given A, using this technique we might be able in some cases to select just some small set I of implications, so that if a column of A violates any implication, then it must necessarily violate an implication in our set I. The number of columns violating any implication will 40then be at most f(t)  jIj. The power of implications comes from the fact that if jIj and f(t) are small enough, even if the total number of implications is large, we can delete every column that violates any implication, thus making the implication pure. The result is that columns previously marked as being \in short supply" are now completely absent. This is a powerful technique, because many times we are able to  nd an upper bound on forb(m; [GjH]) with G;H simple with no columns in common, and we wish to  nd an upper bound on forb(m; [Gjt  H]). Considering a matrix A that has no [Gjt  H], we are able to delete at most f(t)  jIj columns from A, and then this new matrix won’t have [GjH]. So under these assumptions, we are able to conclude that kAk  f(t)  jIj+ forb(m; [GjH]): 41Chapter 3 Computer Program Developed In this chapter we will describe a C++ program we used extensively in the results that follow. We give a description of the algorithms and data structures. Our code computes the following tasks: Sub-Con guration: Answers the question of whether or not a con guration F is a sub- con guration of a con guration A. Conjecture: Determines the asymptotic bound predicted by Conjecture 1.4.1 for a given con guration F . In other words, it  nds X(F ). Classi cation: Once we know how to  nd X(F ), given a certain number of rows, this program  nds all boundary cases. What Is Missing? Given s and a family of con gurations F , this program  nds the list of What Is Missing in each s-tuple of rows if we forbid F . Finding forb: Given F and a small value of m (speci cally, m  5), this program computes forb(m;F). Guessing forb: For m = 6; 7; 8; 9, this program uses local search (in particular, a Genetic Algorithm among others) to guess the extremal matrices ext(m;F). In the subsequent sections we’ll deal with all these problems and describe our own approach to these tasks. All these have been implemented in C++ (and some of them in sage as well) and are available for download from http://www.math.ubc.ca/ anstee/ For some of these tasks, complexity is unfortunately doubly exponential in the number of rows (for What Is Missing, Finding forb, Classi cation), while for others (Sub-Con guration, 42Conjecture) it is merely exponential in the number of rows. In any case, we only deal with these problems for small con gurations, and even for such cases it is often the case we aren’t able to use the information given by the computer to prove results, so better algorithms would not necessarily result in new results in Forbidden Con gurations. 3.1 Representation of a Con guration We are interested in an e cient representation for con gurations, in order to perform the tasks described above. In the progress of our investigations, we have had various versions of the program. We will record here how the program works based on the posted version, and recorded for posterity in the above link. Various things were done in a di erent manner, but we have mainly improved usability and speed with each subsequent version. Each version of the program was tested against many di erent con gurations for correctness, and some tasks were implemented in sage (http://www.sagemath.org/) as well, in order to check the answers when they were too large to check by hand. For the \What Is Missing" calculation it is easy to check that each possibility, if satis ed, indeed implies that the given con gurations are avoided, but there is no easy way to check that the list of possibilities is complete. First of all, we explain how the program stores con gurations. Most of what we want the program to do involves performing a huge number of con guration comparison operations, which is testing whether or not a con guration F is a subcon guration of a con guration A. As a  rst approach it would seem as if, for this task, we would be required to test each row and column permutation of F against each submatrix of A. This is of course a very slow way to do this. A simple trick to speed up the computations is to keep the columns of a con guration always in some canonical order. Then, to test whether or not a con guration F is contained in another A, we just need to permute rows of F and take subsets S of rows of A and place the columns of AjS in canonical order. Most of the tasks we described above involve checking whether or not a given ( xed) con guration F is a subcon guration of a vast number of con gurations A. In particular, any pre-processing we do on F can be considered as almost free. For example,  nding all row permutations of F and storing them would need to be done once for each con guration F , and not at all for con gurations A. After many attempts, it seems that the best (fastest) way to store a con guration that would make many of the other tasks reasonably fast is this: Maintain an array of integers 43where the indices of the array, written in binary, are the columns of the con guration, and the actual numbers of the array represent the number of times a column appears. That is to say, a con guration F in m rows is represented by an array (C++ vector) F of size 2m. For a number  , consider the binary representation of  and consider it as a column with m rows. If necessary, put enough 0’s at the beginning of the binary representation in order to have the required m bits. The number F[ ] (the  -th number of the array) represents the number of times that column  appears in con guration F . In the implementation, we use an array of unsigned characters instead of integers, since we never need a con guration with the same column repeated more than 255 times. An unsigned character consists of 1 byte (8 bits). For example, the array F = [1; 0; 0; 2; 0; 1; 0; 1] represents the following con guration (notice it has 3 rows, since the array has size 8 = 23): F = 2 6 4 0 0 0 1 1 0 1 1 0 1 0 1 1 1 1 3 7 5 : To see this, remember we start from 0. There is a one in position 0 = 000b, meaning the colum (0; 0; 0)T gets repeated one time. A two in position 3 = 011b, a one in position 5 = 101b and a one in position 7 = 111b. The columns of this matrix are the representations of these numbers in binary form. An observant reader might complain that this has the disadvantage that it requires storing 2m bytes, and if F doesn’t have many columns, most of those will be 0’s. But it’s a minor disadvantage, because even at 10 rows we would only need 1024 bytes, and we usually have con gurations for which the number of rows is 5 or less (32 bytes). Perhaps this would become more of an issue with con gurations with a high number of rows, but for those con gurations, most of our tasks would require too much time to be of any practical use. We came to this representation after an implementation which represented columns as an array of bits (C++ bitset) and storing them into an ordered tree-like structure (C++ STL multiset). This might be a more natural implementation, but pro ling the code made clear that the program was spending most of its time counting how many columns of a certain type appeared in a con guration, and was also spending a considerable amount of time navigating the tree. Explicitly storing the number of times each column appears, and making that number instantly accessible by storing it in an array (for random access) gives a very noticeable speedup and allows us to consider larger problems. By representing columns 44as numbers, we can do a lot of preprocessing and compute large tables in which we have almost instant access time. For example, consider the following problem, which has to be done many times for our tasks: Given a column  and a subset S (represented by an integer as written in binary), what column is  S? This is relatively slow to compute, but we can  ll out a table by preprocessing to speed up any further access to it. Since we do this a few million times, the investment is sound. The other advantage is that it becomes immediately clear how to compare two con gu- rations with the same number of rows to see if one is a column-permutation submatrix of the other; check if for any column (index) the integer at position c of the  rst array is bigger than that of the second array. To check if F is a subcon guration of A, we would need to  nd all permutations of the rows of F (which we need to do just once per con guration). 3.2 Subcon gurations Suppose F and A are con gurations and we want to decide if F  A. Then for every s-tuple of rows S (where s is the number of rows of F ), we can extract from A the con guration AjS easily with our pre-stored table of columns and subsets. Once we’ve done this for each column of A and found AjS, then for each permutation of rows of F , we check if every column  in the array corresponding to con guration F appears less than or equal to the corresponding number for column  in the array of AjS. We can check every subset S like this. If at any point this is so, we can return true. There are a few speedups. Sometimes it’s immediately obvious a con guration can’t be contained in another. For example, if there are more 1’s in F than in A, or if F has more columns or rows than A, then F  A. 3.3 Determining X(F) Given a con guration F , we wish to  nd X(F ). In other words, we wish to  nd the conjec- tured asymptotic bound for forb(m;F ). We may make a simpli cation using Lemma 1.4.2 and assume the multiplicity of any column of F is at most 2. First, suppose we wanted to test whether or not there exists r such that con guration F 45is contained in the product Pr(a; b; c) = Ir  ::: Ir| {z } a times  Icr  ::: I c r| {z } b times  Tr  ::: Tr| {z } c times : Building this object with r = R as calculated in Lemma 1.4.2 would be prohibitively slow. Instead, we build a set X of subcon gurations from PR(a; b; c) such that if F  PR(a; b; c), then F  X for some X 2 X . Notice that if F  PR(a; b; c), then the rows of F get \partitioned" into a+ b+ c parts (a part can be empty), where each part belongs to a factor of the product PR(a; b; c). Because of Lemma 1.4.2, we can assume each column appears at most twice. Given s 2 N, consider the following matrices: AI(s) = h 0s 0s Is i ; AIc(s) = h 1s 1s Ics i ; AT (s) = h Ts Ts i : We see that an s-rowed con guration F with each column repeated at most twice is contained in Im for some large m, if and only if F  AI(s). We can then consider all partitions of rows of F and see if each part is contained in the corresponding AI , AIc or AT . For example, to test whether F = 2 6 6 6 6 6 6 4 1 0 1 1 0 1 0 1 1 1 0 0 0 0 1 1 1 0 0 1 3 7 7 7 7 7 7 5 is contained in I  T  T , we would partition the rows of F in three parts. In this case, F has  ve rows, so consider, for example, the following partition of 5: (2; 2; 1), for example. Consider the following representatives: AI(2) = " 0 0 1 0 0 0 0 1 # AT (2) = " 0 0 1 1 1 1 0 0 0 0 1 1 # AT (1) = h 0 0 1 1 i ; and build a 5-rowed matrix A := AI(2) AT (2) AT (1). If F  A, then F  I  T  T . If we do this for every possible partition of 5, we get the desired result. To  nd X(F ) we can build a tree of possibilities of products and return the largest for which F not a subcon guration of a product of this form, observing that if F is a 46subcon guration of a product Pr(a; b; c), then it will be a subcon guration in any product Pr(a0; b0; c0) with a0  a, b0  b and c0  c. 3.4 Boundary Cases to Classify Con gurations We wish to  nd all boundary cases (De nition 1.4.3) for a given number of rows s and a number k using the computer. We use the program described in the previous section to  nd X(F ) for many di erent F . The method we use is very straightforward: start adding columns, one by one. Build a tree of con gurations, where the children of a con guration F are the ones that consist of F plus a column. Find X(F ) for each con guration in the tree. Store all those con gurations F for which X(F ) = k, and then  nd only the maximal and minimal con gurations in the ordering  . If at some point we add a column and the bound jumps to k + 1 or higher, discard and go to the next con guration. Columns will be repeated at most twice because of Lemma 1.4.2. Unfortunately this method is very slow because the same con guration is searched mul- tiple times, since each con guration may have many representatives. To get rid of repetition we might check for equivalence of con gurations against everything we have stored so far. But checking if two con gurations are equivalent is usually slow, so doing it every time is also very slow. What seems to work best is to do check for equivalence, but only up to a point. For example, we can consider all pairs of columns, test those and only take one representative of each equivalence class. Then from each pair, start building the tree as described above. After that, it will be relatively unlikely that two con gurations we search are equivalent, so the amount of repetition will be relatively low. Of course, much repetition will still occur, but much less than in the original tree. Notice that it isn’t as critical that classifying con gurations be a fast calculation. We need to do it once for every s and k, but no more. Once we know the maximal and minimal quadratics for  ve rows, we never need to calculate them again. Other calculations, such as  nding X(F ) or What Is Missing, have to be performed for each con guration (or family) we wish to study, so decreasing the running time is more of a priority in those cases since faster programs allow us to study bigger con gurations. 473.5 Finding What Is Missing and Forbidden We describe the part of the program whose input is a family of con gurations F and a number s, and its output is the list of possibilities for columns absent or in short supply. This computation has been the most useful for the thesis. To  nd a list of possibilities of What Is Missing in s rows after forbidding F 2 F , we present the problem in a complementary way and instead  nd what can be present. To  nd this we may start with St = t  Ks (where t is the maximum multiplicity of a column in any con guration of F) and remove columns from it doing breadth  rst search, to  nd all the maximal con gurations A for which F  A. Then the f0; 1g complements of such A’s are the list of What Is Missing. This is very slow, since the search space could have 22 ts con gurations. But we can speed up the calculations. For example, by carefully choosing which columns we start our search with (i.e. removing unnecessary columns from St). The main speedup for this program comes from choosing a di erent starting point by considering each column sum separately in the following manner. For a sequence (t0; t1; :::; ts) 2 Ns, consider S := [t0  K 0 s j t1  K 1 s j ::: j ts  K s s ]: Instead of making ti = t, we can make ti the biggest number a column of column sum i appears in any con guration F . Of course, since con gurations F 2 F might have less than s rows, each column might count for many di erent column sums, in all the ways it could be  lled. In other words, if F has k rows, we consider the s-rowed con guration F  Ts k and then count how many times each column appears. Then for each column sum take the maximum, and then take the maximum over all con gurations F 2 F . Perhaps an example would be useful. Suppose s = 4 and F consist of the following two con gurations: F1 = 2 6 4 0 0 1 1 0 1 0 1 1 3 7 5 and F2 = 2 6 6 6 6 4 1 1 0 1 1 1 0 0 0 0 1 1 0 0 1 0 3 7 7 7 7 5 We wish to  nd a starting point S for the search. We are searching then for t0; t1; t2; t3 and t4. As instructed above, since F1 only has 3 rows, it acts as F1  T1 = F1  [0 1] for the purpose of considering column sums. We can deduce that t0 = 0, since there are no columns of column sum 0. Since F1  T1 48with a 0 on the bottom contains columns of column-sum 1, but no repeated columns of column sum 1, we can set t1 = 1. F2 has a column of sum two with multiplicity two, so t2 = 2. And because of F1, we have both t3 = 1 and t4 = 1. Then instead of starting with 2  K4, which has 32 columns, we may start with the following matrix: S = [K14 j 2  K 2 4 j K 3 4 j K 4 4 ] = 2 6 6 6 6 4 100011110011000011101 010011001100110011011 001000111100001110111 000100000011111101111 3 7 7 7 7 5 ; which has only 22 columns. The search space has size 222; 1024 times smaller than 232. This alone provides a massive reduction for most families F . Using Lemma 1.4.2, if the multiplicity of a column is 3 or more, we can usually assume it is 2 for our purposes. We only distinguish between columns absent, those in short supply and those in long supply. Once we have the list of subcon gurations A of S such that F  A, we take only the maximals with respect to order  . What Is Missing can be obtained by considering S  A. For the example above, the computer gives us the list of What Is Missing in about one second. P0 = no no no no no no no no 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P1 = no no < 2 no < 2 no no no no 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P2 = no no no no no no no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 49P3 = no < 2 < 2 no < 2 < 2 no no no no 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P4 = no no no no no no no no no 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P5 = no no no no no no no no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P6 = no no no no no no no no no 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P7 = < 2 < 2 < 2 no < 2 < 2 no < 2 no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 50P8 = no no < 2 no no no no < 2 no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P9 = no no no no no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P10 = no no no no no no no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P11 = no no no no no no no no no no 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P12 = no no no no no no no no 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 51P13 = no no no no no no < 2 no < 2 no no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 P14 = no no no no no no no no 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 In practice checking con gurations with s  4 is almost instantaneous, s = 5 takes, depending on the con guration, anywhere from a few minutes to a couple of hours, and with s = 6 it’s typically hopeless, although it can be done if for example the con guration only has columns of column sum 3 and is simple, since 2( 6 3) is still a reasonable number. Finally, note that  nding forb(m;F) is a subproblem of this, when s = m. Start with the complete matrix Km and remove columns one by one until we stop having F 2 F . Perform breadth- rst on the tree of column deletions in order to search for the con guration with the most number of columns for which no F 2 F is a subcon guration. Algorithm 3.5.1 Here is some pseudo-code to  nd What Is Missing (WIM). Input: A family of con gurations F , a number s Output: A list of possibilities for What Is Missing for i = 0 to s do ti := maxf ( ; F  Ts k) :  column of column sum i and F 2 F with k rowsg end for S := [t0  K0s j t1  K 1 s j ::: j ts  K s s ] List := ; for all s-rowed K  S do if F  K then Add K to List end if 52end for List = maximals(List) ListWIM := fS  K : K 2 Listg return ListWIM 3.6 Guessing Forb Consider F a family of (small) con gurations and m a (small)  xed integer. Suppose we wish to to  nd (or rather, guess) forb(m;F) using the computer. This approach has helped considerably with the proofs of 3 exact bounds (described in Section 4.1 and Section 4.3). For these results, once we had an extremal matrix that avoids F it was relatively straightforward to construct a proof of this fact. We defer the details of the proofs to the subsequent sections. For now, we will describe the methods used to provide us with what seems to be a good and dependable guess of forb(m;F). The idea is to consider all the 2m columns in some order and add them one by one into a matrix A, making sure at each step we don’t create any F 2 F as a con guration. The order in which to add columns is what will determine the size of kAk at the end of the process. To do this, enumerate all the columns and for each permutation of [2m]  nd the number of columns that would be added while avoiding all F 2 F if we were to add them in that order. We call the columns that would get added in this procedure good columns and the columns that get discarded bad columns, with respect to the given permutation. We wish then to  nd a permutation that maximizes the number of good columns. So the search space has size (2m)!, which is too big to search exhaustively, but our experiments suggest local search can often  nd the correct answer when m isn’t too large. We’ll describe three methods. Perhaps this explanation could bene t by an example. Suppose we wanted to forbid F = " 1 0 0 1 # 53and suppose m = 3. Consider all eight 3-rowed columns in the following order: 1 2 3 4 5 6 7 8 2 6 4 0 0 0 3 7 5 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 Then consider the permutation (4; 1; 3; 6; 5; 8; 2; 7). We start adding columns in the order given by this permutation. If adding a certain column would rise to a copy of F , then don’t add it and continue to the next one. This is how it would work for this permutation: 1. Add column 4. A = 2 6 4 0 0 1 3 7 5 2. Add column 1: A = 2 6 4 0 0 0 0 1 0 3 7 5 3. Can’t add column 3. 4. Add column 6: A = 2 6 4 0 0 1 0 0 0 1 0 1 3 7 5 5. Can’t add column 5. 6. Add column 8: A = 2 6 4 0 0 1 1 0 0 0 1 1 0 1 1 3 7 5 7. Can’t add column 2. 8. Can’t add column 7. We end up with 4 columns, with the good columns being f4; 1; 6; 8g and the bad columns being f3; 5; 2; 7g: 543.6.1 Brute Force Greedy Search The brute force method does the following:  Choose a permutation at random.  Separate into good and bad columns.  Count the number of good columns.  Repeat, while keeping track of the \best" so far (i.e. the one with the most number of good columns) The strength of this method is that we may do this thousands of times in a relatively short time. This method is good for simple con gurations in which there are many di erent ways to achieve the bound, but in our experience, it fails to  nd the optimal bound in many cases. An easy bound on the probability p of  nding the best con guration in a single try is the following. Let f = forb(m;F). Then p  (2m  f + 1)  2m f  : Indeed, if A 2 ext(m;F) and gi (for i 2 f1; 2; :::; fg) is any permutation of the columns of A and bi (for i 2 f1; 2; :::; (2m  f)g) is a permutation of the columns that are not in A, we see that for any j 2 f0; 1; :::; (2m  f)g any permutation of the form: (g1g2:::gf 1b1; b2; :::; bj; gf ; bj+1; :::; b2m f ) gives rise to a matrix in ext(m;F), since by separating good columns from bad columns, as described in the previous section, we see that at least we are allowed to add g1; g2; :::; gf 1 to the good columns without interruption, and we must be allowed to add to the good columns at least one of b1; b2; :::; bi or gf . The above bound might not be very accurate, as many other permutations might give rise to extremal matrices, but at least it grounds the problem. 3.6.2 Hill Climbing We describe a method that improves over brute force. Start with some permutation and separate the 2m columns into good and bad columns. Then for each column b in bad, form a 55new permutation by putting b at the beginning, but leaving all others in place. This ensures that the chosen column b is selected. From all the possibilities for choice of b, choose the best one, that is, the one that gives the most columns in good. If there are ties, pick one at random. Repeat this process until there is no improvement. In our example, we would have to consider all the following permutations: 1. (3; 4; 1; 6; 5; 8; 2; 7) 2. (5; 4; 1; 3; 6; 8; 2; 7) 3. (2; 4; 1; 3; 6; 5; 8; 7) 4. (7; 4; 1; 3; 6; 5; 8; 2) In this case all of them give size 4, but in general some might be better than others. Local search as described above gets stuck very quickly, but it does perform better than brute force. Algorithm 3.6.1 Here is some pseudo-code to perform hill climbing. This code gives up if it gets stuck. It’s easy to modify it to to be able to \walk" while being stuck, and if there is no improvement after, say, 20 iterations, stop. Input: A family of con gurations F , a number m, a permutation of 2m,  . Output: A con gurations in Avoid(m;F) that is thought to be extremal. stuck := false Find good( ) and bad( ) while not stuck do for all b 2 bad( ) do  b := (b;  n b) Separate  b into good( b) and bad( b). end for  := argmaxfjgood( b)j : b 2 bad( )g if jgood( )j > jgood( )j then    else stuck = true end if 56end while return  3.6.3 Genetic Algorithm An even better way to search the space is using a Genetic Algorithm. The idea is to mimic evolution. We maintain a population of ‘individuals’ and assign ‘ tness’ to them in some way (in our case, the number of ‘good’ columns). Then pick two at random, but giving higher probability to those with higher  tness. Call them father and mother. Then combine them in some way to produce o spring, hopefully with better results than either father or mother. Do this for many generations. There are many ways to combine permutations of course. The one we found that works well is as follows: 1. For both father and mother, separate the columns into good and bad. 2. Take a random number of the good part of mother, and consider all columns from the start up to the chosen random number. 3. Permute the entries of father so that the numbers in the chosen part of mother are in the same order as those in mother. This is better shown with an example. Suppose we had a father and a mother like this: father = (2; 3; 6; 1 j 7; 4; 8; 5) and mother = (5; 3; 8; 1; 4 j 7; 2; 6) with the numbers shown before the vertical line being the good part and the numbers shown after the vertical line being the bad part. Pick a random entry in the good part of mother. For example, pick 1 and look at the numbers that appear (in mother) to the left of the picked number. In this case, (5; 3; 8; 1). In father, select these numbers. father = (2;3; 6;1j7; 4;8;5) Make child by shu ing the selected entries in father so that they match the order of mother. child = (2;5; 6;3; 7; 4;8;1): 57And  nally, separate the columns of child into good and bad. This approach has given very good results in our experience. Algorithm 3.6.2 Here is some pseudo-code to perform genetic search as described in this section. After some empirical experimentation, we  nd that using a population size of 40 and 200 generations are good numbers. Input: A family of con gurations F , a number m Output: A list of con gurations in Avoid(m;F) that are thought to be extremal. De ne Population to be a list 40 random permutations of 2m. for i = 0 to 200 do for all  2 Population do Separate  into good( ) and bad( ) columns. end for De ne NewGeneration :=10 best from Population. while jNewGenerationj < 40 do Pick father and mother from Population fPick random permutations  according to how big good( ) isg Mix father and mother to produce child Add child to NewGeneration end while Population NewGeneration end for 58Chapter 4 Exact Bounds 4.1 Two 3x4 Exact Bounds A list of 3 4 con gurations for which the exact bounds are not known are listed in [AK10]. We solve two of them, and give conjectured bounds for the rest in Section 8.1.3. 4.1.1 Introduction We study the following con gurations V;W (where forb(m;V ) and forb(m;W ) weren’t pre- viously known), and compare to the common submatrix X (where forb(m;X) was found in [ABS11]). X = 2 6 4 1 1 1 1 0 0 3 7 5 ; V = 2 6 4 1 1 0 0 1 1 0 0 0 0 1 1 3 7 5 ; W = 2 6 4 1 1 1 1 1 1 0 0 0 0 1 1 3 7 5 : We used our Genetic Algorithm to seek extremal matrices A 2 ext(m;V ) and A 2 ext(m;W ) for small m. From these examples we guess the structure of extremal matrices in general and then we are subsequently able to prove these matrices are indeed extremal. Guessing such structures would have been challenging without the help of the Genetic Al- gorithm. Theorem 4.1.1 [ABS11] Let m  3. Then forb(m;X) =  m 2  +  m 1  +  m 0  + 1. 59Note that X  V and X  W . We initially thought (before using the Genetic Algorithm) that forb(m;V ) = forb(m;W ) = forb(m;X). Related results are in [ABS11], [AK06]. The structure of the apparently extremal matrices generated by the Genetic Algorithm provided the strategy to tackle the bounds for forb(m;V ) and forb(m;W ). Interestingly, the Genetic Algorithm was used again in each example in the inductive proof to guess forb(m;HV ) and forb(m;HW ). 4.1.2 The Bound for W Recall that W = 2 6 4 1 1 1 1 1 1 0 0 0 0 1 1 3 7 5 : We can use the computer to compute forb(m;W ) for m  5 as described in Section 3.5. We then use the Genetic Algorithm to compute guesses for forb(m;W ) for m = 6; 7; 8 and also guess the structure of the extremal matrices. We then proceed to prove these guesses. Theorem 4.1.2 Let m  2. Then forb(m;W ) =  m 2  +  m 1  +  m 0  +m 2. In order to prove this we proceed using the Standard Induction of Section 2.1: Let A 2 Avoid(m;W ). If we could  nd some row r for which the number of repeated columns kCr(A)k  m+ 1; then we would be done: kAk  m+ 1 +  m 1 2  +  m 1 1  +  m 1 0  + (m 1) 2 =  m 2  +  m 1  +  m 0  +m 2: So we may assume that kCr(A)k  m+ 2 for every r. The minimal inductive children of W are: G1 = 2 6 4 1 1 1 0 0 1 3 7 5 ; G2 = " 1 1 0 0 0 0 1 1 # ; G3 = " 1 1 1 1 1 1 0 0 # : 60Given that Cr(A) 2 Avoid(m  1; fG1; G2; G3g), under these assumptions the following Lemma establishes that kCr(A)k = (m 1) + 3 = m+ 2. Lemma 4.1.3 Let m  4. Then forb(m; fG1; G2; G3g) = m+ 3. Proof: For the lower bound forb(m; fG1; G2; G3g)  m + 3 an example (which was again found using our Local Search strategies) su ces. Consider the matrix A = [0m j K1m j 1m j  ]; where  is any other column. Clearly A 2 Avoid(m; fG1; G2; G3g) and kAk = m + 3. To prove the upper bound, we proceed by induction on m. Let A 2 Avoid(m; fG1; G2; G3g). Then, if we forbid fG1; G2; G3g, below are the 16 possible cases of What Is Missing on each quadruple of rows, found using the program described in Section 3.5. The following list is complete because the search is exhaustive. no no no no no no no no no  1  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no  1  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no  1  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 61no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 62no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 no no no no no no no no no no no no  1  1 > 1 > 1 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no no  1  1 > 1 > 1 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 63no no no no no no no no no no no no  1  1 > 1 > 1 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 We check this list to see that there are at most seven columns present on four rows (at least nine of sixteen are absent) and so forb(4; fG1; G2; G3g) = 7. Now consider m  5. It’s easy to check by considering every quadruple of rows that there is a row and a column we can delete from A and keep the remaining (m 1)-rowed matrix A0 simple. Then by induction, kA0k  (m  1) + 3 = m + 2 and then kAk  kA0k + 1  m + 3. To  nd such a row and column, look at the columns marked  1 and > 1, and see that there is a row we can delete such that the only repeat (if there is one) has one of the columns marked  1. We used the help of the free software sage (http://www.sagemath.org/), but it could also be done by hand. For example, for the  rst possibility,  1  1  1  1  1 > 1 > 1 i j k ‘ 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 ; deleting the row of A corresponding to row ‘ and the column of A corresponding to the  fth column above (0; 0; 0; 1)T (if it exists), keeps the remaining matrix simple. In all cases above one can delete the  nal row (together with at most one column in short supply). Thus kCk  m+ 3.  We need a more detailed lemma about an m (m+3) matrix in Avoid(m; fG1; G2; G3g), one that was predicted using our Genetic Algorithm. Lemma 4.1.4 Let A 2 Avoid(m; fG1; G2; G3g) with m  3 and kAk = m + 3. Then K1m  A. Moreover, the remaining three columns are 0m and two additional columns  ;  with  <  (meaning that on each row for which  has a 1,  also has a 1). Proof: We proceed by induction on m. We checked all cases with m = 3; 4 using a computer. 64Assume A 2 Avoid(m; fG1; G2; G3g) with kAk = m + 3 and m  5. From our proof of Lemma 4.1.3, there is a row and a column from A we can delete to obtain a (m 1) (m+2) simple matrix A0. We may assume K1m 1  A 0 by induction. Assume we deleted the last row from A to obtain A0 and that the deleted column was the last column of A. If we restrict ourselves to the  rst m 1 rows, the deleted column has to be a repeat of one of the columns of A0, else we would have an (m  1)  (m + 3) simple matrix contradicting Lemma 4.1.3. At this point the proof is  nished save for a case analysis on each of the possible columns to repeat. Aside from the column of zeros, there are two columns  ;  which aren’t in K1m 1. Given  <  , we call  the small column and  the big column. We consider 3 di erent types of rows:  Row type 1: Both  ;  has 0 in the row.  Row type 2: Column  has a 1 and  has a 0 in the row.  Row type 3: Both  ;  has 1 in the row. There may not be any rows of type 1, but there has to be at least one row of type 2 (in order to di erentiate between  and  ) and at least two rows of type 3 (in order to di erentiate between  and a column of column-sum 1. Consider the generic rows below. We’ve included the appropriate parts of the copy of K1m 1 and the column of 0’s. The entries marked c1; c2; : : : ; c8; r1; :::; r4 are the entries of the deleted row and column.   0 1 0 0 0 0 0 r1 type 1 0 0 1 0 0 0 1 r2 type 2 0 0 0 1 0 1 1 r3 type 3 0 0 0 0 1 1 1 r4 type 3 c1 c2 c3 c4 c5 c6 c7 c8 : Of course there might be many rows of each of the types 1,2,3, but there is no loss of generality if we focus on these rows. There have to be at least two rows of type 3 so it is possible to have two rows which correspond to the entries r3; r4. We have to be careful because row r1 might not exist. There are some cases for which column gets repeated. In each case we attempt to  nd either G1, G2 or G3. The fact that A has no G1 means A is laminar, as in De nition 1.3.27. 65The following case analysis is probably much easier as an exercise for the reader than it is to either write or read about. Case 1: 0m 1 is the repeated column. Then we have 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 0 c2 c3 c4 c5 c6 c7 1 : So either K1m  A or some of c2; c3; c4; c5 are 1. If c2 = 1 then c6 = 0 and c7 = 0 in order to have a laminar matrix. But then we have G2 in the last and next-to-last rows. So we may assume c2 = 0. If c3 = 1 then c5 = 0, c6 = 0 and c7 = 1 by the laminar property and we have then 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 0 0 1 c4 0 0 1 1 : But then we have G2 in the last two rows. We may assume then that c3 = 0. If c4 = 1 then both c6 and c7 have to be 1, and so we get 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 0 0 0 1 c5 1 1 1 ; which has G3 in the last two rows. This completes Case 1. Case 2: The repeated column has column sum 1. Then there are three sub-cases, de- pending on the position of the 1 in the new column. Let r be the row on which, other than the last row, the new column has a 1. 66Subcase 2a: r is of type 1. We have 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 c1 0 c3 c4 c5 c6 c7 1 ; which contains G2 in the  rst two rows. Subcase 2b: r is of type 2. We have 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 c1 c2 0 c4 c5 c6 c7 1 ; which contains G2 in the second and third rows. Subcase 2c: r is of type 3. We have 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 0 c1 0 c3 c4 c5 c6 c7 1 ; which contains G3 in the third and fourth row. Case 3: The repeated column is the small column  . Then we have this: 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 1 c1 c2 c3 c4 c5 0 c7 1 : So c7 has to be 1 in order to have a laminar matrix. If either c4 or c5 were 0, then we 67would have G3 in the last row together with one of the next-to-last rows. So both have to be 1. But this contradicts the fact that we have a laminar matrix. Case 4: The repeated column is the big column  : Then we have this: 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 1 c1 c2 c3 c4 c5 c6 0 1 : This yields G3 in the second and third rows. This completes all cases. We’ve concluded that K1m  A. We deduce that, apart from the columns of K 1 m and 0m, A has two other columns  and  . Since G2  A we deduce that either  <  or  <  .  Proof of Theorem 4.1.2: We use induction on m. The result is true for m = 2; 3 so we may assume m  4. Let A 2 Avoid(m;W ). Apply the decomposition (2.1.1). If kCr(A)k  m + 1, then we can apply the Standard Induction Section 2.1 to establish the bound for kAk. So assume kCr(A)k  m+2 for all choices of r. By Lemma 4.1.3, we deduce that kCr(A)k = m + 2 for every row r, and by Lemma 4.1.4, we have that K1m 1  Cr(A) also for every row r. Thus K2m  A, since all columns of column sum 1 in Cr(A) appear with a 1 in row r (and this happens for every row r). We also have [K1m j K 0 m]  A. Now in every triple of rows of K2m we have the matrix G1 once in every ordering of the triple. Given that W = 2  G1, the columns of A of column sum at least 3 have no con guration G1. We appeal to Lemma 4.1.5 below to deduce that the number of columns in A of column sum at least 3 is at most m 2. Then kAk   m 2  +  m 1  +  m 0  +m 2; which yields the desired bound.  A quick counting argument reveals the following general result about laminar families which we use in the proof of Theorem 4.1.2. 68Lemma 4.1.5 Let m  3 and let Z be a laminar family of subsets of [m] = f1; 2; :::;mg with the property that for all Z 2 Z we have jZj  3. Then jZj  m 2 and furthermore, this bound is tight (i.e. there exists a family Z for which jZj = m  2). Thus, if A 2 Avoid(m;G1) satis es that all column sums are at least 3, then kAk  m 2. Proof: Let f(x) denote the size of the biggest laminar family of [x] with no sets of size less than or equal to 2. Recall that a laminar family is equivalent to a con guration which avoids G1. Assume A 2 Avoid(m;G1) with the property that all column sums are at least 3. We wish to show that f(m) = m 2. The family f[3]; [4]; : : : ; [m]g has size m  2, which proves f(m)  m  2. We wish to prove f(m)  m 2. Let Z be such that jZj = f(m). We may assume [m] 2 Z: if [m] =2 Z, observe that Z [ f[m]g is also a laminar family of bigger size. Suppose then that the next biggest set Z in Z has size k. We partition [m] into two disjoint sets: Z and [m]nZ. Every set Y 2 Z satis es either Y  Z or Y  [m]nZ or Y = [m]. This gives the recurrence f(m)  1 + f(k) + f(m k): If k 6= m  1, then by induction f(k) = k  2 and f(m  k) = m  k  2, and so f(m)  1+k 2+m k 2 = m 3, a contradiction. When k = m 1 we have f(m)  m 2. Moreover if f(m) = m 2 we observe that Z is \equivalent" to f[3]; [4]; : : : ; [m]g.  4.1.3 The Bound for V Recall that V = 2 6 4 1 1 0 0 1 1 0 0 0 0 1 1 3 7 5 : Using the computer, we can prove by exhaustively looking at all the possibilities that forb(3; V ) = 8, forb(4; V ) = 13, and forb(5; V ) = 18. Using the Genetic Algorithm of Section 3.6, we obtained large matrices with no subcon guration V which suggested to us that forb(6; V ) = 25, forb(7; V ) = 32, forb(8; V ) = 40. For m  6, this suggests forb(m;V ) =  m 2  +  m 1  +  m 0  +3, two more than our  rst guess that forb(m;V ) = forb(m;X). 69Theorem 4.1.6 Let m  6. Then forb(m;V ) =  m 2  +  m 1  +  m 0  + 3. The Genetic Algorithm also predicted the following structure of matrices in ext(m;V ). Recall the product  de ned in De nition 1.3.7. Consider a partition of the m rows into two disjoint sets, say U = f1; 2; : : : ; ug and L = fu+ 1; u+ 2; : : : ;mg. Suppose jU j = u and jLj = ‘ with u+ ‘ = m. Let A have the following structure: A = U L 2 6 4 0u  0‘ 0u  1‘ 0u  K‘ 1‘ 0u  K‘ 2‘ K1u  1‘ K1u  K‘ 1‘ K2u  1‘ 1u  0‘ 1u  1‘ 3 7 5 : (4.1.1) We easily check that for A 2 Avoid(m;V ) and 3  u; ‘  m 3, kAk =  m 2  +  m 1  +  m 0  +3. We will prove that A 2 ext(m;V ) and hence establish Theorem 4.1.6. To prove this, consider A 2 Avoid(m;V ) and apply the standard decomposition (2.1.1). The minimal inductive children of V are: H1 = 2 6 4 1 0 1 0 0 1 3 7 5 ; H2 = " 1 1 0 0 0 0 1 1 # ; H3 = " 1 1 0 0 1 1 0 0 # Thus, Cr(A) 2 Avoid(m  1; fH1; H2; H3g). We used the computer again to conjecture a structure on a matrix in Avoid(m; fH1; H2; H3g). Lemma 4.1.7 Let m  4 and A 2 Avoid(m; fH1; H2; H3g). We have kAk  m+ 2. Proof: Using the program described in Section 3.5, we  nd that one of the following must hold for each quadruple of rows of A: no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 70no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 no no no no no no no no no no  1  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 71no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 72no no no no no no no no no no no  1  1  1 > 1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 no no no no no no no no no no no no  1  1 > 1 > 1 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1  1 > 1 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 no no no no no no no no no no no  1  1  1  1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 no no no no no no no no no no no no  1  1 > 1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 no no no no no no no no no no no no  1  1  1 > 1 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 73no no no no no no no no no no no no  1  1  1 > 1 2 6 6 6 6 4 0 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 This veri es in particular that forb(4; fH1; H2; H3g) = 6 (at least 10 columns are absent in each of the twenty cases). We proceed as we did in Lemma 4.1.3 and verify that each of these possibilities yields one row (the  nal row in each case) and at most one column that we can delete from A, maintaining simplicity. This means that kAk  m + 2 by induction.  Lemma 4.1.8 Assume m  4. Let A 2 Avoid(m; fH1; H2; H3g) with kAk = m + 2. Then there exists a partition U  [m] and L = [m]nU with jU j = u  1 and jLj = ‘ = m u  1 so that if we permute rows, A = U L 2 6 4 0u K1u 0u 0u     1‘ 1‘ K‘ 1‘ 0‘ 3 7 5 or A = U L 2 6 4 0u K1u 0u 1u     1‘ 1‘ K‘ 1‘ 1‘ 3 7 5 (4.1.2) Note that in the former case we must have ‘  2 and in the latter u  2. Proof: We use induction on m. The computer is able to show that the result is true for m = 3. Assume m  4. Let A 2 Avoid(m; fH1; H2; H3g). By our argument in Lemma 4.1.7, there is a row and a single column we can delete, leaving the remainder simple. Let A0 be the resulting simple matrix. We may assume by induction that there exists disjoint sets U 0; L0 such that jU 0j = a  1, jL0j = b  1 where a + b = m 1 so that after permuting rows and columns, A0 = U 0 L0 2 6 4 0a K1a 0a 0a     1b 1b Kb 1b 0b 3 7 5 ; or A0 = U 0 L0 2 6 4 0a K1a 0a 1a     1b 1b Kb 1b 1b 3 7 5 : We will assume b  2. Assume the last column (m + 2) and last row (m) of A is deleted to obtain A0. After deleting row m, the last column of A must be one of the columns of A0 74given that kA0k = forb(m; fH1; H2; H3g). In order to avoid H1; H2 and H3 in A, we can show that we have the desired structure (4.1.1) in A (with either a or b one larger than before). We can make a few general comments about row m. If we have both a 0 and a 1 in row m under the columns containing K1a , then using the two columns containing the 0 and 1 and two rows of the U 0 together with row m, we obtain a copy of H1, a contradiction. Similarly we cannot have both a 0 and a 1 in row m under the columns containing Kb 1b . It is also true that in row m we cannot have a 0 under K1a and a 1 under K b 1 b else we  nd a copy of H1 in two columns containing 0 and 1 in row m and in a row of U 0, a row of L0 and row m. We will consider the two cases that A0 has either the column 0m 1 or 1m 1 together. Note that if a = 1, then A0 has both 0m 1 and 1m 1. As in the proof of Theorem 4.1.2, we must do some case analysis for which column gets repeated. Let  be the last (repeated) column of A. Case 1:  = 0a  1b. We deduce that column 0m 1 of A0 (if present) appears with a 0 in row m else we have the subcon guration H1 in two rows of L0 and row m. Similarly, if column 1m 1 is present and a  2 then 1m 1 appears with a 1 in row m else we have a copy of H1 in two rows of U 0 and row m. Now if we have 1’s in row m under both K1a and K b 1 b , then we have the desired structure with U = U 0 and L = L0 [ fmg. Similarly, if we have 0’s in row m under both K1a and under K b 1 b then we have the desired structure with U = U 0 [ fmg and L = L0. Thus we may assume we have 1’s under K1a and 0’s under K b 1 b . Recall that either A 0 has 0m 1, or A0 has 1m 1 but in that case a  2. We have two subcases. Subcase 1a: A0 has 0m 1 (so A has 0m). If a = 1 we note that A0 has 1m 1 (and therefore we are in Subcase 1b). Else we  nd a copy of H3 in A in a row of L0 and row m in the column 0m of A, a column from 0a Kb 1b  01 with a 0 in the chosen row of L0, the column with 0a  1b  01 and a column from K1a  1b+1. Subcase 1b: A0 has 1m 1 (so A has 1m) and a  2. We have a  2 and  nd a copy of H3 in A in a row of U 0 and row m in the column 1m of A, a column from K1a  1b+1 with a 1 in the chosen row of U 0, the column with 0a  1b  01 and a column from 0a  Kb 1b  01. This completes Case 1. 75Case 2:  = 0a   b, where  b 2 Kb 1b . Given that 0a  b is repeated in A on rows [m 1], we deduce that it appears both with a 1 and with a 0 on row m. Assume any other column 0a  0b of A 0 (with  0b 2 K b 1 b ) has a 0 in row m. Then we  nd a copy of H1 in row m and the two rows of L0 where  b and  0b di er and in the column of A containing 0a  0b and one of the two columns of A containing 0a  b which di ers from the  rst column in row m. Case 3:  =  a  1b, where  a 2 K1a. For a  2 we may follow the argument of the pre- vious case and  nd a copy of H1. Given a = 1, we have that A0 contains 0m 1 as well as  a  1b =  1  1m 2 = 1m 1. We deduce that A has 1m and 1m 1  01. Thus we can  nd a copy of H3 in a row of U 0 together with a row of L0, in the two columns with 1m 1 in A0, the column with 0m 1 in A0 and a column of 01  Kb 1b selected in order to have a 0 on the chosen row of L0. Case 4:  = 0a  0b. We  nd H3 in two rows of L0 since b  2. Case 5:  = 1a  1b. If a  2, we can  nd H3 in two rows of U 0. In case a = 1, we know 0m 1 as well as 1m 1 are in A0. We  nd H3 using a row of U 0 and a row of L0 where we choose a column of 0a  Kb 1b that has a 0 in the chosen row of L 0, plus 0m 1 of A0, 1m 1  0 and 1m 1  1. This completes the proof or Lemma 4.1.8.  Proof of Theorem 4.1.6: We use induction on m for m  6. We established by computer that forb(5; V ) = 18 (which is smaller than the bound of Theorem 4.1.6). Noting that forb(5; fH1; H2; H3g) = 7, we deduce using Section 2.1 that forb(6; V )  forb(5; V ) + forb(5; fH1; H2; H3g) = 18 + 7 = 25 and so forb(6; V ) = 25, because of construction (4.1.1). Thus, we may assume m  7. By induction, assume forb(m 1; V ) =  m 1 2  +  m 1 1  +  m 1 0  + 3. Let A 2 Avoid(m;V ) with kAk = forb(m;V ). Apply the standard decomposition (2.1.1) to A for some row r. If 76kCr(A)k  m, we obtain kAk  kCr(A)k+ forb(m 1; V )  m+  m 1 2  +  m 1 1  +  m 1 0  + 3 =  m 2  +  m 1  +  m 0  + 3 Thus, we may assume that for every r we have kCr(A)k  m+ 1. Using Lemma 4.1.7 (with m replaced by m 1), we may assume kCr(A)k = m+1 for each r. Then using Lemma 4.1.8 we can assume Cr(A) has the structure of (4.1.2) for every r so that every Cr(A) partitions [m]nr rows into sets Ur; Lr with jUrj; jLrj  2. Note also that the only di erence between the two possible structures is a column of 0’s or a column of 1’s neither of which is used in the case analysis below. Furthermore, we will prove that there is a partition of the rows [m] of A into U;L where Ur = Unr and Lr = Lnr. Take two rows, say s and t. Consider Cs(A) and Ct(A) as determined by (2.1.1) by setting r = s and r = t. Applying Lemma 4.1.8 when considering Cs(A) and Ct(A) we obtain the partitions Us, Ls, Ut, Lt (Upper and Lower) of rows. Remember that Cs(A) and Ct(A) both appear twice in A with 0’s and 1’s in rows s and t respectively. We now de ne partitions U 0s = Usnt, L 0 snt, U 0 tns, L 0 tns so that U 0 s [L 0 s = [m]nfs; tg = U 0 t [L 0 t. We assumed m  7 and so j[m]nfs; tgj  5. Hence we may assume that at least one of U 0s and L 0 s has size at least 3. Without loss of generality, assume jU 0sj  3. Let X = 2 6 4 1 1 1 1 0 0 3 7 5 ; Y = 2 6 4 0 0 0 0 1 1 3 7 5 : Consider the following three cases: jU 0s \ L 0 tj  3: We can  nd V in rows U 0 s\L 0 t in A (since A contains two copies of K 1 3 in each triple of rows of U 0s and two copies of K 2 3 in each triple of rows of L 0 t). jU 0s \ L 0 tj = 2: Then jU 0 s\U 0 tj  1, and so we can  nd V in AjU 0s by taking two rows of U 0 s\L 0 t together with any row in the intersection U 0s \ U 0 t . We  nd Y as a submatrix in any row order (A contains two copies of K13 in each triple of rows of U 0 s) and we also have X as a submatrix whose  rst two rows are from U 0s \L 0 t and the last one from U 0 s \U 0 t . 77This yields V . jU 0s \ L 0 tj = 1: We have jU 0 s \ U 0 tj  2, and so we can  nd V in AjU 0s by taking the row of U 0s \ L 0 t together with two rows in the intersection U 0 s \ U 0 t . We  nd Y as a submatrix in any row order (A contains two copies of K13 in each triple of rows of U 0 s) and we also have X as a submatrix whose  rst row is U 0s \ L 0 t and last two rows are from U 0 s \ U 0 t . This yields V . This means that U 0s  U 0 t , but then jU 0 tj  3 and so analogously U 0 t  U 0 s. So U 0 s = U 0 t , and then L0s = L 0 t. The same conclusion will hold if jL 0 sj  3. Thus for all s; t 2 [m], U 0 s = U 0 t , and then L0s = L 0 t. Using m  4, we now may deduce that there is a partition U;L of [m] so that for any r, Ur = Unr and Lr = Lnr This proves that the partition for each Cr is really a global partition. Let jU j = u and jLj = ‘. We may argue u; ‘  2 since for example if jU 0rj  1 and U 0r = fsg then U 0 s [ fsg  U and we have jU j  2. Note that for every row r, we have that [0 1]  Cr(A)  A. We deduce A contains the following columns: B = U L 2 6 4 0u  0‘ 0u  1‘ 0u  K‘ 1‘ 0u  K‘ 2‘ K1u  1‘ K1u  K‘ 1‘ K2u  1‘ 1u  1‘ 3 7 5 : (4.1.3) We have included the column of 0’s and the column of 1’s since such columns can be added to any matrix without creating V . What other columns might we add to this? For u  3, matrix B contains 2 6 4 1 0 0 1 0 0 0 1 1 3 7 5 in any triple of rows or U in any row order. So (A B)jU must not contain the con guration (1; 1; 0)T , else A has subcon guration V . Similarly for ‘  4, (A  B)jL must not contain the con guration (1; 0; 0)T . Thus, for u; ‘  3, all columns of (A  B) are in [0u j K1u j 1u]  [0‘ j K ‘ 1 ‘ j 1‘]. The only column not already in B is 1u  0‘ which is a column of the hypothesized structure (4.1.1). Thus, without loss of generality, we need only consider the case u = 2, ‘  5. Let U = fa; bg and consider any two rows c; d 2 L. We know B contains K1u  [1‘ j K ‘ 1 ‘ ] and 78[0u j K1u] K ‘ 1 ‘ . So B has: a 0 0 1 0 0 1 b 1 1 0 c 1 1 0 0 0 1 d 1 1 0 : Note we can interchange a with b and c with d. To avoid V we must not have columns in (A B) with a 1 1 b 0 c 0 1 d 0 : Thus, the only possible columns of A B are a 0 1 0 1 1 b 0 0 1 1 1 L  1‘ 1‘ 0‘ 1‘ : where  is any column. Recall that since ‘  3, any such  must avoid con guration (1; 0; 0)T . All these columns are already in B, except for 12  0‘, which together with B completes the hypothesized structure (4.1.1). The desired bound follows.  Interestingly, the structure of (4.1.1) falls short of the bound in the case u = 2. 4.2 Exact Bound for Ten Products This section is dedicated to proving an exact bound. The following four matrices are all 2 2 simple matrices (up to row and column permutations). Let I2 = " 1 0 0 1 # T2 = " 1 1 0 1 # U2 = " 0 0 0 1 # V2 = " 1 0 1 0 # : We note forb(m; fI2; T2; U2; V2g) = 1: De ne fI2; T2; U2; V2g  fI2; T2; U2; V2g = fX  Y : X; Y 2 fI2; T2; U2; V2gg as the 10 79possible products of these matrices (note some of the 16 are equivalent). Theorem 4.2.1 We have forb(m; fI2; T2; U2; V2g  fI2; T2; U2; V2g) = m+ 3: Proof: Use the notation F = fI2; T2; U2; V2g fI2; T2; U2; V2g. We establish the lower bound by construction. Let  = 1m 1 01. The m (m+3) matrix A consisting of [0m j I j  j 1m] avoids all con gurations in F , hence forb(m;F)  m+ 3. We use induction on m for the upper bound. We veri ed forb(4;F) = 7 using the computer program described in Section 3.5. To prove the bound for m  5, we will proceed by induction on m. For an m-rowed matrix A that doesn’t contain any con guration in F it su ces by induction to show there exists a row r for which kCrk  1, using the standard decomposition as in (2.1.1). If this were so, we could delete row r and perhaps one column (one instance of the column forming Cr) from A, keeping the remaining matrix simple. This would yield forb(m;F)  1 + forb(m 1;F) = 1 + (m 1) + 3 = m+ 3 as desired. Let us proceed by contradiction. Suppose then that for every row r, kAk  2. We then have at least two columns  and  in C1(A). The matrix A would look like this 1 " 0    0 0 1 1    1     # : But  and  must di er in some row. Without loss of generality, assume they di er on row 2, and suppose  2 = 0 and  2 = 1. We will prove that  and  must be complements of each other. Suppose otherwise and suppose they had something in common, say in row 3. The  rst four rows of A would look like this: 1 2 3 4 2 6 6 6 6 4 0    0 0 1 1    1 0 1 0 1 a a a a b c b c 3 7 7 7 7 5 : for some values of a; b; c (we are using the fact that the matrix has at least 4 rows). Then in 80rows 1 and 3 and rows 2 and 4 we get that this matrix contains 1 3 " 0 1 a a #  2 4 " 0 1 b c # : which is a con guration of F (for any a; b; c), so we conclude  =  . Now C2(A) must have two repeated columns, say  and  . As argued above,  =  . Here is part of the matrix A: 2 6 4 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1         3 7 5 : We have abused notation and refer to  and  on m  2 rows as  and  again. This abuse will continue throughout this proof. Since  and  have to di er somewhere, we can assume  3 = a, and  3 = a. Since  and  must di er somewhere, we can assume  4 = b and  4 = b. Furthermore, since we have at least 5 rows, we can then write the selected columns of A where the columns are given labels below to indicate the source of the column. 2 6 6 6 6 6 6 4 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 a a a a a a a a b b b b b b b b d c d c d c d c 3 7 7 7 7 7 7 5         : There are two cases. Either d = c or d = c. So we either have 81(d = c) : 2 6 6 6 6 6 6 4 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 a a a a a a a a b b b b b b b b d d d d d d d d 3 7 7 7 7 7 7 5         or (d = c) : 2 6 6 6 6 6 6 4 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 a a a a a a a a b b b b b b b b d d d d d d d d 3 7 7 7 7 7 7 5         : These yield the following con gurations in F respectively: (d = c) : 2 4 " 0 1 b b #  3 5 " a a d d # ; (d = c) : 2 3 " 0 1 a a #  4 5 " b b d d # : This is a contradiction to kCrk  2 and hence for m  5, there must be some row r for which kCrk  1. The bound is achieved by induction.  4.3 Critical Substructures In view of Remark 1.3.21, we know that G  F implies forb(m;G)  forb(m;F ). So given F , it is a natural question to try to  nd minimal con gurations G  F such that forb(m;F ) = forb(m;G). In this section we give a conjecture for all such subcon gurations of Kk and prove our conjecture for k = 4. Remember that by Theorem 1.3.29 we have that forb(m;Kk) =  m k 1  +  m k 2  + :::+  m 0  . De nition 4.3.1 Let G be a con guration. We say G  F is a critical substructure if forb(m;G) = forb(m;F ). We give a conjecture that has been veri ed for k = 2; 3 and we verify it for k = 4 in Proposition 4.3.3, Proposition 4.3.7 and Proposition 4.3.9. 82Conjecture 4.3.2 Let k be given. Then the minimal critical substructures of Kk are the k + 3 con gurations 2  1k 1, 2  0k 1 and K‘k for 0  ‘  k. Previous results established this conjecture for k = 3. The fact that all these con gura- tions are critical substructures of Kk isn’t hard to prove. The di culty then lies in proving every critical substructure contains one of these as a con guration. We show that Conjecture 4.3.2 is equivalent to Conjecture 4.3.8 (for any k) and then prove Conjecture 4.3.8 for k = 4. The following proposition was proven by F uredi and Quinn [FQ83], and Gronau [Gro80]. Proposition 4.3.3 [FQ83][Gro80] For all ‘; k with 0  ‘  k we have forb(m;Kk) = forb(m; 2  1k 1) = forb(m; 2  0k 1) = forb(m;K‘k): Notice in particular that this proposition answers the question posed in Example 1.3.15. We have I3 = K13 and so forb(m; I3) = forb(m;K3) =  m 2  +  m 1  +  m 0  . We now consider a shifting argument  rst used by Alon for this context in [Alo83]. Let A be a simple f0; 1g-matrix. We construct a shifted matrix A by performing the following operation repeatedly. Consider an entry of the matrix that has a 1 and make it into a 0 if the resulting matrix is still simple. Keep doing this until there is no entry with a 1 which can be made 0 with the matrix remaining simple. If a column  appears in A, then every column with 0’s in positions where  is 0 (and perhaps others) also appears in A. In other words, the associated family for the shifted matrix A is a downset (if a set is in the family, every subset of that set is in the family as well). Lemma 4.3.4 [Alo83] Let A be a simple f0; 1g-matrix and let A be a shifted version of A. For any set of rows S, we have that the number of di erent columns of AjS is at least the number of di erent columns of AjS. 83Lemma 4.3.5 Let A 2 ext(m;Kk). Then for any set S 2  [m] k  , we have that there is a k  1 column  such that (Kk   )  AjS. Proof: Let A 2 ext(m;Kk). Then for any set S 2  [m] k  , AjS has at most 2k  1 di erent columns since it does not have Kk. Let A be a shifted version of A and so AjS has at most 2k 1 di erent columns. Now the columns of A form a downset and so cannot have a column of sum k or larger. Now kAk is forb(m;Kk) which is equal to the number of columns of sum k 1 or smaller and so A consists of all columns of sum at most k 1. Thus AjS has exactly 2k  1 di erent columns (it must have at least 2k  1 but cannot have 2k since it would contain Kk). Then AjS has exactly 2k 1 di erent columns, establishing our result.  We establish the k-rowed critical substructures of Kk using the following Lemma. Lemma 4.3.6 Let B be an k (k+ 1) matrix consisting of one column of each column sum i for 0  i  k. Let F = Kk  B. Assume m  k. Then forb(m;F ) < forb(m;Kk). Proof: Let A 2 ext(m;F ). Assume forb(m;F ) = forb(m;Kk). Then also A 2 ext(m;Kk). But then by Lemma 4.3.5 we have that there is a k 1 column  so that A has Kk  . Now F is a con guration in Kk   (for any choice of  ), contradicting our assumption. Thus forb(m;F ) < forb(m;Kk).  Proposition 4.3.7 Let F be a minimal k-rowed critical substructure of Kk. Then F = K‘k for some ‘. Proof: Consider any k-rowed critical substructure F of Kk. Then if there exists an ‘ such that K‘k  F , since we know forb(m;K ‘ k) = forb(m;Kk), then F cannot be minimal unless F = K‘k. Consider the case where K ‘ k  F for all ‘. In this case F is contained in Kk B for 84some B a collection of columns with one column of each column sum. Using Lemma 4.3.6, we can conclude forb(m;F )  forb(m;Kk  B) < forb(m;F ): Thus, F is not a critical substructure.  For (k 1)-rowed critical substructures we have more work to do. To prove this conjecture, it would su ce to show that the only (k 1)-rowed minimal critical substructures are 2  0k 1 and 2  1k 1. Given the previous Lemmas, we conclude Conjecture 4.3.2 is equivalent to the following: Conjecture 4.3.8 Let Fk 1 = [0k 1 j 2  K1k 1 j 2  K 2 k 1 j    j 2  K k 2 k 1 j 1k 1]: Then there exists M so that for m  M , forb(m;Fk 1) < forb(m;Kk). Let A 2 ext(m;Kk). A consequence of Lemma 4.3.5 is that for every S 2  [m] k  there is some choice of  for which Kk    AjS. Assume that  has p 1’s and (k  p) 0’s. Then for some T  S with jT j = k 1, we have that for some choice of  , Kk 1   AjT , where  has either p 1’s and k  p 1 0’s or has p 1 1’s and k  p 0’s. Let C be a (k  1)-rowed matrix with at most one column of each column sum and having column of 0’s , column of 1’s and for each 1  i  k  3, we have either a column of sum i or a column of sum i + 1. Then AjT has 2  Kk 1  C. This handles k = 3 for which we can take C to consist of a column of 0’s and a column of 1’s. We can handle the case k = 4 as well. Proposition 4.3.9 Let m  6. Let F3 = [03 j 2  K13 j 2  K 2 3 j 13] as in Conjecture 4.3.8. Then forb(m;F3) < forb(m;K4). Proof: Let A 2 ext(m;F3) and assume ext(m;F3)  ext(m;K4). Consider any column  of [K0m j K 1 m j K m 1 m j K m m ]. By extremality of A, [A j  ] has K4 as a subcon guration, say on rows S. Then for any row j 2 S with Rj := S n fjg, we have that AjRj contains 852  K3   jRj . Selecting j 2 S so that  jRj is either 13 or 03 we have then that A contains F3, a contradiction. So we deduce that [K0m j K 1 m j K m 1 m j K m m ]  A already and hence A contains K3 in every set of 3 rows. Let B consist of the remaining columns of A (not in [K0m j K 1 m j K m 1 m j K m m ]). We deduce that B has no [K13 j K 2 3 ]. Then kBk  forb(m; [K13 j K 2 3 ]) =  m 2  +  m 1  +  m 0  : Thus kAk  2m+ 2 +  m 2  +  m 1  +  m 0  : For m  6, we have 2m+ 2 <  m 3  , a contradiction.  In Section 8.1 we give some further ideas on how to deal with k  5. 86Chapter 5 Three Asymptotic Bounds 5.1 Introduction When we are not able to obtain an exact bound on forb(m;F ) for some F , we might settle for asymptotic bounds. The research in this topic is often guided by Conjecture 1.4.1. In this chapter we prove three main results, each of which completes some part which would be required in order to establish Conjecture 1.4.1. In Section 5.2 we prove the quadratic asymptotic bound predicted by Conjecture 1.4.1 for the con guration F8(t), one of the three maximal 4-rowed con gurations for which the conjecture predicts a quadratic bound. Here are these three con gurations (called F6(t), F7(t), F8(t) for historical reasons). F6(t) = 2 6 6 6 6 4 1 0 1 1 0 1 1 0 0 0 1 1 0 0 0 1 t  2 6 6 6 6 4 1 0 1 1 1 1 0 0 0 1 0 1 0 0 1 0 3 7 7 7 7 5 3 7 7 7 7 5 F7(t) = 2 6 6 6 6 4 1 0 1 1 0 1 1 1 0 0 1 0 0 0 0 1 t  2 6 6 6 6 4 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 0 3 7 7 7 7 5 3 7 7 7 7 5 87F8(t) := 2 6 6 6 6 4 1 0 1 0 0 1 0 1 1 1 0 0 1 1 0 0 t  2 6 6 6 6 4 1 0 0 1 1 1 0 0 3 7 7 7 7 5 3 7 7 7 7 5 : As of this writing, quadratic bounds for F6(t) and F7(t) have not been found. Note that both F6(t) and F7(t) contain the following: t  2 6 6 6 6 4 1 0 1 0 0 1 0 1 3 7 7 7 7 5 : It is not known (but The Conjecture predicts it) whether or not the above matrix has a quadratic bound for t  3, but for t = 2 and t = 1, the quadratic bound was proven by Anstee in [Ans90]. The methods described in the following chapters have so far failed to produce results for these other con gurations. There are some cases in their respective lists of What Is Missing which we don’t know how to deal with. Then in Section 5.3 we prove the bound for F7, one of the nine maximal 5-rowed con g- urations below which are predicted to be quadratic by Conjecture 1.4.1. F3 = 2 6 6 6 6 6 6 4 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F4 = 2 6 6 6 6 6 6 4 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F5 = 2 6 6 6 6 6 6 4 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 1 0 3 7 7 7 7 7 7 5 F6 = 2 6 6 6 6 6 6 4 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F7 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 0 1 0 3 7 7 7 7 7 7 5 F8 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 3 7 7 7 7 7 7 5 88F9 = 2 6 6 6 6 6 6 4 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1 1 0 1 1 1 1 0 0 3 7 7 7 7 7 7 5 F10 = 2 6 6 6 6 6 6 4 1 1 0 1 0 0 1 0 1 1 1 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 1 3 7 7 7 7 7 7 5 F11 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 1 1 3 7 7 7 7 7 7 5 It is not known whether or not any of the other eight boundary con gurations will satisfy the conjecture. Finally, in Section 5.4 we use the result about F7 to classify all 6-rowed quadratic con-  gurations by proving the unique boundary case: G6 3 = 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 3 7 7 7 7 7 7 7 7 7 5 : 5.2 Quadratic Bound for a 4-rowed Con guration Let F8(t) be the following 4 (4 + 2t) con guration (as in the Introduction) F8(t) := 2 6 6 6 6 4 1 0 1 0 0 1 0 1 1 1 0 0 1 1 0 0 t  2 6 6 6 6 4 1 0 0 1 1 1 0 0 3 7 7 7 7 5 3 7 7 7 7 5 : Theorem 5.2.1 Let t  1 be given. Then forb(m;F8(t)) is  (m2). More precisely, forb(m;F8(t))  12tm2. Furthermore, F8(t) is a boundary quadratic case. We have several ingredients to the proof. The  rst is our Standard Induction given in Section 2.1: We  nd the inductive children HF8(t). By Standard Induction, it su ces then to prove that forb(m;HF8(t)) is  (m). We then  nd What Is Missing as in Section 2.2 in a 89triple of rows if HF8(t) is avoided. Lastly, we use Implications, as described in Section 2.3, to delete a linear number of columns from a matrix A 2 Avoid(m;HF8(t)) so that Cr(A) becomes constant. First, let us calculate HF8(t0) for some number t 0. Let A 2 Avoid(m;F8(t0)) and consider Cr(A). The fact that A contains no con guration F8(t0) means Cr(A) doesn’t have any of the following three con gurations, for t = b t 0+1 2 c+ 1: H1(t) = 2 6 4 1 0 0 1 0 0 t  0 B @ 1 0 0 1 1 1 1 C A 3 7 5 ; H2(t) = 2 6 4 1 0 0 1 1 1 t  0 B @ 1 0 0 1 0 0 1 C A 3 7 5 ; H3(t) = 2 6 4 0 1 0 1 1 1 0 0 1 1 0 0 t  0 B @ 0 1 1 1 0 0 1 C A 3 7 5 : So we may conclude that HF8(t) = fH1(t); H2(t); H3(t)g. We now focus our attention to matrices A 2 Avoid(m; fH1(t); H2(t); H3(t)g). We make the following bold claim: Lemma 5.2.2 Let t be given. Then forb(m; fH1(t); H2(t); H3(t)g)  12tm. Note that Theorem 5.2.1 is not equivalent to the previous lemma is false, and this is indeed the case for F6(t) and F7(t), for which we know the bound for the inductive children is quadratic and not linear. The proof of Lemma 5.2.2 appears at the end of this subsection. But by using this lemma we are ready to prove Theorem 5.2.1. Proof of Theorem 5.2.1 : Let A 2 Avoid(m;F8(t)). We simply use induction as in Proposition 2.1.2 (replacing t by t+12 + 1) using Lemma 5.2.2 to deduce kAk is linear and hence forb(m;F8(t)) is O(m2). The fact that F8(t) is a boundary case follows from the constructions in Conjecture 1.4.1. If  2 F8(0), then  has column sum 1 or 3, but then [F8(t) j ]  I  I  I. Consider then the possibilities for  not in F8(1). If  consists of all 1’s or three 1’s or two 1’s except on the  rst two rows then each pair of rows of [F8(1) j ] has a column of two 1’s and so [F8(1) j ] is not in Im=3  Im=3  Im=3. This also handles the complementary case, i.e. where  is all 0’s or one 1 or two 1’s on the last two rows using Icm=3  I c m=3  I c m=3. There are two remaining cases:  having two 1’s in the  rst and fourth rows or in the second and fourth rows. Then [F8(1) j ] has the 2 2 matrix I2 on each pair 90of rows and so [F8(1) j ] is not in Tm=3  Tm=3  Tm=3. Thus forb(m; [F8(1) j ]) is  (m3).  To prove Lemma 5.2.2, we will need some additional Lemmas and properties. 5.2.1 What Is Missing? The following are all the possibilities of columns that are either missing or in short supply on three rows if we forbid H1(t); H2(t); H3(t). This was computed using the C++ program referred to in Chapter 3. Checking that if a triple of rows satis es each Pi, then con gurations H1(t); H2(t); H3(t) are avoided is quite easy, but the computer is used to avoid the enormous amount of work that would be required to establish that the list is complete. Lemma 5.2.3 Let A 2 Avoid(m; fH1(t); H2(t); H3(t)g). For each triple of rows (x; y; z) satis es (at least) one of the 21 cases P1; P2; :::; P21 described below for some ordering of (x; y; z). The cases are: P1 = no no 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 ; P2 = < t < t no no 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 P3 = < t < t no no 2 6 4 0 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 ; P4 = < t < t < t no 2 6 4 1 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 P5 = < t < t < t no 2 6 4 1 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 ; P6 = no < t no 2 6 4 0 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 91P7 = no < t < t no 2 6 4 0 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 ; P8 = no < t < t no 2 6 4 0 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 P9 = < t < t < t < t no 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 ; P10 = < t < t < t < t 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 P11 = no < t < t < t 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 ; P12 = no < t < t < t < t 2 6 4 0 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 P13 = no no no 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 ; P14 = < t no no 2 6 4 1 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 P15 = no no no 2 6 4 0 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 ; P16 = < t no < t no 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 P17 = < t no < t no 2 6 4 1 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 ; P18 = < t < t no < t 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 92P19 = no < t no < t 2 6 4 0 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 ; P20 = no < t no < t 2 6 4 0 0 0 3 7 5 2 6 4 1 0 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 P21 = no no 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 : Proof: By exhaustive computer search as described in Section 3.5.  5.2.2 Case Analysis We now proceed to do some case analysis of the 21 cases mentioned in Lemma 5.2.3. The  rst trick to discard some cases involves attempting to do induction one more time. For some of the 21 cases, if they happened to occur in A we may apply the Standard Induction again. De nition 5.2.4 We say that a row r of A is non-essential if kCr(A)k  4t. The motivation for this de nition is that if we could  nd a non-essential row of A, we could use Standard Induction as in Section 2.1 for the case where we are forbidding fH1(t); H2(t); H3(t)g to prove Lemma 5.2.2 (in fact kCr(A)k  12t would su ce). Lemma 5.2.5 If A 2 Avoid(m; fH1(t); H2(t); H3(t)g) has a triple of rows (i; j; k) that satis es one of P2, P4, P8, P9, P10, P12, P17, P18 and P19, then there is a non-essential row r of A. Proof: If we analyze the columns that could be in long supply for each of the cases, we see in each case that one of the rows of A isn’t necessary to distinguish between columns in long supply. 93Perhaps an example would be useful. The other cases are similar. Suppose A is missing P2 in rows i; j; k, in that order. So A satis es the following from rows i; j; k: P2 = < t < t no no i j k 2 6 4 0 0 1 3 7 5 2 6 4 1 0 1 3 7 5 2 6 4 0 1 1 3 7 5 2 6 4 1 1 1 3 7 5 Let us analyze the columns that could be in long supply (denote l.s.). These columns are: l:s l:s l:s l:s i j k 2 6 4 1 1 0 3 7 5 2 6 4 1 0 0 3 7 5 2 6 4 0 1 0 3 7 5 2 6 4 0 0 0 3 7 5 : We can see that kCk(A)k  2t, and therefore row k is non-essential. The same happens for the other cases: in each of them, there is a row r, like row k, for which kCr(A)k  4t.  If we assume there are no non-essential rows then we may restrict our attention to matrices which for all triples of rows satisfy one of the cases P1, P3, P5, P6, P7, P11, P13, P14, P15, P16, P20 or P21. We will now use the technique of Implications described in Section 2.3 to delete a linear number of columns from A (without deleting any row) in order to obtain a matrix A0 for which kCr(A0)k is bounded by a constant (for any row r). First we give a lemma which is purely a property of directed graphs. Lemma 5.2.6 Let G be a directed graph on m vertices. Then we can colour the edges of G using three colours (blue, red and green) in such a way that for vertices r; a; b we have that G satis es the following properties: (R) There are at most 2m red edges. (B) If r ! a and r ! b are blue, then neither a! b nor b! a (of any colour). (G) If a! b is green, there is a blue-red path from a to b. 94Proof: The idea for this colouring came from an idea  rst introduced by Anstee and Sali in [AS05], although the actual colouring is di erent. We provide an algorithmic proof. 1. Divide G into strongly connected components X1; X2; :::; Xk ordered in a way consistent with the order given by the acyclic ordering (so that if i < j, there might be a path between a vertex of Xi and a vertex of Xj, but there is no path back). 2. Pick a strongly connected component Xi. It is a well known property of directed graphs that there is a strongly connected subgraph Yi of Xi that uses all the vertices of Xi and the number of edges is at most 2jXij. For every edge of Xi, see whether it is in Yi or not. If it is, colour it with red, and if it isn’t, colour it with green. 3. Colour every remaining edge with blue. Notice that currently the only property that may not be satis ed is (B). We will change some of the blue ones to green (leaving the red ones intact) until we get the desired property, but never breaking (R) nor (G). Notice also that red edges always stay in the same strongly connected component, while blue edges always go to a higher level. To make this statement precise, de ne a level function  : V (G)! N as  (v) := i if v 2 Xi: We have the property that if v ! u is a red edge, then  (v) =  (u), and if v ! u is a blue edge, then  (v) <  (u). This property will be preserved during all steps of the colouring algorithm. In particular, it is true when applying steps 1 through 3. 4. Go through all the strongly connected components and for each component go through each vertex. Suppose we are at vertex v. Look at the set of blue edges coming out of v. Say their endpoints are v1; v2; :::; vd. 5. If there is an edge vi ! vj (of any colour) and v ! vi and v ! vj are (still) blue, paint v ! vj with green. It is easy to check that step 5 preserves property (G). We only need to prove that there is a blue-red path from v to vj. We know there is a blue-red path Pvi;vj from vi to vj by (G), and since v ! vi is blue, we can consider the path v ! vi plus Pvi;vj . This can be done as long as Pvi;vj doesn’t contain v ! vj, and indeed it can’t, because red edges always stay in the same connected component, while blue edges always go to a higher level, and since 95v ! vi is blue and the path Pvi;vj is blue-red,  (v) <  (vi)   (vj). This implies that Pvi;vj cannot contain vertex v.  Recall our de nition of implications of Section 2.3. Lemma 5.2.7 Let A 2 Avoid(m; fH1(t); H2(t); H3(t)g) have no non-essential rows. Form the directed graph G on m vertices whose edges are the implications in A and colour G using Lemma 5.2.6. Then we can delete at most 4tm columns from A so that all red implications are pure. Proof: There are at most 2m red edges and there are at most 2t columns that violate any given implication, hence there are at most 4tm columns that violate red implications. We can delete at most 4tm columns to make the red implications pure.  If a column violates a green implication, it must also violate an implication in a blue-red path, so it must violate either a blue or a red implication. So if we manage to purify the blue implications, no column could violate a green implication either. We will devote some time to proving there are at most 4tm columns that violate at least one blue implication. Lemma 5.2.8 Let A 2 Avoid(m; fH1(t); H2(t); H3(t)g) with no non-essential rows (i.e. for every row r, we have kCr(A)k > 4t). Colour the associated implications graph as in Lemma 5.2.6. In any triple of rows r; ri; rj where r ! ri and r ! rj are blue and impure, then either P16 or P21 is satis ed. Proof: By Lemma 5.2.5, we know that in any triple of rows one of P1, P3, P5, P6, P7, P11, P13, P14, P15, P16, P20, P21 has to be satis ed on the triple r; ri; rj. 96Having the implications r ! ri and r ! rj means that the following is satis ed in the triple of rows r; ri; rj for some 0  a < 2t: < 2t a < 2t a = a r ri rj 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 We can go through each of the remaining cases (except P16 and P21) and observe that the implications r ! ri and r ! rj are already not violated. In each case we will  nd a contradiction by  nding either a non-essential row, contradict- ing the hypothesis, or ri ! rj (or rj ! ri), which is a contradiction to the fact that r ! ri and r ! rj are blue. For each case Pi, we number the rows of Pi by 1,2,3 as they appear in our listing of What Is Missing. Note that we can’t have have two implications a ! b and b ! a (yields a non-essential row). There are in fact at most two implication on the three rows r; ri; rj. Here is a quick check for each case: P1: If r corresponds to row 1 of P1, then we get either ri ! rj or rj ! ri, a contradiction. If r corresponds to row 2 we get the same contradiction. And if r corresponds to row 3, then one of the rows ri and row rj is non-essential. P3: Already has the implication 1! 2, so row r must correspond with row 1, but then the row corresponding to row 3 of P3 will be non-essential. P5: Already has 1! 3 and 2! 3, regardless of ordering we’ll have the contradiction. P6: Already has the implication 1! 2. If we set r to correspond to 1, then we also get the implication 2! 3. P7: Already has 2! 3 and 1! 3, so no matter how we set row r we’ll have the contradiction. P11: Already has 2! 3 and 1! 3. P13: The columns involved in the implications already have been marked with ‘no’, so they never get violated anyway. P14: Already has 1 ! 3. Then row r must correspond to row 1 of P14 but then row r becomes non-essential. 97P15: Already has 1 ! 3. Then row r must correspond to row 1 of P15 but then row r becomes non-essential. P20: Already has 1 ! 3. Then row r must correspond to row 1 of P20 but then row r becomes non-essential. Hence we must have either P16 or P21 on any triple r; ri; rj.  Lemma 5.2.9 Let A 2 Avoid(m; fH1(t); H2(t); H3(t)g) with no non-essential rows, whose implication graph is coloured to satisfy the conditions R,B,G of Lemma 5.2.6. We may delete at most 4tm columns from A so that no blue implication is violated in what remains. Proof: We will prove something a bit stronger: for every row r, the number of columns that violate a blue implication coming out of r is bounded by a 4t. Take a row r and consider Bluer, the induced subgraph on the blue children of r. That is, Bluer = fs 2 G : r ! s is blue g: We will assume jBluerj  3. If jBluerj  2, we have at most 4t columns that violate the blue implications out of r. Let Bluer = fr1; :::; r‘g. Notice that if a triple of rows r; ri; rj with r ! ri and r ! rj impure and blue satis es either P16 or P21 then in particular it must satisfy this: < t no < 2t r ri rj 2 6 4 0 1 0 3 7 5 2 6 4 0 0 1 3 7 5 2 6 4 0 1 1 3 7 5 which means that in a column where row r is 0 and row rj is 1, then row ri is 1. Restrict our attention to the submatrix of A given as [BrCr] in (2.1.1). We use the notation row(ri) to denote the set given by row ri considered as an incidence vector (but restricting to the submatrix [BrCr]), we have row(rj)  row(ri). Every pair of rows in Bluer then must have one contained in the other (under the zeros of row r), which means we can order the sets row(r1); row(r2); :::; row(r‘) into an ascending chain. Therefore we can separate the columns that have a zero on row r into three categories. The  rst, C0 consists of the columns with all the entries in rows r1; r2; :::; r‘ being 0. The 98second category, C1 consists of all columns with all the entries in rows r1; r2; :::; r‘ being 1. And the last, C, consists of columns that start with some number of zeros and end with ones, like this: r r1 ... ri ri+1 ... r‘ 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 ... 0 1 ... 1 3 7 7 7 7 7 7 7 7 7 7 7 7 5 : We deal with these three categories separately. Columns in C0: These columns already don’t violate any implication r ! ri. Columns in C: Rows r1 and r‘, in addition to satisfying row(r1)  row(r‘), must also satisfy < t r r1 r‘ 2 6 4 0 0 1 3 7 5 ; which means the number of columns with 0 in row r, 0 in row r1 and 1 in row r‘ is at most t 1. This means jCj < t. Columns in C1: We may use the fact that each triple r; ri; rj satis es < 2t r ri rj 2 6 4 0 1 1 3 7 5 ; so C1 < 2t as well. In conclusion, for each row r, we can delete 3t columns, and then every blue implication r ! ri is pure. We repeat this for every row r, and by deleting at most 3tm columns of A, every blue implication is pure.  99After these column deletions every blue implication is pure and we may now assume every implication is pure: No column violates either blue or red implications, which means no column violates any green implications. We’ve managed to delete only a linear number of columns of A without deleting any row, and now no implication gets violated. 5.2.3 Linear Bound for the Inductive Children We are now ready to prove Lemma 5.2.2. Proof of Lemma 5.2.2 : We will show that forb(m; fH1(t); H2(t); H3(t)g)  12tm. Let A 2 Avoid(m; fH1(t); H2(t); H3(t)g). By Proposition 2.1.2 as described above, we may assume A has no non-essential rows. For every triple of rows of A, we choose Pi if Pi is satis ed in that triple (other Pj might be satis ed as well, but pick one for every triple). This yields a map p : S3(A) ! fPi : i 2 f1; 3; 5; 6; 7; 11; 13; 14; 15; 16; 20; 21gg (by Lemma 5.2.5) where S3(A) is the set of triples of rows of A. Appealing to Lemma 5.2.7 and Lemma 5.2.9 we can delete at most 8tm columns and conclude that all implications associated to one of the Pi in the image of S3(A) are pure. We now do induction again with a new hypothesis. We wish to show that for A 2 Avoid(m; fH1(t); H2(t); H3(t)g) satisfying that each triple of rows has a chosen satis ed condition Pi (for some i 2 f1; 3; 5; 6; 7; 11; 13; 14; 15; 16; 20; 21g) and any implications arising from these chosen conditions is pure, then kAk  4tm. We note that any submatrix of A satis es the same hypotheses. Thus it su ces to appeal to our induction argument Proposition 2.1.2 and show that A has a non-essential row. We will in fact show that row 1 is non-essential. Let A 2 Avoid(m; fH1(t); H2(t); H3(t)g) satisfying that each triple of rows has a chosen satis ed condition Pi (for some i 2 f1; 3; 5; 6; 7; 11; 13; 14; 15; 16; 20; 21g) and any implications arising from these chosen conditions is pure. Consider a triple of rows 1; r; s from A. In this triple, one of the 12 cases will have been chosen to be satis ed, so for the pair r; s in C1(A), two rows corresponding to two rows of one of the 12 Pi’s must be satis ed, since C1(A) consists of the repeated columns. Consider a (0,1)-row with n columns as the incidence vector of a subset of [n] = f1; 2; : : : ; ng. We now use row(r) to denote row r in C1(A), since we’ve restricted our attention to this matrix. We show by the case analysis below that if two columns are absent on a pair of rows r; s, then either row(r) = ; or row(r) = [n], (or row(s) = ; or row(s) = [n]), or row(r) = row(s) or row(r) = row(s)c. If two columns are absent, then only two columns can be present. Thus, 1001. If the two columns absent are no no r s " 1 0 # " 1 1 # or no no r s " 0 1 # " 0 0 # then row(r) = ; (row r is 0’s) in the former, or row(r) = [n] (row r is 1’s) in the latter. 2. If the columns missing are no no r s " 0 1 # " 1 0 # then row(r) = row(s). 3. If the columns missing are no no r s " 0 0 # " 1 1 # then row(r) = row(s)c. We may check each case P1, P3, P5, P6, P7, P11, P13, P14, P15, P16, P20, P21 to  nd that two columns are absent for each pair of rows of C1(A), except for P6 and P14 when row 1 of A corresponds to row 3 of P6 or row 2 of P14, In the case when 1; r; s form a P6 or a P14, and row 1 of A corresponds to row 3 of P6 or row 2 of P14, we have (for some order of r and s) no < t r s " 0 1 # " 1 0 # which means row(s)  row(r), and the di erence row(r) n row(s) is at most t. Construct the following coloured semi-directed graph:  The vertices are the rows r of C1(A) with row(r) 6= ; and row(r) 6= [n].  Place a purple edge between two rows r; s if row(r) = row(s).  Place a yellow edge between two rows r; s if row(r) = row(s)c. 101 Place a directed edge r ! s if row(r)  row(s). If some rows are equal, we will treat them as being just one row. So we can take the quotient of the graph over the purple edges and work in the new graph. If two yellow edges share a vertex, the non-shared vertices must have a purple edge between them, because the complement of the complement is itself. Since we did the quotient over purple edges, we can assume no two yellow edges share a vertex. So we are left with only directed and yellow edges. We will prove there are no yellow edges. We proceed by contradiction. Suppose we have a yellow edge between rows r1 and r2 so that row(r1) = row(r2)c. If there is no other row then the matrix has at most 2 columns and we are done. Assume r is another row, di erent from r1 or r2. Consider the edge between r and r1 and between r and r2. Let us analyze the four possibilities. Clearly it can’t be yellow or purple.  If r ! r1 and r ! r2, then row(r)  row(r1) and row(r)  row(r2) contradicts row(r1) = row(r2)c if row(r) 6= ;. So we conclude row(r) = ;, contradicting our construction.  If r ! r1 and r2 ! r, then row(r2)  row(r1), a contradiction.  If r1 ! r and r ! r2, then row(r1)  row(r2), a contradiction.  If r1 ! r and r2 ! r, then we have that row(r) contains both a set and its complement. This means row(r) = [n] contradicting our construction. Every pair of rows r; s has a directed edge and therefore we have a tournament. We note that the graph has no directed cycles (since a directed edge means containment of rows) and hence it is a transitive tournament. This in particular yields a path that goes through all the vertices (a total ordering of the rows). We have more information: A directed edge only occurs in cases P6 and P14, when row 1 of A corresponds to row 3 of P6 or row 2 of P14. In these two cases, when a row r contains row s, we also have that we get that rows r and s di er in at most t columns. And since the  rst row in the path (the one with the least number of ones) and the last (the one with the most number of ones) have to di er in at most t places, there must be at most t+ 2 columns in C1(A), and so row 1 is non-essential, which proves the lemma.  1025.3 Quadratic Bound for a 5-rowed Con guration Previous work of Ryan, reported in [Ans], computed nine 5-rowed simple matrices F which by Conjecture 1.4.1 should be boundary cases and for which forb(m;F ) should be  (m2). One of them, named F7 in [Ans], is F7 = 2 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 1 1 1 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 0 1 0 3 7 7 7 7 7 7 5 : Note that F7 = F c7 . Theorem 5.3.1 We have that forb(m;F7) is  (m2). Moreover, for any 5  1 f0; 1g- column  , forb(m; [F7 j ]) is  (m3). The proof uses Standard Induction (Section 2.1) and a linear bound for three smaller matrices in Lemma 5.3.2 which in a novel way uses Standard Induction. We give the proof of Theorem 5.3.1 from Lemma 5.3.2 in Section 2.1 and the proof of Lemma 5.3.2 in Section 5.3.3. 5.3.1 Applying Standard Induction Let A 2 Avoid(m;F7) and apply the standard decomposition of (2.1.1) for r = 1. Our goal is to show kAk is quadratic by showing that kC1(A)k is linear. We  nd the inductive children of F7 (basically, delete each row of F7 in turn) and note that C1(A) cannot contain any of the con gurations H1; H2; H3; H4; H5: H1 = 2 6 6 6 6 4 1 0 1 1 1 1 0 1 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 3 7 7 7 7 5 ; H2 = 2 6 6 6 6 4 1 0 1 1 0 0 0 1 0 1 0 1 0 0 1 0 0 0 1 0 3 7 7 7 7 5 ; H3 = 2 6 6 6 6 4 1 1 0 1 0 1 1 1 0 0 1 0 0 0 0 1 3 7 7 7 7 5 ; 103H4 = 2 6 6 6 6 4 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 1 3 7 7 7 7 5 ; H5 = 2 6 6 6 6 4 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 0 1 3 7 7 7 7 5 : We observe that Hc3 = H3, H4 = H c 1, H c 2 = H5. Also H3  H1 (columns 2,3,5,6) and so H3  H4 and so we may ignore H1; H4. Standard Induction (Section 2.1) gives the bound forb(m;F7) < m forb(m; fH2; H3; H5g). The next lemma states that forb(m; fH2; H3; H5g) has a linear bound, which means forb(m;F7) has a quadratic bound. Lemma 5.3.2 We have that forb(m; fH2; H3; H5g) is O(m). We will prove Lemma 5.3.2 in Section 5.3.3. We can now prove that forb(m;F7) is quadratic. Proof of Theorem 5.3.1: The fact that forb(m;F7) is  (m2) comes directly out of the conjecture, as F7  I  I. We show forb(m;F7) is O(m2) using induction on m. Consider A 2 Avoid(m;F7) with kAk = forb(m;F7). Then using Proposition 2.1.2, we have forb(m;F7) = kAk  m 1X i=1 forb(i; fH2; H3; H5g): Given that there is a constant c so that forb(i; fH2; H3; H5g)  ci by Lemma 5.3.2, we deduce the bound forb(m;F7) < c  m(m 1)=2 which is  (m2). Now consider any 5 1 column  . We immediately deduce that forb(m; [F7 j ]) is  (m3) for  having zero, one, four or  ve 1’s, or if  is a column in F7 (considered as a matrix). It is a computational exercise to show that every other  results in forb(m; [F7 j ]) being  (m3) and the computer program described in Section 3.4 indeed gave us this result. We give a proof here for completeness. We need only consider  having two 1’s since F c7 = F7. If  has 0’s on rows 2,3 then [F7 j ]  Ic  Ic  Ic (each pair of rows from the four rows 1,2,3,4 of [F j ] has (0; 0)T ) or two 0’s on rows 1,4 then [F7 j ]  Ic  Ic  Ic (each pair of rows from the four rows 1,3,4,5 has (0; 0)T ) . This only leaves four columns, three of which are already in F7. Only 104 = (0; 0; 1; 1; 0)T is not already in F7, and in such case [F7 j ]  T  T  T since every pair of rows from the four rows 1,2,3,4 has the 2 2 con guration I2.  5.3.2 What Is Missing? Applying the technique of What Is Missing described in Section 2.2 to F = fH2; H3; H5g, we get the following lemma. Lemma 5.3.3 Let A 2 Avoid(m; fH2; H3; H5g). Then there are 13 possibilities Q0, Q1, : : : ; Q12 for What Is Missing on each 4-set of rows. Q0 = no no no no no no 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 Q1 = no no no no no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 Q2 = no no no no no no 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 Q3 = no no no no no no 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 Q4 = no no no no 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 Q5 = no no no no no no no no 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 105Q6 = no no no no no no 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 Q7 = no no no no no no 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 Q8 = no no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 Q9 = no no no no no no 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 Q10 = no no no no no no 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 Q11 = no no no no no no 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 Q12 = no no no no no no no no 2 6 6 6 6 4 1 0 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 Proof of Lemma 5.3.3: An exhaustive computer search by the program described in Section 3.5 yields the result.  1065.3.3 Linear Bound for the Inductive Children The rest of the section is a proof of Lemma 5.3.2. Let A 2 Avoid(m; fH2; H3; H5g). We will use special features of H2; H3; H5 to obtain a linear bound on kAk. The forbidden con guration H3 is used most often in this proof. We will show kAk  7m by induction on m. We analyze the 13 cases of Lemma 5.3.3 one by one and have special arguments for the three troublesome cases Q2; Q3; Q11. Lemma 5.3.4 Let A 2 Avoid(m; fH2; H3; H5g). Consider the standard decomposition (2.1.1) of A based on row r. Let L(r) 6= ; be a minimal set of rows such that CrjL(r) is simple. Then each triple of rows fi; j; kg in L(r) yield a quadruple of rows fr; i; j; kg on which one of the cases Q2; Q3; Q11 occurs, with row r being the  rst row of each of the cases Q2; Q3; Q11 as given in Lemma 5.3.3. Proof: For each Qi we record pairs of rows containing \a copy of K2": namely in the columns marked absent we  nd r i j k no 2 6 6 6 6 4 a e 0 0 3 7 7 7 7 5 ; no 2 6 6 6 6 4 b f 1 0 3 7 7 7 7 5 ; no 2 6 6 6 6 4 c g 0 1 3 7 7 7 7 5 ; no 2 6 6 6 6 4 d h 1 1 3 7 7 7 7 5 : Suppose A had these columns missing on the quadruple of rows r; i; j; k and that rows i; j; k belong to L(r). Then the simple matrix Cr from (2.1.1) has the four 3  1 columns (e; 0; 0)T , (f; 1; 0)T , (g; 0; 1)T and (h; 1; 1)T missing on the triple of rows fi; j; kg. We deduce that row i cannot belong to L(r), a contradiction. By analyzing the cases, we  nd that Q0; Q1; Q5; Q6; Q7; Q8; Q10; Q12 have 3 rows each pair of which have a \K2" and Q4; Q9 have two disjoint pairs of rows each with a \K2". Thus in any of these cases, What Is Missing on a triple of rows in Cr will contain a copy of \K2" and so we can delete a row from Cr without disturbing simplicity of the remainder of Cr. In cases Q2; Q3; Q11, if we choose row r to be any row but the  rst row in each of the cases then there is a \K2" on the remaining triple.  We would like to show that for all A 2 Avoid(m; fH2; H3; H5g) we can choose row r so 107that kCrk  7 as in (2.1.1). Then by Proposition 2.1.2 and induction, kAk  7m. We will assume the contrary, namely that there is A 2 Avoid(m; fH2; H3; H5g) such that for every row r, kCrk  8. In each of the troublesome cases Q2; Q3; Q11, we end up with the following sets of columns missing on a triple of rows in Cr (arising from What Is Missing in A on a quadruple of rows involving r) and we name the cases correspondingly P2; P3; P11. Note the implicatoins arising from P3. P2 : no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 1 3 7 5 no 2 6 4 0 1 1 3 7 5 (5.3.1) P3 : i j k no 2 6 4 1 0 1 3 7 5 no 2 6 4 0 0 1 3 7 5 no 2 6 4 0 1 1 3 7 5 no 2 6 4 0 1 0 3 7 5 yielding i j k no 2 6 4 0 1 3 7 5 no 2 6 4 0 1 3 7 5 no 2 6 4 0 1 3 7 5 (5.3.2) P11 : no 2 6 4 1 0 0 3 7 5 no 2 6 4 0 1 0 3 7 5 no 2 6 4 0 0 1 3 7 5 (5.3.3) Lemma 5.3.5 Let A 2 Avoid(m; fH2; H3; H5g). Consider the standard decomposition (2.1.1) of A based on row r. Let L(r) 6= ; be a minimal set of rows such that CrjL(r) is simple. Then each triple of rows fi; j; kg in L(r) is in one of the cases P2, P3 or P11. Moreover, if any triple in L(r) is in case P2, then all triples of rows of L(r) are in case P2. Similarly if any triple in L(r) is in case P3 (respectively P11), then all triples of rows are in case P3 (resp. P11). Proof: By Lemma 5.3.4, every triple of rows of L(r) satis es one of P2, P3 or P11. A triple of rows fa; b; cg in case P3 can’t overlap with a triple of rows in case P2 (respectively P11) on two rows fa; bg since on the two rows fa; bg What Is Missing (by (5.3.2)) will extend to one 108new column missing on the triple from P2 (resp. P11) yielding a \K2". This would allow us to delete a further row from CrjL(r) while preserving simplicity, a contradiction to the fact that L(r) is minimal with CrjL(r) simple. Thus, if any triple of rows of L(r) is in case P3, then all triples of rows of L(r) are in case P3. Assume all triples of rows are in case P2 or P11. We can’t have a triple of rows in case P2 overlap with a triple of rows in case P11 on two rows as shown below. On the quadruple of rows we have marked ‘OK’ over the columns which can occur on the quadruple of rows. At most 6 columns can be present in CrjL(r) and we note that we can delete the second or third row from CrjL(r) and not a ect simplicity of CrjL(r), a contradiction. Hence such an overlap cannot occur. no no no no no no 1 1 0    1 0 1 1 0 0 0 1 1 0 1 0    0 0 1 OK OK OK OK OK OK 0 1 0 0 1 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 1 1 0 1 Given that each triple of the remaining rows of Cr rows must be in case P2 or P11, we must have all triples satisfy only one of the two.  Lemma 5.3.6 Assume all triples in L(r) are in case P3. Then the rows of Cr can be ordered so that each triple of rows a < b < c corresponds to a = i, b = j, and c = k in P3. Proof: In this case there is an ordering of the rows L(r) so that all triples are consistent with the ordering given. We had noted in (5.3.2) that having P3 on rows i; j; k in that order correspond to three columns, each on two rows, being absent. If we cannot  nd a consistent ordering of the rows of L(r), then on some pair of rows we will be missing two columns and this implies that one of the two rows can be deleted while preserving simplicity of CrjL(r). This contradiction proves the result.  In view of Lemma 5.3.5, we will say L(r) is type i if each triple of rows in L(r) is in case Pi for i = 2; 3 or 11. Recall we assumed kCrk  8. We obtain M(r) from L(r) as follows 109where the type of M(r) is the type of L(r). M(r) = ( L(r) if L(r) is type 2 or 11 L(r)nf rst and last row in orderingg if L(r) is type 3 (5.3.4) Note CrjM(r) need not be simple. Lemma 5.3.7 Let A 2 Avoid(m; fH2; H3; H5g) with (2.1.1) applied for row r and M(r) from (5.3.4). (i) If M(r) is type 2, then CrjM(r) must consist of [0jM(r)j IjM(r)j] and possibly column 1jM(r)j and no other column. Thus kCrk  2  jM(r)j  kCrk  1. In addition, columns of AjM(r) are from [0jM(r)j IjM(r)j 1jM(r)j]. (ii) If M(r) is type 11, then CrjM(r) must consist of [IcjM(r)j 1jM(r)j] and possibly column 0jM(r)j and no other column. Thus kCrk  2  jM(r)j  kCrk  1. In addition columns of AjM(r) are from [0jM(r)j IcjM(r)j 1jM(r)j]. (iii) If M(r) is type 3, then CrjM(r) must consist of [0jM(r)j TjM(r)j 1jM(r)j]. Thus jM(r)j = kCrk  3. In addition, columns of AjM(r) are from TjM(r)j. Proof: For M(r) being type 2, we observe that columns of CrjM(r) must belong to [0jM(r)j IjM(r)j 1jM(r)j]: By minimality of L(r) (which is M(r)), we cannot delete any rows from CrjM(r) and preserve simplicity. Thus all columns of [0jM(r)j IjM(r)j] must be present. A quick count reveals kCrk  2  jM(r)j  kCrk  1: Similarly for M(r) being type 11, CrjM(r) must consist of [IcjM(r)j 1jM(r)j]; and possibly column 0jM(r)j and no other column. For M(r) being type 3 then, with the row ordering of Lemma 5.3.6, CrjL(r) must consist of TjL(r)j. Hence CrjM(r) must consist of [0jM(r)j TjM(r)j 1jM(r)j]; 110and so jM(r)j = kCrk  3. The restricted columns on CrjM(r) extend to restricted columns on AjM(r) as follows. If M(r) is type 2 then for any H  M(r) with jHj = 3, the 6 forbidden columns on rows r[H of Q2 yield the restrictions P2 of 3 forbidden columns on rows H of A. Thus the columns of AjM(r) are all contained in [0jM(r)j IjM(r)j 1jM(r)j]. In a similar way, if M(r) is type 11 then the columns of AjM(r) are all contained in [0jM(r)j IcjM(r)j 1jM(r)j]. If L(r) is type 3 we noted CrjL(r) is TjL(r)j. Indeed, by Lemma 5.3.6, Q3 has each triple i; j; k 2 L(r) ordered consistent with the ordering of the rows of L(r) yielding T . We deduce the following columns are absent in A on rows i < j < k: i j k 2 6 4 1 0 1 3 7 5 i j k 2 6 4 0 1 0 3 7 5 The following two columns are also forbidden on the 4 rows r; i; j; k of A by Q3:  = r i j k 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5  = r i j k 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 Thus, using  , under the 0’s in row r in [BrCr]jL(r) we may only have the columns of TjL(r)j plus one additional column consisting of all 0’s except a 1 in the last row of L(r). Similarly using  , under the 1’s in row r in [CrDr]jL(r) we may only have the columns of TjL(r)j plus one additional column consisting of all 1’s except a 0 in the  rst row of L(r). Thus if M(r) is L(r) with the  rst and last row deleted then CrjM(r) = [0T 1] and the columns of AjM(r) are contained in [TM(r)].  Proof of Lemma 5.3.2: Let A 2 Avoid(m; fH2; H3; H5g). Use the decomposition of A given in (2.1.1). Our procedure is as follows. We use Lemma 5.3.5 to deduce the possible cases we need to consider. Under the assumption that kCrk  8 for all rows r, we will establish by induction an in nite sequence r1; r2; r3; : : : and associated sets of rowsN(r1); N(r2); N(r3); : : : with jN(ri)j  4 for each i. The sets N(r) di er very little from L(r) and M(r). We are able to show that the sets N(r1)nr2, N(r2)nr3, : : :, N(ri)nri+1 are all disjoint (see the beginning of Case 1a) and yet jN(rj)nrj+1j  3. This yields a contradiction (there are only m rows!) and so we may conclude that for some r, kCrk  7. Hence by our induction we deduce that 111kAk  7m. Assume for all rows r that kCrk  8 and hence  nd the sets M(r) with jM(r)j  5 (checking the three cases of Lemma 5.3.7). Let r1 be some row of A. We form M(r1). Note that if M(r1) was type 3 then we have deleted the  rst and last rows (in the ordering) from the originally determined L(r1). We determine the sets N(ri) from M(ri) as follows N(r) = ( M(r) if M(r) is type 2 or 11 M(r)n  rst row in ordering if M(r) is type 3 (5.3.5) Our general step commences with N(ri). We select a row ri+1 2 N(ri), making sure that when N(ri) is of type 3, we select the  rst row in the ordering of Lemma 5.3.6. We obtain M(ri+1) applying Lemma 5.3.4, Lemma 5.3.5, Lemma 5.3.6 and Lemma 5.3.7. Given our assumption that kCrk  8 we have jM(ri+1)j  5. Now by (5.3.5) we deduce jN(ri+1)j  4 in all cases. We hope identifying L(r);M(r); N(r) makes the proof clearer. To show the desired properties of the sets N(ri), we set up an inductive hypothesis concerning the structure of A. In what follows let Z denote a matrix of 0’s (or perhaps a matrix of no columns) and J denote a matrix of 1’s (or perhaps a matrix of no columns). The critical inductive structure is the following. The middle columns correspond to the columns of Cri as shown in (5.3.6). For each p with p < i with N(rp) of type 2 or 3 we have the structure shown in rows N(rp) n rp+1. For each q with q < i with N(rq) of 11 we have the structure shown in rows N(rq) n rq+1. We have three cases depending on the type of N(ri). When N(ri) is type 2: We have S = [0 I] or [0 I 1] and the columns of Ui and Vi are in [0 I 1]. When N(ri) is type 11: We have S = [Ic 1] or [0 Ic 1] and the columns of Ui; Vi are in [0 Ic 1]. When N(ri) is type 3: We have S = [0T 1] and the columns of Ui; Vi are in T . 112A = ri ! ... N(rp)nrp+1 f ... N(rq)nrq+1 f ... N(ri) f ... 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 0    0 0    0 0    0 1    1 1    1 1    1 ... Z W 0p Z Z W 1 p Z ... J W 0q J J W 1 q J ... Ui ZJ S S ZJ Vi ... ... ... ... ... ... 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 |{z} Cri |{z} Cri (5.3.6) We proceed to verify that we have the same inductive structure for ri+1. There will be cases to explore. It is helpful to display representatives of H2; H3; H5 that we will use in our arguments. For M(ri+1) type 2 or 11 we will use H2 = ri+1 s i j 2 6 6 6 6 4 0 0 0 1 1 1 0 0 0 0 1 1 0 1 0 0 0 1 0 1 3 7 7 7 7 5 ; H3 = ri+1 s i j 2 6 6 6 6 4 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 0 3 7 7 7 7 5 (5.3.7) H3 = ri+1 t i j 2 6 6 6 6 4 0 0 0 1 0 1 1 1 1 0 1 1 0 1 0 0 3 7 7 7 7 5 ; H5 = ri+1 t i j 2 6 6 6 6 4 0 0 1 1 1 1 1 1 1 0 0 1 0 1 1 1 0 1 0 0 3 7 7 7 7 5 (5.3.8) For M(ri+1) type 3 we will use H3 = ri+1 s i j 2 6 6 6 6 4 0 1 1 1 1 0 0 0 1 0 1 1 0 0 0 1 3 7 7 7 7 5 ; H3 = ri+1 s i j 2 6 6 6 6 4 0 1 1 1 0 0 0 1 1 0 1 1 1 0 0 0 3 7 7 7 7 5 (5.3.9) 113H3 = ri+1 t i j 2 6 6 6 6 4 0 0 0 1 0 1 1 1 1 0 1 1 1 0 0 0 3 7 7 7 7 5 ; H3 = ri+1 t i j 2 6 6 6 6 4 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 0 3 7 7 7 7 5 (5.3.10) Case 1: N(ri) is type 2. Begin with inductive structure of (5.3.6). Given N(ri) of type 2 we have S = [0 I] or [0 I 1]. Choose a row ri+1 2 N(ri). Now consider the decomposition (2.1.1) applied to A using row r = ri+1. Apply Lemma 5.3.4, Lemma 5.3.5, Lemma 5.3.6 and Lemma 5.3.7 to obtain M(ri+1). Case 1a: M(ri+1) is type 2. The columns of Cri+1 must appear once with a 0 in row ri+1 and once with a 1 in row ri+1. By Lemma 5.3.7 we know that columns of AjN(ri) are contained in [0 I 1]. The only columns of AjN(ri) which di er only in row ri+1 would be the column of 0’s and the column of all 0’s except a 1 in row ri+1. Thus the repeated columns of Cri+1 , when restricted to rows N(ri)nri+1, must be all 0’s. By examining (5.3.6), the only columns of A which on rows N(ri) that have a single 1 (on row ri+1) respectively on the rows N(ri) are the columns which are Z in rows N(rp)nrp+1 for those p < i with N(rp) being type 2 or 3 and J in rows N(rq)nrq+1 for those q < i with N(rq) being type 11. We need to show that N(ri+1) is disjoint from N(rj)nrj+1 for all j < i+ 1. All columns in W 0 or W 1 of (5.3.6) are either all 0’s or all 1’s on the rows of N(ri) and so won’t give rise to columns of Cri+1 . We deduce that the columns of Cri+1 are all 0’s in rows N(rp)nrp+1 for those p < i with N(rp) being type 2 or 3 and all 1’s in rows N(rq)nrq+1 for those q < i with N(rq) being type 11. Recalling that we form L(ri+1) by deleting rows of Cri+1 while preserving simplicity, we deduce that L(ri+1) (and hence M(ri+1) and N(ri+1)) is disjoint from N(rj)nrj+1 for all j < i+ 1. This gives us the structure of Cri+1 given below in (5.3.11) where the two copies of Cri+1 occupy the central columns. To complete (5.3.11) we de ne W 0 and W 1 (likely di erent from those in (5.3.6) in the paragraph above). We choose from the columns of Bri+1 and Dri+1 , all columns which for some ‘ < i, where N(r‘) is type 2 or 3 (and hence rows N(r‘) is Z in Cri+1), have a 1 in some row of N(r‘) or for some ‘ < i, with N(r‘) is type 11 (and hence rows N(r‘) is J in Cri+1), have a 0 in some row of N(r‘). We identify such columns in Bri+1 as W 0 and such columns in Dri+1 as W 1. Moreover let W 0t (respectively W 1 t ) denote the submatrix of W 0 (respectively W 1) in rows N(rt)nrt+1 for t = 1; : : : ; i or in rows M(rt) 114for t = i+ 1. All remaining columns of Bri+1 and Dri+1 are all 0’s on rows of each N(r‘) where N(r‘) is type 2 or 3 and all 1’s on rows of each N(r‘) where N(r‘) is type 11 for ‘ < i. ri+1 ! 0    0 0    0 0    0 1    1 1    1 1    1 ... N(rp)nrp+1 f Z W 0p Z Z W 1 p Z ... N(rq)nrq+1 f J W 0q J J W 1 q J ... N(ri)nri+1 f Z W 0i Z Z W 1 i Z M(ri+1) f Ui+1 W 0i+1 0 I 1 0 I 1 W 1 i+1 Vi+1 ... ... (5.3.11) By Lemma 5.3.7 we know that columns of AjN(ri) are contained in [0 I 1] and so we deduce that columns of Ui+1; Vi+1 are in [0 I 1]. Our remaining goal is to show that W 0i+1 = ZJ and W 1i+1 = ZJ to complete the induction. We will use the four forbidden matrices of (5.3.7), (5.3.8) which have been ordered and labelled to assist the reader in seeing the occurrence of the forbidden objects H2; H3; H5. Assume for some column  of W 0 that  has a 1 in row s 2 N(rp)nrp+1 where N(rp) is type 2 or 3. We will give this  rst case in greater detail. All columns of Cri+1 have 0’s in the rows of N(rp) and in particular in row s. Given that M(ri+1) is type 2 or 11 we deduce Cri+1jM(ri+1) contains either I or I c. Thus each pair of rows i; j 2 M(ri+1) will contain  1 0 0 1  in each copy of Cri+1 . We  nd the following entries in A in the rows r; s; i; j where the left column comes from  and the remaining columns are from the two copies of Cri+1 : ri+1 s i j 2 6 6 6 6 4 0 0 0 1 1 1 0 0 0 0 a 1 0 1 0 b 0 1 0 1 3 7 7 7 7 5 : If  a b  =  1 0  or  0 1  then we have a representative of H2 as noted in the left matrix of (5.3.7). Thus the column  which contains a 1 in some row s of W 0p must either be all 0’s or all 1’s on the rows M(ri+1). Assume for some column  of W 0 that  has a 0 in row t 2 N(rq)nrq+1 where N(rq) is type 11. Using the left matrix of (5.3.8) we may argue as above that column 115 must either be all 0’s or all 1’s on the rows M(ri+1). Given our choice of W 0, this is enough to show that W 0i+1 is ZJ . Assume for some column  of W 1 that  has a 1 in row s 2 N(rp)nrp+1 where N(rp) is type 2 or 3 and hence we  nd 0’s in Cri+1 in row s. Hence by the right matrix in (5.3.7) we cannot have the matrix ij  1 0  in  for any choices i; j 2 M(ri+1). As above, the column  is either all 1’s or all 0’s on the rows of M(ri+1). Similarly, using the right matrix of (5.3.8) , we can show that for any column  of W 1 that has a 0 in row t 2 N(rq)nrq+1 where N(rq) is type 11 that  cannot have the matrix i j  1 0  in  for any choices i; j 2 M(ri+1). Hence  is either all 0’s or all 1’s on the rows of M(ri+1). Thus W 1i+1 = ZJ as desired. Setting N(ri+1) = M(ri+1) results in the same structure of (5.3.6) with ri replaced by ri+1 and S = [0 I] or [0 I 1]. Case 1b: M(ri+1) is type 11. We can use the argument of Case 1a if M(ri+1) is type 11 since any two rows of Ic contain I2 allowing us to use the matrices of (5.3.7),(5.3.8) as above. We would obtain (5.3.6) with ri replaced by ri+1, N(ri+1) = M(ri+1) and S = [Ic 1] or [0 Ic 1]. Case 1c: M(ri+1) is type 3. We follow the argument at the beginning of Case 1a) to obtain most of the structure of (5.3.12). Given that we form L(ri+1) by deleting rows of Cri+1 while preserving simplicity, we deduce that L(ri+1) (and hence M(ri+1)) is disjoint from N(rj)nrj+1 for all j < i + 1. We will use (5.3.9) and (5.3.10) and, arising from the left matrix of (5.3.10), we discover a row of M(ri+1) that must be deleted. ri+1 ! 0    0 0    0 0    0 1    1 1    1 1    1 ... N(rp)nrp+1 f Z W 0p Z Z W 1 p Z ... N(rq)nrq+1 f J W 0q J J W 1 q J ... N(ri)nri+1 f Z W 0i Z Z W 1 i Z M(ri+1) f Ui+1 W 0i+1 0T 1 0T 1 W 1 i+1 Vi+1 (5.3.12) Do not be concerned that Cri+1 as shown is not simple, as we have deleted two rows from L(ri+1) to obtain M(ri+1) which are not displayed here. As before, we note that by Lemma 5.3.7, the columns of Ui+1; Vi+1;W 0i+1;W 1 i+1 are contained in T . Our goal to complete the induction is to show W 0i+1 = ZJ and W 1 i+1 = ZJ . We use the four forbidden matrices of 116(5.3.9),(5.3.10). Given that Cri+1 jM(ri+1) = [0T 1], each pair of rows i; j 2 M(ri+1) with i < j in the special row ordering of M(ri+1) will contain  0 1 1 0 0 1  in each copy of Cri+1 . If we have a column  of W 1 with a 1 in a row s 2 N(rj)nrj+1 where N(rj) is type 2 or 3 and hence we  nd 0’s in columns of Cri+1 in row s. Hence by the right matrix in (5.3.9),  cannot have the submatrix ij  1 0  for each pair of rows i; j 2M(ri+1) with i < j. Given that  jM(ri+1) is a column in T , we deduce that column  is either all 1’s or all 0’s on the rows of M(ri+1). If we have a column  of W 1 with a 0 in a row t 2 N(rj)nrj+1 where N(rj) is type 11, we  nd 1’s in row t of Cri+1 . Hence by the right matrix in (5.3.10),  cannot have the submatrix ij  1 0  for each pair of rows i; j 2M(ri+1) with i < j. As above, the column  is either all 1’s or all 0’s on the rows of M(ri+1). This considers all columns of W 1 and so W 1i+1 = ZJ . If we have a column  of W 0 with a 1 in a row s 2 M(rp)nrp+1 where M(rp) is type 2 or 3, we  nd 0’s in row s of Cri+1 . Hence by the left matrix in (5.3.9),  cannot have the submatrix ij  1 0  for each pair of rows i; j 2 M(ri+1) with i < j and so the column  is either all 1’s or all 0’s on the rows of M(ri+1). If we have a column  of W 0 with a 0 in row t 2 N(rq) where N(rq) is type 11 then we follow a di erent argument that we explain more carefully. For i; j 2M(ri+1) with i < j, we  nd the entries as given below in the rows ri+1; t; i; j in the given column  (the column on the left) and selected columns of Cri+1 (on the right). ri+1 t i j 2 6 6 6 6 4 0 0 0 1 0 1 1 1 a 0 1 1 b 0 0 0 3 7 7 7 7 5 If  a b  =  1 1  then this yields H3 in A as noted in the left matrix in (5.3.10). Now  jM(ri+1) is a column in T and yet cannot have the submatrix  1 1  . Thus  on the rows of M(ri+1) is either all 0’s or possibly the column of all 0’s except a single 1 in the  rst row of M(ri+1). It is for this case that we need to delete the  rst row of M(ri+1) to obtain N(ri+1) (as in (5.3.5)) so that on the rows N(ri+1), the matrix W 0i+1 = ZJ . We now have obtained (5.3.6) with ri replaced by ri+1, and S = T . Case 2: N(ri) is type 11. We use the same argument as Case 1. When N(ri) is type 11 we would have to replace I by Ic in S in (5.3.6) and then proceed to M(ri+1) of type 2 or 11 (essentially Case 1a or 1b) or M(ri+1) of type 3 (essentially Case 1c). 117Case 3: N(ri) is type 3. Begin with inductive structure of (5.3.6) where N(ri) is type 3 and S = [0 0T 1 1]. Now choose the  rst row ri+1 2 N(ri) using the ordering on N(ri). Now consider Standard Induction applied to A using row ri+1. We deduce that in rows N(ri)nri+1, the repeated columns in Cri+1 are Z since for a column to be repeated it extension to row ri+1 with both a 0 and a 1 must be present in Cri+1 . Given that the repeated columns under the 1’s in row ri+1 must correspond to columns of a single 1 on rows N(ri)nri+1 and by (5.3.6) that means we can deduce the structure of the other rows of the columns in Cri+1 . Note that in what follows I have rearranged the columns of Bri+1 and Dri+1 so that we have put in the columns of the Wj’s all columns which either have a 0 in a row of N(rj) where N(rj) is type 2 or 3 (and hence is Z in Cri+1), and all columns which have a 1 in a row of N(rj) where N(rj) is type 11 (and hence is J in Cri+1). This yields (5.3.11) or (5.3.12) . If M(ri+1) is type 2 we follow the same argument as in Case 1a) to deduce that W 0i+1 and W 1i+1 have only constant columns. Similarly the case M(ri+1) is type 11 can use the argument of Case 1b) by switching I with Ic. In either case we set N(ri+1) = M(ri+1). If M(ri+1) is type 3, we follow the same argument as in Case 1c) and again may have to delete the  rst row of M(ri+1) to obtain N(ri+1) and yields (5.3.6) with ri replaces by ri+1. This concludes the induction and so have proven that we can  nd rows r1; r2; r3; : : : and disjoint sets jN(ri)nri+1j  3 yielding a contradiction. As noted this proves the result.  We still have eight 5  6 simple F for which the conjecture predicts they are boundary cases with forb(m;F ) being O(m2). Given the complicated case analysis of this proof, it seems challenging to prove such bounds. One positive observation is that Lemma 5.3.5 may not be necessary. We were only interested in having a large set L(r), say jL(r)j  8, for which each triple is in a given case. We could appeal to Ramsey Theory [Ram30] and given a  nite number of cases, we can identify a large (!) constant c so that if kCrk  c then there are say 8 rows such that every triple is in the same case and in the same row ordering. This would avoid appealing to the particular structures of cases P2; P3; P11 but is not advantageous for our proof. 5.4 Classi cation of 6-rowed Quadratic Bounds This section is devoted to proving a quadratic bound for the 6  3-rowed con guration G6 3. We prove forb(m;G6 3) is  (m2), and furthermore, we prove G6 3 is the only 6-rowed 118boundary quadratic case, therefore classifying all quadratic 6-rowed cases. The following theorem classi es all 6-rowed con gurations F for which forb(m;F ) is  (m2) by giving the unique boundary case. Theorem 5.4.1 Let F be any 6-rowed con guration. Then forb(m;F ) is  (m2) if and only if F is a con guration in G6 3 = 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 3 7 7 7 7 7 7 7 7 7 5 : Furthermore, if F  G6 3, then forb(m;F ) is  (m3). Given the classi cation and Remark 1.3.22, we are not surprised that Gc6 3 = G6 3. Anstee and Keevash [AK06] established the asymptotic bounds for all k 2 con gurations, and in particular concluded that forb  m; 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 0 0 1 0 0 0 0 3 7 7 7 7 7 7 7 7 7 5 ! and forb  m; 2 6 6 6 6 6 6 6 6 6 4 1 1 1 0 1 0 0 1 0 1 0 0 3 7 7 7 7 7 7 7 7 7 5 ! are both  (m2): The proof of the second of these begins to use the full power of the proof in [AK06] and so it is interesting that Theorem 5.4.1 provides a generalization for both of them using an inductive proof (admittedly, a rather complicated one for Theorem 5.3.1) that is quite di erent than that in [AK06]. In order to prove Theorem 5.4.1, we will use three results. First, Lemma 5.4.2 is the \only if" part of the theorem. The second, Lemma 5.4.3, generalizes Lemma 3.2 in [AK06]. Lastly, we will use the main result of the previous chapter, Theorem 5.3.1. 119Lemma 5.4.2 Let F be a 6-rowed con guration such that F  G6 3. Then forb(m;F ) must be  (m3). Proof: We may assume all of F ’s columns have column sum 3, otherwise, if F had a column of column sum 4 or more, then F  I  I  I, and if F had a column sum of 2 or less, then F  Ic  Ic  Ic. Without loss of generality, let the  rst column of F be (1; 1; 1; 0; 0; 0)T . With these assumptions, there are only a few cases left to check, and an exhaustive computer search revealed the lemma to be true. But we present here a more constructive proof, if for no other reason than to check the computer code. Note that the following 2-columned matrices have at least a cubic bound: 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 1 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 5  I  I  I; 2 6 6 6 6 6 6 6 6 6 4 1 0 1 0 1 0 0 1 0 1 0 1 3 7 7 7 7 7 7 7 7 7 5  I  I  T: This means that to in order to have F for which forb(m;F ) is not  (m3), we must put together columns of sum 3 such that for each pair of columns, the number of rows where both columns have 1’s is either one or two. Here are all the possibilities for (the  rst) two columns having 1’s in (the  rst) two rows in common: 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 3 7 7 7 7 7 7 7 7 7 5 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 0 1 0 1 0 1 1 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 5 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 3 7 7 7 7 7 7 7 7 7 5 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 0 1 0 0 0 1 0 0 0 1 0 0 1 3 7 7 7 7 7 7 7 7 7 5 2 6 6 6 6 6 6 6 6 6 4 1 1 0 1 1 0 1 0 1 0 1 1 0 0 1 0 0 0 3 7 7 7 7 7 7 7 7 7 5  Ic  Ic  Ic  I  I  I = G6 3  Ic  Ic  Ic  I  I  I : 120The only other possibility is that each pair of columns has a 1 in only one row in common. 2 6 6 6 6 6 6 6 6 6 4 1 1 0 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 3 7 7 7 7 7 7 7 7 7 5  I  I  T: Thus, the only four-columned matrices F for which forb(m;F ) could be O(m2) have to contain G6 3 in every three-columned subset. The only possibility is then 2 6 6 6 6 6 6 6 6 6 4 1 1 1 0 1 1 0 1 1 0 1 0 0 1 0 1 0 0 1 1 0 0 0 0 3 7 7 7 7 7 7 7 7 7 5  I  I  T ; which means forb(m;F ) is  (m3) for the above matrix F . This concludes the lemma.  The following lemma generalizes Lemma 3.2 in [AK06]. Lemma 5.4.3 Let F = 2 6 4 0    0 1    1 F 0 3 7 5 : Then we can conclude that forb(m;F )  forb  m; " 1    1 F 0 #! + forb  m; " 0    0 F 0 #! : Proof: Let A 2 Avoid(m;F ) with kAk = forb(m;F ). Then permute the columns of A (take 121another representative in the equivalence class) and write it as A = " 0    0 1    1 A0 A00 # : Note that A0 and A00 are simple. Since A0 cannot have " 1    1 F 0 # as a subcon guration, and A00 cannot have " 0    0 F 0 # as a subcon guration, the bound follows.  From the previous lemma, we note that G6 3 has a row of 0’s and a row of 1’s, and therefore the quadratic bound for forb(m;G6 3) would follow from quadratic bounds for forb(m;G) and forb(m;G0), with G and G0 obtained by removing the row of 1’s and the row of 0’s from G6 3 respectively: G = 2 6 6 6 6 6 6 4 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 3 7 7 7 7 7 7 5 and G0 = 2 6 6 6 6 6 6 4 1 1 1 1 1 0 1 0 1 0 1 0 0 0 1 3 7 7 7 7 7 7 5 : We will prove more, as both are contained in the boundary case F7. Observe that G0 = Gc as con gurations. We are now ready to prove Theorem 5.4.1. Proof of Theorem 5.4.1: To prove forb(m;G6 3) is O(m2) we use Lemma 5.4.3 and so we only need to prove forb(m;G) and forb(m;G0) are both O(m2). We check that G  F7 and G0  F7. Now Theorem 5.3.1 shows that forb(m;F7) is O(m2). Applying Remark 1.3.21 yields the bound for G6 3. Lemma 5.4.2 veri es that every con guration F not contained in G6 3 has forb(m;F ) being  (m3).  122Chapter 6 Patterns and Splits 6.1 Patterns and Splits in 2-Dimensions We move away from the world of Forbidden Con gurations to consider another extremal problem which will have applications to Forbidden Con gurations (in Chapter 7), but no knowledge of Forbidden Con gurations is required in order to understand this chapter. Instead of considering the maximum number of columns an m-rowed f0; 1g-matrix can have, we are going to consider the maximum number of 1’s an m n f0; 1g-matrix can have subject to some property. These problems are about geometric patterns in a grid and are close relatives of Zarankiewicz’ problem [KST54],[F ur96] and the investigations of patterns in [FH92],[MT04],[Tar05],[KM07]. Consider the grid [m] [n] in the Euclidean 2-dimensional space. Let A  [m] [n]. Any such subset can be represented by an m  n f0; 1g-matrix A, where A has a 1 in position (x; y) i (x; y) 2 A. In order to give the following de nitions, we  nd it convenient to talk about the subset A instead of the matrix A, but in our applications and proofs it becomes more convenient to talk about the matrix A. Since A and A are equivalent, we will switch back and forth. Sometimes it will be more convenient to use the language of grids and other times to use the language of matrices. Recall that  1(A) denotes the number of 1’s of A. Observe that  1(A) = jAj. We consider horizontal and vertical lines on integer values (i.e. y = ‘ or x = ‘ for some integer k). For a horizontal line y = ‘, we partition the points of the grid into two parts. The  rst part consists of those points which lie in the bottom region of the line (points (x; y) such that y < ‘) and the second part consists of points which lie above the line or in the line itself (points (x; y) with y  ‘). 123Let p and q be numbers. If we consider p 1 horizontal lines and q  1 vertical lines in the plane, we naturally divide the plane into p  q rectangular regions. The horizontal lines can be represented by p  1 numbers h1 < h2 < ::: < hp 1 and the vertical lines by q  1 numbers v1 < v2 < ::: < vq 1. Set h0 = v0 = 1 and set hp = n, vq = m. Then the regions are of the form R(i; j) := [hi; hi+1) [vj; vj+1). De nition 6.1.1 Let p and q be numbers an let A  [m] [n]. We say A has a (p; q)-split if there exist p vertical lines and q horizontal lines such that for every one of the pq regions R(i; j) we have R(i; j) \ A 6= ;. Below is an example of a 3; 3-split in A, where a 1 from each block is indicated 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 1 1 1 1 3 7 7 7 7 7 7 7 7 7 5 A 3,3 split De nition 6.1.2 Let NoSplit(m;n; p; q) denote the maximum number of 1’s in a m  n f0; 1g-matrix that does not have a p; q-split. Here is a de nition that is related to splits, but di ers. Note that row and column order is important. De nition 6.1.3 Let F be a f0; 1g-matrix. We say a f0; 1g matrix A has F as a pattern if there is a submatrix G of F with F  G. That is, if the entry (i; j) is 1 in F then entry (i; j) is also 1. The following theorem was proven by Marcus and Tardos, but using di erent language. Theorem 6.1.4 Marcus and Klazar[KM07], Marcus and Tardos [MT04], Balogh, Bol- lob as and Morris [BBM06]. Let k be given. Then there exists a constant ck such that NoSplit(n; n; k; k)  ckn: 124The result in [MT04] involving forbidden permutation patterns implies the above result by choosing the permutation appropriately. Moreover the proof of [MT04] directly extends to the above result. The papers [KM07], [BBM06] note this as well as deriving higher dimension generalizations. While the constants involved in [MT04] are not optimal (in fact, they are very far from optimal), we can produce best possible constants for small values: Theorem 6.1.5 Let m;n be given with m;n  2. Then NoSplit(m;n; 2; q) = m+ (q  1)  n (q  1) and NoSplit(m;n; 3; q) = 2m+ (q  1)n 2(q  1). Proof: For any m;n; p; q, an m n matrix A with (p 1)m+ (q  1)n (p 1)(q  1) 1’s can be constructed with 1’s in the  rst p 1 rows and the  rst q  1 columns. Then A has no p; q split. We show an example of this for m = 7; n = 6; p = 3 and q = 2. 2 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 5 Of course the matrix shown above has no 3; 2-split. If we take any two horizontal lines and one vertical line, the bottom-right region will have no 1’s. This yields a lower bound for NoSplit(p; q;m;n). For p = 2, consider a matrix A0 by  rst deleting the bottommost 1 from each column. Thus the bottom row has all 0’s. At most m 1’s were deleted. Then delete the q 1 rightmost 1’s from each non-zero row. At most we delete (n 1)(q  1) 1’s. Then we have  1(A)   1(A 0) +m+ (n 1)(q  1) Now, if  1(A0) > 0, pick a 1 from A0. It must have q  1 1’s to its right, and each of those must have a 1 below. This produces a 2; q split, a contradiction. For p = 3 we proceed in a similar way. Form A0 by deleting from A the bottommost 1 from each column. Then delete the topmost 1 from each column. Note that the bottom row as well as the top row cannot have 1’s. Then delete the q 1 rightmost 1’s in each non-zero 125row. So  1(A)   1(A 0) + 2m+ (n 2)(q  1) Now, if  1(A0) > 0, pick a 1 from A0. It must have (q  1) 1’s to its right, and each of those must have a 1 below and a 1 above. This produces a 3; q split, a contradiction.  This proof technique was introduced to the authors by Jozsef Solymosi as a curling technique (the winter sport of curling uses a strategy called ‘peeling’). We do not have good bounds for NoSplit(m;n; p; q) with p; q  4. We conjecture that NoSplit(m;n; 4; 4) = 3m + 3n + min(m;n)  13. We can prove that NoSplit(m;n; 4; 4)  3m+ 3n+ min(m;n) 13 by giving an example with no (4,4)-split. Here is the example for m = n = 8, but it can be easily generalized. 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 The construction above can be generalized for bigger p; q. We believe this to be the best possible construction, but have so far been unable to prove it. Conjecture 6.1.6 Let n; p be given. Then for p  4, NoSplit(n; n; p; p) = (3p 5)n+O(p2) The lower bound is given by a construction like the one given above for p = 4; q = 4. 126We use the following lemma about a particular 3; 3 split in Chapter 7. Lemma 6.1.7 Let A be a f0; 1g-matrix with  1(A) > 4m 4. Then A has a special 3,3 split for which the 1’s in rows are in the same line. In other words, A has nine 1’s that satisfy the following structure, 1            1  ! 1 " 1     1 ! 1 # 1  1        ! 1 : Proof: The proof is the same as Theorem 6.1.5. Form a matrix A0 from A by deleting from A in turn the leftmost and rightmost 1’s in each row and then delete topmost and bottommost 1’s in each column. We have that  1(A)   1(A0) + 4m 4. Hence  1(A0) > 0. Select a 1 in A0. There is a 1 above and a 1 below. Then for each of these three 1’s, there is a 1 to the right and a 1 to the left. This yields the desired structure.  We also study a particular 2; 4 split. Lemma 6.1.8 There exists a f0; 1g-matrix A with  1(A) being  (m (m)), where  (m) is the inverse Ackermann function, and which doesn’t have a 2; 4 split on which the topmost 1’s lie in the same row and the bottommost 1’s lie in the same row as well. In other words, there are no eight 1’s with the following structure: 1     ! 1          ! 1         ! 1 1      ! 1   ! 1          ! 1 The following observation was suggested to us by Tardos. Observe that if A has a special 127split like the one described above, then it must also have the pattern P = " 1 0 1 0 0 1 0 1 # The following proposition of F uredi and Hajnal provides a construction. Proposition 6.1.9 [FH92] There exists a f0; 1g-matrix A with no pattern P for which  1(A) is  (m (m)), where  (m) denotes the inverse Ackermann function. Since the Ackermann function grows very quickly the inverse grows very slowly, but it still is bigger than a constant. This means m (m) is more than linear. 6.2 Patterns and Splits in d-Dimensions The papers [KM07],[BBM06] consider Theorem 6.1.4 generalized to d-dimensional arrays. Unfortunately, the notation for talking about these splits becomes admittedly cumber- some as we consider many dimensions, but the ideas remain simple. We use the following no- tation. Given integers n1; n2; : : : ; nd we can consider the positions Qd i=1[ni] in an n1 n2     nd f0; 1g-array A. Our main interest is in the case n1 = n2 =    = nd. Let p1; p2; : : : ; pd  2 be given. Assume we have d sets of indices I(j) = fr1(j); r2(j); : : : ; rpj 1(j)g for coordinate j, for j = 1; 2; : : : ; d. We can form d sets R1(j); R2(j); : : : ; Rpj(j) with [ pj i=1Ri(j) = [ni] as follows: R1(j) = f1; 2; : : : ; r1(j)g, R2(j) = fr1(j) + 1; r1(j) + 2; : : : ; r2(j)g,..., Rpj(j) = frpj 1(j) + 1; rpj 1(j) + 2; : : : ; njg. We generalize to many dimensions the notion of splits and NoSplit. De nition 6.2.1 We say A has a p1; p2; : : : ; pd split if we can choose the sets as above and for each j 2 [d] and for each possible choice t 2 [pj] with R(j) = Rt(j), the Qd i=1 pj block AjR(1) R(2)     R(d) contains at least one 1. Perhaps an example would be useful to understand this rather cumbersome notation. 128Example 6.2.2 The following matrix R1(1) R2(1) R3(1) R1(2) R2(2) 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 in [8]  [8] has a 3,2 split, where p1 = 3, p2 = 2, and I(1) = f3; 6g and I(2) = f3g, and so R1(1) = f1; 2; 3g, R2(1) = f4; 5; 6g, R3(1) = f7; 8g, R1(2) = f1; 2; 3g, R2(2) = f4; 5; 6; 7; 8g, De nition 6.2.3 Let NoSplit(n1; n2; : : : ; nd; p1; p2; : : : ; pd) be the maximum number of 1’s in n1  n2      nd f0; 1g-array that has no p1; p2; : : : ; pd split. The following describes the asymptotic behavior of NoSplit. Theorem 6.2.4 Klazar and Marcus [KM07], Balogh, Bollab as and Morris [BBM06]. Let k; d be given. Then there exists a constant ck;d so that NoSplit( d z }| { n; : : : ; n; d z }| { k; : : : ; k)  ck;dnd 1: We may also extend the argument in Theorem 6.1.5 to d 1 z }| { 3; 3; : : : ; 3; q splits of d-dimensional arrays. It is surprising that we get exact results here yet the results of [KM07] and [BBM06] do not have a reasonable bound for NoSplit(m;n; 4; 4). Theorem 6.2.5 Let B be the Qd i=1[ni] f0; 1g-array with 1’s in entries whose j coordinate is 1 or 2 for some j = 1; 2; : : : ; d 1 or whose dth coordinate is 1; 2; : : : or q 1. The matrix B has no d 1 z }| { 3; 3; : : : ; 3; q split. Let A be any Qd i=1[ni] f0; 1g-array with no d 1 z }| { 3; 3; : : : ; 3; q split. Then  1(A)   1(B); and hence for q = 3, NoSplit(n1; n2; : : : ; nd; d 1 z }| { 3; 3; : : : ; 3; q) =  1(B). 129Proof: Let A be any Qd i=1[ni] f0; 1g-array with no d 1 z }| { 3; 3; : : : ; 3; q split. For each direction ei with i = 1; 2; : : : ; d 1, remove from A the two 1’s with the smallest and largest coordinate value xi. Finally in the direction ed, remove the largest q  1 points in each line in the direction ed. Let A0 be the resulting matrix. Now if A0 has a 1 in position (y1; y2; : : : ; yd) then we note that A has 1’s in positions (y1; y2; : : : ; yd(j)) for j 2 [q] where yd = yd(1) and yd(1) < yd(2) <    < yd(q). It is now straightforward to show that choosing indices I(1) = fy1  1; y1g, I(2) = fy2  1; y2g, ..., I(d 1) = fyd 1  1; ydg and I(d) = fyd(1); yd(2); : : : ; yd(q  1)g yields a d 1 z }| { 3; 3; : : : ; 3; q split. We can show that  1(B)   1(A)  1(A0), hence if  1(A) >  1(B), then A would have the desired split.  The following notation is helpful. De nition 6.2.6 Thinking of the positions in B as elements of [n]d, we let the coordinates of B be x1; x2; : : : ; xd and for a position y 2 [n] d we de ne xi(y) to be the value of coordinate xi in y. Let proji(B) denote the [n] d 1 (d 1)-dimensional f0; 1g-array C obtained from B by projecting in the direction ei (where ei is the d-dimensional f0; 1g-vector with a single 1 in coordinate xi). For each y 2 [n]d we form y i in [n]d 1 by deleting the ith coordinate of y and placing a 1 in a position y i of C if and only if there is at least one 1 in a position z 2 [n]d of B with z i = y i. We are able to use the proof technique of Theorem 6.1.5 to obtain the following result about arrangements of 1’s. Lemma 6.2.7 Let C be an (n=3) (n=3) (n=3) 3-dimensional f0; 1g-array with more than 6(n=3)2 1’s. Then there are twenty-seven 1’s as follows. There are three values a; b; c for x1 coordinate and each plane x1 = a, x1 = b and x1 = c of C contains nine points. The nine points in each plane form a special 3,3 split as in Lemma 6.1.7 with the central 1’s having the same x2; x3 coordinates in each of the three planes. Proof: Form a matrix C 0 from C by deleting from C in turn the top and bottom 1 in each line in direction x3, the top and bottom 1 in each line in the direction x2 and the top two 1301’s in each line in the direction x1. We obtain  1(C)   1(C 0) + 6(n=3)2: If C 0 has a 1 in position y1, we can  nd twenty-seven 1’s yielding a special 3,3,3 split as follows. There are 2 1’s of C in positions y2;y3 with x1(y1) < x1(y2) < x1(y3), x2(y1) = x2(y2) = x2(y3) and x3(y1) = x3(y2) = x3(y3). Then there are 1’s of C in positions xj, zj for j = 1; 2; 3, where x2(xj) < x2(yj) < x2(zj) and x1(xj) = x1(yj) = x1(zj), x3(xj) = x3(yj) = x3(zj). Now we obtain positions vj;v00j for j = 1; 2; 3 and v = x;y; z with x3(v0j) < x3(vj) < x3(v 00 j ) and x1(v 0 j) = x1(vj) = x1(v 00 j ), x2(v 0 j) = x2(vj) = x2(v 00 j ). In particular there are three planes x1 = a, x1 = b, x1 = c each with nine 1’s and each plane has a special 3,3 split as in Lemma 6.1.7 with the central 1’s of each plane (namely y1;y2;y3) having the same x2; x3 coordinates and the horizontal direction corresponding to the x2 direction.  131Chapter 7 Products In this chapter we will study another related extremal problem related to Conjecture 1.4.1. Instead of considering how many columns out of all possible columns we can have avoiding a con guration F , we now turn our attention to the problem of  nding how many columns we can have out of only a particular, restricted set of columns, while avoiding F . For this we will make heavy use of splits and patterns of Chapter 6. At the end of this section we conclude with forbidden con guration bounds for certain families arising as products. 7.1 Introduction The following are the maximal 2-rowed simple submatrices of the matrices I; T; Ic. Let E1 = " 0 1 0 0 0 1 # ; E2 = " 0 1 1 0 0 1 # ; E3 = " 1 0 1 0 1 1 # Note that E2 = T2. Let T 0m = Tm  0m. That is, T 0 m is the tower matrix Tm except for the column of 00s. We now give an analogous de nition to forb. De nition 7.1.1 Let F be a family of con gurations, and let P be an m-rowed f0; 1g- matrix. We de ne MaxChoiceCols(F ; P ) = maxfkAk : A  P and A 2 Avoid(m;F)g: Notice that according to this de nition, forb(m;F ) = MaxChoiceCols(F;Km). 132Theorem 7.1.2 MaxChoiceCols(E1  E1; Im=2  Im=2) is  (m3=2). Theorem 7.1.3 MaxChoiceCols(E1  E2; Im=2  T 0m=2)  2m. Theorem 7.1.4 MaxChoiceCols(E2  E2; T 0m=2  T 0 m=2)  2m. The bound of Theorem 7.1.2 is perhaps unexpected in view of Conjecture 1.4.1 but it is not a counterexample. The remaining three cases (E1 E3 in Im=2 Icm=2 , E2 E3 in T 0 m=2 I c m=2 and E3 E3 in Icm=2  I c m=2) essentially follow by taking appropriate f0; 1g-complements. The proof of Theorem 7.1.4 is in Section 7.2, the proof of Theorem 7.1.2 is in Section 7.3 and the proof of Theorem 7.1.3 is in Section 7.4. Related results such as Theorem 7.1.5 MaxChoiceCols(E1  E2  E3; Im=3  T 0m=3  I c m=3) is  (m 2). are proved in Section 7.5. A central idea to studying these products is to encode columns of a p-fold product A1 A2     Ap as entries in a p dimensional f0; 1g-array B whose ith coordinate is indexed by the columns of Ai. Then the problem of  nding MaxChoiceCols(F;A1  A2      Ap) can be transformed to the problem of  nding the maximum number of 1’s in a f0; 1g-array that avoids a certain pattern of 1’s. In Chapter 6 we gave results about patterns that we will use in this chapter. The following basic result is proven in Section 7.2. 133Proposition 7.1.6 Let p; q; r; u; v; w be given positive integers. De ne x+ = maxf0; xg. The con guration given by the product F (u; v; w) = u z }| { E1      E1 v z }| { E2      E2 w z }| { E3      E3 is contained in the product Pm(p; q; r) = p z }| { Im      Im q z }| { T 0m      T 0 m r z }| { Icm      I c m for some m if and only if 2  (u p)+ + (v  q)+ + (w  r)+   (p u)+ + (q  v)+ + (r  w)+: For example with u = 2; q = 3 and the rest being 0, Proposition 7.1.6 yields that E1  E1  T 0  T 0  T 0 and hence forb(m;E1  E1) is  (m3). Proof: We note that any row from Ei contains [0 1] and also note that [0 1]  [0 1] = K2. Since none of our product terms I; T 0; Ic contain K2 then two rows of F chosen from two di erent products of the u+v+w 2-rowed products, will necessarily contain K2. This implies that if F is contained in the (p+ q + r)-fold product p z }| { I      I  q z }| { T 0      T 0 r z }| { Ic      Ic; then each product I, T 0, Ic has at most 2 rows of F and if it has two rows then they come from the same 2-rowed product term Ei of F . Of the three matrices I; T 0; Ic, we note that we can  nd E1 only in I, E2 only in T 0 and E3 only in Ic.  We now consider forbidden families of con gurations. It was noted in [AF86] that forb(m; fE1; E2; E3g) = 2. Balogh and Bollob as [BB05] have the much more general re- sult that for a given k, there is a constant ck such that forb(m; fIk; T 0k; I c kg) = ck. Let fE1; E2; E3g  fE1; E2; E3g denote the 6 possible 2-fold products whose terms are chosen from fE1; E2; E3g. We would like to compute forb(m; fE1; E2; E3g  fE1; E2; E3g) but in the interest of a more tractable proof we consider I2 as a replacement for both 134E1 and E3 (Ic2 = I2) and T 0 2 as replacement for E2. It is likely (but unknown) that forb(m; fE1; E2; E3g  fE1; E2; E3g) is O(m3=2). One might ask the relationship of The- orem 7.1.7 to Conjecture 1.4.1. The Conjecture (which applies only for a single forbidden con guration) says that only product constructions are needed for best possible asymptotic bounds, but this case fI2  I2; I2  T 02; T 0 2  T 0 2g are simultaneously missing from all 1-fold products but not simultaneously missing from any 2-fold product of I, Ic and T . In particu- lar I I avoids I2 T 02 and T 0 2 T 0 2 but does not avoid I2 I2 (Proposition 7.1.6). Surprisingly there is an O(m3=2) construction contained in I I and yet avoiding I2 I2 (Theorem 7.1.2) and of course also avoiding I2  T 02, and T 0 2  T 0 2. The other 2-fold products I2  T 0 2 (The- orem 7.1.3) and T 02  T 0 2 (Theorem 7.1.4) behave as the conjecture might suggest. We note forb(m; fI2; T 02g) = 2 and the bounds of Theorem 7.1.2, Theorem 7.1.3 and Theorem 7.1.4 apply. In Section 7.6 we prove a rather surprising result in which the exponent turns out to be fractional, when in most other cases the exponents where always integral: Theorem 7.1.7 forb(m; fI2  I2; I2  T 02; T 0 2  T 0 2g) is  (m 3=2). Let I2  I2 = 2 6 6 6 6 4 1 1 0 0 0 0 1 1 1 0 1 0 0 1 0 1 3 7 7 7 7 5 ; I2  T 0 2 = 2 6 6 6 6 4 1 1 0 0 0 0 1 1 1 1 1 1 0 1 0 1 3 7 7 7 7 5 ; T 02  T 0 2 = 2 6 6 6 6 4 1 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 3 7 7 7 7 5 : (7.1.1) We make use of our Standard Induction (Section 2.1) in a way similar to the way it was used in Section 5.3.3. 7.2 Submatrices of TxT In this section we’ll show how to exploit the results about splits in the context of forbidden con gurations. Proof of Theorem 7.1.4 that MaxChoiceCols(E2  E2; T 0m=2  T 0 m=2)  2m: Let F = E2 E2. Recall the ith column of T 0k is the column with 1’s in rows 1; 2; : : : ; i and 0’s in the 135remaining rows. Let A be an m-rowed submatrix of T 0m=2 T 0 m=2. We can create an m=2 m=2 f0; 1g-matrix B from A by placing a 1 in position r; c if A has the column obtained from the rth column of T 0m=2 placed on top of the cth column of T 0 m=2 namely the column with 1’s only in rows 1; 2; : : : r and m+ 1;m+ 2; : : : ;m+ c. We note that kAk is  1(B). We claim that F  A if and only if B has a 3; 3 split. The only way for a submatrix of T 0m=2 T 0 m=2 to be a row and column permutation of F is to lie in rows r1; r2;m=2+c1;m=2+c2 for some choices 2  r1 < r2  m=2 and 2  c1 < c2  m=2 (using the argument of Proposition 7.1.6 and noting that  rst row of T 0m=2 is 1’s). We have that any two rows of T 0m=2 (not including the  rst) have a copy of E2. We note that the t th column of T 0 m=2 on rows r1; r2 (with r1 < r2) has r1 r2 t " 0 0 # for 1  t < r1; r1 r2 t " 1 0 # for r1  t < r2; and r1 r2 t " 1 1 # for r2  t: Assume A has a copy of F in the 4 rows r1; r2;m=2 + c1;m=2 + c2. We discover that the nine columns of F would correspond to nine 1’s, one 1 in each of the nine blocks in the 3; 3 split of B given by I(1) = fr1  1; r2  1g and I(2) = fc1  1; c2  1g (notation from Chapter 6). Similarly a 3; 3 split of B yields a copy of F in A. We now appeal to the bound in Lemma 6.1.5.  An immediate generalization is the following. Lemma 7.2.1 We have MaxChoiceCols( d z }| { E2  E2      E2; d z }| { T 0m=d  T 0 m=d      T 0 m=d) is  (md 1). Proof: We use the d-dimensional generalization of splits Theorem 6.2.4 where the d-fold product E2  E2      E2 will correspond to a d z }| { 3; 3; : : : ; 3 split. We have an exact bound from Theorem 6.2.5 if needed.  A further generalization considers the matrix Tk. 136Lemma 7.2.2 We have that MaxChoiceCols( d z }| { Tk  Tk      Tk; d z }| { T 0m=d  T 0 m=d      T 0 m=d) is equal to NoSplit(m=d; m=d; : : : ; m=d; k + 1; k + 1; : : : ; k + 1); and so is  (md 1). Proof: We again use the d-dimensional generalization of splits Theorem 6.2.4 where the d-fold product Tk  Tk      Tk will correspond to a d z }| { k + 1; k + 1; : : : ; k + 1 split.  A rather interesting version of Theorem 7.1.4 and Lemma 7.2.1 that uses the idea of ‘peeling’ from Theorem 6.1.5 is the following. Lemma 7.2.3 f(E2  E2; T 0m=p  T 0 m=p      T 0 m=p)  2  4 p 2m. Proof: Let F = E2  E2. We consider A as an (m=p)      (m=p) p-dimensional (0,1)- array B as follows. Let x1; x2; : : : ; xp be the p coordinate directions in B. The entries in coordinate direction xi are indexed by the columns of Tm=p in the given order. We note that jAj =  1(B). We  rst handle the case p = 3. By Theorem 7.1.4, we have that for i = 1; 2; 3,  1(proji(B)) < 2m. In fact if  1(proji(B)) > 2m then we have a 3; 3 split in proji(B) and that yields F in A where no rows of F come from the ith term Tm=3 of the product and 2 rows of F come from another T 0m=3 and the other 2 rows of F come from remaining part T 0m=3. Now proceed to form a matrix B0 from B by deleting from B in turn the top 1 in each line in the direction x3 and then deleting the bottom 1 in each line in the direction x2 and  nally deleting the top two entries in each line in the direction x1. We have  1(B)   1(B 0) +  1(proj3(B)) +  1(proj2(B)) + 2 1(proj1(B))   1(B 0) + 8m: 137Let y1 be a 1 of B0. Then, by our construction, there are 2 further 1’s of B in positions y2;y3 with x1(y1) < x1(y2) < x1(y3), x2(y1) = x2(y2) = x2(y3) and x3(y1) = x3(y2) = x3(y3). For each yj we will have two 1’s in positions y0j;y 00 j of B where y 0 j agrees with yj except in coordinate x2 where x2(y0j) < x2(yj) and y 00 j agrees with yj except in coordinate x3 where x3(yj) < x3(y00j ). Then these nine 1’s in B correspond to a copy of F in A as follows. We choose two coordinates a; b from x1 so that when we consider the columns of A corresponding to y1 (and y01;y 00 1 respectively ), y2 (and y 0 2;y 00 2 resp.), y3 (and y 0 3;y 00 3 resp.) we have y1 y2 y3 a b " 0 0 # " 1 0 # " 1 1 # : For the next part note that column t of T 0m=3 has a 0 in row r if and only if t < r. Noting that x2(y0j) < x2(yj) = x2(yj) and that x3(y 0 j) = x3(yj) < x3(y 00 j ), we choose a value c for x2 and a value d for x3 and so that in A, the columns corresponding to the 1’s yj;y0j;y 00 j have yj y0j y 00 j m=3 + c 2(m=3) + d " 1 0 # " 0 0 # " 1 1 # : This yields a copy of F in A in rows a; b; c; d, a contradiction. We deduce that  1(B0) = 0 and hence  1(B)  8m, concluding the proof for p = 3. For p  4, we proceed in a similar fashion. By induction on p,  1(proji(B)) is at most 2  4p 3m. We form a matrix B0 from B by deleting from B in turn the top 1 in each line in the direction x4 and then deleting the bottom 1 in each line in the direction x3 and then deleting the top 1 in each line in the direction x2  nally deleting the bottom 1 in each line in the direction x1. We have  1(B)   1(B 0) + 4X i=1  1(proji(B))   1(B 0) + 4  2  4p 3m: Let y1 be an 1 of B0. Then, by our construction, there are 2 further 1’s of B in positions y2;y3 with x1(y2) < x1(y1) and xi(y2) = xi(y1) for i 6= 1, and x2(y1) < x2(y3) and xi(y1) = xi(y3) for i 6= 2. For each yj we will have two 1’s in positions y0j;y 00 j of B where y 0 j agrees with yj except in coordinate x3 where x3(y0j) < x3(yj) and y 00 j agrees with yj except in coordinate x4 where x4(yj) < x4(y00j ). Then these nine 1’s in B correspond to a copy of F . In particular 138we can choose coordinates values a; b so that in A the columns contain y2 y1 y3 a b+ (m=4) " 0 0 # " 1 0 # " 1 1 # : As above we can choose values c; d and obtain the copy of F from the nine columns of A given by the nine 1’s of B, a contradiction. We deduce that  1(B0) = 0 and hence  1(B)  2  4p 2m  Some growth in the bound with respect to p is to be expected since forb(m;E2  E2) is  (m3). Note that the proof technique suggests generalizations of Theorem 6.1.4 and Theorem 6.2.4 where we forbid a larger class of arrangements of 1’s in a (m=d)  (m=d)      (m=d) d-dimensional f0; 1g-array and are able to conclude that the array has only a linear number of 1’s. Our geometric argument above has no obvious generalization which would allow us to generalize Theorem 7.2.2. A  rst step would be a simple geometric proof of the linear bound for NoSplit(m=2;m=2; 4; 4) but Lemma 6.1.8 suggests other di culties. 7.3 Submatrices of IxI Proof of Theorem 7.1.2 that MaxChoiceCols(E1  E1; Im=2  Im=2) is  (m3=2). Let F = E1 E1. Let A be a submatrix of Im=2 Im=2 that has no con guration E1 E1. We consider A as an (m=2) (m=2) f0; 1g-matrix B whose rows are indexed by the columns of Im=2 and whose columns are indexed by the columns of Im=2. Then kAk =  1(B). Now 4 rows of A contain F if and only if 2 rows of A chosen from the  rst m=2 contain the  rst two rows of F (and so correspond to one copy of E1) and 2 rows of A chosen from the second m=2 correspond to the second two rows of F (and the other copy of E1). Now the nine columns of A containing F corresponds to B having nine 1’s as follows: a 2 2 submatrix of 1’s and at least one more 1 in each row of the 2 2 submatrix and at least one more 1 in each column of the submatrix and at least one more 1 in neither of the two chosen rows or two chosen columns. To see this, consider the 2 chosen rows of A from the  rst m=2 to be i; j and the two other chosen rows are k+m=2; ‘+m=2. Now Im=2 restricted to rows i; j has i j " 1 0 # in column 139i, has i j " 0 1 # in column j and i j " 0 0 # in all columns not equal to i; j. Thus we get a copy of I2  I2 in rows i; j; k + m=2; ‘ + m=2 of A from 4 1’s in B in rows i; j and columns k; ‘. Similar observations yield the other 5 columns of F from the 5 1’s of B as described. We initially process B by deleting any row or column with at most two 1’s (and hence up to 2m 1’s) repeating the deletion process if necessary so that the resulting matrix B has row and column sums at least 3. We note that  1(B)   1(B) + 2m. We now appeal to K}ovari, S os and Tur an [KST54] for a solution of Zarankiewicz’ problem and deduce that if the number of 1’s in B is  (m3=2), then B has 2  2 block of 1’s and then B has the con guration of nine 1’s yielding E1  E1 in A. Moreover from [KST54] we can point out that a construction using projective planes establishes MaxChoiceCols(E1 E1; Im=2 Im=2) is  (m3=2).  Problem 7.3.1 Determine MaxChoiceCols(E1  E1; Im=3  Im=3  Im=3). We note that MaxChoiceCols(E1 E1; Im=3 Im=3 Im=3) is  (m3=2) by Theorem 7.1.2 and is also O(m7=4). Problem 7.3.2 Determine MaxChoiceCols(E1  E1  E1; Im=3  Im=3  Im=3). The core of this problem would be determining the maximum number of 1’s in a 3-dimensional (m=3) (m=3) (m=3) f0; 1g-matrix which has no 2 2 2 submatrix of eight 1’s. Erd}os [Erd64] has obtained a bound O(m11=4) for this but no matching construction. Note the sharp contrast with results such as Theorem 7.1.4, Lemma 7.2.3, Lemma 7.2.1. 7.4 Submatrices of IxT Proof of Theorem 7.1.3 that MaxChoiceCols(E1 E2; Im=2 T 0m=2)  2m. Let F = E1 E2. Let A be an m-rowed submatrix of Im=2 T 0m=2 with no F . We consider A as an (m=2) (m=2) f0; 1g-matrix B whose rows are indexed by the columns of Im=2 and whose columns are indexed by the columns of T 0m=2 in the usual order. We note that kAk is  1(B). Now A contains F if and only if there are three rows and in each row there are three 1’s such that we can divide the nine 1’s as a sort of \1,3 split", where the three leftmost 1’s in each row are entirely to the left of the middle three 1’s in each row which are entirely to the left of the rightmost three 1’s in each row. By Lemma 6.1.7 we have the desired result. The construction A = Im=2  1m=2 avoids F and is  (m).  140Lemma 7.4.1 MaxChoiceCols(E1  E2; Im=3  T 0m=3  T 0 m=3) is  (m 2). Proof: Let F = E1  E2. Let A be an m-rowed submatrix Im=3  T 0m=3  T 0 m=3 with no F . We consider A as an (m=3) (m=3) (m=3) 3-dimensional f0; 1g-array B with coordinate x1 indexed by the columns of Im=2 and with coordinates x2; x3 indexed by columns of T 0m=3 in the usual order. We note that kAk is  1(B). By Theorem 7.1.3 we have jproj2(B)j and jproj3(B)j being O(m), and of course jproj1(B)j is O(m2). Form a matrix B0 from B by deleting from B in turn the top 1 in each line in direction x3, the bottom 1 in each line in the direction x2 and the top two 1’s in each line in the direction x1. We have that the number of deleted 1’s is at most O(m2). Now assume B0 has a 1 in position y1. Then, by our construction, there are 2 further 1’s of B in positions y2;y3 with x1(y1) < x1(y2) < x1(y3), x2(y1) = x2(y2) = x2(y3) and x3(y1) = x3(y2) = x3(y3). For each yj we will have two 1’s in positions y0j;y 00 j of B where y 0 j agrees with yj except in coordinate x2 where x2(y0j) < x2(yj) and y 00 j agrees with yj except in coordinate x3 where x3(yj) < x2(y00j ). Then these nine 1’s in B correspond to a copy of F = E1 E2. We choose r1; r2 so that in the columns of Im=3 indexed by x1(y1); x1(y2); x1(y3) we  nd E1: r1 r2 " 0 1 0 0 0 1 # . We choose two additional rows b; c following the discussion for p = 3 in Lemma 7.2.3 when looking for E2. Thus  1(B0) = 0 and then  1(B) is O(m2).  Lemma 7.4.2 MaxChoiceCols(E1  E2  E2; Im=3  T 0m=3  T 0 m=3 is  (m 2). Proof: Let A be an m-rowed submatrix Im=3 T 0m=3 T 0 m=3 with no E1 E2 E2. As above, we translate A into the 3-dimensional array B with kAk =  1(B). Now by Lemma 6.2.7, if  1(B) > 6(m=3)2 there will be twenty-seven 1’s in B as described and this will yield a copy of E1  E2  E2 in A. Thus  1(B)  6(m=3)2.  Using an analogous argument one obtains 141Lemma 7.4.3 MaxChoiceCols(E1 p 1 z }| { E2      E2; Im=p p 1 z }| { T 0m=p      T 0 m=p) is  (m p 1).  We note some di culty may arise when considering Tk instead of E2 = T2. Theorem 7.4.4 MaxChoiceCols(T3  I2; T  I) is  (m (m)), where  (m) denotes the inverse Ackermann function. Proof: Form our usual array B and observe that B can’t have the structure described in Lemma 6.1.8.  Initially we had thought MaxChoiceCols(T3  I2; T  I) would be linear, but it isn’t. 7.5 Submatrices of IxTxIc Proof of Theorem 7.1.5 that f(E1  E2  E3; Im=3  T 0m=3  I c m=3) is  (m 2). Let F = E1 E2 E3. Let A be an m-rowed submatrix of Im=3 T 0m=3 I c m=3 with no F . We consider A as an (m=3)  (m=3)  (m=3) 3-dimensional f0; 1g-array B as follows. Let x1; x2; x3 be the three coordinate directions in B. The entries in coordinate direction x1 are indexed by the columns of Im=3, the entries in coordinate direction x2 are indexed by the columns of T 0m=3 in that order and the entries in coordinate direction x3 are indexed by the columns of Icm=3. As before kAk is  1(B). By Lemma 6.2.7, we know that if  1(B) > 6(m=3)2, then there is the con guration of twenty-seven points as described. Then these twenty-seven 1’s in B correspond to a special 3; 3; 3 split that yield copy of F in A following our usual arguments. This contradiction implies that  1(B)  6(m=3)2.  1427.6 Fractional Exponent Bound for a Family of Con-  gurations Proof of Theorem 7.1.7 that forb(m; fI2  I2; I2  T 02; T 0 2  T 0 2g) is  (m 3=2). By Theorem 7.1.2 there exists a matrix A  Im=2  Im=2 with kAk being  (m3=2) and no I2  I2. Because of Proposition 7.1.6, neither I2  T 02  A nor T 0 2  T 0 2  A. Let A 2 Avoid(m; fI2 I2; I2 T 02; T 0 2 T 0 2g). We begin using our usual standard decom- position (2.1.1) on row r A = r ! " 00    0 11    1 Br Cr Cr Dr # We would be done by induction if we could show that kCrk  36m1=2 for some r, so we may assume kCrk  36m1=2 for all r. Our proof will show that we can associate matrix Cr with a certain set of rows M(r) (to be de ned later), where jM(r)j  kCrk=4. Consider a given choice r and set M(r) of rows with jM(r)j  9m1=2. Then for t = 9m1=2 choices r1; r2; : : : ; rt 2M(r) we will show that jM(ri) \M(rj)j  9 (7.6.1) and so we obtain disjoint sets M(r1);M(r2) nM(r1);M(r3) n (M(r1)[M(r2)); : : :M(rt) n (M(r1)[M(r2)[    [M(rt 1)) which yields that M(r1) [M(r2) [M(r3)    is of size 9m1=2 + (9m1=2  9) + (9m1=2  18) +    > m; a contradiction given that we only have m rows. We can  nd the inductive children of fI2  I2; I2  T 02; T 0 2  T 0 2g easily: F4 = 2 6 4 1 1 1 1 0 1 3 7 5 ; F5 = 2 6 4 1 1 1 0 0 1 3 7 5 : (7.6.2) This follows since if Cr has F4, then A has T 02  T 0 2, and if Cr has F5, then A has T 0 2  I2, both forbidden con gurations. Because of F5 we know Cr is a laminar matrix (as in 143De nition 1.3.27). If we consider both forbidden con gurations F4 and F5 we deduce that the columns of Cr of sum at least 2, considered as sets, are disjoint. We need more detailed information and begin by computing what happens on quadru- ples of rows of A in order to avoid the three 4  4 con gurations. There are 11 cases Q0; Q1; : : : ; Q10. These cases were computed using the program described in Chapter 3. Q0 = no no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 ; Q1 = no no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 Q2 = no no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 ; Q3 = no no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 Q4 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 ; Q5 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 Q6 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 0 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 ; Q7 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 144Q8 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 ; Q9 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 0 1 1 1 3 7 7 7 7 5 Q10 = no no no no no 2 6 6 6 6 4 1 1 0 0 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 0 3 7 7 7 7 5 2 6 6 6 6 4 1 1 0 1 3 7 7 7 7 5 2 6 6 6 6 4 1 0 1 1 3 7 7 7 7 5 2 6 6 6 6 4 1 1 1 1 3 7 7 7 7 5 As in the proof given in Lemma 5.3.2, choose L(r) to be a minimal subset of rows of Cr such that CrjL(r) is a simple matrix. Note that CrjL(r) also has the property that the columns of sum at least 2 are disjoint as sets. Now consider the family of sets C corresponding to the columns of Cr. We can have the empty set as well as sets of size 1 and some disjoint family of sets of size at least 2. Let V 2 C be a set of size at least 2 and let U1; U2; : : : ; Ut be the sets of C of size 1 contained in V . Now, if t < jV j and jV j  3, we could delete a row of CrjL(r) and preserve simplicity, a contradiction. As an example, consider deleting the last row of G below. For jV j = 2, we can have t = 1 < jV j and still have no row to delete, as shown for example in H below. G = 2 6 4 0 1 0 1 0 0 1 1 0 0 0 1 3 7 5 ; H = " 0 1 1 0 0 1 # Our choice of L(r) ensures that no row of CrjL(r) can be deleted (while preserving simplicity) and hence CrjL(r) contains the column of 0’s (0jL(r)j) and at least half the columns of IjL(r)j, as well as possibly up to jL(r)j=2 disjoint columns of sum 2 or more. Thus we may estimate jL(r)j  kCrk=2. Thus the number of sets of size 1 in C is at least jL(r)j=2. We choose M(r) to correspond to the sets of size 1, namely CrjM(r) contains IjM(r)j as well as 0jM(r)j. Note that CrjM(r) need not be a simple matrix and may in fact have up to 2 copies of each column from IjM(r)j (imagine deleting the second row of H). We obtain jM(r)j  kCrk=4  9m1=2. 145In what follows we analyze closely each of the 11 cases above to deduce the structure of CrjL(r) in order to prove (7.6.1). It’s possible to determine What Is Missing in CrjL(r) on the triple of rows i; j; k 2 L(r) by considering What Is Missing on the quadruple of rows r; i; j; k, as we did for F7 in Section 5.3. We obtain a contradiction if we  nd a copy of \K2" in What Is Missing, namely if on the triple i; j; k there is a pair of rows i; j with all 4 columns of K2 appearing as follows: i j k no 2 6 4 0 0 a 3 7 5 no 2 6 4 0 1 b 3 7 5 no 2 6 4 1 0 c 3 7 5 no 2 6 4 1 1 d 3 7 5 where a; b; c; d 2 f0; 1g. Perhaps other columns are missing on rows i; j; k. Note that we could delete row k from CrjL(r) and preserve simplicity, contradicting our choice of L(r). The reason for this is that on the three rows i; j; k, the columns present would possibly be i j k 2 6 4 0 0 a 3 7 5 2 6 4 0 1 b 3 7 5 2 6 4 1 0 c 3 7 5 2 6 4 1 1 d 3 7 5 where x denotes the f0; 1g-complement of x. We can see that deleting row k will not result in repeated columns assuming CrjL(r) has no repeated columns. We note that Q0; Q1; Q2 each have 3 rows, each pair of which has a K2 in What Is Missing (rows 1; 2; 3 for Q0, rows 2; 3; 4 for Q1, rows 2,3,4 for Q2) and Q5; Q6 have two disjoint pairs of rows, each of which has K2 (rows 1,2 and rows 3,4 for Q5 and rows 1,3 and rows 2,4 for Q6) so deleting any row of the quadruple will leave a K2 leaving a row to delete from Cr and so by our choice of L(r) we cannot have the cases represented by Q0; Q1; Q2; Q5; Q6 in quadruples of rows consisting of r plus a triple of rows from L(r). For the remaining cases, to avoid leaving a K2 after deleting a row we  nd that we must have deleted particular rows from each quadruple. Note that Q3 and Q8 have K2 on rows 3,4 and so if present, to avoid leaving a copy of K2 row r must either correspond to row 3 or to row 4. The cases Q4; Q7; Q9; Q10 all have K2 on rows 2,4 and rows 3,4 and so row r must be row 4 of such a quadruple. We have used Pi to denote the condition on a triple of rows arising from the quadruple Qi in these ways. 146P3 : no 2 6 4 0 1 0 3 7 5 no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 1 3 7 5 no 2 6 4 1 1 1 3 7 5 or no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 0 3 7 5 no 2 6 4 0 1 1 3 7 5 no 2 6 4 1 1 1 3 7 5 P4 or P9 : no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 1 3 7 5 no 2 6 4 0 1 1 3 7 5 P8 : no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 1 3 7 5 no 2 6 4 0 1 1 3 7 5 no 2 6 4 1 1 1 3 7 5 or no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 0 3 7 5 no 2 6 4 0 1 1 3 7 5 no 2 6 4 1 1 1 3 7 5 P7 or P10 : no 2 6 4 1 1 0 3 7 5 no 2 6 4 1 0 1 3 7 5 no 2 6 4 1 1 1 3 7 5 Note that the presence of IjM(r)j means that on each triple of rows in M(r) we have all three columns of sum 1 present and so cannot have P3 (arising from Q3 with row r being either the third or fourth row) or the second of the two cases for P8 (arising from Q8 with row r being the third row) on triples of rows in M(r) since each forbids a column of sum 1. This careful detail is used in what follows. We now consider what is possible in the full width of A on the rows of M(r). By considering the corresponding Qi’s and the row of Qi that corresponds to r, we  nd that under the 0’s in row r, any triple of rows [BrCr]jM(r) must have two columns of sum 2 absent and hence [BrCr]jM(r) cannot have the con guration F5. Thus the columns form a laminar family on rows M(r), namely, considered as subsets of M(r), any two sets are either disjoint or one is contained in the other. One has to consider Q4; Q8; Q9; Q10 and then in each case the row corresponding to row r must be the fourth row of each such Qi (else a K2 remains on the triple). Similarly under the 1’s in row r, in [CrDr]jM(r) we cannot have the con guration 147F5 and so the columns also form a laminar family on M(r). Had we used row r as the third or fourth row of Q3 then [CrDr]jM(r) might have F5 and had we used row r as the third row of Q8 then [BrCr]jM(r) might have the con guration F5. Fortunately these cases are not possible. Now consider rj 2 M(r) and standard decomposition based on row rj. The columns of Crj , must correspond to columns of A which appear with a 1 and with a 0 in row rj and are the same elsewhere and hence this is also true when restricted to the rows M(r) [ r. We have pairs of columns as follows: rj M(r) n rj f r 2 6 4 0 1   a a 3 7 5 We can show that there are at most three possible choices for pairs of columns and hence at most three choices for  . In fact consider [BrCr] and  6=  be two nonzero choices for  . rj M(r) n rj f r 2 6 4 0 1 0 1     0 0 0 0 3 7 5 The columns of A with 0’s in row r, when restricted to the rows M(r), must form a laminar family. Columns 2 and 4 have 1’s in common on row rj and we deduce, without loss of generality by considering columns 2 and 4, that    . Now considering columns 2,3 and using 0 6=    , we violate the laminar property, a contradiction to  6=  . Thus there is at most one nonzero choice for  in [BrCr] and at most one nonzero choice for  in [CrDr]. Given that there are at most three choices for columns of Crj restricted to the rows M(r) n rj, we can deduce that jL(rj) \M(r)j  2 (we would be able to delete all but two rows of M(r) n rj from Crj jL(rj) without a ecting simplicity). So let M 0(rj) = M(rj) nM(r) and note that jM(rj)j  2  jM 0(rj)j  jM(rj)j. From our previous observations we have that the columns of A indexed by the rows of M 0(rj) consist of the column of 0’s, the columns of the identity matrix, and (possibly) columns of sum at least 2 which are disjoint when considered as sets. We use the following 148notation. Crj = M(r) n rj f M 0(rj) f 2 6 6 4 X Y ... 3 7 7 5 ; A = rj ! M(r) n rj f M 0(rj) f 2 6 6 6 6 4 0    0 0    0 1    1 1    1 X X Y Y ... ... ... ... 3 7 7 7 7 5 : By our previous observations for CrjL(r) and CrjM(r) we know that columns of Y are either columns of 0’s, columns of sum 1 and possibly columns of sum 2 (disjoint as sets). By reducing to M 0(rj) and deleting the possible row of overlap with M(r), we may assume that Y contains IjM 0(rj)j. We can show that two di erent columns of Y of sum 1 (i.e. with 1’s in di erent rows) on rows M 0(rj) cannot lie under two identical nonzero columns (of X) on rows M(r) n rj, else we have I2  T 02 in A as follows. Let i; k correspond to the rows of M 0(rj) which contain the 1’s of the two selected columns of Y and let a 2 M(r) n rj be a row containing 1’s in the repeated nonzero column of X. This gives I2  T 02: rj a i k 2 6 6 6 6 4 0 0 1 1 1 1 1 1 1 0 1 0 0 1 0 1 3 7 7 7 7 5 Recall that there are at most 3 di erent columns in X and Y has IjM 0(rj)j. Let M 00(rj)  M 0(rj) denote the rows corresponding to the 1’s in the columns of the identity IjM 0(rj)j which lie under columns of 0’s in rows of M(r)nrj. We have that jM 0(rj)j 2  jM 00(rj)j  jM 0(rj)j else we create I2  T 02 as above since there are at most two di erent nonzero columns of X. Thus jM(rj)j  4  jM 00(rj)j  jM(rj)j. Now consider two rows rp; rq 2 M(r). Let i; j 2 M 00(rp) \M 00(rq). From the standard decomposition using row rp, we can  nd two columns with 1’s in row rp, 0’s in row rq (and also 0’s in all other rows of M(r)) and I2 on rows i; j. Similarly we can  nd two columns with 1’s in row rq, 0’s in row rp (and all other rows of M(r)) and I2 on rows i; j. Then we can  nd a copy of I2  I2: rp rq i j 2 6 6 6 6 4 1 1 0 0 0 0 1 1 1 0 1 0 0 1 0 1 3 7 7 7 7 5 We deduce that jM 00(rp) \ M 00(rq)j  1, which in turn establishes (7.6.1) that jM(ri) \ 149M(rj)j  9, completing our proof.  7.7 All Pairs of Columns We now establish an asymptotic bound that is a generalization of the exact bound of the bound proved in Section 4.2. Consider all pairs of possible ‘ 1 columns. F‘ := f[  ] :  ;  2 f0; 1g ‘;  6=  g: For example, F2 is the set of matrices considered in Section 4.2, where we proved forb(m;F2  F2) = m+ 3. F2 = (" 0 0 0 1 # ; " 0 1 0 1 # ; " 1 0 0 1 # ; " 1 1 0 1 #) and F3 = 8 >< >: 2 6 4 11 11 10 3 7 5 ; 2 6 4 11 10 10 3 7 5 ; 2 6 4 10 10 10 3 7 5 ; 2 6 4 11 10 01 3 7 5 ; 2 6 4 11 10 00 3 7 5 ; 2 6 4 10 10 01 3 7 5 ; 2 6 4 10 10 00 3 7 5 ; 2 6 4 10 01 00 3 7 5 ; 2 6 4 10 00 00 3 7 5 9 >= >; : Theorem 4.2.1 has an asymptotic version. Observe that forb(m;F‘) = 1 for m  ‘ (and 2m for m < ‘). Theorem 7.7.1 Let T be a family of forbidden con gurations with forb(m; T ) being O(mk). Then for any ‘, we have forb(m;F‘  T ) is O(mk+1). Proof: We consider a matrix A with no con guration in T  F‘. We do an analogous operation to the standard decomposition (2.1.1) on ‘ rows. For any two columns  ;  with  6=  we let A ; denote the (m  ‘) rowed matrix formed as the set of columns on Ajf‘+1;‘+2;:::;mg which appear under both  and  . For example, for ‘ = 2, we would consider 150the following representative A = 2 6 6 4 0 0 1 1 0 1 0 1 A[ 00 ] A[ 10 ] A[ 01 ] A[ 11 ] 3 7 7 5 where A are all simple matrices for  2 f  0 0  ;  1 0  ;  0 1  ;  1 1  g In such matrix, A[ 00 ];[ 1 0 ] would be matrix formed by the columns that are in both A[ 00 ] and A[ 10 ] . We deduce that A ; is a simple matrix with no con guration in T since if F 2 T is in Aj ; then A has [  ] F 2 F‘  T . Hence at Aj ; has at most forb(m ‘; T ) columns. There are at most a constant number of choices for pairs  and  and so after deleting at most O(mk) columns from A to produce A0 we can ensure Ajf‘+1;‘+2;:::;mg is simple and hence, by induction, has at most forb(m ‘; T  F‘) columns. This yields the result.  The O(m) bound of Theorem 4.2.1 would follow from taking T = F2, although to get the exact bound we use the proof given in that section. 151Chapter 8 Conclusions 8.1 Open Problems 8.1.1 Monotonicity It is quite reasonable to conjecture that the function forb(m;F ) is increasing in the  rst variable m, as we know it is in the second variable F (Remark 1.3.21). Problem 8.1.1 Let F be a k-rowed (0,1)-matrix and let m  1. Is it true that forb(m;F )  forb(m+ 1; F )? We can answer positively for single forbidden con gurations with certain properties. We know this is false for families of forbidden con gurations. Proposition 8.1.2 Let F be a con guration that doesn’t have either a row of 0’s, a row of 1’s or a repeated row. Then for all m, forb(m;F )  forb(m+ 1; F ). Proof: Let A 2 ext(m;F ). Then consider the (m+ 1)-rowed matrix A0 = 8 >>>< >>>: A [0] if F doesn’t have a row of 0’s, A [1] if F doesn’t have a row of 1’s,  A r  if F doesn’t have a repeated row, ; 152where r is any row of A. Then clearly A0 2 Avoid(m + 1; F ) and kA0k = kAk, and F  A0.  Note that for families of con guration this is not always the case. Here is an amusing case. Theorem 8.1.3 forb(m; (" 0 0 0 0 # ; " 1 1 1 1 #) ) = 8 >< >: 2 for m = 1 or m  7 6 for m = 3; 4 4 for m = 2; 5; 6 Proof: The result is easy for m = 1; 2. First consider pairs of columns. Let Fabcd denote the (a+ b+ c+ d) 2 matrix of a rows [0 0], b rows [1 0], c rows [0 1] and d rows [1 1]. Consider two columns of A and suppose they are Fabcd. We have a; d  1. We deduce that b; c  2 else with b  3 (F0300), by the pigeonhole principle a third column will create either " 0 0 0 0 # or " 1 1 1 1 # on those 3 rows. For m  7 we deduce then that there are at most 2 columns. For m = 6 we can use this to deduce that every pair of columns must form F1221 and then the 6  4 con guration (K24) T establishes the bound. For m = 5 we can use this to deduce that every pair of columns must form F1211 or F1220 or F0221. We can create 4 columns by deleting one row from (K24) T . Let A 2 ext(m; (" 0 0 0 0 # ; " 1 1 1 1 #) ). Using the technique of What Is Missing, we have that on any pair of rows we have  1 " 0 0 #  1 " 1 1 # For any three rows we  nd that any choice of columns for those three rows will violate one of the restrictions but there are at most 6 restrictions to violate on the three rows so there are at most 6 columns. For m = 3 we may use the construction [K23 j K 1 3 ] and for m = 4 we may use the con guration K24 .  153One could potentially circumvent this example by only seeking monotonicity for m large enough. Conjecture 8.1.4 Let F be a family of forbidden con gurations. Then there exists M for which for all m  M , forb(m;F)  forb(m+ 1;F). 8.1.2 A Common 4-rowed Subcon guration As mentioned in the introduction to Chapter 5, there are two 4-rowed con gurations for which we do not know the asymptotic bound. Both con gurations have the following con guration in common: F (t) = t  2 6 6 6 6 4 1 0 1 0 0 1 0 1 3 7 7 7 7 5 This is the smallest con guration F for which an asymptotic bound on forb(m;F ) is not known. Interestingly, for t = 1 and t = 2 a quadratic bound has been proven. Problem 8.1.5 Prove that forb(m;F (t)) is  (m2) (or  nd a counterexample). 8.1.3 Other 3x4 Exact Bounds In Section 4.1, we found two exact bounds on 3 4 con gurations. There are only 10 others for which the answer isn’t known. We can apply the local search techniques described in Section 3.6 to  nd conjectured extremal matrices. The following table shows the conjectured answers. The entries of the table represent forb(m;Vi). The \?" symbol represents that the computer has found an example (i.e. a lower bound) but we haven’t proven the upper bound. 154m  ! 3 4 5 6 7 8 V1 = 2 6 4 1 1 1 1 1 1 1 1 1 0 0 0 3 7 5 8 16 22 32? 43? 55? V2 = 2 6 4 1 1 1 1 1 1 1 0 1 0 0 1 3 7 5 8 12 18 26? 36? 45? V3 = 2 6 4 1 1 1 1 1 1 1 0 1 0 0 0 3 7 5 8 12 18 26? 36? 45? V4 = 2 6 4 1 1 1 0 1 1 1 0 1 0 0 1 3 7 5 8 12 18 26? 36? 45? V5 = 2 6 4 1 1 1 0 1 1 1 0 1 0 0 0 3 7 5 8 12 18 26? 36? 44? V6 = 2 6 4 1 1 1 0 1 1 1 0 1 0 0 0 3 7 5 8 16 22 29? 37? 45? V7 = 2 6 4 1 1 1 1 1 1 1 0 0 0 0 0 3 7 5 8 16 22 29? 37? 46? V8 = 2 6 4 1 1 1 0 1 1 1 0 0 0 0 1 3 7 5 8 16 22 29? 37? 46? V9 = 2 6 4 1 1 1 0 1 1 1 0 0 0 0 0 3 7 5 8 16 22 29? 37? 46? V10 = 2 6 4 1 1 1 0 1 1 0 1 0 0 1 1 3 7 5 8 12 17 23? 30? 38? Table 8.1: Conjectured values of forb(m;Vi) 155The last numbers on this table (for 8 rows) take a few hours to  nd. The next number, for 9 rows, might take a few weeks of computer time. One could conjecture that forb(m;V10) =  m 2  +  m 1  +  m 0  + 1. 8.1.4 Critical Substructures We give some ideas extending those of Section 4.3. Problem 8.1.6 Prove Conjecture 4.3.2 or the equivalent Conjecture 4.3.8 or give a coun- terexample. What follows are ideas and attempts at proving Conjecture 4.3.2 for k  5. We prove some of the properties required for general k and we prove some for k = 5. Consider applying the idea for the proof of Proposition 4.3.9 for general k to try to show that forb(m;Fk 1) < forb(m;Kk). If we could show this is true for any m, then it will be true for all larger m by induction. One might try for m = k as a base case, but unfortunately we  nd that forb(k; 2  Kk 1k 1 0k 1) = forb(k;Kk) = 2k 1, namely if we delete any column of Kk which has column sum not 0 or k, then the resulting matrix has no 2  Kk 1k 1 0k 1. One might try next to obtain the inequality for m = k + 1. Proceed by contradiction and assume forb(k + 1; Fk 1) = forb(k + 1; Kk). Let A 2 ext(k + 1; Fk 1). As argued before, for any (k + 1)  1 column  which is not a column in A, [Aj ] has a copy of Kk say on rows S. Similarly for a column  not in A, we have that [Aj ] has a copy of Kk say on rows T . Now if S = T , then AjS has the submatrix Kk  jS and also the submatrix Kk  jS. If  S 6=  jS, then we deduce A has Kk, a contradiction. There are at most 2 columns on k+ 1 rows agreeing on a k-set of rows S. Now if S 6= T then jS \ T j = k 1. A similar argument yields a contradiction if  jS\T 6=  jS\T . Table 8.2 shows some computer-found lower bounds (using the algorithms described in Chapter 3). Remember these are conjectures for m  6. We can show forb(k+ 2; Fk) = forb(k+ 2; Kk+1) for k  4 as follows. The k+ 3 columns to delete from Kk+2 are B = 2 6 4 13 Ic3 13    0k 1 0k 1 Ik 1 3 7 5 One can show that this matrix B has the property that for every k-set of rows S, there is a column repeated three times in BjS, while Kk+2 has four copies of each column on a k-set 156m forb(m;F3) forb(m;K4) 4 15 15 5 25 26 6 40(?) 43 7 59(?) 64 m forb(m;F4) forb(m;K5) 5 31 31 6 57 57 7 98(?) 99 Table 8.2: Conjectured values for forb(m;F3) and forb(m;K4) m forb(m; 2  K24) forb(m;K5) 4 16 16 5 30 31 6 53(?) 57 7 86(?) 99 Table 8.3: Conjectured values for forb(m; 2  K24) of rows. We should also expect that forb(m; 2  K24) is much less than forb(m;K5). Several proof techniques will work but the easiest is to consider forb(5; 2 K24) and show that forb(5; 2 K 2 4)  25 2, i.e you must delete at least 2 columns in order to avoid 2  K24 . This is relatively easy. A column of sum 2 can only yield 12  02 on three 4-sets of rows but there are  ve 4-sets of rows. One way to achieve this is to delete from K5, a column of sum 2 and a column of sum 3 that overlap in two rows. Table 8.3 shows some more computer experiments. Lemma 8.1.7 We can establish that forb(m; 2  K24)  forb(m;K5) m+ 4: Proof: We use induction as follows. We  rst note forb(m; [2 K13 j 2 K 2 3 ]) < forb(m;K4). Now applying the standard decomposition (2.1.1) repeatedly we have forb(m;K24)  forb(m  1; K24) + forb(m  1; [2  K 1 3 j 2  K 2 3 ]) and then forb(m;K 2 4)  forb(m  2; K 2 4) + forb(m  2; [2  K13 j 2  K 2 3 ]) + forb(m 1; [2  K 1 3 j 2  K 2 3 ]) and then repeat this m 4 times using that [2  K13 j 2  K 2 3 ] is in F3 and forb(5; F3) = forb(5; K4) 1.  The following idea is probably of no help in establishing these inequalities but it should be noted. 157Proposition 8.1.8 If forb(m;F ) < forb(m;Kk) for m  m0, then forb(m; [0 1] F ) < forb(m;Kk+1) m+m0. Proof: Simply use induction forb(m; [0 1]  F )  forb(m  1; [0 1]  F ) + forb(m  1; F ).  Let forb(m;F4) = forb(m;K5) and let A 2 ext(m;F4)  ext(m;K5). Then on each 5-set of rows exactly one column is missing as described above. So each 5-set contains K5   for some choice of 5 1  . Now if  has column sum 0,1,4,5 then we  nd that K5   has F4. So we deduce that  has column sum 2,3. Thus we deduce that A has no K25K 3 5 on the grounds that one column is missing. Immediately we deduce that A has all columns of sum 0,1,m 1, m (adding these columns can’t create a K5. But then A [K0m j K 1 m j K m 1 m j K m m ] has no [K14 j 2  K 2 4 j K 3 4 ]. Now applying our inductive decomposition on [K 1 4 j 2  K 2 4 j K 3 4 ] yields F3. We would be done if we could show forb(m;F3)  forb(m;K4) 3. Here’s an idea that might not help at all, but gives rise to an interesting problem. Let F be a k-rowed con guration for which forb(m;F ) is known. Let G  F be an ‘-rowed subcon guration. Suppose we wish to  nd out if G is a critical substructure of F . That is, suppose we wished to  nd out if forb(m;G) = forb(m;F ): Proceeding by contradiction, assume this was true and consider a matrix A 2 ext(m;G)  ext(m;F ). If we add a column  to A, by hypothesis we should get that [A j  ] contains F as a con guration. When does this mean A contains G? In particular, this means A contains F   for some column  of F with  =  jS for some subset of the rows S. Note then that G  F   . Thus, any copy of G in [A j  ] must be missing at most 1 column of F . Consider the following reinterpretation of the problem. Lemma 8.1.9 If for every column  of F we have that G  F   , then forb(m;G) < forb(m;F ): 158Proof: If A is in ext(m;G)  ext(m;F ), then by adding any column not already present in A to A we must obtain the con guration F since A 2 ext(m;F ), which means that for some column  of F , we have that F   is present in A, which means G  F   , a contradiction.  8.2 Concluding Remarks We have seen many results concerning a single Forbidden Con guration and just a peek at results concerning multiple Forbidden Con gurations. An observant reader might have noticed that the tools we use to study a single forbidden con guration are mostly the same tools we use to study a family of forbidden con gurations, so, the observant reader might ask, why then have we neglected our families and opted for singles? A perhaps unsatisfying answer could be that single con gurations are hard enough. Imagine for a moment we had an oracle that when given a family of con gurations F would immediately provide the (correct) value of forb(m;F) and perhaps an extremal matrix satisfying the bound. What could we do with such a wonder? In particular, among other things, this oracle would solve all Erd}os-Stone-Simonovits type results, as one could imagine forbidding (1; 1; 1)T together with the incidence matrix of a graph H. When forbidding these con gurations the extremal matrix one would obtain would be [0 j I j G] where G is the incidence matrix of a graph that does not have H as a con guration. Because of this, the idea of having a general theory for families of forbidden con gura- tions looks tantalizing, but perhaps a little daunting. Results about small families of small forbidden con gurations could very well  ourish with a systematic study. If I were to con- tinue research in forbidden con gurations, I would certainly start here, as I am sure there are many new results waiting to be discovered. As of this writing we don’t yet have the tools to study con gurations with 6 rows or more, but we hope that by chipping away at the problem we are making progress toward perhaps a proof or refutation of Conjecture 1.4.1. Proving this pivotal conjecture would indeed be a milestone for the  eld, but it is my personal belief that the conjecture will turn out to be false. The reason for this belief, in the face of so much evidence for it, is its constructive nature. Sure, the product constructions described in this Conjecture 1.4.1 have given the best asymptotic bounds for many (small) con gurations, but these are not, by any stretch, a random sample of con gurations. They are chosen to be small and manageable. As we’ve seen, a very small change in a con guration (for example, substituting a 0 by a 1591 or adding a column) can lead to wildly di erent bounds. I  nd there is not much reason to believe a large, random con guration would satisfy the conjecture that the best (asymptotic) construction will be products of I, Ic and T . On the other hand, the con gurations I; Ic and T do seem to \pop-up" repeatedly while studying con gurations. Perhaps they are in some way special, and we have indeed proven many cases in which the conjecture turns out to be true. 160Bibliography [ABS11] R.P. Anstee, F. Barekat, and A. Sali, Small forbidden con gurations V: Exact bounds for 4 2 cases, Studia. Sci. Math. Hun. 48 (2011), 1{22. ! pages 32, 59, 60 [AF86] R.P. Anstee and Z. F uredi, Forbidden submatrices, Discrete Math. 62 (1986), 225{243. ! pages 15, 18, 134 [AFS01] R.P. Anstee, R. Ferguson, and A. 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Vapnik and A. Ya. Chervonenkis, On the uniform convergence of relative frequencies of events to their probabilities, Soviet Math. Dokl. 9 (1968), 915{918. ! pages 32 [VC71] V.N. Vapnik and A.Ya. Chervonenkis, On the uniform convergence of relative frequencies of events to their probabilities, Theory of Probability and Applica- tions (1971), 264{280. ! pages 15 [Ver05] R. Vershynin, Integer cells in convex sets, Advances in Mathematics 197 (2005), 248{273. ! pages 32 [WD81] R. S. Wencour and R. M. Dudley, Some special Vapnik{Chervonenkis classes, Discrete Mathematics 33(3) (1981), 313{318. ! pages 32 165Index (p; q)-split, 124 f0; 1g-complement, 6 p1; p2; : : : ; pd split, 128 s-boundary case, 18 proji(B), 130 absent, 38 avoids, 11 boundary case, 18 column sum, 5 concatenation, 6 con guration, 10 contain F as a con guration, 11 critical substructure, 82 has no con guration, 11 identity complement Icm, 8 identity matrix Im, 8 implication, 40 impure, 40 inductive child, 35 laminar, 13 long supply, 38 multiplicity, 7 non-essential, 93 pattern, 124 power set, 4 predicted s-boundary case, 18 predicted boundary case, 18 pure, 40 restriction, 7 set system, 4 shifted matrix, 83 short supply, 38 simple, 5 simple con guration, 10 simple hypergraph, 4 standard decomposition, 34 subcon guration, 11 subtraction, 7 tower matrix Tm, 9 violates an implication, 40 166

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