Transversals in Linear Uniform Hypergraphs [1st ed.] 9783030465582, 9783030465599

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Developments in Mathematics

Michael A. Henning Anders Yeo

Transversals in Linear Uniform Hypergraphs

Developments in Mathematics Volume 63

Series Editors Krishnaswami Alladi, Department of Mathematics, University of Florida, Gainesville, FL, USA Pham Huu Tiep, Department of Mathematics, Rutgers University, Piscataway, NJ, USA Loring W. Tu, Department of Mathematics, Tufts University, Medford, MA, USA

Aims and Scope The Developments in Mathematics (DEVM) book series is devoted to publishing well-written monographs within the broad spectrum of pure and applied mathematics. Ideally, each book should be self-contained and fairly comprehensive in treating a particular subject. Topics in the forefront of mathematical research that present new results and/or a unique and engaging approach with a potential relationship to other fields are most welcome. High-quality edited volumes conveying current state-of-the-art research will occasionally also be considered for publication. The DEVM series appeals to a variety of audiences including researchers, postdocs, and advanced graduate students.

More information about this series at http://www.springer.com/series/5834

Michael A. Henning Anders Yeo •

Transversals in Linear Uniform Hypergraphs

123

Michael A. Henning Department of Mathematics and Applied Mathematics University of Johannesburg Auckland Park, South Africa

Anders Yeo Department of Mathematics and Computer Science University of Southern Denmark Odense, Denmark Department of Mathematics and Applied Mathematics University of Johannesburg Auckland Park, South Africa

ISSN 1389-2177 ISSN 2197-795X (electronic) Developments in Mathematics ISBN 978-3-030-46558-2 ISBN 978-3-030-46559-9 (eBook) https://doi.org/10.1007/978-3-030-46559-9 Mathematics Subject Classification: 05C65, 51E10, 51E15 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

We dedicate this book to our wives, Anne and Angela, with heartfelt thanks, for their love, support, encouragement, and patience.

Preface

A subset T of vertices in a hypergraph H is a transversal (also called hitting set or vertex cover or edge cover or blocking set in the literature) if T has a nonempty intersection with every edge of H. The notion of transversal is fundamental in hypergraph theory and has been studied a great deal; a rough estimate says that it occurs in more than 25,000 papers (considering the various names listed above for a transversal). Two of the five chapters in the major monograph of hypergraph theory [C. Berge, Hypergraphs—Combinatorics of Finite Sets. North-Holland, 1989] deal with transversals and their fractional version (real relaxation via Linear Programming). Very few papers give bounds on the transversal number for linear hypergraphs, even though these appear in many applications, as it seems difficult to utilize the linearity in the known techniques. In this book, we nevertheless present bounds on the transversal number for linear hypergraphs. This book is one of the first that give strong nontrivial bounds on the transversal number for linear hypergraphs, which is better than for non-linear hypergraphs. We show, for example, that the best possible upper bound on the transversal number for 4-uniform linear hypergraphs is better than that for 4-uniform non-linear hypergraphs. Key to our proof is the completely new technique of the deficiency of a hypergraph. We have written the book primarily to reach the following audience. The first audience is the graduate student who is interested in exploring the field of transversals in hypergraphs and to “familiarize themselves with the subject, the research techniques and major accomplishments in the field.” It is our hope that such graduate students will find topics and problems that can be developed further. The second audience is the established researcher in hypergraph theory who wishes to have easy access to known results and latest developments in the field of transversals in linear hypergraphs. Chapter 1 introduces graph theory and hypergraph theory concepts fundamental to the chapters that follow. Chapter 2 studies transversals in linear hypergraphs. We show in this chapter that linear intersection hypergraphs have largest possible transversal number among those linear hypergraphs with maximum degree at most two. vii

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Preface

In Chap. 3 we look at finite affine planes. We show a connection between linear uniform hypergraphs with large transversal number and finite affine planes. In Chap. 4 we define the so-called Tuza constants, ck and qk , for the class of k-uniform hypergraphs and for the subclass of k-uniform linear hypergraphs, respectively, for k
2. We determine in this chapter the Tuza constants for small values of k, namely, k ¼ 2 and k ¼ 3. In Chap. 5 we determine the exact value of the Tuza constant c4 . In Chap. 6 we present a result due to Noga Alon that shows that the asymptotic behavior of the Tuza constants, ck , as k grows is of the order lnðkÞ=k. In Chap. 7 we study what we have coined the “West bound” on the transversal number of a k-uniform linear hypergraph motivated by comments by Douglas West. We pose the conjecture that the so-called West bound holds for all k-uniform, linear hypergraphs, and show in this chapter that this is true for small values of k, namely, k ¼ 2 and k ¼ 3. In Chap. 8 we introduce the new technique of the deficiency of a hypergraph. Using this concept of deficiency, we prove a key theorem that we will need in establishing the transversal number of a linear 4-uniform hypergraph, including determining the Tuza constant q4 in the following chapter. In Chap. 9 we illustrate the power of our new technique of the deficiency of a hypergraph discussed in the previous chapter, and we apply this result to obtain best possible upper bounds on the transversal number of a 4-uniform linear hypergraph. In Chap. 10 we show that the asymptotic behaviour of the Tuza constant, qk , as k grows is the same as that of the Tuza constant ck , namely, of the order lnðkÞ=k. We apply this result to projective planes, and show, for example, that when k ¼ 166, the Tuza constant qk [ k þ1 1. In Chap. 11 we discuss the cap set problem and its relation to transversals in 3uniform hypergraphs. We present a brief history of the cap set problem which was considered one of the most intriguing open problems in additive combinatorics and Ramsey theory for the past 20 years. We present a proof of a recent solution to the cap set problem by Ellenberg and Gijswijt. We show that the Ellenberg-Gijswijt result implies that for every e [ 0, there exists a connected 3-uniform linear hypergraph H of sufficiently large order n such that sðHÞ [ ð1  eÞn. In Chap. 12, we consider partial Steiner triple systems; that is, 3-uniform linear hypergraphs. We present here a classical 1986 result due to Phelps and Rödl that establishes an upper bound on the transversal number of a partial Steiner triple system. In Chap. 13 we study upper transversals in linear hypergraphs, where the upper transversal number  ðHÞ of a hypergraph H is the maximum cardinality of a minimal transversal in H. We show that for k 2 f2; 3g and for every given e [ 0, there exists a connected k-uniform linear hypergraph H of sufficiently large order n such that  ðHÞ\e  n. In Chap. 14 we study strong transversals in hypergraphs, also known as double blocking sets in the literature. We show that the set S3 of all pairs ða; bÞ of nonnegative integers a and b for which inequality s2 ðHÞ  anH þ bmH holds for all 3-

Preface

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uniform hypergraphs H is a convex set, where s2 ðHÞ denotes the strong transversal number of H. Further we show that there are infinitely many extreme points of the convex set S3 , and we determine all these extreme points. We also discuss strong transversals in finite projective planes. We close with Chap. 15 which lists some conjectures and open problems that have yet to be settled or solved. We have tried to eliminate errors, but surely several remain. We do welcome any comments the reader may have. A list of typographical errors, corrections, and suggestions can be sent to our email addresses below. Auckland Park, South Africa Odense, Denmark

Michael A. Henning [email protected] Anders Yeo [email protected]

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Graph and Hypergraph Terminology . . . . . . . . . . . . . . . . . . . .

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Linear Intersection Hypergraphs . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Hypergraph Duality . . . . . . . . . . . . . . . . . . . . . 2.3 Properties of Linear Intersecting Hypergraphs . . . 2.4 Linear Hypergraphs with Maximum Degree Two

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Finite Affine Planes and Projective Planes . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Finite Affine Planes and Linear Uniform Hypergraphs . . . . . . .

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The Tuza Constants . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . 4.2 The Tuza Constants c2 and q2 4.3 The Tuza Constants c3 and q3

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The Tuza Constant c4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 4-Uniform Hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Tuza Constant ck for k Large . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 A General Upper Bound for the Tuza Constant ck 6.3 The Asymptotic Behavior of ck as k Grows . . . . .

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The West Bound . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . 7.2 The West bound for 2-Uniform 7.3 The West bound for 3-Uniform

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The West bound for k-Uniform Hypergraphs for Large k . . . . . The Convex Sets Ak and Bk . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 The Tuza Constant qk for Large k . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 The Asymptotic Behavior of the Tuza Constant qk 10.3 Application to Projective Planes . . . . . . . . . . . . . 10.4 Application to the West Bound . . . . . . . . . . . . . .

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11 The Cap Set Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 A Mathematical Formulation . . . . . . . . . . . . . . . . . 11.3 The Relation to Transversals in Linear Hypergraphs

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12 Partial Steiner Triple Systems . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Steiner Triple Systems . . . . . . . . . . . . . . . . . 12.3 Transversals in Partial Steiner Triple Systems .

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13 Upper Transversals in Linear Hypergraphs 13.1 Introduction . . . . . . . . . . . . . . . . . . . . 13.2 Lower Bounds in Terms of Order . . . . 13.2.1 The Constant U2 . . . . . . . . . . 13.2.2 The Constant U3 . . . . . . . . . . 13.3 Upper k-Transversal Number . . . . . . . .

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14 Strong Tranversals in Linear Hypergraphs 14.1 Introduction . . . . . . . . . . . . . . . . . . . . 14.2 A Probabilistic Bound . . . . . . . . . . . . . 14.3 The Convex Set Sk . . . . . . . . . . . . . . . 14.4 The Convex Set S2 . . . . . . . . . . . . . . .

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The Deficiency of a Hypergraph . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . 8.2 Special Hypergraphs . . . . . . . . . . . 8.3 The Deficiency of a Hypergraph . . 8.4 Applying the Deficiency Technique

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The Tuza Constant q4 . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 9.2 The Tuza constant q4 . . . . . . . . . . . . . . . . 9.3 The West Bound when k ¼ 4 . . . . . . . . . . 9.4 Linear Hypergraphs with Maximum Degree 9.5 Application to Total Dominating Sets . . . .

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14.5 The Convex Set S3 . . . . . . . . . . . . . . . . . . . . . . . 14.5.1 Applications to Strong Independence . . . . 14.5.2 Applications to Double Total Domination 14.6 Double Transversals in Finite Projective Planes . . 15 Conjectures and Open Problems . . . . . . 15.1 Introduction . . . . . . . . . . . . . . . . . 15.2 The Tuza Constant c5 . . . . . . . . . . 15.3 The Tuza Constant q5 . . . . . . . . . . 15.4 The Value kmin . . . . . . . . . . . . . . . 15.5 The Values uk and Uk for Large k . 15.6 Partial Steiner Triple Systems . . . . 15.7 Strong Transversals . . . . . . . . . . . . 15.8 Transversals of Triangles in Simple

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Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

List of Figures

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

2.1 3.1 3.2 4.1 4.2 4.3 4.4 5.1 5.2 7.1 7.2 8.1 8.2 8.3 8.4 9.1 9.2 9.3 9.4 9.5

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14.1 14.2 14.3 14.4 14.5 15.1

The hypergraphs L2 , L4 , and L6 . . . . . . . . . . . . . . . . . . . . . . . The affine plane AGð2; 2Þ and the hypergraph F3 . . . . . . . . . . The affine plane AGð2; 3Þ and the hypergraph F8 . . . . . . . . . . The hypergraph Q113 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The hypergraph Q23 that belongs to the family Q2 . . . . . . . . . . The hypergraph Q32;3 that belongs to the family Q3 . . . . . . . . . The hypergraph Q42;2 that belongs to the family Q4 . . . . . . . . . The generalized triangles T4 and T5 . . . . . . . . . . . . . . . . . . . . The generalized triangle T4 . . . . . . . . . . . . . . . . . . . . . . . . . . . The Fano plane F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The first three extreme points of the convex set B3 . . . . . . . . The fifteen special hypergraphs. . . . . . . . . . . . . . . . . . . . . . . . Edges in E  ðXÞ ¼ fe1 ; . . .; e6 g . . . . . . . . . . . . . . . . . . . . . . . . The transformation creating a quadruple . . . . . . . . . . . . . . . . . The graph G if jQj ¼ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The hypergraph H10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Fano plane F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A hypergraph, F6 , in the family F . . . . . . . . . . . . . . . . . . . . The Heawood graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A quadrilateral-free 4-regular graph G30 of order n ¼ 30 with ct ðG30 Þ ¼ 25 n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The hypergraphs E3 , H5 , and the Fano plane . . . . . . . . . . . . . The affine plane AGð2; 3Þ . . . . . . . . . . . . . . . . . . . . . . . . . . . . The convex set S3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The lower part of the convex set S3 . . . . . . . . . . . . . . . . . . . . The Heawood graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 5-uniform 3-regular hypergraph, H, 3 ðmH þ nH Þ . . . . . . . . . . . . . . . . . . . . . . . . with sðHÞ ¼ 3 ¼ 16

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Chapter 1

Introduction

1.1 Introduction In this chapter, we define the necessary concepts needed in the subsequent chapters. We do assume the reader is acquainted with the basic concepts of graph and hypergraph theory. However, since graph and hypergraph terminology sometimes varies, the book is self-contained, and we clarify the terminology that will be adopted in the book in this introductory chapter.

1.2 Graph and Hypergraph Terminology Hypergraphs are systems of sets which are conceived as natural extensions of graphs. A hypergraph H is a finite set V (H ) of elements, called vertices, together with a finite multiset E(H ) of subsets of V , called hyperedges or simply edges. We write H = (V, E) where V = V (H ) and E = E(H ). The order of H is n(H ) = |V |, and the size of H is m(H ) = |E|. For simplicity, we sometimes denote n(H ) and m(H ) by n H and m H , respectively. The size of an edge in H is its cardinality. A k-edge in H is an edge of size k. The hypergraph H is said to be k-uniform if every edge of H is a k-edge. Every (simple) graph is a 2-uniform hypergraph. Thus graphs are special hypergraphs. For i ≥ 2, we denote the number of edges in H of size i by ei (H ). The degree of a vertex v in H , denoted by d H (v), is the number of edges of H which contain v. A degree-k vertex is a vertex of degree k. An isolated vertex of H is a vertex of degree 0. The minimum and maximum degrees among the vertices of H are denoted by δ(H ) and Δ(H ), respectively. Two vertices x and y of H are adjacent if there is an edge e of H such that {x, y} ⊆ e. The neighborhood of a vertex v in H , denoted N H (v) or simply N (v) if H is clear from the context, is the set of all vertices different from v that are © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_1

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1 Introduction

adjacent to v. A vertex in N (v) is a neighbor of v. The closed neighborhood is the set N H (v) = {v} ∪ N H (v). The neighborhood of a set S of vertices of H is the set N (S) = ∪v∈S N H (v), and its closed neighborhood is the set N H (S) = N H (S) ∪ S. The boundary of S is the set ∂(S) = N H (S) \ S. Thus, ∂ H (S) consists of all vertices of H not in S that have a neighbor in S. If H is clear from context, we simply write N (S) and ∂(S) rather than N H (S) and ∂ H (S). Two vertices x and y of H are connected if there is a sequence x = v0 , v1 , v2 . . . , vk = y of vertices of H in which vi−1 is adjacent to vi for i = 1, 2, . . . , k. A connected hypergraph is a hypergraph in which every pair of vertices is connected. A maximal connected subhypergraph of H is a component of H . Thus, no edge in H contains vertices from different components. A component of H isomorphic to a hypergraph F we call an F-component of H . A subset T of vertices in a hypergraph H is a transversal (also called vertex cover or hitting set in many papers) if T has a nonempty intersection with every edge of H . The transversal number τ (H ) of H is the minimum size of a transversal in H . A transversal of size τ (H ) is called a τ -transversal of H . Transversals in hypergraphs are well studied in the literature (see, for example, [2, 21, 24, 25, 28, 51, 52, 54–57, 72, 74, 85, 88]). A hypergraph H is called an intersecting hypergraph if every two distinct edges of H have a nonempty intersection, while H is called a linear hypergraph if every two distinct edges of H intersect in at most one vertex. We say that two edges in H overlap if they intersect in at least two vertices. A linear hypergraph therefore has no overlapping edges. Linear hypergraphs are well studied in the literature (see, for example, [15, 23, 37, 40, 67, 71, 79, 80, 86]), as are uniform hypergraphs (see, for example, [21–23, 40, 44, 57, 59, 74, 77, 79, 80, 86]). A subset of vertices in a hypergraph H is an independent set if it contains no edge of H . Equivalently, a set of vertices S is an independent set in H if and only if V (H ) \ S is a transversal in H . The independence number a(H ) of H is the maximum cardinality of an independent set in H . We note that n(H ) = τ (H ) + α(H ). We state this observation formally as follows. Observation 1.1 If H is a hypergraph of order n H , then n H = τ (H ) + α(H ). A subset of vertices in a hypergraph H is a strongly independent set if it intersects every edge of H in at most one vertex. Independence in hypergraphs is well studied (see, for example, [11, 16, 60, 72], for recent papers on this topic). Given a hypergraph H and subsets X, Y ⊆ V (H ) of vertices, we let H (X, Y ) denote the hypergraph obtained by deleting all vertices in X ∪ Y from H and removing all (hyper)edges containing vertices from X , removing the vertices in Y from any remaining edges, and removing all resulting isolated vertices, if any. If Y = ∅, we simply denote H (X, Y ) by H (X ); that is, H − X denotes the hypergraph obtained from H by removing the vertices X from H , removing all edges that intersect X , and removing all resulting isolated vertices, if any. Further, if X = {x}, we simply write H − x rather than H − X . When we use the definition H (X, Y ) we furthermore assume that no edges of size zero are created. That is, there is no edge e ∈ E(H )

1.2 Graph and Hypergraph Terminology

3

such that V (e) ⊆ Y \ X . In this case we note that if add X to any τ -set of H (X, Y ), then we get a transversal of H , implying that τ (H ) ≤ |X | + τ (H (X, Y )). We will often use this fact throughout the paper. If H is a hypergraph where every edge has size at least 3, then its complement H is the hypergraph on the same vertex set V (H ) and where e is a hyperedge in the complement if and only if V (H ) \ e is a hyperedge in H . If G is a graph, the complement of G, denoted by G , is formed by taking the vertex set of G and joining two vertices by an edge whenever they are not joined in G. An edge coloring of a graph G is an assignment of colors to the edges of G such that adjacent edges receive different colors. The minimum number of colors needed for an edge coloring is called the chromatic index of the graph, denoted χ  (G). Two edges in a graph G are independent if they are not adjacent in G. A matching in a graph G is a set of independent edges in G, while a matching of maximum cardinality is a maximum matching. The number of edges in a maximum matching of G is called the matching number of G which we denote by a  (G). A vertex is covered by a matching M if the vertex is incident with an edge of M. A perfect matching M in G is a matching in G that covers every vertex of G. Thus, G has a perfect matching if and only if α  (G) = |V (G)|/2. The graph G is factor-critical if G − v has a perfect matching for every vertex v in G. A 1-factor is a spanning 1-regular subgraph. Hence the edge set of a 1-factor is a perfect matching. We use the standard notation [k] = {1, 2, . . . , k}.

Chapter 2

Linear Intersection Hypergraphs

2.1 Introduction In this chapter, we look at transversals in linear intersection hypergraphs. We show that linear intersection hypergraphs have largest possible transversal number among those linear hypergraphs with maximum degree at most two. Recall that a hypergraph H is an intersecting hypergraph if every two distinct edges of H have a nonempty intersection. Thus, a linear intersecting hypergraph is a hypergraph in which every two distinct edges of the hypergraph intersect in exactly one vertex. Our aim in this chapter is to establish an upper bound on the transversal number of a k-uniform linear hypergraph with maximum degree 2. For this purpose, for k ≥ 2 we define the following family L of 2-regular kuniform hypergraphs, L k , of size k + 1 inductively as follows. We define L 2 = K 3 and we define L 3 to be the hypergraph with V (L 3 ) = {v1 , v2 , . . . , v6 } and let E(L 3 ) = {e1 , e2 , e3 , e4 }, where e1 = {v1 , v2 , v3 }, e2 = {v1 , v4 , v5 }, e3 = {v2 , v4 , v6 }, and e4 = {v3 , v5 , v6 }. For k ≥ 2, suppose the hypergraph L k has been constructed and that E(L k ) = {e1 , e2 , . . . , ek+1 }. Let L k+2 be the hypergraph of order n(L k ) + 2k + 3 with V (L k+2 ) = V (L k ) ∪ {v} ∪ {u 1 , u 2 , . . . , u k+1 } ∪ {w1 , w2 , . . . , wk+1 } and with edge set E(L k+2 ) = { f 1 , f 2 , . . . , f k+3 }, where f i = ei ∪ {u i , wi } for 1 ≤ i ≤ k + 1 and where f k+2 = {v, u 1 , . . . , u k+1 } and f k+3 = {v, w1 , . . . , wk+1 }. The hypergraphs L 2 , L 4 , and L 6 , for example, are illustrated in Figure 2.1, albeit without their labels. By construction, the hypergraph L k is a k-uniform, 2-regular, linear intersecting hypergraph.

2.2 Hypergraph Duality In this section, we define hypergraph duality. The dual H ∗ of a hypergraph H is a hypergraph whose vertices and edges are interchanged. Thus, given a hypergraph H we form the dual hypergraph H ∗ whose vertices are the edges of H and for every © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_2

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6

2 Linear Intersection Hypergraphs

(a) L2

(b) L4

(c) L6

Fig. 2.1 The hypergraphs L 2 , L 4 , and L 6

vertex v of H , we form a hyperedge in H ∗ consisting of those vertices corresponding with the set of edges of H incident with v. More formally, if V (H ) = {v1 , . . . , vn } and E(H ) = {e1 , . . . .em }, then V (H ∗ ) = E(H ) and E(H ∗ ) = {e1∗ , . . . , en∗ } where ei∗ is the set of edges in H that contain the vertex vi for i ∈ [n]. We note that the operation of taking the dual of a hypergraph is an involution, i.e., (H ∗ )∗ = H .

2.3 Properties of Linear Intersecting Hypergraphs In this section, we establish some properties of linear intersecting hypergraphs. For a hypergraph of H , we denote its order and size by n H = n(H ) and m H = m(H ), respectively. Lemma 2.1 ([28]) For k ≥ 2, there exists a unique k-uniform 2-regular, linear hypergraph, namely, L k . This hypergraph, H , has m H = k + 1 and n H =  intersecting k+1 . 2 Proof For k ≥ 2, let H be a k-uniform, 2-regular, linear intersecting hypergraph. The dual hypergraph H ∗ of H is a k-regular complete 2-uniform hypergraph;   that edges. is, H ∗ is a complete graph K k+1 , and therefore has k + 1 vertices and k+1 2   ∗ Thus, n H = m(H ∗ ) = k+1 and m = n(H ) = k + 1. Further we note that the dual H 2 hypergraph H ∗ is unique, and hence its dual hypergraph (H ∗ )∗ = H is also unique. As observed earlier, for k ≥ 2 the hypergraph L k is a k-uniform, 2-regular, linear intersecting hypergraph. Hence, H = L k ; that is, the hypergraph L k is therefore the unique k-uniform 2-regular, linear intersecting hypergraph. 

2.4 Linear Hypergraphs with Maximum Degree Two

7

2.4 Linear Hypergraphs with Maximum Degree Two In this section, we determine a tight upper bound on the transversal number of a linear hypergraph with maximum degree two. For this purpose, we will use the following result on edge coloring of graphs. Vizing [91] and Gupta [45] independently proved that the edge chromatic index of a graph G always takes one of two values, either Δ(G) or Δ(G) + 1. Theorem 2.1 (Vizing’s Theorem) If G is a graph, then Δ(G) ≤ χ  (G) ≤ Δ(G) + 1. We shall also need the following very special case of the Gallai-Edmonds Structure Theorem [5, 30, 41, 92]. Recall that a graph G is factor-critical if G − v has a perfect matching for every vertex v in G. Theorem 2.2 (Corollary of Gallai-Edmonds Structure Theorem) If G is a graph with the property that no vertex is covered by every maximum matching in G, then every component of G is factor-critical. For k ≥ 2, let E k denote the k-uniform hypergraph on k vertices with exactly one edge. We are now in a position to present a tight upper bound on the transversal number of a linear hypergraph with maximum degree two. Theorem 2.3 ([28]) For k ≥ 2, if H is a k-uniform, linear, connected hypergraph with maximum degree 2, then τ (H ) ≤

1 (n + m H ) k+1 H

with equality if and only if H consists of a single edge E k or k is even and H = L k . Proof For k ≥ 2, let H be a k-uniform, linear connected hypergraph of order n H and size m H satisfying Δ(H ) ≤ 2. If H has an isolated vertex, then by the connectivity of H we have that n H = 1 and m H = 0, implying that (k + 1)τ (H ) = 0 < n H + m H . Hence we may assume that δ(H ) ≥ 1. For i ∈ [2], let n i (H ) be the number of vertices of degree i in H , and so n 1 (H ) + n 2 (H ) = n H . By the k-uniformity of H we have that n 1 (H ) + 2n 2 (H ) = km H , or, equivalently, n 2 (H ) = km H − n H .

(2.1)

If Δ(H ) = 1, then by the connectivity of H we have that H = E k , and so n H = k, m H = 1, τ (H ) = 1 and (k + 1)τ (H ) = n H + m H . Hence, we may assume that Δ(H ) = 2, for otherwise the desired result follows. We now form a graph G whose vertices are the edges of H , with edges of H adjacent in G if they intersect in H (and therefore have at least one common vertex). Thus, the edges of G correspond to the vertices of degree at least 2 in H : if a vertex of

8

2 Linear Intersection Hypergraphs

H is contained in the edges e and f of H , then the corresponding edge of the graph G joins vertices e and f of G. We call the resulting graph G the intersection graph of H (also called the line graph of H in the literature). Since H is k-uniform and Δ(H ) ≤ 2, the maximum degree, Δ(G), in G is at most k. By construction we have n(G) = m H and m(G) = n 2 (H ) where n 2 (H ) is the number of vertices of degree 2 in H . Further since H is connected, so too is G. Let C be an arbitrary edge coloring of the edges of G using χ  (G) colors. The matching number of G is at least the cardinality of a maximum edge color class in C , and so, by Vizing’s Theorem, α  (G) ≥

m(G) m(G) n 2 (H ) m(G) ≥ ≥ = . χ  (G) Δ(G) + 1 k+1 k+1

(2.2)

Let M be a maximum matching in G. Let T denote the set of vertices in H corresponding to the edges in M. Then, |T | = |M| = α  (G) and by Eqs. (2.1) and (2.2), we have that n 2 (H ) km H − n H |T | ≥ = . (2.3) k+1 k+1 Now, T covers 2α  (G) edges in H , noting that each vertex in T has degree 2 in H and the set T is an independent set in H . Add to T a vertex from each edge of H not covered by T , yielding a transversal T ∗ of H . Hence, the lower bound on |T | in Equation (2.1) yields τ (H ) ≤ |T ∗ | = |T | + (m H − 2|T |) = m H − |T | ≤ mH − =

km H − n H k+1

1 (m + n H ), k+1 H

which establishes the desired upper bound. 1 Suppose next that τ (H ) = k+1 (n H + m H ) (and still Δ(H ) = 2). Then we must have equality throughout the above inequality chain, implying in particular that we have equality throughout in Eqs. (2.2) and (2.3). Thus, Δ(G) = k, χ  (G) = k + 1 and α  (G) = m(G)/(k + 1), implying that each color class of C must be a maximum matching. Since Δ(G) = k and there are k + 1 color classes, this in turn implies that no vertex is covered by every maximum matching in G. Therefore by the corollary of the Gallai-Edmonds Structure Theorem, the graph G, which by assumption is connected, has odd order and is factor-critical. Therefore, α  (G) = (n(G) − 1)/2. Since Δ(G) = k, we note that 2m(G) ≤ kn(G). Hence, kn(G) n(G) − 1 m(G) = α  (G) = ≤ , 2 k+1 2(k + 1)

(2.4)

2.4 Linear Hypergraphs with Maximum Degree Two

9

implying that n(G) ≤ k + 1. However, n(G) ≥ Δ(G) + 1 = k + 1. Consequently, n(G) = k + 1, α  (G) = k/2, and k is even. Since each of the k + 1 color classes is a maximum matching, we therefore have that m(G) = (k + 1)α  (G) = k(k + 1)/2 = k+1 , implying that G is the complete graph, K k+1 , on k + 1 vertices. 2 We show next that H = L k ; that is, H is the unique k-uniform, 2-regular, linear intersecting hypergraph. The hypergraph H has size m H =n(G) = k + 1 and . In particular, we order n H = n 1 (H ) + n 2 (H ) = n 1 (H ) + m(G) = n 1 (H ) + k+1 2 note that 2n 2 (H ) = k(k + 1). We therefore have that k(k + 1) = km H = n 1 (H ) + 2n 2 (H ) = n 1 (H ) + k(k + 1), implying that n 1 (H ) = 0 and therefore that H is 2-regular. Thus, H is a k-uniform,   . 2-regular, linear connected hypergraph of size m H = k + 1 and order n H = k+1 2 If H is not an intersecting hypergraph and e and f are two edges in H that do not intersect, then the two vertices in the intersection graph G of H corresponding to the edges e and f would not be adjacent, contradicting the fact that G is a complete graph. Hence, H is an intersecting hypergraph, implying by Lemma 2.1 that H = L k , as desired. This completes the proof of Theorem 2.3.  Note that the bound in Theorem 2.3 is not true if we relax the linearity constraint. For example, the unique 2-regular 4-uniform hypergraph containing three edges and six vertices has transversal number two and therefore would not satisfy Theorem 2.3.

Chapter 3

Finite Affine Planes and Projective Planes

3.1 Introduction In this chapter, we look at finite affine planes and finite projective planes. We show a connection between linear uniform hypergraphs with large transversal number and finite affine planes. In geometry, a finite affine plane is a system of points and lines that satisfy the following rules: • Any two distinct points lie on a unique line. • Each line has at least two points. • Given a point and a line, there is a unique line which contains the point and is parallel to the line, where two lines are called parallel if they are equal or disjoint. • There exist three non-collinear points (points not on a single line). A finite affine plane of order q ≥ 2 is a collection of q 2 points and q 2 + q lines, such that each line contains q points and each point is contained in q + 1 lines. A finite projective plane is a system of points and lines that satisfy the following rules, where we say that a point P is incident with a line  in the plane if the line  contains the point P. • Any two distinct points lie on a unique line. • Given any two distinct lines, there is exactly one point incident with both of them. • There are four points such that no line is incident with more than two of them. Special kind of projective planes, called Desarguesian planes, named after Girard Desargues, are constructed from a three-dimensional vector space over a skew field. We will not in this book describe the construction, but just note that Desarguesian planes are denoted by P G(2, q), where P G stands for “projective geometry,” the “2” is the dimension and q is called the order of the projective plane, which is one less than the number of points on any line. When the projective plane is constructed from a finite field then they are also referred to as Galois planes. We note that there © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_3

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3 Finite Affine Planes and Projective Planes

also do exist non-Desarguesian planes. For example, when the order is 9, there exist three non-isomorphic non-Desarguesian planes as well as the Desarguesian plane P G(2, 9). If q = p n for some prime p and integer n ≥ 1, then there exists a Desarguesian plane with q 2 + q + 1 points, which as mentioned above is denoted by P G(2, q). A projective plane may be considered as a hypergraph, whose vertex set is the set of points and whose edge set is the set of lines of the plane. The hypergraph associated with a projective plane P G(2, q) is a (q + 1)-uniform, linear hypergraph on q 2 + q + 1 vertices. An affine plane can be obtained from any projective plane by removing a line and all the points on it, and conversely any affine plane can be used to construct a projective plane by adding a line at infinity, each of whose points is that point at infinity where an equivalence class of parallel lines meets. Deleting one of the lines and all its incident points from a Desarguesian plane P G(2, q) produces a finite affine plane, denoted by AG(2, q), of order q. We note that the order q of the resulting finite affine plane is the number of points on any of its lines, which is the order of the projective plane from which it comes. An affine plane of order q exists if and only if a projective plane of order q exists. When there is only one affine plane of order q there is only one projective plane of order q, but the converse is not true. Jamison [66] and Brouwer and Schrijver [20] proved that the minimum cardinality of a subset of AG(2, k) which intersects all lines is 2(k − 1) + 1. Theorem 3.1 ([20, 66]) The minimum cardinality of a subset of AG(2, k) which intersects all lines is 2(k − 1) + 1. One of the most fundamental results in combinatorics is the result due to Bose [18] that there are k − 1 mutually orthogonal Latin squares if and only if there is an affine plane of order k. The prime power conjecture for affine and projective planes states that there is an affine plane of order k if only if k is a prime power. Veblen and Wedderburn [90] proved that there exist non-Desarguesian finite projective planes of small orders. In 1989 Lam, Thiel and Swiercz [76] established the non-existence of a finite projective plane of order 10 using a computer-based proof. Lam [75] gave an expository article describing the history of the problem and how computers were used to solve it. The affine plane AG(2, 2) of dimension 2 and order 2 is illustrated in Figure 3.1a. This is equivalent to a linear 2-uniform 3-regular hypergraph F4 (the complete graph K 4 on four vertices), where the lines of AG(2, 2) correspond to the 2-edges of F4 . Deleting any vertex from F4 yields the 2-uniform 2-regular hypergraph F3 (the complete graph K 3 ) illustrated in Figure 3.1b, where the lines correspond to the 2edges of F3 . Further, deleting any vertex from F3 yields a linear 2-uniform hypergraph F2 (the complete graph K 2 , which we also denote by E 2 ). The affine plane AG(2, 3) of dimension 2 and order 3 is illustrated in Figure 3.2a. This is equivalent to a linear 3-uniform 4-regular hypergraph F9 on nine vertices, where the lines of AG(2, 3) correspond to the 3-edges of F9 . Deleting any vertex

3.1 Introduction

13

(a) F4 = AG(2, 2)

(b) F3

Fig. 3.1 The affine plane AG(2, 2) and the hypergraph F3

(a) F9 = AG(2, 3)

(b) F8

Fig. 3.2 The affine plane AG(2, 3) and the hypergraph F8

from F9 yields the hypergraph F8 (see Figure 3.2b, where the lines correspond to the 3-edges of F8 ). Further, deleting any vertex from F8 yields a linear 3-uniform hypergraph F7 . Also, F9 − e = F9 − f for any two distinct edges e and f in F9 . Let xi j be the vertex of F9 in the ith row and jth column for i ∈ [3] and j ∈ [3]. Thus, F8 and F9 can be defined in terms of the following 3-uniform hypergraphs. Definition 3.1 Let F8 be the hypergraph with vertex set V (F8 ) = {x11 , x12 , x13 , x21 , x22 , x23 , x31 , x32 } and the following edge set. E(F8 ) = {{x11 , x12 , x13 }, {x21 , x22 , x23 }, {x11 , x21 , x31 }, {x12 , x22 , x32 }, {x11 , x23 , x32 }, {x12 , x23 , x31 }, {x13 , x22 , x31 }, {x13 , x21 , x32 }}. Definition 3.2 Let F9 be obtained from the hypergraph F8 of Definition 3.1 by adding the vertex x33 and the edges {x31 , x32 , x33 }, {x12 , x21 , x33 }, {x11 , x22 , x33 }, {x13 , x23 , x33 }. Definition 3.3 Let F7 be obtained from the hypergraph F8 of Definition 3.1 by deleting the vertex x32 . As observed in [56], any set of five vertices in F9 contains an edge, due to the following. Let X ⊆ V (F9 ) and |X | = 5 and suppose, to the contrary, that X does not

14

3 Finite Affine Planes and Projective Planes

contain an edge of F9 . Due to the three horizontal edges {xi1 , xi2 , xi3 } for i ∈ [3] and due to symmetry, we may assume without loss of generality that X contains two vertices in the first row, two vertices in the second row, and one vertex from the last row. Further we may assume that x11 , x21 ∈ X . The presence of the edge {x11 , x21 , x31 } / X , and so either x32 ∈ X or x33 ∈ X . By symmetry we therefore implies that x31 ∈ may assume that x32 ∈ X . Due to the edges {x11 , x23 , x32 } and {x13 , x21 , x32 } we / X and x23 ∈ / X , implying that {x12 , x22 , x32 } ⊆ X , a contradiction as note that x13 ∈ {x12 , x22 , x32 } is an edge in F9 . Therefore, τ (F9 ) ≥ 5. Since {x11 , x21 , x23 , x32 , x33 } hits every edge, τ (F9 ) ≤ 5. Consequently, τ (F9 ) = 5 which is shown, for example, in [20, 66]. As observed in [56], since F8 keeps all edges on its eight vertices, it remains true that any set of five vertices there also contains an edge, implying that τ (F8 ) ≥ 4. Since {x11 , x21 , x23 , x32 } hits every edge, τ (F8 ) ≤ 4. Consequently, τ (F8 ) = 4 which is shown, for example, in [54]. Since F7 keeps all edges on its seven vertices, it remains true that any set of five vertices there also contains an edge, implying that τ (F7 ) ≥ 3. Since {x11 , x21 , x23 } hits every edge of F7 , τ (F7 ) ≤ 3. Consequently, τ (F8 ) = 3. We state these results formally as follows. Observation 3.4 τ (F7 ) = 3, τ (F8 ) = 4 and τ (F9 ) = 5.

3.2 Finite Affine Planes and Linear Uniform Hypergraphs In this section, we show that there is a connection between linear uniform hypergraphs with large transversal number and affine planes and projective planes, as shown by the following result. As before, we denote the order and size of a hypergraph H by n H = n(H ) and m H = m(H ), respectively. We are now in a position to present the following result due to the authors [64]. Theorem 3.2 ([64]) Let Fk 2 be the linear, k-uniform, (k + 1)-regular hypergraph of order k 2 which is equivalent to the affine plane AG(2, k) of order k for some k ≥ 2. Let e ∈ E(Fk 2 ) be an arbitrary edge in Fk 2 and let X ⊆ V (e) be any nonempty subset of vertices belonging to the edge e. If H = Fk 2 (X ) is the linear, k-uniform hypergraph obtained from Fk 2 by deleting the vertex set X and all edges intersecting X , then 1 (n + m H ). τ (H ) = k+1 H Proof Let Fk 2 , e, and X be defined as in the statement of the theorem. For notational simplicity, as k is fixed, denote Fk 2 by F. By Theorem 3.1, the transversal number of F is 2(k − 1) + 1; that is, τ (F) = 2(k − 1) + 1. Let e ∈ E(F) \ {e} be an arbitrary edge intersecting e. Since F is equivalent to an affine plane, every edge in E(F) \ {e, e } will intersect e or e (or both e and e ). Thus, V (e) ∪ V (e ) is a transversal in F of size 2(k − 1) + 1 = 2k − 1 as F is k-uniform. Let H = Fk 2 (X ) = F − X ; that is,

3.2 Finite Affine Planes and Linear Uniform Hypergraphs

15

H is obtained from F by deleting X and all edges incident with X . If TX is a transversal in H , then TX ∪ X is a transversal in F, which implies that τ (H ) ≥ 2k − 1 − |X |. However, (V (e) ∪ V (e )) \ X is a transversal in H of size 2k − 1 − |X |, and so τ (H ) ≤ 2k − 1 − |X |. Consequently, τ (H ) = 2k − 1 − |X |. We will now compute the order and size of H . The order of H is n H = n F − |X | = k 2 − |X |. Let x be an arbitrary vertex in X . As F − x is k-regular and has order k 2 − 1, the hypergraph F − x contains k 2 − 1 edges. Further since F is linear and e is not an edge in F − x, no edge in F − x contains more than one vertex from X \ {x}. Therefore, we remove k edges from F − x for every vertex in X \ {x} we remove when constructing H . Thus, the size of H is m H = k 2 − 1 − (|X | − 1)k = k 2 + k − 1 − k|X |. Therefore, τ (H ) = 2k − 1 − |X | 1 ((k 2 − |X |) + (k 2 + k − 1 − k|X |)) = k+1 1 (n + m H ). = k+1 H



We close this section with the following remarks. Let Lk denote the class of 1 (n H + m H ) for some k ≥ 2 and for all k-uniform linear hypergraphs. If τ (H ) ≤ k+1 hypergraphs H ∈ Lk , then the hypergraphs H that achieve equality in this bound cannot have average degree greater than k as removing any vertex, x, of degree more than k from H and applying the bound on τ (H − x) gives us a transversal in H of size at most the following: τ (H − x) + |{x}| ≤

n + mH (n H − 1) + (m H − (k + 1)) +1< H , k+1 k+1

1 a contradiction. Similarly, if τ (H ) ≤ k+1 (n H + m H ) for some k ≥ 2 and for all hypergraphs H ∈ Lk , then the hypergraphs H ∈ Lk that achieve equality in this bound cannot have average degree less than 1 as then there are isolated vertices that can be removed. As a consequence of Theorem 3.2 we therefore have the following result.

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3 Finite Affine Planes and Projective Planes

1 Theorem 3.3 Suppose that τ (H ) ≤ k+1 (n H + m H ) for some k ≥ 2 and for all hypergraphs H ∈ Lk . If the affine plane AG(2, k) of order k exists, then this bound is tight for average degree 1 and average degree k and for a number of average degrees in the interval from 1 to k.

Thus by Theorem 3.3, if the affine plane AG(2, k) of order k exists for some k ≥ 2, then there are a large number of densities of hypergraphs H ∈ Lk such that τ (H ) =

1 (n + m H ). k+1 H

We remark that Theorem 3.3 is somewhat surprising as there are no similar kinds of bounds which hold for a wide variety of average degrees if we consider non-linear hypergraphs.

Chapter 4

The Tuza Constants

4.1 Introduction In this chapter, we define the so-called Tuza constants, ck and qk , for the class of k-uniform hypergraphs and for the subclass of k-uniform linear hypergraphs, respectively. We determine in this chapter the Tuza constants for small values of k, namely, k = 2 and k = 3. Let H be a hypergraph of order n H = n(H ) and size m H = m(H ). For k ≥ 2, let Hk denote the class of all k-uniform hypergraphs. Tuza [88] (pictured below) proposed the problem of determining or estimating the best possible constants ck (which depends only on k) such that τ (H ) ≤ ck (n H + m H ) for all H ∈ Hk . These constants are given by τ (H ) . ck = sup n + mH H ∈H k H

Zsolt Tuza

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_4

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4 The Tuza Constants

We call the constants ck the Tuza constants. Very few papers give bounds on the transversal number for linear hypergraphs, even though these appear in many applications, as it seems difficult to utilize the linearity in the known techniques. In this chapter, we nevertheless consider the class of k-uniform linear hypergraphs, which we denote by Lk . Motivated by Tuza [88], we propose an analogous problem of determining or estimating the best possible constants qk (which depends only on k) such that τ (H ) ≤ qk (n H + m H ) for all H ∈ Lk . These constants are given by qk = sup

H ∈L k

τ (H ) . nH + mH

We also refer to the constants qk as the Tuza constants. For k ≥ 3, the family Lk is a proper subfamily of Hk , implying that qk ≤ ck and that strict inequality may be possible. For k ≥ 2, let E k denote the k-uniform hypergraph on k vertices with exactly one edge. If H = E k , then H ∈ Lk and τ (H )/(n H + m H ) = 1/(k + 1), implying that qk ≥ 1/(k + 1). We state these observations formally as follows. Observation 4.1 For k ≥ 2, the Tuza constants satisfy ck ≥ qk ≥

1 . k+1

For k ≥ 2 and Δ ≥ 1, let Lk,Δ denote the subclass of Lk consisting of all kuniform linear hypergraphs of maximum degree Δ. Further, let qk,Δ = sup

H ∈L k,Δ

τ (H ) . nH + mH

As an immediate consequence of Theorem 2.3, we have the following result. Theorem 4.2 For all k ≥ 2, the Tuza constant qk,2 =

1 . k+1

4.2 The Tuza Constants c2 and q2 In this section, we determine the Tuza constants c2 and q2 . Thus, we restrict our attention in this simplest case when k = 2 to 2-uniform hypergraphs, which are graphs (possibly with multiple edges). It is a simple exercise to show that the Tuza constant c2 = 13 as first shown by Erd˝os (pictured below) and Tuza [32, p. 1180], and subsequently by Thomassé and Yeo [85].

4.2 The Tuza Constants c2 and q2

19

Paul Erd˝os As a gentle introduction to the Tuza constants, we prove that c2 = q2 = characterize the hypergraphs H in H2 satisfying τ (H ) = c2 (n H + m H ).

1 3

and we

Theorem 4.3 If H ∈ H2 , then τ (H ) ≤ 13 (n H + m H ), with equality if and only if every component of H is a single edge E 2 or a copy of K 3 . Proof. Let H ∈ H2 , and so H is a 2-uniform hypergraph of order n H and size m H . We proceed by induction on the number n H ≥ 1 of vertices to show that τ (H ) ≤ (n H + m H )/3, with equality if and only if every component of H is a single edge E 2 or a copy of K 3 . If m H = 0, then τ (H ) = 0 < (n H + m H )/3 and the result is immediate. Hence we may assume that m H ≥ 1, and so n H ≥ 2. If n H = 2, then H = E 2 and τ (H ) = 1 and n H + m H = 2 + 1 = 3 = 3τ (H ), as desired. This establishes the base case. Assume, then, that n H ≥ 3 and that all 2-uniform hypergraphs H  of order n H and size m H where n H < n H satisfy τ (H  ) ≤ (n H + m H )/3, with equality if and only if every component of H  is a single edge E 2 or a copy of K 3 . Let H be a 2-uniform hypergraph of order n H and size m H . In what follows, we denote the order and size of a hypergraph H  by n H and  m H , respectively. Further if n H < n H , then applying the inductive hypothesis to H  , we let T  denote a minimum transversal in H  such that |T  | ≤ (n H + m H )/3. By linearity we may assume that H is connected, since the induction hypothesis applies to each component when H is disconnected. Since n H ≥ 3, this assumption implies that Δ(H ) ≥ 2. Let v be a vertex of maximum degree in H , and consider the hypergraph H  = H − v. We note that n H = n H − 1 and m H = m H − d H (v) = m H − Δ(H ) ≤ m H − 2. Thus, |T  | ≤ (n H + m H )/3 ≤ (n H + m H )/3 − 1. Since T  ∪ {v} is a transversal of H , we have τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/3. Further if Δ(H ) ≥ 3, then m H ≤ m H − 3, implying that τ (H ) < (n H + m H )/3. Hence, we may assume that Δ(H ) = 2, implying that H is a path or a cycle.

20

4 The Tuza Constants

If H is a path (and still n H ≥ 3), then we choose v to be a support vertex on the path and let u be a leaf neighbor of v. We now consider the hypergraph H  = H − {u, v}. We note that n H = n H − 2 and m H = m H − 2. Thus, |T  | ≤ (n H + m H )/3 ≤ (n H + m H )/3 − 4/3. Since T  ∪ {v} is a transversal of H , we have τ (H ) ≤ |T  | + 1 < (n H + m H )/3. Hence, we may assume that H is a cycle. Let v be an arbitrary vertex of the cycle H , and consider the hypergraph H  = H − v. We note that n H = n H − 1 and size m H = m H − 2. Further, we note that H  is a path. If n H ≥ 3, then as shown above |T  | < (n H + m H )/3. Since T  ∪ {v} is a transversal of H , we have τ (H ) ≤ |T  | + 1 < (n H + m H )/3. Hence, we may assume  that n H + 3, and so H = K 3 , as desired. Adopting our notation in Chapter 3.1, recall that F2 and F3 denote the complete graphs K 2 and K 3 , respectively. Further, recall that F3 and F2 are obtained from the affine plane AG(2, 2) of order 2 by deleting one and two vertices, respectively. Hence, Theorem 4.3 can be restated as follows. Theorem 4.4 If H ∈ H2 , then τ (H ) ≤ 13 (n H + m H ), with equality if and only if every component of H is obtained from the affine plane AG(2, 2) of order 2 by deleting one or two vertices; that is, every component of H is F2 or F3 . As an immediate consequence of Theorem 4.4, we have that c2 = sup

H ∈H 2

τ (H ) 1 ≤ . nH + mH 3

By Observation 4.1, we note that c2 ≥ 13 . Consequently, the Tuza constant c2 = 13 . We note that the family L2 is precisely the family H2 , and so q2 = c2 . We state these results, which determine precisely the values of the Tuza constants c2 and q2 , formally as follows. Theorem 4.5 ([32, p. 1180],[85]) The Tuza constants q2 = c2 = 13 .

4.3 The Tuza Constants c3 and q3 In this section, we consider 3-uniform hypergraphs and determine the Tuza constants c3 and q3 . Chvátal and McDiarmid [24] (pictured below) and Tuza [88] independently established that the Tuza constant c3 = 41 . The proof we present of this result is from [53].

4.3 The Tuza Constants c3 and q3

Vašek Chvátal

21

Colin McDiarmid

Theorem 4.6 ([24, 88]) If H ∈ H3 , then τ (H ) ≤ 14 (n H + m H ). Proof. Let H be a 3-uniform hypergraph of order n H and size m H . We proceed by induction on the number n H of vertices to show that τ (H ) ≤ (n H + m H )/4. If m H = 0, then τ (H ) = 0 and the result follows. Hence we may assume that m H ≥ 1, and so n H ≥ 3. If n H = 3, then H = E 3 and τ (H ) = 1 and n H + m H = 3 + 1 = 4 = 3τ (H ), as desired. This establishes the base case. Assume, then, that n H ≥ 4 and that all 3-uniform hypergraphs H  of order n H and size m H where n H < n H satisfy τ (H  ) ≤ (n H + m H )/4. Let H be a 3-uniform hypergraph of order n H and size m H . In what follows, we denote the order and size of a hypergraph H  by n H and size m H , respectively. Further if n H < n H , then applying the inductive hypothesis to H  , we let T  denote a transversal in H  such that |T  | ≤ (n H + m H )/4. If H has an isolated vertex x, then we apply the induction to the hypergraph H  = H − x of order n H = n H − 1 and size m H = m H . In this case, τ (H ) = τ (H  ) ≤ (n H + m H )/3 < (n H + m H )/3. Hence, we may assume that H has no isolated vertex, implying that δ(H ) ≥ 1. Suppose that Δ(H ) ≥ 3. Let v be a vertex of degree at least 3 in H . We now consider the hypergraph H  = H − v and note that n H = n H − 1 and size m H = m H − d H (v) ≤ m H − 3. Thus, |T  | ≤ (n H + m H )/4 ≤ (n H + m H )/4 − 1. Since T  ∪ {v} is a transversal of H , we have τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/4, as desired. Hence, we may assume that Δ(H ) ≤ 2. Suppose that H contains an edge e such that all three vertices in the edge e have degree 1 in H . In this case, we consider the hypergraph H = H  − V (e) of order n H = n H − 3 and size m H = m H − 1. Thus, |T  | ≤ (n H + m H )/4 ≤ (n H + m H )/4 − 1. Since T  ∪ {v} is a transversal of H where v is an arbitrary vertex in e, we have τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/4, as desired. Hence, we may assume that every edge of H contains a vertex of degree 2.

22

4 The Tuza Constants

Suppose that H contains an edge e that contains a vertex x of degree 1 and a vertex y of degree 2. In this case, we consider the hypergraph H  = H − {x, y} of order n H = n H − 2 and size m H = m H − 2. Thus, |T  | ≤ (n H + m H )/4 ≤ (n H + m H )/4 − 1. Since T  ∪ {y} is a transversal of H , we have τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/4, as desired. Hence, we may assume that H is 2-regular; that is, d H (v) = 2 for every vertex v in H . If some edge e of H appears twice in H , then we let H  = H − e. In this case, n H = n H and m H = m H − 1. Further, τ (H ) = τ (H  ) ≤ (n H + m H )/3 < (n H + m H )/3. Hence, we may assume that H has no duplicated edges. Suppose two edges e and f of H intersect in two vertices, say a and b. Let e = {a, b, c} and f = {a, b, d}, and consider the hypergraph H  = H − {a, b} of order n H = n H − 2 and size m H = m H − 1. Thus, |T  | ≤ (n H + m H )/4 ≤ (n H + m H )/4 − 2. Since T  ∪ {a} is a transversal of H , we have τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/4, as desired. Hence, we may assume that H is a linear hypergraph. Let e = {x1 , x2 , x3 } be any edge in H . Let ei be the edge in H different from e that contains the vertex xi for i ∈ [3]. From our earlier assumptions, we note that |(ei ∪ e j ) \ {x1 , x2 , x3 }| ≥ 3 for 1 ≤ i < j ≤ 3, and we let ei j be any 3-element subset of (ei ∪ e j ) \ {x1 , x2 , x3 }. Let H  be obtained from the hypergraph H − {x1 , x2 , x3 } by adding the edges e12 , e13 and e23 . We note that H  is a 3-uniform hypergraph of order n H = n H − 3 and size m H = m H − 1. If T  contains at least one vertex from each of the deleted edges e1 , e2 , and e3 , then T  ∪ {x1 } is a transversal of H , implying that τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/4 + 1 = (n H + m H )/4. Hence we may assume that at least one of e1 , e2 , and e3 , say e1 , is not covered by T  in H , and so T  contains no vertex in e1 . This implies that T  ∩ e12 ⊆ e2 and T  ∩ e13 ⊆ e3 . Thus, e2 and e3 are covered by T  in H . Therefore, T = T  ∪ {x1 } is a transversal in H , implying that τ (H ) ≤ |T  | + 1 ≤ (n H + m H )/4 + 1 = (n H + m H )/4, as desired. This completes the inductive proof.  As an immediate consequence of Theorem 4.4, we have that c3 = sup

H ∈H 3

τ (H ) 1 ≤ . nH + mH 4

By Observation 4.1, we note that c3 ≥ 41 . Consequently, the Tuza constant c3 = 14 . By Observation 4.1, we note that 14 ≤ q3 ≤ c3 = 41 , implying that q3 = 41 . We state this formally as follows. Theorem 4.7 ([24, 88]) The Tuza constants q3 = c3 = 41 . We define next an infinite family of 3-regular, connected hypergraphs H ∈ H3 that satisfy τ (H ) = 41 (n H + m H ).

4.3 The Tuza Constants c3 and q3

23

Definition 4.1 Let i ≥ 1 be an arbitrary integer. Let Q i1 be the 3-uniform hypergraph defined as follows. Let V (Q i1 ) = {x0 , x1 , . . . , xi , y0 , y1 , . . . , yi }. Let the edge set of Q i1 be defined as follows: E(Q i1 ) = {{xi , x0 , y0 }, {yi , x0 , y0 }} ∪ (

i 

{{xa−1 , xa , ya }, {ya−1 , xa , ya }}).

a=1

Let Q1 = {Q 11 , Q 12 , Q 13 , . . .}. The hypergraph Q 113 is shown in Figure 4.1, albeit without labels. The authors showed in [54] that if H ∈ Q1 , then τ (H ) = 41 (n H + m H ). Further, they define three other infinite families, which we call Q2 , Q3 , and Q4 , of connected hypergraphs H , where all edges in H have size at least 3 and where Δ(H ) ≤ 3, for which equality holds. Definition 4.2 Let i ≥ 0 be an arbitrary integer. Let Q i2 be the 3-uniform hypergraph defined as follows. Let V (Q i1 ) = {u, x0 , x1 , . . . , xi , y0 , y1 , . . . , yi }. Let the edge set of Q i2 be defined as follows: E(Q i2 ) = {{u, x0 , y0 }} ∪ (

i 

{{xa−1 , xa , ya }, {ya−1 , xa , ya }}).

a=1

Let Q2 = {Q 20 , Q 21 , Q 22 , . . .}. The hypergraph Q 23 that belongs to the family Q2 is shown in Figure 4.2.

Fig. 4.1 The hypergraph Q 113

24

4 The Tuza Constants

Fig. 4.2 The hypergraph Q 23 that belongs to the family Q2

Definition 4.3 Let i, j ≥ 1 be arbitrary integers. Let Q i,3 j be the hypergraph defined as follows: V (Q i,3 j ) = {t1 , t2 , t3 , x1 , x2 , . . . , xi , y1 , y2 , . . . , yi , w1 , w2 , . . . , w j , z 1 , z 2 , . . . , z j }, E 1 = {{t1 , x1 , y1 }, {t2 , x1 , y1 }} ∪ (

i 

{{xa−1 , xa , ya }, {ya−1 , xa , ya }} ),

a=2

E 2 = {{t1 , w1 , z 1 }, {t3 , w1 , z 1 }} ∪ (

j 

{{wb−1 , wb , z b }, {z b−1 , wb , z b }}),

b=2

E(Q i,3 j ) = {{t1 , t2 , t3 }} ∪ E 1 ∪ E 2 . Let Q3 =



{Q i,3 j }.

i≥1 j≥1

The hypergraph Q 32,3 that belongs to the family Q3 is shown in Figure 4.3.

Fig. 4.3 The hypergraph Q 32,3 that belongs to the family Q3

4.3 The Tuza Constants c3 and q3

25

Fig. 4.4 The hypergraph Q 42,2 that belongs to the family Q4

Definition 4.4 Let i, j ≥ 1 be arbitrary integers. Let Q i,4 j be the hypergraph defined as follows: V (Q i,4 j ) = {u, x0 , x1 , . . . , xi , y0 , y1 , . . . , yi , w0 , w1 , . . . , w j , z 0 , z 1 , . . . , z j }, E1 =

i 

{{xa−1 , xa , ya }, {ya−1 , xa , ya }},

a=1

E2 =

j 

{{wb−1 , wb , z b }, {z b−1 , wb , z b }},

b=1

E(Q i,4 j ) = {{u, x0 , y0 }, {u, w0 , z 0 }, {x0 , y0 , z 0 , w0 }} ∪ E 1 ∪ E 2 . Let Q4 =



{Q i,4 j }.

i≥0 j≥0

The hypergraph Q 42,2 that belongs to the family Q4 is shown in Figure 4.4. Recall that the hypergraphs F8 and F7 defined in Section 3.1 are obtained from the affine plane AG(2, 3) of order 3 by deleting one and two vertices, respectively. By Theorem 4.7, q3 = c3 = 41 . The authors showed in [54] that if H ∈ Q1 , then τ (H ) = c3 (n H + m H ). Further, they showed that the only connected hypergraphs H where every edge has size at least 3 and that satisfy τ (H ) = c3 (n H + m H ) belong to one of these four infinite families Q1 , Q2 , Q3 , or Q4 or are the two exceptional hypergraphs F7 and F8 . We state this result formally as follows. Theorem 4.8 ([54]) If H is a connected hypergraph in which every edge has size at least 3, then 1 τ (H ) ≤ (n H + m H ), 4 with equality if and only H belongs to one of the four families Q1 , Q2 , Q3 , or Q4 or H ∈ {F7 , F8 }. As a consequence of Theorem 4.10, we have the following result.

26

4 The Tuza Constants

Theorem 4.9 ([54]) If H ∈ H3 is 3-regular, then τ (H ) ≤

1 (n + m H ) 4 H

with equality if and only if every component of H belongs to the family Q1 or is obtained from the affine plane AG(2, 3) of order 3 by deleting one vertex; that is, every component of H belongs to Q1 or is F8 . Since every hypergraph in one of the four infinite families Q1 , Q2 , Q3 , or Q4 is non-linear (that is, has edges that intersect in at least two vertices), except for the hypergraph E 3 consisting of a single 3-edge, we have the following characterization of hypergraphs H in the family L3 that satisfy τ (H ) = q3 (n H + m H ). Theorem 4.10 ([54]) If H ∈ L3 , then τ (H ) ≤

1 (n + m H ), 4 H

with equality if and only if every component of H is a single edge E 3 or is obtained from the affine plane AG(2, 3) of order 3 by deleting one or two vertices; that is, every component of H is E 3 or F7 or F8 . We summarize the results of Theorems 4.4 and 4.10 as follows. Theorem 4.11 For k ∈ {2, 3} if H ∈ Lk , then τ (H ) ≤

1 (n + m H ) k+1 H

with equality if and only if every component of H consists of a single edge E k or is obtained from the affine plane AG(2, k) of order k by deleting one or two vertices; that is, every component of H is E k or Fk 2 −2 or Fk 2 −1 .

Chapter 5

The Tuza Constant c4

5.1 Introduction In this chapter, we consider 4-uniform hypergraphs and determine the Tuza constant C4 . This establishes an upper bound for the Tuza constant q4 , which we discuss in detail in Chapter 9.

5.2 4-Uniform Hypergraphs For k ≥ 2 a generalized triangle Tk is defined as follows. Let A, B, C, and D be vertexdisjoint sets of vertices with |A| = k/2, |B| = |C| = k/2 and |D| = k/2 − k/2. In particular, if k is even, the set D = ∅, while if k is odd, the set D consists of a singleton vertex. Let Tk denote the k-uniform hypergraph with V (Tk ) = A ∪ B ∪ C ∪ D and with E(Tk ) = {e1 , e2 , e3 }, where V (e1 ) = A ∪ B, V (e2 ) = A ∪ C, and V (e3 ) = B ∪ C ∪ D. The hypergraphs T4 and T5 are illustrated in Figure 5.1. If H = T4 , then H ∈ H4 and τ (H ) = 2 =

2 2 (6 + 3) = (n H + m H ), 9 9

implying that the Tuza constant c4 ≥ 29 . We show next that this lower bound of 29 on the Tuza constant c4 is in fact also an upper bound. Recall that if H is a hypergraph and X and Y are subset of vertices in H , then H (X, Y ) denotes the hypergraph obtained from H by deleting all vertices in X ∪ Y from H and removing all (hyper)edges containing vertices from X , removing the vertices in Y from any remaining edges, and removing all resulting isolated vertices, if any. In particular, if Y = ∅, then we denote H (X, Y ) by H − X . If X = {x}, we denote H − X simply by H − x. For

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_5

27

28

5 The Tuza Constant c4

(a) T4

(b) T5

Fig. 5.1 The generalized triangles T4 and T5

i ∈ {3, 4}, let ei (H ) denote the number of edges in H of size i. Recall that n(H ) = n H and m(H ) = m H denote the order and size of H , respectively. Theorem 5.1 If H is a hypergraph with all edges of size 3 or 4, then 9τ (H ) ≤ 2n(H ) + 2m 4 (H ) + 3m 3 (H ). Proof. Let Φ(H ) = 2n(H ) + 2m 4 (H ) + 3m 3 (H ). We proceed by induction on the order n(H ) of H to show that 9τ (H ) ≤ Φ(H ). Let H be a hypergraph of order n(H ) with all edges of size 3 or 4. If n(H ) = 1 or n(H ) = 2, then τ (H ) = 0 and the desired result is immediate, which establishes the base cases. Suppose that n(H ) ≥ 3 and that if H  is a hypergraph with n(H  ) < n(H ) and with all edges of size 3 or 4, then 9τ (H  ) ≤ Φ(H  ). If H is disconnected, then we apply the inductive hypothesis to each component of H to yield the desired result. Hence, we may assume that H is connected. Claim 5.2 If Δ(H ) ≥ 4, then 9τ (H ) ≤ Φ(H ). Proof. Suppose that Δ(H ) ≥ 4. Let x be a vertex of maximum degree in H , and so d H (x) = Δ(H ) ≥ 4. Let H  = H − x. We note that n(H  ) ≤ n(H ) − 1 and m(H  ) = m(H ) − d H (x) ≤ m(H ) − 4. Further, 2m 4 (H  ) + 3m 3 (H  ) ≤ 2m 4 (H ) + 3m 3 (H ) − 2 · d H (x) ≤ 2m 4 (H ) + 3m 3 (H ) − 8. Every τ -transversal of H  can be extended to a transversal in H by adding to it the vertex x. Hence applying the inductive hypothesis to H  , we have 9τ (H ) ≤ 9(τ (H  ) + 1) ≤ (2n(H  ) + 2m 4 (H  ) + 3m 3 (H  )) + 9 ≤ 2(n(H ) − 1) + (2m 4 (H ) + 3m 3 (H ) − 8) + 9 < Φ(H ).  By Claim 5.2, we may assume that Δ(H ) ≤ 3, for otherwise the desired result follows. Claim 5.3 If H contains a vertex x such that d H (x) ≤ 2 and x only intersects 4edges, then 9τ (H ) ≤ Φ(H ). Proof. Suppose that H contains a vertex x such that d H (x) ≤ 2 and x only intersects 4-edges. In this case, we consider the hypergraph H  = H (∅, {x}) obtained from H by deleting the vertex x and removing the vertex x from all edges containing x, and

5.2 4-Uniform Hypergraphs

29

removing all resulting isolated vertices, if any. We note that n(H  ) ≤ n(H ) − 1. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Every τ transversal in H  is a transversal in H , and so τ (H ) ≤ τ (H  ). By supposition, d H (x) = i where i ∈ [2]. Thus, m 4 (H  ) = m 4 (H ) − i and m 3 (H  ) = m 3 (H ) + i. Since i ≤ 2, this implies that 9τ (H ) ≤ 9τ (H  ) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) ≤ 2(n(H ) − 1) + 2(m 4 (H ) − i) + 3(m 3 (H ) + i) = Φ(H ) + i − 2 ≤ Φ(H ).  By Claim 5.3, we may assume that every vertex of degree at most 2 in H is contained in a 3-edge. By our earlier assumptions, H is connected, n(H ) ≥ 3, and Δ(H ) ≤ 3. Claim 5.4 If H ∈ / H4 , then 9τ (H ) ≤ Φ(H ). Proof. Suppose that H ∈ / H4 , and so H contains 3-edges. Let e = {x1 , x2 , x3 } be an arbitrary 3-edge in H , where d H (x1 ) ≤ d H (x2 ) ≤ d H (x3 ). If d H (x3 ) = 3, then let H  = H − x3 . In this case, n(H  ) ≤ n(H ) − 1 and m(H  ) = m(H ) − 3. Since we delete at least one 3-edge incident with x3 , we note that 2m 4 (H  ) + 3m 3 (H  ) ≤ 2m 4 (H ) + 3m 3 (H ) − 2 · 2 − 1 · 3 = 2m 4 (H ) + 3m 3 (H ) − 7. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Every τ -transversal in H  can be extended to a transversal in H by adding to it the vertex x3 , and so 9τ (H ) ≤ 9(τ (H  ) + 1) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9 ≤ 2(n(H ) − 1) + (2m 4 (H ) + 3m 3 (H ) − 7) + 9 = Φ(H ). Hence, we may assume that d H (x3 ) ≤ 2, for otherwise the desired result holds. If n(H ) = 3, then 9τ (H ) = 9 = 2 · 3 + 0 + 3 ≤ 2n(H ) + 2m 4 (H ) + 3m 3 (H ) = Φ(H ), so we may assume that n(H ) ≥ 4 and as H is connected, d H (x3 ) = 2. Suppose that d H (x1 ) = 1. In this case, we let H  = H − x3 , and note that when x3 is deleted from H , the vertex x1 is isolated and is therefore deleted when constructing H  . Thus, n(H  ) ≤ n(H ) − 2. Further, m(H  ) = m(H ) − 2. Since we delete at least one 3-edge incident with x3 , we note that 2m 4 (H  ) + 3m 3 (H  ) ≤ 2m 4 (H ) + 3m 3 (H ) − 2 · 1 − 1 · 3 = 2m 4 (H ) + 3m 3 (H ) − 5. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Every τ -transversal in H  can be extended to a transversal in H by adding to it the vertex x3 , and so 9τ (H ) ≤ 9(τ (H  ) + 1) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9 ≤ 2(n(H ) − 2) + (2m 4 (H ) + 3m 3 (H ) − 5) + 9 = Φ(H ). Hence, we may assume that d H (x1 ) = 2, for otherwise the desired result holds. Therefore, d H (x1 ) = d H (x2 ) = d H (x3 ) = 2. For i ∈ [3], let ei be the edge different from e that contains xi . Suppose that ei = e j for some i and j where 1 ≤ i < j ≤ 3. Renaming vertices if necessary, we may assume that e1 = e2 . Let H  = H − x1 , and note that x2 is deleted when constructing H  . Thus, n(H  ) ≤ n(H ) − 2. Further, m(H  ) = m(H ) − 2. Analogously as in the preceding paragraph, 9τ (H ) ≤ Φ(H ). Hence, we may assume that e1 , e2 , and e3 are distinct, for otherwise the desired result holds. Suppose that e1 , e2 , and e3 are all 4-edges. In this case, we let X = {x1 } and Y = {x2 , x3 }, and consider the hypergraph H  = H (X, Y ). We note that n(H  ) = n(H ) − 3, m 4 (H  ) = m 4 (H ) − 3 and m 3 (H  ) = m 3 (H ) + 1. Every τ -transversal in H  can be extended to a transversal in H by adding to it the vertex x1 ,

30

5 The Tuza Constant c4

and so 9τ (H ) ≤ 9(τ (H  ) + 1) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9 ≤ 2(n(H ) − 3) + 2(m 4 (H ) − 3) + 3(m 3 (H ) + 1) + 9 = Φ(H ). Hence renaming vertices if necessary, we may assume that e1 is a 3-edge. Suppose that e2 or e3 is a 4-edge, say e2 . In this case, we let X = {x1 } and Y = {x2 }, and consider the hypergraph H  = H (X, Y ). We note that n(H  ) = n(H ) − 2, m 4 (H  ) = m 4 (H ) − 1, and m 3 (H  ) = m 3 (H ) − 1. Every τ -transversal in H  can be extended to a transversal in H by adding to it the vertex x1 , and so 9τ (H ) ≤ 9(τ (H  ) + 1) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9 ≤ 2(n(H ) − 2) + 2(m 4 (H ) − 1) + 3(m 3 (H ) − 1) + 9 = Φ(H ). Hence, we may assume that e1 , e2 , and e3 are all 3-edges. By our earlier arguments, we may assume that no two 3-edges overlap, for otherwise the desired result holds. Further we may assume that no 3-edge intersects a 4-edge. This implies that H is a 2-regular, 3-uniform, linear hypergraph. Thus, m 4 (H ) = 0 and 3m 3 (H ) = 3m(H ) =



d H (v) = 2n(H ).

(5.1)

v∈V (H )

We could now complete the proof of Claim 5.4 using Theorem 2.3, which states that 4τ (H ) ≤ n(H ) + m(H ), as this implies by Equation (5.1) and m 4 (H ) = 0 that 36τ (H ) ≤ 9n(H ) + 9m 3 (H ) = 7n(H ) + 12m 3 (H ) < 8n(H ) + 12m 3 (H ) = 4Φ(H ), or, equivalently, 9τ (H ) < Φ(H ). However we will also give a direct proof for the sake of completeness. Recall that e = {x1 , x2 , x3 } and for i ∈ [3], ei is the edge different from e that contains xi . By our earlier assumptions, e1 , e2 , e3 are distinct edges. Let Vi, j = (V (ei ) ∪ V (e j )) \ V (e), for all 1 ≤ i < j ≤ 3 and note that 3 ≤ |Vi, j | ≤ 4 by the linearity of H . Let ei, j be a (new) edge containing three vertices from Vi, j and define H  as follows. Let V (H  ) = V (H ) \ {x1 , x2 , x3 } and let E(H  ) = E(H ) ∪ {e1,2 , e1,3 , e2,3 } \ {e, e1 , e2 , e3 }. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Let T  be a τ -transversal in H  . If T  does not contain vertices that belong to the deleted edges ei and e j for some i and j where 1 ≤ i < j ≤ 3, then the edge ei, j is not covered by T  , a contradiction. This implies that T  contains vertices from at least two of the edges e1 , e2 , e3 . Renaming the vertices x1 , x2 and x3 if necessary, we may assume that e2 and e3 are covered by T  , which implies that T  ∪ {x1 } is a transversal in H . Therefore, τ (H ) ≤ τ (H  ) + 1. Furthermore, we note that n(H  ) = n(H ) − 3, m 4 (H  ) = m 4 (H ) = 0 and m 3 (H  ) = m 3 (H ) + 3 − 4 = m 3 (H ) − 1, and so

5.2 4-Uniform Hypergraphs

9τ (H ) ≤ ≤ = =

31

9(τ (H  ) + 1) 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9 2(n(H ) − 3) + 2m 4 (H ) + 3(m 3 (H ) − 1) + 9 Φ(H ).

This completes the proof of Claim 5.4.  By Claim 5.4, we may assume that H ∈ H4 , and so H contains no 3-edge; that is, m(H ) = m 4 (H ) and m 3 (H ) = 0. By Claims 5.2 and 5.3, H is a 3-regular, 4-uniform hypergraph. Claim 5.5 If H has overlapping edges, then 9τ (H ) ≤ Φ(H ). Proof. Suppose that H has overlapping edges, say e and f . Let u and v be two vertices that belong to the intersection of e and f , and so {u, v} ⊆ e ∩ f . Let e and f  be edges containing u and v, respectively, different from e and f . We now consider the hypergraph H  = H ({u}, {v}) of order n(H  ) ≤ n(H ) − 2. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Every τ -transversal in H  can be extended to a transversal in H by adding to it the vertex u, and so τ (H ) ≤ τ (H  ) + 1. If e = f  , then H  ∈ H4 and m 4 (H  ) = m 4 (H ) − 3, and so 9τ (H ) ≤ ≤ ≤ =

9(τ (H  ) + 1) 2n(H  ) + 2m 4 (H  ) + 9 2(n(H ) − 2) + 2(m 4 (H ) − 3) + 9 Φ(H ) − 1.

If e = f  , then m 4 (H  ) = m 4 (H ) − 4 and m 3 (H  ) = 1, and so 9τ (H ) ≤ 9(τ (H  ) + 1) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9 ≤ 2(n(H ) − 2) + 2(m 4 (H ) − 4) + 3 + 9 = Φ(H ). This completes the proof of Claim 5.5.  By Claim 5.5, we may assume that H is a 3-regular, 4-uniform, linear hypergraph, for otherwise the desired result holds. Thus, m 3 (H ) = 0 and 4m 4 (H ) = 4m(H ) =



d H (v) = 3n(H ).

(5.2)

v∈V (H )

We could now complete the proof using Theorem 5.12, which states that 21τ (H ) ≤ 5n(H ) + 4m(H ), as this implies by Equation (5.2) and m 3 (H ) = 0 that

32

5 The Tuza Constant c4

63τ (H ) ≤ 15n(H ) + 12m 4 (H ) = 14n(H ) + 14m 4 (H ) + n(H ) − 2m 4 (H ) = 14n(H ) + 21m 3 (H ) − 21 n(H ) < 14n(H ) + 21m 3 (H ) = 7Φ(H ), or, equivalently, 9τ (H ) < Φ(H ). However for the sake of completeness, we provide a direct proof that doesn’t depend on other results. Let C : x1 x2 . . . x x1 be a shortest cycle in H , and let f i be the edge of the cycle C that contains xi and xi+1 for i ∈ [], where addition is taken modulo . Since H is a linear hypergraph, we note that  ≥ 3. Let f i be an edge different from f i−1 and f i that contains the vertex xi . Let x be a vertex different from x−1 and x that belongs to the edge f −1 , and let e and f be the two edges different from f −1 that contains the vertex x. By the linearity of H and the minimality of the cycle C, the edges f 1 , . . . , f  , f 1 , . . . , f  , e, f are all distinct. Let Vodd be the set of vertices xi on C with i odd, and let Veven be the set of vertices xi on C with i even. Suppose that  is even. In this case, we let X = Vodd and Y = Veven . We now consider the hypergraph H  = H (X, Y ). We note that |X | = |Y | = /2, n(H  ) ≤ n(H ) − , m 4 (H  ) = m 4 (H ) − 2 and m 3 (H  ) = /2. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Every τ -transversal in H  can be extended to a transversal in H by adding to it the set X , and so τ (H ) ≤ τ (H  ) + |X |. Thus, 9τ (H ) ≤ 9(τ (H  ) + |X |) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + ≤ 2(n(H ) − ) + 2(m 4 (H ) − 2) + = Φ(H ).

9 2 3 2

+

9 2

Suppose that  is odd. In this case, we let X = (Vodd \ {x }) ∪ {x} and Y = Veven ∪ {x }. We now consider the hypergraph H  = H (X, Y ). We note that |X | = |Y | = ( + 1)/2, n(H  ) ≤ n(H ) −  − 1, m 4 (H  ) = m 4 (H ) − 2 − 2, and m 3 (H  ) = ( + 1)/2. Applying the inductive hypothesis to H  , we have 9τ (H  ) ≤ Φ(H  ). Every τ transversal in H  can be extended to a transversal in H by adding to it the set X , and so τ (H ) ≤ τ (H  ) + |X |. Thus, 9τ (H ) ≤ 9(τ (H  ) + |X |) ≤ 2n(H  ) + 2m 4 (H  ) + 3m 3 (H  ) + 9(+1) 2 ≤ 2(n(H ) −  − 1) + 2(m 4 (H ) − 2 − 2) + = Φ(H ).

3(+1) 2

+

9(+1) 2

This completes the proof of Theorem 5.1.  As an immediate consequence of Theorem 5.1, we have the following result due to Lai and Chang [74] (pictured below). We remark that Feng-Chu Lai (now known as Tina Lai) was a student of Professor Gerard Jennhwa Chang at National Chiao

5.2 4-Uniform Hypergraphs

33

Tung University, and the beautiful result in Theorem 5.6 formed part of her master thesis. Theorem 5.6 ([74]) If H ∈ H4 , then τ (H ) ≤ 29 (n H + m H ).

Feng-Chu Lai By Theorem 5.6,

Gerard Jennhwa Chang

c4 = sup

H ∈H 4

τ (H ) 2 ≤ . nH + mH 9

As observed earlier, c4 ≥ 29 . Consequently, the exact value of the Tuza constant c4 is determined. Theorem 5.7 ([74]) The Tuza constant c4 = 29 . As an immediate consequence of Observation 4.1 and Theorem 5.6, we have the following upper bound on the Tuza constant q4 . Corollary 5.1 The Tuza constant q4 ≤ 29 . In order to characterize the hypergraphs achieving equality in the Lai-Chang Theorem 5.6, we present a series of results that culminate in its characterization. Since the main focus in this book is on linear hypergraphs, we present a summary of these important results without proof. Chvátal and McDiarmid [24] established the following upper bound on the transversal number of a uniform hypergraph in terms of its order and size. Theorem 5.8 ([24]) For k ≥ 2, if H ∈ Hk , then   n H + k2 m H  3k  τ (H ) ≤ . 2

As a special case of the Chvátal-McDiarmid Theorem 5.8 when k = 4, we have the following result. Theorem 5.9 ([24]) If H ∈ H4 , then τ (H ) ≤ 16 n H + 13 m H .

34

5 The Tuza Constant c4

The hypergraphs achieving equality in the Chvátal-McDiarmid Theorem 5.8 when k = 3 were characterized by Henning and Löwenstein [52]. (Recall that the characterization in the case when k = 3 is given by Theorem 4.8.) Theorem 5.10 ([52]) For k = 2 or k ≥ 4, if H ∈ Hk is connected and   n H + 2k m H  3k  τ (H ) = 2

then H = E k consists of a single edge or H = Tk is the generalized triangle. As a special case of Theorem 5.10 when k = 4, we have the following result. Theorem 5.11 ([52]) If H ∈ Hk is connected and τ (H ) = 16 n H + 13 m H , then H consists of a single edge E 4 or H is the generalized triangle T4 . Thomassé (pictured below) and Yeo [85] proved the following upper bound on the transversal number of a 4-uniform hypergraph. Theorem 5.12 ([85]) If H ∈ Hk , then τ (H ) ≤

5 n 21 H

+

4 m . 21 H

We are now in a position to characterize the hypergraphs achieving equality in the Lai-Chang Theorem 5.6. Let H ∈ H4 be a connected hypergraph satisfying τ (H ) = 29 (n H + m H ). By Theorem 5.9, 1 1 2 (n H + m H ) = τ (H ) ≤ n H + m H , 9 6 3 implying that n H ≤ 2m H . By Theorem 5.12,

5.2 4-Uniform Hypergraphs

35

Fig. 5.2 The generalized triangle T4

2 5 4 (n H + m H ) = τ (H ) ≤ nH + mH , 9 21 21 implying that n H ≥ 2m H . Consequently, n H = 2m H and τ (H ) = 16 n H + 13 m H , implying by Theorem 5.11 that H is the generalized triangle T4 , shown in Figure 5.2. We summarize the results in this section as follows. Theorem 5.13 ([51]) The Tuza constant c4 = 29 . Moreover if H ∈ H4 and τ (H ) =

2 (n + m H ), 9 H

then every component of H is the generalized triangle T4 .

Chapter 6

The Tuza Constant ck for k Large

6.1 Introduction In this chapter, we determine the asymptotic behavior of the Tuza constants, ck , ck as k grows.

6.2 A General Upper Bound for the Tuza Constant ck Applying probabilistic arguments, Noga Alon [2] (pictured below) determined a general upper bound for the Tuza constant ck .

n  n  Recall that 1 − n1 < 1e for all n ≥ 1 and limn→∞ 1 − n1 = 1e , where e is the base of the natural logarithm. Hence we have the following well-known result.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_6

37

38

6 The Tuza Constant ck for k Large

Lemma 6.1 The following holds.  x (a) For all real x ≥ 1, we have 1 − x1 < 1e . x  (b) Given ε, where 0 < ε < 1, there exists a constant Cε such that 1 − x1 > when x ≥ Cε .

1−ε , e

Lemma 6.2 For all 0 ≤ x ≤ 21 , we have ln(1 − x) ≥ −x − x 2 . Proof Using the Taylor expansion of ln(1 − x), and noting that 0 ≤ x ≤ 21 , we obtain the following which proves the lemma: ∞  ∞  1 xi 2 i−2 ln(1 − x) = − = −x − x ·x i i i=1 i=2 ∞   1  1 i−2 2 · ≥ −x − x 2 2 i=2 ∞     1 i−1 = −x − x 2 2 i=2 = −x − x 2 .



In order to establish a general upper bound on the Tuza constant ck , Alon [2] first proved the following proposition, where ln denotes the natural logarithm to the base e. Proposition 6.1 ([2]) For k > 1 an integer, if H is a k-uniform hypergraph with n vertices and m edges, then for any positive real number α, we have  τ (H ) ≤

α ln k k



 n+

1 kα

 m.

Proof Since τ (H ) ≤ n, we may assume that α lnk k ≤ 1, for otherwise the desired result is immediate. Applying probabilistic arguments, we choose, randomly and independently, each vertex of H with probability p where p=

ln k α α ln k = . k k

Let T be the (random) set of vertices picked, and let F be the set of all edges e ∈ E(H ) that do not intersect the set T . For a fixed edge e ∈ E(H ), we note that by Lemma 6.1(a)

6.2 A General Upper Bound for the Tuza Constant ck

39

Pr(e ∈ F) = (1 − p)k k  = 1 − α lnk k 

lnkk α ln k α 1 = 1− k ( ln k α )  1 ln k α ≤ e =

1 . kα

By the linearity of expectation, the expected value of the quantity |T | + |F| is at most n · p + k1α · m, that is, E[|T | + |F|] ≤ pn +

m = kα



α ln k k



 n+

1 kα

 m.

Thus, there is at least one choice of the set T ⊆ V (H ) such that  |T | + |F| ≤

α ln k k



 n+

1 kα

 m.

By adding to T , arbitrarily, each edge of F, we obtain a set  one vertex  from  of size at most |T | + |F| ≤ α lnk k n + k1α m that intersects every edge. Hence,     τ (H ) ≤ α lnk k n + k1α m.  As a consequence of Proposition 6.1, Alon [2] established the following result. Theorem 6.2 ([2]) For k > 1 an integer, if H is a k-uniform hypergraph, then τ (H ) ≤

ln k (n H + m H ). k

Proof When k = 2, we have by Theorem 4.5 that τ (H ) ≤ 13 (n H + m H ). Since 13 < ln k ≈ 0.3465, the desired result follows for k = 2. Hence we may assume that k ≥ 3. k By substituting α = 1 in Proposition 6.1 where here n H = n and m H = m, we obtain  τ (H ) ≤ or, equivalently, τ (H ) ≤

ln k (n H k

ln k k



  1 nH + mH , k

+ m H ). 

As an immediate consequence of Theorem 6.2, we have the following upper bound on the Tuza constant ck for all k ≥ 2. Theorem 6.3 ([2]) For k > 1 an integer, ck ≤

ln k . k

40

6 The Tuza Constant ck for k Large

6.3 The Asymptotic Behavior of ck as k Grows We next study the asymptotic behavior of ck as k grows. Alon [2] proved that when k is sufficiently large, the upper bound in Theorem 6.2 is essentially best possible. Theorem 6.4 ([2]) For sufficiently large k, the Tuza constant ck ≥ (1 − o(1)) lnkk . Proof Let n = k ln(k) and let m = k. Let H = (V, E) be a random k-uniform hypergraph with vertex set V on n vertices and with m (not necessarily distinct) edges, constructed by choosing each edge randomly and independently according to a uniform distribution on the k-element subsets of V . Let e1 , e2 , . . . , ek denote the resulting edges of H . We show next that with high probability τ (H ) > ln2 (k) − 10 ln(k) · ln(ln(k)) for k sufficiently large. For k sufficiently large, let X be an arbitrary (fixed) set of vertices of H of cardinality |X | ≤ ln2 (k) − 10 ln(k) · ln(ln(k)).

(6.1)

For each i ∈ [k], the probability that the edge ei does not intersect X is given by n−|X | Pr(ei ∩ X = ∅) =

nk  k

≥

(n−|X |−k)k k! (n−k)k k!

 =

n − |X | − k n−k

k .

By Inequality (6.1) and since n ≥ k ln(k), we therefore have that  Pr(e ∩ X = ∅) ≥

k ln(k) − ln2 (k) + 10 ln(k) · ln(ln(k)) − k k ln(k) − k

where θ=

k = (1 − θ )k (6.2)

ln(k) − 10 ln(ln(k)) . k k − ln(k)

We prove next that (1 − θ )k ≥

ln9 (k) . k

We note that kθ =

ln(k) − 10 ln(ln(k)) ln(k) − 10 ln(ln(k)) = ln(k) − 10 ln(ln(k)) + 1 ln(k) − 1 1 − ln(k)

and kθ 2 =

(ln(k) − 10 ln(ln(k)))2 , which tends to zero when k gets large.

2 1 k 1 − ln(k)

(6.3)

6.3 The Asymptotic Behavior of ck as k Grows

41

Further, we note that ln(k) − 10 ln(ln(k)) tends to one when k gets large. ln(k) − 1 Hence for all sufficiently large k, we have 0 ≥ − ln(ln(k)) +

ln(k)−10 ln(ln(k)) ln(k)−1

+ kθ 2

= 9 ln(ln(k)) − ln(k) + ln(k) − 10 ln(ln(k)) +

9 + kθ + kθ 2 , = ln (ln(k)) k or, equivalently,



(ln(k))9 k(−θ − θ ) ≥ ln k

ln(k)−10 ln(ln(k)) ln(k)−1

+ kθ 2

 .

2

For k sufficiently large, we have 0 ≤ θ ≤ 21 , implying by Lemma 6.2 that 

(ln(k))9 k ln(1 − θ ) ≥ ln k

 ,

which in turn implies that Inequality (6.3) holds. By Inequalities (6.2) and (6.3), we have (ln(k))9 . (6.4) Pr(e ∩ X = ∅) ≥ k Since the m = k edges are chosen independently, Inequality (6.4) implies, using Lemma 6.1, that Pr(X is a transversal) =

k 

Pr(ei ∩ X = ∅)

i=1

=

k  i=1

≤

k 

(1 − Pr(ei ∩ X = ∅)) 

i=1



=

1−

ln9 (k) 1− k ln9 (k) k

⎡  = ⎣ 1−



k

1 k/ ln9 (k)

9 ≤ e− ln (k) .

⎤ 9 k/ ln9 (k) ln (k) ⎦

42

6 The Tuza Constant ck for k Large

For k large enough, the number of choices for X is bounded by the following, as we ln2 (k) times pick a vertex to add to X or choose not to add a vertex to X . (n + 1)ln

2

(k)

< (k ln(k) + 2)ln

2

(k)

= eln

2

(k) ln(k ln(k)+2)

< eln

4

(k)

.

Hence for k sufficiently large, we have Pr(τ (H ) ≤ ln2 (k) − 10 ln(k) · ln(ln(k))) ≤ eln

4

(k)

· e− ln

9

(k)

=

eln

4

(k)

eln

9

(k)

< 1.

Hence, for k sufficiently large there is at least one k-uniform hypergraph H on n = k ln(k) vertices and m = k edges such that τ (H ) > ln2 (k) − 10 ln(k) · ln(ln(k))   2 ln (k) − 10 ln(k) · ln(ln(k)) ( k ln(k) + k) = k ln(k) + k  2  ln (k) − 10 ln(k) · ln(ln(k)) ≥ (n H + m H ) k ln(k) + k + 1   ln(k) ln(k) − 10 ln(ln(k)) = (n H + m H ) k ln(k) + 1 + k1 = (1 − o(1))

ln(k) (n H + m H ). k

Thus for sufficiently large k, we have ck ≥ (1 − o(1)) lnkk .  As a consequence of Theorem 6.3 and 6.4, the asymptotic behavior of the Tuza constant ck as k grows is determined.   ln(k) as k → ∞. Theorem 6.5 ([2]) Tuza constant ck = (1 − o(1)) k

Chapter 7

The West Bound

7.1 Introduction In this chapter, motivated by comments by Douglas West, we study what we have coined the “West bound” on the transversal number of a k-uniform linear hypergraph. We pose a conjecture that the so-called West bound holds for all k-uniform, linear hypergraphs, and show here that this is true for small values of k, namely, k = 2 and k = 3. West [93] (pictured below) observed that the Chvátal-McDiarmid Theorem (see Theorem 5.8) can be used to obtain a lower bound on the covering number C(n, s, t) which is defined as the minimum number of s-sets in [n] needed to cover every t-set in [n]; that is, each t-set must be contained in some s-set in the family. Coverings have been studied, for example, in [1, 33, 83].

Douglas West © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_7

43

44

7 The West Bound

When n is large compared to n − s and t, the covering problem can be translated into a transversal problem for uniform hypergraphs. Let k = n − s. As remarked by West [93], a family S of s-sets covers all the t-sets if and only if every t-set is disjoint from the complement of some member of S . Thus, C(n, n − k, t) is the minimum size of a k-uniform hypergraph on n vertices such that no t-set intersects all edges. Equivalently, C(n, n − k, t) is the minimum number of edges in a k-uniform hypergraph on n vertices with transversal number t + 1. West posed the problem of proving better lower bounds on C(n, n − k, t) when n < 43 k(t + 1). As observed by West [93], “proving lower bounds in this range is equivalent to improving the upper bound on transversal numbers in the ChvátalMcDiarmid Theorem when n is small relative to k(t + 1). For even k, the constructions show that the Chvátal-McDiarmid Theorem is optimal when n exceeds about 3 kτ (H ) (recall that t + 1 = τ (H )). To obtain optimality in the next range of cov4 ering numbers when n < 34 kτ (H ), it would suffice to show that the upper bound in the Chvátal-McDiarmid Theorem can be improved to τ (H ) ≤ k1 n H + 16 m H in this range.” Motivated by these comments of West, if τ (H ) ≤ k1 n H + 16 m H holds for a hypergraph H ∈ Hk , then we say that H satisfies the West bound of k1 n H + 16 m H . The authors in [56] pose the following conjecture. Conjecture 7.1 ([56]) For k ≥ 2, if H ∈ Lk , then H satisfies the West bound; that is, τ (H ) ≤ k1 n H + 16 m H . We remark that the linearity constraint in the statement of Conjecture 7.1 is essential. Indeed for k ≥ 4, if H is non-linear, then it is not necessarily true that τ (H ) ≤ k1 n H + 16 m H . For example, let H = F be the complement of the Fano plane F, where the Fano plane is shown in Figure 7.1 and where its complement F is the hypergraph on the same vertex set V (F) and where e is a hyperedge in the complement if and only if V (F) \ e is a hyperedge in F. We note that H = F is a 4-uniform, non-linear hypergraph; that is, H ∈ H4 \ L4 . Further, H has order n H = 7 and size m H = 7. Any two vertices lie in one common edge e in F. The edge of H that is the complement of e is not covered by these two vertices. Therefore, τ (H ) ≥ 3. Any

Fig. 7.1 The Fano plane F

7.1 Introduction

45

three vertices that do not form an edge in F are not the complement of any edge in H , so equality holds. Hence, τ (H ) = 3 > 47 + 76 = k1 n H + 16 m H .

7.2 The West bound for 2-Uniform Hypergraphs In this section, we show that the West bound holds for 2-uniform hypergraphs, even without the linearity constraint. In particular, we prove that Conjecture 7.1 is true when k = 2. As observed earlier, a 2-uniform hypergraph is a graph, with possibly multiple edges (if edges of the hypergraph are duplicated). Theorem 7.1 ([56]) If H ∈ H2 , then τ (H ) ≤ 21 n H + 16 m H , with equality if and only every component of H is K 3 or K 4 . Proof. We proceed by induction on n H ; the statement is trivial when n H ≤ 2. For n H ≥ 3, the induction hypothesis applies to each component when H is disconnected, so we may assume that H is connected. We may also exclude multiple edges, since they only increase m H without changing τ (H ). Suppose that Δ(H ) ≤ 2. In this case, H is a path of a cycle. If H is a path, then m H = n H − 1 and τ (H ) =  21 n H  < 21 n H + 16 m H . If H is a cycle, then m H = n H and  τ (H ) =

1 n 2 H

 ≤

1 2 1 1 (n H + 1) ≤ n H = n H + m H , 2 3 2 6

with equality only if n = 3, where in fact τ (H ) = 21 n H + 16 m H . Hence, we may assume that Δ(G) ≥ 3, for otherwise the desired result follows. Suppose that Δ(G) ≥ 4. Let v be a vertex of maximum degree in H , and consider the hypergraph H  = H − v. We denote the order and size of a hypergraph H  by n H and m H , respectively. We note that n H = n H − 1 and m H = m H − d H (v) = m H − Δ(H ) ≤ m H − 4. Let T  denote a minimum transversal in H  . Applying the inductive hypothesis to H  , |T  | ≤

1  1 1 1 7 n H + m H ≤ n H + m H − . 2 6 2 6 6

Since T  ∪ {v} is a transversal of H , we have τ (H ) ≤ |T  | + 1 < 21 n H + 16 m H . Hence, we may assume Δ(H ) = 3, for otherwise the desired result follows. Let T be a largest independent set among the set of vertices of degree 3 in H , and let t = |T |. Let H  = H − T , and define n H and m H as above. By the choice of T , we have Δ(H  ) ≤ 2. Thus, τ (H  ) ≤ 21 n H + 16 m H , with equality if and only if every component of H  is K 3 . Every minimum transversal in H  can be extended to a transversal in H by adding to it the set T  , implying that

46

7 The West Bound

τ (H ) ≤ τ (H  ) + |T | ≤ ( 21 n H + 16 m H ) + t ≤ 21 (n H − t) + 16 (m H − 3t) + t = 21 n H + 16 m H . Furthermore, equality requires that every component of H  is a triangle, and adding T to any choice of two vertices from each component of H  completes a smallest transversal of H . If some vertex v in T has neighbors in more than one component of H  , then the transversal in H  can be chosen to include all neighbors of v, in which case v can be dropped. Hence equality (when H is connected) requires H = K 4 .  Adopting our notation in Chapter 3.1, recall that F4 denotes the complete graph K 4 , and is equivalent to the affine plane AG(2, 2) of order 2. Hence, Theorem 7.1 can be restated as follows. Theorem 7.2 ([56]) If H ∈ H2 , then τ (H ) ≤ 21 n H + 16 m H , with equality if and only every component of H is the affine plane AG(2, 2) of order 2 or is obtained from the affine plane AG(2, 2) of order 2 by deleting one vertex; that is, every component of H is F3 or F4 .

7.3 The West bound for 3-Uniform Hypergraphs In this section, we show that the West bound holds for 3-uniform hypergraphs, even if we relax the linearity constraint. In particular, we prove that Conjecture 7.1 is true when k = 3. Recall that a τ -transversal of a hypergraph H is a minimum transversal of H of size τ (H ). Theorem 7.3 ([56]) If H ∈ H3 is connected, then τ (H ) ≤

1 1 n + m , 3 H 6 H

with equality if and only H ∈ {F8 , F9 } ∪ Q. Proof. Let H ∈ H3 be connected and have order n H and size m H . We proceed by induction on m H . If m H = 0, then τ (H ) = 0 and n H = 1, and so τ (H ) < 13 n H + 1 m . Hence we may assume that m H ≥ 1, and therefore n H ≥ 3. If n H = 3, then 6 H τ (H ) < 13 n H + 16 m H . Hence we may assume m H ≥ 2 and n H ≥ 4. Let Q ∗ = {F8 } ∪ Q. We note that every hypergraph in Q ∗ is 3-regular. In what follows, we denote the order and size of a hypergraph H  by n H and size  m H , respectively. Further if m H < m H , then applying the inductive hypothesis to H  , we let T  denote a transversal in H  such that |T  | ≤ 13 n H + 16 m H . Further by the inductive hypothesis, if |T  | = 13 n H + 16 m H , then H  ∈ {F8 , F9 } ∪ Q.

7.3 The West bound for 3-Uniform Hypergraphs

47

Suppose that Δ(H ) ≥ 5. Let v be a vertex of maximum degree in H . We now consider the hypergraph H  = H − v and note that n H = n H − 1 and m H = m H − d H (v) ≤ m H − 5. Thus, |T  | ≤ 13 n H + 16 m H < 13 n H + 16 m H − 1. Since T  ∪ {v} is a transversal of H , we have τ (H ) ≤ |T  | + 1 < 13 n H + 16 m H , as desired. Hence, we may assume that Δ(H ) ≤ 4.  Suppose that Δ(H ) ≤ 3. In this case, m H = 13 v∈V (H ) d H (v) ≤ n H . By Theorem 4.6, we therefore have τ (H ) ≤ 41 (n H + m H ) ≤ 13 n H + 16 m H . Equality requires τ (H ) = 41 (n H + m H ) and n H = m H , implying that H is 3-regular. Thus Theorem 4.9 applies, and the only hypergraphs achieving equality are those in Q ∗ = {F8 } ∪ Q. Hence, we may assume that Δ(H ) = 4. With this assumption, we show that H is the affine plane AG(2, 3) of order 3; that is, H = F9 . Let T be a largest independent set among the set of vertices of degree 4. We now consider the hypergraph H  = H − T , and we note that n H = n H − |T | and m H = m H − 4|T |. Thus, |T  | ≤ 13 n H + 16 m H = 13 (n H − |T |) + 16 (m H − 4|T |) = 13 n H + 1 m − |T |. Since T  ∪ T is a transversal of H , we have τ (H ) ≤ |T  | + |T | ≤ 6 H 1 n + 16 m H , which proves the desired upper bound. 3 H Suppose that equality holds for H . This requires τ (H ) = |T  | + |T | and τ (H  ) = |T  | = 13 n H + 16 m H . By the inductive hypothesis, every component of H  is in {F8 , F9 } ∪ Q. By our choice of T , Δ(H  ) ≤ 3. Thus since F9 is 4-regular, every component of H  is therefore in Q ∗ . We proceed further with the following claim. Claim 7.4 No vertex v of degree 4 lies in two edges with another vertex w. Proof. Suppose, to the contrary, that such a pair occurs. Since the set T is maximal and not necessarily maximum, we may assume that v ∈ T . Thus since T is an independent set, we have w ∈ / T and therefore w ∈ V (H  ). As observed earlier, every  ∗ component of H is in Q and every hypergraph in Q ∗ is 3-regular. In particular, the component of H  containing w is in Q ∗ . Thus, we have d H  (w) = 3, implying that d H (w) ≥ 5 since two edges incident with w were removed when constructing H  , a contradiction.  We now return to the proof of Theorem 7.3. Let v be an arbitrary vertex in T , and let e1 , e2 , e3 , e4 be the edges incident with v. For i ∈ [4], let ei be a 2-element subset of V (ei ) \ {v}. Let M = {e1 , e2 , e3 , e4 }, and let VM =

4 

V (ei ) ⊆ V (H  ).

i=1

Since all components of H  are in Q ∗ , we have d H  (w) = 3 and d H (w) = 4 for every vertex w ∈ VM . Since each edge ei contains the vertex v, Claim 7.4 now implies that M is a matching and that no edge in M is contained in an edge of H  . Let R be the subhypergraph of H  induced by VM .

48

7 The West Bound

Claim 7.5 R ∼ = F8 . Proof. Suppose, to the contrary, that R  F8 . In this case, we obtain a τ -transversal of H  that intersects each edge of M. This makes v unnecessary, yielding τ (H ) ≤ τ (H  ) + |T \ {v}| < τ (H  ) + |T | = |T  | + |T | = τ (H ), a contradiction. First suppose that all four edges of M intersect some component Q of H  that is isomorphic to F8 . At least one edge of M, say ej , has only one vertex of Q, since otherwise R = Q. Any three vertices of F8 belong to a τ -transversal of F8 , so we can select a τ -transversal of Q that intersects the other three edges of M. The vertex of ej outside V (Q) belongs to a τ -transversal of its component of H  , since every hypergraph H ∗ ∈ Q ∗ has the property that any specified vertex belongs to a τ -transversal of H ∗ (in fact, as shown in [54] any two arbitrary vertices in H ∗ ∈ Q ∗ belong to a common τ -transversal of H ∗ ). Combining the two transversals yields a τ -transversal of H  intersecting each edge of M, which we have forbidden. Thus at most three edges of M intersect any copy of F8 in H  . Let F be the union of all components of H  that are copies of F8 . Since any three vertices in each component can be used in a τ -transversal of F, there is a τ -transversal S of F that intersects all edges of M intersecting F. Since no τ -transversal of H  intersects all of M, we have F = H  . Let M  be the set of edges in M that do not intersect copies of F8 in H  . Since F = H  and every component of H  lies in Q ∗ , there is a component Q of  H that is isomorphic to Q d for some d. Under the vertex labeling in Definition 4.1, let M Q consist of {x0 , y0 }, {x1 , y1 }, . . . , {xd , yd }. Let M ∗ be the union of M Q for all such components Q in H  . Since M ∗ and M  are matchings, their union consists of disjoint paths and cycles. Hence these edges have a transversal of size at most |V (M ∗ ∪ M  )|/2, which we call S  . Now S ∪ S  is a τ -transversal of H  that intersects all edges of M, again a contradiction.  By Claim 7.5, R ∼ = F8 . Let the vertices and edges of R be indexed as in Definition 3.1 defined in Chapter 3. Since the only pairs of nonadjacent vertices in R are {x1 , x8 }, {x2 , x7 }, {x3 , x4 }, and {x5 , x6 }, Claim 7.4 implies that these are the edges of M. Hence, the set VM ∪ {v} induces a copy of F9 . Since F9 is 4-regular and H is  connected, H = F9 . This completes the proof of Theorem 7.3. As observed in the previous chapter, namely, Chapter 4, every hypergraph in the family Q is non-linear. Recall that the hypergraph F9 defined in Chapter 3.1 is the affine plane AG(2, 3) of order 3. Hence, as an immediate consequence of Theorem 7.3, we have the following result. Theorem 7.6 ([56]) If H ∈ L3 , then τ (H ) ≤

1 1 n + m , 3 H 6 H

with equality if and only every component of H is the affine plane AG(2, 3) of order 3 or is obtained from the affine plane AG(2, 3) of order 3 by deleting one vertex; that is, every component of H is F8 or F9 .

7.3 The West bound for 3-Uniform Hypergraphs

49

We summarize the results of Theorems 7.2 and 7.6 as follows. Theorem 7.7 For k ∈ {2, 3} if H ∈ Lk , then τ (H ) ≤

1 1 n + m k H 6 H

with equality if and only if every component of H is the affine plane AG(2, k) of order k or is obtained from the affine plane AG(2, k) of order k by deleting one vertex; that is, every component of H is Fk 2 −1 or Fk 2 . If H ∈ Hk consists only of isolated vertices, then τ (H ) = m H = 0. If Δ(H ) ≥ 1, let v be a vertex of maximum degree Δ(H ) in H and let Tv be the set consisting of the vertex v and one vertex from every edge that does not contain v. Since Tv is a transversal in H , we note that τ (H ) ≤ |Tv | ≤ 1 + (m − Δ(H )) ≤ m, with strict inequality if Δ(H ) ≥ 2. Hence we have the following trivial bound on the transversal number, where E k denotes the k-uniform hypergraph on k vertices with exactly one edge. Observation 7.8 If H ∈ Hk is connected, then τ (H ) ≤ m H with equality if and only if H = E k . The following result summarizes the bounds given in Theorems 4.10, 7.7, and Observation 7.8. Recall that F9 is the affine plane AG(2, 3). Further, F8 and F7 are obtained from the affine plane AG(2, 3) by deleting one and two vertices, respectively. Theorem 7.9 If H ∈ L3 is connected, then the following holds. (a) τ (H ) ≤ m H with equality if and only if H = E 3 . (b) τ (H ) ≤ 41 n H + 41 m H with equality if and only if H ∈ {E 3 , F7 , F8 }. (c) τ (H ) ≤ 13 n H + 16 m H with equality if and only if H ∈ {F8 , F9 }. If we relax the linearity constraint, then the upper bounds in Theorem 7.9 still hold. However in this case when H ∈ H3 , there are four infinite families achieving equality in the lower bound τ (H ) ≤ 41 n H + 41 m H (see, Theorem 4.8) and one infinite family achieving equality in the lower bound τ (H ) ≤ 13 n H + 16 m H (see, Theorem 7.3).

7.4 The West bound for k-Uniform Hypergraphs for Large k As remarked earlier, the linearity constraint in the statement of Conjecture 7.1 is essential when k ≥ 4 since there exists hypergraphs H ∈ H4 \ L4 satisfying τ (H ) > 1 n + 16 m H . k H

50

7 The West Bound

In Chapter 9, we show (see Theorem 9.2) that the West bound also holds for 4-uniform, linear hypergraphs; that is, we prove that Conjecture 7.1 is true when k = 4. In Chapter 10.4, we show (see Theorem 10.9) that the West bound holds for large uniformity and for sufficiently large average degree.

7.5 The Convex Sets Ak and Bk For k ≥ 2, let Ak be the set of all pairs (ak , bk ) of nonnegative integers ak and bk for which the inequality τ (H ) ≤ ak n H + bk m H holds for all k-uniform hypergraphs H ; that is, for all H ∈ Hk . Chvátal and McDiarmid [24] proved that the set Ak is a convex set and has infinitely many extreme points. Theorem 7.10 ([24]) For each k ≥ 2, the set Ak is a convex set and has infinitely many extreme points. By Theorem 14.5, the set Ak is a convex set for every k ≥ 3. It remains however an open problem to determine the extreme points of the convex set Ak for k ≥ 3. For k ≥ 3, let Bk be the set of all pairs (ak , bk ) of nonnegative integers ak and bk for which the inequality τ (H ) ≤ ak n H + bk m H holds for all k-uniform linear hypergraphs H ; that is, for all H ∈ Lk . We note that the set Bk is a convex set for every k ≥ 3. We say that a point (α, β) of nonnegative integers α and β is tight in the convex set Bk if the point (α, β) satisfies τ (H ) ≤ αn H + βm H

(7.1)

for all H ∈ Lk and achieves equality in this bound for some H ∈ Lk . As a consequence of Theorem 7.9, we have the following result. Corollary 7.1 The points (0, 1), ( 41 , 14 ) and ( 16 , 13 ) are tight for the convex set B3 . With the extreme points of B3 ordered by the first coordinate, we determine the first three points of the convex set B3 . Theorem 7.11 The first extreme point of B3 is the point (0, 1), which is tight only for the hypergraph E 3 . Proof. By Corollary 7.1, the points (0, 1) and ( 41 , 41 ) are tight for the convex set B3 , and both points are tight at the hypergraph H = E 3 . Moreover, if H = E 3 , then n H = 3m H , implying that  τ (H ) =

   1 1 − ε nH + + 3ε m H 4 4

7.5 The Convex Sets Ak and Bk

51

for all ε where 0 ≤ ε ≤ 41 . Hence, every point on the line segment joining the points (0, 1) and ( 41 , 41 ) belongs to the set B3 and is tight. Therefore when the extreme points are ordered by their first coordinate, the first extreme point of B3 is the point (0, 1),  which by Corollary 7.1 is tight only for the hypergraph E 3 . Theorem 7.12 The second extreme point of B3 is the point ( 41 , 14 ), which is tight only for the hypergraphs E 3 , F7 , and F8 . Proof. By Corollary 7.1, the points ( 41 , 41 ) and ( 31 , 16 ) are tight for the convex set B3 , and both points are tight at the hypergraph H = F8 . Moreover, if H = F8 , then n H = m H , implying that  τ (H ) =

   1 1 − ε nH + + ε mH 3 6

1 for all ε where 0 ≤ ε ≤ 12 . Hence, every point on the line segment joining the points 1 1 1 1 ( 4 , 4 ) and ( 6 , 3 ) belongs to the set B3 and is tight. Therefore when the extreme points are ordered by their first coordinate, the second extreme point of B3 is the point ( 41 , 14 ), which by Corollary 7.1 is tight only for the hypergraphs E 3 , F7 , and  F8 .

Theorem 7.13 The third extreme point of B3 is the point ( 13 , 16 ), which is tight only for the hypergraphs F8 and F9 . Proof. By Corollary 7.1, the point ( 13 , 16 ) is tight for the hypergraph F9 . Moreover, if H = F9 , then 3m H = 4n H , implying that

1

(0, 1)

B3

1 4 1 6

( 14 , 14 ) ( 13 , 16 )

1 4

Fig. 7.2 The first three extreme points of the convex set B3

1 3

52

7 The West Bound

 τ2 (H ) =

   1 1 + 4ε n H + − 3ε m H 3 6

for all ε ≥ 0. Hence, every point ( 16 + 4ε, 13 − 3ε) on the line segment starting at the point ( 13 , 16 ) is tight for the hypergraph H = F9 . Further, no point on this line segment different from the point ( 13 , 16 ) is tight for the hypergraph F8 . Hence when the extreme points are ordered by their first coordinate, the third extreme point of B3 is the point ( 13 , 16 ).  We summarize the above results as follows. Theorem 7.14 With the extreme points of the convex set B3 ordered by the first coordinate, the first three extreme points of B3 are (0, 1), ( 41 , 41 ), and ( 13 , 16 ). The first three extreme points of the convex set B3 are illustrated in Figure 7.2.

Chapter 8

The Deficiency of a Hypergraph

8.1 Introduction In this chapter, we introduce the completely new technique of the deficiency of a hypergraph. Using this concept of deficiency, we prove here a key theorem that we will need in establishing the transversal number of a linear 4-uniform hypergraph, including determining the Tuza constant q4 in the next chapter. We begin this chapter by defining a few so-called special hypergraphs. Throughout this section, we let L4,3 be the class of all 4-uniform, linear hypergraphs with maximum degree at most 3. Thus, L4,3 is a subclass of the class L4 of all 4-uniform, linear hypergraphs.

8.2 Special Hypergraphs In this section, we define fifteen special hypergraphs, which are shown in Fig. 8.1. We define H14 = {H14,1 , H14,2 , H14,3 , H14,4 , H14,5 , H14,6 } and H21 = {H21,1 , H21,2 , H21,3 , H21,4 , H21,5 , H21,6 }. The following properties of special hypergraphs will prove to be useful. We remark that these properties are straightforward, albeit tedious to check “by hand.” However, we have also verified these properties by computer.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_8

53

54

8 The Deficiency of a Hypergraph

(b) H10

(a) H4

(d) H14,1

(f) H14,3

(h) H14,5 Fig. 8.1 The fifteen special hypergraphs

(c) H11

(e) H14,2

(g) H14,4

(i) H14,6

8.2 Special Hypergraphs

55

(k) H21,2

(j) H21,1

() H21,3

(n) H21,5

(m) H21,4

(o) H21,6

Fig. 8.1 (continued)

56

8 The Deficiency of a Hypergraph

Observation 8.1 Let H be a special hypergraph of order n H and size m H , and let u and v be two arbitrary distinct vertices in H . Then the following hold. (a) (b) (c) (d) (e) (f) (g) (h) (i)

(j) (k)

(l)

(m)

(n)

(o)

(p)

If H = H4 , then n H = 4, m H = 1 and τ (H ) = 1. If H = H10 , then n H = 10, m H = 5 and τ (H ) = 3. If H = H11 , then n H = 11, m H = 5 and τ (H ) = 3. If H ∈ H14 , then n H = 14, m H = 7 and τ (H ) = 4. If H ∈ H21 , then n H = 21, m H = 11 and τ (H ) = 6. If H ∈ {H10 , H14,5 , H14,6 }, then H is 2-regular. Any given vertex in H belongs to some τ -transversal of H . If H ∈ {H10 , H14,6 }, then there exists a τ -transversal of H that contains both u and v. If H = H4 , T ⊂ V (H ) and |T | = 3, then there exists a τ -transversal of H that contains at least two vertices of T , unless H = H11 and T is the set of three vertices in H that have no neighbor of degree 1 in H . If H = H4 , T ⊂ V (H ) and |T | = 4, then there exists a τ -transversal of H that contains at least two vertices of T . If H = H4 and T1 and T2 are vertex-disjoint subsets of H such that |T1 | = |T2 | = 2, then there exists a τ -transversal of H that contains a vertex from T1 and a vertex from T2 . If T1 and T2 are vertex-disjoint subsets of H such that |T1 | = 3 and |T2 | = 1 and T1 contains two vertices that are not adjacent in H , then there exists a τ -transversal of H that contains a vertex from T1 and a vertex from T2 . If T1 and T2 are vertex-disjoint subsets of H such that |T1 | = 1 and |T2 | = 2 where T2 contains two vertices that are not adjacent in H , then there exists a τ -transversal of H that contains a vertex from T1 and a vertex from T2 , except if H = H11 , and one degree-1 vertex in H belongs to T1 and the other degree-1 vertex to T2 , and the second vertex of T2 is adjacent to the vertex of T1 . If T1 , T2 , and T3 are vertex-disjoint subsets of H such that |T1 | = |T2 | = 3 and |T3 | ≥ 2, and T1 , T2 , and T3 are strongly independent sets in H , then there exists a τ -transversal of H that contains a vertex from each of T1 , T2 , and T3 . If H is a special subhypergraph of a 4-uniform linear hypergraph F where Δ(F) ≤ 3, and if there are three edges of F each of which intersect H in two vertices, then there is a τ -transversal of H that covers one of these edges and any specified edge of H . If H = H11 or if H ∈ H14 ∪ H21 , and if v is a vertex of degree 2 in H , then either H − v is connected or H − v is disconnected with exactly two components, one of which consists of an isolated vertex. Further, H − y does not contain two vertex-disjoint copies of H4 that are both intersected by a common edge and such that both copies of H4 have three vertices of degree 1 and one vertex of degree 2.

8.3 The Deficiency of a Hypergraph

57

8.3 The Deficiency of a Hypergraph Let H be a 4-uniform hypergraph. A set X is a special H -set if it consists of subhypergraphs of H with the property that every subhypergraph in X is a special hypergraph and further these special hypergraphs are pairwise vertex-disjoint. For notational simplicity, we write V (X ) and E(X ) to denote the set of all vertices and edges, respectively, in H that belong to a subhypergraph H  ∈ X in the special H -set X . Let X be an arbitrary special H -set. A set T of vertices in V (X ) is an X -transversal if T is a minimum set of vertices that intersects every edge from every subhypergraph in X . We define E ∗H (X ) to be the set of all edges in H that do not belong to a subhypergraph in X but which intersect at / E(H  ) for every subhyleast one subhypergraph in X . Hence if e ∈ E ∗ (X ), then e ∈   pergraph H ∈ X but V (e) ∩ V (H ) = ∅ for at least one subhypergraph H  ∈ X . If the hypergraph H is clear from context, we simply write E ∗ (X ) rather than E ∗H (X ). To illustrate the definition of the set E ∗ (X ), suppose that X = {S1 , S2 , . . . , S6 } is a special H -set, and so every subhypergraph Si in X for i ∈ [6] is a special hypergraph and these special hypergraphs are pairwise vertex-disjoint. If e1 , e2 , . . . , e6 represent the edges in H that do not belong to a subhypergraph in X but which intersect at least one subhypergraph in X (as illustrated in Fig. 8.2), then E ∗ (X ) = {e1 , . . . , e6 }). We associate with the set X a bipartite graph, which we denote by G X , with partite sets X and E ∗ (X ), where an edge joins e ∈ E ∗ (X ) and H  ∈ X in G X if and only if the edge e intersects the subhypergraph H  of X in H . We define a weak partition of X = (X 4 , X 10 , X 11 , X 14 , X 21 ) (where a weak partition is a partition in which some of the sets may be empty) where X i ⊆ X consists of all subhypergraphs in X of order i, i ∈ {4, 10, 11, 14, 21}. Thus, X =

e6 e1 e3

S1

S2

S3

S4

S5

e4 e2 Fig. 8.2 Edges in E ∗ (X ) = {e1 , . . . , e6 }

e5

S6

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8 The Deficiency of a Hypergraph

X 4 ∪ X 10 ∪ X 11 ∪ X 14 ∪ X 21 and |X | = |X 4 | + |X 10 | + |X 11 | + |X 14 | + |X 21 |. As an immediate consequence of Observation 8.1(a)–(e), we have the following result. Observation 8.2 If X is a special H -set and T is an X -transversal, then |T | = |X 4 | + 3|X 10 | + 3|X 11 | + 4|X 12 | + 6|X 21 |. We define the deficiency of X in H as def H (X ) = 10|X 10 | + 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | − 13|E ∗ (X )|. We define the deficiency of H by def(H ) = max def H (X ) where the maximum is taken over all special H -sets X . We note that taking X = ∅, we have def(H ) ≥ 0.

8.4 Applying the Deficiency Technique Recall that L4,3 is the class of all 4-uniform linear hypergraphs with maximum degree at most 3. In this section, we apply the technique of deficiency of a hypergraph to prove the following powerful result that we will need when establishing upper bounds on the transversal number of a 4-uniform linear hypergraph in the following Chap. 9. For this purpose, we shall need the following theorem of Berge [13] about the matching number of a graph, which is sometimes referred to as the Tutte-Berge formulation for the matching number. Theorem 8.3 (Tutte-Berge Formula) For every graph G, α  (G) = min

X ⊆V (G)

1 (|V (G)| + |X | − oc(G − X )) , 2

where oc(G − X ) denotes the number of odd components of G − X . We shall also rely heavily on the following well-known theorem due to König [70] and Hall [46] in 1935. Theorem 8.4 (Hall’s Theorem) Let G be a bipartite graph with partite sets X and Y . Then X can be matched to a subset of Y if and only if |N (S)| ≥ |S| for every nonempty subset S of X .

8.4 Applying the Deficiency Technique

59

We are now ready to state and prove a key theorem, namely, Theorem 8.5. A summary of the key proof ideas used to prove this theorem appears in [63]. Here we provide a comprehensive proof which was first posted by the authors on arXiv on February 6, 2018 (see, [62]). Recall in this chapter that a subset of vertices in a hypergraph H is a strongly independent set if it intersects every edge of H in at most one vertex. Theorem 8.5 ([63]) If H ∈ L4,3 , then 45τ (H ) ≤ 6n(H ) + 13m(H ) + def(H ). Proof For a 4-uniform hypergraph H , let ξ(H ) = 45τ (H ) − 6n(H ) − 13m(H ) − def(H ). We wish to show that if H ∈ L4,3 , then ξ(H ) ≤ 0. Suppose, to the contrary, that the theorem is false and that H ∈ L4,3 is a counterexample with minimum value of n(H ) + m(H ). Thus, ξ(H ) > 0 but every hypergraph H  ∈ L4,3 with n(H  ) + m(H  ) < n(H ) + m(H ) satisfies ξ(H  ) ≤ 0. We will now prove a number of claims, where the last claim completes the proof of the theorem. Claim A: The hypergraph H is connected and δ(H ) ≥ 1. Proof of Claim A: If H is disconnected, then by the minimality of H we have that the theorem holds for all components of H and therefore also for H , a contradiction. Therefore, H is connected. If H has an isolated vertex, then by the connectivity of H we have that n(H ) = 1 and m(H ) = 0, implying that 45τ (H ) = 0 < 6 = 6n(H ) + 13m(H ) + def(H ), a contradiction. Hence, δ(H ) ≥ 1. () Claim B: Given a special H -set, X , there is no X -transversal, T , such that |T | = |X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + |X 21 | and T intersects every edge in E ∗ (X ). Proof of Claim B: Suppose that there does exist an X -transversal, T , such that |T | = |X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + |X 21 | and T intersects every edge in E ∗ (X ). Let H  be obtained from H by removing all vertices in V (X ) and removing all edges of H that intersect V (X ). Since T is an X -transversal and T intersects every edge in E ∗ (X ), we remark that H  = H (T, V (X ) \ T ). Then, H  is a 4-uniform hypergraph with maximum degree Δ(H ) ≤ 3. Further, n(H  ) = n(H ) − 4|X 4 | − 10|X 10 | − 11|X 11 | − 14|X 14 | − 21|X 21 |, m(H  ) = m(H ) − |X 4 | − 5|X 10 | − 5|X 11 | − 7|X 14 | − 11|X 21 | − |E ∗ (X )|, and τ (H ) ≤ |T | + τ (H  ). Let def(H  ) = def H  (X  ) for some special H  -set, X  . Let X ∗ = X ∪ X  . Then, def(H  ) = def H (X ∗ ) − def H (X ). By the minimality of n(H ) + m(H ), H  is not a counterexample to our theorem, and so 45τ (H  ) ≤ 6n(H  ) + 13m(H  ) + def(H  ). Hence,

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8 The Deficiency of a Hypergraph

45τ (H ) ≤ 45τ (H  ) + 45|T | ≤ (6n(H  ) + 13m(H  ) + def(H  ))+ 45(|X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + 6|X 21 |) ≤ 6(n(H ) − 4|X 4 | − 10|X 10 | − 11|X 11 | − 14|X 14 | − 21|X 21 |)+ 13(m(H ) − |X 4 | − 5|X 10 | − 5|X 11 | − 7|X 14 | − 11|X 21 | − |E ∗ (X )|)+ def H (X ∗ ) − def H (X ) + 45(|X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + 6|X 21 |) = 6n(H ) + 13m(H ) + 8|X 4 | + 10|X 10 | + 4|X 11 | + 5|X 14 | + |X 21 | − 13|E ∗ (X )|+ def H (X ∗ ) − def H (X ) = 6n(H ) + 13m(H ) + def H (X ∗ ) ≤ 6n(H ) + 13m(H ) + def(H ),

contradicting the fact that H is a counterexample. () As an immediate consequence of Claim A and Claim B, H is not a special hypergraph. We state this formally as follows. Claim C: H is not a special hypergraph. Among all special nonempty H -sets, let X be chosen so that (1) |E ∗ (X )| − |X | is minimum. (2) Subject to (1), |X | is maximum. Claim D: |E ∗ (X )| ≥ |X | + 1. Proof of Claim D: Suppose, to the contrary, that |E ∗ (X )| ≤ |X |. Let G X be the bipartite graph associated with the special H -set X . Suppose there exists a matching, M, in G X that matches E ∗ (X ) to a subset of X . Then, by Observation 8.1(g), there exists a minimum X -transversal, T , that intersects every edge in E ∗ (X ), contradicting Claim B. Therefore, no matching in G X matches E ∗ (X ) to a subset of X . By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗ (X ) such that |NG X (S)| < |S|. We now consider the special H -set, X  = X \ NG X (S). Then, |X  | = |X | − |NG X (S)| and |E ∗ (X  )| = |E ∗ (X )| − |S|. Thus, |E ∗ (X  )| − |X  | = (|E ∗ (X )| − |S|) − (|X | − |NG X (S)|) = (|E ∗ (X )| − |X |) + (|NG X (S)| − |S|) < |E ∗ (X )| − |X |, contradicting our choice of the special H -set X .

()

As an immediate consequence of Claim D and by our choice of the special H -set X , we have the following claim. Claim E: If X  = ∅ is a special H -set, then |E ∗ (X  )| ≥ |X  | + 1. Claim F: def(H ) = 0. Proof of Claim F: Let X ∗ be a special H -set. If X ∗ = ∅, then by Claim E, |E (X ∗ )| ≥ |X ∗ | + 1, implying that def H (X ∗ ) ≤ 10|X ∗ | − 13|E ∗ (X ∗ )| < 0. However, taking X ∗ = ∅, we note that def H (X ∗ ) = 0. Consequently, def(H ) = 0. () ∗

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For a hypergraph H  ∈ L4,3 , let Φ( H  ) = ξ(H  ) − ξ(H ). Claim G: If H  ∈ L4,3 satisfies n(H  ) + m(H  ) < n(H ) + m(H ), then Φ(H  ) < 0. Proof of Claim G: Since H is a counterexample with minimum value of n(H ) + m(H ), and since n(H  ) + m(H  ) < n(H ) + m(H ), we note that ξ(H  ) ≤ 0. Thus, 45τ (H ) = ξ(H ) + 6n(H ) + 13m(H ) + def(H ) = ξ(H  ) − Φ(H  ) + 6n(H ) + 13m(H ) + def(H ) ≤ −Φ(H  ) + 6n(H ) + 13m(H ) + def(H ), implying that Φ(H  ) ≤ 6n(H ) + 13m(H ) + def(H ) − 45τ (H ) = −ξ(H ) < 0.

()

∗

Claim H: |E (X )| ≥ |X | + 2. Proof of Claim H: Suppose, to the contrary, that |E ∗ (X )| ≤ |X | + 1. Then, by Claim D, |E ∗ (X )| = |X | + 1. We define an edge e ∈ E ∗ (X ) to be X -universal if there exists a special subhypergraph F ∈ X , such that e intersects F and for every other edge, say f , that intersects F there exists a τ -transversal of F that covers both e and f . We proceed further with the following series of subclaims. Claim H.1: If e ∈ E ∗ (X ), then there exists an X -transversal that intersects every edge in E ∗ (X ) \ {e}. Furthermore, no edge in E ∗ (X ) is X -universal. Proof of Claim H.1: Let G X be the bipartite graph associated with the special H -set X and consider the graph G X − e, where e is an arbitrary vertex in E ∗ (X ). We note that the partite sets of G X − e are E ∗ (X ) \ {e} and X and these sets have equal cardinalities. Suppose that G X − e does not have a perfect matching. By Hall’s Theorem, there is a nonempty subset S ⊆ X such that in the graph G X − e, we have |N (S)| < |S|. (We remark that here N (S) ⊂ E ∗ (X ) \ {e}.) Since the edge e may possibly intersect a subhypergraph of S in H , we note that in the graph G X , |N (S)| ≤ |S|. However, in the graph G X , |E ∗ (S)| = |N (S)|, implying that |E ∗ (S)| ≤ |S|. By our choice of the special H -set X , |E ∗ (X )| − |X | ≤ |E ∗ (S)| − |S| ≤ 0, contradicting Claim D. Therefore, G X − e has a perfect matching. By Observation 8.1(g), this implies the existence of a minimum X -transversal, T , that intersects every edge in E ∗ (X ) \ {e}. This proves the first part of Claim H.1. For the sake of contradiction, suppose that e ∈ E ∗ (X ) is a universal edge. Let F ∈ X be a special subhypergraph of X that is intersected by e, such that for every f ∈ E ∗ (X ) intersecting F there exists a τ -transversal of F that covers both e and f . By the above argument there exists a perfect matching, M, in G X − e. Let g be the edge that is matched to F in M. By definition there is a τ -transversal of F intersecting both e and g. Due to the matching M \ {g F} and Observation 8.1(g), we therefore obtain an X -transversal, T , that intersects every edge in E ∗ (X ), a contradiction to Claim B. This proves the second part of Claim H.1. ()

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Claim H.2: |X 10 | = 0. Proof of Claim H.2: Suppose, to the contrary, that |X 10 | > 0. Let H  be a copy of H10 that belongs to the special H -set X . By Claim E, at least two edges in E ∗ (X ) intersect H  in H . Let e be such an edge of E ∗ (X ) that intersects H  . Let G X be the bipartite graph associated with X and consider the graph G X − e. We note that the partite sets of G X − e are E ∗ (X ) \ {e} and X and these sets have equal cardinalities. Suppose that G X − e does not have a perfect matching. By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗ (X ) \ {e} such that in the graph G X − e (and in the graph G X ), we have |N (S)| < |S|. We remark that here N (S) ⊂ X . We now consider the special H -set, X  = X \ N (S). Then, X  = ∅ and |E ∗ (X  )| ≤ ≤ = =

|E ∗ (X )| − |S| |E ∗ (X )| − (|N (S)| + 1) |X | + 1 − |N (S)| − 1 |X  |,

contradicting Claim E. Therefore, G X − e has a perfect matching, M say. Let e be the vertex of E ∗ (X ) that is M-matched to H  in G X − e. Let u and v be vertices in the subhypergraph H  that belong to the edges e and e , respectively, in H . By Observation 8.1(h), there exists a τ -transversal of H  that contains both u and v. If H  is a subhypergraph in X different from H  and if e is the vertex of E ∗ (X ) that is M-matched to H  in G X − e, then by Observation 8.1(g) there exists a τ -transversal of H  that contains a vertex of H  that belongs to e in H . This implies the existence of a minimum X -transversal, T , that intersects every edge in E ∗ (X ), contradicting () Claim B. Recall that the boundary of a set S of vertices in H , denoted ∂ H (S) or simply ∂(S) if H is clear from context, is the set N (S) \ S. Abusing notation, we write ∂(X ), rather than ∂(V (X )), as the boundary of the set V (X ). Let X  be a set of new vertices (not in H ), where |X  | = max(0, 4 − |∂(X )|). We note that |X  ∪ ∂(X )| ≥ 4. Further if |∂(X )| ≥ 4, then X  = ∅. Let H  be the hypergraph obtained from H − V (X ) by adding the set X  of new vertices and adding a 4-edge, e, containing four vertices in X  ∪ ∂(X ). Note that H  may not be linear as the edge e may overlap other edges in H  . Claim H.3: Either def(H  ) = 0 or def(H  ) = def H  (Y ) where |Y | = 1 and e is an edge of the hypergraph in the special H  -set Y . Proof of Claim H.3: Suppose that def(H  ) > 0. Let Y be a special H  -set such that def(H  ) = def H  (Y ). Since def(H  ) > 0, we note that in H  , |E ∗ (Y )| < |Y |. Suppose that e ∈ / E(Y ). We now consider the special H -set Q = X ∪ Y . Then in H , |E ∗ (Q)| ≤ |E ∗ (X )| + |E ∗ (Y )| ≤ (|X | + 1) + (|Y | − 1) = |X | + |Y | = |Q|, contradicting Claim E. Therefore, e ∈ E(Y ). Let e ∈ E(R), where R ∈ Y . Suppose that |Y | ≥ 2. In this case, we consider the special H -set Q = X ∪ (Y \ {R}). Then in H , |E ∗ (Q)| ≤ |E ∗ (X )| + |E ∗ (Y )| ≤ (|X | + 1) + (|Y | − 1) = |X | + |Y | = |Q| + 1. By Claim E, |E ∗ (Q)| ≥ |Q| + 1. Consequently, |E ∗ (Q)| = |Q| + 1.

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However since |Y | ≥ 2, we note that |Q| > |X |, contradicting our choice of the spe() cial H -set X . Claim H.4: Φ(H  ) ≥ −8|X 4 | − 5|X 14 | − 4|X 11 | − |X 21 | − 6|X  | + 13|X | − def(H  ). Furthermore if H  is linear, then both of the following statements hold. (a): 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − def(H  ) ≥ 13|X | − 6|X  | − 10. (b): If X  = ∅, then 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − 8. Proof of Claim H.4: By Claim F, def(H ) = 0. Thus, Φ(H  ) = 45(τ (H  ) − τ (H )) − 6(n(H  ) − n(H )) − 13(m(H  ) − m(H )) − def(H  ).

We will first show that τ (H ) ≤ τ (H  ) + |X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + 6|X 21 |. Let T  be a minimum transversal in H  . If some vertex of X  belongs to T  , then we can simply replace such a vertex in T  with a vertex from V (e) ∩ ∂(X ) since a vertex in X  belongs to the edge e but no other edge of H  . We may therefore assume that some vertex in ∂(X ) belongs to T  , and hence T  covers at least one edge from E ∗ (X ). By Claim H.1, we can now obtain a transversal of H of size τ (H  ) + |X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + 6|X 21 |, which implies that τ (H ) ≤ τ (H  ) + |X 4 | + 3|X 10 | + 3|X 11 | + 4|X 14 | + 6|X 21 | as desired. Let F ∈ X . By Claim H.2, X 10 = ∅. By Observation 8.1, if F ∈ X 4 , then F contributes 4 to the sum n(H ) − n(H  ), 1 to the sum m(H ) − m(H  ), and at most 1 to the sum τ (H ) − τ (H  ) (due to the above bound on τ (H )), and therefore contributes at least 45 · (−1) − 6 · (−4) − 13 · (−1) = −8 to Φ(H  ). Similar arguments show that if F ∈ X 14 , then F contributes at least −5 to Φ(H  ). If F ∈ X 11 , then F contributes at least −4 to Φ(H  ), while if F ∈ X 21 , then F contributes at least −1 to Φ(H  ). The edges in E ∗ (X ) contribute |E ∗ (X )| = |X | + 1 to the sum m(H ) − m(H  ) and therefore 13(|X | + 1) to Φ(H  ), while the added edge e contributes 1 to the sum m(H  ) − m(H ) and therefore −13 to Φ(H  ). If X  = ∅, then each vertex in X  contributes 1 to the sum n(H  ) − n(H ), and therefore the vertices in X  contribute −6|X  | to Φ(H  ). This proves the first part of Claim H.4. Suppose, next, that H  is linear. Then, by Claim G, Φ(H  ) < 0. Thus, from our previous inequality established in the first part of Claim H.4, 0 > Φ(H  ) ≥ −8|X 4 | − 5|X 14 | − 4|X 11 | − |X 21 | − 6|X  | + 13|X | − def(H  ). This immediately implies part (a) as def(H  ) ≤ 10 by Claim H.3. If X  = ∅, then the edge e contains a degree-1 vertex and is therefore not part of a H10 subhypergraph in H  . Part (b) now () again follows from Claim H.3, noting that in this case def(H  ) ≤ 8. Claim H.5: Suppose the added edge e overlaps with some other edge in H  . (a): If e contains a vertex of degree 1 in H  , then both of the following statements hold. (i): Φ(H  ) ≤ 3 − def(H  ). (ii): 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≥ 13|X | − 6|X  | − 3.

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(b): If e contains no vertex of degree 1 in H  , then both of the following statements hold. (i): Φ(H  ) ≤ 7 − def(H  ). (ii): 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≥ 13|X | − 6|X  | − 7. Proof of Claim H.5: Suppose that the added edge e overlaps with some other edge e in H  . We will first show the following: Φ(H  ) ≤



3 − def(H  ) if e contains a vertex of degree 1 in H  7 − def(H  ) ife contains no vertex of degree 1 in H  .

Suppose, to the contrary, that Φ(H  ) ≥ 4 − def(H  ) if e contains a vertex of degree 1 in H  and that Φ(H  ) ≥ 8 − def(H  ) if e contains no vertex of degree 1 in H  . Let {x, y} ⊆ e ∩ e . Suppose firstly that d H  (x) = 3. Let e be the edge incident with x different from e and e , and consider the hypergraph H = H  − x. Since H ∈ L4,3 , by construction, H ∈ L4,3 . Then, m(H ) = m(H  ) − 3 and τ (H  ) ≤ τ (H ) + 1. Further, if e contains a vertex of degree 1 in H  , then n(H ) ≤ n(H  ) − 2, while if e contains no vertex of degree 1 in H  , then n(H ) ≤ n(H  ) − 1. Recall that Φ(H  ) = ξ(H  ) − ξ(H ) and Φ(H ) = ξ(H ) − ξ(H ). By Claim G, Φ(H ) < 0. Thus, Φ(H ) ξ(H ) − ξ(H ) ξ(H ) + Φ(H  ) − ξ(H  ) Φ(H  ) − 45(τ (H  ) − τ (H )) + 6(n(H  ) − n(H ))+ 13(m(H  ) − m(H )) + (def(H  ) − def(H )) ≥ Φ(H  ) − 45 + 6(n(H  ) − n(H )) + 13 · 3 + def(H  ) − def(H ) ≥ Φ(H  ) − 6 + 6(n(H  ) − n(H )) + def(H  ) − def(H ).

0> = = =

Claim H.5.1: If e contains a vertex of degree 1 in H  , then def(H ) ≥ 11. Proof of Claim H.5.1: If e contains a vertex of degree 1 in H  , then n(H  ) − n(H ) ≥ 2 and, by supposition, Φ(H  ) ≥ 4 − def(H  ). Thus, in this case, the above inequality chain simplifies to 0 > Φ(H ) ≥ (4 − def(H  )) − 6 + 6 · 2 + def(H  ) − () def(H ) = 10 − def(H ), and so, def(H ) > 10. Claim H.5.2: If e contains no vertex of degree 1 in H  , then def(H ) ≥ 9. Proof of Claim H.5.2: If e contains no vertex of degree 1 in H  , then n(H  ) − n(H ) ≥ 1 and, by supposition, Φ(H  ) ≥ 8 − def(H  ). Thus, in this case, our inequality chain simplifies to 0 > Φ(H ) ≥ (8 − def(H  )) − 6 + 6 · 1 + def(H  ) − () def(H ) = 8 − def(H ), and so, def(H ) > 8. By Claim H.5.1 and Claim H.5.2, def(H ) ≥ 9. Let Y be a special H -set such that def(H ) = def H (Y ).

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65

Claim H.5.3: |Y | = 1. Proof of Claim H.5.3: Suppose, to the contrary, that |Y | ≥ 2. If |E ∗ (Y )| ≥ |Y | − 1, then def H (Y ) ≤ 10|Y | − 13|E ∗ (Y )| ≤ 10|Y | − 13(|Y | − 1) = −3|Y | + 13 ≤ 7, a contradiction. Hence, |E ∗ (Y )| ≤ |Y | − 2. We now consider the special H -set Q = X ∪ Y . Then in H , |E ∗ (Q)| ≤ |E ∗ (X )| + |E ∗ (Y )| + |{e , e }| ≤ (|X | + 1) + (|Y | − 2) + 2 = |X | + |Y | + 1 = |Q| + 1. By Claim E, |E ∗ (Q)| ≥ |Q| + 1. Consequently, |E ∗ (Q)| = |Q| + 1. However |Q| > |X |, contradicting our choice of the () special H -set X . Therefore, |Y | = 1. By Claim H.5.3, |Y | = 1. Let Y = {R}. If R = H10 , then def H (Y ) ≤ 8|Y | − 13|E ∗ (Y )| ≤ 8, a contradiction. Hence, R = H10 . If |E ∗ (Y )| ≥ 1, then def H (Y ) = 10 − 13|E ∗ (Y )| ≤ −3, a contradiction. Therefore, R is a component of H and def H (Y ) = 10. In particular, we note that the edge e therefore contains no vertex of degree 1 in H  . By Observation 8.1(f), R is 2-regular. Recall that {x, y} ⊆ e ∩ e , implying that d H  (y) ≥ 2. If d H  (y) = 2, then both x and y are removed from H  when constructing H , implying that n(H  ) − n(H ) ≥ 2. In this case, our inequality chain in the proof of Claim H.5.2 simplifies to 0 > Φ(H ) ≥ (8 − def(H  )) − 6 + 6 · 2 + def(H  ) − def(H ) = 14 − def(H ), and so, def(H ) > 14, a contradiction. Therefore, d H  (y) = 3, implying that d H (y) = 1 and y ∈ / V (R). Let e be the edge incident with y different from e and e . If e intersects R, then since R is a component of H  , this would imply y ∈ V (R), a contradiction. Therefore, e does not intersect R. Interchanging the roles of x and y, and considering the hypergraph H = H  − y (instead of H = H  − x), there exists an H10 -component, say R  , in H . Anal/ V (R  ) and e ogous arguments as employed earlier show that d H (x) = 1, x ∈ does not intersect R  . Thus, R and R  are H10 -components in the hypergraph, H  − {e, e , e , e }, obtained from H  by deleting the edges e, e , e , e . If R = R  , then neither e nor e intersects R. Hence, |E ∗ (X ∪ R)| ≤ |E ∗ (X )| +  |{e }| ≤ (|X | + 1) + 1 = |X ∪ R| + 1. By Claim E, |E ∗ (X ∪ R)| ≥ |X ∪ R| + 1. Consequently, |E ∗ (X ∪ R)| = |X ∪ R| + 1. However |X ∪ R| > |X |, contradicting our choice of the special H -set X . Therefore, R = R  , implying that R and R  are distinct and vertex-disjoint components of H  − {e, e , e , e }. As observed earlier, e does not intersect R. Suppose that e does not intersect R. Then, |E ∗ (X ∪ R)| ≤ |E ∗ (X )| + |{e }| ≤ (|X | + 1) + 1 = |X ∪ R| + 1, a contradiction. Therefore, e intersects R. Analogously, e intersects R  . Let r  ∈ e ∩ V (R) and let r  ∈ e ∩ V (R  ). If R and R  contain no vertex in ∂(X ), then |E ∗ (R ∪ R  )| ≤ |{e , e , e }| = 3 = |R ∪ R  | + 1, implying that E ∗ (R ∪ R  ) = {e , e , e } and |E ∗ (R ∪ R  )| = |R ∪ R  | + 1. Let r  ∈ e ∩ V (R) ∪ V (R  ). Without loss of generality, we may assume that r  ∈ V (R). By Observation 8.1(h), there exists a τ -transversal of R that contains both r  and r  , and a τ -transversal of R  that contains r  . Thus, given the special H set, {R, R  }, we can find an {R, R  }-transversal that covers every edge in E ∗ (R ∪ R  ), contradicting Claim B. Therefore, R or R  (or both R and R  ) contain a vertex in ∂(X ). Let z be such a vertex and let ez be an edge of E ∗ (X ) that contains z. Without loss of generality, we may assume that r  ∈ V (R). Applying Observation 8.1(h)

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and Claim H.1, we can find an (X ∪ {R, R  })-transversal that covers every edge in E ∗ (X ∪ R ∪ R  ), contradicting Claim B. We deduce, therefore, that d H  (x) = 2 (recall that in H  , {x, y} ⊆ e ∩ e ). Analogously, all vertices in e ∩ e have degree 2 in H  . In particular, d H  (y) = 2. We once again consider the hypergraph H = H − x, noting that here both x and y are removed from H  when constructing H . Since H ∈ L4,3 , as before def(H ) ≥ 0. Further, m(H ) = m(H  ) − 2 and τ (H  ) ≤ τ (H ) + 1. If e contains a vertex of degree 1 in H  , then n(H  ) − n(H ) ≥ 3 and, by supposition, Φ(H  ) ≥ 4 − def(H  ). Thus, in this case, our inequality chain simplifies to 0 > Φ(H ) ≥ (4 − def(H  )) − 45 + 6 · 3 + 13 · 2 + def(H  ) − def(H ) = 3 − def(H ), and so, def(H ) ≥ 4. If e contains no vertex of degree 1 in H  , then n(H  ) − n(H ) ≥ 2 and, by supposition, Φ(H  ) ≥ 8 − def(H  ). Thus, in this case, our inequality chain simplifies to 0 > Φ(H ) ≥ (8 − def(H  )) − 45 + 6 · 2 + 13 · 2 + def(H  ) − def(H ) = 1 − def(H ), and so, def(H ) ≥ 2. In both cases, def(H ) > 0. Let Y be a special H -set such that def(H ) = def H (Y ). Since def H (Y ) > 0, we note that |E ∗ (Y )| ≤ |Y | − 1. We now consider the special H -set X ∪ Y . In H , |E ∗ (X ∪ Y )| ≤ |E ∗ (X )| + |E ∗ (Y )| + |{e }| ≤ (|X | + 1) + (|Y | − 1) + 1 = |X | + |Y | + 1 = |X ∪ Y | + 1, contradicting our choice of the special H -set X . This completes the proof of the following:  3 − def(H  ) if e contains a vertex of degree 1 in H   Φ(H ) ≤ 7 − def(H  ) if e contains no vertex of degree 1 in H  , thereby establishing part (a)(i) and part (b)(i). By Claim H.4, Φ(H  ) ≥ −8|X 4 | − 5|X 14 | − 4|X 11 | − |X 21 | − 6|X  | + 13|X | − def(H  ). Thus, if e contains a vertex of degree 1 in H  , then 3 − def(H  ) ≥ Φ(H  ) ≥ −8|X 4 | − 5|X 14 | − 4|X 11 | − |X 21 | − 6|X  | + 13|X | − def(H  ).

This immediately completes the proof of part (a)(ii), as def(H  ) cancels out. Part (b) () is easily proved analogously. Claim H.6: The case |X | ≥ 2 and |∂(X )| ≥ 4 cannot occur. Proof of Claim H.6: Note that 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≤ 8|X | ≤ 13|X | − 6|X  | − 10, as |X | ≥ 2 and |X  | = 0. If H  is linear, then we are obtain a contradiction to Claim H.4(a), and if H  is not linear we obtain a contradiction to Claim H.5. () Claim H.7: The case |X | ≥ 3 and |∂(X )| = 3 cannot occur. Proof of Claim H.7: Note that 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≤ 8|X | ≤ 13|X | − 6|X  | − 9, as |X | ≥ 3 and |X  | = 1. If H  is linear, then we obtain a contradiction to Claim H.4(b), and if H  is not linear we obtain a contradiction to () Claim H.5. Claim H.8: The case |X | = 2 and |∂(X )| = 3 cannot occur. Proof of Claim H.8: Suppose, to the contrary, that |X | = 2 and |∂(X )| = 3. Then, |E ∗ (X )| = 3 and |X  | = 1. If H  is not linear, then we obtain a contradiction by

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Claim H.5(a), as 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≤ 16 and 13|X | − 6|X  | − 3 = 17. Therefore, H  is linear. Claim H.8.1: |X 4 | ≤ 1. Proof of Claim H.8.1: Suppose, to the contrary, that X consists of two (vertexdisjoint) copies of H4 . By Claim H.4(a), 16 = 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − def(H  ) = 26 − 6 − def(H  ), implying that def(H  ) > 4. If e is a H4 -component in H  , then, by the connectivity and linearity of H , we note that H = H11 , contradicting Claim C. Hence, H  = H4 , implying by Claim H.3 and the fact that e contains a degree-1 vertex (and so, Y = H10 ) that Y ∈ X 14 and def(H  ) = def(Y ) = 5. Since e ∈ E(Y ) and e contains a vertex of degree 1 in H  , we note that H  = H14,5 and H  = H14,6 . Therefore, H  = H14,i for some i ∈ [4], implying that () H = H21,i , contradicting Claim C. Claim H.8.2: H  = H4 , |X 4 | = 1 and |X 14 | = 1. Proof of Claim H.8.2: Recall that H  is linear and |X  | = 1. By Claim H.4(a), 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − def(H  ) = 20 − def(H  ). By Claim H.8.1, |X 4 | ≤ 1. By Claim H.3 and the fact that e contains a degree-1 vertex, def(H  ) ≤ 8. Thus, 8 + 5 ≥ 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 20 − def(H  ), () implying that |X 4 | = 1, |X 14 | = 1 and def(H  ) = 8, as desired. By Claim H.8.2, H  = H4 , |X 4 | = 1, and |X 14 | = 1. Let E ∗ (X ) = {e1 , e2 , e3 }. Let F1 and F2 be the (vertex-disjoint) hypergraphs that belong to X , where F1 = H14 and F2 = H4 . Further, let E(F2 ) = { f 2 }. Let ∂(X ) = {x, y, z}. Claim H.8.3: All three edges in E ∗ (X ) intersect F1 . Proof of Claim H.8.3: Suppose, to the contrary, that at most two edges in E ∗ (X ) intersect F1 . Therefore, by Claim E, exactly two edges in E ∗ (X ) intersect F1 . Renaming edges in E ∗ (X ) if necessary, we may assume that e1 and e2 intersect F1 , and that e3 does not intersect F1 . By the linearity of H , the edge e3 contains one vertex of F2 and all three vertices of ∂(X ). This in turn implies by the linearity of H that both edges e1 and e2 intersect F1 in at least two vertices. Let T1 = e1 ∩ V (F1 ) and T2 = e2 ∩ V (F1 ). Thus, |T1 | ≥ 2 and |T2 | ≥ 2. If e1 and e2 intersect in a common vertex of F1 , then, by Observation 8.1(g), we can cover e1 and e2 by a τ -transversal of F1 and we can cover e3 by a τ -transversal of F2 , implying that there is an X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Hence, T1 ∩ T2 = ∅. By Observation 8.1(l), there exists a τ -transversal of F1 that contains one vertex from T1 and one vertex from T2 . Once again, this implies that there is an X -transversal () that intersects every edge in E ∗ (X ), contradicting Claim B. By Claim H.8.3, all three edges in E ∗ (X ) intersect F1 . If two edges of E ∗ (X ) intersect in a common vertex of F1 , then we can cover these two edges by a τ transversal of F1 . If no two edges of E ∗ (X ) intersect in a common vertex of F1 , then by Observation 8.1(i), we can cover two edges of E ∗ (X ) by a τ -transversal of F1 . In both cases, two edges of E ∗ (X ) can be covered by a τ -transversal of F1 . Renaming edges in E ∗ (X ) if necessary, we may assume that e1 and e2 can be

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covered by a τ -transversal of F1 . If e3 intersects F2 , then there is an X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Hence, e3 does not intersect F2 , implying that both e1 and e2 intersect F2 . If |e3 ∩ V (F1 )| ≥ 3, then by Observation 8.1(l), we can cover e1 and e3 by a τ -transversal of F1 . Since we can cover e2 by a τ -transversal of F2 , there is an X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Hence, |e3 ∩ V (F1 )| ≤ 2. If |(e1 ∪ e2 ) ∩ V (F1 )| = 1, then both e1 and e2 contain one vertex from F1 , one vertex from F2 and two vertices from ∂(X ). However since |∂(X )| = 3, e1 and e2 would then overlap, a contradiction. Hence, |(e1 ∪ e2 ) ∩ V (F1 )| ≥ 2. Thus, if |e3 ∩ V (F1 )| = 2, then by Observation 8.1(k), there exists a τ -transversal of F1 that contains a vertex from e3 ∩ V (F1 ) and a vertex from (e1 ∪ e2 ) ∩ V (F1 ), implying that we can cover e3 and one of e1 and e2 by a τ -transversal of F1 and we can cover the remaining edge in E ∗ (X ) by a τ -transversal of F2 , once again contradicting Claim B. Therefore, |e3 ∩ V (F1 )| = 1, implying that the edge e3 contains one vertex of F1 and all three vertices in ∂(X ), namely, x, y, and z. Thus, both e1 and e2 contain one vertex from F2 , one vertex from ∂(X ), and two vertices from F1 . In particular, we note that e1 ∩ V (F1 ) is a strongly independent set in F1 , as is e2 ∩ V (F1 ). Further, since e1 and e2 do not overlap, |(e1 ∪ e2 ) ∩ V (F1 )| ≥ 3. Thus, by Observation 8.1(l), there exists a τ -transversal of F1 that contains a vertex from e3 ∩ V (F1 ) and a vertex from (e1 ∪ e2 ) ∩ V (F1 ), implying that we can cover e3 and one of e1 and e2 by a τ -transversal of F1 and we can cover the remaining edge in E ∗ (X ) by a τ -transversal of F2 , once again contradicting Claim B. This completes the proof of Claim H.8. () Claim H.9: If |X | ≥ 2 and |∂(X )| = 2, then H  ∈ L4,3 . Proof of Claim H.9: Let |X | ≥ 2 and |∂(X )| = 2 and suppose, to the contrary, that H ∈ / L4,3 . Thus, H  is not a linear hypergraph, implying that the edge e overlaps in  H with some other edge, e say. Since e contains two vertices of degree 1, namely, the two vertices in X  , we note that in H  , e ∩ e = ∂(X ). We now consider the hypergraph H  = H − e . Since H  has no overlapping edges, H  ∈ L4,3 , and so by Claim G, Φ(H  ) < 0. By Claim H.2, |X 10 | = 0. An analogous proof to that of Claim H.4 shows that Φ(H  ) ≥ −8|X 4 | − 5|X 14 | − 4|X 11 | − |X 21 | − 6|X  | + 13(|X | + 1) − def(H  ), noting that the deleted edge e contributes 1 to the sum m(H ) − m(H  ). Thus, since |X  | = 2 and |X | = 2, Φ(H  ) ≥ −8|X | − 12 + 13(|X | + 1) − def(H  ) = 5|X | + 1 − def(H  ) = 11 − def(H  ). If def(H  ) ≤ 11, then Φ(H  ) ≥ 0, a contradiction. Hence, def(H  ) ≥ 12. Let Y be a special H  -set such that def(H  ) = def H  (Y ). If |Y | = 1, then def H  (Y ) ≤ 10, a contradiction. Hence, |Y | ≥ 2. If |E ∗ (Y )| ≥ |Y | − 1 in H  , then def H (Y ) ≤ 10|Y | − 13|E ∗ (Y )| ≤ 10|Y | − 13(|Y | − 1) = −3|Y | + 13 < 12, a contradiction. Therefore, |E ∗ (Y )| ≤ |Y | − 2 in H  . We now consider the special H -set X ∪ Y . Suppose that e ∈ / E(Y ). Then in H , |E ∗ (X ∪ Y )| ≤ |E ∗ (X )| + |E ∗ (Y )| + |{e }| ≤

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(|X | + 1) + (|Y | − 2) + 1 = |X | + |Y |, contradicting Claim E. Therefore, e ∈ E(Y ). Let e ∈ E(R), where R ∈ Y . We consider the special H -set Q = X ∪ (Y \ {R}). Then in H , |E ∗ (Q)| ≤ |E ∗ (X )| + |E ∗ (Y )| + |{e }| ≤ (|X | + 1) + (|Y | − 2) + 1 = |X | + |Y | = |Q| + 1. By Claim E, |E ∗ (Q)| ≥ |Q| + 1. Consequently, |E ∗ (Q)| = |Q| + 1. However since |Y | ≥ 2, we note that |Q| > |X |, contradicting our choice () of the special H -set X . Claim H.10: The case |X | ≥ 4 and |∂(X )| = 2 cannot occur. Proof of Claim H.10: Note that 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≤ 8|X | ≤ 13|X | − 6|X  | − 8, as |X | ≥ 4 and |X  | = 2. As H  is linear by Claim H.9, this () contradicts Claim H.4(b). Claim H.11: The case |X | = 3 and |∂(X )| = 2 cannot occur. Proof of Claim H.11: Suppose, to the contrary, that |X | = 3 and |∂(X )| = 2, which implies that |X  | = 2. By Claim H.9, H  ∈ L4,3 , and so, by Claim H.4(a), we have 24 = 8|X | ≥ 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − def(H  ) = 39 − 12 − def(H  ), and so def(H  ) > 3. As e contains two degree-1 vertices, we must have Y = H4 and def(H  ) = def(Y ) = 8, by Claim H.3. Therefore, by Claim H.4(a), 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 39 − 12 − def(H  ) = 19, implying that |X 4 | ≥ 2. By the connectivity of H , we note therefore that V (H ) = V (X ) ∪ ∂(X ). Let E ∗ (X ) = {e1 , e2 , e3 , e4 }. Let F1 , F2 and F3 be the (vertex-disjoint) hypergraphs that belong to X . By Claim H.2, |X 10 | = 0. Claim H.11.1: |X 4 | = 2. Proof of Claim H.11.1: As observed earlier, |X 4 | ≥ 2. Suppose, to the contrary, that |X 4 | ≥ 3, implying that |X | = |X 4 | = 3. Thus, every hypergraph in X is a copy of H4 . Let ∂(X ) = {x, y}. Thus, F1 , F2 , and F3 are all copies of H4 . Let E(Fi ) = { f i } for i ∈ [3]. Further, V (H ) = V (X ) ∪ {x, y} and E(H ) = E(X ) ∪ E ∗ (X ). In particular, n(H ) = 14 and m(H ) = 6. Renaming x and y, if necessary, we may assume that d H (x) ≥ d H (y). Suppose that every vertex in V (X ) has degree at most 2 in H . Then, by the linearity of H , there are three cases to consider. If d H (x) = 3 and d H (y) = 2, then H = H14,1 . If d H (x) = 3 and d H (y) = 1, then H = H14,2 . If d H (x) = 2, then d H (y) = 2 and H = H14,5 . In all three cases, we contradict Claim C. Therefore, some vertex in V (X ) has degree 3. We may assume that F1 contains a vertex, z 1 say, of degree 3 in H and that e1 and e4 contain z. By Claim E, at least three edges in E ∗ (X ) intersect F2 ∪ F3 . Renaming e2 and e3 , if necessary, we may assume that e2 intersects F2 . If e3 intersects F3 , then in this case letting z 2 ∈ e2 ∩ f 2 and z 3 ∈ e3 ∩ f 3 , the set {z 1 , z 2 , z 3 } is an X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Hence, e3 does not intersect F3 , implying by the linearity of H that e3 contains both x and y, and intersects both F1 and F2 . This in turn implies that each of e1 , e2 , and e3 contain exactly one vertex in {x, y} and therefore exactly one vertex of Fi for each i ∈ [3]. In this case, letting z 2 ∈ e3 ∩ f 2 and z 3 ∈ e2 ∩ f 3 , the set {z 1 , z 2 , z 3 } is an X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Therefore, |X | = 2. ()

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By Claim H.11.1, |X 4 | = 2. We may assume that F1 = H4 . Thus, both F2 and F3 are copies of H4 . Let E(F2 ) = { f 2 } and E(F3 ) = { f 3 }. Claim H.11.2: All four edges in E ∗ (X ) intersect F1 . Proof of Claim H.11.2: Suppose, to the contrary, that at most three edges in E ∗ (X ) intersect F1 . If two edges in E ∗ (X ) do not intersect F1 , then these two edges would both contain x and y, contradicting the linearity of H . Hence, at least three edges in E ∗ (X ) intersect F1 . Consequently, exactly three edges in E ∗ (X ) intersect F1 . Renaming edges in E ∗ (X ) if necessary, we may assume that e1 , e2 , and e3 intersect F1 . By the linearity of H , the edge e4 contains both x and y, and intersects both F2 and F3 . Suppose that two of the edges e1 , e2 , and e3 , say e1 and e2 , can be covered by a minimum transversal, T1 say, in F1 . If e3 intersects F2 ∪ F3 , say e3 intersects F2 , then we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it the vertex in e3 ∩ f 2 and the vertex in e4 ∩ f 3 , contradicting Claim B. Hence, e3 does not intersect F2 ∪ F3 , implying that e3 contains at most one of x and y and at least three vertices of F1 . Further this implies that both e1 and e2 intersect F2 ∪ F3 since by Claim E, at least three edges in E ∗ (X ) intersect F2 ∪ F3 . By Observation 8.1(l), there exists a τ -transversal of F1 that covers e3 and one of e1 and e2 , say e1 . We can then cover e2 and e4 from F2 ∪ F3 , once again contradicting Claim B. Therefore, we can only cover at most one of the three edges e1 , e2 , and e3 by a minimum transversal in F1 . By Observation 8.1(i), this implies that F1 = H11 . Further, for i ∈ [3] the edge ei intersects F1 in exactly one vertex and this vertex has no neighbor of degree 1 in F1 . Let ei ∩ V (F1 ) = {u i } for i ∈ [3]. Thus, U = {u 1 , u 2 , u 3 } is the set of three vertices in F1 all of whose neighbors in F1 have 2 in F1 . Let H  be the hypergraph obtained from H by deleting the eight vertices in V (F1 ) \ U (and their incident edges), adding a new vertex u  , and then adding the edge f  = {u  , u 1 , u 2 , u 3 }. Then, H  has order n(H  ) = 14 and size m(H  ) = 7. We note that E(H  ) = {e1 , e2 , e3 , e4 , f  , f 2 , f 3 }. Recall that the edge e4 contains both x and y, and intersects both F2 and F3 . Therefore, the edge ei contains exactly one vertex from each of F2 and F3 , and exactly one of x and y, for each i ∈ [3]. If every vertex in F2 ∪ F3 has degree at most 2 in H  , then by the linearity of H , H  = H14,1 , implying that H = H21,1 , contradicting Claim C. Therefore, some vertex in F2 ∪ F3 has degree 3 in H  . We may assume that z 2 is such a vertex and that z 2 ∈ V (F2 ). Further, we may assume that e1 and e2 contain the vertex z 2 . Let z 1 and z 3 be the vertices of e3 and e4 , respectively, that belongs to F1 and F3 . Then, {z 1 , z 2 , z 3 } is a τ -transversal of H  that can be extended to a τ -transversal of H by adding to it two additional vertices of V (H ) \ V (H  ). This produces an X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Therefore, all four edges in E ∗ (X ) () intersect F1 . By Claim H.11.2, all four edges in E ∗ (X ) intersect F1 . Claim H.11.3: All four edges in E ∗ (X ) intersect F2 ∪ F3 . Proof of Claim H.11.2: Suppose, to the contrary, that at most three edges in E ∗ (X ) intersect F2 ∪ F3 . By Claim E, at least three edges in E ∗ (X ) intersect F2 ∪ F3 .

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Therefore, exactly three edges in E ∗ (X ) intersect F2 ∪ F3 . We may assume that e1 does not intersect F2 ∪ F3 . Thus, e2 , e3 , e4 all intersect F2 ∪ F3 . We note that |e1 ∩ V (F1 )| ≥ 2. Further since Δ(H ) ≤ 3, we note that |(e2 ∪ e3 ∪ e4 ) ∩ V (F1 )| ≥ 2. Thus, by Observation 8.1(k), there is a τ -transversal of F1 , T1 say, that covers e1 and covers one of the edges e2 , e3 , e4 , say e2 . Renaming F2 and F3 , if necessary, we may assume that e3 intersects F2 . If e4 intersects F3 , then we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it the vertex in e3 ∩ f 2 and the vertex in e4 ∩ f 3 , contradicting Claim B. Therefore, e4 does not intersect F3 , implying that e4 intersects F2 . If e3 intersects F3 , then analogously we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ), a contradiction. Therefore, neither e3 nor e4 intersects F3 , implying that e2 is the only possible edge in E ∗ (X ) () that intersect F3 , contradicting Claim E. By Claim H.11.3, all four edges in E ∗ (X ) intersect F2 ∪ F3 . Claim H.11.4: At least three edges in E ∗ (X ) intersect F3 . Proof of Claim H.11.4: Suppose, to the contrary, that at most two edges in E ∗ (X ) intersect F3 . Then, by Claim E, exactly two edges in E ∗ (X ) intersect F3 . We may assume that e1 and e2 intersect F3 , and therefore e3 and e4 do not intersect F3 . By the linearity of H , this implies that at least one of e3 and e4 intersects F1 in at least two vertices. Renaming e3 and e4 if necessary, we may assume that |e3 ∩ V (F1 )| ≥ 2. If |(e1 ∪ e2 ) ∩ V (F1 )| ≥ 2, then by Observation 8.1(k), there is a τ -transversal of F1 , T1 say, that covers e3 and covers one of the edges e1 and e2 , say e1 . We can now extend T1 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it the vertex in e2 ∩ f 3 and the vertex in e4 ∩ f 2 , contradicting Claim B. Therefore, |(e1 ∪ e2 ) ∩ V (F1 )| = 1. By the linearity of H , this implies that both e1 and e2 intersect F2 . Since Δ(H ) ≤ 3, we note that |(e1 ∪ e2 ∪ e4 ) ∩ V (F1 )| ≥ 2. Thus, considering the sets |e3 ∩ V (F1 )| ≥ 2 and the set |(e1 ∪ e2 ∪ e4 ) ∩ V (F1 )| ≥ 2, by Observation 8.1(k), there is a τ -transversal of F1 , T1 say, that covers e3 and covers one of the edges e1 , e2 , e4 . If T1 covers e1 (and therefore also e2 ), then we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it the vertex in e4 ∩ f 2 and any vertex in f 3 , contradicting Claim B. Therefore, T1 covers e4 . Since both e1 and e2 intersect both F2 and F3 , we can once again extend T1 to an () X -transversal that intersects every edge in E ∗ (X ), a contradiction. By Claim H.11.4, at least three edges in E ∗ (X ) intersect F3 . Analogously, at least three edges in E ∗ (X ) intersect F2 . Recall that by Claim H.11.2, all four edges in E ∗ (X ) intersect F1 . On the one hand, if two edges of E ∗ (X ) intersect F1 in a common vertex, then there is a τ -transversal of F1 that covers two edges of E ∗ (X ). On the other hand, if no two edges of E ∗ (X ) intersect F1 in a common vertex, then by Observation 8.1(k), we can once again find a τ -transversal of F1 that covers two edges of E ∗ (X ) (by considering, for example, the set |(e1 ∪ e2 ) ∩ V (F1 )| ≥ 2 and the set |(e1 ∪ e2 ) ∩ V (F1 )| ≥ 2). In both cases, there is a τ -transversal of F1 , T1 , that covers two edges of E ∗ (X ). Renaming edges in E ∗ (X ) if necessary, we may assume that T1 covers e1 and e2 . Recall that by Claim H.11.3, all four edges in

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E ∗ (X ) intersect F2 ∪ F3 . In particular, both e3 and e4 intersect F2 ∪ F3 . Renaming F2 and F3 if necessary, we may assume that e3 intersects F2 . If e4 intersects F3 , then we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it the vertex in e3 ∩ f 2 and the vertex in e4 ∩ f 3 , contradicting Claim B. Therefore, e4 does not intersect F3 , implying that e4 intersects F2 . If e3 intersects F3 , then analogously we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ), a contradiction. Therefore, neither e3 nor e4 intersects F3 , implying that e1 and e2 are the only possible edge in E ∗ (X ) that intersect F3 , contradicting () Claim H.11.4. This completes the proof of Claim H.11. Claim H.12: The case |X | = 2 and |∂(X )| = 2 cannot occur. Proof of Claim H.12: Suppose, to the contrary, that |X | = 2 and |∂(X )| = 2. We note that |E ∗ (X )| = 3 and |X  | = 2. Let E ∗ (X ) = {e1 , e2 , e3 }. By Claim H.9, H  ∈ L4,3 , and so by Claim G, Φ(H  ) < 0. Since at most one of the edges in E ∗ (X ) can contain both vertices in ∂(X ), we note that 3 

|ei ∩ V (F1 )| +

i=1

3 

|ei ∩ V (F2 )| ≥ 8.

(8.1)

i=1

Suppose that at most two edges in E ∗ (X ) intersect F1 . Then, by Claim E, exactly two edges in E ∗ (X ) intersect F1 . We may assume that both e1 and e2 intersect F1 , and therefore that e3 does not intersect F1 . Thus, |e3 ∩ V (F1 )| = 0 and |e3 ∩ V (F2 ) ≥ 2. If |e1 ∩ V (F1 )| + |e2 ∩ V (F1 )| ≥ 4, then by Observation 8.1, we can find a τ transversal of F1 , T1 , that covers both e1 and e2 . In this case, we can extend T1 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it a τ -transversal of F2 that covers e3 , a contradiction. Therefore, |e1 ∩ V (F1 )| + |e2 ∩ V (F1 )| ≤ 3, implying by Inequality (8.1) that 3 

|ei ∩ V (F2 )| ≥ 5.

(8.2)

i=1

Therefore, by Inequality (8.2), we note that (|e1 ∩ V (F2 )| + |e3 ∩ V (F2 )|) + (|e2 ∩ V (F2 )| + |e3 ∩ V (F2 )|) ≥ 5 + |e3 ∩ V (F2 )| ≥ 7. Renaming e1 and e2 if necessary, we may assume that |e1 ∩ V (F2 )| + |e3 ∩ V (F2 )| ≥ 4. By Observation 8.1, we can find a τ -transversal of F2 , T2 , that covers both e1 and e3 . In this case, we can extend T3 to an X -transversal that intersects every edge in E ∗ (X ) by adding to it a τ -transversal of F1 that covers e2 , a contradiction. Hence, all three edges in E ∗ (X ) intersect F1 . Analogously, all three edges in E ∗ (X ) intersect 3 F2 . Renaming F1 and |ei ∩ V (F1 )| ≥ 4. By F2 if necessary, we may assume by Inequality (8.1) that i=1 Observation 8.1, we can find a τ -transversal of F1 , T1 , that covers two of the edges in E ∗ (X ). The third edge in E ∗ (X ) can be covered by a τ -transversal of F2 noting that all three edges in E ∗ (X ) intersect F2 . This produces an X -transversal that intersects () every edge in E ∗ (X ), a contradiction.

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We proceed further with some additional notation. We associate with the set X a bipartite multigraph, which we denote by M X , with partite sets X and E ∗ (X ) as follows. If an edge e ∈ E ∗ (X ) intersects a subhypergraph H  ∈ X in H in k vertices, then we add k multiple edges joining e ∈ E ∗ (X ) and H  ∈ X in M X . Two multiple edges (also called parallel edges in the literature) joining two vertices in M X we call double edges, while three multiple edges joining two vertices in M X we call triple edges. An edge that is not a multiple edge we call a single edge. By supposition in our proof of Claim H, |E ∗ (X )| = |X | + 1. We say that a pair (e1 , e2 ) of edges in E ∗ (X ) forms an E ∗ (X )-pair if they intersect a common subhypergraph, F, of X in H , and, further, there exists a τ -transversal of F that covers both e1 and e2 in H . We call F a subhypergraph of X associated with the E ∗ (X )-pair, (e1 , e2 ). Claim H.13: |E(M X )| ≥ 4|E ∗ (X )| − 3|∂(X )| = 4(|X | + 1) − 3|∂(X )|. Furthermore, the following holds. (a) M X does not contain triple edges. (b) No F ∈ X has double edges to three distinct vertices of E ∗ (X ) in M X . (c) Every e ∈ E ∗ (X ) has degree at least 4 − |∂(X )| in M X . (d) Every F ∈ X has degree at least 2 in M X . Proof of Claim H.13: Consider the bipartite multigraph, M X , which is identical to M X , except we add a new vertex b to M X and for each vertex e ∈ E ∗ (X ) we add r multiple edges joining e and b in M X if e intersects ∂(X ) in r vertices in H . In M X , every vertex in E ∗ (X ) has degree 4. Further, since Δ(H ) ≤ 3, the vertex b has degree at most 3|∂(X )|. Therefore, M X = M X − b contains at least 4|E ∗ (X )| − 3|∂(X )| edges. This proves the first part of Claim H.13. To prove part (a), for the sake of contradiction, suppose that M X does contain triple edges that join vertices e ∈ E ∗ (X ) and F ∈ X . By Observation 8.1( ), we note that e is an X -universal edge, a contradiction to Claim H.1. This proves part (a). To prove part (b), for the sake of contradiction, suppose that some F ∈ X has double edges to three distinct vertices, say e1 , e2 , e3 , in E ∗ (X ) in M X . By Observation 8.1(o), we note that at least one of e1 , e2 , or e3 is an X -universal edge, a contradiction to Claim H.1. This proves part (b). By the construction of M X , every vertex e ∈ E ∗ (X ) has degree 4 − |V (e) ∩ ∂(X )| ≥ 4 − |∂(X )| in M X . By Claim E (used on F), we note that at least two edges in E ∗ (X ) intersect F in H , and therefore the degree of F in M X is at least 2. () This proves part (c) and part (d). Claim H.14: The case |X | ≥ 5 and |∂(X )| = 1 cannot occur. Proof of Claim H.14: Suppose, to the contrary, that |X | ≥ 5 and |∂(X )| = 1. We note that in this case |X  | = 3 and H  is linear. Let ∂(X ) = {y}. Claim H.14.1: |X | = 5 and X = X 4 . Proof of Claim H.14.1: By Claim H.4(b), 8|X | ≥ 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − 8 = 8|X | + 5|X | − 26, and so 5|X | < 26, implying that

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|X | ≤ 5. This clearly implies that |X | = 5. By supposition, |X | ≥ 5. Consequently, |X | = 5, implying that 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 8|X | − 1, which in turn () implies that X = X 4 . By Claim H.14.1, X = X 4 and |X 4 | = 5. Let X = {x1 , x2 , . . . , x5 } and let E ∗ (X ) = {e1 , e2 , . . . , e6 }. We now consider the bipartite multigraph M X defined earlier. Since X = X 4 and H is linear, the multigraph M X is in this case a graph. By Claim H.13(c), since here |∂(X )| = 1, d M X (e) ≥ 3 for each e ∈ E ∗ (X ). By Claim H.13(d), d M X (F) ≥ 2 for each F ∈ X . Also by Claim H.13, we note that  F∈X d M X (F) = |E(M X )| ≥ 4(|X | + 1) − 3|∂(X )| = 24 − 3 = 21. Claim H.14.2: Suppose the edges ei and e j form an E ∗ (X )-pair and that x is the vertex in X corresponding to a copy of H4 of X that contains a vertex covering both e1 and e2 . Then there exists a vertex xk ∈ X \ {x } in M X such that ei and e j are both neighbors of xk in M X and one of the following holds. (a) d M X (xk ) = 2. (b) d M X (xk ) = 3 and N M X (xk  ) ⊆ N M X (xk ) for some vertex xk  ∈ X \ {xk , x }. Proof of Claim H.14.2: For notational convenience, we may assume that i = 1 and j = 2. Thus, (e1 , e2 ) is an E ∗ (X )-pair. Further, we may assume that x1 is the vertex in X corresponding to a copy of H4 of X that contains a vertex covering both e1 and e2 . We now consider the bipartite graph M X = M X − {e1 , e2 , x1 } with partite sets X  = X \ {x1 } and E X = E ∗ (X ) \ {e1 , e2 }. If there exists a matching in M X that matches E X to X  , then there exists a minimum X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Therefore, no matching in M X matches E X to X  . By Hall’s Theorem, there is a nonempty subset S ⊆ E X such that in M X , |N (S)| < |S|. If S = E X , then let xk ∈ X  \ N M X (S). In this case, the vertices e1 and e2 are the only possible neighbors of xk in M X , implying that N M X (xk ) = {e1 , e2 } and d M X (xk ) = 2. Thus, Part (a) holds, as desired. Hence, we may assume that |S| ≤ 3 and that no vertex in X  \ {x1 } has degree 2 with e1 and e2 as its neighbors. Since |S| ≤ 3, we note that |N M X (S)| ≤ 2. Since every vertex in E ∗ (X ) has degree at least 3 in M X , we note that |N M X (S)| ≥ 2. Consequently, |N M X (S)| = 2, implying that |S| = 3. Renaming vertices if necessary, we may assume that S = {e3 , e4 , e5 } and that N M X (S) = {x2 , x3 }. Since d M X (e) ≥ 3 for each e ∈ E ∗ (X ), this implies that for each vertex e ∈ S, we have N M X (e) = {x1 , x2 , x3 }. We note that the only possible neighbors of x4 and x5 in M X are e1 , e2 , and e6 , and so d M X (x4 ) ≤ 3 and d M X (x5 ) ≤ 3. We show next that x4 or x5 dominate both e1 and e2 in M X . Suppose, to the contrary, that neither x4 nor x5 dominates both e1 and e2 in M X . Since both x4 and x5 have degree at least 2 in M X , this implies that d M X (x4 ) = d M X (x5 ) = 2 and that we may assume e6 is adjacent to both x4 and x5 . Renaming e1 and e2 , if necessary,  that e2 is adjacent to x5 , and so N M X (x5 ) = {e2 , e6 }. Since x∈X d M X (x) ≥ 21, the degree sequence of vertices of X in M X is therefore either 2, 2, 5, 6, 6 or 2, 2, 6, 6, 6. If the degree sequence is given by 2, 2, 6, 6, 6, then e6 would be adjacent to every vertex of X in M X , which is not possible since d M X (e) ≤ 4 for each e ∈ E ∗ (X ). Thus, the degree sequence of vertices of X in M X is 2, 2, 5, 6, 6. In particular, this implies that d M X (e6 ) = 4, which in turn implies that x2 and x3 are both adjacent to

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e1 and e2 in M X . Thus, M X − {e6 , x4 , x5 } = K 3,5 . By assumption, e2 is adjacent to x5 . Thus, since d M X (e2 ) = 4, we note that N M X (e2 ) = {x1 , x2 , x3 , x5 } and therefore N M X (x4 ) = {e1 , e6 }. The graph M X is therefore determined. We note that the two vertices of X of degree 6 in M X both give rise to at least two E ∗ (X )-pairs, while the vertex of X of degree 5 in M X gives rise to at least one E ∗ (X )-pairs. Further since there are no overlapping edges, these five E ∗ (X )pairs are distinct. Therefore, there are at least five distinct E ∗ (X )-pairs. Recall that N M X (x4 ) = {e1 , e6 } and N M X (x5 ) = {e2 , e6 }. Suppose that there is a vertex in the copy of H4 in H corresponding to the vertex x4 that contains a vertex covering both e1 and e6 . Then the structure of M X implies that there exists a minimum X -transversal that intersects every edge in E ∗ (X ), contradicting Claim B. Analogously, if there is a vertex in the copy of H4 in H corresponding to the vertex x5 that contains a vertex covering both e2 and e6 , we produce a contradiction. Therefore, since e6 has only two neighbors in M X different from x4 and x5 , at most two E ∗ (X )-pairs contain e6 . Since there are at least five E ∗ (X )-pairs, there exists an E ∗ (X )-pair, (ei1 , ei2 ) say, / {ei1 , ei2 }. that is not the pair (e1 , e2 ) and such that e6 ∈ / {e1 , e2 }. Thus, ei1 ∈ Renaming i 1 and i 2 , if necessary, we may assume that ei1 ∈ {e3 , e4 , e5 }. Let x j be the vertex in X corresponding to a copy of H4 of X that contains a vertex covering both ei1 and ei2 . Since ei1 is adjacent to neither x4 nor x5 , we note that j ∈ {1, 2, 3}. We now proceed analogously as we did with the E ∗ (X )pair (e1 , e2 ). We consider the bipartite graph M X = M X − {ei1 , ei2 , x j } with partite sets X  = X \ {x j } and E X = E ∗ (X ) \ {ei1 , ei2 }. No matching in M X matches E X to X  . By Hall’s Theorem, there is a nonempty subset R ⊆ E X such that in M X , |N (R)| < |R|. If R = E X , then let x  ∈ X \ N M X (R). The only possible neighbors of x  are the vertices ei1 and ei2 . Since d M X (x  ) = 2, we note that x  ∈ {x4 , x5 }. However, x  is not adjacent to ei1 , and so d M X (x  ) ≤ 1, a contradiction. Therefore, |R| = 3 and in M X , |N (R)| = 2. We note that both vertices in N (R) are adjacent to all three vertices in R, implying that the two vertices of X not in R ∪ {x j } are the / R and e6 is the two vertices of X of degree 2 in M X , namely, x4 and x5 . Further, e6 ∈ only common neighbor of x4 and x5 . If i 2 = 1, then x5 is adjacent to neither ei1 nor ei2 , implying that d M X (x5 ) = 1. If i 2 = 1, then x4 is adjacent to neither ei1 nor ei2 , implying that d M X (x4 ) = 1. Both cases produce a contradiction. Therefore, at least one of x4 and x5 dominate both e1 and e2 in M X . Renaming x4 and x5 , if necessary, we may assume that x4 dominates both e1 and e2 in M X . By our earlier assumption, no vertex in X  \ {x1 } has degree 2 with e1 and e2 as its neighbors. Hence, N M X (x4 ) = {e1 , e2 , e6 }. As observed earlier, the only possible neighbors of x5 in M X are e1 , e2 and e6 , and so N M X (x5 ) ⊆ N M X (x4 ). Taking xk = x4 and xk  = x5 , Part (b) holds. This completes the proof of Claim H.14.2. () We now consider the degree sequence of vertices of X in M X . Let this degree sequence, in nondecreasing order, be given 5by s : d1 , d2 , d3 , d4 , d5 , and so 2 ≤ d1 ≤ di ≥ 21. By Claim H.14.2, d1 = 2 or d2 ≤ · · · ≤ d5 ≤ 6. As observed earlier, i=1 d1 = d2 = 3.

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Claim H.14.3: There are at most three E ∗ (X )-pairs. Proof of Claim H.14.3: Suppose, to the contrary, that there are at least four E ∗ (X )pairs. Renaming vertices of E ∗ (X ), if necessary, we may assume that (e1 , e2 ) and (e3 , e4 ) are E ∗ (X )-pairs. Let x12 (respectively, x34 ) be the vertex in X corresponding to a copy of H4 of X that contains a vertex covering both e1 and e2 (respectively, e3 and e4 ). Possibly, x12 = x34 . Let x1 ∈ X \ {x12 } be adjacent to both e1 and e2 in M X and such that either d M X (x1 ) = 2 or d M X (x1 ) = 3 and there exists a vertex x1 ∈ X \ {x1 , x12 } with N M X (x1 ) ⊆ N M X (x1 ). Further, let x2 ∈ X \ {x34 } be adjacent to both e3 and e4 in M X and such that either d M X (x2 ) = 2 or d M X (x2 ) = 3 and there exists a vertex x2 ∈ X \ {x2 , x34 } with N M X (x2 ) ⊆ N M X (x2 ). We note that x1 and x2 exists by Claim H.14.2. Further, x12 ∈ / {x1 , x2 }, x34 ∈ / {x1 , x2 }, and x1 = x2 . We may assume that d M X (x1 ) ≤ d M X (x2 ). Suppose that d M X (x1 ) = 2, and so N M X (x1 ) = {e1 , e2 }. Then, d1 = 2. Since d2 ≤ 5 3 and i=1 di ≥ 21, this implies that d3 ≥ 4, and so x1 and x2 are the only vertices in X of degree at most 3 in M X . If d M X (x2 ) = 3, then by Claim H.14.2, N M X (x1 ) ⊆ N M X (x2 ), a contradiction since at least one of e1 and e2 is not adjacent to x2 . Thus, d M X (x2 ) = 2, and so d2 = 2 and N M X (x2 ) = {e3 , e4 }. Let (ei , e j ) be an E ∗ (X )-pair different from (e1 , e2 ) and (e3 , e4 ). By Claim H.14.2, there exists a vertex in X of degree at most 3 adjacent to both ei and e j . The vertices x1 and x2 are the only two to vertices in X of degree at most 3 in M X . However, neither x1 nor x2 is adjacent 5 di ≥ both ei and e j , a contradiction. Therefore, d M X (x1 ) = d M X (x2 ) = 3. Since i=1 21, we note that d3 ≥ 3. This implies that N M X (x1 ) = N M X (x1 ) and N M X (x2 ) = N M X (x2 ). Suppose d3 = 3. Then, s is given by 3, 3, 3, 6, 6. Let x3 be the vertex of X \ {x1 , x2 } of degree 3 in M X . Since N M X (x1 ) = N M X (x2 ), we note that x1 = x3 and x2 = x3 . But then {e1 , e2 , e3 , e4 } ⊆ N M X (x3 ), and so d M X (x3 ) ≥ 4, a contradiction. Therefore, d3 ≥ 4. Thus, x1 and x2 are the only vertices in X of degree at most 3 in M X . This implies that x1 = x2 (and x2 = x1 ). But then {e1 , e2 , e3 , e4 } ⊆ N M X (x2 ), () and so d M X (x2 ) ≥ 4, a contradiction. Claim H.14.4: The degree sequence s is given by 3, 3, 5, 5, 5. Proof of Claim H.14.4: We show firstly that d1 = 3. Suppose, to the contrary, that d1 = 2. If d2 = 2, then s is given by 2, 2, 5, 6, 6 or 2, 2, 6, 6, 6, implying that there are at least five E ∗ (X )-pairs. If d2 = 3, then s is given by 2, 3, 4, 6, 6 or 2, 3, 5, 5, 6 or 2, 3, 5, 6, 6 or 2, 3, 6, 6, 6, implying that there are at least four E ∗ (X )-pairs. In both cases, we contradict Claim H.14.3. Therefore, d1 = d2 = 3. We show next that d6 ≤ 5. Suppose, to the contrary, that d6 = 6. If d3 = 3, then s is given by 3, 3, 3, 6, 6, implying that there are at least four E ∗ (X )-pairs, a contradiction. Hence, d3 ≥ 4. Let x1 and x2 be the two vertices of degree 3 in M X . By Claim H.14.2, N M X (x1 ) = N M X (x2 ). We may assume that e1 , e2 , and e3 are the three neighbors of x1 and x2 . By Claim H.14.2, if (ei , e j ) is an E ∗ (X )-pair, then 1 ≤ i, j ≤ 3. We may assume that x4 is a vertex of degree 6 in M X . Then, x4 gives rise to two E ∗ (X )-pairs that comprise four distinct vertices. At least one such pair is distinct from (e1 , e2 ), (e1 , e3 ), and (e2 , e3 ), a contradiction. Therefore, d3 ≤ 5. Since 5 d () i=1 i ≥ 21, the degree sequence s is given by 3, 3, 5, 5, 5.

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Renaming vertices in X if necessary, we may assume that d M X (xi ) = di . By Claim H.14.4, x1 and x2 have degree 3 in M X , while x3 , x4 , and x5 have degree 3 in M X . By Claim H.14.2, N M X (x1 ) = N M X (x2 ). We may assume that e1 , e2 , and e3 are the three neighbors of x1 and x2 . Further, we may assume that e1 is not adjacent to x3 , e2 is not adjacent to x4 , and e3 is not adjacent to x5 . By Claim H.14.2, the three E ∗ (X )-pairs are (e1 , e2 ), (e1 , e3 ), and (e2 , e3 ). The graph M X is completely determined. Let Fi be the copy of H4 ∈ X 4 associated with the vertex xi in M X for i ∈ [5]. We note that e4 , e5 , and e6 all have degree 3 in M X and are all adjacent in M X to x3 , x4 , and x5 . Thus, in H , each of the edges e4 , e5 , and e6 contain the vertex y and one vertex from each copy of H4 associated with x3 , x4 , and x5 . In H , each of the edges e1 , e2 , and e3 contain one vertex from each copy of H4 associated with x3 , x4 , and x5 . The copy of H4 associated with x3 contains a vertex covering both e2 and e3 . The copy of H4 associated with x4 contains a vertex covering both e1 and e3 . The copy of H4 associated with x5 contains a vertex covering both e1 and e2 . By the linearity of H , and since (e1 , e2 ), (e1 , e3 ), and (e2 , e3 ) are the only three E ∗ (X )-pairs, the graph H is now completely determined, and H = H21,2 . By Observation 8.1, ξ(H ) = 0, contradicting the fact that H is a counterexample to the theorem. This completes the () proof of Claim H.14. Claim H.15: The case |X | = 4, and |∂(X )| = 1 cannot occur. Proof of Claim H.15: By Claim H.13, there are no triple edges in M X . By Claim H.4(b) we note that 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − 8 = 26, as |X | = 4 and |X  | = 3. This implies that |X 4 | ≥ 3. Since |∂(X )| = 1, by Claim H.13(c), each vertex of E ∗ (X ) has degree at least 3 in M X . By Claim H.13(d), each vertex of X has degree at least 2 in M X . We now prove the following subclaims. Claim H.15.1: There does not exist a double edge between e ∈ E ∗ (X ) and F ∈ X in M X , such that e belongs to a E ∗ (X )-pair associated with F. Proof of Claim H.15.1: For the sake of contradiction, suppose that there is a double edge between e ∈ E ∗ (X ) and F ∈ X in M X , such that e belongs to a E ∗ (X )-pair, say (e, e ), associated with F. As there is a double edge incident with F in M X we note that F ∈ / X 4 , implying that |X 4 | = 3. We now consider the bipartite multigraph M X∗ , where M X∗ is defined as follows. Let M X∗ be obtained from M X by removing the vertex e ∈ E ∗ (X ) and removing all edges (e , F), where (e, e ) is not a E ∗ (X )-pair associated with F. We note that (e , F) is an edge in M X∗ and, therefore, d M X∗ (F) ≥ 1. First suppose that there is a perfect matching, M, in M X∗ . Let (e∗ , F) be an edge of the matching incident with F. By the definition of M X∗ , we note that (e∗ , e) is a E ∗ (X )-pair associated with F. By Observation 8.1(g), we can find a X -transversal covering E ∗ (X ), contradicting Claim B. Therefore, there is no perfect matching in M X∗ . By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗ (X ) \ {e} such that |N M X∗ (S)| < |S|. Let X ∗ = X \ N M X∗ (S). / X ∗, We will show that F ∈ X ∗ . For the sake of contradiction, suppose that F ∈ ∗ ∗ ∗ / X , the neighbors of a vertex of X in M X remain and so F ∈ N M X (S). Since F ∈ unchanged in M X∗ . We note that no vertex in X ∗ has a neighbor that belongs to S

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in M X∗ , and therefore also no neighbor that belongs to S in M X . This implies that in M X the vertices in X ∗ are adjacent to at most |E ∗ (X ) \ S| = |E ∗ (X )| − |S| = |X | + 1 − |S| ≤ |X ∗ | vertices in E ∗ (X ). Thus, |E ∗ (X ∗ )| ≤ |X ∗ | in H , contradicting Claim D. Therefore, F ∈ X ∗ . Since all three subhypergraphs in X \ {F} belong to X 4 , all edges in M X from S to vertices in N M X∗ (S) are therefore single edges. If |S| = 4, then the vertices in X ∗ have no edge to S = E ∗ (X ) \ {e} in M X∗ . In particular, the vertex F ∈ X ∗ is isolated in M X∗ , contradicting our earlier observation that d M X∗ (F) ≥ 1. Therefore, |S| ≤ 3. Suppose that |S| = 3, and let S = {e1 , e2 , e3 }. As observed earlier, all edges in M X from S to vertices in N M X∗ (S) are single edges. Since each vertex of S has degree at least 3 in M X and |N M X∗ (S)| ≤ 2, each vertex of S therefore has F as a neighbor in M X , but not in M X∗ since F ∈ X ∗ . Since Δ(H ) ≤ 3, this implies that |(V (e1 ) ∪ V (e2 ) ∪ V (e3 )) ∩ V (F)| ≥ 2. By Observation 8.1(k), either (e, e1 ) or (e, e2 ) or (e, e3 ) is a E ∗ (X )-pair associated with F. Renaming edges in E ∗ (X ) if necessary, we may assume that (e, e1 ) is a E ∗ (X )-pair associated with F. However, since the edge (e1 , F) was removed from M X when constructing M X∗ , this implies that (e, e1 ) is not a E ∗ (X )-pair associated with F, a contradiction. Therefore, |S| ≤ 2. Suppose that |S| = 2, and let S = {e1 , e2 }. We note that there is no edge in M X∗ joining e1 and F. If e1 has a double edge to F in M X , then (e1 , e) would be a E ∗ (X )pair associated with F, by Observation 8.1(k), and therefore the edge (e1 , F) would still exist in M X∗ , a contradiction. Hence, either e1 has a single edge to F in M X or is not adjacent to F in M X . This implies that the degree of e1 is at most 2 in M X since it can only be adjacent to the vertex in N M X∗ (S) and to F. This contradicts Claim H.13. Therefore, |S| = 1. However if |S| = 1, we analogously get a contradiction as the vertex in S has degree at most 1 in M X . This completes the proof of Claim H.15.1. () Claim H.15.2: There does not exist a double edge in M X . Proof of Claim H.15.2: Suppose, to the contrary, that there is a double edge between e ∈ E ∗ (X ) and F ∈ X in M X . By Claim H.15.1, for every edge (e , F) in M X , the pair (e, e ) is not a E ∗ (X )-pair associated with F. If there is a double edge between some vertex e ∈ E ∗ (X ) \ {e} and F in M X , then (e, e ) would be a E ∗ (X )-pair associated with F, a contradiction. Therefore, there is no other double edge incident to F, except for the double edge that joins it to e ∈ E ∗ (X ). By Claim D, the set N M X (F) contains at least two vertices. If the set N M X (F) contains four or more vertices, then by Observation 8.1(k), we would get a E ∗ (X )-pair associated with F containing e, a contradiction. Therefore, the set N M X (F) contains at most three vertices. That is, |N M X (F)| ∈ {2, 3}. / X 4 , implyAs there is a double edge incident with F in M X , we note that F ∈ ing that |X 4 | = 3 and X \ {F} = X 4 . By Claim H.13, |E(M X )| ≥ 4(|X | + 1) − 3|∂(X )| = 4 × 5 − 3 = 17. We first consider the case when |N M X (F)| = 2, and let N M X (F) = {e, e }. That is d M X (F) = 3, since the double edge between e and F counts 2 to the degree of F in M X . Since X \ {F} = X 4 , we have no double edges incident with a vertex in X \ {F} in M X . Further, as |E(M X )| ≥ 17, the degree sequence of the four vertices of X in M X is either (3, 4, 5, 5) or (3, 5, 5, 5). Let F1 and F2 be two vertices of degree 5

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in M X that belong to X , and let F3 be the remaining vertex in X \ {F}. We note that in H , both F1 and F2 belong to X 4 , and have an associated E ∗ (X )-pair in M X . Since H is linear, there are in fact distinct E ∗ (X )-pairs associated with F1 and F2 . Renaming F1 and F2 , if necessary, we may assume that the E ∗ (X )-pair, say (e1 , e2 ), associated with F1 is distinct from {e, e }. Thus, there is a vertex in F1 that covers both e1 and e2 , and we can cover a vertex in {e, e } \ {e1 , e2 } using a τ -transversal of F. Let e3 be such a vertex covered from F. The vertex F3 in M X has degree at least 4 in M X , and can be used to cover a vertex in E ∗ (X ) \ {e1 , e2 , e3 }, say e4 . Finally, the vertex F2 in M X which has degree 5 in M X can be used to cover the vertex in E ∗ (X ) \ {e1 , e2 , e3 , e4 }. We thereby obtain a contradiction to Claim B. We next consider the case when |N M X (F)| = 3. Let N M X (F) = {e1 , e2 , e3 }, where e = e1 , and let E ∗ (X ) \ {e1 , e2 , e3 } = {e4 , e5 }. Since there is no τ -transversal of F covering e1 and a vertex in V (e2 ) ∪ V (e3 ), we note that e2 and e3 intersect F in the same vertex (by Observation 8.1(k)). Therefore, (e2 , e3 ) is a E ∗ (X )-pair associated with F. Since d M X (e1 ) ≥ 3, and since there are no triple edges in M X , there is a vertex F  in X \ {F} adjacent to e1 in M X . Since neither e4 nor e5 has an edge to F, but they do have degree at least 3 in M X , both e4 and e5 must be adjacent in M X to all three vertices in X \ {F}. We can therefore cover the edge e1 from F  , cover both edges e2 and e3 from F, and cover the remaining two edges, e4 and e5 , from X \ {F, F  }, thereby obtaining a X -transversal covering E ∗ (X ), contradicting () Claim B. This completes the proof of Claim H.15.2. We now return to the proof of Claim H.15. By Claim H.15.2, there does not exist a double edge in M X , implying that M X is therefore a graph. As observed earlier, |X 4 | ≥ 3. By Claim H.13, |E(M X )| ≥ 4(|X | + 1) − 3|∂(X )| = 17. Recall that each vertex of X has degree at least 2 in M X . If some vertex of X has degree 2 in M X , then the degree sequence of X in M X is (2, 5, 5, 5), implying that there is a unique vertex of degree 2 in M X . Since |X 4 | ≥ 3, there exist two distinct vertices F, F  ∈ X 4 of degree 5 in M X . Both F and F  are associated with E ∗ (X )-pairs. Further, since H is linear, the E ∗ (X )-pairs associated with F and F  are distinct. Renaming F and F  , if necessary, we may assume that F has an associated E ∗ (X )-pair, say (e1 , e2 ), such that the neighborhood of the degree-2 vertex in X is not the set {e1 , e2 }. If every vertex of X has degree at least 3 in M X , then we must still have a vertex, F, in X of degree 5 in the graph M X , and such a vertex is necessarily associated with an E ∗ (X )-pair, say (e1 , e2 ). In both cases, we have therefore determined a vertex F ∈ X of degree 5 in M X with an associated E ∗ (X )-pair, (e1 , e2 ), such that the neighborhood of the degree-2 vertex in X , if it exists, is not the set {e1 , e2 }. We now consider the bipartite graph M X = M X − {e1 , e2 , F} with partite sets  X = X \ {F} and E X = E ∗ (X ) \ {e1 , e2 }. We note that |E X | = 3, and that there is no matching in M X that matches E X to X  , by Claim B. By Hall’s Theorem, there is a nonempty subset S ⊆ E X such that in M X , |N (S)| < |S|. Since every vertex in E ∗ (X ) has degree at least 3 in M X , we note that |N M X (S)| ≥ 2. Consequently, |S| = 3 and |N M X (S)| = 2. Thus, S = E X . Let F  be the vertex in X that is adjacent to no vertex of S in M X . Therefore, d M X (F  ) = 2 and N M X (F  ) = {e1 , e2 }. However, this is a contradiction to our choice of the E ∗ (X )-pair (e1 , e2 ), which was chosen so

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that the neighborhood of the degree-2 vertex in X , if it exists, is not the set {e1 , e2 }. () This completes the proof of Claim H.15. Claim H.16: The case |X | = 3 and |∂(X )| = 1 cannot occur. Proof of Claim H.16: Suppose, to the contrary, that |X | = 3 and |∂(X )| = 1. By Claim H.13, there are no triple edges in M X . Also, by Claim H.13(b), every vertex F ∈ X has double edges to at most two vertices in E ∗ (X ) in M X . Since |∂(X )| = 1, by Claim H.13(c), each vertex of E ∗ (X ) has degree at least 3 in M X . By Claim H.13(d), each vertex of X has degree at least 2 in M X . Suppose that some vertex F ∈ X has double edges to two distinct vertices, say {e1 , e2 }, of E ∗ (X ) in M X . Let E ∗ (X ) \ {e1 , e2 } = {e3 , e4 }. If there exists a perfect matching in M X − {e1 , e2 , F}, then we obtain a X -transversal covering all edges of E ∗ (X ) (by Observation 8.1(g) and 8.1(k)), contradicting Claim B. Therefore, there does not exist a perfect matching in M X − {e1 , e2 , F}. By Hall’s Theorem, there is a nonempty subset S ⊆ {e3 , e4 } such that |N M X (S) \ {F}| < |S|. If |S| = 1, say S = {e3 }, then e3 has degree at most 1 in M X , a contradiction. Therefore, |S| = 2 and S = {e3 , e4 }. Let F  ∈ X \ {F} be defined such that N M X (S) \ {F} = {F  }. Since both e3 and e4 have degree at least 3 in M X , we must have an edge from e3 to F and two edges from e3 to F  . Analogously, there is one edge from e4 to F and two edges from e4 to F  . Therefore, by Observation 8.1(k), there exists a τ -transversal of F  covering both e3 and e4 . Since e1 and e2 can be covered by a τ -transversal of F, we can therefore cover all edges of E ∗ (X ) with a X -transversal, a contradiction to Claim B. Hence, every F ∈ X has double edges to at most one vertex of E ∗ (X ) in M X . By Claim H.13, |E(M X )| ≥ 4(|X | + 1) − 3|∂(X )| = 16 − 3 = 13. By the Pigeonhole Principle, there is therefore a vertex F of X of degree at least 5 in M X . By Observation 8.1 this implies that there exists a E ∗ (X )-pair, say (e1 , e2 ), associated with F. Let E ∗ (X ) \ {e1 , e2 } = {e3 , e4 }. If there exists a perfect matching in M X − {e1 , e2 , F}, then, as before, we obtain a X -transversal covering all edges of E ∗ (X ). Therefore, there is no perfect matching in M X − {e1 , e2 , F}. By Hall’s Theorem, there is a nonempty subset S ⊆ {e1 , e2 } such that |N M X (S) \ {F}| < |S|. If |S| = 1, say S = {e3 }, then e3 has degree at most 2 in M X , a contradiction. Therefore, |S| = 2 and S = {e3 , e4 }. Let F  ∈ X \ {F} be defined such that N M X (S) \ {F} = {F  }. Suppose firstly that the E ∗ (X )-pair, say (e1 , e2 ), can be chosen so that either e1 or e2 has a double edge to F in M X . In this case, both e3 and e4 have at most one edge to F. Since both e3 and e4 have degree at least 3 in M X , this implies that both e3 and e4 have double edges to F  , contradicting the fact that every vertex in X has double edges to at most one vertex of E ∗ (X ) in M X . Therefore, all E ∗ (X )-pairs, (e1 , e2 ), have only single edges to F. Renaming e3 and e4 , if necessary, we may assume that e3 has a single edge to F  in M X . Thus, e3 has a double edge to F in M X , which in turn implies that e4 has a single edge to F and a double edge to F  in M X . However, this implies that F has a double edge to e3 , and single edges to e1 , e2 , and e4 . Therefore, by Observation 8.1(k), there exists a E ∗ (X )-pair containing e3

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and one of e1 , e2 , and e4 , contradicting the fact that all E ∗ (X )-pairs have only single () edges to F. This completes the proof of Claim H.16. Claim H.17: The case |X | = 2 and |∂(X )| = 1 cannot occur. Proof of Claim H.17: Suppose, to the contrary, that |X | = 2 and |∂(X )| = 1. By Claim H.13, there are no triple edges in M X , and |E(M X )| ≥ 4(|X | + 1) − 3|∂(X )| = 12 − 3 = 9. By the Pigeonhole Principle, there is therefore a vertex F of X of degree at least 5 in M X . By Observation 8.1 this implies that there exists a E ∗ (X )pair, say (e1 , e2 ), associated with F. Let E ∗ (X ) = {e1 , e2 , e3 } and X = {F, F  }. If there is an edge from e3 to F  in M X we obtain a contradiction to Claim B analogously to the above cases. If there is no edge from e3 to F  in M X we obtain a contradiction to e3 having degree at least three in M X . This contradiction completes the proof of () Claim H.17. Claim H.18: If |∂(X )| = 0, then |X | ≤ 6 and the following holds. (a) If |X | = 6, then X = X 4 . (b) If |X | = 5, then |X 4 | ≥ 3. (c) All vertices in E ∗ (X ) have degree 4 in M X and |E(M X )| = 4|X | + 4. (d) M X does not contain triple edges. (e) No F ∈ X has double edges to three distinct vertices of E ∗ (X ) in M X . Proof of Claim H.18: Suppose that ∂(X ) = ∅, and so V (H ) = V (X ). Further, H  is obtained from H − V (X ) by adding the set X  of four new vertices and adding a 4-edge, e, containing these four vertices. Thus, H  = H4 and |X  | = 4. By Claim H.4(b), we have 8|X | ≥ 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − 8 = 8|X | + 5|X | − 32. In particular, 5|X | ≤ 32, and so |X | ≤ 6. Furthermore if |X | = 6, then we must have X = X 4 , while if |X | = 5, then |X 4 | ≥ 3. The last three () statements in Claim H.18 follow directly from Claim H.13. Claim H.19: If |∂(X )| = 0, then there does not exist a double edge between e ∈ E ∗ (X ) and F ∈ X in M X , such that e belongs to a E ∗ (X )-pair associated with F. Proof of Claim H.19: For the sake of contradiction, suppose that there is a double edge between e ∈ E ∗ (X ) and F ∈ X in M X , such that e belongs to a E ∗ (X )-pair, say (e, e ), associated with F. We assume, further, that e and F are chosen so that the number of E ∗ (X )-pairs, (e, e ), associated with F, where e ∈ E ∗ (X ) \ {e, e }, is / X 4. maximized. As there is a double edge incident with F in M X , we note that F ∈ We now consider the bipartite multigraph M X∗ , where M X∗ is defined as follows. Let M X∗ be obtained from M X by removing the vertex e ∈ E ∗ (X ) and removing all edges (e , F), where (e, e ) is not a E ∗ (X )-pair associated with F. We note that (e , F) is an edge in M X∗ and, therefore, d M X∗ (F) ≥ 1. Also due to Observation 8.1(k), we note that at most two edges (and no double edges) have been removed from F to E ∗ (X ) \ {e} when constructing M X∗ . First suppose that there is a perfect matching, M, in M X∗ . Let (e∗ , F) be an edge of the matching incident with F. By the definition of M X∗ , we note that (e∗ , e) is a

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E ∗ (X )-pair associated with F. By Observation 8.1(g), we can find a X -transversal covering E ∗ (X ), contradicting Claim B. Therefore, there is no perfect matching in M X∗ . By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗ (X ) \ {e} such that |N M X∗ (S)| < |S|. Let X ∗ = X \ N M X∗ (S). / X ∗, We will show that F ∈ X ∗ . For the sake of contradiction, suppose that F ∈ ∗ ∗ / X , the neighbors of a vertex of X in M X remain and so F ∈ N M X∗ (S). Since F ∈ unchanged in M X∗ . We note that no vertex in X ∗ has a neighbor that belongs to S in M X∗ , and therefore also no neighbor that belongs to S in M X . This implies that in M X the vertices in X ∗ are adjacent to at most |E ∗ (X ) \ S| = |E ∗ (X )| − |S| = |X | + 1 − |S| ≤ |X ∗ | vertices in E ∗ (X ). Thus, |E ∗ (X ∗ )| ≤ |X ∗ | in H , contradicting / X 4, Claim D. Therefore, F ∈ X ∗ . By Claim H.18, |X | ≤ 6. As observed earlier, F ∈ which implies that |X | = 6, by Claim H.18. Thus, |X | ≤ 5. Suppose that |X | = 5. If |S| = 5, then since F is adjacent to e in M X∗ , this would imply that F ∈ N M X∗ (S), contradicting the fact that F ∈ X ∗ . Hence, |S| ≤ 4. Suppose that |S| = 4 and S = {e1 , e2 , e3 , e4 }. We note that in this case, E ∗ (X ) \ S = {e, e } and that (e, e ) is the only E ∗ (X )-pair associated with F. As observed earlier, at most two edges (and no double edges) were removed from F to E ∗ (X ) \ {e} when constructing M X∗ . Thus, at least two vertices in S did not have any edges to F in M X . Renaming vertices of S, if necessary, we may assume that neither e1 nor e2 has an edge to F in M X . By Claim H.18(c), all vertices in E ∗ (X ) have degree 4 in M X , implying that both e1 and e2 have a double edge to a vertex in N M X∗ (S). Since |X 4 | ≥ 3 and F∈ / X 4 , e1 and e2 both have double edges to the same vertex F  ∈ N M X∗ (S). Thus, X \ {F, F  } = X 4 . Furthermore, as e3 and e4 have degree 4 in M X , and therefore degree at least 3 in M X∗ , both e3 and e4 are adjacent to F  in M X∗ . Therefore, by Observation 8.1(k) and 8.1( ), there is a E ∗ (X )-pair containing one vertex from {e1 , e2 } and one vertex from {e3 , e4 } associated with F  . Renaming vertices of S, if necessary, we may assume that (e1 , e3 ) is a E ∗ (X )-pair associated with F  . If we had used this E ∗ (X )-pair instead of (e, e ), and F  instead of F, then we note that e1 is a double edge to F  and both (e1 , e2 ) and (e1 , e3 ) are E ∗ (X )-pairs associated with F  , while e is a double edge to F but there is only one E ∗ (X )-pair of the form (e, e ), where e ∈ E ∗ (X ) \ {e, e } (namely, the pair (e, e )) that is associated with F. This contradicts our choice of e and F. Therefore, |S| ≤ 3. Suppose that |S| = 3 and S = {e1 , e2 , e3 }. Since at most two edges (and no double edges) were removed from F to E ∗ (X ) \ {e} when constructing M X∗ , at least one vertex in S, say e1 , has the same degree in M X∗ as in M X . Thus, e1 has degree 4 in M X∗ , implying that e1 has double edges to both vertices in N M X∗ (S). However, this implies that no vertex in / X 4 . This implies that |X 4 | ≤ 2, a N M X∗ (S) belongs to X 4 . As observed earlier, F ∈ contradiction. Therefore, |S| ≤ 2. However in this case the vertices in S cannot have degree 4 in M X , a contradiction. This completes the case when |X | = 5. Suppose that |X | = 4. If |S| = 4, then since F is adjacent to e in M X∗ , this would imply that F ∈ N M X∗ (S), contradicting the fact that F ∈ X ∗ . Hence, |S| ≤ 3. Suppose that |S| = 3 and S = {e1 , e2 , e3 }. We note that in this case, E ∗ (X ) \ S = {e, e }. Since at most two edges (and no double edges) were removed from F to E ∗ (X ) \ {e} when constructing M X∗ , at least one vertex in S, say e3 , has the same degree in M X∗ as in M X . Thus, e3 has degree 4 in M X∗ , implying that e3 has double edges to both vertices

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in N M X∗ (S) = {F1 , F2 }. Since both e1 and e2 have degree 4 in M X , and degree at least 3 in M X∗ , both e1 and e2 have a double edge to F1 or F2 . They cannot have a double edge to the same vertex in {F1 , F2 }, for otherwise such a vertex would then have double edges to all three vertices in S, contradicting Claim H.18(e). Renaming e1 and e2 , if necessary, we may assume that e1 has a double edge to F1 and e2 has a double edge to F2 in M X∗ (and in M X ). We can now cover {e, e } from F, {e1 , e3 } from F1 , and e2 from F2 , contradicting Claim B. Therefore, |S| ≤ 2. However in this case the vertices in S cannot have degree 4 in M X , a contradiction. This completes the case when |X | = 4. Consider the case when |X | = 3. If |S| = 3, then we contradict the fact that F ∈ X ∗ . Hence, |S| ≤ 2. However in this case the vertices in S cannot have degree 4 in M X , a contradiction. This completes the case when |X | = 3. Analogously, we get () a contradiction when |X | = 2. This completes the proof of Claim H.19. Claim H.20: The case |X | ≥ 6 and |∂(X )| = 0 cannot occur. Proof of Claim H.20: Suppose, to the contrary, that |X | ≥ 6 and |∂(X )| = 0. By Claim H.18, |X | = 6, X = X 4 , and |E(M X )| = 4|X | + 4 = 28. Since X = X 4 , the multigraph M X is a graph. By the Pigeonhole Principle, there is a vertex F of X of degree at least 5 in M X . By Observation 8.1 this implies that there exists a E ∗ (X )pair, say (e1 , e2 ), associated with F. Among all such E ∗ (X )-pairs, we choose the pair (e1 , e2 ) so that the number of neighbors of e1 and e2 in M X of degree less than 4 is a minimum. Since F ∈ X 4 , we note that in H , the edges e1 and e2 intersect F in the same vertex. We now consider the bipartite multigraph M X = M X − {e1 , e2 , F} with partite sets X  = X \ {F} and E X = E ∗ (X ) \ {e1 , e2 }. If there exists a perfect matching in M X , then we can find a X -transversal covering E ∗ (X ), contradicting Claim B. Therefore, there is no perfect matching in M X . By Hall’s Theorem, there is a nonempty subset S ⊆ E X such that in M X , |N (S)| < |S|. By Claim H.18(c), all vertices in E ∗ (X ) have degree 4 in M X . Thus, each vertex of E X has degree at least 3 in M X , implying that in M X , |S| > |N (S)| ≥ 3. Hence, |S| ∈ {4, 5}. Suppose that |S| = 5. If |N M X (S)| ≤ 3, then let X ∗ = X \ N M X (S) and note that ∗ |E (X ∗ )| ≤ |{e1 , e2 }| ≤ |X ∗ |, a contradiction to Claim D. Therefore, |N M X (S)| = 4 and let F(e1 , e2 ) denote the vertex in X \ (N M X (S) ∪ {F}). The degree of every vertex of X is at least 2 in M X , implying that the vertex F(e1 , e2 ) is adjacent in M X to both e1 and e2 (and to no other vertex of E ∗ (X )), and therefore has degree 2 in M X . Hence, in this case when |S| = 5, the vertices e1 and e2 have a neighbor of degree 2. Before completing our proof of this case when |S| = 5, we first consider the case when |S| = 4. Suppose that |S| = 4. Since every vertex of E X has degree at least 3 in M X , we note that in M X , |N (S)| = 3. Further, in M X , every vertex of S is adjacent to every vertex of N (S). This implies that in M X , the vertex F is adjacent to all four vertices of S, and therefore F has degree at least 6 in M X . Let S = {e3 , e4 , e5 , e6 } and let N M X (S) = {F1 , F2 , F3 }. Let X \ {F1 , F2 , F3 } = {F, F5 , F6 }. We note that F5 (respectively, F6 ) has degree 2 or 3 in M X , and is adjacent to at least one of e1 and

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e2 . Hence, in this case when |S| = 4, there are two vertices of degree less than 4 adjacent to e1 or e2 . Recall that in H , the edges e1 and e2 intersect F in the same vertex, implying that in H , two edges in {e3 , e4 , e5 , e6 } intersect F in the same vertex. Renaming vertices of S, if necessary, we may assume that in H , the edges e3 and e4 both intersect F in the same vertex, and therefore form a E ∗ (X )-pair associated with F. If we had considered the E ∗ (X )-pair (e3 , e4 ) instead of the E ∗ (X )-pair (e1 , e2 ), then we note that every neighbor of e3 and e4 has degree at least 4 in M X . This contradicts our choice of the E ∗ (X )-pair, (e1 , e2 ). Therefore, |S| = 4 is not possible. We now return to the case when |S| = 5. Given a E ∗ (X )-pair, (e1 , e2 ), we showed that there exists a vertex F(e1 , e2 ) in X \ {F} that has degree 2 in M X , and has e1 and e2 as its neighbors. Since |S| ≤ 4 is not possible, this implies that for every E ∗ (X )pair, (e, e ), associated with some vertex F  ∈ X , there exists a vertex F(e, e ) in X \ {F  } that has degree 2 in M X , and has e and e as its neighbors. We note that if a vertex F ∗ ∈ X has degree 4 + i in M X , then it is associated with at least i E ∗ (X )pairs. Thus, since the vertex F(e1 , e2 ) has degree 2 in M X , and since |E(M X )| = 28, there are at least six E ∗ (X )-pairs, which are all distinct. To each one we associated a vertex of degree 2 in M X that belongs to X . However this implies that there are only 12 edges in M X , a contradiction to |E(M X )| = 28. This completes the proof of () Claim H.20. Claim H.21: The case |X | = 5 and |∂(X )| = 0 cannot occur. Proof of Claim H.21: Suppose, to the contrary, that |X | = 5 and |∂(X )| = 0. By Claim H.18, |X 4 | ≥ 3 and |E(M X )| = 4|X | + 4 = 24. Also, by Claim H.18, M X does not contain triple edges. By Claim H.19 and Observation 8.1(k), no F ∈ X has double edges to two distinct vertices of E ∗ (X ) in M X . By Claim H.18(c), all vertices in E ∗ (X ) have degree 4 in M X . Since |E(M X )| = 24 and |X | = 5, there is a vertex F of X of degree at least 5 in M X . By Observation 8.1 this implies that there exists a E ∗ (X )-pair, say (e1 , e2 ), associated with F. We now consider the bipartite multigraph M X = M X − {e1 , e2 , F} with partite sets X  = X \ {F} and E X = E ∗ (X ) \ {e1 , e2 }. If there exists a perfect matching in M X , then we can find a X -transversal covering E ∗ (X ), contradicting Claim B. Therefore, there is no perfect matching in M X . By Hall’s Theorem, there is a nonempty subset S ⊆ E X such that in M X , |N (S)| < |S|. We note that 1 ≤ |S| ≤ 4. If |S| = 1, then the vertex in S has degree at most 2 in M X , a contradiction. If |S| = 2, then the two vertices in S must have double edges to F (and to the vertex in N M X (S)), contradicting the fact that no vertex in X has double edges to two distinct vertices of E ∗ (X ) in M X . Therefore, |S| ≥ 3. Suppose that |S| = 3, and let S = {e3 , e4 , e5 }. For each i ∈ [3], the vertex ei ∈ S has degree 4 in M X and N M X (ei ) ⊆ {F} ∪ N M X (S), implying that |N M X (S)| = 2 and that ei ∈ S is adjacent to a double edge. Let N M X (S) = {F1 , F2 }. Since no vertex in X is adjacent to two double edges, we may assume, renaming vertices if necessary, that e3 has a double edge to F, e4 has a double edge to F1 , and e5 has a double edge to F2 . However, this implies that none of F, F1 , or F2 belongs to X 4 , contradicting the fact that |X 4 | ≥ 3. Therefore, |S| ≥ 4.

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We have shown that |S| = 4. Let S = {e3 , e4 , e5 , e6 }. If |N M X (S)| ≤ 2, then let X ∗ = X \ (N M X (S) ∪ {F}) and note that |E ∗ (X ∗ )| ≤ |{e1 , e2 }| ≤ |X ∗ |, a contradiction to Claim D. Therefore, |N M X (S)| = 3. Let N M X (S) = {F1 , F2 , F3 } and let X \ N M X (S) = {F, F(e1 , e2 )}. We note that N M X (F(e1 , e2 )) ⊆ {e1 , e2 }. By Claim E, |N M X (F(e1 , e2 ))| ≥ 2, implying that N M X (F(e1 , e2 )) = {e1 , e2 }. Since no vertex of X has double edges in M X to two distinct vertices in E ∗ (X ), we note that d M X (F(e1 , e2 )) ≤ 3. Thus, for each E ∗ (X )-pair, (e , e ) say, there is a vertex F  ∈ X such that N M X (F(e , e )) = {e , e } and d M X (F(e , e )) ≤ 3. We show next that if F  ∈ X has degree 4 + i in M X , then it is associated with / X 4 , for otherwise this propat least i E ∗ (X )-pairs. We may assume that F  ∈ erty clearly holds if F  ∈ X 4 . If there are three neighbors { f 1 , f 2 , f 3 } of F  in M X , no two of which form a E ∗ (X )-pair associated with F  , then every vertex in N M X (F  ) \ { f 1 , f 2 , f 3 } forms a E ∗ (X )-pair with one of the vertices in { f 1 , f 2 , f 3 }, by Observation 8.1(j). This gives us at least |N M X (F  )| − 3 distinct E ∗ (X )-pairs associated with F  . Since no vertex in X has double edges to two distinct vertices in E ∗ (X ), we note that |N M X (F  )| ≥ d M X (F  ) − 1 = 3 + i and the result follows. Therefore, we may assume that every set of three neighbors of F  in M X yields at least one E ∗ (X )-pair associated with F  . If every two neighbors of F  in M X form a E ∗ (X )pair associated with F  , then we clearly have enough E ∗ (X )-pairs, so we assume that there are two neighbors, f 1 and f 2 say, of F  in M X which do not form a E ∗ (X )-pair associated with F  . However, since every set of three neighbors of F  in M X contains a E ∗ (X )-pair, for every f  ∈ N M X (F  ) \ { f 1 , f 2 } at least one of ( f  , f 1 ) and ( f  , f 2 ) is a E ∗ (X )-pair associated with F  , yielding at least d M X (F  ) − 2 = 2 + i E ∗ (X )pairs associated with F  . Therefore, we have shown that if F  ∈ X has degree 4 + i in M X , then it is associated with at least i E ∗ (X )-pairs. Recall that d M X (F(e1 , e2 )) ≤ 3 and |E(M X )| = 24. Thus, there are at least five E ∗ (X )-pairs. If (e, e ) is an arbitrary E ∗ (X )-pair, then, by the linearity of H , (e, e ) can be associated with at most one vertex from X 4 . Hence, since |X \ X 4 | ≤ 2, the E ∗ (X )pair can be associated with at most three vertices of X . Since there are at least five E ∗ (X )-pairs, there are therefore at least 53  = 2 distinct E ∗ (X )-pairs associated with distinct vertices of X . Let (e , e ) and ( f  , f  ) be two such E ∗ (X )-pairs. Thus, there exist two vertices in X , say F  and F  , of degree at most 3 in M X , where F  has neighborhood {e , e } and F  has neighborhood { f  , f  }, in M X . Suppose that both F  and F  have degree 3 in M X . Then, both F  and F  are incident with double edges in M X and therefore do not belong to X 4 . Since |X 4 | ≥ 3, this implies that X \ {F  , F  } = X 4 . Hence, no vertex in X \ {F  , F  } is incident with a double edge in M X , and therefore has degree at most |E ∗ (X )| = 6 in M X . Since |E(M X )| = 24, the degree sequence of the vertices in X must be (3, 3, 6, 6, 6) in M X . However in this case, each of the three vertices of X of degree 6 belongs to X 4 , and is therefore intersected by six edges in H , which gives rise to at least two associated E ∗ (X )-pairs. Further, by the linearity of H , the E ∗ (X )-pairs associated with two distinct vertices of X that belong to X 4 are distinct. Hence, there are at least six distinct E ∗ (X )-pairs associated with distinct vertices of X . However, as observed earlier, for each E ∗ (X )-pair, (e, f ), there is a vertex in X whose neighborhood is

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precisely {e, f }, implying that |X | ≥ 6, a contradiction. Therefore, at least one of F  and F  has degree 2 in M X . Since |E(M X )| = 24, and since the sum of the degrees of F  and F  in M X is at most 5, the sum of the degrees of the three vertices of X \ {F  , F  } in M X is at least 19. As observed earlier, if F  ∈ X has degree 4 + i in M X , then it is associated with at least i E ∗ (X )-pairs. Hence there are at least seven E ∗ (X )-pairs. Since each E ∗ (X )-pair can be associated with at most three vertices of X , there are therefore at least 73  = 3 distinct E ∗ (X )-pairs associated with distinct vertices of X . Thus, there exist at least three vertices in X of degree at most 3 in M X . Since |E(M X )| = 24, the sum of the remaining two vertices of X in M X is at least 15, implying that X has a vertex of degree at least 8 in M X . Such a vertex has triple edges to a vertex of E ∗ (X ) or double edges to two distinct vertices in E ∗ (X ), a contradiction. This completes () the proof of Claim H.21. Claim H.22: The case |X | = 4 and |∂(X )| = 0 cannot occur. Proof of Claim H.22: Suppose, to the contrary, that |X | = 4 and |∂(X )| = 0. By Claim H.18, |E(M X )| = 4|X | + 4 = 20. Also, by Claim H.18, M X does not contain triple edges. By Claim H.19 and Observation 8.1(k), no F ∈ X has double edges to two distinct vertices of E ∗ (X ) in M X . By Claim H.18(c), all vertices in E ∗ (X ) have degree 4 in M X . Since |E(M X )| = 20 and |X | = 4, there is a vertex F of X of degree at least 5 in M X . By Observation 8.1 this implies that there exists a E ∗ (X )-pair associated with F. Among all such E ∗ (X )-pairs, we choose the pair (e1 , e2 ) to maximize the number of edges between {e1 , e2 } and F in M X . Since F has double edges to at most one vertex of E ∗ (X ) in M X , we note that there are at most three edges between {e1 , e2 } and F in M X . We now consider the bipartite multigraph M X = M X − {e1 , e2 , F} with partite sets X  = X \ {F} and E X = E ∗ (X ) \ {e1 , e2 }. If there exists a perfect matching in M X , then we can find a X -transversal covering E ∗ (X ), contradicting Claim B. Therefore, there is no perfect matching in M X . By Hall’s Theorem, there is a nonempty subset S ⊆ E X such that in M X , |N (S)| < |S|. We note that 1 ≤ |S| ≤ 3. If |S| = 1, then the vertex in S has degree at most 2 in M X , a contradiction. If |S| = 2, then the two vertices in S must have double edges to F (and to the vertex in N M X (S)), contradicting the fact that no vertex in X has double edges to two distinct vertices of E ∗ (X ) in M X . Therefore, |S| = 3. Let S = {e3 , e4 , e5 }. For each i ∈ [3], the vertex ei ∈ S has degree 4 in M X and N M X (ei ) ⊆ {F} ∪ N M X (S), implying that |N M X (S)| = 2 and that ei ∈ S is adjacent to a double edge. Let N M X (S) = {F1 , F2 }. Since no vertex in X is adjacent to two double edges, we may assume, renaming vertices if necessary, that e3 has a double edge to F, e4 has a double edge to F1 , and e5 has a double edge to F2 . Furthermore, e4 and e5 also have (single) edges to F. Thus, F has degree 6 in M X , with a single edge to each of e1 , e2 , e4 , e5 and a double edge to e3 . By Observation 8.1 this implies that there exists a E ∗ (X )-pair, (e3 , e), associated with F, for some e ∈ {e1 , e2 , e4 , e5 }. The number of edges between {e1 , e2 } and F in M X is 2, while the number of edges

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between {e3 , e} and F in M X is 3. This contradicts our choice of the E ∗ (X )-pair () {e1 , e2 }, and completes the proof of Claim H.22. Claim H.23: The case |∂(X )| = 0 cannot occur. Proof of Claim H.23: Suppose, to the contrary, that |∂(X )| = 0. By Claim H.20, Claim H.21, and Claim H.22 we must have |X | ≤ 3. By Claim H.18, M X does not contain triple edges. By Claim H.19 and Observation 8.1(k), no F ∈ X has double edges to two distinct vertices of E ∗ (X ) in M X . Since every vertex of E ∗ (X ) has degree 4 in M X , we note that |X | ≤ 2 is therefore impossible. Therefore, |X | = 3. Let E ∗ (X ) = {e1 , e2 , e3 , e4 }. Since every vertex of E ∗ (X ) = {e1 , e2 , e3 , e4 } has a double edge to a vertex of X in M X , by the Pigeonhole Principle, some vertex in X has double edges to two distinct vertices in E ∗ (X ), a contradiction. This completes () the proof of Claim H.23. Claim H.24: |X | = 1. Proof of Claim H.24: By Claim H.10, Claim H.14, Claim H.15, and Claim H.23, the case |X | ≥ 4 cannot occur. By Claim H.11, Claim H.16, Claim H.17, and Claim H.23, the case |X | = 3 cannot occur. By Claim H.6, Claim H.8, Claim H.12, Claim H.17, and Claim H.23, the case |X | = 2 cannot occur. Therefore, |X | = 1. () Throughout the remaining subclaims of Claim H, we implicitly use the fact that |X | = 1, and we let e1 and e2 be the two edges of E ∗ (X ) that intersect X in H . Claim H.25: |∂(X )| ≥ 4. Proof of Claim H.25: Suppose, to the contrary, that |∂(X )| ≤ 3. Recall that e1 and e2 were the edges intersecting X in H . If |V (e1 ) ∩ V (X )| = 1, then ∂(X ) ⊆ V (e1 ) and therefore |V (e2 ) ∩ V (X )| ≥ 3 as H is linear. Therefore, by Observation 8.1( ), we can cover E ∗ (X ) with a X -transversal, a contradiction to Claim B. If |V (e1 ) ∩ V (X )| ≥ 3, we analogously get a contradiction using Observation 8.1( ) and Claim B. If |V (e1 ) ∩ V (X )| = 2, then |V (e2 ) ∩ V (X )| ≥ 2, as |∂(X )| ≤ 3. We now obtain a contradiction to Claim B using Observation 8.1(k). Therefore, () |∂(X )| ≥ 3. Claim H.26: If H  is linear, then one of the following hold. (a) def(H  ) ∈ {8, 10}. (b) If def(H  ) = 8, then X = X 4 . Proof of Claim H.26: By Claim H.4(a), we note that 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | > 13|X | − 6|X  | − def(H  ). As |X | = 1 and |X  | = 0, we note that def(H  ) ≥ 6. By Claim H.3, def(H  ) = def H  (Y ), where |Y | = 1 and e is an edge of the hypergraph in the special H  -set Y . Thus, either Y is an H4 -component of H  , in which case def(H  ) = 8, or Y is an H10 -component of H  , in which case def(H  ) = 10. This proves Part (a). Suppose that def(H  ) = 8, and so Y is an H4 -component of H  consisting of the edge e. We now let H ∗ = H − V (X ) − V (e). Equivalently, H ∗ is obtained from H  by deleting the H4 -component Y . If def(H ∗ ) > 0 and Y ∗ is a special

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H ∗ -set such that def(H ∗ ) = def H ∗ (Y ∗ ) > 0, then def(H  ) ≥ def H  (Y ∪ Y ∗ ) > 8, a contradiction. Hence, def(H ∗ ) = 0. By Observation 8.1(e), we can choose a X transversal, TX , to contain a vertex of e1 . Every τ -transversal of H ∗ can be extended to a transversal of H by adding to the set TX an arbitrary vertex of e2 , implying that τ (H ) ≤ τ (H ∗ ) + τ (X ) + 1. We note that n(H ∗ ) = n(H ) − n(X ) − 4 and m(H ∗ ) = m(H ) − m(H ) − 2. We show that X = X 4 . Suppose, to the contrary, X = X 4 . Thus, X = X i for some i ∈ {11, 14, 21}. If X = X 11 , then 45τ (X ) = 6n(X ) + 13m(X ) + 4. If X = X 14 , then 45τ (X ) = 6n(X ) + 13m(X ) + 5. If X = X 21 , then 45τ (X ) = 6n(X ) + 13m(X ) + 1. In all cases, 45τ (X ) ≤ 6n(X ) + 13m(X ) + 5. Thus, Φ(H ∗ ) = ξ(H ∗ ) − ξ(H ) = 45τ (H ∗ ) − 6n(H ∗ ) − 13m(H ∗ ) − def(H ∗ ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − τ (X ) − 1) − 6(n(H ) − n(X ) − 4) −13(m(H ) − m(H ) − 2) −45τ (H ) + 6n(H ) + 13m(H ) = −45τ (X ) + 6n(X ) + 13m(X ) − 45 + 24 + 26 ≥ −5 − 45 + 24 + 26 = 0, a contradiction to Claim G. This proves Part (b) and completes the proof of () Claim H.26. Claim H.27: def(H  ) ≤ 8. Proof of Claim H.27: By Claim H.3, either def(H  ) = 0 or def(H  ) = def H  (Y ) where |Y | = 1 and e is an edge of the hypergraph in the special H  -set Y . For the sake of contradiction suppose that def(H  ) > 8, which implies that def(H  ) = def H  (Y ) where |Y | = 1 and Y is an H10 -component, F, of H  and e is an edge of F. By Claim H.25, |∂(X )| ≥ 4. Suppose that |∂(X )| = 4. If |V (e1 ) ∩ V (X )| = 1 and |V (e2 ) ∩ V (X )| = 1, then the linearity of H implies that |∂(X )| ≥ 5, a contradiction. Renaming the edges e1 and e2 , we may assume that |V (e2 ) ∩ V (X )| ≥ 2. By Claim B, there is no X -transversal covering both e1 and e2 . Therefore, by Observation 8.1(k) and 8.1( ), we note that |V (e1 ) ∩ V (X )| = 1 and |V (e2 ) ∩ V (X )| = 2. Thus, by Observation 8.1(m), we have X = X 11 . Let T1 = V (e1 ) ∩ V (X ) and let T2 = V (e2 ) ∩ V (X ), and so T1 and T2 are vertex-disjoint subsets of X such that |T1 | = 1 and |T2 | = 2 where T2 contains two vertices that are not adjacent in X . Further by Observation 8.1(m), one degree-1 vertex in X = X 11 belongs to T1 and the other degree-1 vertex to T2 , and the second vertex of T2 is adjacent to the vertex of T1 . As |∂(X )| = 4, the edges e1 and e2 intersect in a vertex in ∂(X ). As every edge in H10 is equivalent, the above observations imply that we get that H = H21,6 , contradicting Claim C. Hence, |∂(X )| ≥ 5. Since e is an edge of F, we note that |∂(X ) ∩ V (F)| ≥ 4. Suppose that |∂(X ) ∩ V (F)| ≥ 5. Let z ∈ (∂(X ) ∩ V (F)) \ V (e) be arbitrary. Removing from F the edge

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e and all edges incident to z, we are left with only two edges. Further, these two remaining edges intersect in a vertex, say z  . Let H  be obtained from H  by removing the two vertices z and z  (and all edges incident with z and z  ) and removing the edge e; that is, H  = H  − {z, z  } − e. Equivalently, H  is obtained from H  by deleting the H10 -component F. If def(H  ) > 0 and Y  is a special H  -set such that def(H  ) = def H  (Y  ) > 0, then def(H  ) ≥ def H  (Y ∪ Y  ) > def(H  ), a contradiction. Hence, def(H  ) = 0. Every τ -transversal of H  can be extended to a transversal of H by adding to a X -transversal the two vertices {z, z  }, implying that τ (H ) ≤ τ (H  ) + τ (X ) + 2. We note that 45τ (X ) = 6n(X ) + 13m(X ) + def(X ) ≤ 6n(X ) + 13m(X ) + 10. Thus, Φ(H  ) = ξ(H  ) − ξ(H ) = 45τ (H  ) − 6n(H  ) − 13m(H  ) − def(H  ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − τ (X ) − 2) − 6(n(H ) − n(X ) − 10) −13(m(H ) − m(X ) − 6) −45τ (H ) + 6n(H ) + 13m(H ) = −45τ (X ) + 6n(X ) + 13m(X ) − 90 + 60 + 78 ≥ −10 − 90 + 60 + 78 > 0, a contradiction to Claim G. Hence, |∂(X ) ∩ V (F)| = 4, implying that ∂(X ) ∩ V (F) = V (e). Let w ∈ ∂(X ) \ V (F) be arbitrary. Assume we had picked the edge e to contain three vertices from V (F) and the vertex w. In this case, H  would be linear. Further, def(H  ) = def H  (Y  ), where |Y  | = 1 and e is an edge of the hypergraph in the special H  -set Y  . However, Y is neither an H4 -component nor an H10 -component / {8, 10}, contradicting Claim H.26. This completes of H  , implying that def(H  ) ∈ () the proof of Claim H.27. Claim H.28: H  is not linear. Proof of Claim H.28: Suppose, to the contrary, that H  is linear. By Claim H.26 and Claim H.27, X = X 4 and def(H  ) = 8. As X = X 4 and H is linear, we note that |∂(X )| ≥ 5. By Claim H.3, the edge e is an isolated edge in H  and therefore there are at least four isolated vertices in ∂(X ) in H  − e. Suppose that there is a non-isolated vertex in ∂(X ) in H − V (X ). In this case, choosing the edge e to contain this vertex together with three isolated vertices of H − V (X ) that belong to ∂(X ) yields a new linear hypergraph H  in which the newly chosen edge e does not belong to a H4 - or H10 -component, contradicting Claim H.3 and Claim H.26. Therefore, all vertices in ∂(X ) are isolated vertices in H − V (X ), implying that |V (H )| ∈ {9, 10}, |E(H )| = 3 and τ (H ) = 2, which implies that the theorem holds for H , a contradiction. Thus, ξ(H ) = 45τ (H ) − 6n(H ) − 13m(H ) − def(H ) ≤ 90 − 54 − () 39 = −3, contradicting the fact that ξ(H ) > 0. Claim H.29: The following holds.

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(a) X = X 4 . (b) The edge e contains no vertex of degree 1 in H  . (c) Every edge in H  − e intersects ∂(X ) in at most two vertices. Proof of Claim H.29: By Claim H.28, H  is not linear, implying that the edge e overlaps some other edge in H  . If the edge e contains a degree-1 vertex in H  , then, by Claim H.5(a), 8 ≥ 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≥ 13|X | − 6|X  | − 3 = 10, a contradiction. Therefore, the edge e does not contain a degree-1 vertex in H  . Hence, by Claim H.5(b), 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | ≥ 13|X | − 6|X  | − 7 = 6, which implies that X = X 4 . This completes the proof of Part (a) and Part (b). To prove Part (c), suppose to the contrary that some edge, say f , in H  − e intersects ∂(X ) in at least three vertices. In this case, the edge f must intersect e1 or e2 in at () least two vertices, a contradiction to H being linear. Claim H.30: There exists two distinct edges in H  − e that both intersect ∂(X ) in exactly two vertices. Proof of Claim H.30: By Claim H.29, X = X 4 . The linearity of H implies that |∂(X )| ≥ 5. Suppose that at most one edge in H  − e intersects ∂(X ) in more than one vertex. By Claim H.29, such an edge intersects ∂(X ) in exactly two vertices, implying that the edge e could have be chosen so that H  is linear, contradicting Claim H.28. Therefore there are at least two distinct edges in H  − e that intersects ∂(X ) in more than one vertex. By Claim H.29, they each intersect ∂(X ) in exactly () two vertices, as claimed. By Claim H.30, there exists two distinct edges in H  − e that both intersect ∂(X ) in exactly two vertices. Let f 1 and f 2 be two such edges in H  − e. Thus, f 1 and f 2 are edges in H − V (X ) and |V ( f 1 ) ∩ ∂(X )| = |V ( f 2 ) ∩ ∂(X )| = 2. Claim H.31: The following holds. (a) Every vertex in ∂(X ) ∩ (V ( f 1 ) ∪ V ( f 2 )) has degree 1 in H  − e. (b) Every vertex in ∂(X ) belongs to at most one of f 1 and f 2 . Proof of Claim H.31: Suppose, to the contrary, that x ∈ ∂(X ) ∩ (V ( f 1 ) ∪ V ( f 2 )) and d H  −e (x) = 1. If d H  −e (x) = 0, then we could have chosen the edge e to contain the vertex x, implying that e would contain a vertex of degree 1 in H  − e, contradicting Claim H.29(b). Therefore, d H  −e (x) = 2 noting that Δ(H ) ≤ 3 and x ∈ ∂(X ). Renaming f 1 and f 2 , if necessary, we may assume that x ∈ V ( f 1 ). Let y be the vertex in V ( f 1 ) ∩ ∂(X ) different from x. Let H ∗ be obtained from H  − e − f 1 by adding two new vertices, say s1 and s2 , and the edge s = {x, y, s1 , s2 }. We show that τ (H ) ≤ τ (H ∗ ) + 1. Let T ∗ be a minimum transversal in H ∗ . As T ∗ covers the edge {x, y, s1 , s2 } and s1 and s2 have degree 1 in H ∗ , we can choose T ∗ to contain x or y, implying that T ∗ covers the edge f 1 . Further since {x, y} ⊂ ∂(X ), the set T ∗ covers at least one of e1 and e2 in H , say e1 . We can therefore add to T ∗ a vertex from X ∩ V (e2 ) in order to cover the edge in X and the edge e2 , thereby getting a transversal for H . This proves that τ (H ) ≤ |T ∗ | + 1 = τ (H ∗ ) + 1. We note that H ∗ is linear, and so H ∗ ∈ L4,3 . Further, n(H ) = n(H ∗ ) + 2 and m(H ) = m(H ∗ ) + 3. By Claim G, we have Φ(H ∗ ) < 0. Thus,

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0 > Φ(H ∗ ) = ξ(H ∗ ) − ξ(H ) = 45τ (H ∗ ) − 6n(H ∗ ) − 13m(H ∗ ) − def(H ∗ ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − 1) − 6(n(H ) − 2) − 13(m(H ) − 3) −def(H ∗ ) − 45τ (H ) + 6n(H ) + 13m(H ) = −45 + 12 + 39 − def(H ∗ ) = 6 − def(H ∗ ), and so def(H ∗ ) ≥ 7. Let Y ⊆ H ∗ be a special H ∗ -set such that def H ∗ (Y ) = def(H ∗ ) ≥ 7. If |E ∗H ∗ (Y )| ≥ |Y |, then def H ∗ (Y ) ≤ 10|Y | − 13|E ∗H ∗ (Y )| ≤ 10|Y | − / E(Y ), then 13|Y | < 0, a contradiction. Hence, |E ∗H ∗ (Y )| ≤ |Y | − 1. If s ∈ |E ∗H (X ∪ Y )| ≤ |E ∗H ∗ (Y )| + |{e1 , e2 , f 1 }| ≤ (|Y | − 1) + 3 = |X ∪ Y | + 1, contradicting the maximality of |X |. Therefore, s ∈ E(Y ). Thus, since the edge s contains two degree-1 vertices, namely, s1 and s2 , and at least one vertex of degree 2, namely, x, in H ∗ , we note that |Y | ≥ 2 and that s is a H4 -component in Y . If |E ∗H ∗ (Y )| = |Y | − 1, then def H ∗ (Y ) ≤ 10(|Y | − 1) + 8 − 13|E ∗H ∗ (Y )| = 10|Y | − 2 − 13(|Y | − 1) = −3|Y | + 11 ≤ 5, a contradiction. Hence, |E ∗H ∗ (Y )| ≤ |Y | − 2. Therefore, |E ∗H (X ∪ (Y \ {s})| ≤ |E ∗H ∗ (Y )| + |{e1 , e2 , f 1 }| ≤ (|Y | − 2) + 3 = |X ∪ (Y \ {s})| + 1,

contradicting the maximality of |X |. This completes the proof of Part (a). Part (b) () follows directly from Part (a). Recall that |V ( f 1 ) ∩ ∂(X )| = |V ( f 2 ) ∩ ∂(X )| = 2. By Claim H.31, every vertex in ∂(X ) belongs to at most one of f 1 and f 2 . We now choose the edge e to contain the four vertices in ∂(X ) ∩ (V ( f 1 ) ∪ V ( f 2 )). We next define a new hypergraph H f as follows. Let H  be constructed from H by removing the four vertices in X and the four vertices in e and removing the edge in E(X ) and the four edges e1 , e2 , f 1 , f 2 . We now define the edge f as follows. If f 1 and f 2 have no common neighbor in V (H  ) \ ∂(X ), then |(V ( f 1 ) ∪ V ( f 2 )) \ V (e)| = 4 and, in this case, we let f contain these four vertices. If f 1 and f 2 have a common neighbor in V (H  ) \ ∂(X ), then |(V ( f 1 ) ∪ V ( f 2 )) \ V (e)| = 3 and, in this case, we let f contain these three vertices as well as a vertex from ∂(X ) \ V (e). Let H f be obtained from H  by adding to it the edge f ; that is, H f = H  ∪ f . Claim H.31: τ (H ) ≤ τ (H f ) + 2. Proof of Claim H.31: Let T f be a minimum transversal in H f . In order to cover the edge f , there is a vertex z in T f that belongs to f . Suppose that z ∈ V ( f 1 ) ∪ V ( f 2 ). Thus, at least one of the edges in { f 1 , f 2 }, say f 2 , is covered by T f . Let w1 be any vertex in V (e) ∩ V ( f 2 ). We note that w1 covers the edge f 2 and at least one of the vertices in {e1 , e2 }, say e1 . Let w2 be the vertex in V (X ) ∩ V (e2 ). We note that w2 covers

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the edge in X and the edge e2 . Hence, T f ∪ {w1 , w2 } is a transversal in H , which / V ( f 1 ) ∪ V ( f 2 ). This implies that implies that τ (H ) ≤ τ (H f ) + 2. Suppose that z ∈ |(V ( f 1 ) ∪ V ( f 2 )) \ V (e)| = 3 and z ∈ ∂(X ) \ V (e). We note that z covers at least one of the vertices in {e1 , e2 }, say e1 . Let z 2 be the vertex in V (X ) ∩ V (e2 ). We note that w2 covers the edge in X and the edge e2 . Let z 1 be the vertex common to f 1 and f 2 . Hence, T f ∪ {z 1 , z 2 } is a transversal in H , which implies that τ (H ) ≤ τ (H f ) + 2. () Claim H.32: If Y is a special H f -set and f ∈ / E(Y ), then |E ∗H f (Y )| ≥ |Y | + 1. Proof of Claim H.32: Suppose, to the contrary, that |E ∗H f (Y )| ≤ |Y |. We now prove the following claims. Claim H.32.1: f ∈ E ∗H f (Y ). Proof of Claim H.32.1: If f ∈ / E ∗H f (Y ), then |E ∗H (X ∪ Y )| ≤ |E ∗H f (Y )| + () |{e1 , e2 }| ≤ |Y | + 2 = |X ∪ Y | + 1, contradicting the maximality of |X |. Claim H.32.2: |E ∗H f (Y )| = |Y |. Proof of Claim H.32.2: If |E ∗H f (Y )| < |Y |, then |E ∗H (X ∪ Y )| ≤ |E ∗H f (Y )| − |{ f }| + |{ f 1 , f 2 , e1 , e2 }| ≤ (|Y | − 1) − 1 + 4 = |X ∪ Y | + 1, contradicting the () maximality of |X |. We construct a bipartite graph G Y , with partite sets Y ∪ Y10 (that is, there are two copies of every element in Y10 ) and E ∗H f (Y ), where an edge joins g ∈ E ∗H f (Y ) and F ∈ Y ∪ Y10 in G Y if and only if the edge g intersects the subhypergraph F of Y in H f . Claim H.32.3: There is no matching in G Y saturating every vertex in E ∗H f (Y ). Proof of Claim H.32.3: Suppose, to the contrary, that there is a matching, M, in G Y saturating every vertex in E ∗H f (Y ). In this case, by Observation 8.1(g) and 8.1(h), using the matching M we can find a Y -transversal in H f covering all edges in E(Y ) ∪ E ∗H f (Y ). Let H ∗ = H f − V (Y ); that is, H ∗ is obtained from H f by removing all vertices in Y and all edges in E(Y ) ∪ E ∗H f (Y ). We show that def(H ∗ ) = 0. Suppose, to the contrary, that def(H ∗ ) > 0. Let Y ∗ be a special H ∗ -set with def H ∗ (Y ∗ ) > 0. This implies that |E ∗H ∗ (Y ∗ )| < |Y ∗ |. By Claim H.32.2, |E ∗H f (Y )| = |Y |. Thus, |E ∗H f (Y ∪ Y ∗ )| = |E ∗H f (Y )| + |E ∗H ∗ (Y ∗ )| < |Y | + |Y ∗ |. Hence we have shown the existence of a special H f -set, Y ∪ Y ∗ , such that f ∈ / E(Y ∪ Y ∗ ) and |E ∗H f (Y ∪ Y ∗ )| < |Y | + |Y ∗ |, contradicting Claim H.32.2. Hence, def(H ∗ ) = 0. / E(H ∗ ). Therefore, H ∗ is linear. Recall that As f ∈ E ∗H f (Y ), we note that f ∈ for any given special hypergraph, F, we have def(F) = 45τ (F) − 6|V (F)| − 13|E(F)|. Thus, def(H10 ) = 10, def(H4 ) = 8, def(H14 ) = 5, def(H11 ) = 4, and def(H21 ) = 1. In particular, def(F) ≤ 10 for any given special hypergraph, F. By Claim H.31, τ (H ) ≤ τ (H f ) + 2. Thus, since τ (H f ) ≤ τ (H ∗ ) + τ (Y ), we have τ (H ) ≤ τ (H ∗ ) + τ (Y ) + 2. We note that n(H ) = n(H ∗ ) + n(Y ) + 8 and m(H ) = m(H ∗ ) + m(Y ) + |E ∗H f (Y )| + 4. Thus,

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Φ(H ∗ ) = ξ(H ∗ ) − ξ(H ) = 45τ (H ∗ ) − 6n(H ∗ ) − 13m(H ∗ ) − def(H ∗ ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − τ (Y ) − 2) − 6(n(H ) − n(Y ) − 8) −13(m(H ) − m(Y ) − |E ∗H f (Y )| − 4) −45τ (H ) + 6n(H ) + 13m(H ) = 10 + 13|E ∗H f (Y )| − 45τ (Y ) + 6n(Y ) + 13m(Y )  = 10 + 13|Y | − def(F) = 10 +



F∈Y

(13 − def(F))

F∈Y

> 10, a contradiction to Claim G. This completes the proof of Claim H.32.3.

()

By Claim H.32.3, there is no matching in G Y saturating every vertex in E ∗H f (Y ). By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗H f (Y ) such that |NG Y (S)| < |S|. We now consider the bipartite graph G Y , with partite sets Y and E ∗H f (Y ), where an edge joins g ∈ E ∗H f (Y ) and F ∈ Y in G Y if and only if the edge g intersects the subhypergraph F of Y in H f . Thus, G Y is obtained from G Y by deleting all duplicated vertices associated with copies of H10 in Y . We note that |NG Y  (S)| ≤ |NG Y (S)| < |S|. We now consider the special H -set, Y  = Y \ NG Y  (S). Recall that by Claim H.32.2, |E ∗H f (Y )| = |Y |. Thus, |E ∗H f (Y  )| = |E ∗H f (Y )| − |S| = |Y | − |S| = |Y  | + |NG Y  (S)| − |S| < |Y  |, contradicting Claim H.32.2. This com() pletes the proof of Claim H.32. Claim H.33: Let Y be a special H f -set. If f ∈ / E(Y ) and f ∈ E ∗H f (Y ) and |Y10 | > ∗ 0, then |E H f (Y )| ≥ |Y | + 2. Proof of Claim H.33: Suppose, to the contrary, that |E ∗H f (Y )| ≤ |Y | + 1. By Claim H.32, |E ∗H f (Y )| ≥ |Y | + 1. Consequently, |E ∗H f (Y )| = |Y | + 1. Let G Y be the bipartite graph as defined in the proof of Claim H.32. Claim H.33.1: There is no matching in G Y saturating every vertex in E ∗H f (Y ). Proof of Claim H.33.1: Suppose, to the contrary, that there is a matching, M, in G Y saturating every vertex in E ∗H f (Y ). We proceed now analogously as in the proof of Claim H.32.3. If Y ∗ is a special H ∗ -set with def H ∗ (Y ∗ ) > 0, then |E ∗H ∗ (Y ∗ )| < |Y ∗ | and |E ∗H f (Y ∪ Y ∗ )| = |E ∗H f (Y )| + |E ∗H ∗ (Y ∗ )| ≤ (|Y | + 1) + (|Y ∗ | − 1) = |Y | + |Y ∗ |, contradicting Claim H.32. Hence, def(H ∗ ) = 0. Proceeding now exactly as in the proof of Claim H.32.3, we show that Φ(H ∗ ) > 0, contra() dicting Claim G. By Claim H.32.3, there is no matching in G Y saturating every vertex in E ∗H f (Y ). By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗H f (Y ) such that |NG Y (S)| < |S|.

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If S = E ∗H f (Y ), then NG Y (S) = Y ∪ Y10 . However, by assumption |Y10 | > 0, and so |NG Y (S)| ≥ |Y | + 1 = |E ∗H f (Y )| = |S|, a contradiction. Hence, S is a proper subset of E ∗H f (Y ). Let G Y be the bipartite graph defined in the proof of Claim H.32. We note that |NG Y  (S)| ≤ |NG Y (S)| < |S|. We now consider the special H -set, Y  = Y \ NG Y  (S). If Y  = ∅, then, |NG Y  (S)| = |Y | = |E ∗H f (Y )| − 1 ≥ |S|, a contradiction. Hence, Y  = ∅. Further, |E ∗H f (Y  )| = |E ∗H f (Y )| − |S| = (|Y | + 1) − |S| = |Y  | + 1 + |NG Y  (S)| − |S| < |Y  | + 1, and so |E ∗H f (Y  )| ≤ |Y  |, contradicting Claim H.32. () This completes the proof of Claim H.33. Claim H.34: If Y is a special H f -set and f ∈ E(Y ), then |E ∗H f (Y )| ≥ |Y | − 1. Proof of Claim H.34: Suppose, to the contrary, that |E ∗H f (Y )| < |Y | − 1. This implies that |Y | ≥ 2. Let F ∈ Y be the special subhypergraph containing f and let Y  = Y \ F. In this case, |E ∗H f (Y  )| ≤ |E ∗H f (Y )| < |Y | − 1 = |Y  |. Therefore, Y  is a / Y  satisfying |E ∗H f (Y  )| ≤ |Y  |, contradicting Claim H.32. () special H f -set and f ∈ Claim H.35: H f is not linear. Proof of Claim H.35: Suppose, to the contrary, that H f is linear. Thus, by Claim G and Claim H.31, we have 0 > Φ(H f ) = ξ(H f ) − ξ(H ) = 45τ (H f ) − 6n(H f ) − 13m(H f ) − def(H f ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − 2) − 6(n(H ) − 8) − 13(m(H ) − 4) −45τ (H ) + 6n(H ) + 13m(H ) = −45 · 2 + 6 · 8 + 13 · 4 − def(H f ) = 10 − def(H f ), and so, def(H f ) ≥ 11. Let Y be a special H f -set such that def(H f ) = def H f (Y ). As def H f (Y ) ≥ 11, we note that |E ∗H f (Y )| ≤ |Y | − 2. This is a contradiction to Claim H.32, Claim H.33, and Claim H.34, no matter whether f ∈ / E(Y ) or f ∈ () E(Y ). We now return to the proof of Claim H one final time. By Claim H.35, there exists an edge g in H f that overlaps f . Let {u, v} ⊆ V ( f ) ∩ V (g). Let H ∗f be the hypergraph obtained from H f by removing the edges f and g, adding two new vertices z 1 and z 2 , and adding a new edge e f g = {u, v, z 1 , z 2 }. Since H f − f is linear, we note that H ∗f is linear. Let T f∗ be a minimum transversal of H ∗f . If z 1 or z 2 belong to T f∗ , we can replace it with u or v, implying that T f∗ is a transversal of H f . Thus, by Claim H.31, τ (T f∗ ) ≥ τ (H f ) ≥ τ (H ) − 2. Thus, by Claim G, we have

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0 > Φ(H ∗f ) = ξ(H ∗f ) − ξ(H ) = 45τ (H ∗f ) − 6n(H ∗f ) − 13m(H ∗f ) − def(H ∗f ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − 2) − 6(n(H ) − 6) − 13(m(H ) − 5) −45τ (H ) + 6n(H ) + 13m(H ) = −45 · 2 + 6 · 6 + 13 · 5 − def(H ∗f ) = 11 − def(H ∗f ), and so, def(H ∗f ) ≥ 12. Let Y be a special H ∗f -set such that def(H ∗f ) = def H ∗f (Y ). As / E(Y ), then |E ∗H f (Y )| ≤ def H ∗f (Y ) ≥ 12 we note that |E ∗H ∗ (Y )| ≤ |Y | − 2. If e f g ∈ f |Y | − 2 + |{ f, g}| = |Y |, contradicting Claim H.32. Therefore, e f g ∈ E(Y ). Let F f g ∈ Y be the special subhypergraph containing the edge e f g . Since e f g contains at least two degree-1 vertices in H ∗f , we note that F f g ∈ Y4 . Let Y f g = Y \ {F f g }, and note that f ∈ / E(Y f g ). Further, |E ∗H f (Y f g )| ≤ |E ∗H ∗f (Y )| + |{ f, g}| = |E ∗H ∗f (Y )| + 2. As observed earlier, |E ∗H ∗ (Y )| ≤ |Y | − 2. Suppose that |E ∗H ∗ (Y )| = |Y | − 2. f f In this case, since def H ∗f (Y ) ≥ 12, we note that |Y10 | > 0. Since F f g ∈ Y4 , this implies that |(Y f g )10 | > 0. If f ∈ E ∗H f (Y f g ), then |E ∗H f (Y f g )| ≤ |E ∗H ∗ (Y )| + 2 = f / E ∗H f (Y f g ), then |E ∗H f (Y f g )| ≤ |Y | = |Y f g | + 1, contradicting Claim H.33. If f ∈ |E ∗H ∗ (Y )| + |{g}| = |E ∗H ∗ (Y )| + 1 = |Y | − 1 = |Y f g |, contradicting Claim H.32. f f Therefore, |E ∗H ∗ (Y )| ≤ |Y | − 3, implying that |E ∗H f (Y f g )| ≤ |E ∗H ∗ (Y )| + 2 ≤ |Y | − f f 1 = |Y f g |, once again contradicting Claim H.32. This completes the proof of () Claim H. Claim I: If Y be a special H -set, then the following holds. (a) |E ∗ (Y )| ≥ |Y | + 2. (b) If |Y10 | ≥ 2, then |E ∗ (Y )| ≥ |Y | + 3. Proof of Claim I: Part (a) follows immediately from our choice of the H -special set X and by Claim H. To prove Part (b), suppose, to the contrary, that |Y10 | ≥ 2 and |E ∗ (Y )| ≤ |Y | + 2. By Part (a), |E ∗H (Y )| ≥ |Y | + 2. Consequently, |E ∗ (Y )| = |Y | + 2. We construct a bipartite graph G Y , with partite sets Y ∪ Y10 (that is, there are two copies of every element in Y10 ) and E ∗ (Y ), where an edge joins e ∈ E ∗ (Y ) and F ∈ Y ∪ Y10 in G Y if and only if the edge e intersects the subhypergraph F of Y in H . Suppose that there is a matching, M, in G Y saturating every vertex in E ∗ (Y ). By Observation 8.1(g) and 8.1(h), using the matching M we can find a Y -transversal in H covering all edges in E(Y ) ∪ E ∗ (Y ), contradicting Claim B. Hence, there is no matching in G Y saturating every vertex in E ∗ (Y ). By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗ (Y ) such that |NG Y (S)| < |S|. If S = E ∗ (Y ), then NG Y (S) = Y ∪ Y10 . However, by assumption |Y10 | ≥ 2, and

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so |NG Y (S)| = |Y | + |Y10 | ≥ |Y | + 2 = |E ∗ (Y )| = |S|, a contradiction. Hence, S is a proper subset of E ∗ (Y ). We now consider the special H -set, Y  , obtained from Y by deleting every element in Y that is a neighbor of S in H . We note that at most |NG Y (S)| such elements have been deleted from Y , possibly fewer since some elements F ∈ Y10 may get deleted twice, and so |Y  | ≥ |Y | − |NG Y (S)|. Therefore, |E ∗ (Y  )| ≤ |E ∗ (Y )| − |S| < (|Y | + 2) − |NG Y (S)| ≤ |Y  | + 2. Thus, |E ∗ (Y  )| − |Y  | < 2. We therefore have a contradiction to the choice of the special H -set X which was chosen () so that |E ∗ (X )| − |X | is minimum. In particular, Claim I implies that if |X 10 | ≥ 2, then |E ∗ (X )| ≥ |X | + 3. Claim J: There is no H10 -subhypergraph in H . Proof of Claim J: Suppose, to the contrary, that F is a H10 -subhypergraph in H . By Claim C, F is not a component of H , and so there exists an edge e ∈ E ∗ (F). We choose such an edge e so that |V (e) ∩ V (F)| is maximum possible. Since H is linear, we note that |V (e) ∩ V (F)| ≤ 2. Let x ∈ V (e) ∩ V (F) be arbitrary. Let E(F) = {e1 , e2 , e3 , e4 , e5 }, where e1 and e2 are the two edge of F that contain x. We note that d H (x) = 3, and e, e1 , e2 are the three edges of H that contains x. Let H  = H − x. Let E  = {e3 , e4 , e5 } and note that E  ⊆ E(H  ) and every pair of edges in E  intersect in H  . Every τ -transversal of H  can be extended to a transversal of H by adding to it the vertex x, and so τ (H ) ≤ τ (H  ) + 1. By Claim G, 0 > Φ(H  ) = ξ(H  ) − ξ(H ) = 45τ (H  ) − 6n(H  ) − 13m(H  ) − def(H  ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − 1) − 6(n(H ) − 1) − 13(m(H ) − 3) −45τ (H ) + 6n(H ) + 13m(H ) = −45 + 6 · 1 + 13 · 3 − def(H  ) = −def(H  ), and so, def(H  ) > 0. Let Y be a special H  -set such that def(H  ) = def H  (Y ). By Claim I(a), |E ∗H (Y )| ≥ |Y | + 2. Claim J.1: |Y | = 2, |E ∗H  (Y )| = 1, and 0 < def H  (Y ) ≤ 5. Proof of Claim J.1: If |E ∗H  (Y )| ≤ |Y | − 2, then |E ∗H (Y )| ≤ |E ∗H  (Y )| + |{e, e1 , e2 }| ≤ |Y | + 1, a contradiction. Therefore, |E ∗H  (Y )| ≥ |Y | − 1. As def H  (Y ) ≥ 1, we note that |E ∗H  (Y )| ≤ |Y | − 1. Consequently, |E ∗H  (Y )| = |Y | − 1. If |Y | ≥ 5, then def(H  ) ≤ 10|Y | − 13|E ∗H  (Y )| = 10|Y | − 13(|Y | − 1) = −3|Y | + 13 < 0, a contradiction. Hence, |Y | ≤ 4. Suppose that |Y | ≥ 3. If |Y10 | ≤ 1, then def(H  ) ≤ 10 + 8(|Y | − 1) − 13|E ∗H  (Y )| = 8|Y | + 2 − 13(|Y | − 1) = −5|Y | + 15 ≤ 0, a contradiction. Hence, |Y10 | ≥ 2. However, |E ∗H (Y )| ≤ |E ∗H  (Y )| + |{e, e1 , e2 }| = |Y | + 2, contradicting Claim I(b). Hence, |Y | ≤ 2. Suppose that |Y | = 1, and let Y = {F1 }. By Claim I(a), |Y | + 2 ≤ |E ∗H (Y )| ≤ ∗ |E H  (Y )| + |{e, e1 , e2 }| = |Y | + 2, implying that |E ∗H (Y )| = |Y | + 2 and E ∗ (F1 ) =

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{e, e1 , e2 }. Therefore, E  ⊆ E(F1 ), which implies that V (F) \ {x} ⊆ V (F1 ). Therefore, the edges e1 and e2 intersect F1 in three vertices, while the edge e intersects F1 in at least one vertex. If e intersects F1 in at least two vertices, then by Observation 8.1(n) we can cover E ∗ (F1 ) with a τ -transversal of F1 in H , a contradiction to Claim B. Therefore, we may assume that the edge e intersects F1 in only one vertex. By Observation 8.1( ), there exists a τ -transversal of F1 , T1 , of F1 covering both e1 and e2 . Let H1 = H − T1 . We will now show that def(H1 ) ≤ 8. Suppose, to the contrary, that def(H1 ) ≥ 9 and let Y1 be a special H1 -set such that def(H1 ) = def H1 (Y1 ). If the edge e does not belong to any special subhypergraph in Y1 , then def H  (Y1 ∪ F1 ) > def H  (F1 ) = def(H  ), a contradiction. Therefore, e ∈ E(Fe ) for some special subhypergraph in Y1 . As the edge e has at least two vertices of degree 1 in H1 (namely, the vertex x and the vertex in V (F1 ) ∪ V (e)), the subhypergraph Fe is an H4 -component. However since def H1 (Y1 ) ≥ 9, the special set {F1 } ∪ (Y1 \ {Fe }) has higher deficiency than F1 in H  , a contradiction. Therefore, def(H1 ) ≤ 8, as desired. As there is an edge, e1 (and e2 ), that intersects F1 in three vertices we note that the deficiency of F1 is at most 5. Note that we have removed the edges e1 and e2 as well as all edges in F1 to get from H to H1 as well as all vertices in F1 except the vertex in V (e) ∩ V (F1 ). Therefore, n(H1 ) = n(H ) − (n(F1 ) − 1), m(H1 ) = m(H ) − m(F1 ) − |{e1 , e2 }|, and τ (H ) ≤ τ (H1 ) + τ (F1 ). As observed earlier, def(F1 ) ≤ 5. Thus, 45τ (F1 ) ≤ 6n(F1 ) + 13m(F1 ) + 5, and therefore 0 > Φ(H1 ) = ξ(H1 ) − ξ(H ) = 45τ (H1 ) − 6n(H1 ) − 13m(H1 ) − def(H1 ) −45τ (H ) + 6n(H ) + 13m(H ) + def(H ) ≥ 45(τ (H ) − τ (F1 )) − 6(n(H ) − n(F1 ) + 1) − 13(m(H ) − m(F1 ) − 2) − 8 −45τ (H ) + 6n(H ) + 13m(H ) = −45τ (F1 ) + 6 · (n(F1 ) − 1) + 13 · (m(F1 ) + 2) − 8 ≥ −5 − 6 · 1 + 13 · 2 − 8 > 0,

a contradiction. Therefore, |Y | = 2 and |E ∗H  (Y )| = |Y | − 1 = 1. As observed earlier, if |Y10 | ≥ 2, then |E ∗H (Y )| ≤ |Y | + 2, contradicting Claim I(b). Hence, |Y10 | ≤ 1, implying that def(H  ) = def H  (Y ) ≤ 10 + 8 − 13|E ∗H  (Y )| = 18 − 13 = 5. This () completes the proof of Claim J.1. By Claim J.1, we have |Y | = 2, |E ∗H  (Y )| = 1, and 0 < def H  (Y ) ≤ 5. Let Y = {F1 , F2 }. Since F1 and F2 are vertex-disjoint and every pair of edges in E  intersect in H  , we note that E(F1 ) ∩ E  = ∅ or E(F2 ) ∩ E  = ∅. Renaming F1 and F2 if necessary, we may assume that E(F2 ) ∩ E  = ∅. If neither e1 nor e2 intersects F2 in H , then E ∗H (F2 ) ⊆ {e} ∪ E ∗H  (Y ), implying that |E ∗H (F2 )| ≤ 1 + |E ∗H  (Y )| = 2. However, by Claim I(a), |E ∗H (F2 )| ≥ |F2 | + 2 = 3, a contradiction. Therefore, e1 or e2 intersects F2 in H . Renaming e1 and e2 if necessary, we may assume that e1 intersects F2 in H .

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Claim J.2: F2 ∈ / Y10 . Proof of Claim J.2: Suppose, to the contrary, that F2 ∈ Y10 . Let z ∈ V (e1 ) ∩ V (F2 ) be arbitrary. The vertex z is incident with two edges from F and two edges from F2 in H and these four edges are distinct, contradicting the fact that the maximum / Y10 . () degree in H is three. Therefore, F2 ∈ Claim J.3: F2 ∈ / Y4 . Proof of Claim J.3: Suppose, to the contrary, that F2 ∈ Y4 . Since H is linear and F2 ∈ Y4 , every edge in E ∗H (F2 ) intersects F2 in exactly one vertex. Let V (e1 ) ∩ V (F2 ) = {z 1 }. The vertex z 1 is incident with two edges from F and the one edge from F2 in H , and so d H (z) = 3. Suppose that both e and e2 intersect F2 in H . Let V (e2 ) ∩ V (F2 ) = {z 2 }. Since x is the only vertex common to both e1 and e2 , we note that z 1 = z 2 . Let f be the edge of F2 . By the linearity of H and since F is a H10 -subhypergraph in H , we note that V ( f ) ∩ V (F) = {z 1 , z 2 }. Thus, the edge f ∈ E ∗H (F) and |V ( f ) ∩ V (F)| = 2, implying by our choice of the edge e that |V (e) ∩ V (F)| = 2. This in turn implies by the linearity of H and the structure of F that the edge e does not intersect F2 , a contradiction. As observed above, at least one of e and e2 does not intersect F2 in H . Thus, |E ∗H (F2 )| ≤ 2 + |E ∗H  (Y )| = 3, implying that there exists a vertex q in F2 that has degree 1 in H . If we had chosen to use the vertex z 1 instead of x when constructing H  , the vertex q would become isolated and therefore deleted from H  , implying that n(H  ) ≤ n(H ) − 2. Thus, 0 > Φ(H  ) ≥ −45 + 6 · 2 + 13 · 3 − def(H  ) = 6 − def(H  ). Hence, def(H  ) > 6, which is a contradiction to Claim J.1. Therefore, / F4 . () F2 ∈ Claim J.4: F1 ∈ / Y10 . Proof of Claim J.4: Suppose, to the contrary, that F1 ∈ Y10 . Suppose that |E(F1 ) ∩ E  | = 0. By Claim I(a), at least two of the edges in {e, e1 , e2 } intersect F1 in H , implying that there is a vertex q ∈ V (F2 ) ∩ V (F). The vertex q is incident with two edges from F and two edges from F2 in H and these four edges are distinct, contradicting the fact that the maximum degree in H is three. Therefore, |E(F1 ) ∩ E  | ≥ 1. If |E(F1 ) ∩ E  | = 1, then |E ∗H  (Y )| ≥ 2, as all edges in E  intersect each other, contradicting Claim J.1. Therefore, |E(F1 ) ∩ E  | ≥ 2. If |E(F1 ) ∩ E  | = 3 (and so, E  ⊂ E(F1 )), then, since F1 and F2 are vertex-disjoint, the edges e1 and e2 do not intersect F2 , implying that E ∗H (F2 ) ⊆ {e} ∪ E ∗H  (Y ) and hence that |E ∗H (F2 )| ≤ 1 + |E ∗H  (Y )| = 2, contradicting Claim I(a). Therefore, |E(F1 ) ∩ E  | = 2. Renaming the edges in E  if necessary, we may assume that e3 , e4 ∈ E(F1 ), which implies that E ∗H  (Y ) = {e5 }. As e5 intersects both e3 and e4 , the edge e5 intersects F1 in at least two vertices; that is, e5 ∈ E ∗H (F1 ) and |V (e5 ) ∩ V (F1 )| ≥ 2 where recall that F1 is a H10 -subhypergraph in H . By the maximality of |V (e) ∩ V (F)|, we may therefore assume that |V (e) ∩ V (F)| ≥ 2. Let {x, y} ⊆ V (e) ∩ V (F), where recall

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that H  = H − x. By the linearity of H and since F is a H10 -subhypergraph in H , we note that V (e) ∩ V (F) = {x, y} and y ∈ / V (e1 ) ∪ V (e2 ), which implies that y belongs to two of the edges in {e3 , e4 , e5 }. Therefore, y ∈ V (F1 ) and all four edges e, e1 , e2 , e5 intersect F1 . Recall that E ∗H  (Y ) = {e5 }. If the edge e5 does not intersect F2 , then neither do the edges e1 and e2 , implying that |E ∗H (F2 )| ≤ |{e}| = 1, contradicting Claim I(a). Therefore, V (e5 ) ∩ V (F2 ) = ∅. This implies that one of the edges in {e1 , e2 , e3 , e4 } intersects F2 as every vertex of e5 intersects one of the edges in {e1 , e2 , e3 , e4 }. Since F1 and F2 are vertex-disjoint and e3 , e4 ∈ E(F1 ), the edges e3 and e4 do not intersect F2 . Thus, the common vertex of e1 and e5 belongs to F2 or the common vertex of e2 and e5 belongs to F2 . Renaming e1 and e2 if necessary, we may assume that the common vertex of e1 and e5 belongs to F2 , and therefore we can cover e1 and e5 with a τ -transversal of F2 . As F1 is a H10 subhypergraph in H we can cover e and e2 with a τ -transversal of F1 by Observation 8.1(h). Therefore, we can cover E ∗ (Y ) in H with a Y -transversal in H , contradicting Claim B. This completes the proof of () Claim J.4. We now return to the proof of Claim J. Recall that Y = {F1 , F2 } and that by / Y4 ∪ Y10 . By Claim J.4, F1 ∈ / Claim J.1, |E ∗H  (Y )| = 1. By Claims J.2 and J.3, F2 ∈ F10 . Thus, def(H  ) = def H  (Y ) ≤ 8 + 5 − 13 = 0, a contradiction to def(Y ) > 0. () This completes the proof of Claim J. Recall that the boundary of a set Z of vertices in a hypergraph H is the set N H (Z ) \ Z , denoted ∂ H (Z ) or simply ∂(Z ) if H is clear from context. Claim K: Let Z ⊆ V (H ) be an arbitrary nonempty set of vertices and let H  = H − Z . Then either |E ∗H  (Y )| ≥ |Y | for all special H  -sets Y in H  (and therefore def(H  ) = 0) or there exists a transversal T  in H  , such that 45|T  | ≤ 6n(H  ) + 13m(H  ) + def(H  )

and

T  ∩ ∂(Z ) = ∅.

Proof of Claim K: Suppose that |E ∗H  (Y )| < |Y | for some special H  -set, Y , in H  , as otherwise the claim holds. Among all such special H  -sets Y , let Y be chosen so that (1) |Y | − |E ∗H  (Y )| is maximum. (2) Subject to (1), |Y | is minimum. If ∂(Z ) ∩ V (Y ) = ∅, then |E ∗H (Y )| < |Y |, a contradiction to Claim I(a). Therefore, ∂(Z ) ∩ V (Y ) = ∅ and let q be any vertex in ∂(Z ) ∩ V (Y ). We now consider the bipartite graph, G Y , with partite sets Y and E ∗H  (Y ) ∪ {q}, where an edge joins e ∈ E ∗H  (Y ) and F ∈ Y in G Y if and only if the edge e intersects the subhypergraph F of Y in H  and q is joined to F ∈ Y in G Y if and only if q ∈ V (F). Claim K.1: If there is a matching in G Y saturating every vertex in E ∗H  (Y ) ∪ {q}, then Claim K holds.

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Proof of Claim K.1: Suppose that there exists a matching in G Y that saturates every vertex in E ∗H  (Y ) ∪ {q}. This implies that there exists a Y -transversal, TY , in H  covering E ∗H  (Y ) and with q ∈ TY . Let HY be obtained from H  by deleting the vertices V (Y ) and edges E(Y ) ∪ E ∗H  (Y ); that is, HY = H  − TY . Suppose that def(HY ) > 0, and let Y  be a special HY -set in HY such that def HY (Y  ) = def(HY ) > 0. Since |E ∗HY (Y  )| ≤ |Y  | − 1, we note that |E ∗H  (Y ∪ Y  )| = |E ∗H  (Y )| + |E ∗HY (Y  )| < |Y | + |Y  | and |Y ∪ Y  | − |E ∗H  (Y ∪ Y  )| = |Y | + |Y  | − (|E ∗H  (Y )| + |E ∗HY (Y  )|) ≥ |Y | − |E ∗H  (Y )| + 1,

contradicting the choice of the special H  -set Y . Therefore, def(HY ) = 0. Since H is a counterexample with minimum value of n(H ) + m(H ), and since n(HY ) + m(HY ) < n(H ) + m(H ), we note that 45τ (HY ) ≤ 6n(HY ) + 13m(HY ) + def(HY ) = 6n(HY ) + 13m(HY ). As 45|TY | = 6n(Y ) + 13m(Y ) + 13|E ∗H  (Y )| + def H  (Y ) we get the following: 45|T  | ≤ 45|TY | + 45τ (HY ) ≤ 6n(Y ) + 13m(Y ) + 13|E ∗H  (Y )| + def H  (Y ) + 6n(HY ) + 13m(HY ) ≤ 6n(H  ) + 13m(H  ) + def H  (Y ). As q belonged to TY this proves Claim K in this case.

()

By Claim K.1, we may consider the case when there is no perfect matching in G Y saturating every vertex in E ∗H  (Y ) ∪ {q}. By Hall’s Theorem, there is therefore a nonempty subset S ⊆ E ∗H  (Y ) ∪ {q} such that |NG Y (S)| < |S|. We now consider the special H -set Y  = Y \ NG Y (S). Since |Y  | = |Y | − |NG Y (S)| and |E ∗H  (Y  )| ≤ |(E ∗H  (Y ) ∪ {q}) \ S| = |E ∗H  (Y )| + 1 − |S| ≤ |E ∗H  (Y )| − |NG Y (S)|, we note that |Y  | − |E ∗H  (Y  )| ≥ |Y | − |E ∗H  (Y )| and |Y  | < |Y |, contradicting our choice of Y . () This completes the proof of Claim K. Now let f be the function defined in Table 8.1. Claim L: Let Z ⊆ V (H ) be an arbitrary nonempty set of vertices that intersects at least two edges of H , and let H  = H − Z . If def(H  ) ≤ 21 and |∂(Z )| ≥ 1, then there exists a transversal, T  , in H  , such that T  ∩ ∂(Z ) = ∅ and the following holds. (a) 45|T  | ≤ 6n(H  ) + 13m(H  ) + f (|∂(Z )|). (b) If |∂(Z )| ≥ 5 and H  does not contain two intersecting edges e and f , such that (i) ∂(Z ) ⊆ (V (e) ∪ V ( f )) \ (V (e) ∩ V ( f )),

Table 8.1 The function f i f (i)

1 39

2 33

3 27

4 23

≥5 22

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101

(ii) e contains three degree-1 vertices, and (iii) |∂(Z ) ∩ V (e)|, |∂(Z ) ∩ V ( f )| ≥ 2, then 21.

45|T  | ≤ 6n(H  ) + 13m(H  ) + f (|∂(Z )|) − 1 = 6n(H  ) + 13m(H  ) +

Proof of Claim L: Suppose, to the contrary, that there exists a set Z ⊆ V (H ), such that H  = H − Z with def(H  ) ≤ 21 and |∂(Z )| ≥ 1, but there exists no transversal T  in H  satisfying (a) or (b) in the statement of the claim. Among all such sets, let Z be chosen so that |Z | is as large as possible. Let T  be a smallest possible transversal in H  containing a vertex from ∂(Z ). Claim L.1: For any special H  -set, Y , in H  we have |E ∗H  (Y )| ≥ |Y |. Furthermore def(H  ) = 0. Proof of Claim L.1: If |E ∗H  (Y )| < |Y | for some special H  -set, Y , in H  , then by Claim K there exists a transversal T  in H  , such that 45|T  | ≤ 6n(H  ) + 13m(H  ) + def(H  ) and T  ∩ ∂(Z ) = ∅. Since def(H  ) ≤ 21, this implies that 45|T  | ≤ 6n(H  ) + 13m(H  ) + 21 < 6n(H  ) + 13m(H  ) + f (|∂(Z )|), contradicting our definition of Z . Therefore |E ∗H  (Y )| ≥ |Y | for all H  -sets Y , which furthermore implies that def(H  ) = 0. () By supposition, |∂(Z )| ≥ 1. Let Z  be a set of new vertices (not in H ), where |Z | = max(0, 4 − |∂(Z )|). We note that |Z  ∪ ∂(Z )| ≥ 4 and 0 ≤ |Z  | ≤ 3. Further if |∂(Z )| ≥ 4, then Z  = ∅. Let He be the hypergraph obtained from H  by adding the set Z  of new vertices and adding a 4-edge, e, containing four vertices in Z  ∪ ∂(Z ). Note that He may not be linear as the edge e may overlap other edges in He . Every τ -transversal of He is a transversal of H  . Further, if Z  = ∅, then in order to cover the edge e every τ -transversal of He contains a vertex of ∂(Z ), while if Z  = ∅, then the edge e contains at least one vertex of ∂(Z ) and we can choose a τ -transversal of He to contain such a vertex of ∂(Z ). Therefore, |T  | ≤ τ (He ), which recall that T  is a smallest possible transversal in H  containing a vertex from ∂(Z ). 

Claim L.2: He is not linear and |∂(Z )| ≥ 2. Proof of Claim L.2: Suppose, to the contrary, that He is linear. By construction, n(He ) ≤ n(H ) and m(He ) = m(H  ) + 1 < m(H ). Therefore, n(He ) + m(He ) < n(H ) + m(H ). Since H is a counterexample with minimum value of n(H ) + m(H ), we note that 45τ (He ) ≤ 6n(He ) + 13m(He ) + def(He ). Claim L.2.1: def(He ) ≥ 9. Proof of Claim L.2.1: Suppose, to the contrary, that def(He ) ≤ 8. Recall that m(He ) = m(H  ) + 1. If Z  = ∅, then n(He ) = n(H  ) and either |∂(Z )| = 4 and f (|∂(Z )|) = 23 or |∂(Z )| ≥ 5 and f (|∂(Z )|) = 22. In this case, 45|T  | ≤ 45τ (He ) ≤ 6n(He ) + 13m(He ) + def(He ) = 6n(H  ) + 13m(H  )| + 13 + 8 ≤ 6n(H  ) + 13m(H  ) + f (|∂(Z )|) − 1,

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a contradiction. If |Z  | = 1, then n(He ) = n(H  ) + 1, |∂(Z )| = 3 and f (|∂(Z )|) = 27, implying that 45|T  | ≤ 6n(H  ) + 13m(H  )| + 6 + 13 + 8 = 6n(H  ) + 13m(H  ) + f (|∂(Z )|), a contradiction. If |Z  | = 2, then n(He ) = n(H  ) + 2, |∂(Z )| = 2 and f (|∂(Z )|) = 33, implying that 45|T  | ≤ 6n(H  ) + 13m(H  )| + 12 + 13 + 8 = 6n(H  ) + 13m(H  ) + f (|∂(Z )|), a contradiction. If |Z  | = 3, then n(He ) = n(H  ) + 3, |∂(Z )| = 1 and f (|∂(Z )|) = 39, implying that 45|T  | ≤ 6n(H  ) + 13m(H  )| + 18 + 13 + 8 = 6n(H  ) + 13m(H  ) + f (|∂(Z )|), () a contradiction. This completes the proof of Claim L.2.1. By Claim L.2.1, def(He ) ≥ 9. Let Ye be a special He -set in He such that def He (Ye ) = def(Ye ) ≥ 9. Suppose that |Ye | ≥ 2. In this case |E ∗He (Ye )| ≤ |Ye | − 2, for otherwise def(He ) < 9. Let Ye be equal to Ye , except we remove any special subhypergraph from Ye if it contains the edge e. We note that |E H  (Ye )| ≤ |E ∗He (Ye )| ≤ |Ye | − 2 ≤ |Ye | − 1, contradicting Claim L.1. Therefore, |Ye | = 1. Since def(He ) ≥ 9 and |Ye | = 1, the special He -set consists of one H10 -component in He and def(He ) = 10. By Claim J, this H10 -component, Ye , in He contains the edge e. If |∂(Z )| ≤ 3, then Z  = ∅ and the edge e would contain a vertex of degree 1 in He , implying that Ye would contain a vertex of degree 1. However, H10 is 2-regular, a contradiction. Hence, |∂(Z )| ≥ 4. Thus, Z  = ∅ and n(He ) = n(H  ). If |∂(Z )| = 4, then f (|∂(Z )|) = 23, implying that 45|T  | ≤ 45τ (He ) ≤ 6n(He ) + 13m(He ) + def(He ) = 6n(H  ) + 13m(H  )| + 13 + 10 = 6n(H  ) + 13m(H  ) + f (|∂(Z )|), a contradiction. Hence, |∂(Z )| ≥ 5, and so f (|∂(Z )|) = 22. If there is a vertex v ∈ ∂(Z ) \ V (Ye ), then we can change e by removing from it an arbitrary vertex and adding to it the vertex v instead. With this new choice of the added edge e, we note that He is once again linear but now the component in He containing the edge e contains at least 11 vertices, implying that there would be no H10 -component in He containing the edge e, contradicting our earlier arguments. Therefore, ∂(Z ) ⊆ V (Ye ). Since H is connected, this implies that H  = Ye − e. Let u ∈ ∂(Z ) \ V (e) be arbitrary. We note that u exists since |∂(Z )| ≥ 5 and |V (e)| = 4. We note further that there are exactly two edges in H  − u and these two edges intersect in a vertex, say w. Hence, {u, w} is a transversal in H  containing a vertex from ∂(Z ), implying that 45|T  | ≤ 90 < 60 + 52 + 21 = 6n(H  ) + 13m(H  ) + f (|∂(Z )|) − 1, a contradiction. Therefore, He cannot be lin() ear. This implies that |∂(Z )| ≥ 2, which completes the proof of Claim L.2. Claim L.3: For any nonempty special H  -set, Y , in H  we have |E ∗H  (Y )| ≥ |Y | + 1 or |∂(Z ) \ V (Y )| ≤ 1. Proof of Claim L.3: Suppose, to the contrary, that there is a nonempty special H  -set, Y in H  where |E ∗H  (Y )| ≤ |Y | and |∂(Z ) \ V (Y )| ≥ 2. Assume that |Y | is minimum possible with the above property. By Claim L.1, |E ∗H  (Y )| = |Y |. By Claim E, we note that ∂(Z ) ∩ V (Y ) = ∅. Let Z ∗ = Z ∪ V (Y ) and let H ∗ = H − Z ∗ . Further, let Q ∗ = ∂(Z ∗ ), and so Q ∗ = N H (Z ∗ ) \ Z ∗ . Since |∂(Z ) \ V (Y )| ≥ 2, we have |Q ∗ | ≥ 2.

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Claim L.3.1: def(H ∗ ) ≤ 21. Proof of Claim L.3.1: Suppose, to the contrary, that def(H ∗ ) ≥ 22. Let Y ∗ be a special H ∗ -set with def H ∗ (Y ∗ ) ≥ 22. By Claim J, there is no H10 -subhypergraph in H , implying that def H ∗ (Y ∗ ) ≤ 8|Y ∗ | − 13|E ∗H ∗ (Y ∗ )|. If |E ∗H ∗ (Y ∗ )| ≥ |Y ∗ | − 2, then 22 ≤ def H ∗ (Y ∗ ) ≤ 8|Y ∗ | − 13|Y ∗ | + 26 = −5|Y ∗ | + 26, implying that |Y ∗ | = 0, a contradiction. Thus, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Hence, |E ∗H  (Y ∪ Y ∗ )| ≤ |E ∗H  (Y )| + () |E ∗H ∗ (Y ∗ )| ≤ |Y | + |Y ∗ | − 3 < |Y ∪ Y ∗ |, contradicting Claim L.1. Let T ∗ be a minimum transversal in H ∗ that contains a vertex from Q ∗ . By the maximality of |Z |, we note that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|Q ∗ |). Claim L.3.2: |T  | ≤ |T ∗ | + τ (Y ). Proof of Claim L.3.2: By construction of T ∗ , either ∂(Z ) ∩ T ∗ = ∅ or T ∗ intersects an edge in E ∗H  (Y ). We now consider the bipartite graph, G Y , with partite sets Y and E ∗H  (Y ) ∪ {q}, where an edge joins e ∈ E ∗H  (Y ) and F ∈ Y in G Y if and only if the edge e intersects the subhypergraph F of Y in H  and q is joined to F ∈ Y in G Y if and only if ∂(Z ) ∩ V (F) = ∅. We now construct G Y from G Y by removing either q if ∂(Z ) ∩ T ∗ = ∅ or by removing some e ∈ E ∗H  (Y ) where T ∗ intersects the edge e . We note that since |E ∗H  (Y )| = |Y |, the partite sets in G Y have the same size. Suppose there is no perfect matching in G Y . By Hall’s Theorem, there is a nonempty subset S ⊆ V (G Y ) \ Y such that |NG Y (S)| < |S|. We now consider the special H  -set, Y  = Y \ NG Y (S). Then, |Y  | = |Y | − |NG Y (S)|. Since possibly q ∈ S, we therefore have |E ∗H  (Y  )| ≤ |E ∗H  (Y )| − (|S| − 1) ≤ |Y | − |NG Y (S)| = |Y  |. Thus, Y  is a nonempty special H  -set such that |E ∗H  (Y  )| ≤ |Y  | and |Y  | < |Y |. Further, since Y  ⊂ Y and |∂(Z ) \ V (Y )| ≥ 2, we note that |∂(Z ) \ V (Y  )| ≥ 2. This contradicts our choice of Y . Therefore, there is a perfect matching in G Y , implying that we can find a Y -transversal that together with T ∗ intersects all the edges in E ∗H  (Y ) and intersects ∂(Z ). This implies that |T  | ≤ |T ∗ | + τ (Y ), as desired. () As observed earlier, 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|Q ∗ |). Since Y is a nonempty special H  -set, we note that 45τ (Y ) = 6n(Y ) + 13m(Y ) + def H  (Y ) ≤ 6n(Y ) + 13m(Y ) + def H  (Y ) + 13|E ∗H  (Y )|. By Claim L.3.2, |T  | ≤ |T ∗ | + τ (Y ). Therefore, 45|T  | ≤ 45|T ∗ | + 45τ (Y ) ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|Q ∗ |) + 6n(Y ) + 13m(Y ) + 13|E ∗H  (Y )| + def H  (Y )

= 6n(H  ) + 13m(H  ) + f (|Q ∗ |) + def H  (Y ).

As observed earlier, |E ∗H  (Y )| = |Y |, implying that def H  (Y ) < 0. In fact, by Claim J, def H  (Y ) = 8|Y4 | + 5|Y14 | + 4|Y11 | + |Y21 | − 13|Y | = −5|Y4 | − 8|Y14 | − 9|Y11 | − 12|Y21 |. Since |Q ∗ | ≥ 2, we note that f (|Q ∗ |) ≤ 33. Suppose that |Y | ≥ 2. In this case, def H  (Y ) ≤ −10, implying that f (|Q ∗ |) + def H  (Y ) ≤ 33 − 10 = 22 ≤ f (|∂(Z )|). By supposition, the transversal T  in H  does not satisfy (a) or (b) in the statement of the claim. Hence we immediately obtain a contradiction unless f (|Q ∗ |) = 33, def H  (Y ) = −10 and |∂(Z )| ≥ 5. This

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in turn implies that |Y | = 2, |Y4 | = 2, and |Q ∗ | = 2. However if |Y | = |Y4 | = 2 and |E H  (Y )| = 2, then by the linearity of H the two edges intersecting Y both have two vertices not in Y , implying that |Q ∗ | ≥ 3, a contradiction to |Q ∗ | = 2. Therefore, |Y | = 1. Let E ∗H  (Y ) = { f }. Recall that by supposition, |∂(Z ) \ V (Y )| ≥ 2. Suppose that |V ( f ) ∩ V (Y )| ≥ 2. By the linearity of H we note that Y4 = ∅. Thus, def H  (Y ) = −8|Y14 | − 9|Y11 | − 12|Y21 | ≤ −8, and so 45τ (Y ) = 6n(Y ) + 13m(Y ) + def H  (Y ) ≤ 6n(Y ) + 13m(Y ) − 8. By Observation 8.1(k) we can find a Y -transversal, TY , that contains a vertex in ∂(Z ) and a vertex in f . Let H  be the hypergraph obtained from H  by removing the vertices V (Y ) and edges E(Y ) ∪ E ∗H  (Y ); that is, H  = H  − TY . If def(H  ) > 0, then letting Y  be a special H  -set with def H  (Y  ) > 0 we note that |E ∗H  (Y ∪ Y  )| ≤ |E ∗H  (Y )| + |E ∗H  (Y  )| ≤ 1 + (|Y  | − 1) < |Y  ∪ Y |, contradicting Claim L.1. Therefore, def(H  ) = 0, and so 45τ (H  ) ≤ 6n(H  ) + 13m(H  ) + def(H  ) = 6n(H  ) + 13m(H  ). Hence, 45|T  | ≤ ≤ =
9 and let Ye f be a special He f -set with def He f (Ye f ) > 9. By Claim J, (Ye f )10 = ∅, implying that |E ∗He f (Ye f )| ≤ |Ye f | − 2. If g∈ / E(Ye f ), then |E H  (Ye f )| ≤ |E He f (Ye f )| + |{ f }| ≤ (|Ye f | − 2) + 1 < |Ye f |, contradicting Claim L.1. Therefore, g ∈ E(Ye f ). Since g has two degree-1 vertices in He f , we note that g is the edge of a H4 -hypergraph, Rg , in Ye f . In this case, |E ∗H  (Ye f \ {Rg })| ≤ |E ∗He f (Ye f )| + |{ f }| ≤ (|Ye f | − 2) + 1 = |Ye f \ {Rg }|. As {u, v} ⊆ ∂(Z ) and u and v do not belong to Ye f \ {Rg } (as they belong to V (Rg )) we obtain a contradiction to Claim L.3. Therefore, def(He f ) ≤ 9. Thus, by the minimality of n(H ) + m(H ), 45|T  | ≤ 45τ (He f ) ≤ ≤ = ≤ ≤

6n(He f ) + 13m(He f ) + def(He f ) 6(n(H  ) + 2) + 13m(H  ) + def(He f ) 6n(H  ) + 13m(H  ) + 12 + def(He f ) 6n(H  ) + 13m(H  ) + 21 6n(H  ) + 13m(H  ) + f (|∂(Z )|) − 1,

() a contradiction. This completes the proof of Claim L. We call a component of a 4-uniform, linear hypergraph that contains two vertexdisjoint copies of H4 that are both intersected by a common edge and such that each copy of H4 has three vertices of degree 1 and one vertex of degree 2 a double-H4 -component. We call these two copies of H4 the H4 -pair of the double-H4 component, and the edge that intersects them the linking edge. We note that a doubleH4 -component contains at least ten vertices, namely, eight vertices from the H4 -pair and at least two additional vertices that belong to the linking edge.

Claim M: If x is an arbitrary vertex of H of degree 3, then one of the following holds. (a) def(H − x) = 8 and the hypergraph H − x contains an H4 -component that is intersected by all three edges incident with x. (b) def(H − x) = 3 and the hypergraph H − x contains a double-H4 -component. Further, the H4 -pair in this component is intersected by all three edges incident with x. Proof of Claim M: Let x be an arbitrary vertex of H of degree 3 and let e1 , e2 , and e3 be the three edges incident with x. By the linearity of H , the vertex x is the only common vertex in ei and e j for 1 ≤ i < j ≤ 3. Let H  = H − x. We note that n(H  ) = n(H ) − 1 and m(H  ) = m(H ) − 3. Every transversal in H  can be extended to a transversal in H by adding to it the vertex x. Hence, since H  is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H  ) + 1) ≤

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6n(H  ) + 13m(H  ) + def(H  ) + 45 ≤ 6n(H ) + 13m(H ) + def(H  ). If def(H  ) = 0, then 45τ (H ) ≤ 6n(H ) + 13m(H ), implying that ξ(H ) ≤ 0, contradicting the fact that H is a counterexample to the theorem. Hence, def(H  ) > 0. Let X be a special H  -set satisfying def(H  ) = def H  (X ). Claim M.1: |E ∗H  (X )| = |X | − 1. Further, all three edges incident with the vertex x intersect X . Proof of Claim M.1: Since def(H  ) > 0, we note that |E ∗H  (X )| ≤ |X | − 1. By Claim I and since E ∗H (X ) \ E ∗H  (X ) ⊆ {e1 , e2 , e3 }, we note that |X | + 2 ≤ |E ∗H (X )| ≤ |E ∗H  (X )| + 3, implying that |E ∗H  (X )| ≥ |X | − 1. Consequently, |E ∗H  (X )| = () |X | − 1 and all three edges incident with the vertex x intersect X . Claim M.2: |X | = 1 or |X | = 2 and X = X 4 . Proof of Claim M.2: By Claim M.1, |E ∗H  (X )| = |X | − 1 and all three edges incident with the vertex x intersect X . By Claim J, there is no H10 -subhypergraph in H . Thus, def H  (X ) = 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | − 13|E ∗H  (X )| = 8|X 4 | + 5|X 14 | + 4|X 11 | + |X 21 | − 13(|X | − 1) = 13 − 5|X 4 | − 8|X 14 | − 9|X 11 | − 12|X 21 |. Since def H  (X ) > 0, either |X | = 1 or |X | = 2 and X = X 4 .

()

Claim M.3: If |X | = 2, then H − x contains a double-H4 -component and def(H  ) = 3. Proof of Claim M.3: Suppose that |X | = 2. By Claim M.2, X = X 4 , and so def H  (X ) = 3. By Claim M.1, |E ∗H  (X )| = 1. Let e be the edge in E ∗H  (X ). If e intersects only one of the copies of H4 in X , then the other copy of H4 is an H4 -component in H − x implying that def(H  ) ≥ 8, contradicting the fact that def(H  ) = def H  (X ) = 3. Therefore, e intersects both copies of H4 in X . By the linearity of H the edge e intersects each copy in one vertex, implying that H − x () contains a double-H4 -component. Claim M.4: If |X | = 1, then H − x contains an H4 -component and def(H  ) = 8. Proof of Claim M.4: Suppose that |X | = 1, and let R be the special subhypergraph in X . By Claim M.1 we note that |E ∗H  (X )| = 0, and so R is a component of H − x. Suppose, to the contrary, that |X 4 | = 0. Among all degree-3 vertices, we choose the vertex x so that X = X j where j is a maximum; that is, X is chosen to contain a special subhypergraph of maximum possible size j. By supposition, j = 4, and so j ∈ {11, 14, 21}. Suppose that each edge incident with x intersects R in at most two vertices. By Claim M.1, all three edges incident with x intersect R in at least one vertex, implying that in this case at least three neighbors of x in H belong to R. Since every special subhypergraph different from H4 contains at most two vertices of degree 1, we can choose a neighbor y of x in R such that d H  (y) ≥ 2. Since every neighbor of x

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in R has degree at most 2 in R since H has maximum degree 3, this implies that d H  (y) = 2, and so d H (y) = 3. This implies by Observation 8.1(p) that H  − y is connected or is disconnected with exactly two components, one of which consists of an isolated vertex. In both cases, since all three edges incident with x intersect R we note that either H − y is connected and has cardinality at least n(R) + 3 or H − y is disconnected with two components, one of which consists of an isolated vertex with the other component of cardinality at least n(R) + 2. Analogously as with the vertex x, there is a special (H − y)-set, Y , with def(H − y) = def H −y (Y ) satisfying |Y | = 1 or |Y | = 2 and Y = Y4 . If |Y | = 1, then by our earlier observations, Y consists of a special subhypergraph of cardinality at least n(R) + 2, contradicting the maximality of R. Hence, |Y | = 2 and Y = Y4 . Analogously as with the vertex x (see Claim M.3), H − y contains a double-H4 -component. Let R1 and R2 be the H4 -pair of this double-H4 -component, and let h ∗ be the linking edge that intersects them. We note that Y = {R1 , R2 }. By Observation 8.1(p), R − y does not contain a double-H4 -component. Further, R − y does not contain a component of order 4. Suppose that the edge of R1 belongs to R. If the edge h ∗ does not belong to R, then R1 is a component in R − y (of order 4), a contradiction. Hence, the edge h ∗ also belongs to R. If the edge in R2 belongs to R, then R − y would contain a double-H4 -component, a contradiction. Hence, the edge in R2 does not belong to R and must therefore contain the vertex x. However in this case R2 is intersected by at least two edges in R − y, namely, h ∗ and the edge incident with x that intersects R2 . Therefore, the pair R1 and R2 is not the H4 -pair of a double-H4 -component in H − y, a contradiction. Therefore, the edge in R1 does not belong to R. Analogously, the edge in R2 does not belong to R. Since the H4 -pair, R1 and R2 , in this double-H4 -component is intersected by all three edges incident with y, we note that both R1 and R2 intersect V (R). However as they do not belong to R, they must both contain the vertex x, which implies that they are not vertex-disjoint, a contradiction. Therefore, at least one edge incident with x intersects R in three vertices. Renaming edges if necessary, we may assume that the edge e1 intersects R in three vertices. From the structure of special subhypergraphs, this implies that we can choose a vertex y ∈ V (e1 ) ∩ V (R) so that R − y is connected (of cardinality n(R) − 1). If at least one neighbor of x does not belong to R, then analogously as before H − y is a connected hypergraph of cardinality at least n(R) + 1, contradicting the maximality of R. Hence, letting Ti = V (ei ) ∩ V (R), we note that |Ti | = 3 for all i ∈ [3]. Further, by the linearity of H , the sets T1 , T2 , and T3 are vertex-disjoint strongly independent sets in R. By Observation 8.1(n) there exists a τ -transversal of R, T ∗ say, that contains a vertex from each of the sets Ti for i ∈ [3]. Since H is connected and all neighbors of x belong to R, we note that H  = R, V (H ) = V (R) ∪ {x}, and E(H ) = E(R) ∪ {e1 , e2 , e3 }. Thus, T ∗ is a transversal of H , implying that 45τ (H ) ≤ 45|T ∗ | = 45τ (H  ) ≤ 6n(H  ) + 13m(H  ) + def(H  ) = 6n(H ) + 13m(H ) + def(H  ) − 45. By Claim M.2, def(H  ) ≤ 8, and so 45τ (H ) < 6n(H ) + 13m(H ), a contradiction. Therefore, X = X 4 , implying that R = H4 is an H4 -component of H − x and def(H  ) = 13 − 5|X 4 | = 8. ()

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By Claims M.2, M.3, and M.4, H − x contains an H4 -component or a double() H4 -component. This completes the proof of Claim M. Claim N: No edge in H contains two degree-1 vertices of H . Proof of Claim N: Suppose, to the contrary, that e = {v1 , v2 , v3 , v4 } is an edge in H , where 1 = d H (v1 ) = d H (v2 ) ≤ d H (v3 ) ≤ d H (v4 ). Suppose that d H (v4 ) = 3. Let H  be obtained from H − v4 by deleting all resulting isolated vertices (including v1 and v2 ). By Claim M, def(H  ) ≤ 8. We note that n(H  ) ≤ n(H ) − 3 and m(H  ) = m(H ) − 3. Since H  is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H  ) + 1) ≤ 6n(H  ) + 13m(H  ) + def(H  ) + 45 ≤ 6n(H ) − 3 × 6 + 13m(H ) − 3 × 13 + def(H  ) + 45 ≤ 6n(H ) + 13m(H ) − 18 − 39 + 8 + 45 < 6n(H ) + 13m(H ), a contradiction. Therefore, d H (v4 ) ≤ 2, implying that at most two edges of H intersect the edge e. Taking Y to the special H -set consisting only of the special subhypergraph H4 with e as its edge, we have () |E ∗ (Y )| ≤ 2 = |Y | + 1, contradicting Claim I(a). Claim O: Let H  be a 4-uniform, linear hypergraph with no H10 -subhypergraph satisfying def(H  ) > 0. If Y is a special H  -set satisfying def(H  ) = def H  (Y ), then |E ∗H  (Y )| = |Y | − i for some i ≥ 1 and def(H  ) ≤ 13i − 5|Y4 | − 8|Y14 | − 9|Y11 | − 12|Y21 |. In particular, def(H  ) ≤ 13i − 5|Y |. Further, if def(H  ) > 8 j for some j ≥ 0, then |E ∗H  (Y )| ≤ |Y | − ( j + 1). Proof of Claim O: Since def H  (Y ) > 0, we note that |E ∗H  (Y )| = |Y | − i for some i ≥ 1. Thus, def H  (Y ) = 8|Y4 | + 5|Y14 | + 4|Y11 | + |Y21 | − 13|E ∗H  (Y )| = 8|Y4 | + 5|Y14 | + 4|Y11 | + |Y21 | − 13(|Y | − i) = 13i − 5|Y4 | − 8|Y14 | − 9|Y11 | − 12|Y21 |, and so def H  (Y ) ≤ 13i − 5|Y |. Further, let def(H  ) > 8 j for some j ≥ 0 and suppose, to the contrary, that ∗ |E H  (Y )| ≥ |Y | − j. Hence, |E ∗H  (Y )| = |Y | − k where k ≤ j. Therefore, def(H  ) ≤ 13k − 5|Y | ≤ 13k − 5k = 8k ≤ 8 j, a contradiction. Therefore, |E ∗H  (Y )| ≤ |Y | − () ( j + 1). We show next that the removal of any specified vertex of degree 3 from H produces an H4 -component. In what follows, we use the following notation for simplicity. If e and f are intersecting edges of H , then we denote the vertex in this intersection by (e f ); that is, e ∩ f = {(e f )}. Claim P: If x is an arbitrary vertex of H of degree 3, then H − x contains an H4 -component. Proof of Claim P: Let x be an arbitrary vertex of H of degree 3. Let e1 , e2 , and e3 be the three edges incident with x and let H  = H − x. By the linearity of H , the vertex x is the only common vertex in ei and e j for 1 ≤ i < j ≤ 3. Suppose, to the contrary, that H  does not contain an H4 -component. By Claim M, H  therefore

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contains a double-H4 -component, say C  , and def(H  ) = 3. Let f 1 and f 2 be the two edges belonging to the H4 -pair in C  and let h be the linking edge in H  that intersects f 1 and f 2 . By definition of a double-H4 -component, we note that f i has three vertices of degree 1 and one vertex of degree 2 in H  for i ∈ [2]. We proceed further with the following series of subclaims. Claim P.1: 2 ≤ | f i ∩ N H (x)| ≤ 3 for i ∈ [2]. Proof of Claim P.1: By the linearity of H , | f i ∩ e j | ≤ 1 for every j ∈ [3], and so | f i ∩ N H (x)| ≤ 3. If | f i ∩ N H (x)| ≤ 1, then the edge f i contains at least two () vertices of degree 1 in H , contradicting Claim N. Claim P.2: Any two edges among e1 , e2 , and e3 can be covered by two vertices one of which belongs to f 1 and the other to f 2 . Proof of Claim P.2: Renaming edges if necessary, it suffices to consider the two edges e1 and e2 . By Claim P.1, the edge f 1 intersects at least one of e1 and e2 , say e1 . If f 2 intersects e2 , then we can cover e1 by the vertex (e1 f 1 ) and we can cover e2 by the vertex (e2 f 2 ). If f 2 does not intersect e2 , then by Claim M and Claim P.1, the edge f 1 intersects e2 and the edge f 2 intersects e1 , and we can therefore cover e1 by () the vertex (e1 f 2 ) and we can cover e2 by the vertex (e2 f 1 ). Let h \ ( f 1 ∪ f 2 ) = {a, b}, and let h ∩ f 1 = {c} and h ∩ f 2 = {d}. Thus, h = {a, b, c, d}. By definition of a double-H4 -component, we note that d H (c) = d H  (c) = 2 and d H (d) = d H  (d) = 2. Claim P.3: d H  (a) ≥ 2 and d H  (b) ≥ 2. Proof of Claim P.3: Suppose, to the contrary, that d H  (a) = 1 or d H  (b) = 1. Interchanging the names of a and b if necessary, we may assume that d H  (a) = 1. Let H  be obtained from H − {x, c, d} by deleting all resulting isolated vertices (including the vertex a, as well as three vertices of degree 1 in each of f 1 and f 2 ). We note that n(H  ) ≤ n(H  ) − 10. Further, the edges e1 , e2 , e3 , f 1 , f 2 , and h are deleted from H when constructing H  , and so m(H  ) = m(H ) − 6. Since H  is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H  ) + 3) ≤ 6n(H  ) + 13m(H  ) + def(H  ) + 135 ≤ 6n(H ) − 6 × 10 + 13m(H ) − 13 × 6 + def(H  ) + 135 = 6n(H ) + 13m(H ) + def(H  ) − 3. If def(H  ) ≤ 3, then 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, def(H  ) ≥ 4. Let Y  be a special H  -set satisfying def H  (Y ) = def(H  ) > 0. By Claim O, |E ∗H  (Y  )| ≤ |Y  | − 1. Let Yi be the H4 -subhypergraph of H with E(Yi ) = { f i } for i ∈ [2], and consider the special H -set Y = Y  ∪ {Y1 , Y2 }. We note that |E ∗H (Y )| ≤ |E ∗H  (Y  )| + |{e1 , e2 , e3 , h}| ≤ (|Y  | − 1) + 4 = |Y  | + 3 = |Y | + 1, () contradicting Claim I(a). As observed earlier, the edge f i intersects the edge e j in at most one vertex for every i ∈ [2] and j ∈ [3]. For j ∈ [3], let s j be a vertex different from x that belongs to e j but not to f 1 ∪ f 2 .

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Claim P.4: The following holds. (a) d H (si ) ≥ 2 for all i ∈ [3]. (b) If d H (si ) = 3 for some i ∈ [3], then ei ∩ f 1 = ∅ or ei ∩ f 2 = ∅. (c) d H (si ) = 2 for at least one i ∈ [3]. Proof of Claim P.4: (a) Suppose, to the contrary, that the vertex si has degree 1 in H for some i ∈ [3], and so ei is the only edge in H containing si . Letting H  = H − {x, si }, we note that n(H  ) = n(H ) − 2, m(H  ) = m(H ) − 3 and def(H  ) = def(H  ) = 3. Every transversal in H  can be extended to a transversal in H by adding to it the vertex x. Hence, since H  is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H  ) + 1) ≤ 6n(H  ) + 13m(H  ) + def(H  ) + 45 ≤ 6n(H )+ 3m(H ) − 6 · 2 − 13 · 3 + 3 + 45 = 6n(H ) + 13m(H ) − 3, a contradiction. Therefore, d H (si ) ≥ 2 for all i ∈ [3]. (b) Let d H (si ) = 3 for some i ∈ [3], and suppose, to the contrary, both edges f 1 and f 2 intersect ei . Renaming indices if necessary, we may assume i = 1. By Claim M, H − s1 either contains an H4 -component that is intersected by the edge e1 or a double-H4 -component in which the H4 -pair is intersected by the edge e1 . However such a component of H − s1 would contain the vertex x and the four edges e2 , e3 , f 1 , f 2 , which is not possible. (c) Suppose that d H (s1 ) = d H (s2 ) = d H (s3 ) = 3. By Claim P.1, ei ∩ f 1 = ∅ and () ei ∩ f 2 = ∅ for some i ∈ [3]. This, however, contradicts part (b) above. We show next that the edge h contains no neighbor of x. Claim P.5: N H (x) ∩ V (h) = ∅. Proof of Claim P.5: Suppose, to the contrary, that y ∈ N H (x) ∩ V (h). Claim P.5.1: y ∈ / f1 ∪ f2 . Proof of Claim P.5: Suppose, to the contrary, that y ∈ f 1 ∪ f 2 . Renaming edges if necessary, we may assume that y ∈ f 1 ∩ e3 . Since y belongs to the three edges e3 , f 1 , h, we note that d H (y) = 3. Further since f 1 ∩ h = {y}, we note that the edge f 1 contains a vertex of degree 1 in H that is not a neighbor of x. By Claim N, no edge contains two degree-1 vertices, implying that in this case f 1 intersects all three edges e1 , e2 , and e3 . Let vi = f 1 ∩ ei for i ∈ [3]. By Claim M, H − y contains an H4 -component or a double-H4 -component. If H − y contains an H4 -component R, then by Claim M the component R intersects all three edges incident with y. In particular, R intersects the edge f 1 and therefore contains the vertex v1 or v2 . In both cases, the component R contains the vertex x and both edges e1 and e2 , a contradiction. Hence, H − y contains a double-H4 -component R  . By Claim M, all three edges incident with y intersect the H4 -pair in R  . In particular, the edge f 1 intersects the H4 -pair in R  , implying that e1 or e2 belongs to the H4 -pair in R  . In both cases, the component R  contains the vertex x and both edges e1 and e2 . Renaming e1 and e2 if necessary, we may assume that e1 belongs to the H4 -pair in R  , and so e1 contains three vertices of degree 1 and one vertex, namely, x, of degree 2 in H − y. Let e1 = {x, v1 , u, v}. We note that neither u nor v belongs to f 1 or

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e3 , and at most one of u and v belong to h. Thus, at least one of u and v, say v, has degree 1 in H . Thus, v is isolated in H  = H − x. Hence, letting H  = H − {x, y}, we note that n(H  ) = n(H ) − 2, m(H  ) = m(H ) − 3 and def(H  ) = def(H  ) = 3. Every transversal in H  can be extended to a transversal in H by adding to it the vertex x. Hence, since H  is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H  ) + 1) ≤ 6n(H  ) + 13m(H  ) + def(H  ) + 45 ≤ 6n(H ) − 6 · 2 + 13m(H ) − 13 · 3 + def(H  ) + 45 = 6n(H ) + 13m(H ) − 3, () a contradiction. By Claim P.5.1, y ∈ / f 1 ∪ f 2 . Thus, y ∈ {a, d}. Recall that y ∈ N H (x), and so y is incident with e1 , e2 , or e3 . Thus, by Claim P.3, d H (y) = d H  (y) + 1 ≥ 3. Consequently, d H (y) = 3. Renaming edges if necessary, we may assume that y is incident with e1 . Let g be the third edge (different from e1 and h) incident with y. By Claim P.2, the edges e2 and e3 can be covered by two vertices one of which belongs to f 1 and the other to f 2 . Thus, e2 ∩ f i = ∅ and e3 ∩ f 3−i = ∅ for some i ∈ [2]. Let Z = f 1 ∪ f 2 ∪ {x, y} and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 10. The edges e1 , e2 , e3 , f 1 , f 2 , g, h are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 7. Every transversal in H ∗ can be extended to a transversal in H by adding to it the vertices y, (e2 f i ) and (e3 f 3−i ). Hence, since H ∗ is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H ∗ ) + 3) ≤ 6n(H ∗ ) + 13m(H ∗ ) + def(H ∗ ) + 45 ≤ 6n(H ) + 13m(H ) − 6 · 10 − 13 · 7 + def(H ∗ ) + 45 × 3 = 6n(H ) + 13m(H )+def(H ∗ ) + 16. If def(H ∗ ) ≤ 16, then 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, def(H ∗ ) ≥ 17. Recall that H  = H − x and note that H ∗ is obtained from H  by deleting the edges f 1 , f 2 , g, h and the resulting isolated vertex y. We now consider the hypergraph H  obtained from H  by deleting the edges f 1 , f 2 , h; that is, H  = H  − f 1 − f 2 − h. We note that def(H  ) ≥ def(H ∗ ) − 13 ≥ 4 since the edge g, which contributes at most 13 to def(H ∗ ), has been added back to H ∗ (along with the vertex y). Let Y  be a special H  -set satisfying def H  (Y  ) = def(H  ). Let Yi be the H4 -subhypergraph of H with E(Yi ) = { f i } for i ∈ [2], and consider the special H  -set Y = Y  ∪ {Y1 , Y2 }. We note that each of Y1 and Y2 contribute 8 to def H  (Y ), and the edge h contributes −13 to def H  (Y ), implying that def H  (Y ) = def H  (Y  ) + 8 + 8 − 13 ≥ 4 + 8 + 8 − 13 = 7, contradicting () the fact that def(H  ) = 3. This completes the proof of Claim P.5. For i ∈ [3], let si be a vertex in ei \ {x} that does not belong to f 1 ∪ f 2 . By Claim P.5, the edge h contains no neighbor of x in H . In particular, h ∩ {s1 , s2 , s3 } = ∅. Further, we note by Claim P.5 that every vertex that belongs to the edge h has the same degree in H and in H  , that is, d H (v) = d H  (v) for all v ∈ {a, b, c, d} = V (h). Recall that d H (c) = d H (d) = 2 and by Claim P.3, d H (a) ≥ 2 and d H (b) ≥ 2. Claim P.6: d H (a) = 2 and d H (b) = 2. Proof of Claim P.6: Suppose, to the contrary, that d H (a) = 3 or d H (b) = 3. Interchanging the names of a and b if necessary, we may assume that d H (a) = 3. Let h, h 1 , and h 2 be the three edges incident with a. By Claim M, def(H − a) = 3 or def(H − a) = 8. We consider the two possibilities in turn and show that neither case can occur.

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Claim P.6.1: The case def(H − a) = 3 cannot occur. Proof of Claim P.6.1: Suppose that def(H − a) = 3. By Claim M, H − a therefore contains a double-H4 -component, say C ∗ . Let g1 and g2 be the two edges belonging to the H4 -pair in C ∗ and let h ∗ be the linking edge in H − a that intersects g1 and g2 . By definition of a double-H4 -component, we note that gi has three vertices of degree 1 and one vertex of degree 2 in H − a for i ∈ [2]. By Claim M, the edges g1 and g2 combined are intersected by all three edges h, h 1 , h 2 incident with a. Analogously as in the proof of Claim P.1, 2 ≤ |gi ∩ N H (a)| ≤ 3 for i ∈ [2]. By Claim P.1 and P.5, we note that the edges g1 and g2 are distinct from the edges f 1 and f 2 . Further, we note that the vertices c and d both have degree 1 in H − a and belong to the edges f 1 and f 2 , respectively, in H − a. Thus, neither g1 nor g2 contains the vertex c or d, while at least one of g1 and g2 contains the vertex b. Renaming g1 and g2 if necessary, we may assume that the vertex b belongs to the edge g1 . Thus, b has degree 1 in H − a. Let Z = f 1 ∪ f 2 ∪ g1 ∪ g2 ∪ {x, a} and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 18. The edges e1 , e2 , e3 , f 1 , f 2 , g1 , g2 , h, h 1 , h 2 , h ∗ are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 11. We show firstly that def(H ∗ ) ≤ 21. Suppose, to the contrary, that def(H ∗ ) ≥ 22. Let Y ∗ be a special H ∗ -set satisfying def H ∗ (Y ∗ ) = def(H ∗ ). By Claim O, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Let Y1 , Y2 , Y3 , Y4 be the H4 -subhypergraph of H with edge set f 1 , f 2 , g1 , g2 , respectively, and consider the special H -set Y = Y ∗ ∪ {Y1 , Y2 , Y3 , Y4 }. We note that |E ∗H (Y )| ≤ |E ∗H ∗ (Y ∗ )| + |{e1 , e2 , e3 , h, h 1 , h 2 , h ∗ }| ≤ (|Y ∗ | − 3)+7 = |Y ∗ | + 4 = |Y |, contradicting Claim I(a). Therefore, def(H ∗ ) ≤ 21. We show next that every transversal in H ∗ that contains a vertex in ∂(Z ) can be extended to a transversal in H by adding to it five vertices. We note that if ∂(Z ) contains a vertex from some edge e of H , then e ∈ {e1 , e2 , e3 , h 1 , h 2 , h ∗ }. Let T ∗ be a transversal in H ∗ that contains a vertex in ∂(Z ). Suppose that T ∗ contains a vertex in e1 , e2 , or e3 . Renaming edges if necessary, we may assume T ∗ contains a vertex in e1 . By Claim P.2, the edges e2 and e3 can be covered by two vertices one of which belongs to f 1 and the other to f 2 . Thus, e2 ∩ f i = ∅ and e3 ∩ f 3−i = ∅ for some i ∈ [2]. In this case, T ∗ ∪ {(e2 f i ), (e3 f 3−i ), a, (g1 h ∗ ), (g2 h ∗ )} is a transversal in H of size |T ∗ | + 5. Suppose that T ∗ contains a vertex in h 1 or h 2 , say in h 1 . We note that h 2 ∩ g j = ∅ for some j ∈ [2]. In this case, T ∗ ∪ {x, c, d, (g j h 2 ), (g3− j h ∗ )} is a transversal in H of size |T ∗ | + 5. Suppose that T ∗ contains a vertex in h ∗ . We note that h 1 ∩ gi = ∅ and h 2 ∩ g3−i = ∅ for some i ∈ [2]. In this case, T ∗ ∪ {x, c, d, (gi h 1 ), (g3−i h 2 )} is a transversal in H of size |T ∗ | + 5. In all cases, T ∗ can be extended to a transversal in H by adding to it five vertices. For i ∈ [2], let θi be a vertex in h i that does not belong to g1 ∪ g2 . Let θ3 and θ4 be the two vertices in h ∗ that do not belong to g1 or g2 . Analogously as in Claim P.5, we note that N H (a) ∩ h ∗ = ∅. In particular, the vertices θ1 , θ2 , θ3 , and θ4 are distinct, implying that |∂(Z )| ≥ 4. Recall that def(H ∗ ) ≤ 21. By Claim L, there exists a transversal, T ∗ , in H ∗ , that contains a vertex in ∂(Z ) such that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|∂(Z )|) ≤ 6n(H ∗ ) + 13m(H ∗ ) + 23. As observed earlier, T ∗ can be extended to a transversal in H by adding to it five vertices. Hence, 45τ (H ) ≤

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45(|T ∗ | + 5) ≤ (6n(H ∗ ) + 13m(H ∗ ) + 23) + 225 = 6n(H ) + 13m(H ) − 6 · 18 − 13 · 11 + 23 + 225 = 6n(H ) + 13m(H ) − 3, a contradiction. Hence the case () def(H − a) = 3 cannot occur. Claim P.6.2: The case def(H − a) = 8 cannot occur. Proof of Claim P.6.2: Suppose that def(H − a) = 8. By Claim M, H − a therefore contains an H4 -component. Let g be the edge belonging in this H4 -component. By Claim M, the edge g is intersected by all three edges h, h 1 , h 2 incident with the vertex a. We note that the edge g is distinct from f 1 and f 2 , and contains the vertex b which therefore has degree 1 in H − a. For i ∈ [2], let z i1 and z i2 be the two vertices in h i \ {a} that do not belong to the edge g, and let z i3 be the vertex (h i g) that is common to h i and g. Thus, h i = {a, z i1 , z i2 , z i3 }. Let Z = f 1 ∪ f 2 ∪ g ∪ {x, a} and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 14. The edges e1 , e2 , e3 , f 1 , f 2 , g, h, h 1 , h 2 are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 9. We show firstly that def(H ∗ ) ≤ 21. Suppose, to the contrary, that def(H ∗ ) ≥ 22. Let Y ∗ be a special H ∗ -set satisfying def H ∗ (Y ∗ ) = def(H ∗ ). By Claim O, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Let R1 , R2 , R3 be the H4 -subhypergraphs of H with edge set f 1 , f 2 , g, respectively, and consider the special H -set Y = Y ∗ ∪ {R1 , R2 , R3 }. We note that |E ∗H (Y )| ≤ |E ∗H ∗ (Y ∗ )| + |{e1 , e2 , e3 , h 1 , h 2 }| ≤ (|Y ∗ | − 3) + 6 = |Y ∗ | + 3 = |Y |, contradicting Claim I(a). Therefore, def(H ∗ ) ≤ 21. We show next that every transversal in H ∗ that contains a vertex in ∂(Z ) can be extended to a transversal in H by adding to it four vertices. We note that if ∂(Z ) contains a vertex from some edge e of H , then e ∈ {e1 , e2 , e3 , h 1 , h 2 }. Let T ∗ be a transversal in H ∗ that contains a vertex in ∂(Z ). Suppose that T ∗ contains a vertex in e1 , e2 , or e3 . Renaming edges if necessary, we may assume T ∗ contains a vertex in e1 . By Claim P.2, e2 ∩ f i = ∅ and e3 ∩ f 3−i = ∅ for some i ∈ [2], implying that T ∗ ∪ {(e2 f i ), (e3 f 3−i ), a, b} is a transversal in H of size |T ∗ | + 4. If T ∗ contains a vertex in h 1 or h 2 , say in h 1 , then T ∗ ∪ {(gh 2 ), x, c, d} is a transversal in H of size |T ∗ | + 4. In all cases, T ∗ can be extended to a transversal in H by adding to it four vertices. Recall that si is a vertex in ei \ {x} that does not belong to f 1 or f 2 . Thus, s1 , s2 , s3 are distinct vertices in ∂(Z ). Recall that def(H ∗ ) ≤ 21. As observed earlier, T ∗ can be extended to a transversal in H by adding to it four vertices, and so τ (H ) ≤ |T ∗ | + 4. If H ∗ contains two intersecting edges e and f , such that ∂(Z ) ⊆ (V (e) ∪ V ( f )) \ (V (e) ∩ V ( f )), then |∂(Z )| ≤ 6. Hence if |∂(Z )| ≥ 7, then by Claim L(b) there exists a transversal, T ∗ , in H ∗ , that contains a vertex in ∂(Z ) such that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + 21. Thus, 45τ (H ) ≤ 45(|T ∗ | + 4) ≤ (6n(H ∗ ) + 13m(H ∗ ) + 21) + 180 = 6n(H ) + 13m(H ) − 6 · 14 − 13 · 9 + 21 + 180 = 6n(H ) + 13m(H ), a contradiction. Hence, |∂(Z )| ≤ 6. Renaming vertices if necessary, we may assume that s1 = z 11 . If d H (s1 ) = 2, then the vertex s1 is isolated in H ∗ . In this case, adding s1 to the set Z we note that n(H ∗ ) = n(H ) − 15. Further z 12 , z 21 , z 22 are distinct vertices in ∂(Z ), and so |∂(Z )| ≥ 3 and f (|∂(Z )|) ≤ 27. By Claim L, there exists a transversal, T ∗ , in H ∗ , that contains a vertex in ∂(Z ) such that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) +

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f (|∂(Z )|) ≤ 6n(H ) + 13m(H ) + 27 − 6 · 15 − 13 · 9 = 6n(H ) + 13m(H ) − 180, implying that 45τ (H ) ≤ 45(|T ∗ | + 4) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, d H (s1 ) = 3. Since z 11 , z 12 , z 21 , z 22 are distinct vertices in ∂(Z ), we note that |∂(Z )| ≥ 4. If |∂(Z )| = 4, then {s1 , s2 , s3 } ⊂ {z 11 , z 12 , z 21 , z 22 }, implying analogously as above that d H (si ) = 3 for all i ∈ [3], contradicting Claim P.5. Therefore, |∂(Z )| ≥ 5, and so f (|∂(Z )|) = 22. If the statement of Claim L(b) holds, then 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + 21, implying as before that 45τ (H ) ≤ 45(|T ∗ | + 4) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, H ∗ contains two intersecting edges e and f having properties (i), (ii), and (iii) in Claim L(b). Recall that (e f ) denotes the vertex in the intersection of e and f . Thus, ∂(Z ) ⊆ (V (e) ∪ V ( f )) \ {(e f )}, e contains three degree-1 vertices, and |∂(Z ) ∩ V (e)|, |∂(Z ) ∩ V ( f )| ≥ 2. Renaming vertices if necessary, we may assume that s1 = z 11 . If z 11 ∈ e, let z ∈ ∂(Z ) ∩ V ( f ), while if z 11 ∈ f , let z ∈ ∂(Z ) ∩ V (e). Let Z • = Z ∪ (e \ {(e f )}) ∪ {z} and let H • = H − Z • . We note that n(H • ) = n(H ) − 14 − 4 = n(H ) − 18. The edges e, e1 , e2 , e3 , f, f 1 , f 2 , g, h, h 1 , h 2 are deleted from H when constructing H • , and so m(H • ) = m(H ) − 11. By Claim P.2, e2 ∩ f i = ∅ and e3 ∩ f 3−i = ∅ for some i ∈ [2]. Every transversal in H • can be extended to a transversal in H by adding to it the five vertices in the set {b, (e2 f i ), (e3 f 3−i ), s1 , z}, and so τ (H ) ≤ τ (H • ) + 5. Since H • is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H • ) + 5) ≤ 6n(H • ) + 13m(H • ) + def(H • ) + 225 ≤ 6n(H ) + 13m(H ) + def(H • ) − 6 × 18 − 13 × 11 + 225 ≤ 6n(H ) + 13m(H ) + def(H • ) − 26. If def(H • ) ≤ 26, then 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, def(H • ) ≥ 27. Let Y • be a special H • -set satisfying def(H • ) = def H • (Y • ). By Claim O, |E ∗H • (Y • )| ≤ |Y ∗ | − 4. Let Y1 , Y2 , Y3 be the H4 -subhypergraphs of H with edge set f 1 , f 2 , g, respectively, and consider the special H -set Y = Y • ∪ {Y1 , Y2 , Y3 }. We note that |E ∗H (Y )| ≤ |E ∗H • (Y • )| + |{e, e1 , e2 , e3 , f, h, h 1 , h 2 }| ≤ (|Y • | − 4) + 8 = |Y • | + 4 = |Y | + 1, contradicting Claim I(a). Hence the case def(H − a) = 8 () cannot occur. Since neither the case def(H − a) = 3 nor the case def(H − a) = 8 can occur, () this completes the proof of Claim P.6. Recall that si is a vertex in V (ei ) \ {x} that does not belong to f 1 or to f 2 , and that by Claim P.4, d H (si ) ≥ 2. Claim P.7: If d H (si ) = 3 and d H (s j ) = 2 for some i and j where 1 ≤ i, j ≤ 3, then the vertices si and s j do not belong to a common edge. Proof of Claim P.7: Renaming edges e1 , e2 , e3 if necessary, let d H (s1 ) = 3 and d H (s2 ) = 2 and suppose, to the contrary, that there is an edge g containing both s1 and s2 . Let g • be the third edge containing s1 different from e1 and g. By Claim M, def(H − s1 ) = 3 or def(H − s1 ) = 8. We consider the two possibilities in turn and show that neither case can occur. Claim P.7.1: The case def(H − s1 ) = 3 cannot occur. Proof of Claim P.7.1: Suppose that def(H − s1 ) = 3. By Claim M, H − s1 therefore contains a double-H4 -component, say F. Let q1 and q2 be the two edges belong-

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ing to the H4 -pair in F and let q • be the linking edge in H − s1 that intersects q1 and q2 . Recall that V (h) = {a, b, c, d}, where the edges f 1 and h intersect in the vertex c and where the edges f 2 and h intersect in the vertex d. By Claim P.6 and our earlier observations, d H (v) = 2 for all v ∈ V (h). Let q • ∩ q1 = {cq } and q • ∩ q2 = {dq }. Let Z = ( f 1 ∪ f 2 ∪ q1 ∪ q2 ∪ {x, s1 , s2 }) \ {c, d, cq , dq } and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 15. The edges e1 , e2 , e3 , f 1 , f 2 , g, g • , q1 , q2 are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 9. Claim P.7.1.1: def(H ∗ ) ≤ 21. Proof of Claim P.7.1.1: Suppose, to the contrary, that def(H ∗ ) ≥ 22. Let Y ∗ be a special H ∗ -set satisfying def H ∗ (Y ∗ ) = def(H ∗ ). By Claim O, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Let F1 , F2 , F3 , F4 be the H4 -subhypergraphs of H with edge set f 1 , f 2 , q1 , q2 , respectively. Let Y ∗∗ be the special H ∗ -set obtained from Y ∗ by removing all special subhypergraphs in Y ∗ , if any, that contain the edge h or the edge q • , and consider the special H -set Y = Y ∗∗ ∪ {F1 , F2 , F3 , F4 }. Since we remove at most two special subhypergraphs from Y ∗ when constructing Y , we note that |Y | ≥ |Y ∗ | − 2. Further, we note that |E ∗H ∗ (Y ∗∗ )| ≤ |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3, and |E ∗H (Y )| ≤ |E ∗H ∗ (Y ∗∗ )| + |{e1 , e2 , e3 , g, g • , h, q • }| ≤ (|Y ∗ | − 3) + 7 = |Y ∗ | + 4. If |Y | ≥ |Y ∗ | − 1 or if |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 4, then |E ∗H (Y )| ≤ |Y | + 1, contradicting Claim I(a). Therefore, |Y | = |Y ∗ | − 2 and |E ∗H ∗ (Y ∗ )| = |Y ∗ | − 3. In particular, since |E ∗H ∗ (Y ∗ )| ≥ 0, we note that |Y ∗ | ≥ 3. By Claim O, 22 ≤ ∗ ∗ ∗ | − 9|Y11 | − 12|Y21 |, implying that |Y ∗ | = |Y4∗ | = def(H ∗ ) ≤ 13 · 3 − 5|Y4∗ | − 8|Y14 ∗ ∗ ∗ 3 and |E H ∗ (Y )| = 0. Let Y = {R1 , R2 , R3 }, where R1 and R2 are the special subhypergraphs in Y ∗ containing the edges h and q • , respectively. We note that all three subhypergraphs R1 , R2 , and R3 are H4 -components in H ∗ . In particular, E(R1 ) = {h} and E(R2 ) = {q • }. Let E(R3 ) = {θ }. If q1 or q2 intersects the edge h, then h = q • , a contradiction. Hence, neither q1 nor q2 intersects h. By Claim P.6, d H (a) = 2 and d H (b) = 2, implying by the linearity of H that both edges g and g • intersect h. Analogously, both edges e2 and e3 intersect the edge q • . Thus, no vertex in N H [x] belongs to the edge θ , and no vertex in N H [s1 ] belongs to the edge θ . Thus, no edge in H intersects the edge θ , implying that R3 is an H4 -component in H , a contradiction. This completes the proof () of Claim P.7.1.1. Claim P.7.1.2: Every transversal in H ∗ that contains a vertex in ∂(Z ) can be extended to a transversal in H by adding to it four vertices. Proof of Claim P.7.1.2: We note that if ∂(Z ) contains a vertex from some edge e of H , then e ∈ {e2 , e3 , f 1 , f 2 , g, g • , q1 , q2 }. Let T ∗ be a transversal in H ∗ that contains a vertex in ∂(Z ). By Claim P.2, any two edges among e1 , e2 , and e3 (incident with x) can be covered by two vertices one of which belongs to f 1 and the other to f 2 . Analogously, any two edges among e1 , g, and g • (incident with s1 ) can be covered by two vertices one of which belongs to q1 and the other to q2 . If T ∗ contains a vertex in f 1 or f 2 , say in f 1 , then T ∗ ∪ {( f 2 h), (gqi ), (g • q3−i ), x} is a transversal in H of size |T ∗ | + 4 for some i ∈ [2]. If T ∗ contains a vertex in q1 or q2 , say in q1 , then T ∗ ∪ {(e2 f i ), (e3 f 3−i ), s1 , v2 } is a transversal in H of size |T ∗ | + 4 for some i ∈ [2],

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where v2 is an arbitrary vertex in q2 . If T ∗ contains a vertex in e2 or e3 , say in e2 , then T ∗ ∪ {(e1 f i ), (e3 f 3−i ), (gq j ), (g • q3− j )} is a transversal in H of size |T ∗ | + 4 for some i ∈ [2] and j ∈ [2]. If T ∗ contains a vertex in g or g • , say in g, then T ∗ ∪ {(g • qi ), (e1 q3−i ), (gq j ), (e2 f j ), (e3 f 3− j )} is a transversal in H of size |T ∗ | + 4 for some i ∈ [2] and j ∈ [2]. In all cases, T ∗ can be extended to a transversal in H () by adding to it four vertices. This completes the proof of Claim P.7.1.2. By Claim P.5, the vertex s3 does not belong to the edge h, implying that {a, b, s3 } ⊆ ∂(Z ), and so |∂(Z )| ≥ 3 and f (|∂(Z )|) ≤ 27. By Claim P.7.1.1, def(H ∗ ) ≤ 21. By Claim L, there exists a transversal, T ∗ , in H ∗ , that contains a vertex in ∂(Z ) such that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|∂(Z )|) ≤ 6n(H ∗ ) + 13m(H ∗ ) + 27. By Claim P.7.1.2, T ∗ can be extended to a transversal in H by adding to it four vertices. Hence, 45τ (H ) ≤ 45(|T ∗ | + 4) ≤ (6n(H ∗ ) + 13m(H ∗ ) + 27) + 180 = 6n(H ) + 13m(H ) − 6 · 15 − 13 · 9 + 27 + 180 = 6n(H ) + 13m(H ), a contradic() tion. Hence the case def(H − s1 ) = 3 cannot occur. Claim P.7.2: The case def(H − s1 ) = 8 cannot occur. Proof of Claim P.7.2: Suppose that def(H − s1 ) = 8. By Claim M, H − a therefore contains an H4 -component. Let q be the edge belonging in this H4 -component. By Claim M, the edge q is intersected by all three edges e1 , g, g • incident with the vertex s1 . Let Z = f 1 ∪ f 2 ∪ q ∪ {x, s1 , s2 } and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 15. The edges e1 , e2 , e3 , f 1 , f 2 , g, g • , h, q are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 9. We show firstly that def(H ∗ ) ≤ 21. Suppose, to the contrary, that def(H ∗ ) ≥ 22. Let Y ∗ be a special H ∗ -set satisfying def H ∗ (Y ∗ ) = def(H ∗ ). By Claim O, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Let F1 , F2 , F3 be the H4 -subhypergraphs of H with edge set f 1 , f 2 , q, respectively, and consider the special H -set Y = Y ∗ ∪ {F1 , F2 , F3 }. We note that |E ∗H (Y )| ≤ |E ∗H ∗ (Y ∗ )| + |{e1 , e2 , e3 , g, g • , h}| ≤ (|Y ∗ | − 3) + 6 = |Y ∗ | + 3 = |Y |, contradicting Claim I(a). Therefore, def(H ∗ ) ≤ 21. We show next that every transversal in H ∗ that contains a vertex in ∂(Z ) can be extended to a transversal in H by adding to it four vertices. We note that if ∂(Z ) contains a vertex from some edge e of H , then e ∈ {e2 , e3 , g, g • , h}. Let T ∗ be a transversal in H ∗ that contains a vertex in ∂(Z ). If T ∗ contains a vertex in e2 or e3 , say in e2 , then T ∗ ∪ {(e3 f i ), (h f 3−i ), s1 , w} is a transversal in H of size |T ∗ | + 4 for some i ∈ [2] where w is an arbitrary vertex in the edge q. If T ∗ contains a vertex in h, then T ∗ ∪ {(e2 f i ), (e3 f 3−i ), s1 , w} is a transversal in H of size |T ∗ | + 4 for some i ∈ [2] where w is an arbitrary vertex in the edge q. If T ∗ contains a vertex in g or g • , say in g, then T ∗ ∪ {(g • q), c, d, x} is a transversal in H of size |T ∗ | + 4. In all cases, T ∗ can be extended to a transversal in H by adding to it four vertices. By Claim P.5, the vertex s3 does not belong to the edge h, implying that {a, b, s3 } ⊆ ∂(Z ), and so |∂(Z )| ≥ 3 and f (|∂(Z )|) ≤ 27. An identical argument as in the last paragraph of the proof of Claim P.7.1 yields the contradiction 45τ (H ) ≤ 6n(H ) + () 13m(H ). Hence the case def(H − s1 ) = 8 cannot occur. Since neither the case def(H − s1 ) = 3 nor the case def(H − s1 ) = 8 can occur, () this completes the proof of Claim P.7.

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Claim P.8: There is no edge containing all of s1 , s2 , and s3 . Proof of Claim P.8: Suppose, to the contrary, that there is an edge, g say, in H containing all of s1 , s2 , and s3 . By Claim P.4, d H (si ) ≥ 2 for all i ∈ [3], and d H (si ) = 2 for at least one i ∈ [3]. Thus, by Claim P.7, d H (s1 ) = d H (s2 ) = d H (s3 ) = 2. Let Z = f 1 ∪ f 2 ∪ {x, s1 , s2 , s3 } and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 12. The edges e1 , e2 , e3 , f 1 , f 2 , g, h are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 7. If def(H ∗ ) ≥ 22, then let Y ∗ be a special H ∗ -set satisfying def H ∗ (Y ∗ ) = def(H ∗ ). By Claim O, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Let F1 , F2 be the H4 -subhypergraphs of H with edge set f 1 , f 2 , respectively, and consider the special H -set Y = Y ∗ ∪ {F1 , F2 }. We note that |E ∗H (Y )| ≤ |E ∗H ∗ (Y ∗ )| + |{e1 , e2 , e3 , g, h}| ≤ (|Y ∗ | − 3) + 5 = |Y ∗ | + 2 = |Y |, contradicting Claim I(a). Therefore, def(H ∗ ) ≤ 21. We show next that every transversal in H ∗ that contains a vertex in ∂(Z ) can be extended to a transversal in H by adding to it three vertices. We note that if ∂(Z ) contains a vertex from some edge e of H , then e ∈ {e1 , e2 , e3 , g, h}. Let T ∗ be a transversal in H ∗ that contains a vertex in ∂(Z ). If T ∗ contains a vertex in e1 , e2 , or e3 , say in e1 , then T ∗ ∪ {s2 , (e3 f i ), (h f 3−i )} is a transversal in H of size |T ∗ | + 3 for some i ∈ [2]. If T ∗ contains a vertex in h, then T ∗ ∪ {s1 , (e2 f i ), (e3 f 3−i )} is a transversal in H of size |T ∗ | + 3 for some i ∈ [2]. If T ∗ contains a vertex in g, then T ∗ ∪ {c, d, x} is a transversal in H of size |T ∗ | + 3. In all cases, T ∗ can be extended to a transversal in H by adding to it three vertices. As observed earlier, def(H ∗ ) ≤ 21. By Claim L, there exists a transversal, T ∗ , in H ∗ , that contains a vertex in ∂(Z ) such that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|∂(Z )|). As observed earlier, T ∗ can be extended to a transversal in H by adding to it three vertices. Hence, 45τ (H ) ≤ 45(|T ∗ | + 3) ≤ (6n(H ∗ ) + 13m(H ∗ ) + f (|∂(Z )|)) + 135 = 6n(H ) + 13m(H ) + f (|∂(Z )|) − 6 · 12 − 13 · 7 + 135 = 6n(H ) + 13m(H ) + f (|∂(Z )|)) − 28. If |∂(Z )| ≥ 3, then f (|∂(Z )|) ≤ 27 and 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, |∂(Z )| ≤ 2. Since {a, b} ⊆ ∂(Z ), we note that |∂(Z )| ≥ 2. Consequently, |∂(Z )| = 2 and ∂(Z ) = {a, b}. This implies that f i intersects each of the edges e1 , e2 , e3 for i ∈ [2] and that either a or b belongs to the edge g. Renaming the vertices a and b if necessary, we may assume that g = {a, s1 , s2 , s3 }. This implies that the only edges in H that intersect the edges f 1 , f 2 , and g are e1 , e2 , e3 , h. We now let F1 , F2 , F3 be the H4 -subhypergraphs of H with edge set f 1 , f 2 , g, respectively, and consider the special H -set Y = {F1 , F2 , F3 }. We note that |E ∗H (Y )| ≤ |{e1 , e2 , e3 , h}| = 4 = |Y | + 1, contradicting Claim I(a). This completes () the proof of Claim P.8. Claim P.9: If d H (si ) = d H (s j ) = 3 for some i and j where 1 ≤ i, j ≤ 3 and i = j, then the vertices si and s j do not belong to a common edge. Proof of Claim P.9: Renaming edges e1 , e2 , e3 if necessary, let d H (s1 ) = d H (s2 ) = 3 and suppose, to the contrary, that there is an edge g containing both s1 and s2 . By Claim P.8, the edge g does not contain the vertex s3 . By Claim P.4, we may assume, renaming the edges f 1 and f 2 if necessary, that e1 ∩ f 1 = ∅, e1 ∩ f 2 = ∅, and that

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e2 ∩ f 1 = ∅ and e2 ∩ f 2 = ∅. Thus, e1 intersects f 1 but not f 2 , and e2 intersects f 2 but not f 1 . By Claim P.1, this implies that e3 intersects both f 1 and f 2 ; that is, e3 ∩ f 1 = ∅ and e3 ∩ f 2 = ∅. For i ∈ [2], let wi be the vertex in ei different from (ei f i ), x and si , and so V (ei ) = {x, si , wi , (ei f i )}. We note that the vertex (ei f i ) has degree 2 in H for i ∈ [2]. Claim P.9.1: There is no edge containing both w1 and w2 . Proof of Claim P.9.1: Suppose, to the contrary, that there is an edge f  that contains both w1 and w2 . We now consider the hypergraph H − s1 . By Claim M, either f  belongs to an H4 -component or to a double-H4 -component. Since f  is intersected by the edge e2 in H − s1 , the edge f  cannot belong to an H4 -component. Hence, f  belongs to a double-H4 -component. Let f  be the second edge belonging to the H4 pair in this double-H4 -component. We note that e2 is the linking edge in H − s1 that intersects f  and f  . Analogously as in Claim P.5, N H (s1 ) ∩ V (e2 ) = ∅. However, () s2 ∈ N H (s1 ) ∩ V (e2 ), a contradiction. Claim P.9.2: There is no edge containing s3 and one of w1 or w2 . Proof of Claim P.9.2: Suppose, to the contrary, that there is an edge f  that contains s3 and one of w1 or w2 . By symmetry, we may assume that w1 ∈ V ( f  ). We now consider the hypergraph H − s1 . By Claim M, either f  belongs to an H4 -component or to a double-H4 -component. Since f  is intersected by the edge e3 in H − s1 , the edge f  cannot belong to an H4 -component. Hence, f  belongs to a double-H4 component. Let f  be the second edge belonging to the H4 -pair in this double-H4 component. We note that e3 is the linking edge in H − s1 that intersects f  and f  . Analogously as in Claim P.5, N H (s1 ) ∩ V (e3 ) = ∅. However, x ∈ N H (s1 ) ∩ V (e3 ), () a contradiction. By Claim P.9.1 and P.9.2, {w1 , w2 , s3 } is a strongly independent set. Recall that h = {a, b, c, d}, where the edges f 1 and h intersect in the vertex c and where the edges f 2 and h intersect in the vertex d. Further, d H (c) = d H  (c) = 2 and d H (d) = d H  (d) = 2. In particular, we note that {d, w1 , w2 , s3 } is a strongly independent set. Let Z = ( f 1 ∪ f 2 ∪ {x}) \ {c, d} and let H • be obtained from H − Z by adding to it the edge f • = {d, w1 , w2 , s3 }. We note that n(H • ) = n(H ) − 7 and m(H • ) = m(H ) − 4. Further since {d, w1 , w2 , s3 } is a strongly independent set in H , the hypergraph H • is linear. We show next that τ (H ) ≤ τ (H • ) + 2. Let T • be a τ (H • )-transversal. In order to cover the edge f • , the transversal T • contains at least one vertex in f • . If w1 ∈ T • , let T = T • ∪ {(e2 f 2 ), (e3 f 1 )}. If w2 ∈ T • , let T = T • ∪ {(e1 f 1 ), (e3 f 2 )}. If s3 ∈ T • , let T = T • ∪ {(e1 f 1 ), (e2 f 2 )}. If d ∈ T • , let T = T • ∪ {x, c}. In all cases, T is a transversal in H of size |T • | + 2, and so τ (H ) ≤ |T • | + 2 = τ (H • ) + 2. since H • is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45 (τ (H • ) + 2) ≤ 6n(H • ) + 13m(H • ) + def(H • ) + 90 ≤ 6n(H ) + 13m(H ) + def (H • ) − 7 × 6 − 4 × 13 + 90 = 6n(H ) + 13m(H ) + def(H • ) − 4. If def(H • ) ≤ 4, then 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, def(H • ) ≥ 5. Let Y • be a special H • -set satisfying def H • (Y • ) = def(H • ). By Claim O, |E ∗H • (Y • )| ≤ |Y • | − 1.

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Claim P.9.3: f • ∈ Y • . Proof of Claim P.9.3: Suppose, to the contrary, that f • ∈ / Y • . If f • ∈ / E ∗H • (Y • ), ∗ ∗ • • • • then h ∈ / Y , implying that |E H (Y )| ≤ |E H • (Y )| + |{e1 , e2 }| ≤ (|Y | − 1) + 2 = |Y • | + 1, contradicting Claim I. Hence, f • ∈ E ∗H • (Y • ). Since H − Z = H • − f • , this implies that def(H − Z ) = def(H • − f • ) = def(H • ) + 13 ≥ 18. Let Y be a / Y , then neither special (H − Z )-set satisfying def H −Z (Y ) = def(H − Z ). If h ∈ f 1 nor f 2 intersects any special subhypergraphs in Y , implying that def H −x (Y ) = def H −Z (Y ) ≥ 18, contradicting Claim M. Hence, h ∈ Y . Let R1 be the special subhypergraph in Y that contains the edge h. We note that both c and d have degree 1 in H − Z , implying that the subhypergraph R1 is an H4 -component in H − Z and therefore contributes 8 to the deficiency of Y in H − Z . Since neither f 1 nor f 2 intersects any special subhypergraph in Y different from R1 , this implies that () def H −x (Y \ {R1 }) ≥ 18 − 8 = 10, once again contradicting Claim M. By Claim P.9.3, f • ∈ Y • . Let R1 be the special subhypergraph in Y • that contains the edge f • . Suppose that |Y • | ≥ 2. Let |E ∗H • (Y • )| = |Y • | − i for some i ≥ 1. By Claim O and by our earlier observations, 5 ≤ def(H • ) ≤ 13i − 5|Y • | ≤ 13i − 10, implying that i ≥ 2; that is, |E ∗H • (Y • )| ≤ |Y • | − 2. Let R2 and R3 be the H4 subhypergraphs of H with edge set f 1 and f 2 , respectively. Let Y ∗ = (Y • \ {R1 }) ∪ {R2 , R3 }, and so |Y ∗ | = |Y • | + 1. Thus, |E ∗H (Y ∗ )| ≤ |E ∗H • (Y • )| + |{e1 , e2 , e3 , h}| ≤ (|Y • | − 2) + 4 = |Y ∗ | + 1, contradicting Claim I. Hence, |Y • | = 1; that is, Y • = {R1 }. Recall that def(H • ) ≥ 5 and |E ∗H • (Y • )| ≤ |Y • | − 1. Thus since |Y • | = 1, we note that |E ∗H • (Y • )| = 0, implying that R1 is a component in H • . Since the vertex s3 belongs to R1 and is contained in both edges f • and h, the component R1 is not an H4 -component in H • , implying that R1 is an H14 -component in H • and def(H • ) = 5. We now consider the hypergraph H − s1 . By Claim M, either H − s1 contains an H4 -component or a double-H4 -component. We note that in either case, the vertex w1 belongs to such a component. If H − s1 contains an H4 -component, then this component contains a vertex of degree 1 in H different from w1 . Suppose that H − s1 contains a double-H4 -component. In this case, let q1 and q2 be the two edges belonging to the H4 -pair in this component and let q • be the linking edge in H − s1 that intersects q1 and q2 . Renaming q1 and q2 if necessary, we may assume that w1 ∈ V (q1 ). Thus, q1 ∩ V (e1 ) = ∅, implying that q2 contains a vertex of degree 1 in H . In both cases, there is a vertex of degree 1 in H that belongs to the component of H − s1 that contains w1 , implying that the component R1 of H • contains a vertex of degree 1 in H . Further, the component R1 of H • contains the vertex c which has degree 1 in H • and degree 2 in H . Thus, R1 has at least two vertices of degree 1 in H • , implying that R1 is an H14,4 -component. We now consider the edge h in R1 that contains the vertex c of degree 1 in H • . All three edges intersecting the edge h in R1 ∼ = H14,4 have at least one vertex of degree 3. Since f • is one of the three edges that intersect h in R1 , this implies in particular that the edge f • contains at least one vertex of degree 3 in R1 . However, no vertex of f • has degree 3 in H • , a () contradiction. This completes the proof of Claim P.9.

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Claim P.10: Every edge contains at most one vertex from the set {s1 , s2 , s3 }. Proof of Claim P.10: Suppose, to the contrary, that there is an edge, g say, in H containing si and s j where 1 ≤ i < j ≤ 3. Renaming the edges e1 , e2 , e3 if necessary, we may assume that {s1 , s2 } ⊂ V (g). Let g = {s1 , s2 , u, v}. By Claim P.8, the edge / {u, v}. By Claim P.4, Claim P.7, and g does not contain the vertex s3 , and so s3 ∈ Claim P.9, d H (s1 ) = d H (s2 ) = 2. Let Z = f 1 ∪ f 2 ∪ {s1 , s2 , x} and let H ∗ = H − Z . We note that n(H ∗ ) = n(H ) − 11. The edges e1 , e2 , e3 , f 1 , f 2 , g, h are deleted from H when constructing H ∗ , and so m(H ∗ ) = m(H ) − 7. We show firstly that def(H ∗ ) ≤ 21. Suppose, to the contrary, that def(H ∗ ) ≥ 22. Let Y ∗ be a special H ∗ -set satisfying def H ∗ (Y ∗ ) = def(H ∗ ). By Claim O, |E ∗H ∗ (Y ∗ )| ≤ |Y ∗ | − 3. Let R1 and R2 be the H4 -subhypergraphs of H with edges f 1 and f 2 , respectively, and consider the special H -set Y = Y ∗ ∪ {R1 , R2 }. We note that |E ∗H (Y )| ≤ |E ∗H ∗ (Y ∗ )| + |{e1 , e2 , e3 , g, h}| ≤ (|Y ∗ | − 3) + 5 = |Y ∗ | + 2 = |Y |, contradicting Claim I(a). Therefore, def(H ∗ ) ≤ 21. We show next that every transversal in H ∗ that contains a vertex in ∂(Z ) can be extended to a transversal in H by adding to it three vertices. We note that if ∂(Z ) contains a vertex from some edge e of H , then e ∈ {e1 , e2 , e3 , g, h}. Let T ∗ be a transversal in H ∗ that contains a vertex in ∂(Z ). If T ∗ contains a vertex in e1 or in e2 , say in e1 by symmetry, then T ∗ ∪ {(e2 g), (e3 f i ), (h f 3−i )} is a transversal in H for some i ∈ [2]. If T ∗ contains a vertex in g, then T ∗ ∪ {c, d, x} is a transversal in H . If T ∗ contains a vertex in h, then T ∗ ∪ {(e1 g), (e2 f i ), (e3 f 3−i )} is a transversal in H for some i ∈ [2]. In all cases, T ∗ can be extended to a transversal in H by adding to it three vertices. By Claim L, there exists a transversal, T ∗ , in H ∗ , that contains a vertex in ∂(Z ) such that 45|T ∗ | ≤ 6n(H ∗ ) + 13m(H ∗ ) + f (|∂(Z )|) = 6n(H ) + 13m(H ) + f (|∂(Z )|) − 11 · 6 − 7 · 13 = 6n(H ) + 13m(H ) + f (|∂(Z )|) − 157. As observed earlier, T ∗ can be extended to a transversal in H by adding to it three vertices. Hence, 45τ (H ) ≤ 45(|T ∗ | + 3) ≤ 6n(H ) + 13m(H ) + f (|∂(Z )|) − 22. If |∂(Z )| ≥ 5, then f (|∂(Z )|) = 22, implying that 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, |∂(Z )| ≤ 4. We note that {a, b} ⊆ ∂(Z ), {u, v} ⊆ ∂(Z ), and s3 ∈ ∂(Z ). By Claim P.5, the / {a, b}, and so {a, b, s3 } ⊆ ∂(Z ). edge h contains no neighbor of x. In particular, s3 ∈ / {u, v}. By the linearity of H , we note that {a, b} = {u, v}. As observed earlier, s3 ∈ Since |∂(Z )| ≤ 4, this implies that |{a, b} ∩ {u, v}| = 1 and |∂(Z )| = 4. Renaming vertices if necessary, we may assume that b = v. By Claim P.6, d H (b) = 2. Since b = v, the vertex b is therefore incident only with the edge g and h in H , implying that d H ∗ (b) = 0. We now add the vertex b to the set Z . With this addition of b to Z , we note that n(H ∗ ) = n(H ) − 12, m(H ∗ ) = m(H ) − 7, and 45|T ∗ | ≤ 6n(H ) + 13m(H ) + f (|∂(Z )|) − 157 − 6 = 6n(H ) + 13m(H ) + f (|∂(Z )|) − 163. Further since {a, s3 , u} ⊆ ∂(Z ), we note that |∂(Z )| ≥ 3 and f (|∂(Z )|) ≤ 27. Thus, 45τ (H ) ≤ 45(|T ∗ | + 3) ≤ 6n(H ) + 13m(H ) + f (|∂(Z )|) − 28 < 6n(H ) + 13m(H ), a con() tradiction. This completes the proof of Claim P.10. We now continue with our proof of Claim P. By Claim P.10, {s1 , s2 , s3 } is a strongly independent set. Recall that h = {a, b, c, d}, where the edges f 1 and h

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intersect in the vertex c and where the edges f 2 and h intersect in the vertex d. Further, d H (c) = d H  (c) = 2 and d H (d) = d H  (d) = 2. In particular, we note that {d, s1 , s2 , s3 } is a strongly independent set. Let Z = ( f 1 ∪ f 2 ∪ {x}) \ {c, d} and let H • be obtained from H − Z by adding to it the edge f • = {d, s1 , s2 , s3 }. We note that n(H • ) = n(H ) − 7 and m(H • ) = m(H ) − 4. Further since {d, s1 , s2 , s3 } is a strongly independent set in H , the hypergraph H • is linear. By Claim P.1, at least one of the edges e1 , e2 , e3 intersects both f 1 and f 2 . Renaming edges if necessary, we may assume that e3 intersects both f 1 and f 2 ; that is, e3 ∩ f 1 = ∅ and e3 ∩ f 2 = ∅. Let Y • be a special H • -set satisfying def H • (Y • ) = def(H • ). Claim P.11: The following properties hold in the hypergraph H • . (a) (b) (c) (d) (e)

τ (H ) ≤ τ (H • ) + 2. def(H • ) ≥ 5 and |E ∗H • (Y • )| ≤ |Y • | − 1. There is no H10 -subhypergraph in H • . f • ∈ Y •. |Y • | = 1.

Proof of Claim P.11: (a) Let T • be a τ (H • )-transversal. In order to cover the edge f , the transversal T • contains at least one vertex in f • . Suppose that s j ∈ T • for some j ∈ [3]. Renaming edges e1 , e2 , e3 if necessary, we may assume that s1 ∈ T • . In this case, T • ∪ {(e2 f i ), (e3 f 3−1 )} is a transversal of H for some i ∈ [2]. If d ∈ T • , then T • ∪ {x, c} is a transversal of H . In both cases, we produce a transversal in H of size |T • | + 2, and so τ (H ) ≤ |T • | + 2 = τ (H • ) + 2. (b) Since H • is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H • ) + 2) ≤ 6n(H • ) + 13m(H • ) + def(H • ) + 90 ≤ 6n(H )+ 13m(H ) + def(H • ) − 7 × 6 − 4 × 13 + 90 = 6n(H ) + 13m(H ) + def(H • ) − 4. If def(H • ) ≤ 4, then 45τ (H ) ≤ 6n(H ) + 13m(H ), a contradiction. Hence, def(H • ) ≥ 5. By Claim O, |E ∗H • (Y • )| ≤ |Y • | − 1. (c) Suppose, to the contrary, that there is a H10 -subhypergraph, say F, in H • . By Claim J, F is not a subhypergraph of H , implying that the added edge f • is an edge of F and therefore the vertex d belongs to F. We note that every vertex of F has degree 2 in F and therefore degree at least 2 in H • . Since the vertex d has degree 2 in H • and is contained in the edges f • and h in H • , the edge h must belong to F. This implies that the vertex c belongs to F. However the vertex c has degree 1 in H • , and therefore degree 1 in F, a contradiction. / Y • . Recall that the edge e3 intersects both (d) Suppose, to the contrary, that f • ∈ ∗ • • / E H • (Y ), then h ∈ / Y • , implying that |E ∗H (Y • )| ≤ |E ∗H • (Y • )| + f 1 and f 2 . If f ∈ • • |{e1 , e2 }| ≤ (|Y | − 1) + 2 = |Y | + 1, contradicting Claim I. Hence, f • ∈ E ∗H • (Y • ). If h ∈ / Y , then neither f 1 nor f 2 intersects any special subhypergraphs in Y • , implying that |E ∗H (Y • )| ≤ |E ∗H • (Y • )| + |{e1 , e2 , e3 }| − |{ f • | ≤ (|Y • | − 1) + 3 − 1 = |Y • | + 1, contradicting Claim I. Hence, h ∈ Y . Let R1 be the special subhypergraph in Y that contains the edge h. We note that both c and d have degree 1 in R1 , implying that the subhypergraph R1 is an H4 -component in H • . By Claim P.5 and Claim P.6, both vertices a and b have degree 2 in H • . By the linearity of H , the edge different from h that contains a and the edge different from h that contains b are distinct. These •

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two edges, together with the edge f • , all intersect the H4 -component R1 , implying that |E ∗H (Y • )| ≥ 3. Thus, by Claim O, |E ∗H (Y • )| ≤ |Y • | − 3. Hence, |E ∗H (Y • )| ≤ |E ∗H • (Y • )| + |{e1 , e2 , e3 , f 1 , f 2 }| − |{ f • | ≤ (|Y • | − 3) + 5 − 1 = |Y • | + 1, contradicting Claim I. (e) By Part (d), f • ∈ Y • . Let R1 be the special subhypergraph in Y • that contains the edge f • . If |Y • | ≥ 3, then by Claim O and by our earlier observations, |E ∗H • (Y • )| ≤ |Y • | − 2. Let R2 and R3 be the H4 -subhypergraphs of H with edge set f 1 and f 2 , respectively. Let Y ∗ = (Y • \ {R1 }) ∪ {R2 , R3 }, and so |Y ∗ | = |Y • | + 1. Thus, |E ∗H (Y ∗ )| ≤ |E ∗H • (Y • )| + |{e1 , e2 , e3 , h}| ≤ (|Y • | − 2) + 4 = |Y ∗ | + 1, contradicting Claim I. Hence, |Y • | ≤ 2. Suppose that |Y • | = 2. Let R2 be the special subhypergraph in Y • different from R1 , where recall that f • ∈ E(R1 ). As observed earlier, def(H • ) ≥ 5 and there is no H10 -subhypergraph in H • . Thus, 5 ≤ def(H • ) ≤ 8|Y • | − 13|E ∗H • (Y • )| = 16 − 13|E ∗H • (Y • )|, implying that E ∗H • (Y • ) = ∅. Letting Y ∗ = {R2 }, this in turn implies that E ∗H (Y ∗ ) = ∅ and therefore that def(H ) > 0, () a contradiction. Hence, |Y • | = 1. This completes the proof of Claim P.11. By Claim P.11(d), f • ∈ Y • . By Claim P.11(e), |Y • | = 1. Let R1 be the special subhypergraph in Y • that contains the edge f • ; that is, Y • = {R1 }. By Claim P.11, we note that 5 ≤ def(H • ) ≤ 8|Y • | − 13|E ∗H • (Y • )| = 8 − 13|E ∗H • (Y • )|, implying that E ∗H • (Y • ) = ∅; that is, R1 is a component in H • . By our way in which H • is constructed, this implies that R1 = H • . Thus, n(H ) = n(H • ) + 7 = 14 + 7 = 21. Since the vertex d belongs to R1 and is contained in both edges f • and h, the component R1 is not an H4 -component in H • , implying that R1 is an H14 -component in H • and def(H • ) = 5. Since R1 contains the edge h, the component R1 has at least one vertex of degree 1, namely, the vertex c which has degree 1 in H • . This implies that / {H14,5 , H14,6 }; that is, R1 ∼ R1 ∈ = H14,i for some i ∈ [4]. Suppose that R1 = H14,1 . In this case, h is the edge in R1 that contains the (unique) vertex of degree 1 in R1 . Further, the edge h is intersected by three edges in R1 , one of which is the edge f • . The structure of H14,1 implies that we can choose a τ -transversal of H • , T • , to contain three vertices of f • , one of which is the vertex d (that belongs to both f • and h). Renaming the edges e1 , e2 , e3 if necessary, we may assume that {e2 , e3 } ⊂ T • . If the edges e1 and f 1 intersect, let T = T • ∪ {(e1 f 1 )}. If the edges e1 and f 1 do not intersect, then the edges e1 and f 2 intersect and we let T = (T • \ {d}) ∪ {c, (e1 f 2 )}. In both cases, the set T is a transversal in H of size |T • | + 1, and so τ (H ) ≤ τ (H • ) + 1. Thus, 45τ (H ) ≤ 45(τ (H • ) + 1) ≤ 6n(H • ) + 13m(H • ) + def(H • ) + 45 ≤ 6n(H ) + 13m(H ) + 5 + 45 − 7 × 6 − 4 × 13 < 6n(H ) + 13m (H ), a contradiction. Suppose that R1 = H14,3 . In this case, h is the edge in R1 that contains the (unique) vertex of degree 1 in R1 . Further, the edge h is intersected by three edges in R1 , one of which is the edge f • . Using analogous arguments as in the previous case, due to the structure of H14,3 we can choose a τ -transversal of H • , T • , to contain three vertices of f • , one of which is the vertex d, implying as before that τ (H ) ≤ τ (H • ) + 1, producing a contradiction. Suppose that R1 = H14,4 . In this case, R1 contains two vertices of degree 1, and h is one of the two edges in R1 that contain a vertex of degree 1 in R1 . Further, the edge

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h is intersected by three edges in R1 , one of which is the edge f • . Using analogous arguments as in the previous two cases, due to the structure of H14,4 we can choose a τ -transversal of H • , T • , to contain at least three vertices of f • , one of which is the vertex d, implying as before that τ (H ) ≤ τ (H • ) + 1, producing a contradiction. By the above, we must have R1 = H14,2 . In this case, h is the edge in R1 that contains the (unique) vertex of degree 1 in R1 . Further, the edge h is intersected by three edges in R1 , one of which is the edge f • . We note that in this case, each vertex si has degree 2 in R1 and therefore degree 2 in H . Suppose that f 1 or f 2 (or both f 1 and f 2 ) does not intersect one of the edges e1 , e2 , e3 . Renaming vertices and edges, if necessary, we may assume that f 1 does not intersect the edge e2 . By Claim P.1, the edge f 1 therefore intersects both e1 and e3 . Let x2 be the vertex in the edge e2 different from x, s2 , and (e2 f 2 ). We note that we could have chosen s2 to be the vertex x2 , implying by our earlier observations that d H (x2 ) = 2 and {s1 , x2 , s3 } is a strongly independent set. If the vertex x2 belongs to the set V (R1 ), then x2 would be adjacent to s1 or s3 in H or x2 and s2 would be adjacent / V (R1 ). Interchanging the roles of x2 and s2 in H − e2 , a contradiction. Hence, x2 ∈ in our earlier arguments and letting f • = {d, s1 , x2 , s3 }, the special subhypergraph, say R1• , in Y • that contains the edge f • is a H14,2 -component of H • that does not contain the vertex s2 . This, however, is a contradiction since the vertex of degree 3 in R1 is also the degree 3 vertex in R1• , implying that s2 ∈ V (R1• ), a contradiction. Therefore, the edge f i intersects all three edges e1 , e2 , e3 for i ∈ [2]. The hyper() graph H is now determined and H = H21,5 , contradicting Claim C. Claim Q: No edge in H contains two degree-3 vertices. Proof of Claim Q: We first need the following subclaim. Claim Q.1: There does not exist three degree-3 vertices in H that are pairwise adjacent. Proof of Claim Q.1: For the sake of contradiction, suppose to the contrary that x1 , x2 , and x3 are pairwise adjacent degree-3 vertices in H . First consider the case when some edge, e, in H contains all three vertices. Let V (e) = {x1 , x2 , x3 , x4 }. By Claim P, the hypergraph H − xi contains an H4 -component, Ri , for all i ∈ [3]. Further, let f i be the edge in Ri . By Claim M, the edge f i is intersected by all three edges incident with xi . Since every vertex in f i has degree at most 2 in H and d(x1 ) = d(x2 ) = d(x3 ) = 3, we note that x4 ∈ V ( f i ) for all i ∈ [3], which implies that f 1 = f 2 = f 3 , which is impossible as the two edges different from e containing x1 intersect f 1 and therefore R1 is not an H4 -component in H − x2 . Therefore there exists three distinct edges e12 , e13 , and e23 , such that ei j contains xi and x j for all 1 ≤ i < j ≤ 3. Let Ri be the H4 -component in H − xi for i ∈ [3], and let h i be the edge in Ri . Note that V (h i ) ∩ {x1 , x2 , x3 } = ∅ for all i ∈ [3] as no vertex in h i has degree 3. If h = h 1 = h 2 , then the edge h intersects all edges containing x1 (as h = h 1 ) and all edges intersecting x2 (as h = h 2 ), but then h does not belong to an H4 component in H − x1 (or in H − x2 ), a contradiction. Hence, h 1 , h 2 , and h 3 are distinct edges. Furthermore, if V (h 1 ) ∩ V (h 2 ) = ∅ and y ∈ V (h 1 ) ∩ V (h 2 ), then y is contained in the three edges e12 , h 1 , and h 2 , and so d(y) = 3, a contradiction

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to all vertices in h 1 having degree at most two. Therefore, h 1 , h 2 , and h 3 are nonintersecting edges. For each i ∈ [3], let f i be the edge incident with xi , different from e12 , e13 , and e23 , which exists as d(xi ) = 3 and two of the three edges e12 , e13 , and e23 contain xi . By the above definitions, we note that h 1 intersects the three edges e12 , e13 , and f 1 , which all contain x1 . Also, h 2 intersects the three edges e12 , e23 , and f 2 , which all contain x2 . And finally, h 3 intersects the three edges e13 , e23 , and f 3 , which all contain x3 . Let Z = V (e12 ) ∪ V (e13 ) ∪ V (e23 ) ∪ V (h 1 ) ∪ V (h 2 ) ∪ V (h 3 ) and note that |Z | = 15, as |V (e12 ) ∪ V (e13 ) ∪ V (e23 )| = 9 and each h i contains two vertices not in V (e12 ) ∪ V (e13 ) ∪ V (e23 ). Also note that the only edges in H intersecting Z are e12 , e13 , e23 , h 1 , h 2 , h 3 , f 1 , f 2 , f 3 . Let H  = H \ Z and note that n(H  ) = n(H ) − 15 and m(H  ) = m(H ) − 9. Further, we note that |V ( f i ) \ Z | = 2 for i ∈ [3]. Moreover since H is linear, V ( f 1 ) \ Z and V ( f 2 ) \ Z can have at most one vertex in common, implying that |∂(Z )| ≥ 3. We wish to use Claim L, so first we need to show that def(H  ) ≤ 21. Suppose to the contrary that def(H  ) ≥ 22. Let Y  be a special H  -sets in H  with def H  (Y  ) = def(H  ) ≥ 22. By Claim O, |E ∗H  (Y  )| ≤ |Y  | − 3. This implies that |E ∗H (Y  )| ≤ |E ∗H  (Y  )| + |{ f 1 , f 2 , f 3 }| ≤ (|Y  | − 3) + 3 = |Y  |, a contradiction to Claim I(a). Therefore, def(H  ) ≤ 21. As observed earlier, |∂(Z )| ≥ 3, and so f (|∂(Z )|) ≤ 27. By Claim L, there exists a transversal, T  , in H  , that contains a vertex in ∂(Z ) such that 45|T  | ≤ 6n(H  ) + 13m(H  ) + f (|∂(Z )|) ≤ 6(n(H ) − 15) + 13(m(H ) − 9) + 27 = 6n(H ) + 13m(H ) − 180. Since T  contains a vertex in ∂(Z ), the transversal T  contains a vertex from V ( f 1 ), V ( f 2 ), or V ( f 3 ). Without loss of generality, T  contains a vertex from V ( f 1 ). With this assumption, the set T = T  ∪ {x3 , (e12 h 1 ), ( f 2 h 2 ), w} where w is an arbitrary vertex from V (h 3 ) is a transversal in H of size |T  | + 4. Hence, 45τ (H ) ≤ 45|T | = 45(|T  | + 4) ≤ 6n(H ) + 13m(H ) − 180 + 4 · 45 = () 6n(H ) + 13m(H ), a contradiction. This proves Claim Q.1 We now return to the proof of Claim Q. Suppose, to the contrary, that there exists an edge e = {x1 , x2 , x3 , x4 }, such that d(x1 ) = d(x2 ) = 3 in H . Let e, f 1 , and f 2 be the three edges in H containing x1 and let e, g1 , and g2 be the three edges in H containing x2 . Let Ri be the H4 -component in H − xi and let h i be the edge of Ri for i ∈ [2], and so E(Ri ) = {h i }. Renaming vertices if necessary, we may assume that x3 ∈ V (h 1 ) and x4 ∈ V (h 2 ), as h i intersects all three edges incident with xi for i ∈ [2] and h 1 and h 2 are non-intersecting edges. Let Z  = V (h 1 ) ∪ V (h 2 ) ∪ {x1 , x2 } and let Q 1 = V (g1 ) \ {(g1 h 2 ), x2 }, Q 2 = V (g2 ) \ {(g2 h 2 ), x2 }, Q 4 = V ( f 2 ) \ {( f 2 h 1 ), x1 }. Q 3 = V ( f 1 ) \ {( f 1 h 1 ), x1 }, Let Q  = Q 1 ∪ Q 2 ∪ Q 3 ∪ Q 4 and note that Q  = ∂(Z  ) = N H (Z  ) \ Z  . Let H  = H − Z  . We note that n(H  ) = n(H ) − 10 and m(H  ) = m(H ) − |{e, f 1 , f 2 , g1 , g2 , h 1 , h 2 }| = m(H ) − 7. Let R1 = Q 1 ∪ Q 2 and R2 = Q 3 ∪ Q 4 . As |R1 | = |R2 | = 4 and Q  = R1 ∪ R2 we note that 4 ≤ |Q  | ≤ 8.

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Claim Q.2: If there is a transversal, T  , in H  containing a vertex from Q  , then τ (H ) ≤ |T  | + 3. Furthermore, if T  contains a vertex from three of the sets Q 1 , Q 2 , Q 3 , Q 4 , then τ (H ) ≤ |T  | + 2. Proof of Claim Q.2: Let T  be a transversal in H  containing a vertex from Q  . Renaming vertices if necessary, we may assume that T  intersects Q 1 , and therefore covers the edge g1 . With this assumption, the set T = T  ∪ {x1 , x3 , (g2 h 2 )} is a transversal in H of size |T  | + 3, where we note that the vertex x3 covers h 1 , the vertex x1 covers e, f 1 and f 2 , and the vertex (g2 h 2 ) covers g2 and h 2 . Hence, τ (H ) ≤ |T | = |T  | + 3. Furthermore, suppose that the transversal T  of H  contains a vertex from three of the sets Q 1 , Q 2 , Q 3 , Q 4 . Without loss of generality, we may assume that T  intersects Q 1 , Q 2 , and Q 3 and therefore covers the edges g1 , g2 , and f 1 . In this case, the set T = T  ∪ {(eh 2 ), ( f 2 h 1 )} is a transversal in H of size |T  | + 2, implying that τ (H ) ≤ |T | = |T  | + 2. This completes the proof of Claim Q.2 () Claim Q.3: Every vertex in R1 ∩ R2 has degree 2 in H . Proof of Claim Q.3: Every vertex in R1 ∩ R2 is intersected by one of g1 or g2 and one of f 1 or f 2 and therefore has degree at least 2. By Claim Q.1, such a vertex in () R1 ∩ R2 cannot have degree 3, which completes the proof of Claim Q.3 Claim Q.4: |Q  | ≥ 5. Proof of Claim Q.4: For the sake of contradiction, suppose that |Q  | ≤ 4. As observed earlier, |Q  | ≥ 4. Consequently, |Q  | = 4 and Q  = R1 = R2 , implying by Claim Q.3 that every vertex in Q has degree 2. The hypergraph H is now determined and by linearity we note that H is isomorphic to H14,4 , a contradiction. This proves () Claim Q.4 Claim Q.5: |E ∗H  (Y )| ≥ |Y | for all special H  -sets Y in H  and therefore def(H  ) = 0. Proof of Claim Q.5: We first prove that def(H  ) ≤ 16. Suppose to the contrary that def(H  ) ≥ 17. Let Y be a special H  -sets with def H  (Y ) = def(H  ) ≥ 17. By Claim O, |E ∗H  (Y )| ≤ |Y | − 3. Let R1 and R2 be the H4 -subhypergraphs of H with edges h 1 and h 2 , respectively. Let Y ∗ = Y ∪ {R1 , R2 }, and so |Y ∗ | = |Y | + 2. Thus, |E ∗H (Y ∗ )| ≤ |E ∗H  (Y )| + |{e, f 1 , f 2 , g1 , g2 }| ≤ |Y | − 3 + 5 = |Y ∗ |, contradicting Claim I(a). Therefore def(H  ) ≤ 16. For the sake of contradiction suppose that |E ∗H  (Y )| < |Y | for some special  H -sets Y in H  . By Claim K, there exists a transversal T  in H  , such that 45|T  | ≤ 6n(H  ) + 13m(H  ) + def(H  ) and T  ∩ Q  = ∅. By Claim Q.2 we note that τ (H ) ≤ |T  | + 3. Recall that n(H  ) = n(H ) − 10 and m(H  ) = m(H ) − 7. Thus, 45τ (H ) ≤ 45(|T  | + 3) ≤ 6n(H  ) + 13m(H  ) + def(H  ) + 135 ≤ 6(n(H )−10)+ 13(m(H ) − 7) + 16 + 135 = 6n(H ) + 13m(H ), a contradiction. Therefore, |E ∗H  (Y )| ≥ |Y | for all special H  -sets Y in H  , and so def(H  ) = 0 which proves () Claim Q.5 Claim Q.6: |Q  | = 8.

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Proof of Claim Q.6: For the sake of contradiction suppose that |Q  | = 8. By Claim Q.4 and the fact that |Q  | ≤ 8 we obtain 5 ≤ |Q  | ≤ 7. Therefore, R = R1 ∩ R2 is nonempty and by Claim Q.3 every vertex in R has degree 2 in H and therefore degree zero in H  . Let Z  = Z  ∪ R, let Q  = Q  \ R and note that Q  = ∂(Z  ). Let H  = H − Z  . By Claim Q.5 we note that def(H  ) = 0 and therefore def(H  ) = 0 also holds (as we only removed isolated vertices). Note that n(H  ) = n(H ) − 10 − |R| and m(H  ) = m(H  ) = m(H ) − 7. Further, we note that |Q  | = 8 − |R| and |Q  | = |Q  | − |R| = 8 − 2|R|. By Claim L, there exists a transversal, T  , in H  containing a vertex from Q  , such that 45|T  | ≤ 6n(H  ) + m(H  ) + f (|Q  |) ≤ 6(n(H ) − 10 − |R|) + 13(m(H ) − 7) + f (8 − 2|R|) = 6n(H ) + 13m(H ) − 3 · 45 − 16 − 6|R| + f (8 − 2|R|). If |R| = 1, then f (8 − 2|R|) − 16 − 6|R| = 22 − 22 = 0. If |R| = 2, then f (8 − 2|R|) − 16 − 6|R| = 23 − 28 = −6. If |R| = 3, then f (8 − 2|R|) − 16 − 6|R| = 33 − 34 = −1. So in all cases, f (8 − 2|R|) − 16 − 6|R| ≤ 0, implying that 45|T  | ≤ 6n(H ) + 13m(H ) − 3 · 45. Thus by Claim Q.2, 45τ (H ) ≤ 45(|T  | + 3) ≤ () 6n(H ) + 13m(H ), a contradiction. This completes the proof of Claim Q.6 By Claim Q.6 we note that Q 1 , Q 2 , Q 3 , and Q 4 are vertex-disjoint. We will now define the following four different hypergraphs. Let H1∗ be the hypergraph obtained from H  by adding three vertices x23 , x24 , and x34 and three hyperedges h ∗2 = Q 2 ∪ {x23 , x24 }, h ∗3 = Q 3 ∪ {x23 , x34 }, and h ∗4 = Q 4 ∪ {x24 , x34 }. Analogously let H2∗ be the obtained from H  by adding three vertices y13 , y14 , and y34 and three hyperedges e1∗ = Q 1 ∪ {y13 , y14 }, e3∗ = Q 3 ∪ {y13 , y34 }, and e4∗ = Q 4 ∪ {y14 , y34 }. Let H3∗ be the obtained from H  by adding three vertices z 13 , z 14 , and z 34 and three hyperedges f 1∗ = Q 1 ∪ {z 12 , z 14 }, f 2∗ = Q 3 ∪ {z 12 , z 24 }, and f 4∗ = Q 4 ∪ {z 14 , z 24 }. Finally let H4∗ be the obtained from H  by adding three vertices w12 , w13 , and w23 and three hyperedges g1∗ = Q 1 ∪ {w12 , w13 }, g2∗ = Q 2 ∪ {w12 , w23 }, and g3∗ = Q 3 ∪ {w13 , w23 }. Claim Q.7: τ (H ) ≤ τ (Hi∗ ) + 2, for i ∈ [4]. Proof of Claim Q.7: Let T ∗ be a minimum transversal in H1∗ . If T ∗ ∩ {x23 , x24 , x34 } = ∅, then T ∗ covers Q 2 , Q 3 , and Q 4 , and by Claim Q.2, τ (H ) ≤ |T ∗ | + 2. We may therefore, without loss of generality, assume that x23 ∈ T ∗ . We may also assume that T ∗ ∩ {x24 , x34 } = ∅, since otherwise we could have picked a vertex from Q 4 instead of this vertex (or vertices). Therefore, Q 4 is covered by T ∗ \ {x23 }. By Claim Q.2, τ (H ) ≤ |T ∗ \ {x23 }| + 3 = |T ∗ | + 2, which completes the proof for H1∗ . () Analogously, the claim holds for Hi∗ , with i = 2, 3, 4.

8.4 Applying the Deficiency Technique

127

Claim Q.8: def(Hi∗ ) ≥ 5, for i ∈ [4]. Proof of Claim Q.8: For the sake of contradiction, suppose to the contrary that def(H1∗ ) ≤ 4 and let T ∗ be a minimum transversal in H1∗ . Since H  is not a counterexample to our theorem, we have by Claim Q.7 that 45τ (H ) ≤ 45(|T ∗ | + 2) ≤ 6n(H1∗ ) + 13m(H1∗ ) + def(H1∗ ) + 90 ≤ 6(n(H ) − 7) + 13(m(H ) − 4) + 4 + 90 = 6n(H ) + 13m(H ), a contradiction. Hence, def(H1∗ ) ≥ 5. Analogously, def(Hi∗ ) ≥ () 5, for i = 2, 3, 4. Let Yi∗ be a special Hi∗ -set satisfying def(Hi∗ ) = def Hi∗ (Yi∗ ) for i ∈ [4]. Claim Q.9: |Y1∗ | = 1 and Y1∗ is an H14 -component in H1∗ containing the edges

h ∗2 , h ∗3 , h ∗4 .

Proof of Claim Q.9: If none of h ∗2 , h ∗3 , or h ∗4 belongs to E(Y1∗ ), then in this case |E ∗ (Y1∗ )| ≤ |Y1∗ | − 1 in H1∗ , which implies that |E ∗ (Y1∗ )| ≤ |Y1∗ | − 1 in H  , contradicting Claim Q.5. Therefore, without loss of generality, we may assume that h ∗2 ∈ E(Y1∗ ). Suppose that there is an H4 -component, R ∗ , in Y1∗ that contains the edge h ∗2 . Thus E(R ∗ ) = {h ∗2 }. In this case, we note that h ∗3 , h ∗4 ∈ E ∗ (Y1∗ ) in H1∗ . As def(H1∗ ) ≥ 5 and |E ∗ (Y1∗ )| ≥ 2, we have |E ∗ (Y1∗ )| ≤ |Y1∗ | − 2 in H1∗ . Let Y1 = Y1∗ \ {R1 }. Since h ∗3 , h ∗4 ∈ E ∗ (Y1∗ ), we note that |E ∗ (Y1 )| ≤ |Y1∗ | − 2 − 2 = |Y1 | − 3 in H  , contradicting Claim Q.5. Therefore, R ∗ is not an H4 -component in Y1∗ . Analogously, there is no H4 -component in Y1∗ that contains the edge h ∗3 or h ∗4 . If two of the three edges h ∗2 , h ∗3 , h ∗4 belong to E(Y1∗ ) in H1∗ , then these two edges, which intersect, contain a degree-1 vertex in Y1∗ . However, there is no special hypergraph with two intersecting edges both containing a degree-1 vertex, a contradiction. Therefore, we must have all three edges h ∗2 , h ∗3 , h ∗4 belonging to E(Y1∗ ). Now consider the case when |Y1∗ | ≥ 2. Let Y1∗ = {R1∗ , R2∗ , . . . , R ∗ }, where ≥ 2. Renaming the elements of Y1∗ , we may assume that {h ∗2 , h ∗3 , h ∗4 } ⊆ E(R1∗ ). As def(H1∗ ) ≥ 5, we must have |E ∗ (Y1∗ )| ≤ |Y1∗ | − 2 in H1∗ . This implies that in H we have |E ∗ (Y1∗ \ {R1∗ })| ≤ |Y1∗ | − 2 + |{g1 }| = |Y1∗ \ {R1∗ }|, contradicting Claim I(a). Therefore, |Y1∗ | = 1 and E ∗ (Y1∗ ) = ∅ in H ∗ . As all three edges h ∗2 , h ∗3 , h ∗4 belong to E(Y1∗ ), we note that Y1∗ is not an H4 -component. As def(H1∗ ) ≥ 5 we must therefore () have that Y1∗ is an H14 -component, completing the proof of Claim Q.9. We note that an analogous claim to Claim Q.9 also holds for H2∗ , H3∗ , and H4∗ . Claim Q.10: Both vertices in at least two of the sets Q 1 , Q 2 , Q 3 , Q 4 have degree 1 in H  . Proof of Claim Q.10: For the sake of contradiction, suppose to the contrary that Q 2 , Q 3 , and Q 4 contain a vertex which does not have degree 1 in H  . If it is an isolated vertex, then without loss of generality assume that y ∈ Q 2 is isolated in H  . Consider the hypergraph H  = H − y and let Z  = Z  ∪ {y}. By Claim Q.5 we note that def(H  ) = def(H  ) = 0. Furthermore, H  = H − Z  . Let Q  = N (Z  ) \ Z  , and so Q  = ∂(Z  ). By Claim Q.6, we note that |Q  | = |Q  \ {y}| = 7. By Claim L, there exists a transversal, T  , in H  containing a vertex from Q  , such that the following holds:

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8 The Deficiency of a Hypergraph

45|T  | ≤ ≤ = =

6n(H  ) + m(H  ) + f (|Q  |) 6(n(H ) − 11) + 13(m(H ) − 7) + f (7) 6n(H ) + 13m(H ) − 66 − 91 + 22 6n(H ) + 13m(H ) − 3 × 45.

By Claim Q.2, this implies that 45τ (H ) ≤ 45(|T  | + 3) ≤ 6n(H ) + 13m(H ), a contradiction. Therefore, Q 2 , Q 3 , and Q 4 do not contain isolated vertices in H  and they therefore all contain a vertex of degree at least 2 (in fact, equal to 2), which implies that they have degree 3 in H1∗ . As all these vertices belong to Y1∗ , we get a contradiction to the fact that no H14 -component contains three degree-3 vertices. () This completes the proof of Claim Q.10 By Claim Q.10, we assume without loss of generality that both vertices in Q 1 and in Q 2 have degree 1 in H  , which implies that the vertices in Q 1 have degree 1 in H1∗ and all vertices in the edge h ∗2 (which contains Q 2 ) have degree 2 in H1∗ . Claim Q.11: Q  ⊆ V (Y1∗ ). Proof of Claim Q.11: By Claim Q.9 we note that Q i ⊆ V (Y1∗ ) for i = 2, 3, 4. For the sake of contradiction, suppose to the contrary that Q 1  V (Y1∗ ). As Y1∗ is an H14 -component and all vertices in h ∗2 have degree 2 in H1∗ and therefore also in Y1∗ , we note that Y1∗ − h ∗2 is a component containing fourteen vertices. This implies that the component containing {x23 , x24 , x34 } in H2∗ has more than fourteen vertices as it contains all vertices of V (Y1∗ ) as well as Q 1 . This contradiction to Claim Q.9 () completes the proof of Claim Q.11. By Claim Q.11 we note that Y1∗ contains two degree-1 vertices (namely, the two vertices in Q 1 ) and an edge consisting of only degree-2 vertices (namely, the edge h ∗2 ). However no H14 -component has these properties, a contradiction. This completes () the proof of Claim Q. By Claim M and Claim P, if x is an arbitrary vertex of H of degree 3, then H − x contains an H4 -component that is intersected by all three edges incident with x, and def(H − x) = 8. By Claim Q, no edge in H contains two degree-3 vertices. Claim R: Every degree-3 vertex in H has at most one neighbor of degree 1. Proof of Claim R: Suppose, to the contrary, that there exists a vertex x of degree 3 in H with at least two degree-1 neighbors, say y and z. Let H  = H − {x, y, z}, and note that n(H  ) = n(H ) − 3 and m(H  ) = m(H ) − 3. Let e be the edge in the H4 -component of H − x. By Claim M and Claim P, the edge e is intersected by all three edges incident with x and contains a vertex (of degree 1 in H ) not adjacent to x. Further by Claim M, def(H − x) = 8. We note that def(H  ) = def(H − x) since H  is obtained from H − x by deleting the two isolated vertices y and z in H − x. Every transversal in H  can be extended to a transversal in H by adding to it the vertex x, and so τ (H ) ≤ τ (H  ) + 1. since H  is not a counterexample to our theorem, we have that 45τ (H ) ≤ 45(τ (H  ) + 1) ≤ 6n(H  ) + 13m(H  ) + def(H  ) + 45 = 6(n(H ) − 3) + 13(m(H ) − 3) + 8 + 45 = 6n(H ) + 13m(H ) − 4 () < 6n(H ) + 13m(H ), a contradiction.

8.4 Applying the Deficiency Technique

129

By Claim R, every degree-3 vertex in H has at most one neighbor of degree 1. We now define the operation of duplicating a degree-3 vertex x as follows. Let e1 , e2 , and e3 be the three edges incident with x. By Claim Q and Claim R, every neighbor of x has degree 2, except possibly for one vertex which has degree 1. Renaming edges if necessary, we may assume that the edge e1 contains no vertex of degree 1, and therefore every vertex in e1 different from x has degree 2. We now delete the edge e1 from H , and add a new vertex x  and a new edge e1 = (V (e1 ) \ {x}) ∪ {x  } to H . We note that in the resulting hypergraph the vertex x now has degree 2 (and is incident with the edges e2 and e3 ) and the new vertex x  has degree 1 with all its three neighbors of degree 2. We call x  the vertex duplicated copy of x. Let H  be obtained from H by duplicating every degree-3 vertex as described above. By construction, H  is a linear 4-uniform connected hypergraph with minimum degree δ(H  ) ≥ 1 and maximum degree Δ(H  ) ≤ 2. For i ∈ [2], let n i (H  ) be the number of vertices of degree i in H  . Then, n(H  ) = n 1 (H  ) + n 2 (H  ) and 4m(H  ) = 2n 2 (H  ) + n 1 (H  ). We proceed further with the following properties of the hypergraph H  . Claim S: The following properties hold in the hypergraph H  . (a) τ (H ) = τ (H  ). (b) def(H  ) = 0. Proof of Claim S: In order to show that τ (H ) = τ (H  ) we show that the operation of duplicating a degree-3 vertex x leaves the transversal number unchanged. Let x  , e1 , e2 , and e3 be defined as in the description of duplication. Let ex be the edge in the H4 -component in H − x and let Hx be the hypergraph obtained from H by duplicating x. Let x1∗ be the vertex (e1 ex ) in V (ex ) ∩ V (e1 ). / Tx . Now Let Tx be a transversal in Hx . As d H  (x  ) = 1, we may assume that x  ∈ we note that Tx is also a transversal in H and therefore τ (H ) ≤ τ (Hx ). Conversely assume that T is a transversal in H . If x ∈ / T , let T  = T . If x ∈ T , then since T is a transversal in H , there exists a vertex y ∈ V (ex ) ∩ T . In this case, we let T  = (T \ {y}) ∪ {x1∗ }. In both cases, |T  | = |T | and T  is a transversal in Hx , implying that τ (Hx ) ≤ |T  | = |T | = τ (H ). Consequently, τ (H ) = τ (Hx ), which implies that τ (H ) = τ (H  ) and (a) holds. In order to prove part (b) we, for the sake of contradiction, suppose that def(H  ) > 0. Let Y be a special H  -set satisfying def H  (Y ) = def(H  ) > 0. By Claim N, no edge in H contains two vertices of degree 1. We note that this is still the case after every duplication and therefore no edge in H  contains two vertices of degree 1. This implies that every H4 -component in Y4 is intersected by at least three edges of E ∗ (Y ). If Y \ Y4 = ∅, let R ∈ Y \ Y4 be arbitrary. Since Δ(H  ) ≤ 2, we note that R ∈ {H10 , H11 , H14,5 , H14,6 }. Let x be a vertex of degree 3 in H and let x  be the duplicated copy of x in H  . Furthermore let e1 be the edge containing x  in H  and let ex be the edge in the H4 -component in H − x. Finally let e2 and e3 be the edges containing x in H  . Since ex and e1 both contain a degree-1 vertex and intersect

130

8 The Deficiency of a Hypergraph

each other we note that they both cannot belong to R. If e1 ∈ E(R), then it would therefore contain two degree-1 vertices in R (namely, x  and the vertex (ex e1 ) in / E(R). Analogously ex ∈ / E(R). V (ex ) ∩ V (e1 )), a contradiction. Therefore, e1 ∈ The edges e2 and e3 cannot both belong to R as they intersect and would both contain degree-1 vertices in R (namely, the vertices (e2 ex ) and (e3 ex ) in V (e2 ) ∩ V (ex ) and V (e3 ) ∩ V (ex ), respectively). However if e2 ∈ E(R), then e2 would contain two degree-1 vertices in R (namely, x and the vertex (e2 ex ) in V (e2 ) ∩ V (ex )), a / R and analogously e3 ∈ / R. This implies that R was contradiction. Therefore, e2 ∈ also a special hypergraph in H . By Claim J, R is not isomorphic to H10 , and so R ∈ {H11 , H14,5 , H14,6 }. Furthermore if R is intersected by k edges in H  , then it is also intersected by k edges in H , and so by Claim I(a) we note that at least three edges intersect R in H and therefore also in H  . Therefore, all components in Y are intersected by at least three edges from |E ∗ (Y )|. This implies that 4|E ∗ (Y )| ≥



|V (e) ∩ V (Y )| ≥ 3|Y |,

e∈E ∗ (Y )

|Y | > 8|Y |. As observed earlier, Y10 = ∅. Hence, def(Y ) ≤ and so 13|E ∗ (Y )| ≥ 39 4 8|Y | − 13|E ∗ (Y )| < 0, a contradiction. Therefore def(H  ) = 0. This completes the () proof of Claim S. We now consider the multigraph G whose vertices are the edges of H  and whose edges correspond to the n 2 (H  ) vertices of degree 2 in H  : if a vertex of H  is contained in the edges e and f of H  , then the corresponding edge of the multigraph G joins vertices e and f of G. By the linearity of H  , the multigraph G is in fact a graph, called the dual of H  . We shall need the following properties about the dual G of the hypergraph H  . Claim T: The following properties hold in the dual, G, of the hypergraph H  . (a) G is connected, n(G) = m(H  ) and m(G) = n 2 (H  ). (b) Δ(G) ≤ 4, and so m(G) ≤ 2n(G) and 8n(G) + 6m(G) ≤ 20n(G). (c) τ (H  ) = m(H  ) − α  (G). Proof of Claim T: Since H  is connected, so too is G by construction. Further by construction, n(G) = m(H  ) and m(G) = n 2 (H  ). Since H  is 4-uniform and Δ(H  ) ≤ 2, we see that Δ(G) ≤ 4, implying that m(G) ≤ 2n(G) and 8n(G) + 6m(G) ≤ 20n(G). This establishes Part (a) and Part (b). Part (c) is well known () (see, for example, [28, 60]), so we omit the details of this property. Suppose that x is a vertex of degree 3 in H , and let x  be the vertex duplicated from x when constructing H . Adopting our earlier notation, let e1 , e2 , and e3 be the three edges incident with x. Further, let e be the edge in the H4 -component of H − x, and let y be the vertex of degree 1 in e. Let yi be the vertex common to e and ei for i ∈ [3], and so yi = (eei ) and V (e) = {y, y1 , y2 , y3 }. We note that in the graph G, which is the dual of H  , the vertex e has degree 3 and is adjacent to the vertices e1 , e2 , and e3 . Further, we note that in the graph G, the vertex e1 has degree 3, while the vertices e2 and e3 are adjacent and have degree at most 4. Further, the edge yi in G

8.4 Applying the Deficiency Technique x

131

x

y1 y1

y2

y3

y

e e1

a

a

e

y3

y2 e2

b

x

e3

b

e1

e2

e3

Fig. 8.3 The transformation creating a quadruple

is the edge eei , while the edge x in G is the edge e2 e3 . The vertex e1 is adjacent in G to neither e2 nor e3 . This set of four vertices {e, e1 , e2 , e3 } in the graph G we call a quadruple in G. We illustrate this quadruple in G in Fig. 8.3. We denote the number of (vertex-disjoint) quadruples in G by Q. We shall need the following additional property about the dual G of the hypergraph H  . Claim U: If G is the dual of the hypergraph H  , then 45τ (H  ) ≤ 6n(H  ) + 13m(H  ) − 6|Q| ⇔ 45α  (G) ≥ 8n(G) + 6m(G) + 6|Q|. Proof of Claim U: Recall that n(H  ) = n 1 (H  ) + n 2 (H  ) and 4m(H  ) = 2n 2 (H  ) + n 1 (H  ). The following holds by Claim T: 45τ (H  ) ≤ ⇔ 45(m(H ) − α  (G)) ≤ ⇔ 45α  (G) ≥ ⇔ 45α  (G) ≥ ⇔ 45α  (G) ≥ ⇔ 45α  (G) ≥ ⇔ 45α  (G) ≥ 

6n(H  ) + 13m(H  ) − 6|Q| 6n(H  ) + 13m(H  ) − 6|Q| 32m(H  ) − 6n(H  ) + 6|Q| (16n 2 (H  ) + 8n 1 (H  )) − (6n 2 (H  ) + 6n 1 (H  )) + 6|Q| 10n 2 (H  ) + 2n 1 (H  ) + 6|Q| 8m(H  ) + 6n 2 (H  ) + 6|Q| 8n(G) + 6m(G) + 6|Q|.

This completes the proof of Claim U.

()

Let S be a set of vertices in G such that (n(G) + |S| − oc(G − S))/2 is minimum. By the Tutte-Berge Formula, α  (G) =

1 (n(G) + |S| − oc(G − S)) . 2

We now consider two cases, depending on whether S = ∅ or S = ∅. Claim V: If S = ∅, then 45τ (H  ) ≤ 6n(H  ) + 13m(H  ) − 6|Q|.

(8.3)

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8 The Deficiency of a Hypergraph

Proof of Claim V: Suppose that S = ∅. For i ≥ 1, let n i (G − S) denote the number of components on G − S of order i. Let n 15 (G − S) be the number of components of G − S isomorphic to K 5 − e and let n 25 (G − S) denote all remaining components of G − S on five vertices (with at most eight edges), and so n 5 (G − S) = n 15 (G − S) + n 25 (G − S). For notational convenience, let n = n(G), m = m(G), n 15 = n 15 (G − S), n 25 = n 25 (G − S), and n i = n i (G − S) for i ≥ 1. Let ZZ + denote the set of all + + and ZZ odd denote the set of all even and odd integers, positive integers, and let ZZ even j + respectively, in ZZ . Further for a fixed j ∈ ZZ + , let ZZ ≥ j = {i ∈ ZZ | i ≥ j}, ZZ even = j {i ∈ ZZ ≥ j | i even}, and ZZ odd = {i ∈ ZZ ≥ j | i odd}. We note that n = |S| +



i · ni .

(8.4)

i∈ZZ +

By Equation (8.3) and Equation (8.4), and since oc(G − S) =



ni ,

+ i∈ZZ odd

we have the equation ⎞ ⎛   45 ⎝( 45α  (G) = 45|S| + (i − 1) · n i ) + i · ni ⎠ . 2 3 2 i∈ZZ odd

(8.5)

i∈ZZ even

Claim V.1: m ≤ 4|S| + n 2 + 3n 3 + 6n 4 + 9n 15 + 8n 25 +



(2i − 1)n i − |Q|.

i∈ZZ 6

Proof of Claim V.1: Since G is connected and Δ(G) ≤ 4, we note that if F is a component of G − S of order i, then m(F) ≤ 2i − 1. Further, every component of G − S of order 5 is either isomorphic to K 5 − e or contains at most eight edges, while every component of G − S of order 2, 3, and 4 contains at most 1, 3, and 6 edges, respectively. The above observations imply that m ≤ 4|S| + n 2 + 3n 3 + 6n 4 + 9n 15 + 8n 25 +



(2i − 1)n i .

i∈ZZ 6

We show next that each quadruple in the graph G decreases the count on the right-hand side expression of the above inequality by at least 1. Adopting our earlier notation, consider a quadruple {e, e1 , e2 , e3 }. Recall that the vertices e and e1 both have degree 3 in G, and there is no vertex in G that is adjacent to both e and e1 . Further, recall that the vertices e2 and e3 are adjacent in G. If e or e1 or if both e2 and e3 belong to the set S, then the quadruple decreases the count 4|S| by at least 1. Hence, we may assume that e, e1 , and e2 all belong to a component, C say, of G − S. In particular, we note that C has order at least 3. Abusing notation, we

8.4 Applying the Deficiency Technique

133

say that the component C contains the quadruple {e, e1 , e2 , e3 }, although possibly the vertex e3 may belong to S. Since no vertex in G is adjacent to both e and e1 , we note that if the component C has order 3, 4, or 5, then it contains at most 2, 4, and 7 edges, respectively. Further, since we define a component to contain a quadruple if it contains at least three of the four vertices in the quadruple, we note in this case when the component C has order at most 5 that it contains exactly one quadruple. Further, this quadruple decreases the count 3n 3 + 6n 4 + 9n 15 + 8n 25 by at least 1. It remains for us to consider a component F of G − S of order i ≥ 6 that contains q quadruples, and to show that these q quadruples decrease the count 2i − 1 by at least q. We note that each quadruple contains a pair of adjacent vertices of degree 3 in G. Further, at least one vertex v in F is joined to at least one vertex of S in G, implying that d F (v) < dG (v). These observations imply that 2m(F) =  v∈V (F) d F (v) ≤ 4n(F) − 2q − 1 = 4i − 2q − 1, and therefore that m(F) ≤ 2i − q − 1. Hence, these q quadruples contained in F combined decrease the count 2i − 1 () by at least q. This completes the proof of Claim V.1. By Claim V.1 and by Eq. (8.4), we have 8n + 6m + 6|Q| ≤ 32|S| + 8n 1 + 22n 2 + 42n 3 + 68n 4  + 88n 25 + 94n 15 + (20i − 6)n i .

(8.6)

i∈ZZ 6

Let Σeven =

 5 i + 6 · ni 2 6

and

Σodd =

i∈ZZ even

 1 (5i − 33) · n i . 2 7

i∈ZZ odd

We note that every (odd) component in G isomorphic to K 1 corresponds to a subhypergraph H4 in H  , while every (odd) component in G isomorphic to K 5 − e corresponds to a subhypergraph H11 in H  . Hence the odd components of G isomorphic to K 1 or isomorphic to K 5 − e correspond to a special H  -set, X say, where |X | = |X 4 | + |X 11 |, |X 4 | = n 1 and |X 11 | = n 15 . Further, the set S of vertices in G correspond to the set E ∗ (X ) of edges in H  , and so |E ∗ (X )| ≤ |S|. Thus, def H  (X ) = 8|X 4 | + 4|X 11 | − 13|E ∗ (X )| ≥ 8n 1 + 4n 15 − 13|S|. By Claim S(b), def H  (X ) ≤ def(H  ) = 0, and therefore we have that 13|S| ≥ 8n 1 + 4n 15 .

(8.7)

By Equation (8.5), and by Inequalities (8.6) and (8.7), and noting that Σeven ≥ 0 and Σodd ≥ 0, the following now holds:

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8 The Deficiency of a Hypergraph



(8.5)

45α  (G) = 32|S| + 8n 1 + 22n 2 + 42n 3 + 68n 4 + 88n 25 + 94n 15 +

(20i − 6)n i

i∈ZZ 6

+13|S| − 8n 1 + 23n 2 + 3n 3 + 22n 4 + 2n 25 − 4n 15 + Σeven + Σodd (8.6)

≥ (8n + 6m + 6|Q|) + (13|S| − 8n 1 − 4n 15 )

(8.7)

≥ 8n + 6m + 6|Q|.

Claim V now follows from Claim U.

()

Claim W: If S = ∅, then 45τ (H ) ≤ 6n(H ) + 13m(H ) − 6|Q|. Proof of Claim W: Suppose that S = ∅. Then, α  (G) = (n(G) − oc(G))/2. Since G is connected by Claim T, we have the following: 1 n(G) if n(G) is even 2  α (G) = 1 (n(G) − 1) if n(G) is odd. 2 By Claim T(b), Δ(G) ≤ 4. As every quadruple in G contains two vertices of degree 3,  dG (v) ≤ 4n(G) − 2|Q|, 2m(G) = v∈G

implying that 12n(G) ≥ 6m(G) + 6|Q|. If n(G) is even, then α  (G) = n(G)/2, and so 45 45α  (G) = n(G) > 8n(G) + 12n(G) ≥ 8n(G) + 6m(G) + 6|Q|. 2 This completes the case when n(G) is even by Claim U. Suppose next that n(G) is odd. In this case 45α  (G) = 45(n(G) − 1)/2, and so 90α  (G) = 45n(G) − 45 = 21n(G) + 24n(G) − 45 ≥ 21n(G) + 12m(G) + 12|Q| − 45.

If 5n(G) ≥ 45, then 90α  (G) ≥ 16n(G) + 12m(G) + 12|Q|, which completes the proof by Claim U. We may therefore assume that 5n(G) < 45, implying that n(G) ∈ {1, 3, 5, 7}. Since every quadruple contains four vertices, we must therefore have |Q| ≤ 1. We first consider the case when |Q| = 1. In this case n(G) ≥ 6, as the quadruple contains four vertices and one vertex (called e1 in the definition of a quadruple) has two neighbors outside the quadruple. Therefore, n(G) = 7. Since two vertices in the quadruple have degree at most 3, 2m(G) = v∈V (G) d(v) ≤ 4n(G) − 2 = 26, and so m(G) ≤ 13. If m(G) ≤ 12, then 45α  (G) = 45 · 3 > 8 · 7 + 6 · 12 + 6 ≥ 8n(G) + 6m(G) + 6|Q|, and the desired result follows from Claim U. Therefore, we may assume that m(G) = 13, for otherwise the case is complete. In this case, all vertices in G have degree 4 except for two vertices in the quadruple which have

8.4 Applying the Deficiency Technique

135

Fig. 8.4 The graph G if |Q| = 1

e e1

e2

e3

u1

u2

w

degree 3. Let {e, e1 , e2 , e3 } be the vertices in the quadruple in G, such that d(e) = d(e1 ) = 3 and e, e2 , and e3 form a 3-cycle in G. Define u 1 and u 2 such that N (e1 ) = {e, u 1 , u 2 } and define w, such that V (G) = {e, e1 , e2 , e3 , u 1 , u 2 , w}. As d(w) = 4 and w is not adjacent to e or e1 we have N (w) = {e2 , e3 , u 1 , u 2 }. Therefore, e2 is adjacent to e, e3 , and w. Its fourth neighbor is either u 1 or u 2 . Renaming u 1 and u 2 if necessary, we may assume that N (e2 ) = {e, e3 , w, u 2 }. This implies that u 1 must be adjacent to u 2 and e3 and G is the graph shown in Fig. 8.4. If we draw the corresponding hypergraph whose dual is the graph G, we note that it is obtained by duplicating the degree-3 vertex in H14,3 . However, H is not equal to H14,3 by Claim C, implying that G cannot be the graph in Fig. 8.4, a contradiction. This completes the case when |Q| = 1. Therefore, we may assume that |Q| = 0. If n(G) = 1, then H = H4 , contradicting Claim C. Hence, n(G) ∈ {3, 5, 7}. Suppose that n(G) = 3. Then, α  (G) = 1 and m(G) ≤ 3. In this case, 8n(G) + 6m(G) ≤ 8 · 3 + 6 · 3 = 42 < 45 = 45α  (G), which by Claim U completes the proof. Hence we may assume that n(G) = 5 or n(G) = 7. Suppose that n(G) = 5. Then, α  (G) = 2 and by Claim T(b), m(G) ≤ 10. If m(G) = 10, then G = K 5 . In this case, H is a 4-uniform 2-regular linear intersecting hypergraph. However, H10 is the unique such hypergraph as shown, for example, in [28, 60]. Thus if m(G) = 10, then H = H10 , contradicting Claim C. Hence, m(G) ≤ 9. If m(G) = 9, then G = K 5 − e, where e denotes an arbitrary edge in K 5 . In this case, H = H11 , contradicting Claim C. Hence, m(G) ≤ 8. Thus, 8n(G) + 6m(G) ≤ 8 · 5 + 6 · 8 = 88 < 90 = 45α  (G), which by Claim U completes the proof in this case. Finally suppose that n(G) = 7. Then, α  (G) = 3 and by Claim T(b), m(G) ≤ 14. Suppose that m(G) = 14. Then, G is a 4-regular graph of order 7. Equivalently, the complement, G, of G is a 2-regular graph of order 7. If G = C3 ∪ C4 , then H = H14,2 . If G = C7 , then H = H14,4 . Both cases contradict Claim C. Hence, m(G) ≤ 13. Thus, 8n(G) + 6m(G) ≤ 8 · 7 + 6 · 13 = 134 < 135 = 45α  (G), which by () Claim U completes the proof. Recall that n(H  ) = n(H ) + |Q| and m(H  ) = m(H ). By Claim S(a), Claim V, and Claim W, 45τ (H ) = 45τ (H  ) ≤ 6n(H  ) + 13m(H  ) − 6|Q| = 6n(H ) + 13m(H ), a contradiction. This completes the proof of Theorem 8.5. 

Chapter 9

The Tuza Constant q4

9.1 Introduction In this chapter, we illustrate the power of Theorem 8.5 proven in the previous chapter using our new technique of the deficiency of a hypergraph. We apply this result to obtain best possible upper bounds on the transversal number of a 4-uniform linear hypergraph, and determine the Tuza constant q4 . By Theorem 5.13, the Tuza constant c4 = 29 . Moreover if H ∈ H4 and τ (H ) = c4 (n H + m H ), then every component of H is the generalized triangle T4 illustrated in Figure 5.1(a). Although the generalized triangle T4 is not linear, it is still possible that q4 ≤ 29 since q4 is the supremum of the ratio τ (H )/(n H + m H ) over all hypergraphs H ∈ L4 . In this chapter, we determine the exact value of the Tuza constant q4 .

9.2 The Tuza constant q4 In this section, we show that as a consequence of Theorem 8.5, the Tuza constant q4 can be readily determined. By Observation 4.1, q4 ≥ 15 and if H ∈ L4 consists of a single edge E 4 (also named H4 in Figure 8.1(a)), then τ (H ) = 1 = 15 (4 + 1) = 1 (n H + m H ). If H ∈ L4 is the 2-regular, linear hypergraph shown in Figure 8.1(b), 5 which we redraw in Figure 9.1, then τ (H ) = 3 = 15 (10 + 5) = 15 (n H + m H ), once again showing that q4 ≥ 15 . We remark that the hypergraph H10 is precisely the linear intersection hypergraph L 4 defined in Chapter 2.1 (and illustrated in Figure 2.1). As an application of Theorem 8.5, we show that this lower bound of 15 on the Tuza constant q4 is in fact also an upper bound. Theorem 9.1 ([63]) The Tuza constant q4 = 15 . Proof. Let H ∈ L4 have order n H = n(H ) and size m H = m(H ) edges. We show that τ (H ) ≤ 15 (n H + m H ). We proceed by induction on n H . If n H = 4, then H con© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_9

137

138

9 The Tuza Constant q4

Fig. 9.1 The hypergraph H10

sists of a single edge E 4 , and τ (H ) = 1 = 15 (4 + 1) = 15 (n H + m H ). Let n H ≥ 5 and suppose that the result holds for all hypergraphs in L4 on fewer than n H vertices. Let H ∈ L4 have order n H and size m H edges. Suppose that Δ(H ) ≥ 4. Let v be a vertex of maximum degree in H , and consider the 4-uniform, linear hypergraph H  = H − v. Let H  have order n H and size m H . We note that n H = n H − 1 and m H = m H − Δ(H ) ≤ m H − 4. Every transversal in H  can be extended to a transversal in H by adding to it the vertex v. Hence, applying the inductive hypothesis to H  , we have that τ (H ) ≤ τ (H  ) + 1 ≤ 15 (n H + m H ) + 1 ≤ 1 (n H + m H − 5) + 1 = 15 (n H + m H ). Hence, we may assume that Δ(H ) ≤ 3, for 5 otherwise the desired result follows. Thus, H ∈ L4,3 where recall that L4,3 is the class of all 4-uniform, linear hypergraphs with maximum degree at most 3. We note that in this case, 4m H ≤ 3n H . Applying Theorem 8.5 to the hypergraph H , we have 45τ (H ) ≤ 6n H + 13m H + def(H ). If def(H ) = 0, then 45τ (H ) ≤ 6n H + 13m H = (9n H + 9m H ) + (4m H − 3n H ) ≤ 9(n H + m H ), and so τ (H ) ≤ 15 (n H + m H ). Hence, we may assume that def(H ) > 0, for otherwise the desired result follows. Among all special nonempty H -sets, let X be chosen so that |E ∗ (X )| − |X | is minimum. We note that since def(H ) > 0, |E ∗ (X )| − |X | < 0. As in Section 8.3, we associate with the set X a bipartite graph, G X , with partite sets X and E ∗ (X ), where an edge joins e ∈ E ∗ (X ) and H  ∈ X in G X if and only if the edge e intersects the subhypergraph H  of X in H . Suppose that there is no matching in G X that matches E ∗ (X ) to a subset of X . By Hall’s Theorem, there is a nonempty subset S ⊆ E ∗ (X ) such that |NG X (S)| < |S|. We now consider the special H -set, X  = X \ NG X (S), and note that |X  | = |X | − |NG X (S)| > |E ∗ (X )| − |S| ≥ 0 and |E ∗ (X  )| = |E ∗ (X )| − |S|. Thus, X  is a special nonempty H -set satisfying |E ∗ (X  )| − |X  | = (|E ∗ (X )| − |S|) − (|X | − |NG X (S)|) = (|E ∗ (X )| − |X |) + (|NG X (S)| − |S|) < |E ∗ (X )| − |X |, contradicting our choice of the special H -set X . Hence, there exists a matching in G X that matches E ∗ (X ) to a subset of X . By Observation 8.1(f), there exists a minimum

9.2 The Tuza constant q4

139

Table 9.1 Hypergraphs H ∈ L4 satisfying τ (H ) = 15 (n H + m H ) H = F16 (X ) τ (H )

nH

mH

1 5 (n H

|X | = 1 |X | = 2

6 5

15 14

15 11

6 5

|X | = 3

4

13

7

4

|X | = 4

3

12

3

3

+ mH )

Average degree 60/15 = 4 44/14 ≈ 3.14 . . . 28/13 ≈ 2.15 . . . 12/12 = 1

X -transversal, TX , that intersects every edge in E ∗ (X ). By Observation 8.1, every special hypergraph F satisfies τ (F) ≤ 15 (n F + m F ) where F has order n F and size m F . Hence, letting n(X ) =



nF

and

F∈X

m(X ) =



mF ,

F∈X

we note that |TX | =

 F∈X

τ (F) ≤

1 1 (n F + m F ) = (n(X ) + m(X )). 5 5 F∈X

We now consider the 4-uniform, linear hypergraph H  = H − V (X ). Let H  have order n H and size m H . We note that n H = n H − n(X ) and m H = m H − m(X ) − |E ∗ (X )| ≤ m H − m(X ). Every transversal in H  can be extended to a transversal in H by adding to it the set TX . Hence, applying the inductive hypothesis to H  , we have that τ (H ) ≤ τ (H  ) + |TX | ≤ 15 (n H + m H ) + |TX | ≤ 15 (n H + m H ) − 15 (n(X ) + m(X )) + |TX | ≤ 15 (n H + m H ). Thus, q4 ≤ 15 . By Observation 4.1, q4 ≥ 15 . Consequently, q4 = 15 .



As further examples showing hypergraphs H ∈ L4 for which τ (H ) = 15 (n H + m H ), we illustrate Theorem 3.2 presented in Chapter 3 in the case when k = 4. When k = 4 we know that the affine plane AG(2, 4) exists. Therefore by Theorem 3.2, the following hypergraphs H in Table 9.1 satisfy τ (H ) = 15 (n H + m H ), where H = F16 (X ) is defined as in the statement of Theorem 3.2. By Theorem 9.1, τ (H ) ≤ 15 (n H + m H ) for every hypergraph H ∈ L4 . We remark that this bound is not necessary true for the non-linear hypergraph case; that is, for hypergraphs H ∈ H4 \ L4 . As observed in Chapter 4, if H is the generalized

140

9 The Tuza Constant q4

Fig. 9.2 The Fano plane F

triangle T4 shown in Figure 5.1(a), then H ∈ H4 \ L4 and τ (H ) = 29 (n H + m H ) > 1 (n H + m H ). 5 As a further example, let H = F be the complement of the Fano plane F, where the Fano plane is shown in Figure 9.2 and where its complement F is the hypergraph on the same vertex set V (F) and where e is a hyperedge in the complement if and only if V (F) \ e is a hyperedge in F. In this case, H = F ∈ H4 \ L4 and 3 (n H + m H ) > 15 (n H + m H ). τ (H ) = 3 = 14 If H is a 3-regular, 4-uniform, linear hypergraph of order n H and size m H , then m H = 43 n H , and so, by Theorem 9.1, τ (H ) ≤

1 3 7 1 (n H + m H ) = (n H + n H ) = n . 5 5 4 20 H

Hence, as an immediate corollary of Theorem 9.1, we have the following result. Corollary 9.1 If H ∈ L4 is 3-regular, then τ (H ) ≤

7 n 20 H

= 0.35n H .

9.3 The West Bound when k = 4 In this section, we show that the West bound holds for 4-uniform, linear hypergraphs; that is, we prove that Conjecture 7.1 is true when k = 4. Theorem 9.2 ([63]) If H ∈ L4 , then τ (H ) < 14 n H + 16 m H . Proof. Let H ∈ L4 have order n H and size m H . We show that τ (H ) ≤ 41 n H + 16 m H . We proceed by induction on n H . If n H = 4, then H consists of a single edge E 4 , and τ (H ) = 1 < 41 n H + 16 m H . Let n H ≥ 5 and suppose that the result holds for all 4-uniform, linear hypergraphs on fewer than n H vertices. Let H ∈ L4 have order n H and size m H . Suppose that Δ(H ) ≥ 6. Let v be a vertex of maximum degree in H , and consider the 4-uniform, linear hypergraph H  = H − v. Let H  have order n H and size m H .

9.3 The West Bound when k = 4

141

We note that n H = n H − 1 and m H = m H − Δ(H ) ≤ m H − 6. Every transversal in H  can be extended to a transversal in H by adding to it the vertex v. Hence, applying the inductive hypothesis to H  , we have that τ (H ) ≤ 1 + τ (H  ) < 1 + 41 n H + 16 m H ≤ 1 + 41 (n H − 1) + 16 (m H − 6) = 41 n H + 16 m H . Hence, we may assume that Δ(H ) < 6, for otherwise τ (H ) < 41 n H + 16 m H . In this case, 2m H < 3n H . By Theorem 9.1, 60τ (H ) ≤ 12n H + 12m H = 15n H + 10m H + 2m H − 3n H < 15n H + 10m H , or, equivalently, τ (H ) < 41 n H + 16 m H .



9.4 Linear Hypergraphs with Maximum Degree Two In the special case of Theorem 2.3 when k = 4, we have that if H ∈ L4 and Δ(H ) ≤ 2, then τ (H ) ≤ 15 (n H + m H ) with equality if and only if every component of H consists of a single edge E k or every component of H is the linear intersecting hypergraph H10 (also called L 4 in Chapter 2). We show in this section that the 1 3 1 (n H + m H ) bound can be further improved to 16 (n H + m H ) + 16 if we exclude the 5 special hypergraph H10 . For this purpose, we construct a family, F , of 4-uniform, connected, linear hypergraphs with maximum degree Δ(H ) = 2 as follows. Let F0 be the hypergraph with one edge (illustrated in Figure 8.1(a), but with a different name, H4 ). For i ≥ 1, we now build a hypergraph Fi inductively as follows. Let Fi be obtained from Fi−1 by adding 12 new vertices, adding three new edges so that each new vertex belongs to exactly one of these added edges, and adding one further edge that contains a vertex in V (Fi−1 ) and three additional vertices, one from each of the three newly added edges, in such a way that Δ(Fi ) = 2. Let F be the family of all such hypergraphs, Fi , where i ≥ 0. A hypergraph, F6 , in the family F is illustrated in Figure 9.3. We proceed further with the following lemma. Let c(H ) denote the number of components of a hypergraph H . Recall that if X is a special H -set, we write E ∗ (X ) to denote the set E ∗H (X ) if the hypergraph H is clear from context. Lemma 9.1 If H ∈ L4 and X is a special H -set, then 3|E ∗H (X )| ≥ |X | − c(H ).

142

9 The Tuza Constant q4

Fig. 9.3 A hypergraph, F6 , in the family F

Proof. We proceed by induction on |E ∗H (X )| = k ≥ 0. If k = 0, then c(H ) ≥ |X |, and so 3|E ∗ (X )| = 0 ≥ |X | − c(H ). This establishes the base case. Suppose k ≥ 1 and the result holds for special H -sets, X , such that |E ∗H (X )| < k. Let X be a special H -set satisfying |E ∗H (X )| = k. Let e ∈ E ∗H (X ) and consider the hypergraph H  = H − e. We note that H  ∈ L4 , and that c(H  ) ≤ c(H ) + 3. Further, the set X is a special H  -set satisfying |E ∗H  (X )| = k − 1. Applying the inductive hypothesis to the hypergraph H  ∈ L4,3 and to the special H  -set, X , we have 3(|E ∗H (X )| − 1) = 3|E ∗H  (X )| ≥ |X | − c(H  ) ≥ |X | − (c(H ) + 3), implying that 3|E ∗H (X )| ≥ |X | − c(H ).  We are now in a position to state the following result, where H10 , H14,5 , and H14,6 are the 4-uniform, linear hypergraphs shown in Figure 8.1(b), 8.1(h) and 8.1(i), respectively. As observed earlier, H4 = F0 , and so H4 ∈ F . Theorem 9.3 ([63]) Let H ∈ L4 be a connected hypergraph such that Δ(H ) ≤ 2 3 1 (n H + m H ) + 16 , with equality if and only if H ∈ and H = H10 . Then, τ (H ) ≤ 16 {H14,5 , H14,6 } or H ∈ F . Proof. Let H ∈ L4 be a connected hypergraph such that Δ(H ) ≤ 2 and H = H10 . Let H have order n H and size m H . Suppose firstly that H is a special hypergraph. By assumption, H = H10 . Since Δ(H ) ≤ 2, we note that H ∈ {H4 , H11 , H14,5 , H14,6 }. 3 (n H + m H ). If H ∈ {H4 , H14,5 , If H = H11 , then by Observation 8.1(c), τ (H ) = 16 3 1 H14,6 }, then by Observation 8.1(a) and (d), τ (H ) = 16 (n H + m H ) + 16 . Hence, we may assume that H is not a special hypergraph, for otherwise the desired result holds, noting that H4 ∈ F . Since Δ(H ) ≤ 2, we observe that m H ≤ 21 n H . By Theorem 8.5, 45τ (H ) ≤ 6n H + 13m H + def(H ). If def(H ) = 0, then, since 0 ≤ 21 n H − m H , we have

9.4 Linear Hypergraphs with Maximum Degree Two

143

45τ (H ) ≤ 6n H + 13m H ≤ 6n H + 13m H + = (6 + =

14 )n H 6

25 (n H 3

+

14 1 ( n − mH ) 3 2 H (13 − 14 )m H 3

+ m H ),

5 3 or, equivalently, τ (H ) ≤ 27 (n H + m H ) < 16 (n H + m H ). Hence, we may assume that def(H ) > 0, for otherwise the desired result follows. Let X be a special H -set such that def(H ) = def H (X ). If H10 belongs to X , then, since Δ(H ) ≤ 2 and H is connected, H = H10 , a contradiction. If H14,5 or H14,6 belong to X , then, analogously, H ∈ {H14,5 , H14,6 }, contradicting our assumption that H is not a special hypergraph. Thus, if F ∈ X , then F ∈ {H4 , H11 }, noting that Δ(H ) ≤ 2. Recall that if F is a hypergraph, we denote by n 1 (F) the number of vertices of degree 1 in H . Let

n 1 (X ) =



n 1 (F),

F∈X

and note that n 1 (H ) ≥ n 1 (X ) − 4|E ∗ (X )|, since every edge in E ∗ (X ) contains at most four vertices whose degree is 1 in some subhypergraph F ∈ X . Since Δ(H ) ≤ 2 and H is 4-uniform, we note that 4m H = 2n H − n 1 (H ), or, equivalently, . If F = H4 , then n 1 (F) = 4 and def(F) = 8 = n 1 (H ) = 2n H − 4m H . Let β = 73 64 (8 − 4β) + 4β = (8 − 4β) + n 1 (F) · β. If F = H11 , then n 1 (F) = 2 and def(F) = 4 < 5.71875 = 8 − 2β = (8 − 4β) + n 1 (F) · β. Hence, if F ∈ X , then def(F) ≤ (8 − 4β) + n 1 (F) · β, with strict inequality if F = H11 . Therefore, def(H ) = 8|X 4 | + 4|X 11 | − 13|E ∗ (X )|    = def(F) − 13|E ∗ (X )| F∈X

≤ (8 − 4β)|X | + n 1 (X ) · β − 13|E ∗ (X )|, with strict inequality if X = X 4 . By Lemma 9.1, |E ∗ (X )| ≥ 13 (|X | − 1). We also note that n H ≥ 4|X 4 | + 11|X 11 | ≥ 4|X |, and so |X | ≤ 14 n H . Thus by our previous observations, 45τ (H ) ≤ 6n H + 13m H + def(H )

≤ 6n H + 13m H + (8 − 4β)|X | + n 1 (X ) · β − 13|E ∗ (X )|

≤ 6n H + 13m H + (8 − 4β)|X | + (n 1 (H ) + 4|E ∗ (X )|) · β − 13|E ∗ (X )|

= 6n H + 13m H + (8 − 4β)|X | + n 1 (H ) · β − |E ∗ (X )|(13 − 4β)

≤ 6n H + 13m H + (8 − 4β)|X | + n 1 (H ) · β − 13 (|X | − 1)(13 − 4β) = 6n H + 13m H + (8 − 4β − 13 (13 − 4β))|X | + (2n H − 4m H ) · β + 13 (13 − 4β)

n ≤ 6n H + 13m H + (8 − 4β − 13 (13 − 4β)) 4H + (2n H − 4m H ) · β + 13 (13 − 4β)

83 + 4 β)n + (13 − 4β)m + 1 (13 − 4β) = ( 12 H H 3 3

144

9 The Tuza Constant q4

= (13 − 4β)(n H + m H ) + 13 (13 − 4β) 135 = 135 16 (n H + m H ) + 48 , 3 1 or, equivalently, τ (H ) ≤ 16 (n H + m H ) + 16 . This establishes the desired upper bound. Recall that H ∈ L4 is a connected hypergraph such that Δ(H ) ≤ 2 and H = H10 . 3 1 (n H + m H ) + 16 . Then we must have equality throughSuppose that τ (H ) = 16 out the above inequality chain. This implies that X = X 4 , V (H ) = V (X ), n H = 4|X |, E(H ) = E(X ) ∪ E ∗ (X ), and |E ∗ (X )| = 13 (|X | − 1). We show by induction on n H ≥ 4 that these conditions imply that H ∈ F . When n H = 4, |X | = 1 and |E ∗ (X )| = 0, and so H = H4 ∈ F . This establishes the base case. Suppose that n H > 4. Thus, |X | ≥ 2 and n H = 4|X | ≥ 8. We now consider the bipartite graph, G X , with partite sets X and E ∗ (X ), where an edge joins e ∈ E ∗ (X ) and H  ∈ X in G X if and only if the edge e intersects the subhypergraph H  of X in H . Since H is 4-uniform and linear, each vertex in E ∗ (X ) has degree 4 in G X . Let n 1 = n 1 (G), and so n 1 is the number of vertices of degree 1 in G. Counting the edges in G, we note that 43 (|X | − 1) = 4|E ∗ (X )| = m(G) ≥ n 1 + 2(|X | − n 1 ), implying that n 1 ≥ 13 (2|X | + 4). By the Pigeonhole Principle, there is a vertex of E ∗ (X ) adjacent in G to at least   2|X |+4 3 6 n1 2|X | + 4  = =  =2+ |X |−1 |E ∗ (X )| |X | − 1 |X | − 1 3

vertices of degree 1 (that belong to X ). Thus, since |X | ≥ 2 here, some vertex e ∈ E ∗ (X ) in G X is adjacent to three vertices of degree 1, say x1 , x2 , and x3 . Let x4 be the remaining neighbor of e in G X . We now consider the hypergraph H  obtained from H by deleting the 12 vertices from the three special H4 -subhypergraphs, say F1 , F2 , and F3 , corresponding to x1 , x2 , and x3 , respectively, and deleting the hyperedge corresponding to e. Since H is connected and linear, so too is H  . Let X  = X \ {F1 , F2 , F3 }, and so |X  | = |X | − 3. We note that |E ∗H  (X  )| = |E ∗H (X )| − 1 = 13 (|X | − 1) − 1 = 13 (|X  | − 1). Further, X  = X 4 , V (H  ) = V (X  ), n  = 4|X  |, and E(H  ) = E(X  ) ∪ E ∗H  (X  ). Applying the inductive hypothesis to H  , we deduce that H  ∈ F . The original hypergraph H can now be reconstructed from H  by adding back the three deleted edges and 12 deleted vertices in F1 ∪ F2 ∪ F3 , and adding back the deleted edge e that contains the vertex x4 ∈ V (H  ) and contains one vertex from each edge in F1 , F2 , and F3 . Thus, H ∈ F . This completes the proof of Theorem 9.3. 

9.5 Application to Total Dominating Sets

145

9.5 Application to Total Dominating Sets A total dominating set, also called a TD-set, of a graph G with no isolated vertex is a set S of vertices of G such that every vertex is adjacent to a vertex in S. The total domination number of G, denoted by γt (G), is the minimum cardinality of a TD-set of G. Total domination in graphs is now well studied in graph theory. The literature on this subject has been surveyed and detailed in a recent book on this topic that can be found in [58]. The Heawood graph, shown in Figure 9.4, is the unique 6-cage. The bipartite complement of the Heawood graph is the bipartite graph formed by taking the two partite sets of the Heawood graph and joining a vertex from one partite set to a vertex from the other partite set by an edge whenever they are not joined in the Heawood graph. The bipartite complement of the Heawood graph can also be seen as the incidence bipartite graph of the complement of the Fano plane. Thomassé and Yeo [85] established the following upper bound on the total domination number of a graph with minimum degree at least 4. Recall that δ(G) denotes the minimum degree of a graph G. Theorem 9.4 ([85]) If G is a graph of order n with δ(G) ≥ 4, then > (G) ≤ 37 n. The extremal graphs achieving equality in the Thomassé-Yeo bound of Theorem 9.4 are given by the following result. Theorem 9.5 ([56, 58]) If G is a connected graph of order n with δ(G) ≥ 4 that satisfies γt (G) = 37 n, then G is the bipartite complement of the Heawood Graph. We remark that every vertex in the bipartite complement of the Heawood Graph belongs to a 4-cycle. It is therefore a natural question to ask whether the ThomasséYeo upper bound of 37 n can be improved if we restrict G to contain no 4-cycles. As a consequence of Theorem 9.1, this question can now be answered in the affirmative. For a graph G, the open neighborhood hypergraph, abbreviated ONH, of G is the hypergraph HG with vertex set V (HG ) = V (G) and with edge set E(HG ) = {NG (x) | x ∈ V (G)} consisting of the open neighborhoods of vertices

Fig. 9.4 The Heawood graph

146

9 The Tuza Constant q4

Fig. 9.5 A quadrilateral-free 4-regular graph G 30 of order n = 30 with γt (G 30 ) = 25 n

in G. As first observed in [85] (see also [58]), the transversal number of the ONH of a graph is precisely the total domination number of the graph; that is, for a graph G, we have γt (G) = τ (HG ). As an application of Theorem 9.1, we have the following result, which significantly improves the upper bound of Theorem 9.4 when the graph G contains no 4-cycle. Theorem 9.6 ([63]) If G is a quadrilateral-free graph of order n with δ(G) ≥ 4, then γt (G) ≤ 25 n. Proof. Let G be a quadrilateral-free graph of order n with δ(G) ≥ 4 and let HG be the ONH of G. Then, each edge of HG has size at least 4. Since G contains no 4cycle, the hypergraph HG contains no overlapping edges and is therefore linear. Let H be obtained from HG by shrinking all edges of HG , if necessary, to edges of size 4. Then, H is a 4-uniform linear hypergraph of order n H = n and size m H = n; that is, n H = m H = n(G) = n. By Theorem 9.1 we note that τ (H ) ≤ 15 (n H + m H ) = 25 n.  This completes the proof of the theorem since γt (G) = τ (HG ) ≤ τ (H ). That the bound in Theorem 9.6 is best possible may be seen by taking, for example, the 4-regular bipartite quadrilateral-free graph G 30 of order n = 30 illustrated in Figure 9.5 satisfying γt (G 30 ) = 12 = 25 n. We note that the graph G 30 is the incidence bipartite graph of the linear 4-uniform hypergraph obtained by removing an arbitrary vertex from the affine plane AG(2, 4) of order 4.

Chapter 10

The Tuza Constant qk for Large k

10.1 Introduction In this chapter, we determine the asymptotic behavior of the Tuza constant, qk , as k grows. Recall that in Chapter 6 we presented results due to Alon [2] who determined the asymptotic behavior of the Tuza constant ck as k grows, and showed that 

ln(k) ck = (1 + o(1)) k

 as k → ∞.

In this chapter, we apply probabilistic arguments to determine the asymptotic behavior of the Tuza constant, qk , as k grows.

10.2 The Asymptotic Behavior of the Tuza Constant qk In this section, we show that qk approaches ck as k gets large. For this purpose, we shall prove a result that was motivated by an important paper by Alon, Kalai, Matousek and Meshulam [3] (pictured below) on transversal numbers for hypergraphs arising in geometry where each edge in a projective plane is shrunk randomly.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_10

147

148

10 The Tuza Constant qk for Large k

Gil Kalai

Jiri Matousek

Roy Meshulam

However, we remark that although the ingenious construction by Alon et al. is applied to projective planes, our result, namely, Theorem 10.1, can be used on general linear hypergraphs H which are not necessarily projective planes. We shall need the following well-known lemma. x x−1   . Lemma 10.1 For all x > 1, we have 1 − x1 < e−1 < 1 − x1 The proof of the following result due to the authors can be found in [64]. For completeness, we present the proof in this chapter. Theorem 10.1 ([64]) For k ≥ 2, let H be an arbitrary 2k-uniform 2k-regular hypergraph on n vertices and let 0



 c ln(k) n. k

10.2 The Asymptotic Behavior of the Tuza Constant qk

149

Proof. Let H and H  be defined as in the statement of Theorem 10.1. Let E(H ) = {e1 , e2 , . . . , en } and let ei be the edge in H  obtained by picking k vertices at random from ei for i ∈ [n]. Let T be a random set of vertices in H  where  |T | =

c ln(k) n k



and let ti = |T ∩ V (ei )| for each i ∈ [n]. The probability, Pr(ei not covered), that the edge ei is not covered by T (that is, V (ei ) ∩ T = ∅) is given by Pr(ei

2k−ti  not covered) = 2kk  . k

Given the values ti for all i ∈ [n], we note that the probability of an edge ei being covered by T is independent of an edge e j being covered by T where j ∈ [n] \ {i}. Therefore, the probability that T is a transversal of H  is given by Pr(T is a transversal of H  ) =

n    1 − Pr(ei not covered) i=1

=

n 



2k−ti 

1 − 2kk 

i=1

.

(10.1)

k

As every vertex in T belongs to 2k edges of H , we note that n 

ti = 2k|T |

i=1

by double counting. Let s1 , s2 , . . . , sn be nonnegative integers, such that the expression 2k−si 

n  (10.2) 1 − 2kk  i=1

is maximized, where

n 

k

si = 2k|T |.

i=1

We proceed further with the following series of claims.

(10.3)

150

10 The Tuza Constant qk for Large k

Claim 10.2 |si − s j | ≤ 1 for all i, j ∈ [n]. Proof. For the sake of contradiction, suppose that si ≤ s j − 2 for some i and j where i, j ∈ [n]. We will now show that by increasing si by one and decreasing s j by one we increase the value of the maximized expression in (10.2), a contradiction. Hence it suffices for us to show that Inequality (10.4) holds: 1−

2k−(si +1) 2kk  k

2k−(s j −1)

1−

> 1−

2kk  k

2k−si  2kk  k

2k−s j 

1 − 2kk 

. (10.4)

k

We remark that Inequality (10.4) is equivalent to the following: 

               2k 2k − s j 2k 2k − si − 1 2k 2k − s j + 1 2k 2k − si − . − − > − k k k k k k k k

Defining Ai , A j , Bi , B j , Ci , C j as follows,         j i Ai = 2kk − 2k−s A j = 2kk − 2k−s k k     j +1 i −1 Bi = 2k−s B j = 2k−s k−1 k−1         Ci = 2kk − 2k−sk i −1 C j = 2kk − 2k−sk j +1 , we note that Inequality (10.4) is equivalent to C i C j > Ai A j . By Pascal’s rule, which states that Ci = Cj =

2k  k

−

2k  k

n  k

  2k−si

−

k



=

−

2k−s j  k

n−1 k−1

(10.5) +

2k−si −1

+

k−1

2k−s j  k−1

n−1 , we obtain the following: k = Ai + Bi = Aj − Bj.

Since si ≤ s j − 2, we note that A j > Ci = Ai + Bi and Bi ≥ B j , implying that Ci C j = (Ai + Bi )(A j − B j ) ≥ (Ai + Bi )(A j − Bi ) = Ai A j + Bi (A j − (Ai + Bi )) > Ai A j . Thus, Inequality (10.5) holds, producing the desired contradiction. This completes () the proof of Claim 10.2.

10.2 The Asymptotic Behavior of the Tuza Constant qk

151

Let Pr(H ) denote the probability that τ (H  ) ≤ integer, we note that Pr(H ) = Pr(τ (H  ) ≤ |T |). 2k−si 

  n n Claim 10.3 Pr(H ) ≤ 1 − 2kk  . |T | i=1 k



c ln(k) k

n. Since τ (H  ) is an

  Proof. As there are |Tn | possible ways of choosing the set T , Equation (10.1) implies that the probability that there exists a transversal of H  of size at most |T | is bounded as follows:   n  × Pr(T is a transversal of H  ) Pr(τ (H ) ≤ |T |) ≤ |T | 2k−ti 

n  |T | ≤n · 1 − 2kk  i=1

≤ n |T | ·

n 



k

2k−si 

1 − 2kk 

i=1

.

k

This completes the proof of Claim 10.3. () Let sav denote the average value of the integers s1 , s2 , . . . , sn , and let s ∗ = 2c ln(k). Claim 10.4 sav ≤ s ∗ . Proof. Since

n 

si = 2k|T |, the average value of s1 , s2 , . . . , sn is bounded as

i=1

follows: sav =

2k c ln(k) n 2k|T | k = ≤ 2c ln(k) = s ∗ . n n

This completed the proof of Claim 10.4. 2k−si    1 1 si k Claim 10.5 2k  > . 5 2 k

()

Proof. We note that 2k−si  k 2k  k

=

k(k − 1) · · · (k − si + 1) (2k − si )! k!k! = ≥ k!(k − si )!) (2k)! (2k)(2k − 1) · · · (2k − si + 1)



 k − si + 1 si . 2k − si + 1

152

10 The Tuza Constant qk for Large k

Let θ (si ) = Since θ (si )si + 1 = 

2k−si +1 , si −1

k − si + 1 2k − si + 1

si

2k − 2si + 2 . si (si − 1)

we note by Lemma 10.1 that the following holds:

 1 2k − 2si + 2 si = × 2 2k − si + 1 si  si  1 si − 1 = × 1− 2 2k − si + 1

si  si 1 1 = × 1 − 2k−si +1 2 s −1 

i

θ(si )si θ (s1i )  si  1 1 1− = × 2 θ (si )si + 1  si   1 1 > × e−1 θ (si ) . 2 In order to prove our desired result, it therefore suffices for us to prove that e

− θ (s1

i)

≥

1 . 5

By Claims 10.2 and 10.4, we note that si ≤ s ∗ + 1 for all i ∈ [n]. As θ (si ) is a decreasing function, the function e−1/θ(si ) is also a decreasing function in si . Hence ∗ it suffices for us to show that e−1/θ(s +1) ≥ 1/5. We note that

−

e

1 (s ∗ + 1)s ∗ (1 + 2c ln(k))2c ln(k) . − − + 1) = e 2k − 2(s ∗ + 1) + 2 = e 2k − 4c ln(k)

θ (s ∗

(10.6)

Since c < 1/ ln(4),

1 + 2 ln(k) 2 ln(k) ln(4) (1 + 2c ln(k))2c ln(k) ≤ . 2k − 4c ln(k) 2 ln(4)k − 4 ln(k) The maximum value of the function on the right-hand side of the above inequality is approximately 1.5037 obtained at k = 3.753, which is always less than ln(5) ≈ 1.6094. Thus, (1 + 2c ln(k))2c ln(k) < ln(5), 2k − 4c ln(k)

10.2 The Asymptotic Behavior of the Tuza Constant qk

153

implying by (10.6) that 1 1 ≥ ∗ + 1) θ (s 5 e . −

This completes the proof of Claim 10.5.   ∗

n   n 1 1 s Claim 10.6 Pr(H ) ≤ . 1− 5 2 |T |

()

Proof. For a nonnegative real number s, let f (s) = 1 −

1 5

 s 1 . 2

We show firstly that f (x − y) f (x + y) ≤ f (x)2

(10.7)

holds for all 0 ≤ y ≤ x. This is the case due to the following, where Y = (1/2) y and X = (1/2)x : 0 ≤ (Y − 1)2



2Y ≤ Y 2 + 1



2X ≤ X Y +



− 15 X Y −



1−

  

1 X 5Y

(1 − 1−

1 5

− 15 X Y + 1 X )(1 5Y

1 X 5Y

1 X2 25

X Y

≤ − 25 X ≤ 1 − 25 X +

1 X2 25

− 15 X Y ) ≤ (1 − 15 X )2

 1 x−y     x 2 1 1 x+y 1 − ≤ 1 − 15 21 2 5 2 f (x − y) f (x + y) ≤ f (x)2 .

Therefore the product f (s1 ) f (s2 ) · · · f (sn ), where si can be reals and not just n si = 2k|T | is maximized when all si have the same value. Furtherintegers and i=1 more, note that as f (x) is an increasing function the following holds by Lemma 10.1, Claims 10.3 and 10.5.

154

10 The Tuza Constant qk for Large k

2k−si 

 n n  Pr(H ) ≤ 1 − 2kk  |T | i=1 k 

    n  1 1 si n  1− ≤ 5 2 |T | i=1   av n 1 1 s |T |



    c ln(k) c ln(k)  ∗ n ∗H = n H + m ∗H . k 2 k

Example 1 Let H be the hypergraph associated with a projective plane of order p = 5507 (which is prime). Thus, H is a 2k-regular, 2k-uniform, linear hypergraph with k = 2754. Taking c = 0.25244 in the statement of Theorem 10.1 yields a k-uniform hypergraph, H ∗ , of order n ∗H = p 2 + p + 1 and size m ∗H = p 2 + p + 1 satisfying

156

10 The Tuza Constant qk for Large k

τ (H ∗ ) > and so q2754 > 0.12622 ×





ln(k) k

  0.12622 × ln(k)  ∗ n H + m ∗H , k

.

Example 2 Let H be the hypergraph associated with a projective plane of order p = 331. Thus, H is a 2k-regular, 2k-uniform, linear hypergraph with k = 166. We will now show that using Claim 10.3 in the proof of Theorem 10.1 it is possible to prove that the probability that H  (as defined in the proof of Claim 10.3) has transversal 1 (n H + m H ) is strictly less than one, where as before n H and m H number less that k+1 denote the order and size of H  , respectively. Recall that in the proof of Theorem 10.1 we look at the probability of a set T of size n · c · ln(k)/k being a transversal in the hypergraph H on n vertices. Now we want to consider sets T of size (n + m)/(k + 1) in H , where m is the number of edges in H . However we know that m = n in this example, so we will choose c such that n · c · ln(k)/k = (n + n)/(k + 1). That is, c=

2 · 166 2k = ≈ 0.3888945. (k + 1) ln(k) 167 ln(166)

Now |T | = n · c · ln(k)/k = 2n/(k + 1), and so letting t = 2/(k + 1), we note that |T | = t · n and 2 2 t= = ≈ 0.0119760479. k+1 167 Since |T | = t · n, Equation (10.3) in the proof of Theorem 10.1 can be written as n 

si = 2k|T | = 2ktn.

i=1

Claim 10.2 states that all si where i ∈ [n] differ in value by at most one. As 2kt ≈ 3.9760479 we note that all si ∈ {3, 4} for all i ∈ [n]. Furthermore, (4 − 2kt)n ≈ 0.0239521n si ’s have value three and (2kt − 3)n ≈ 0.9760479n have value four. Therefore, Claim 10.3 implies the following: 2k−si 

  n n Pr(H ) ≤ 1 − 2kk  |T | i=1 k 332−4 0.9760479n 332−3 0.0239521n   n 166 166 ≈ . 1 − 332 1 − 332  tn 166 166 Using Stirling’s approximation, we note that   n en tn e tn ≤ = . tn tn t

10.3 Application to Projective Planes

157

We therefore get the following bound:

n    e tn  166 · 165 · 164 · 163 0.9760479 166 · 165 · 164 0.0239521 1− 1− Pr(H ) ≤ . t 332 · 331 · 330 332 · 331 · 330 · 329

Computing the above values we get the following:  n Pr(H ) ≤ (1.0671251139558)n (0.87613293)0.0239521 (0.9386312)0.9760479 = 0.99998518n < 1. Therefore, there must exist a linear k-uniform hypergraph, H ∗ , with k = 166 of order n ∗H and size m ∗H satisfying τ (H ∗ ) >

1 (n ∗ + m ∗H ). k+1 H

Thus when k = 166, we note that qk > follows.

1 . k+1

We state this observation formally as

Observation 10.8 When k = 166, we have qk >

1 . k+1

We remark that the authors have subsequently proven that when k = 60, we have 1 . However, it remains an open problem to determine the smallest k for which qk > k+1 1 qk > k+1 .

10.4 Application to the West Bound In this section, we apply the result of Theorem 10.1 to show that the West bound holds for large uniformity and for sufficiently large average degree. Recall the statement of Conjecture 7.1. Conjecture 7.1. For k ≥ 2, if H ∈ Lk , then H satisfies the West bound; that is, τ (H ) ≤ k1 n H + 16 m H . As remarked in Chapter 7, the linearity constraint in the statement of Conjecture 7.1 is essential, even when k = 4. By Theorem 10.7, qk is bounded from above by a constant times ln(k)/k for k sufficiently large. Thus there exists a constant C, such that   ln(k) qk ≤ C × k for k sufficiently large. Hence, for all hypergraphs H ∈ Hk , τ (H ) ≤

1 · C · ln(k)(n H + m H ). k

158

10 The Tuza Constant qk for Large k

In what follows, let dav (H ) denote the average degree of the graph H , and so n H · dav (H ) =



d H (v) = k · m H .

v∈V (H )

Theorem 10.9 For k sufficiently large, if H ∈ Hk and dav (H ) ≥ 6 · C · ln(k), then H satisfies the West bound; that is, τ (H ) ≤ k1 n H + 16 m H . Proof. For k sufficiently large, let H ∈ Hk have order n = n H , size m = m H , and average degree d = dav (H ). Thus, n · d = k · m and C · ln(k) (n + m) k     C · ln(k) 1 n m C · ln(k) − 1 n+ m = + + − k 6 k k 6     n m C · ln(k) − 1 km 6 · C · ln(k) − k = + + × + m k 6 k d 6k   m C · ln(k) − 1 6 · C · ln(k) − k n + m. = + + k 6 d 6k

τ (H ) ≤

Therefore, τ (H ) ≤ k1 n + 16 m holds if the following holds:



C · ln(k) − 1 6 · C · ln(k) − k + ≤0 d 6k C · ln(k) − 1 k − 6 · C · ln(k) ≤ d 6k



6k(C · ln(k) − 1) ≤ d. k − 6 · C · ln(k)

Further, we note that

  

6k(C · ln(k) − 1) ≤ 6 · C · ln(k) k − 6C · ln(k) 6k(C · ln(k) − 1) ≤ (k − 6C · ln(k))(6 · C · ln(k)) −6k ≤ −62 C 2 ln2 (k) 6C 2 ln2 (k) ≤ k.

For k sufficiently large, we note that 6C 2 ln2 (k) ≤ k

10.4 Application to the West Bound

159

holds. This, together with our supposition that d ≥ 6 · C · ln(k), implies that 6k(C · ln(k) − 1) ≤ 6 · C · ln(k) ≤ d. k − 6C · ln(k) This in turn implies that τ (H ) ≤ k1 n + 16 m.



We remark that Theorem 10.9 holds for general hypergraphs H ∈ Hk (even without the linearity constraint). In particular, for k sufficiently large, if H ∈ Lk and dav (H ) ≥ 6 · C · ln(k), then Conjecture 7.1 holds.

Chapter 11

The Cap Set Problem

11.1 Introduction In this chapter, we discuss the cap set problem and its relation to transversals in 3-uniform hypergraphs. The card game SET is played with a deck of 3n cards. On each card is printed n symbols or features. For example, if n = 4, then these four features are color, number, shape, and shading. Each card is unique and the 3n cards are created by choosing one of three possible values for each of the n features. Three cards form a set if each of the n features, considered individually, is either the same on each card or different on each card. A question that arises is “how many cards must be dealt to guarantee the presence of a set?” or, equivalently, “How large can a hand be if it does not contain a set?” For example, if n = 4, then Pellegrino [81] showed a collection of 20 cards (from the pack of 81 cards) that does not contain a set, but proved that any collection of 21 cards will always contain a set.

11.2 A Mathematical Formulation Using F3 = {0, 1, 2} to denote the field with three elements, each SET card a may be associated with an element a = (a1 , a2 , . . . , an ) in Fn3 , and so ai ∈ {0, 1, 2} for each i ∈ [n]. The three cards a, b, and c where a = (a1 , a2 , . . . , an ), b = (b1 , b2 , . . . , bn ), and c = (c1 , c2 , . . . , cn ) form a set if and only if a + b + c = 0. That is, if and only if either ai = bi = ci or {ai , bi , ci } = {0, 1, 2} for each i ∈ [n]. The question about the game of SET can be translated as follows: “How large can a subset A ⊂ Fn3 be if there are no solutions to the equation a + b + c = 0 with a, b and c in A distinct?” Such a subset A is called an n-cap. Equivalently, a cap set is a subset A of Fn3 in which no three terms are in an arithmetic progression. This is also © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_11

161

162

11 The Cap Set Problem

equivalent to a subset A of the n-dimensional affine space Fn3 in which no three points are on a common affine line, implying the condition that for all λ ∈ F3 = {0, 1, 2} and for all three distinct a, b, c ∈ A, we have b − a = λ(c − b). Let Cn denote the maximum cardinality of an n-cap. For small n, one can show that C2 = 4 and C3 = 9, while Pellegrino’s result shows that C4 = 20. One way to construct a large cap set would be to choose the subset A of Fn3 so that each a = (a1 , a2 , . . . , an ) ∈ A satisfies ai ∈ {0, 1} for each i ∈ [n]. The resulting set A is a cap set of size 2n . As observed earlier, Pellegrino’s solution for the 4dimensional cap set problem (when n = 4) shows a larger cap set than size 24 . In 2004, Edel [29] improved the trivial lower bound of 2n for Cn to 2.2174n . Theorem 11.1 (Edel [29]) Cn ≥ 2.2174n . In 1987 Frankl, Graham, and Rödl [38] posed the question of whether there exists a constant c < 3 such that for all sufficiently large values of n, any cap set in ZZ 3n has size at most cn , implying that any set in ZZ 3n of size exceeding cn contains an affine line. This question was once again posed by Alon and Dubiner [4] in 1995. In 1995, Meshulam [78] established the best known upper bound at that time and proved, primarily using Fourier-analytic arguments, that the size of a cap set does not exceed n1 (2 · 3n ); that is, Cn ≤ n1 (2 · 3n ). In 2012, Bateman and Katz [12] improved this upper bound further to O(3n/n 1+ε ) for some fixed ε, using more sophisticated Fourier/spectral arguments. For the past 20 years, the problem of determining whether Meshulam’s bound can be improved to cn with c < 3 was “considered one of the most intriguing open problems in additive combinatorics and Ramsey theory,” and attracted the attention of Fields medalists including Timothy Gowers and Terence Tao. So captivated were the world leading Combinatorialists by this question that Terence Tao referred to this problem in his blog of February 23, 2007 entitled “open questions: best bounds for cap sets” as “perhaps, my favorite open problem.” Tao was of the opinion that the Fourier-type arguments used to establish the best upper bounds up to that time by Meshulam [78] and Bateman and Katz [12] had reached their natural limit, and that any breakthrough would have to be “a radically new idea.” The breakthrough came on May 5, 2016 when Croot, Lev, and Pach announced on arXiv that they had drastically improved the best known bounds. Their paper [27] was published in 2017, and in it they use the polynomial method to show that a subset of the cyclic group ZZ 4n with no three-term arithmetic progression has size at most cn for some c ≈ 3.61. Within a few weeks of the breakthrough result announced by Croot, Lev, and Pach, on May 27, 2016, Ellenberg and Gijswijt (pictured below) independently extended their ideas to vector spaces over a general finite field and proved the upper bound of 2.756n on the cap set problem. The resulting paper of Ellenberg and Gijswijt [31] appeared in the same issue of the prestigious journal, Annals of Mathematics, as the paper by Croot, Lev, and Pach [27]. We state their result formally as follows. Theorem 11.2 (Ellenberg, Gijswijt [31]) Cn ≤ 2.756n .

11.2 A Mathematical Formulation

163

Jordan Ellenberg

Dion Gijswijt

Recall that the n-cap problem is to determine how large can a subset A ⊂ Fn3 be if there are no solutions to the equation a + b + c = 0 with a, b, and c in A distinct. Since we are working in F3 where 1 + 2 = 0, we note that a + b + c = 0 is equivalent to a + c = −b = 2b, and so a − 2b + c = 0 and b = (a + c)/2 is the arithmetic mean of a and c. Thus, a, b, and c are consecutive terms of an arithmetic progression. Our problem can be generalized by allowing q possibilities for each of the features on a SET card where q is an odd prime by asking “How large can a subset A ⊂ Fqn be for which the equation a − 2b + c = 0 has no solutions with a, b and c in A distinct?” Solutions to this equation form consecutive terms a, b, and c in an arithmetic progression in Fqn and we call such a set A having no distinct solutions progression-free. In order to prove Theorem 11.2, Ellenberg and Gijswijt [31] prove a more general in Fqn and establish an upper bound on sets A ⊂ Fqn that avoid progressions of the form αa + βb + γ c with α, β, γ ∈ Fq = {0, 1, . . . , q − 1} fixed, satisfying α + β + γ = 0 with a, b, and c in A. Thus taking α = γ = 1 and β = −2 gives the desired result for the special case of progression-free subsets of Fn3 (where q = 3, which is our ultimate focus). We are now in a position to present the proof of Theorem 11.2 given by Ellenberg and Gijswijt [31], which is beautifully explained and illustrated by David Austin [6] in his feature column in the August 2016 issue of the American Mathematical Society to make the proof more accessible to a more general audience, including interested graduate students. Recall the statement of Theorem 11.2. Theorem 11.2 (Ellenberg, Gijswijt [31]) Cn = O(2.756n ). Proof. Let Mn be the set of monomials x1α1 x2α2 . . . xnαn ,

164

11 The Cap Set Problem

where the degree αi in each variable xi satisfies 0 ≤ αi ≤ q − 1. Let Sn be the span of Mn over Fqn ; that is, Sn is the set of polynomials with coefficients in Fq spanned by Mn . We note that the dimension of Sn is q n . If P ∈ Sn , then x = (x1 , x2 , . . . , xn ) ∈ Fqn we have  Cα1 ,α2 ,...,αn x1α1 x2α2 . . . xnαn P(x) = α1 ,α2 ,...,αn

for some constant Cα1 ,α2 ,...,αn (depending only on α1 , α2 , . . . , αn ). The polynomial P defines an Fq -valued function in the set Fqn that assigns to each a = (a1 , a2 , . . . , an ) ∈ Fqn a value in Fq , namely, P(a) =



Cα1 ,α2 ,...,αn a1α1 a2α2 . . . anαn .

α1 ,α2 ,...,αn

Let Map(Fqn , Fq ) denote the set of Fq -valued functions on Fqn . The evaluation map e : Sn → Map(Fqn , Fq ) is an isomorphism since Map(Fqn , Fq ) is an Fq -vector space having the same dimension as Sn , namely, q n , and every function in Map(Fqn , Fq ) is given by evaluating a polynomial in Sn . Let Mnd be the set of monomials with total degree at most d. Thus a monomial α1 α2 x1 x2 . . . xnαn belongs to Mnd if 0 ≤ αi ≤ q − 1 and α1 + α2 + · · · + αn ≤ d. Let Snd be the subspace spanned by these monomials, and let m d denote the dimension of Snd . Thus, m d = dimFq (Snd ). We note that there is an involution of monomials in Mn given by q−1−α1 q−1−α2 x2 . . . xnq−1−αn x1α1 x2α2 . . . xnαn → x1 that maps a monomial of degree d into a monomial of degree (q − 1)n − d. This involution is a bijection between the set of monomials of degree greater than d and the set of monomials whose degree is less than (q − 1)n − d, implying that q n − m d = m (q−1)n−d−1 ≤ m (q−1)n−d .

(11.1)

Recall that our immediate aim is to establish an upper bound on a progression-free subset A ⊂ Fqn . For such a subset A the equation a − 2b + c = 0 has no solutions with a, b, and c in A distinct. Given a subset A ⊂ Fqn , we define the set B and C in Fqn by B = {2a | a ∈ A} and C = {a + a | a, a ∈ A, a = a }. We note that A is progression-free if and only if B and C are disjoint. Further since q is odd, we note that |A| = |B|. We now consider polynomials P whose support, Supp(P), is contained in B, where the support of P is the set of points in Fqn for which P is nonzero. We proceed further with the following two claims. Claim 11.3 If P ∈ Snd is a polynomial of degree d such that P(c) = 0 for all c ∈ C, then the number of points in B for which P(b) = 0 is at most 2m d/2 ; that is,

11.2 A Mathematical Formulation

165

|{b ∈ B | P(b) = 0}| ≤ 2m d/2 . Proof Let R be the |A| × |A| matrix with a, a entry Ra,a given by P(a + a ). If a = a , then a + a ∈ C and P(a + a ) = 0 by our assumption that P(c) = 0 for all c ∈ C. The only nonzero elements in R are therefore diagonal entries Ra,a = P(2a). Hence, the rank of the diagonal matrix R equals the number of elements of A for which P(2a) = 0, or, equivalently, the number of elements of B for which P(b) = 0. We next study properties of the matrix R by considering P(x + y) where x, y ∈ Fqn . Let x = (x1 , x2 , . . . , xn ) ∈ Fqn and y = (y1 , y2 , . . . , yn ) ∈ Fqn . Expanding P(x + y) and letting α = α1 + α2 + · · · + αn , we have P(x + y) =



cα (x1 + y1 )α1 (x2 + y2 )α2 . . . (xn + yn )αn

|α|≤d

=



cα





β

α −β1 α2 −β2 y2

· · · ynαn −βn

β

|α|≤d

=

β

cβ x1 1 x2 2 · · · xnβn y1 1

cm,m m(x)m (y).

∈Mnd

m,m mm )≤d

Since the product of monomials m(x)m (y) in each term of the summand of the above expression has degree at most d, at least one of the monomials m or m has degree at most d/2. We can therefore write P(x + y) as P(x + y) =





m(x)Fm (y) +

d/2

G m (x)m (y)

d/2

m ∈Mn

m∈Mn

for some family of polynomials Fm and G m indexed by m ∈ Mn and m ∈ Mn , respectively. We note that the above decomposition of P(x + y) is not necessarily unique. Since d/2



Ra,a = P(a + a ) =

m(a)Fm (a ) +

d/2



G m (a)m (a )

d/2

m ∈Mn

m∈Mn

we can write the matrix R as R=





Sm +

d/2 m∈Mn

Tm ,

d/2 m ∈Mn

where the matrices Sm and Tm are defined by (Sm )a,a = m(a)Fm (a )

and

(Tm )a,a = G m (a)m (a ).

d/2

166

11 The Cap Set Problem

Thus, the matrix R is the sum of 2m d/2 matrices. We note that every row of Sm is a scalar multiple of Fm (a ), implying that Sm has rank 1. Analogously, every column of Tm is a scalar multiple of G m (a) and therefore Tm has rank 1. The rank of R is therefore at most 2m d/2 . As observed earlier, the rank of R equals the number of elements of B for which P(b) = 0, implying that the number of elements of B for  which P(b) = 0 is at most 2m d/2 . This completes the proof of Claim 11.3. Claim 11.4 |A| ≤ 3m (q−1)n/3 . Proof Let W be the vector space of polynomials whose support is contained in B. Since B and C are disjoint, the polynomials P in W satisfy P(c) = 0 for all c ∈ C. Let V be the space of polynomials in W of degree at most d, and so V = Snd ∩ W . As observed earlier, m d = dimFq (Snd ) and the evaluation map is an isomorphism, we note that dimFq (W ) = |B| = |A|. Thus, dimFq (V ) ≥ dimFq (Snd ) + dimFq (W ) − dimFq (Sn ) = m d + |A| − q n .

(11.2)

For d sufficiently large, we have dimFq (V ) > 0, implying that V is not empty. Among all polynomials in V , let P be chosen so that the support Supp(P) is maximum. We claim that |Supp(P)| ≥ dimFq (V ). Suppose, to the contrary, that |Supp(P)| < dimFq (V ). Let X be a the subspace of polynomials in V with support contained in Supp(P). Since |Supp(P)| < dimFq (V ), there exists a nontrivial subspace X so that V = X ⊕ X . This implies that there exists a nonzero polynomial Q in the subspace X that vanishes on Supp(P); that is, such that Q(x) = 0 for all x ∈ Supp(P). However, P + Q is then a polynomial in V with strictly larger support than P, contradicting our choice of P. Hence, |Supp(P)| ≥ dimFq (V ). Since P ∈ Snd is a polynomial of degree d such that P(c) = 0 for all c ∈ C, by Claim 11.3 the number of points in B for which P(b) = 0 is at most 2m d/2 , and so |Supp(P)| ≤ 2m d/2 . Thus by Inequality (11.2), we have 2m d/2 ≥ |Supp(P)| ≥ dimFq (V ) ≥ m d + |A| − q n , implying that |A| ≤ 2m d/2 + q n − m d .

(11.3)

Applying Inequality (11.1) when d = 2(q − 1)n/3, we have q n − m 2(q−1)n/3 ≤ m (q−1)n/3 .

(11.4)

Inequality (11.3) and Inequality (11.4) imply that |A| ≤ 2m 2(q−1)n/3 + q n − m 2(q−1)n/3 ≤ 3m (q−1)n/3 . This completes the proof of Claim 11.4.



By Claim 11.4, we have bound the size of the progression-free subset A ⊂ Fqn by the dimension of a specific space of polynomials. Recall that m d is the number of monomials of degree at most d, and the total number of monomials in Mn is q n .

11.2 A Mathematical Formulation

167

We wish to find the fraction of monomials in Mn whose degree is at most d; that is, we wish to determine m d /q n . For this purpose, we use probabilistic arguments. We select the exponents α1 , α2 , . . . , αn of the monomial x1α1 x2α2 . . . xnαn where 0 ≤ αi ≤ q − 1 randomly and with equal probability. The ratio m d /q n is the probability that α1 + α2 + · · · + αn ≤ d, or, equivalently, the probability that the mean αn =

d α1 + α2 + · · · + αn ≤ . n n

We now use the theory of large deviations from probability in estimating the ratio   md d = Pr(α1 + α2 + · · · + αn ≤ d). = Pr α ≤ n qn n

(11.5)

In order to apply Markov’s inequality to the right-hand side of Equation (11.5), we multiply by a negative constant θ < 0 to reverse the inequality and exponentiate to obtain a positive random variable. By Markov’s inequality and the moment generating function, we obtain md = Pr (θ (α1 + α2 + · · · + αn ) ≥ θ d) qn  n  = Pr eθ i=1 αi ≥ eθd n

E[eθ i=1 αi ] ≤ eθd  n θα

E i=1 e i = eθd n  E[eθα1 ] ≤ eθd/n 

n 1 + eθ + e2θ + · · · + e(q−1)θ /q ≤ . eθd/n Our ultimate focus is when q = 3 and when d = 2(q − 1)n/3 = 2n/3. From the above, it suffices for us in this case to minimize the function f (θ ) =

(1 + eθ + e2θ )/3 e2θ/3

over the set of real number θ < 0. This is equivalent to minimizing 2 h(θ ) = ln( f (θ )) = ln((1 + eθ + e2θ )/3) − θ. 3

168

11 The Cap Set Problem

√ Applying calculus, the function h has a minimum when eθ = ( 33 − 1)/8 and  √  33 − 1 f ln ≈ 0.918368, 8 implying that

m 2n/3 ≤ 0.918368n , 3n

and therefore that m 2n/3 ≤ (3 × 0.918368)n ≤ 2.756n . Hence by Claim 11.4, the progression-free subset A ⊂ Fqn has size |A| ≤ 3m 2n/3 ≤ 3 · 2.756n . Hence, the maximum cardinality of an n-cap is given by |Cn | = O(2.756n ). A similar analysis shows that for odd primes q, there is a constant c < q such that a progression-free subset A ⊂ Fqn satisfies |A| = O(cn ). This completes the proof of Theorem 11.2. 

11.3 The Relation to Transversals in Linear Hypergraphs The cap set problem can be translated to the problem of determining lower and upper bounds on the transversal number of a 3-uniform linear hypergraph in terms of its order. For n ≥ 1, let Rn be the 3-uniform hypergraph whose vertex set, V (Rn ), consists of all n-tuples of the form a = (a1 , a2 , . . . , an ) where each ai ∈ {0, 1, 2}. Further, the edge set, E(Rn ), of Rn consists of all 3-edges of the form {a, b, c} where a = (a1 , a2 , . . . , an ), b = (b1 , b2 , . . . , bn ), and c = (c1 , c2 , . . . , cn ) and where either ai = bi = ci or {ai , bi , ci } = {0, 1, 2} for each i ∈ [n]. We note that R1 is the hypergraph E 3 consisting of a single edge, and R2 is the affine plane AG(2, 3) (of dimension 2 and order 3) which we call F9 in Figure 3.2a. We note that the hypergraph Rn has order n R = 3n . If some edge e = {a, b, c} ∈ E(Rn ) contains the two vertices a = (a1 , a2 , . . . , an ) and b = (b1 , b2 , . . . , bn ), then the third vertex, c = (c1 , c2 , . . . , cn ), in e is completely defined: if ai = bi , then ci = ai = bi , while if ai = bi , then ci = 3 − ai − bi for all i ∈ [n], implying that Rn is a linear hypergraph. Further, this implies that every two vertices in Rn belong to some edge, and that Rn is a 21 (3n − 1)-regular hypergraph of size m R = 16 (3n (3n − 1)). Recall that a subset of vertices in a hypergraph H is an independent set if it contains no edge of H . Equivalently, a set of vertices S is an independent set in H

11.3 The Relation to Transversals in Linear Hypergraphs

169

if and only if V (H ) \ S is a transversal in H . Further, recall that the independence number, α(H ), of H is the maximum cardinality of an independent set in H , and that by Observation 1.1 we have n(H ) = τ (H ) + α(H ). We propose here the problem of determining or estimating the best possible constants u k (which depends only on k) such that τ (H ) ≤ u k · n H for all H ∈ Lk and k ≥ 2, where we recall that Lk is the class of k-uniform linear hypergraphs. These constants are given by τ (H ) . u k = sup nH H ∈L k The case when k = 2 is trivial, since we note that the 2-uniform complete hypergraph H = K n satisfies n H = n and τ (H ) = n H − 1, implying that u 2 = 1. When k = 3, we note that an independent set in the hypergraph Rn constructed earlier corresponds to a cap set in Fn3 (in which no three points belong to a common affine line). Thus, α(Rn ) = Cn . Hence, by Theorem 11.2 and Observation 1.1, we have that τ (Rn ) = n(Rn ) − α(Rn ) = 3n − Cn ≥ 3n − 2.756n . Taking H to be the hypergraph Rn , we therefore have that τ (H ) 3n − 2.756n ≥ =1− nH 3n



2.756 3

n > 1 − (0.918667)n .

The existence of the hypergraph Rn constructed above implies the following results. Theorem 11.5 For every given ε > 0, there exists a connected hypergraph H ∈ L3 of sufficiently large order such that τ (H ) > (1 − ε)n H . Theorem 11.6 u 3 = 1. The precise value of u k has yet to be determined for any values of k with k ≥ 4. Given the considerable interest in the cap set problem and the breakthrough papers by Croot et al. [27] and Ellenberg and Gijswijt [31] that show that u 3 = 1, the immediate challenge is to determine if u 4 = 1 also holds or whether u 4 < 1. Indeed, it is even possible that u k = 1 for all k ≥ 2.

Chapter 12

Partial Steiner Triple Systems

12.1 Introduction In this chapter, we consider 3-uniform linear hypergraphs which are more commonly known as partial Steiner triple systems. A partial Steiner (n, 3, 2)-system, is a 3uniform hypergraph on n vertices in which every pair of its vertices is contained in at most edge, implying that the hypergraph is linear. We present here a classical 1986 result due to Phelps and Rödl that establishes an upper bound on the transversal number of a partial Steiner triple system. Before we do so, we briefly discuss Steiner triple systems.

12.2 Steiner Triple Systems A Steiner triple (n, 3, 2)-system is a partial Steiner (n, 3, 2)-system with the added property that every pair of its vertices is contained in a unique edge (or triple). Equivalently, a Steiner triple (n, 3, 2)-system consists of a set S with n elements and a set T consisting of triples of S (called blocks) such that every pair of elements of S appears together in a unique triple of T . Hence, a “triple” in this context corresponds to a 3-edge in the associated linear hypergraph. We note that a partial Steiner triple (n, 3, 2)-system differs from a Steiner triple (n, 3, 2)-system in that every 2-element subset (pair) is contained in at most one triple. We abbreviate a Steiner triple (n, 3, 2)-system by STS(n) and refer to it as a Steiner triple system of order n. For example, the projective plane of order 2 (also known as the Fano plane, illustrated in Fig. 7.1) is an STS(7), while the affine plane of order 3 (illustrated in Fig. 3.2(a)) is an STS(9). We abbreviate a partial Steiner triple (n, 3, 2)-system by PSTS(n).

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_12

171

172

12 Partial Steiner Triple Systems

Let H be the 3-uniform, linear, hypergraph associated with a Steiner triple system of order n. The total number of pairs of vertices in H is n(n − 1)/2 and every three vertices appear in a unique edge (triple), implying that the total number of edges is n(n − 1)/6. For every vertex v in H , if we remove the vertex v from all the edges that contain it we form a partition of V (H ) \ {v} into pairs, implying that n − 1 is even, and so n is odd. Therefore, n ≡ 1, 3, 5 (mod 6). However, if n = 6k + 5 for some integer k ≥ 0, then the number of edges/triples is given by n(n − 1)/6 = (6k + 5)(6k + 4)/6 = (36k 2 + 54k + 20)/6, which is not an integer. Hence, n ≡ 5 (mod 6), implying that n ≡ 1, 3 (mod 6). This necessary condition for the existence of a Steiner triple system of order n is also sufficient since there exist constructions of such systems for all n ≡ 1, 3 (mod 6) as first proven in 1847 by Rev. Kirkman [68]. Theorem 12.1 ([68]) A Steiner triple system of order n exists if and only if n ≡ 1, 3 (mod 6).

12.3 Transversals in Partial Steiner Triple Systems Recall that a subset of vertices in a hypergraph H is an independent set if it contains no edge of H , and the independence number, α(H ), of H is the maximum cardinality of an independent set in H . As observed in Chapter 11.3, a set of vertices S is an independent set in H if and only if V (H ) \ S is a transversal in H . By Observation 1.1, if H is a hypergraph of order n H , then n H = τ (H ) + α(H ). Hence in order to find a lower bound on the transversal number of a partial Steiner triple systems it suffices to find an upper bound on its independence number. A cycle of length k in a hypergraph H with vertex set V (H ) = V and edge set E(H ) = E is a sequence of distinct vertices and edges v1 , e1 , v2 , e2 , . . . , vk , ek , v1 such that {vi , vi+1 } ⊆ ei , where vi ∈ V and ei ∈ E for i ∈ [k] and addition is taken modulo k. For g ≥ 3, the hypergraph H has girth at least g if it has no cycles of length 2, . . . , g − 1. In particular, the girth of H is at least 3 if H is a linear hypergraph. In 1971, Harary and Manvel [49] gave formulae for the number of cycles of lengths 3 and 4 in a graph in terms of the trace of the adjacency matrix of the graph (where we recall that the trace of a matrix is the sum of the diagonal entries of the matrix). In particular, they showed that if G is a graph with adjacency matrix A, then the number of 3-cycles in G is given by 16 tr(A3 ), while the number of 4-cycles in G is bounded above by 18 tr(A4 ). Below we will need the following weaker bounds for the number of 3-cycles and 4-cycles of a graph G. Since a 3-cycle is uniquely  determined by the three vertices in the 3-cycle there are at most n3 < 16 n 3 3-cycles in G. Analogously since there are  at most three 4-cycles on any set of 4-vertices, we note that there are at most 3 n4 < 18 n 4 4-cycles in G. Analogously, in a linear hypergraph, a 3-cycle, v1 e1 v2 e2 v3 e3 v1 , is uniquely determined by the three vertices, v1 , v2 , v3 , in the 3-cycle (as there is at most one edge containing any two vertices)

12.3 Transversals in Partial Steiner Triple Systems

173

and there are at most three 4-cycles on any set of 4-vertices. Therefore the following observation holds. Observation 12.2 The number of 3-cycles and 4-cycles in a 3-uniform linear hypergraph of order n is bounded from above by 16 n 3 and 18 n 4 , respectively. If H is a 3-uniform, linear hypergraph of order n H = n, then every vertex in H belongs to at most (n − 1)/2 edges; that is, the maximum degree of H is (H ) ≤ (n − 1)/2. Thus, the number of edges in H is mH =

1 1  1 d H (v) ≤ n(n − 1) ≤ n 2 . 3 v∈V (H ) 6 6

We state this formally as follows. Observation 12.3 If H is a 3-uniform, linear hypergraph, then m H ≤ 16 n 2H . Komlós et al. [69] proved the following lower bound on the transversal number of a 3-uniform hypergraph with girth at least 5. Lemma 12.1 ([69]) Let H = (Q, E Q ) be a 3-uniform, linear hypergraph with vertex set Q and edge set E Q and with girth at least 5. For |Q| = q and t > 0 such that t < q 0.1 and 3|E Q | < t 2 q both hold, the following holds: α(H ) > c∗ ·

q   · log t t

for some absolute constant c∗ independent of q. We are now in a position to present the 1986 result due to Phelps and Rödl [82] (pictured below) that establishes a lower bound on the independence number of a partial Steiner triple system of order n (or, equivalently, a 3-uniform linear hypergraph on n vertices). Their proof utilizes the probabilistic method.

174

12 Partial Steiner Triple Systems

Theorem 12.4 ([82]) If H is a 3-uniform linear hypergraph of order n, then  α(H ) ≥ C n log n for some absolute constant C > 0 independent of n. Proof Let H be a 3-uniform linear hypergraph of order n H = n and size m H = m. For each subset X of vertices of H , let e X be the number of edges of H that are subsets of X . Further, let c3 (X ) and c4 (X ) be the number of cycles of length 3 or 4, respectively, whose edges are subsets of X . Let p=

1 −4 n 9 10

and form a random subset X ⊆ V (H ) where a vertex is chosen to be in X with probability p and independently of the choice for any other vertex. The expected value of |X | is E[|X |] = np. The random variable e X can be written as the sum of m indicator random variables Ye where e ∈ E(H ) and where Ye (X ) = 1 if e ⊆ X and Ye (X ) = 0 otherwise. For each edge e ∈ E(H ), we have E[Ye ] = p 3 . By linearity of expectation and by Observation 12.3, 

E[e X ] =

E[Ye ] =

e∈E(H )



p 3 = mp 3 ≤

e∈E(H )

1 3 2 p n , 6

implying by Markov’s inequality that  E[e X ]  1 Pr e X ≥ p 3 n 2 ≤ 3 2 ≤ . p n 6

(12.1)

Let C3 (H ) be the set of all 3-cycles in H . By Observation 12.2, we have |C3 (H )| ≤ The random variable c3 (X ) can be written as the sum of |C3 (H )| indicator random variables WC where C ∈ C3 (H ) and where WC (X ) = 1 if V (C) ⊆ X and WC (X ) = 0 otherwise. For each 3-cycle C in C3 (H ), we have |V (C)| = 6 (that is, if C = v1 e1 v2 e2 v3 e3 v1 , then |V (e1 ) ∪ V (e2 ) ∪ V (e3 )| = 6), and E[WC ] = p 6 . By linearity of expectation, 1 3 n . 6

E[c3 (X )] =



E[WC ] =

C∈C 3 (H )



p 6 = |C3 (H )| · p 6 ≤

C∈C 3 (H )

1 6 3 p n , 6

implying by Markov’s inequality that   E[c3 (X )] 1 Pr c3 (X ) ≥ p 6 n 3 ≤ ≤ . 6 3 p n 6

(12.2)

12.3 Transversals in Partial Steiner Triple Systems

175

Let C4 (H ) be the set of all 4-cycles in H . By Observation 12.2, we have |C4 (H )| ≤ The random variable c4 (X ) can be written as the sum of |C4 (H )| indicator random variables Z C where C ∈ C4 (H ) and where Z C (X ) = 1 if V (C) ⊆ X and Z C (X ) = 0 otherwise. For each 4-cycle C in C4 (H ), we have |V (C)| = 8 and E[Z C ] = p 8 . By linearity of expectation, 1 4 n . 8

E[c4 (X )] =



E[Z C ] =

C∈C 4 (H )



p 8 = |C4 (H )| · p 8 ≤

C∈C 4 (H )

1 8 4 p n , 8

implying by Markov’s inequality that   E[c4 (X )] 1 Pr c4 (X ) ≥ p 8 n 4 ≤ ≤ . 8 4 p n 8

(12.3)

1 1/9 Note that np 2 = 100 n , which can be made arbitrary big by increasing n. Let N be chosen such that np 2 > 1.4 > 2 ln(2) for all n > N . We may assume that n > N , as otherwise the theorem holds by letting C be large enough. By Hoeffding’s inequality we now note that the following holds, when k = 21 np:

 (np−k)2 np 2 1 1 Pr |X | ≤ pn = Pr (|X | ≤ k) ≤ e−2 n = e− 2 < e− ln(2) = . 2 2

(12.4)

By Inequalities (12.1), (12.2), (12.3), and (12.4), the probability that one of the 23 < 1: below statements fail is at most 61 + 16 + 18 + 21 = 24 e X ≤ p3 n 2 ,

c3 (X ) ≤ p 6 n 3 ,

c4 (X ) ≤ p 8 n 4

and

|X | ≥

1 pn. 2

Therefore there exists an X satisfying all of the above. Note that for i ∈ [4] we 1 1 have p( p 2 n)i = 102i+1 n (i−4)/9 < 10 , which implies that 8 p 6 n 3 = pn · 8 p( p 2 n)2 < 8 4 2 3 pn and 8 p n = pn · 8 p( p n) < pn. From each 3-cycle in H [X ] and from each 4-cycle in H [X ], we select an arbitrary vertex and denote the resulting set by X . Thus, |X | ≤ p 6 n 3 + p 8 n 4 < 18 pn + 18 pn = 41 pn, and so |X \ X | > 21 pn − 41 pn = 41 pn. Removing the vertices in X from H [X ] results in a 3-uniform, linear hypergraph that contains no 3-cycles or 4-cycles; that is, the 3-uniform, linear hypergraph H [X \ X ] has girth at least 5. Let q be an integer chosen such that 14 pn < q ≤ |X \ X | and let Q ⊆ X \ X

be chosen so that |Q| = q. We will now consider the subhypergraph H = H [Q] induced by the set Q. Thus, H has vertex set V (H ) = Q. Let H have edge set E(H ) = E Q . We note that 3|E Q | ≤ 3e X ≤ 3 p 3 n 2 =

3n 2/3 . 1000

(12.5)

176

12 Partial Steiner Triple Systems

Letting t = ( 41 pn)0.1 , our earlier observations imply that  t q≥ 2

1 pn 4

0.2

1 · pn = 4



1 pn 4

65

 =

1 5 n9 40

65

=

n 2/3 3n 2/3 . > 406/5 1000

(12.6)

Thus by Inequalities (12.5) and (12.6), we have 3|E Q | < t 2 q. Thus, H = (Q, E Q ) is a 3-uniform, linear hypergraph with girth at least 5 and with vertex set Q and edge set E Q satisfying |Q| = q and 3|E Q | < t 2 q where t = ( 41 pn)0.1 < q 0.1 . 1 5/9 Let r = 14 pn = 40 n . By Lemma 12.1, for some absolute constant c∗ independent of q we have the following as t = r 0.1 and r < q:   √ α(H ) > c∗ · qt · log t  > c∗ · r 0.9 · log r 0.1   1 5/9 0.9

1 = c∗ · 40 n · 0.1 · log 40 n 5/9

   1 0.9 1/2 √ 1 = c∗ · 40 + 59 log(n) · n · 0.1 · log 40 √ > C · n log n,

for √ some absolute constant C independent of n. Thus, α(H ) ≥ α(H ) ≥ C n log n. 

Brandes et al. [19] established an upper bound on the minimum independence number of a partial Steiner triple system of order n. Their results, together with the results due to Phelps and Rödl [82] given in Theorem 12.4, imply the following result. Recall that we abbreviate a partial Steiner triple (n, 3, 2)-system of order n by PSTS(n). Theorem 12.5 ([82]) For a partial Steiner triple system of order n, we have  c1 n log n ≤

inf

H ∈PSTS(n)

 α(H ) ≤ c2 n log n

for some absolute constants c1 and c2 independent √ of n; that is, if H is a partial Steiner triple system of order n, then α(H ) = ( n log n). This beautiful 1986 result due to Phelps and Rödl [82] was improved in 2013 by Eustis and Verstaëte [34] (pictured below) who proved a more general result about partial Steiner (n, r, )-systems for integers r and  with r ≥ 2 − 1 ≥ 3, where a partial Steiner (n, r, )-system is r -uniform hypergraphs on n vertices in which every set of  vertices is contained in at most one edge. In the special case when r = 3 and  = 2, we have the following result on a partial Steiner (n, 3, 2)-system, where if f and g are positive-valued functions of n, then we write f  g if and only if lim supn→∞ f (n)/g(n) ≤ 1.

12.3 Transversals in Partial Steiner Triple Systems

177

Theorem 12.6 ([34]) There exists a partial Steiner triple system H of order n such that  α(H )  3n log n as n → ∞. As an immediate consequence of Observation 1.1 and Theorem 12.6, we have the following result. Theorem 12.7 ([34]) There exists a partial Steiner triple system H of order n such that  τ (H )  n − 3n log n as n → ∞. We remark that Theorem 12.7 implies Theorem 11.5 in a stronger form.

Chapter 13

Upper Transversals in Linear Hypergraphs

13.1 Introduction In this chapter, we study upper transversals in linear hypergraphs. The upper transversal number Υ (H ) of a hypergraph H is the maximum cardinality of a minimal transversal in H . For k ≥ 1, a transversal T in a hypergraph H is a k-transversal if it intersects every edge of H in at least k vertices; that is, |T ∩ e| ≥ k for every edge e in H . The upper k-transversal number Υk (H ) of H is the maximum cardinality of a minimal k-transversal in H . When k = 1, the upper k-transversal number is the upper transversal number, and so Υ (H ) = Υ1 (H ). In this chapter we study the upper k-transversal number Υk (H ) of an r -uniform, linear hypergraph H for all r ≥ k ≥ 1 where r ≥ 2.

13.2 Lower Bounds in Terms of Order Let Lk,≥1 be the class of k-uniform, connected, linear hypergraphs with minimum degree of at least 1. We propose here the problem of determining or estimating the best possible constants Uk (which depends only on k) such that Υ (H ) ≥ Uk · n H for all H ∈ Lk,≥1 and k ≥ 2. These constants are given by Uk =

inf

H ∈L k,≥1

Υ (H ) . nH

13.2.1 The Constant U2 In this section, we first consider the case when k = 2. In this case, H is a 2uniform, connected hypergraph with minimum degree at least 1. Equivalently, H is a © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_13

179

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13 Upper Transversals in Linear Hypergraphs

connected, isolate-free graph, where an isolate-free graph is a graph that contains no isolated vertex. We proceed further by recalling that an independent dominating set of a graph G is a set S of vertices of G that is both independent and dominating. Equivalently, no two vertices of S are adjacent, and every other vertex is adjacent to at least one vertex of S. The independent domination number, i(G), of G is the minimum cardinality of an independent dominating set of G. In order to determine the constant U2 , we first establish a relationship between the upper transversal number and independent domination number of a graph. Theorem 13.1 ([61]) If G is an isolate-free graph, then i(G) + Υ (G) = n(G). Proof. Let G be an isolate-free graph of order n. Let S be an independent dominating set in G of minimum cardinality, and so |S| = i(G). Let T = V (G) \ S and note that T is a transversal in G as S is an independent set. Since every vertex in T has a neighbor in S, we furthermore note that T is a minimal transversal, which implies that Υ (G) ≥ |T | = n − |S| = n − i(G). Conversely, let T be a minimal transversal in G of maximum cardinality, and so |T | = Υ (G). Let S = V (G) \ T and note that S is an independent set as T is a transversal. Since T is a minimal transversal, every vertex in T has a neighbor in S, implying that S is an independent dominating set in G. Therefore, i(G) ≤ |S| = n − |T | = n − Υ (G). Consequently, i(G) + Υ (G) = n.  Favaron [35] was the first to prove the following upper bound on the independent domination of a graph with √ no isolated vertex: If G is an isolate-free graph of order n, then i(G) ≤ n + 2 − 2 n. We remark that this result also follows from a result due to Bollobás and Cockayne [17] (and was also proved in [42]). Sun and Wang [84] proved the following more general result, which was originally posed as a conjecture by Favaron [35] and was proved for δ = 2 by Glebov and Kostochka [43]. Theorem √ 13.2 ([84]) If G is a graph of order n with δ(G) ≥ δ, then i(G) ≤ n + 2δ − 2 δn. As an immediate consequence of Theorems 13.1 and 13.2, we have the following result. Theorem 13.3 If H is a connected graph with δ(H ) ≥ δ, then  Υ (H ) ≥ 2 δn H − 2δ, and this bound is sharp. That this bound is sharp follows from a result of Favaron [35] who showed that for every positive integer δ, the bound in Theorem 13.2 is attained for infinitely many graphs. The same graphs achieve equality in the bound for Theorem 13.3. For example, for c ≥ 2, let G c be the connected graph constructed as follows. Let Fc be the complete graph of order c, and so Fc ∼ = K c . For every vertex v of Fc , add c − 1 new vertices v1 , . . . , vc−1 and add the c − 1 edges vvi for all i ∈ [c − 1]. Let G = G c

13.2 Lower Bounds in Terms of Order

181

denote the resulting graph of order n = c2 . We note that all the new vertices added to Fc have degree 1 in G c . Every transversal in G c must contain all except possibly one vertex of Fc in order to cover all the edges of Fc . If a minimal transversal in G c contains exactly c − 1 vertices of Fc , say all vertices of Fc except for v, then the transversal contains exactly c − 1 vertices not in Fc , namely, v1 , . . . , vc−1 , in order to cover the edges vvi for all i ∈ [c − 1]. Such a minimal transversal therefore has size exactly 2(c − 1). If a minimal transversal in G c contains all c vertices of Fc , then it contains no other vertex of G c and therefore has size exactly c. Therefore, the connected graph G of order n = c2 satisfies Υ (G) = max{2(c − 1), c} = 2(c − √ 1) = 2 n − 2, noting that c ≥ 2. Thus, the bound of Theorem 13.3 when δ = 1 is sharp. For every δ ≥ 2 one can similarly show that Theorem 13.3 is tight. Note that the following holds for G: √ Υ (G) = 2 n − 2
0, there exists a connected hypergraph H ∈ L2 of sufficiently large order such that Υ (H ) < ε · n H . As a special case of Theorem √ 13.3, if H is a connected 2-uniform hypergraph √ of order n ≥ 2, then Υ (H ) ≥ 2 n − 2. When n ≥ 3, we √ observe that 2 n − 2 ≥ √ n/2. Further, we observe that when n = 2, Υ (H ) = 1 = n/2. Thus, as an immediate consequence of Theorem 13.3 we observe that if H is a connected graph, then  Υ (H ) ≥ n H /2, implying the following.  Υ (H ) ≥

1 2n H

 nH .

Theorem 13.5 U2 = 0.

13.2.2 The Constant U3 In this section, we consider the case when k = 3. Let H1 and H2 be two vertex-disjoint copies of a linear 3-uniform hypergraph Rn constructed in Chapter 10 corresponding to a cap set in Fn3 . As shown in Section 11.3, the hypergraph Rn has order 3n and τ (Rn ) = n(Rn ) − α(Rn ) = 3n − Cn , where Cn ≤ 2.756n . Thus, n(H1 ) = n(H2 ) = 3n

and

τ (H1 ) = τ (H2 ) = 3n − Cn .

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13 Upper Transversals in Linear Hypergraphs

Let H be the hypergraph obtained from the union of H1 and H2 as follows: For every vertex xi ∈ V (H1 ) and every vertex y j ∈ V (H2 ), add a vertex wi j and the edge ei j = {xi , y j , wi j } for all i, j ∈ [3n ]. The resulting hypergraph H is a 3-uniform, connected, linear hypergraphs with minimum degree at least 1; that is, H ∈ L3,≥1 . We note that H has order n H = n(H1 ) + n(H2 ) + n(H1 ) · n(H2 ) = 2 · 3n + 32n . Let T be a minimal transversal in H , and let T1 and T2 be the restriction of T to H1 and H2 , respectively; that is, Ti = T ∩ V (Hi ) for i ∈ [2]. Let n i = |Ti | for i ∈ [2]. Since T is a transversal in H , we note that Ti is a transversal in Hi for i ∈ [2], implying that n i ≥ τ (Hi ) = 3n − Cn . Let T12 consist of the set of all vertices of the form wi j where the vertex xi ∈ V (H1 ) \ T1 and the vertex y j ∈ V (H2 ) \ T2 for some i ∈ [3n ] and j ∈ [3n ]. In order to cover all the edges ei j where i, j ∈ [3n ], we note that T = T1 ∪ T2 ∪ T12 and |T | = n 1 + n 2 + (3n − n 1 )(3n − n 2 ). This above expression is maximized if n 1 and n 2 are minimized; that is, when n i = τ (Hi ) = 3n − Cn for i ∈ [2], implying that Υ (H ) = (3n − Cn ) + (3n − Cn ) + Cn · Cn = 2 · 3n − 2Cn + Cn2 = 2 · 3n − 2(2.756)n + (2.756)2n . Hence, 2 · 3n − 2(2.756)n + (2.756)2n Υ (H ) = → nH 2 · 3n + 32n



2.7562 9

n < (0.8439484444445)n

as n → ∞. The existence of the above hypergraph H implies the following result. Theorem 13.6 For every given ε > 0, there exists a connected hypergraph H ∈ L3 of sufficiently large order such that Υ (H ) < ε · n H . Theorem 13.7 U3 = 0. The precise value of Uk has yet to be determined for any values of k with k ≥ 4. The immediate challenge is to determine if U4 = 0 also holds or whether U4 > 0. Indeed, it is even possible that Uk = 0 for all k ≥ 2.

13.3 Upper k-Transversal Number

183

13.3 Upper k-Transversal Number In this section, we study the upper k-transversal number of linear hypergraphs. Our aim is twofold. First, to obtain an upper bound on Υk (H ) in terms of the order of H . Secondly, to obtain an upper bound on Υk (H ) in terms of the order and size of H . Theorem 13.8 For r ≥ 2 an integer and for every integer k ∈ [r ], if H is an r uniform, linear hypergraph with maximum degree Δ, then the following holds:  (a)

Υk (H ) ≤ 

(b) Υk (H ) ≤

k·Δ k(Δ − 1) + r

 nH

k·Δ Δ(k + 1) + r − k

 (n H + m H ).

Further, both bounds are tight. Proof. Let r ≥ 2 and let H be a r -uniform hypergraph. For k ∈ [r ], let Tk be a minimal k-transversal in H of maximum cardinality, and so |Tk | = Υk (H ). Let Tk = {v1 , v2 , . . . , vtk }, where tk = |Tk |. By the minimality of the k-transversal Tk , for each i ∈ [tk ] the set Tk \ {vi } is not a k-transversal in H , implying that there exists an edge that contains the vertex vi and contains exactly k vertices of Tk . Among all such edges, we select one such edge ei that we associate with the vertex vi . Thus, |ei ∩ Tk | = k and vi ∈ ei . Let ei be the set of vertices in ei that do not belong to Tk , and so ei = ei \ Tk and |ei | = |ei | − k = r − k. We call the edge ei associated with the vertex vi a critical edge. Further, we call each of the r − k vertices in ei a vertex associated with the vertex vi . Thus, there are r − k vertices associated with each vertex in Tk . Let t be the number of vertices in V (H ) \ Tk that belong to at least one critical edge. Thus,   t k     t =  ei  .   i=1

The sum of the number of vertices in a critical edge that do not belong to Tk over all critical edges, allowing for repetition and multiple counting of vertices, is given by tk

|ei | = tk (r − k). i=1

Since the maximum degree in H is Δ, each vertex not in Tk belongs to at most Δ critical edges. Further, since every critical edge contains exactly k vertices of Tk , each vertex not in Tk is associated with at most k · Δ vertices of Tk . Thus, each vertex tk |ei | is counted at most k · Δ times, implying that in the sum i=1 t ≥

tk (r − k) . k·Δ

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13 Upper Transversals in Linear Hypergraphs

Therefore,

n H = |Tk | + |V (H ) \ Tk | ≥ tk +  t ≥ tk +  =

or, equivalently,

 tk ≤

tk (r − k) k·Δ

k(Δ − 1) + r k·Δ

k·Δ k(Δ − 1) + r

 tk ,

 nH .

(13.1)

As observed earlier, every critical edge contains exactly k vertices of Tk , implying that there are at least tk /k critical edges. Since m H is at least the number of critical edges in H , we note that m H ≥ tk /k, implying that     k(Δ − 1) + r Δ tk + tk , nH + mH ≥ k·Δ k·Δ or, equivalently,

 tk ≤

k·Δ Δ(k + 1) + r − k

 (n H + m H ).

(13.2)

Since Υk (H ) = tk , the upper bound in (a) follows from Inequality (13.1), while the upper bound in (b) follows from Inequality (13.2). We note that the above proof holds even if we relax the linearity constraint and allow non-linear hypergraphs. We show next that the bounds are tight, even for linear hypergraphs. Suppose firstly that k = r . In this case, H is a k-uniform hypergraph and every k-transversal in H contains all k vertices from each edge. Thus, if H is a k-uniform linear hypergraph that contains no isolated vertex, then every k-transversal in H is precisely the set V (H ), implying that  Υk (H ) = n H =

k·Δ k(Δ − 1) + r

 nH .

Thus, the bound in (a) is tight in this case when k = r . We note further that if k = r and we take H to be the hypergraph consisting of a singleton edge of size r and no isolated vertex, then m H = 1, n H = r , Δ = 1, and  Υk (H ) = r =

k·Δ Δ(k + 1) + r − k

 (n H + m H ).

13.3 Upper k-Transversal Number

185

More generally, for 1 ≤ Δ ≤ k + 1, if k = r and H is obtained from the disjoint union of Δ edges of size r by identifying one vertex from every edge, then m H = Δ, n H = Δ(r − 1) + 1, and  Υk (H ) = Δ(r − 1) + 1 =

k·Δ Δ(k + 1) + r − k



(n H + m H ) ,

noting that (Δ − 1)/(k + 1) = 0. Thus, the bound in (b) is in a sense best possible in this case when k = r . Suppose next that k = r − 1. For Δ ≥ 1, let f i = {v1 , u i1 , . . . , u ik } for each i ∈ [Δ]. We note that f i ∩ f j = {v1 } for all i and j, where 1 ≤ i < j ≤ Δ. Let H be the r -uniform hypergraph with edge set E(H ) = { f 1 , . . . , f Δ } and with Δ  V (H ) = {v1 } ∪ {u i1 , . . . , u ik } . i=1

Thus, H has exactly Δ edges and every two edges of H have one vertex in common, namely, the vertex v1 . Further, H has n H = kΔ + 1 = k(Δ − 1) + r vertices, noting that here k = r − 1. Every minimal k-transversal in H of maximum cardinality consists precisely of the set V (H ) \ {v1 }. Thus, Υk (H ) = n H − 1 = k · Δ. Hence,   k·Δ nH . Υk (H ) = k · Δ = k(Δ − 1) + r Further, since H has m H = Δ edges, we note that  Υk (H ) = k · Δ =

k·Δ Δ(k + 1) + r − k

 (n H + m H ).

Thus, the bounds in (a) and (b) are tight in this case when k = r − 1. Suppose, finally, that k ≤ r − 2. Let H  be an arbitrary Δ-regular (r − k)-uniform linear hypergraph, and let H be the r -uniform hypergraph obtained from H  by adding to each edge of H  , k new distinct vertices (of degree 1). We note that m H = m H  and n H = n H  + k · m H  . Let T = V (H ) \ V (H  ) denote the set of newly added vertices that belong to H but not to H  , and so |T | = k · m H  = k · m H . Since H  is a Δ-regular (r − k)-uniform hypergraph, (r − k)m H = (r − k)m H  = Δn H  , and so n H  /(k · m H ) = (r − k)/(k · Δ). Thus, since T is a minimal k-transversal in H ,

186

13 Upper Transversals in Linear Hypergraphs

Υk (H ) nH + mH

≥ = = = = = T heor em 13.8(b)

≥

|T | nH + mH k · mH k · m H + n H + m H 1 1+

nH k·m H

1+

r −k k·Δ

+

1 k

1 +

1 k

k·Δ k ·Δ+r −k +Δ k·Δ Δ(k + 1) + r − k Υk (H ) . nH + mH

Consequently, we must have equality throughout the above inequality chain, implying that  Υk (H ) = |T | =

k·Δ Δ(k + 1) + r − k

 (n H + m H ).

Moreover, since (r − k)m H = Δn H  = Δ(n H − k · m H ), we note that  mH =

Δ k(Δ − 1) + r 

and so Υk (H ) = |T | = k · m H =

 nH ,

kΔ k(Δ − 1) + r

 nH .

Thus, the bounds in (a) and (b) are also tight in this case when k ≤ r − 2. This completes the proof of Theorem 13.8. 

Chapter 14

Strong Tranversals in Linear Hypergraphs

14.1 Introduction A set S of vertices in a hypergraph H is a strong independent set (sometimes simply called independent in the literature) if no two vertices in S belong to a common edge. The strong independence number αs (H ) is the maximum size of a strong independent set in H . Strong independence in hypergraphs is well studied in the literature (see, for example, [11, 60, 72], for recent papers on this topic). In this chapter, we study the complement of a strong independent set in a hypergraph H . Such sets are called strong transversals of H . Formally, a strong transversal in a hypergraph H is a transversal in H that intersect every edge e of H in at least |e| − 1 vertices; that is, if T is a strong transversal in H , then |T ∩ e| ≥ |e| − 1 for every edge e ∈ E(H ). The strong transversal number of H is the minimum size of a strong transversal in H . A transversal of size τs (H ) is called a τs -transversal of H . The complement of a strong transversal in a hypergraph H is a strong independent set of H and vice versa. Hence by definition, we have the following relationship between strong transversals and strong independent sets. Observation 14.1 τs (H ) + αs (H ) = n(H ) for every hypergraph H . If H is a hypergraph in which every edge has size at least 2, then a transversal in H that intersect every edge e of H in at least two vertices is called a double transversal in H , also known as a double blocking set in the literature. Hence if T is a double transversal in H , then |T ∩ e| ≥ 2 for every edge e ∈ E(H ). The double transversal number τ2 (H ) of H is the minimum size of a double transversal in H . A transversal of size τ2 (H ) is called a τ2 -transversal of H . If H is a 3-uniform hypergraph, then we note that a double transversal is a strong transversal and vice versa. In this case, τ2 (H ) = τs (H ). In this chapter, we present results on the strong transversal number in linear hypergraphs. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_14

187

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14 Strong Tranversals in Linear Hypergraphs

14.2 A Probabilistic Bound Let k ≥ 2 be an integer. There are many pairs of numbers (ak , bk ) such that the inequality τs (H ) ≤ ak n H + bk m H holds for all k-uniform hypergraph. The set of all vertices of H is a strong transversal in a k-uniform hypergraph, H , and so we can choose (ak , bk ) = (1, 0). On the other hand, choosing k − 1 vertices from every edge of H produces a strong transversal in H , and so we can choose (ak , bk ) = (0, k − 1). In this section, we give a trivial probabilistic upper bound on the strong transversal number of H in terms of its order n H and size m H . For a positive integer k ≥ 1, we use the standard notation [k] = {1, . . . , k} and [k]0 = {0, 1, . . . , k}. Theorem 14.2 ([65]) For k ≥ 2, if  (ak , bk ) =

k−2    k (k − 1 − i) pi (1 − p)k−i p, i i=0

 (14.1)

for any value of p with 0 ≤ p ≤ 1, then τs (H ) ≤ ak n H + bk m H

(14.2)

holds for every k-uniform hypergraph H . Proof. Applying probabilistic arguments similar to the ones uded by Alon [2], we choose, randomly and independently, each vertex of H with probability p. Let T be the (random) set of vertices picked, and let E i be the set of all edges e ∈ E(H ) that intersect T in exactly i vertices for i ∈ [k − 2]0 . For a fixed edge e ∈ E(H ) and i ∈ [k − 2]0 , we note that Pr (e ∈ E i ) =

  k i p (1 − p)k−i . i

Letting n = n H and m = m H , the expected value of the quantity |T | is np, while the expected value of the quantity |E i | for i ∈ [k − 2]0 is   k i p (1 − p)k−i m. i By adding to T , arbitrarily, k − 1 − i vertices from each edge of E i for i ∈ [k − 2]0 , we obtain a set of size at most |T | +

k−2  (k − 1 − i)|E i | i=0

14.2 A Probabilistic Bound

189

that intersects every edge e of H in at least |e| − 1 = k − 1 vertices. By the linearity of expectation, the expected value of such a set is at most k−2    k np + (k − 1 − i) pi (1 − p)k−i m. i i=0

Thus,

 k−2     k i k−i (k − 1 − i) p (1 − p) m τs (H ) ≤ pn + i i=0

for all 0 ≤ p ≤ 1.



In the special cases of Theorem 14.2 when k = 2 and k = 3, we have the following probabilistic upper bounds on the strong transversal number of 2-uniform and 3-uniform hypergraph, respectively. Theorem 14.3 ([65]) If (a2 , b2 ) = ( p, (1 − p)2 ) for any value of p with 0 ≤ p ≤ 1, then τs (H ) ≤ a2 n H + b2 m H holds for every 2-uniform hypergraph H . Theorem 14.4 ([65]) If (a3 , b3 ) = ( p, 2(1 − p)3 + 3 p(1 − p)2 ) for any value of p with 0 ≤ p ≤ 1, then τs (H ) ≤ a3 n H + b3 m H holds for every 3-uniform hypergraph H . In Chapters 14.4 and 14.5, we improve the bounds of Theorem 14.3 and 14.4, respectively, by determining all (ak , bk ) of nonnegative integers ak and bk such that the inequality τs (H ) ≤ ak n H + bk m H holds for k ∈ {2, 3}.

14.3 The Convex Set Sk For k ≥ 2, let Sk be the set of all pairs (ak , bk ) of nonnegative integers ak and bk for which the inequality τs (H ) ≤ ak n H + bk m H holds for all H ∈ Hk . We show that the set Sk is a convex set. Theorem 14.5 ([65]) For k ≥ 2, the set Sk is a convex set. Proof. For k ≥ 2, let (α1 , β1 ) and (α2 , β2 ) be two arbitrary points in the set Sk . Let H ∈ Hk and note that τs (H ) ≤ α1 n H + β1 m H and τs (H ) ≤ α2 n H + β2 m H . We show that the line segment L joining the points (α1 , β1 ) and (α2 , β2 ) lies completely within the set Sk . Let (α, β) be an arbitrary point on the line segment L. Thus, (α, β) = ε(α1 , β1 ) + (1 − ε)(α2 , β2 ) = (εα1 + (1 − ε)α2 , εβ1 + (1 − ε)β2 )

190

14 Strong Tranversals in Linear Hypergraphs

for some real number ε where 0 ≤ ε ≤ 1. The point (α, β) satisfies αn H + βm H = (εα1 + (1 − ε)α2 )n H + (εβ1 + (1 − ε)β2 )m H = ε(α1 n H + β1 m H ) + (1 − ε)(α2 n H + β2 m H ) ≥ ετs (H ) + (1 − ε)τs (H ) = τs (H ), implying that the point (α, β) belongs to the set Sk .



By Theorem 14.5, the set Sk is a convex set for every k ≥ 2. In Chapter 14.4, we determine all extreme points of the convex set S2 . In the following Chapter 14.5, we determine all extreme points of the convex set S3 . We show that there are infinitely many extreme points in S3 , and that the line segment between two consecutive extreme points of the convex set S3 corresponds to a Steiner triple system STS(n) where n ≥ 3 and n (mod 6) ∈ {1, 3}. For this purpose, for r ≥ 2 we say that a point (α, β) of nonnegative integers α and β is tight in the convex set Sr if the point (α, β) belongs to the set Sr and τs (H ) = αn H + βm H holds for some r -uniform hypergraph H . Further, if (α, β) is tight in the convex set Sr and H is a r -uniform hypergraph for which τs (H ) = αn H + βm H holds, then we call H a realizable hypergraph of (α, β). The following lemma will prove to be useful. Lemma 14.1 For r ≥ 2, if (ai , bi ) and (ai+1 , bi+1 ) are both tight in the convex set Sr for some r -uniform hypergraph F, then all points ε(ai , bi ) + (1 − ε)(ai+1 , bi+1 ) where 0 ≤ ε ≤ 1 are also tight in Sr for F. Proof. Since (ai , bi ) is tight for the r -uniform hypergraph F, the point (ai , bi ) belongs to the convex set Sr , and so τs (H ) ≤ ai n H + bi m H for all r -uniform hypergraphs H . Analogously, we have τs (H ) ≤ ai+1 n H + bi+1 m H for all r -uniform hypergraphs H . Thus for all r -uniform hypergraphs H , the following now holds. τs (H ) = ετs (H ) + (1 − ε)τs (H ) ≤ ε(ai n H + bi m H ) + (1 − ε)(ai+1 n H + bi+1 m H ) = (εai + (1 − ε)ai+1 )n H + (εbi + (1 − ε)bi+1 )m H . Therefore the point ε(ai , bi ) + (1 − ε)(ai+1 , bi+1 ) lies in the convex set Sr . Furthermore since F is a realizable hypergraph for both points (ai , bi ) and (ai+1 , bi+1 ), we have ai n F + bi m F = τs (F) = ai+1 n F + bi+1 m F . Therefore if we considered F instead of H above we would have equality everywhere, implying that the point ε(ai , bi ) + (1 − ε)(ai+1 , bi+1 ) is also tight in Sr , and F is a realizable hypergraph for this point. 

14.4 The Convex Set S2

191

14.4 The Convex Set S2 We note that if H is a 2-uniform hypergraph, then H is a graph. Further in this case, a transversal of H is a strong transversal of H and vice versa. In particular, τ (H ) = τs (H ). The following theorem was first proved by Chvátal and McDiarmid in [24]. We give an alternative proof below. Theorem 14.6 ([24]) For every integer k ≥ 1, if H is a graph, then k(k + 1)τ (H ) ≤ k(k − 1)n H + 2m H , with equality if and only if H is a complete graph of order k or k + 1. Proof. We will prove the theorem by induction on the order of H . If n H = 0, then m H = 0 which implies that τ (H ) = 0 and the theorem holds. If n H = 1, then m H = 0 and again the theorem holds. We may therefore assume n H ≥ 2 and that the theorem holds for all graphs of smaller order than H . Further we may also assume that H is connected as otherwise we use induction on each component of H . Let x be a vertex of minimum degree in H and let Q = N H [x]. Let H  be obtained from H by deleting all vertices of Q and all edges incident with vertices of Q; that is, H  = H − Q. Let H  have order n H  and size m H  . We note that possibly V (H ) = Q, in which case n H  = m H  = τ (H  ) = 0. Applying the inductive hypothesis to H  , the following holds (even in the case when V (H ) = Q): k(k + 1)τ (H  ) ≤ k(k − 1)n H  + 2m H  . We note that n H − n H  = |Q| = d H (x) + 1. Further since d H (x) = δ(H ), we have m H − m H ≥

1 1 1 d H (q) ≥ |Q| · d H (x) = (d H (x) + 1)d H (x). 2 q∈Q 2 2

(14.3)

Every transversal in H  can be extended to a transversal in H by adding to it the set N H (x) = Q \ {x}, implying that τ (H ) ≤ τ (H  ) + d H (x). We therefore obtain the following noting that d H (x) and k are integers: k(k + 1)τ (H ) ≤ k(k + 1)τ (H  ) + k(k + 1)d H (x) ≤ k(k − 1)n H  + 2m H  + k(k + 1)d H (x) ≤ k(k − 1)(n H − d H (x) − 1)+   2 m H − 21 (d H (x) + 1)d H (x) + k(k + 1)d H (x) = k(k − 1)n H + 2m H + 2k · d H (x) − k(k − 1) − d H (x)(d H (x) + 1) = k(k − 1)n H + 2m H − ((d H (x))2 − (2k − 1)d H (x) + k(k − 1)) = k(k − 1)n H + 2m H − (d H (x) − k)(d H (x) − (k − 1)) ≤ k(k − 1)n H + 2m H .

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14 Strong Tranversals in Linear Hypergraphs

We have therefore shown that k(k + 1)τ (H ) ≤ k(k − 1)n H + 2m H , as desired. Suppose next that k(k + 1)τ (H ) = k(k − 1)n H + 2m H . This implies that all the above inequalities must be equalities. In particular, d H (x) = k or d H (x) = k − 1. Further, equality throughout Inequality (14.3) implies that Q = V (H ) and that all vertices q ∈ Q have degree d H (q) = d H (x). This in turn implies H is a complete graph of order k or k + 1.  Theorem 14.6 implies the following bounds when k ∈ [4]: τ (H ) ≤ m H τ (H ) ≤ 21 n H + 16 m H

τ (H ) ≤ 13 n H + 13 m H 1 τ (H ) ≤ 35 n H + 10 mH .

We are now in a position to determine all extreme points in the convex set S2 . Let  



2 k−1

k = 1, 2, 3 . . . ∪ {(1, 0)}. , EP2 = k + 1 k(k + 1)

We show that the set EP2 determines all extreme points for the convex set S2 . Theorem 14.7 ([24]) The set EP2 is the set of extreme points of the convex set S2 . Proof. We note that the (1, 0) is an accumulation point (also called a limit point, in the literature). For k ≥ 1, let  (ak , bk ) =

 2 k−1 , . k + 1 k(k + 1)

By Theorem 14.6, the points (ak , bk ) and (ak+1 , bk+1 ) are both tight for the complete graph of order k + 1. Hence by Lemma 14.1 (with r = 2), every point on the line segment joining the points (ak , bk ) and (ak+1 , bk+1 ) is tight in S2 (for the complete graph of order k + 1). 

14.5 The Convex Set S3 As observed earlier, if H is a 3-uniform hypergraph, then a double transversal in H is a strong transversal in H and vice versa. In this case, τ2 (H ) = τs (H ). By Theorem 14.5, the set S3 consisting of all pairs (α, β) of nonnegative integers α and β for which inequality τs (H ) ≤ αn H + βm H holds is a convex set. In this section, we show that there are infinitely many extreme points of the convex set S3 , and we determine all these extreme points. With the extreme points of S3 ordered by the first coordinate, Theorem 11.5 (and Theorem 12.7) implies that (1, 0) is the last extreme point of S3 and this point is an accumulation point (also called a limit point, in the literature). In order to determine

14.5 The Convex Set S3

193

the infinitely many extreme points of the convex set S3 , we prove two more general results in Theorems 14.8 and 14.15. For this purpose, we introduce some additional notation. Let H2,3 be the class of hypergraphs in which every edge is a 2-edge or a 3-edge. By definition, a strong transversal of a hypergraph H ∈ H2,3 is a transversal of H that intersects every 2-edge in at least one vertex and intersects every 3-edge in at least two vertices. For a hypergraph H ∈ H2,3 , let m 2 (H ) and m 3 (H ) denote the number of 2-edges and 3-edges in H . Further for a vertex v ∈ V (H ), we denote by d2 (v, H ) and d3 (v, H ) the number of 2-edges and 3-edges, respectively, in H that contain the vertex v. Thus, if H ∈ H2,3 , then m H = m 2 (H ) + m 3 (H ) and d H (v) = d2 (v, H ) + d3 (v, H ). If the hypergraph H is clear from the context, we simply write d2 (v) and d3 (v) rather than d2 (v, H ) and d3 (v, H ), respectively. Let E 3 denote the 3-uniform hypergraph on three vertices containing exactly one edge, which is also shown in Figure 14.1(a). Theorem 14.8 ([65]) For every integer k ≥ 1, if H ∈ H2,3 is connected hypergraph, then (2k + 1)(2k + 3)τs (H ) ≤ 4k(k + 1)n H + 6m 3 (H ) + 3m 2 (H ), with equality if and only if the following holds. (a) H ∈ L3 ; that is, H is a 3-uniform linear hypergraph. (b) H is k-regular of order 2k + 1 or H is (k + 1)-regular of order 2k + 3. (c) Every two vertices of H belong to a common edge. Proof. Let k ≥ 1 be an arbitrary positive integer, and let H ∈ H2,3 . We define the weight of H to be w(H ) = 6m 3 (H ) + 3m 2 (H ). Further, we define ak = 4k(k + 1) and ck = (2k + 1)(2k + 3), and φk (H ) = ck τs (H )

and

Φk (H ) = ak n H + w(H ).

We proceed by induction on the order n H ≥ 2 of a hypergraph H in the class H2,3 to show that φk (H ) ≤ Φk (H ). If n H = 2, then m H = m 2 (H ), w(H ) ≥ 3 and τs (H ) = 1, implying that φk (H ) = 4k 2 + 8k + 3 and Φk (H ) ≥ 8k 2 + 8k + 3, and so φk (H ) < Φk (H ). If n H = 3 and m 3 (H ) ≥ 1, then w(H ) ≥ 6 and τs (H ) = 2, implying that φk (H ) = 8k 2 + 16k + 6 and Φk (H ) ≥ 12k 2 + 12k + 6, and so φk (H ) < Φk (H ) unless k = 1 and m H = m 3 (H ) = 1, in which case H = E 3 and φk (H ) = Φk (H ). If n H = 3 and m H = m 2 (H ), then either m H = 2, in which case w(H ) = 6 and τs (H ) = 1, or m H ≥ 3, in which case w(H ) ≥ 9 and τs (H ) ≤ 2. In both cases, φk (H ) < Φk (H ). This establishes the base cases. Let n H ≥ 4 and

194

14 Strong Tranversals in Linear Hypergraphs

assume the desired result hold for all hypergraphs H  ∈ H2,3 of order n H  < n H . Let H ∈ H2,3 have order n H . For each vertex v ∈ V (H ), let 3 θ H (v) = 2d3 (v) + d2 (v) 2

and

θ (H ) = min θ H (v). v∈V (H )

Let x be a vertex of H such that θ H (x) = θ (H ). Let Q = N H [x] and let H  be the hypergraph obtained from H by deleting all vertices from the set Q and all edges incident with vertices in Q, and for each 3-edge e of H that intersect Q in exactly one vertex, say v, we add to H  the edge e = e \ {v}. Let H  have order n H  and  size m H  . We proceed further with the following series of claims. Claim 14.9 The following holds: 1 |Q| ≤

+ θ (H ) if |Q| is even 1 + θ (H ) if |Q| is odd. 2

Proof. By our choice of the vertex x and definition of θ H (x), we have |Q| ≤ 1 + 2d3 (x) + d2 (x) = 1 + θ H (x) − 21 d2 (x) = 1 + θ (H ) − 21 d2 (x). Since d2 (x) ≥ 0, we have |Q| ≤ 1 + θ (H ). If, however, |Q| is even, then d2 (x) ≥ 1 or two distinct edges that contain the vertex x intersect in at least two vertices, that is, x is incident with two overlapping edges. Further if d2 (x) ≥ 1, then |Q| ≤ 21 + θ (H ), while if x is incident with two overlapping edges, then |Q| ≤ 2d3 (x) + d2 (x), () implying that |Q| ≤ θ (H ) − 21 d2 (x) ≤ θ (H ).  Claim 14.10 w(H  ) ≤ w(H ) − θ H (q), with strict inequality if V (H ) = Q. q∈Q

Proof. Let e be an arbitrary edge intersecting Q. We now consider the following cases. Case 14.10.1. The edge e is a 2-edge and the edge e

|e ∩ Q| = 1. In this case, 3 3 to θ (q), and therefore to the term contributes 3 to the weight w(H ) and H q∈Q 2 2

w(H ) − q∈Q θ H (q). Since the edge e is deleted from H when constructing H  , it contributes 0 to w(H  ). Case 14.10.2. The edge e is a 2-edge and |e ∩ Q| = 2. In this case, the edge e contributes 3 to the weight w(H ) and 3 to q∈Q θ H (q) (it contributes 23 to two θ H (q)values), implying that the edge e contributes 0 to the term w(H ) − q∈Q θ H (q) and contributes 0 to w(H  ).

14.5 The Convex Set S3

195

Case 14.10.3. The edge e is a 3-edge and |e ∩ Q| = 1. In this case, the edge constructing H  , e is deleted from H and a new edge e = e \ {q} is added when implying that the edge e contributes

6 to the weight w(H ) and 2 to q∈Q θ H (q), and therefore 4 to the term w(H ) − q∈Q θ H (q), while the edge e contributes 3 to the weight w(H  ). Case 14.10.4. The edge e is a 3-edge and

|e ∩ Q| = 2. In this case, the edge e contributes 6 to the weight w(H ) and 4 to q∈Q θ H (q), implying that the edge e contributes 2 to the term w(H ) − q∈Q θ H (q) and contributes 0 to w(H  ). Case 14.10.5. The edge e is a 3-edge and

|e ∩ Q| = 3. In this case, the edge e contributes 6 to the weight w(H ) and 6 to q∈Q θ H (q), implying that the edge e contributes 0 to the term w(H ) − q∈Q θ H (q) and contributes 0 to w(H  ). By the above five cases, if e ⊆ Q, then Case 14.10.2 or Case 14.10.5 applies, and in both cases the edge e contributes 0 to the term w(H ) − q∈Q θ H (q) and contributes 0 to w(H  ). However if e  Q (and so, |e ∩ Q| < |e|), then Case 14.10.1, Case 14.10.3, or Case 14.10.4

applies, and in all three cases the contribution of the edge e to the term w(H ) − q∈Q θ H (q) exceeds its contribution to w(H  ). () Applying the inductive hypothesis to each component of H  , we have φk (H  ) ≤ Φk (H  ), or, equivalently, ck τs (H  ) ≤ ak n H  + w(H  ). Thus by Claim 14.10 and noting that n H  = n H − |Q|, we have ⎛ ck τs (H  ) ≤ ak n H + w(H ) − ⎝ak |Q| +



⎞ θ H (q)⎠ .

(14.4)

q∈Q

Every strong transversal in H  can be extended to a strong transversal in H by adding to it the set Q \ {x}, implying that τs (H ) ≤ τs (H  ) + |Q| − 1. Hence by Inequality (14.4), we have ⎛ ck τs (H ) ≤ ak n H + w(H ) − ⎝ak |Q| − ck |Q| + ck +



⎞ θ H (q)⎠ .

(14.5)

q∈Q

By Inequality (14.5), it suffices for us to show that ak |Q| − ck |Q| + ck +



θ H (q) ≥ 0.

(14.6)

q∈Q

By definition of θ (H ), we note that θ (H ) ≤ θ H (q) for every vertex q ∈ Q. Thus by definition of ak and ck , we have

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14 Strong Tranversals in Linear Hypergraphs

ak |Q| − ck |Q| + ck +

θ H (q) ≥ 0

q∈Q

ck − (ck − ak )|Q| +



θ H (q) ≥ 0

q∈Q

ck − (4k + 3)|Q| + ⇑





θ H (q) ≥ 0

q∈Q

ck − (4k + 3)|Q| + |Q| · θ (H ) ≥ 0.

In order to prove Inequality (14.6), it therefore suffices for us to prove that f (Q) := ck − (4k + 3 − θ (H ))|Q| ≥ 0.

(14.7)

Claim 14.11 If |Q| = 2k + 1 or |Q| = 2k + 3, then f (Q) ≥ 0. Proof. If |Q| = 2k + 1, then by Claim 14.9, θ (H ) ≥ |Q| − 1 = 2k, implying that f (Q) ≥ ck − (4k + 3 − 2k)|Q| = (2k + 1)(2k + 3) − (2k + 3)(2k + 1) = 0. If |Q| = 2k + 3, then by Claim 14.9, θ (H ) ≥ |Q| − 1 = 2k + 2, implying that f (Q) ≥ ck − (4k + 3 − 2k − 2)|Q|=(2k + 1)(2k + 3)−(2k + 1)(2k + 3)=0. () Claim 14.12 If |Q| ∈ / {2k + 1, 2k + 3}, then f (Q) > 0. Proof. If |Q| = 2k + 2, then by Claim 14.9, θ (H ) ≥ |Q| − 21 = 2k + 23 ,    implying that f (Q) ≥ ck − 4k + 3 − 2k + 23 |Q| = (2k + 1)(2k + 3) − (2k + 3 )(2k + 2) = k > 0. We may therefore assume that |Q| < 2k + 1 or |Q| > 2k + 3, 2 for otherwise the desired result, namely, f (Q) > 0, follows. By Claim 14.9, θ (H ) ≥ |Q| − 1, implying that f (Q) = ck − (4k + 3 − θ (H ))|Q| ≥ ck − (4k + 3 − |Q| + 1)|Q| = ck + |Q|2 − 4|Q|(k + 1). Therefore, f  (Q) = 2|Q| − 4(k + 1). If |Q| < 2k + 1, then f  (Q) < 2(2k + 1) − 4(k + 1) < 0. Hence, the function f (Q) is a decreasing function on the interval [0, 2k + 1]. Since f (2k + 1) ≥ 0, this implies that f (Q) > 0 for |Q| ≤ 2k. If |Q| > 2k + 3, then f  (Q) > 2(2k + 3) − 4(k + 1) > 0. Hence, the function f (Q) is an increasing function on the interval [2k + 3, ∞). Since f (2k + 3) ≥ 0, this implies that () f (Q) > 0 for |Q| ≥ 2k + 4. By Claims 14.11 and 14.12, Inequality (14.6) holds. This completes the proof of the desired upper bound, namely, ck τs (H ) ≤ ak n H + w(H ), or, equivalently, (2k + 1)(2k + 3)τs (H ) ≤ 4k(k + 1)n H + 6m 3 (H ) + 3m 2 (H ).

(14.8)

Suppose, next, that we have equality in the above Inequality (14.8). In this case, all the above inequalities are equalities. In particular, we have equality in Claim 14.11, implying that |Q| = 2k + 1 or |Q| = 2k + 3. This in turn implies that |Q| = θ (H ) + 1, and therefore by the proof of Claim 14.9, we have d2 (x) = 0 and

14.5 The Convex Set S3

(a) E3

197

(c) The Fano plane

(b) H5

Fig. 14.1 The hypergraphs E 3 , H5 , and the Fano plane

there are no overlapping edges that contain the vertex x. Thus, every edge incident with x is a 3-edge, and d H (x) = d3 (x) and |Q| = |N H [x]| = 1 + 2d3 (x). We also must have θ (H ) = θ H (q) for every vertex q ∈ Q. Moreover, equality in Inequality (14.8) implies that we must have equality in Claim 14.10, which implies that V (H ) = Q. From these observations, properties (a), (b), and (c) in the statement of the theorem readily follow. This completes the proof of Theorem 14.8.  Before proving our second main theorem of this chapter we need the following upper bound on the strong transversal number given in [55]. Let H5 be the hypergraph illustrated in Figure 14.1(b). Recall that in finite geometry the Fano plane, illustrated in Figure 14.1(c), is the finite projective plane P G(2, 2) of order 2 (where PG stands for “projective geometry”). Theorem 14.13 ([55]) If H is a connected 3-uniform hypergraph, then τs (H ) ≤

4 2 nH + mH , 7 7

with equality if and only if H = E 3 or H = H5 or H is the Fano plane. In particular, if H is linear, then equality holds if and only if H = E 3 or H is the Fano plane. We shall also need the following result by Feder and Subi [36] on packing edgedisjoint triangles in graphs. Theorem 14.14 ([36]) Let n = 6 j + i, where i and j are integers and 0 ≤ i ≤ 5. The maximum number ofedges  in the complete graph K n that can be covered by edge-disjoint triangles is n2 − k /3, where k is defined as follows: ⎧ ⎪ ⎪ ⎪ ⎨

0 4

if i if i k= n ⎪ 2 if i ⎪ ⎪ ⎩n + 1 if i 2

∈ {1, 3} =5 ∈ {0, 2} =4

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14 Strong Tranversals in Linear Hypergraphs

Our next result extends Theorem 14.13. In fact, Theorem 14.13 shows that Theorem 14.15 can be extended to k ≥ 2 where k ≡ 2 (mod 3) when there is no 2-edge in the hypergraph. Theorem 14.15 ([65]) For every integer k ≥ 5 where k ≡ 2 (mod 3), if H ∈ H2,3 is a connected hypergraph, then (2k − 1)(2k + 3)τs (H ) ≤ 4(k − 1)(k + 1)n H + 6m 3 (H ) + 3m 2 (H ). Furthermore if H is 3-uniform and linear, then we have equality if and only if every two vertices of H belong to a common edge and H is either (k − 1)-regular of order 2k − 1 or H is (k + 1)-regular of order 2k + 3. Proof. Let k ≥ 5 be an arbitrary positive integer such that k ≡ 2 (mod 3), and let H ∈ H2,3 be a connected hypergraph. We first prove the main part of the theorem. Thereafter we will prove that if H is 3-uniform and linear, then we have equality if and only if every two vertices of H belong to a common edge and H is either (k − 1)-regular of order 2k − 1 or H is (k + 1)-regular of order 2k + 3. We define the weight of H to be w(H ) = 6m 3 (H ) + 3m 2 (H ). Further, we define ak = 4k 2 − 4 and ck = (2k − 1)(2k + 3), and φk (H ) = ck τs (H )

and

Φk (H ) = ak n H + w(H ).

We proceed by induction on the order n H of a hypergraph H in the class H2,3 to show that φk (H ) ≤ Φk (H ). If n H = 0 the theorem vacuously holds, and if n H = 1, then m H = 0 and τs (H ) = 0 and the theorem holds. This establishes the base case. Let n H ≥ 2 and assume the desired result hold for all hypergraphs H  ∈ H2,3 of order n H  < n H . Let H ∈ H2,3 have order n H . We first prove the following claim. Claim 14.16 We may assume that H is linear. Proof. Suppose that H is not linear and let e1 , e2 ∈ E(H ) be two overlapping edges in H . We will now produce a new hypergraph H  , such that the following holds. (i): (ii): (iii): (iv):

φk (H  ) = φk (H ). Φk (H  ) ≤ Φk (H ). n(H ) = n(H  ). 3m 3 (H  ) + m 2 (H  ) < 3m 3 (H ) + m 2 (H ).

Repeating this process as long as there are overlapping edges, which is a finite process due to (iv), we obtain a linear hypergraph H ∗ , such that φk (H ∗ ) = φk (H ) and Φk (H ∗ ) ≤ Φk (H ). So if we can prove the theorem for H ∗ , then it would also hold for H . It therefore suffices for us to prove that a hypergraph H  satisfying (i)–(iv) exists.

14.5 The Convex Set S3

199

As the edges e1 and e2 overlap, we have |V (e1 ) ∩ V (e2 )| ≥ 2. If V (e2 ) ⊆ V (e1 ), then we delete e2 from H in order to obtain a hypergraph H  ∈ H2,3 . If |V (e1 )| = |V (e2 )|, then clearly H  satisfies (i)–(iv). So assume that e2 is a 2-edge and e1 is a 3-edge. In this case a strong transversal in H  will contain at least two vertices in V (e1 ) and therefore also at least one vertex in V (e2 ). Therefore, φk (H  ) = φk (H ) and properties (i)–(iv) are also satisfied in this case. We may therefore assume that V (e2 )  V (e1 ) and analogously that V (e1 )  V (e2 ). This implies that e1 and e2 are both 3-edges and |V (e1 ) ∩ V (e2 )| = 2. Let V (e1 ) = {x, u 1 , u 2 } and V (e2 ) = {y, u 1 , u 2 } and construct H  from H by deleting e2 and adding the two 2-edges f 1 = {y, u 1 } and f 2 = {y, u 2 }. We note that the connected hypergraph H  ∈ H2,3 satisfies n H  = n H , w(H  ) = w(H ) and Φk (H  ) = Φk (H ). Furthermore, if T  is a strong transversal in H  , then it must contain at least two vertices from e2 , due to the following. If y ∈ T  , then either u 1 or u 2 is in T  due to the / T  , then {u 1 , u 2 } ⊆ T  due to the edges f 1 and f 2 . Hence, T  is also edge e1 and if y ∈ a strong transversal in H , implying that τs (H ) ≤ τs (H  ). Analogous arguments show that every strong transversal in H is a strong transversal in H  , and so τs (H  ) ≤ τs (H ). Consequently, τs (H  ) = τs (H ) and (i)–(iv) are satisfied, completing the proof of the () claim. For each vertex v ∈ V (H ), let 3 θ H (v) = 2d3 (v) + d2 (v) 2

and

θ (H ) = min θ H (v). v∈V (H )

Let Y be an arbitrary set of vertices in H such that no two vertices in Y are adjacent in H . Thus, every edge of H intersects Y in at most one vertex. Let QY =



N H [y]

y∈Y

and let HY be the hypergraph obtained from H by deleting all vertices from the set Q Y and all edges incident with vertices in Q Y , and for each 3-edge e of H that intersect Q Y in exactly one vertex, say v, we add to HY the edge e = e \ {v}. We define the boundary, ∂(Q Y ), of the set Q Y as the set of all vertices not in Q Y that have a neighbor in Q Y ; that is, ∂(Q Y ) = N (Q Y ) \ Q Y . We now define  H (Y ) by  H (Y ) =

 1 |∂(Q Y )| + θ H (q). 2 q∈Q

(14.9)

Y

Claim 14.17 w(HY ) ≤ w(H ) −  H (Y ). Proof. Let e be an arbitrary edge intersecting Q Y . We now consider the following cases. Case 14.17.1. The edge e is a 2-edge and |e ∩ Q Y | = 1. In this case,

the edge e contributes 3 to the weight w(H ), at most 21 to 21 |∂(Q Y )|, and 23 to q∈Q Y θ H (q),

200

14 Strong Tranversals in Linear Hypergraphs

and therefore at least 1 to the term w(H ) −  H (Y ). Since the edge e is deleted from H when constructing H  , it contributes 0 to w(H  ). Case 14.17.2. The edge e is a 2-edge and |e ∩ Q Y | =

2. In this case, the edge e contributes 3 to the weight w(H ), 0 to 21 |∂(Q Y )|, and 3 to q∈Q Y θ H (q) (it contributes 3 to two θ H (q)-values), implying that the edge e contributes 0 to the term w(H ) − 2  H (Y ). Since the edge e is deleted from H when constructing H  , it contributes 0 to w(H  ). Case 14.17.3. The edge e is a 3-edge and |e ∩ Q Y | = 1. In this case, the edge e is deleted from H and a new edge e = e \ {v} is added when constructing HY , where e ∩ Q Y = {v}. This implies that the edge e contributes 6 to the weight w(H ), at most 1 to 21 |∂(Q Y )|, and 2 to q∈Q Y θ H (q), and therefore at least 3 to the term w(H ) −  H (Y ). Moreover, the edge e contributes 3 to the weight w(H  ). Case 14.17.4. The edge e is a 3-edge and |e ∩ Q Y | = 2. In this case,

the edge e contributes 6 to the weight w(H ), at most 21 to 21 |∂(Q Y )|, and 4 to q∈Q Y θ H (q), and therefore at least 23 to the term w(H ) −  H (Y ). Since the edge e is deleted from H when constructing H  , it contributes 0 to w(H  ). Case 14.17.5. The edge e is a 3-edge and |e ∩ Q Y | = 3. In this case, the edge e contributes 6 to the weight w(H ), 0 to 21 |∂(Q Y )|, and 6 to q∈Q Y θ H (q), and therefore 0 to the term w(H ) −  H (Y ). Since the edge e is deleted from H when constructing () H  , it contributes 0 to w(H  ). Let x be a vertex of H such that θ H (x) = θ (H ) and let H (Y ) be defined by H (Y ) =

 1 (θ H (q) − θ H (x)), |∂(Q Y )| + 2 q∈Q

(14.10)

Y

and so, by Equation (14.9) and noting that

q∈Q Y

θ H (x) = |Q Y | · θ H (x), we have

H (Y ) =  H (Y ) − |Q Y | · θ H (x).

(14.11)

We prove next the following claim. Claim 14.18 If |Q Y |(θ H (x) − 4k − 1) + ck |Y | + H (Y ) ≥ 0

(14.12)

holds for some nonempty set Y where no two vertices in Y are adjacent in H , then the desired result, namely, φk (H ) ≤ Φk (H ), holds. Proof. Suppose that Inequality (14.12) holds for some nonempty set Y where no two vertices in Y are adjacent in H . Let H  = HY where H  has order n H  . Applying the inductive hypothesis to each component of H  , we have φk (H  ) ≤ Φk (H  ), or, equivalently, ck τs (H  ) ≤ ak n H  + w(H  ). Furthermore every strong transversal in H  can be extended to a strong transversal in H by adding to it the set Q Y \ Y , implying that τs (H ) ≤ τs (H  ) + |Q Y | − |Y |. Thus by Claim 14.17 and noting that n H  = n H − |Q Y |, we have

14.5 The Convex Set S3

201

ck τs (H ) ≤ ck (τs (H  ) + |Q Y | − |Y |) ≤ ak n H  + w(H  ) + ck (|Q Y | − |Y |) ≤ ak (n H − |Q Y |) + (w(H ) −  H (Y )) + ck (|Q Y | − |Y |) = ak n H + w(H ) − (ak |Q Y | − ck |Q Y | + ck |Y | +  H (Y )) . Therefore it suffices for us to show that ak |Q Y | − ck |Q Y | + ck |Y | +  H (Y ) ≥ 0.

(14.13)

By Equation (14.11) and noting that ck − ak = 4k + 1, Inequality (14.13) is equivalent to the following: |Q Y |(−4k − 1) + ck |Y | + H (Y ) + |Q Y | · θ H (x) ≥ 0.

(14.14)

However, Inequality (14.14) is a restatement of Inequality (14.12), which by the () supposition in the statement of our claim is assumed to hold. Recall that x is chosen to be a vertex of H such that θ H (x) = θ (H ). We now consider the case when Y = {x}. Abusing notation we write Q x = Q {x} (= N H [x])  . and Hx = H{x} Claim 14.19 The following holds: 1 |Q x | ≤

+ θ (H ) if |Q x | is even 1 + θ (H ) if |Q x | is odd. 2

Proof. By our choice of the vertex x and definition of θ H (x), we have |Q x | ≤ 1 + 2d3 (x) + d2 (x) = 1 + θ H (x) − 21 d2 (x) = 1 + θ (H ) − 21 d2 (x). Since d2 (x) ≥ 0, we have |Q x | ≤ 1 + θ (H ). If, however, |Q x | is even, then d2 (x) ≥ 1 or two distinct edges that contain the vertex x intersect in at least two vertices, that is, x is incident with two overlapping edges. Further if d2 (x) ≥ 1, then |Q x | ≤ 21 + θ (H ), while if x is incident with two overlapping edges, then |Q x | ≤ 2d3 (x) + d2 (x), implying that |Q x | ≤ θ (H ) − 21 d2 (x) ≤ θ (H ). () By Equation (14.10) and noting that here Y = {x}, we have H ({x}) ≥ 1 |∂(Q x )| ≥ 0. We prove next the following claim. 2

202

14 Strong Tranversals in Linear Hypergraphs

Claim 14.20 We may assume that the following three conditions all hold, for otherwise the desired result holds. (a) |Q x | = 2k + 1. (b) θ H (x) = |Q x | − 1 = 2k. (c) H ({x}) < 4. Proof. Let r = |Q x | − 2k − 1 and let p = θ H (x) − |Q x | + 1 and note that r is an integer and p ≥ 0 by Claim 14.19. Furthermore p ≥ 21 if |Q x | is even which is equivalent to r being odd. By Claim 14.18 we would be done if the following holds: 0 ≤ |Q x |(( p + |Q x | − 1) − 4k − 1) + ck |{x}| + H ({x}) = (2k + 1 + r )( p + 2k + 1 + r − 4k − 2) + ck + H ({x}) = (2k + 1 + r )(−(2k + 1) + p + r ) + (2k − 1)(2k + 3) + H ({x}) = −(2k + 1)2 + r (−(2k + 1) + p + r ) + (2k + 1)( p + r ) +(2k − 1)(2k + 3) + H ({x}) = −(4k 2 + 4k + 1) + r ( p + r ) + (2k + 1) p + (4k 2 + 4k − 3) + H ({x}) = −4 + r 2 + p(r + 2k + 1) + H ({x}). Let ξ(x) = −4 + r 2 + p(r + 2k + 1) + H ({x}). As shown above, it suffices for us to show that ξ(x) ≥ 0. As observed earlier, p ≥ 0 and H ({x}) ≥ 0. Further, r + 2k + 1 = |Q x | ≥ 1, and so p(r + 2k + 1) ≥ 0. Hence, ξ(x) = −4 + r 2 + p(r + 2k + 1) + H ({x}) ≥ −4 + r 2 . Thus if r ≤ 2 or if r ≥ 2, then ξ(x) ≥ 0, and we are done. Hence we may assume that r ∈ {−1, 0, 1}. If r ∈ {−1, 1}, then |Q x | is even, and so by Claim 14.19, |Q x | ≤ 21 + θ (H ), implying that p = θ H (x) − |Q x | + 1 ≥ 21 . Hence since in this case, r 2 = 1 and p(r + 2k + 1) ≥ 21 (2k) = k ≥ 5, we have ξ(x) ≥ −4 + r 2 + p(r + 2k + 1) ≥ −4 + 1 + 5 ≥ 2. We may therefore assume that r = 0. Thus, |Q| = 2k + 1. If p > 0, then p ≥ 21 and p(r + 2k + 1) > 5, so we have ξ(x) ≥ −4 + r 2 + p(r + 2k + 1) > −4 + 0 + 5 = 1. Hence we may assume that p = 0, for otherwise we are done. Thus, θ H (x) = |Q x | − 1 = 2k. By our earlier assumption, we have r = 0 and p = 0. Thus in this case, ξ(x) = −4 + H ({x}). If H ({x}) ≥ 4, then we are also done, so we must have H ({x}) < 4. Hence, the three conditions (a), (b), and (c) in the statement of the claim must () hold, for otherwise ξ(x) ≥ 0 and the desired result follows. Claim 14.21 Every two vertices in Q x are adjacent. Proof. Suppose, to the contrary, that y1 , y2 ∈ Q x are nonadjacent and let Y = {y1 , y2 }. By Claim 14.18, we may assume that |Q Y |(θ H (x) − 4k − 1) + ck |Y | + H (Y ) < 0,

(14.15)

14.5 The Convex Set S3

203

for otherwise the desired result follows and we would be done. By Claim 14.20(b) we note that θ H (x) = 2k. Thus recalling that ck = (2k − 1)(2k + 3), Inequality (14.15) simplifies to the following inequality: (2k − 1)(2k + 3)|Y | − |Q Y |(2k + 1) + H (Y ) < 0.

(14.16)

By Claim 14.20(a), we have |Q x | = 2k + 1. By Claim 14.20(c), we have 4 > H ({x}) ≥ 21 |∂(Q x )|, implying that |∂(Q x )| ≤ 7. We note that Q Y ⊆ Q x ∪ ∂(Q x ), and therefore we have |Q Y | ≤ |Q x | + |∂(Q x )| ≤ (2k + 1) + 7 = 2k + 8. Since |Y | = 2, Inequality (14.16) is therefore equivalent to the following inequality: 2(2k − 1)(2k + 3) + H (Y ) < |Q Y |(2k + 1) ≤ (2k + 8)(2k + 1).

(14.17)

Since 2(2k − 1)(2k + 3) = 8k 2 + 8k − 6 and (2k + 8)(2k + 1) = 4k 2 + 18k + 8, Inequality (14.17) is in turn equivalent to the following inequality: 4k 2 − 10k − 14 + H (Y ) < 0.

(14.18)

However, H (Y ) ≥ 21 |∂(Q Y )| ≥ 0 and by supposition, k ≥ 5, implying that 4k 2 − () 10k − 14 + H (Y ) > 0, a contradiction. We now return to the proof of Theorem 14.15. By Claim 14.16 we may assume that H is linear. By Claim 14.20 we have |Q x | = 2k + 1, θ H (x) = 2k and H ({x}) < 4. By Claim 14.21 every two vertices in Q x are adjacent in H . Therefore for every pair of vertices u, v ∈ Q x there exists a unique edge in H containing u and v. We denote this edge by e(u, v). Recall that a 2-section graph, (F)2 , of a hypergraph F is defined as a graph (F)2 with the same vertex set as F and in which two vertices are adjacent in the graph (F)2 if and only if they belong to a common edge in F. In our case, the subgraph of the 2-section graph (H )2 induced by the set Q x is a complete graph, which we denote by G x . In particular, V (G x ) = Q x . Since |Q x | = 2k + 1 and k ≡ 2 (mod 3), this implies that |Q k | ≡ 5 (mod 6). Thus, every 3-edge of H which lies entirely within Q x can be thought of as a 3cycle in the complete graph G x . Therefore, by Theorem 14.14 we must have at least four pairs of vertices {u i , vi } where i ∈ [] ( ≥ 4) such that e(u i , vi ) is not a 3-edge of H that lies completely within Q x . Therefore, e(u i , vi ) is either a 2-edge or a 3-edge containing one vertex outside Q x for each i ∈ []. Let Rx =

  {u i , vi }. i=1

Claim 14.22 If z ∈ Rx , then θ H (z) ≥ 2k + 1. Proof. Since H is a linear hypergraph, |Q x | = 2k + 1 is odd and every two vertices in Q x are adjacent in H , we note that every vertex y ∈ Q x lies in an even number, say 2n y , of pairs {u i , vi } where i ∈ []. Let z ∈ Rx be arbitrary and note that z belongs

204

14 Strong Tranversals in Linear Hypergraphs

to 2n z pairs {u i , vi } (i ∈ []) and (k − n z ) 3-edges that connect z to the 2k − 2n z vertices of Q x not contained in these 2n z pairs. Thus, θ H (z) ≥ 2(k − n z ) + 2n z × 3 = 2k + n z ≥ 2k + 1. () 2 As  ≥ 4, we note that there are at least four vertices in Rx . By Equation (14.10) and Claim 14.22, we therefore have that H ({x}) =

 1 |∂(Q x )| + (θ H (q) − θ H (x)) 2 q∈Q x



1 (θ H (z) − θ H (x)) ·0+ 2 z∈Rx  ((2k + 1) − 2k) ≥

≥

z∈Rx

= |Rx | ≥ 4, contradicting Claim 14.20(c). This completes the main part of the proof of Theorem 14.15. In order to prove the last part of the theorem, assume that the hypergraph H is 3-uniform and linear and tight for Theorem 14.15. We first note that in Claim 14.18 we must have equality in Inequality (14.12). We note next that in Claim 14.20 H can only be tight if ξ(x) = 0, which implies that r ∈ {−2, 0, 2}. If r ∈ {−2, 2}, then we must have p = 0 and H ({x}) = 0, which implies that V (H ) = Q x and θ H (u) = |Q x | − 1 for all u ∈ V (H ). This implies that if H is 3-uniform and linear and tight for Theorem 14.15, then every two vertices of H belong to a common edge and H is either (k − 1)-regular of order 2k − 1 or H is (k + 1)-regular of order 2k + 3, as desired. We therefore only need to consider the case when r = 0. In this case we note that H ({x}) = 4, and the proof of Claim 14.21 still holds (the final equation would be 4k 2 − 12k − 15 > 0, which is true). However, since H is 3-uniform and linear, this implies that the  edges e(u i , vi ) where i ∈ [] are 3-edges with exactly two vertices in Q x . However in Case 4 of Claim 14.17 we note that these  edges each contribute at least 3/2 to the term w(H ) −  H (Y ) but zero to w(H  ). Therefore, H cannot be tight for Theorem 14.15, when r = 0 in Claim 14.20. We have now shown that if H is 3-uniform and linear and is tight for Theorem 14.15, then it has the desired property. It is easy to see that if it has the desired property, then it is tight. This completes the proof of Theorem 14.15.  In 1847 Kirkman [68] proved that a complete graph K n on n ≥ 3 vertices has a triangle decomposition if and only if n (mod 6) ∈ {1, 3} (that is, n is 1 or 3 modulo 6); equivalently, there is a Steiner Triple System on n ≥ 3 vertices if n (mod 6) ∈ {1, 3}, where we recall that a Steiner triple system STS(n) of order n is a 3-uniform linear

14.5 The Convex Set S3

205

Fig. 14.2 The affine plane AG(2, 3)

hypergraph with the property that every pair of vertices is contained in a unique edge (or triple) as defined in Chapter 12. This result also follows from Theorem 14.14. If n ≡ 1 (mod 6), then let n = 2k + 1 and note that the 3-uniform linear hypergraph H associated with the Steiner triple system STS(n) is k-regular where k ≡ 0 (mod 3). If n ≡ 3 (mod 6), then the 3-uniform linear hypergraph H associated with the Steiner triple system STS(n) is k-regular where n = 2k + 1 and k ≡ 1 (mod 3). Thus, we have the following observation. Observation 14.23 For every integer k ≥ 1 where k (mod 3) ∈ {0, 1}, there exists a 3-uniform k-regular linear hypergraph H on 2k + 1 vertices, such that every two vertices belong to a common edge. Further, all such hypergraphs H are precisely the (3-uniform linear) hypergraphs associated with the Steiner triple system STS(n) where n ≥ 3 and n (mod 6) ∈ {1, 3}. Recall from Chapter 3 that the affine plane AG(2,3) of dimension 2 and order 3, illustrated in Figure 14.2, is equivalent to a linear 3-uniform 4-regular hypergraph F9 on nine vertices, where the lines of AG(2, 3) correspond to the 3-edges of F9 . As a consequence of Theorem 14.8 and Observation 14.23, we have the following upper bound on the strong transversal number. Theorem 14.24 ([65]) If H is a connected 3-uniform linear hypergraph, then τs (H ) ≤

16 2 nH + mH 21 21

with equality if and only if H is the Fano plane or H is the affine plane AF(2, 3). 2 Proof. The upper bound τs (H ) ≤ 16 n + 21 m H is a special case of Theorem 14.8 21 H when k = 3. Suppose that we have equality in this bound. Then by Theorem 14.8, the hypergraph H ∈ L3 , and so H is a 3-uniform linear hypergraph. Further, H is either 3-regular of order 7 or 4-regular of 9. In both cases, every two vertices of H belong to a common edge. Hence, H is either a Steiner triple system STS(7) of order 7 or

206

14 Strong Tranversals in Linear Hypergraphs

a Steiner triple system STS(9) of order 9. It is well known (see, for example, [26], p. 60) that STS(7) is unique and is the Fano plane, and that STS(9) is unique and is the affine plane AF(2, 3) of order 3.  We are now in a position to determine all extreme points in the convex set S3 . We define the set EP3 , EP3 , and EP as follows. Let EP3

 =

4k(k + 1) 6 , (2k + 1)(2k + 3) (2k + 1)(2k + 3)

 



k ≥ 3 and k ≡ 0 (mod 3)

and EP3

 =

Further let

6 4(k − 1)(k + 1) , (2k − 1)(2k + 3) (2k − 1)(2k + 3)

 



k ≥ 5 and k ≡ 2 (mod 3) .

  4 2 EP3 = (0, 2), (1, 0), , ∪ EP3 ∪ EP3 . 7 7

We will show that the (infinite) set of extreme points of the convex S3 is precisely the set EP3 . For this purpose, we define ai and bi , such that EP3 = {(a1 , b1 ), (a2 , b2 ), (a3 , b3 ), . . .} and such that a1 < a2 < a3 < · · · . The first six pairs in EP3 are given below: (a1 , b1 ) = (0, 2)   (a4 , b5 ) = 32 , 2 39 39

(a2 , b2 ) = (a5 , b5 ) =

4



7

, 27

65

2 , 65

 56



(a3 , b3 ) = (a6 , b6 ) =

 16 21

2 , 21

95

2 , 95

 84

 

We will show that the convex set S3 is in fact the gray area depicted in Figure 14.3, where the line passing through S3 is the bound we found in Theorem 14.2 (with k = 3) using trivial probabilistic bounds. If we zoom in to the lower part of the graph we get the illustration of the convex set S3 depicted in Figure 14.4. Recall that a point (α, β) of nonnegative integers α and β is tight in the convex set S3 if the point (α, β) belongs to the set S3 and τs (H ) = αn H + βm H holds for some 3-uniform hypergraph H . In order to prove that EP3 contains all extreme points for S3 we need the following lemma which is obtained from Lemma 14.1 by letting r = 3. Lemma 14.2 If (ai , bi ) and (ai+1 , bi+1 ) are both tight for some 3-uniform hypergraph F, then all points ε(ai , bi ) + (1 − ε)(ai+1 , bi+1 ) where 0 ≤ ε ≤ 1 are also tight for F. We are now in a position to prove that the set EP3 determines all extreme points for S3 .

14.5 The Convex Set S3

207

(0, 2) 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4



0.2

0.20

4 2 7, 7





0.40

16 , 2 21 21



0.80

0.60

1.00

Fig. 14.3 The convex set S3 

32 2 39 , 39



0.050 0.045 0.040 0.035 0.030



0.025

56 , 2 65 65





84 2 95 , 95



0.020 0.015



0.010

120 , 2 133 133



0.005

0.84

0.86

0.88

0.90

0.92

0.94

0.96

0.98

1.00

Fig. 14.4 The lower part of the convex set S3

Theorem 14.25 ([65]) The set EP3 is the set of extreme points of the convex set S3 .   Proof. The points (a1 , b1 ) = (0, 2) and (a2 , b2 ) = 47 , 27 are both tight for E 3 due to  4 2  Theorem 14.13. The points (a2 , b2 ) = 7 , 7 and (a3 , b3 ) = 16 , 2 are both tight 21 21 for the Fano plane due to Theorems 14.13 and 14.8 (when k = 3). By Observation 14.23 there exists a 3-uniform k-regular hypergraph Hk on 2k + 1 vertices, such that every two vertices belong to a common edge whenever k ≡ 0 (mod 3) or k ≡ 1 (mod 3). We note that (a3 , b3 ) and (a4 , b4 ) are both tight for

208

14 Strong Tranversals in Linear Hypergraphs

H4 defined above. The points (a4 , b4 ) and (a5 , b5 ) are both tight for H6 . Continuing this process we note that (a2r −1 , b2r −1 ) and (a2r , b2r ) are tight for H3r −2 and (a2r , b2r ) and (a2r +1 , b2r +1 ) are tight for H3r , for all r ≥ 2. This completes the proof due to Lemma 14.2.  By Observation 14.23 and our earlier results, including Theorems 14.13, 14.24 and 14.25, we note that the first three extreme points of S3 are tight for the following hypergraphs. Observation 14.26 With the extreme points of the convex set S3 ordered by the first coordinate, the first extreme points of S3 are the points (a1 , b1 ) = (0, 2), (a2 , b2 ) = 16 2 , 21 ). Further the following holds. ( 47 , 27 ) and (a3 , b3 ) = ( 21 (a) The extreme point (0, 2) is tight only for the hypergraph E 3 . (b) The extreme point ( 47 , 27 ) is tight only for the hypergraphs E 3 , H5 and the Fano plane P G(2, 2). , 2 ) is tight only for the Fano plane P G(2, 2) and the (c) The extreme point ( 16 21 21 affine plane AF(2, 3). By Theorems 14.8, 14.15, and 14.25, we have proven in this section that the set EP3 determines all extreme points for the convex set S3 , and shown that the linear hypergraphs that are tight for these extreme points correspond exactly with the Steiner triple systems STS(n) (where n ≥ 3 and n (mod 6) ∈ {1, 3}). We note that the last extreme point, namely, (1, 0), which is an accumulation point of S3 , is not actually tight for any hypergraph as picking all vertices of a hypergraph except one is a strong transversal. However all other extreme points for S3 are tight for some Steiner triple systems. We state our main result formally as follows. Theorem 14.27 ([65]) The set EP3 is the (infinite) set of extreme points of the convex set S3 . Furthermore, the linear hypergraphs that are tight for these extreme points correspond exactly with the Steiner triple systems STS(n).

14.5.1 Applications to Strong Independence Let I3 be the set of all pairs (γ , λ) of nonnegative integers γ and λ for which the inequality αs (H ) ≥ γ n H − λm H holds for all 3-uniform hypergraphs H . As a consequence of our main result summarized in Theorem 14.27, we find all extreme points for the convex set I3 noting that for a 3-uniform hypergraph H , the inequality τs (H ) ≤ αn H + βm H holds if and only if the inequality αs (H ) ≥ (1 − α)n H − βm H holds.

14.5 The Convex Set S3

209

14.5.2 Applications to Double Total Domination Double transversals in hypergraphs can be used, for example, to obtain results on double total domination in graphs, where a double total dominating set (also called a 2-tuple total dominating set in the literature), abbreviated DTD-set, in a graph G is a set S of vertices such that every vertex of G is adjacent to at least two vertices in S and the double total domination number γ×2,t (G) is the minimum cardinality of a DTDset in G. There is a natural interplay between double transversals in hypergraphs and double total domination in graphs, as first observed in [55]. Observation 14.28 ([55]) If G is a graph with δ(G) ≥ 2, then γ×2,t (G) = τ2 (H ), where H is the open neighborhood hypergraph of G. As a consequence of Observation 14.28 and the observation that the open neighborhood hypergraph H of a graph G has size m H equal to its order n H , the inequality τ2 (H ) ≤ ak n H + bk m H implies that if G is a graph of order n with δ(G) ≥ k, then γ×2,t (G) ≤ (ak + bk )n. Thus in some applications of double transversals we wish to choose ak and bk so that their sum is as small as possible. This is the same approach used for total domination discussed in Section 9.5. We therefore wish to minimize ak + bk and to decide which values of ak and bk to use. Note that if G is a graph with δ(G) ≥ 3 then the open neighborhood hypergraph, H , of G has edge sizes at least three. We can make H 3-uniform by shrinking the edges of H and as a consequence of Theorem 14.13 and Observation 14.28, we have the following upper bound on the double transversal number of a graph, where the Heawood graph is shown in Figure 14.5. Corollary 14.1 ([55]) If G is a connected graph of order n with δ(G) ≥ 3, then γ×2,t (G) ≤ 67 n with equality if and only if G is the Heawood graph.

Fig. 14.5 The Heawood graph

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14 Strong Tranversals in Linear Hypergraphs

14.6 Double Transversals in Finite Projective Planes Very interesting work on double transversals in a finite projective plane is presented by Bacsó, Héger and Sz˝onyi [7]. As remarked earlier, a double transversal is also called a twofold blocking set or double blocking set in the literature. The authors in [7] establish a close relation between double transversals and the upper chromatic number of C -hypergraphs, where a C -hypergraph H = (X, C ) has an underlying vertex set X and set system C over X . A proper vertex coloring of H is defined in [7] as a mapping ϕ from X to a set of colors, such that each C -edge C ∈ C has at least two vertices with a common color. A strict k-coloring is proper vertex coloring ϕ : X → [k] that uses each of the k colors on at least one vertex. The upper chromatic number of H , denoted by χ(H ), is the largest k admitting a strict k-coloring. Let Πq denote an arbitrary finite projective plane of order q, and let P G(2, q) denote the projective plane of order q over the finite field G F(q) of q elements. As remarked earlier, a projective plane may be considered as a hypergraph, whose vertex set is the set of points and whose edge set is the set of lines of the plane. Considering every line as C -edge in the associated hypergraph, our earlier definitions imply that a coloring of the points of a finite projective plane Π is proper if every line contains at least two points of the same color, and the upper chromatic number χ (Π ) is the maximum number of colors in a proper coloring. Bacsó and Tuza [8] gave the following general upper bound on the upper chromatic number for any projective plane, as a function of the order. Theorem 14.29 ([8]) As q → ∞, every projective plane Πq of order q satisfies χ (Πq ) ≤ q 2 − q −

1√ √ q + o( q). 2

If T is a double transversal in Πq , then one can obtain a proper coloring of Πq by coloring all the points in T with the same color, and then assigning a different color to each of the remaining points that do not belong to T . The resulting proper coloring uses n − |T | + 1 = n − τ2 (Πq ) + 1 colors, where n = q 2 + q + 1 denotes the number of points in Πq . Thus such a coloring of Πq contains a monochromatic double transversal of size τ2 (Πq ), and every other color class consists of a single point. We call such a coloring of Πq a trivial coloring. Hence, we have the following trivial lower bound on the upper chromatic number of Πq . Observation 14.30 For every projective plane Πq of order q on n points, we have χ(Πq ) ≥ n − τ2 (Πq ) + 1. Bacsó, Héger and Sz˝onyi [7] establish the following upper bound on the double transversal number of a projective plane.

14.6 Double Transversals in Finite Projective Planes

211

Theorem 14.31 ([7]) If h ≥ 3 is an odd integer, α ≥ 1 an integer, p an odd prime, r = p α , and q = r h , then there exist two disjoint transversals of size q + (q − 1)/(r − 1) in P G(2, q), implying that   q −1 . τ2 (P G(2, q) ≤ 2 q + r −1 Ball and Blokhuis [10] established the following upper bound on the double transversal number of a projective plane P G(2, q) when q is a square. Theorem 14.32 ([10]) If q is a square, then   √ τ2 (P G(2, q) ≤ 2 q + q + 1 . As a consequence of Theorems 14.31 and 14.32, we have the following result of Bacsó, Héger and Sz˝onyi [7]. Theorem 14.33 ([7]) If r denotes the order of the largest proper subfield of G F(q) where q is odd and not a prime, then   q −1 τ2 (P G(2, q) ≤ 2 q + . r −1 Recall that by Observation 14.30, every projective plane Πq of order q on n points satisfies χ (Πq ) ≥ n − τ2 (Πq ) + 1. Using the above results on the double transversal number, Bacsó, Héger and Sz˝onyi [7] prove that equality holds in this lower bound if q and p are large enough, where q = p h and p is prime. We state their result formally as follows. Theorem 14.34 ([7]) Let n = q 2 + q + 1 where q = p h and p prime. If q > 256 is a square or if p ≥ 29 and h ≥ 3 is odd, then χ(Πq ) = n − τ2 (Πq ) + 1, and equality is achieved only by trivial proper colorings.

Chapter 15

Conjectures and Open Problems

15.1 Introduction In this chapter, we list some conjectures and open problems that have yet to be settled or solved.

15.2 The Tuza Constant c5 As shown in Chapter 4, c2 = 13 and c3 = 41 , while in Chapter 5, we show that c4 = 29 . As remarked by Alon [2], “it would be extremely interesting to determine precisely the value of ck for every k. The considerable effort made in [74] to show that c4 = 29 suggests that this may be difficult.” Indeed, the precise value of ck has yet to be determined for any values of k with k ≥ 5. If we take H to be the generalize triangle T5 defined in Chapter 5 (and illustrated in Figure 5.1(b)), then H has order n H = 8, size m H = 3 and transversal number 2 2 (n H + m H ), implying that c5 ≥ 11 . τ (H ) = 2 = 11 Using a construction due to Thomassé and Yeo [85] we can improve this lower 2 3 and show that c5 ≥ 16 . Thomassé and Yeo construct a 5-uniform bound of 11 hypergraph which we call R as follows. Let R be the hypergraph with vertex set V (R) = {0, 1, . . . , 10} and edge set E(R) = {e0 , e1 , . . . , e10 }, where the hyperedge ei = Q + i for i = 0, 1, . . . , 10 and where Q = {1, 3, 4, 5, 9} is the set of nonzero quadratic residues modulo 11. The hypergraph R is a 5-uniform, 5-regular hypergraph of order n R = 11 and size m R = 11 and with transversal number τ (R) = 4. Let H be obtained from the hypergraph R by deleting an arbitrary vertex of R and its five incident edges. We illustrate a drawing of the resulting hypergraph H in Figure 15.1, where in this diagram V (H ) = [10] and E(H ) = {{1, 3, 4, 5, 9}, {2, 4, 5, 6, 10}, {4, 6, 7, 8, 1},

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9_15

213

214

15 Conjectures and Open Problems

4

1

4

3

6

1

3

6

10

10 9

8 7

9

8 7

5

5

2

2

(a) Hyperedges not containing vertex 10.

(b) Hyperedges containing vertex 10.

Fig. 15.1 A 5-uniform 3-regular hypergraph, H , with τ (H ) = 3 =

3 16 (m H

+ nH )

{5, 7, 8, 9, 2}, {6, 8, 9, 10, 3}, {10, 1, 2, 3, 7}}. The hypergraph H is a 5-uniform, 3-regular hypergraph of order n H = 10 and size m H = 6. The three hyperedges of H not containing the vertex named 10 are drawn in Figure 15.1(a), and the remaining three hyperedges of H that contain the vertex named 10 are drawn in Figure 15.1(b). We note that the set {1, 2, 10}, for example, is a transversal of H and that no two vertices of H will cover all hyperedges of H . Thus, H has transversal number τ (H ) = 3. Therefore, τ (H ) = 3 =

3 3 (10 + 6) = (n + m H ). 16 16 H

The existence of the above hypergraph H yields the following lower bound on c5 . Observation 15.1 c5 ≥

3 . 16

3 We conjecture that the lower bound of 16 on the Tuza constant c5 is also an upper bound. That is, we pose the following conjecture.

Conjecture 15.1 c5 =

3 . 16

15.3 The Tuza Constant q5 As shown in Chapter 4, q2 = 13 and q3 = 14 . Moreover as shown in Chapter 9, c4 = 15 . 1 Thus for small k ∈ {2, 3, 4}, we know that qk = k+1 . Applying probabilistic arguments, we show in this paper that the asymptotic behavior of qk as k grows is the same as that of ck , namely, of the order ln(k)/k.

15.3 The Tuza Constant q5

215

It would be extremely interesting to determine precisely the value of qk for every k. It remains an open problem, even for k = 5, to determine the precise value of qk for k ≥ 5. The amount of work to show that q4 = 15 suggests that this may be difficult. We pose the following conjecture on the value of the Tuza constant q5 . Conjecture 15.2 q5 = 16 .

15.4 The Value kmin 1 As mentioned earlier, for small k ∈ {2, 3, 4}, we know that qk = k+1 , while the asymptotic behavior of qk as k grows is of the order ln(k)/k. It would be interesting 1 holds. Recently the to determine the smallest value, kmin , of k for which qk > k+1 1 authors have shown that when k = 60 we have qk > k+1 . Therefore, 5 ≤ kmin ≤ 60. However, it remains an open problem to determine the exact value of kmin . Conjecture 15.2 can be restated as follows.

Conjecture 15.3 kmin ≥ 6.

15.5 The Values u k and Uk for Large k As shown in Chapters 11 and 13, for k ∈ {2, 3}, we have u k = 1 and Uk = 0. We pose the following problem. Problem 15.2 Determine the value of u k for all k ≥ 4. In particular, is it true that u k = 1 for all k ≥ 4? Problem 15.3 Determine the value of Uk for all k ≥ 4. In particular, is it true that Uk = 0 for all k ≥ 4?

15.6 Partial Steiner Triple Systems As shown in Chapter 12, Eustis and Verstaëte [34] proved √ that there exists a partial Steiner triple system H of order n such that α(H )  3n log n as n → ∞. The following conjecture was posed by Eustis and Verstaëte [34] in terms of partial Steiner (n, r, )-systems for integers r and  with r ≥ 2 − 1 ≥ 3. We state their conjecture in the special case when r = 3 and  = 2.

216

15 Conjectures and Open Problems

Conjecture 15.4 ([34]) For every partial Steiner √ triple system on n vertices, the independence number is at least asymptotic to 3n log n as n → ∞. The current best lower bounds are due to Kostochka, Mubayi and Verstaëte [72] who proved that for a partial Steiner triple system H of order n, we have α(H )  √ 0.458 n log n as n → ∞. By Observation 1.1, the above Conjecture 15.4 can be restated in terms of transversals as follows: For every partial Steiner √ triple system on n vertices, the transversal number is at least asymptotic to n − 3n log n as n → ∞.

15.7 Strong Transversals For k ≥ 2, the set Sk is defined in Chapter 14 to be the set of all pairs (ak , bk ) of nonnegative integers ak and bk for which inequality τs (H ) ≤ ak n H + bk m H holds. By Theorem 14.5, the set Sk is a convex set. In Theorem 14.7, the convex set S3 is fully described. Further, it is shown that the line segment between two consecutive extreme points of the convex set S3 corresponds to a Steiner triple system STS(n) where n ≥ 3 and n (mod 6) ∈ {1, 3}. However, for all k ≥ 4, it remains an open problem to determine all the extreme points of the convex set Sk . We pose this problem formally as follows. Problem 15.4 Determine the extreme points of the convex set S4 . More generally, determine the extreme points of the convex set Sk for all k ≥ 4.

15.8 Transversals of Triangles in Simple Graphs An intriguing conjecture of Tuza [87] from 1981 states that if a graph G (with no multiple edges or loops) does not contain more than k pairwise edge-disjoint triangles (subgraphs isomorphic to K 3 ), then there exists a set of at most 2k edges that meet all triangles of G. We note that the conjecture, if true, is achieved, for example, for the complete graphs of orders 4 and 5. Baron and Kahn [9] showed that Tuza’s conjecture is asymptotically tight for dense graphs. This conjecture due to Tuza can be posed in the language of transversals in linear hypergraphs noting that any two triangles of a graph share at most one edge. In order to state the equivalent hypergraph conjecture, we define a triangle hypergraph as a hypergraph H that can be obtained from a graph G by taking the edges of G as vertices of H , and the triplets of edges of the triangles of G as hyperedges in H . We note that a triangle hypergraph is a 3-uniform linear hypergraph. A set of edges in a hypergraph H is independent if the edges in the set are pairwise vertex-disjoint. The matching number, denoted α  (H ), of a hypergraph H is the maximum cardinality of an independent set of edges in H .

15.8 Transversals of Triangles in Simple Graphs

217

Conjecture 15.5 ([87]) If H is a hypergraph obtained from a graph G by taking the edges of G as vertices of H , and the triplets of edges of the triangles of G as hyperedges in H , then the resulting 3-uniform linear hypergraph H satisfies τ (H ) ≤ 2α  (H ). We observe that if M is a maximum matching in H (of size α  (H )), then the set of all vertices that belong to edges in M forms a transversal in H , and so trivially τ (H ) ≤ 3|M| = 3α  (H ). Conjecture 15.5 has attracted considerable attention since it was raised by Tuza some 38 years ago. Although the conjecture has been resolved for various classes of graphs G, the conjecture has yet to be solved in general. For example, Tuza [89] proved the conjecture for planar graphs and for chordal graphs, 7 2 n edges. Krivelevich [73] proved the and for graphs with n vertices and at least 16 conjecture for graphs H without homeomorphic copies of K 3,3 , thereby improving Tuza’s result that the conjecture holds for planar graphs. Tuza [89] proved that if G is a tripartite graph, that is, a graph whose vertex set can be partitioned into three disjoint independent sets, then the associated triangle hypergraph H of G satisfies τ (H ) ≤ 73 α  (H ). This bound was subsequently improved by Haxell and Kohayakawa [47] who showed that for tripartite graphs G, the associated triangle hypergraph H satisfies τ (H ) ≤ (2 − )α  (H ) where  > 0.044. The best general result to date is that of Haxell [48] who showed that a triangle α  (H ). Thus although many partial results exist, hypergraph H satisfies τ (H ) ≤ 66 23 Conjecture 15.5 in general remains open.

Glossary

AG(2, 2): AG(2, 3): AG(2, 4): AG(2, q): a(H ): ∂ H (S): c2 : c3 : c4 : c5 : ck : C(n, n − k, t): C(n, s, t): τ2 (H ): χ  (G): E(H ): E ∗H (X ): Ek : G: H2 : H3 : H4 : H: Hk :

Finite affine plane of order 2. 12, 19 Finite affine plane of order 3. 12, 24, 196 Finite affine plane of order 4. 133 Finite affine plane of order q ≥ 2 12 Maximum cardinality of an independent set in H . 2 The boundary of the set S. 1 The Tuza constant c2 . 18 The Tuza constant c3 . 20 The Tuza constant c4 . 27, 32 The Tuza constant c5 . 204 The Tuza constant ck . 17, 35, 203 The minimum number of edges in a k-uniform hypergraph on n vertices with transversal number t + 1. 42 The minimum number of s-sets in [n] needed to cover every t-set in [n]. 41 The double transversal number of H . 179 Cardinality of a minimum edge coloring in G. 3 The edge set of H . 1 Set of edges, not in X , but intersecting V (X ). 55 The k-uniform hypergraph on k vertices with exactly one edge. 18 The complement of a graph. 3 The class of all 2-uniform hypergraphs. 19 The class of all 3-uniform hypergraphs. 20 The class of all 4-uniform hypergraphs. 32, 33 The complement of a hypergraph. 3 The class of all k-uniform hypergraphs. 17

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9

219

220

H − X: H (X, Y ):

i(G): kmin : L2 : L3 : L4 : L4,3 : Hk : Lk,≥1 : Lk,Δ : τ2 -transversal of H : m(H ): α  (G): τs -transversal of H : τ -transversal of H : MX : n(H ): N H (S): N H [S]: N H (v): N H [v]: P G(2, q): Φ(H  ): q2 : q3 : q4 : q5 : qk : qk,Δ : τs (H ): γt (G): τ (H ): U2 :

Glossary

Hypergraph obtained from H by removing the vertices X from H , removing all edges that intersect X , and removing all resulting isolated vertices, if any 2 The hypergraph obtained by deleting all vertices in X ∪ Y from H and removing all edges containing vertices from X , removing the vertices in Y from any remaining edges, and removing all resulting isolated vertices, if any. 2, 28 The independent domination number of G. 172 1 The smallest value of k for which qk > k+1 holds. 205 The class of all 2-uniform linear hypergraphs. 20 The class of all 3-uniform linear hypergraphs. 25, 46 The class of all 4-uniform linear hypergraphs. 51, 131, 133– 135 The subclass of hypergraphs in L4 with maximum degree 3. 51, 56, 132, 136 The class of all k-uniform linear hypergraphs. 18, 42, 150 The class of hypergraphs in Lk with minimum degree at least 1. 171 The subclass of hypergraphs in Lk with maximum degree Δ. 18 Double transversal of H of minimum size. 179 The size of H . 1 Cardinality of a maximum matching in G. 3 Strong transversal of H of minimum size. 179 Transversal of H of minimum size. 2 Bipartite graph with partite sets X and E ∗ (X ). 70 The order of H . 1 The open neighborhood of the set S. 1 The closed neighborhood of the set S. 1 The open neighborhood of v. 1 The closed neighborhood of v. 1 Finite projective plane of order q ≥ 2. 11 Equal ξ(H  ) − ξ(H ). 58 The Tuza constant q2 . 18 The Tuza constant q3 . 20 The Tuza constant q4 . 27, 33, 51 The Tuza constant q5 . 204 The Tuza constant qk . 18, 204 The Tuza constant qk restricted to hypergraphs H ∈ Lk,Δ . 18 The strong transversal number of H . 179 The total domination number of G. 138 The transversal number of H . 2 The best possible constant U2 such that Υ (H ) ≥ U2 · n H for all H ∈ L2,≥1 . 172, 173

Glossary

u2: U3 : u3: U4 : u4: Uk : uk : Υ (H ): Υk (H ): V (H ):

221

The best possible constant u 2 such that τ (H ) ≤ u 2 · n H all H ∈ L2 . 160 The best possible constant U3 such that Υ (H ) ≥ U3 · n H all H ∈ L3,≥1 . 174 The best possible constant u 3 such that τ (H ) ≤ u 3 · n H all H ∈ L3 . 161 The best possible constant U4 such that Υ (H ) ≥ U4 · n H all H ∈ L4,≥1 . 174 The best possible constant u 4 such that τ (H ) ≤ u 4 · n H all H ∈ L4 . 161 The best possible constant Uk such that Υ (H ) ≥ Uk · n H all H ∈ Lk,≥1 . 171, 174, 205 The best possible constant u k such that τ (H ) ≤ u k · n H all H ∈ Lk 160, 161, 205 The upper transversal number of H . 171 The upper k-transversal number of H . 171, 174 The vertex set of H . 1

for for for for for for for

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Index

A Adjacent, 1

B Boundary, ∂ H (S), 2

C Cap set problem, 161 Chromatic index, 3 Closed neighborhood, N H [S], 2 Closed neighborhood, N H [v], 2 Colin McDiarmid, 21 Complement of a graph, G, 3 Complement of a hypergraph, H , 3 Component, 2 Connected, 2 Connected hypergraph, 2 Convex set, 189 S2 , 191 S3 , 192 Sk , 189 Covering number, C(n, s, t), 43

D Deficiency, 57 def(H ), 58 def H (X ), 58 Degree, d H (v), 1 Dion Gijswijt, 163 Double transversal, see transversal Double transversal number, τ2 (H ), 187 Douglas West, 43

E Edge coloring, 3

F Factor-critical, 7 Fano plane, 44, 140, 145 F-component, 2 Feng-Chu Lai, 32 Finite affine plane, 11 Finite projective plane, 11

G Generalized triangle T4 , 27, 35 T5 , 27 Tk , 27 Gerard Jennhwa Chang, 32 Gil Kalai, 148

H Hall’s Theorem, 58 Heawood graph, 145, 209 Hitting set, see transversal Hyperedge, 1 k-edge, 1 Hypergraph, 1 2-regular, 6, 9 2-uniform, 1, 18, 45 3-regular, 12 3-uniform, 20, 46 4-regular, 12 4-uniform, 27, 34, 53 Δ(H ) ≤ 2, 141 k-regular, 14

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 M. A. Henning and A. Yeo, Transversals in Linear Uniform Hypergraphs, Developments in Mathematics 63, https://doi.org/10.1007/978-3-030-46559-9

227

228 k-uniform, 1, 49 k-uniform k-regular, 148 duality, 5

I Independence number, 2 Independent dominating set, 180 Independent domination number, 180 Independent edges, 3 Independent set of vertices, 2 Intersecting hypergraph, 2 Isolated vertex, 1

J Jacques Verstaëte, 176 Jiri Matousek, 148 Jordan Ellenberg, 163

K Kevin T. Phelps, 173

L Latin squares, 12 Linear hypergraph, 2 Linear intersecting hypergraph, 5

M Matching, 3 Matching number, 3 Maximum degree, Δ(H ), 1 Maximum matching, 3 Minimum degree, δ(H ), 1

N Neighbor, 2 Neighborhood, N H (S), 2 Neighborhood, N H (v), 1 Noga Alon, 37

O Open Neighborhood Hypergraph (ONH), 145 Order, n(H ), 1 Overlapping edges, 2

Index P Partial Steiner, 171 Paul Erd˝os, 19 Perfect matching, 3 Projective plane, 147, 155

Q Quadrilateral-free, 146

S Size, m(H ), 1 Special hypergraphs, 53 H4 , H10 , H11 , 54 H14,1 , H14,2 , H14,3 , H14,4 , H14,5 , H14,6 , 54 H21,1 , H21,2 , H21,3 , H21,4 , H21,5 , H21,6 , 54 H14 , H21 , 53 Steiner system, 171, 215 Stephan Thomassé, 34 Strong transversal, see transversal Strong transversal number, τs (H ), 187 Strongly independent set, 2

T Total dominating set, 145 Total domination number, 145 Transversal, 2 number, τ (H ), 2 Tutte-Berge Formula, 58 Tuza constant c2 , 18, 20 c3 , 20, 22 c4 , 27, 33 c5 , 214 ck , 17, 37, 213 kmin , 215 q2 , 18, 20 q3 , 20, 22 q4 , 27, 33, 53 q5 , 215 qk , 18 U Upper k-transversal number, 183 Upper transversal number, 179 U2 , 179 U3 , 181 U4 , 182 Uk , 179, 215

Index V Vašek Chvátal, 21 Vertex cover, see transversal Vizing’s Theorem, 7 Vojtech Rödl, 173 W West bound, 43, 44, 46, 49, 50, 157

229 2-uniform, 45 3-uniform, 46 k-uniform, 49

Z Zsolt Tuza, 17