Combinatorial Nullstellensatz [1 ed.] 0367686945, 9780367686949

Table of contents :
Cover
Half Title
Title Page
Copyright Page
Dedication
Contents
Preface
Authors
Acknowledgements
1. Some Definitions and Notations
2. Combinatorial Nullstellensatz
2.1. Introduction
2.2. An application of CNS to additive number theory
2.3. Application of CNS to geometry
2.4. CNS and a subgraph problem
2.5. 0-1 vectors in a hyperplane
3. Alon–Tarsi Theorem and Its Applications
3.1. Alon-Tarsi theorem
3.2. Bipartite graphs and acyclic orientations
3.3. The Cartesian product of a path and an odd cycle
3.4. A solution to a problem of Erd}os
3.5. Bound for AT(G) in terms of degree
3.6. Planar graphs
3.7. Planar graph minus a matching
3.8. Discharging method
3.9. Hypergraph colouring
3.10. Paintability of graphs
4. Generalizations of CNS and Applications
4.1. Number of non-zero points
4.2. Multisets
4.3. Coefficient of a highest degree monomial
4.4. Calculation of NS(a)
4.5. Alon-Tarsi number of K 2*n and cycle powers
4.6. Alon-Tarsi numbers of toroidal grids
4.7. List colouring of line graphs
4.8. r-regular planar graphs
4.9. Complete graphs K p+1 for odd prime p
4.10. Jaeger's conjecture
5. Permanent and Vertex-edge Weighting
5.1. Permanent as the coefficient
5.2. Edge weighting and total weighting
5.3. Polynomial associated to total weighting
5.4. Permanent index
5.5. Trees with an even number of edges
5.6. Complete graphs
5.7. Every graph is (2,3)-choosable
Bibliography
Index

Citation preview

Combinatorial Nullstellensatz

Combinatorial Nullstellensatz With Applications to Graph Colouring

Xuding Zhu R. Balakrishnan

First edition published 2022 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2022 Xuding Zhu and R. Balakrishnan CRC Press is an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data ISBN: [978-0-367-68694-9] (hbk) ISBN: [978-1-003-13867-9] (ebk) ISBN: [978-0-367-68701-4] (pbk) Typeset in CMR10 by KnowledgeWorks Global Ltd.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page v — #1

The first author (XZ) dedicates this book to his wife Tong Zhang. The second author (RB) dedicates this book to Professor N. Ramanujam, former Head of the Department of Mathematics, Bharathidasan University.

“2020-12-10” — 2021/5/8 — 18:39 — page vii — #3

i

i

Contents

Preface

ix

Authors

xiii

Acknowledgements

xv

1 Some Definitions and Notations 2

3

Combinatorial Nullstellensatz 2.1 Introduction . . . . . . . . . . . . . . . . 2.2 An application of CNS to additive number 2.3 Application of CNS to geometry . . . . . 2.4 CNS and a subgraph problem . . . . . . 2.5 0–1 vectors in a hyperplane . . . . . . . .

1

. . .

7 7 9 11 12 13

Alon–Tarsi Theorem and Its Applications 3.1 Alon–Tarsi theorem . . . . . . . . . . . . . . . . . 3.2 Bipartite graphs and acyclic orientations . . . . . 3.3 The Cartesian product of a path and an odd cycle 3.4 A solution to a problem of Erd˝ os . . . . . . . . . . 3.5 Bound for AT (G) in terms of degree . . . . . . . . 3.6 Planar graphs . . . . . . . . . . . . . . . . . . . . 3.7 Planar graph minus a matching . . . . . . . . . . 3.8 Discharging method . . . . . . . . . . . . . . . . . 3.9 Hypergraph colouring . . . . . . . . . . . . . . . . 3.10 Paintability of graphs . . . . . . . . . . . . . . . .

17 17 26 27 30 35 37 41 47 56 61

. . . . theory . . . . . . . . . . . .

4 Generalizations of CNS and Applications 4.1 Number of non-zero points . . . . . . . . . . 4.2 Multisets . . . . . . . . . . . . . . . . . . . . 4.3 Coefficient of a highest degree monomial . . 4.4 Calculation of NS (a) . . . . . . . . . . . . . 4.5 Alon–Tarsi number of K2?n and cycle powers

. . . . .

. . . . .

.

. . . . .

69 69 71 75 77 80 vii

“2020-12-10” — 2021/5/8 — 18:39 — page viii — #4

i

i

viii

Contents 4.6 4.7 4.8 4.9 4.10

Alon–Tarsi numbers of toroidal grids . List colouring of line graphs . . . . . . r-regular planar graphs . . . . . . . . . Complete graphs Kp+1 for odd prime p Jaeger’s conjecture . . . . . . . . . . .

5 Permanent and Vertex-edge Weighting 5.1 Permanent as the coefficient . . . . . . 5.2 Edge weighting and total weighting . . 5.3 Polynomial associated to total weighting 5.4 Permanent index . . . . . . . . . . . . . 5.5 Trees with an even number of edges . . 5.6 Complete graphs . . . . . . . . . . . . . 5.7 Every graph is (2,3)-choosable . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

84 89 92 94 103

. . . . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

107 107 110 113 116 119 122 125

Bibliography

127

Index

133

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page ix — #5

Preface

Combinatorial Nullstellensatz (abbreviated as CNS), first introduced by Noga Alon in 1990, is a celebrated theorem in Algebra which has been used to tackle combinatorial problems in diverse areas of mathematics. Suppose f (x) ∈ F[x] is a polynomial with coefficients in a field F. asserts that if the coefficient of a highQCNS n est degree monomial i=1 xtii in the expansion of f (x) is nonzero, then any grid S1 × S2 × . . . × Sn with Si ⊆ F and |Si | = ti + 1, contains a nonzero point of f (x). To apply CNS to solve a combinatorial problem, we construct an appropriate polynomial f (x) so that the solutions of the combinatorial problem correspond to nonzero points of the polynomial. Then we show that the coefficient of a certain monomial of f (x) is non-zero and invoke CNS to conclude the existence of a solution to the combinatorial problem. Chapter 1 introduces some basic notations. Chapter 2 presents some examples of applications of CNS to show the varieties of problems where CNS is applied. The main part of this book is devoted to the application of CNS to graph colouring and labeling problems. For these applications, the step of constructing a polynomial for the combinatorial problems is easy. We concentrate on methods used to show that the coefficient of a certain monomial in the expansion of a polynomial is nonzero. In general, it is a difficult problem to determine if the coefficient of a monomial of a polymonial is nonzero or not. Nevertheless, some methods are developed for this purpose for some classes of special polynomials. One of the methods is the so called Alon–Tarsi orientation. An orientation D of a graph G is an Alon–Tarsi orientation if the number of even Eulerian subdigraphs of D differs from the number of odd Eulerian subdigraphs of D. For the polynomial fG (x) corresponding to proper colouring of a graph G, Alon–Tarsi’s Theorem asserts that the coefficient of a given monomial in the expansion of fG is nonzero if and only if a corresponding orientation D of G is Alon– Tarsi. So the task reduces to finding an appropriate orientation of a given graph. In general, it is difficult to determine if an orientaix

“2020-12-10” — 2021/5/8 — 18:39 — page x — #6

i

i

x

Preface

tion is Alon–Tarsi. Nevertheless, this method is successfully used to solve some interesting list colouring problems. In Chapter 3, we present a family of such results. For example, if a graph D is bipartite, then any orientation D is an Alon–Tarsi orientation, as D has no odd Eulerian subdigraph. A consequence of this observation is that bipartite planar graphs are 3-choosable. Two other easy observations are that if the total number of Eulerian subdigraphs of D is odd, then the number of even Eulerian subdigraphs is certainly different from the number of odd Eulerian subdigraphs. So D is an Alon–Tarsi orientation. If the total number of Eulerian subdigraphs of D is congruent to 2 modulo 4, and the number of even Eulerian subdigraphs of D is even, then again D is an Alon–Tarsi orientation. The former is used in showing that any grid graph is 3-choosable, and the latter is used to solve an open problem of Erd˝os that every 4-regular graph that decomposes into a Hamiltonian cycle and a family of triangles is 3-colourable. Induction is also frequently used in proving the existence of an appropriate Alon–Tarsi orientation. Chapter 3 presents a few such proofs that solve list colouring problems for planar graphs. In particular, this method is used to show that every planar graph G has a matching M such that G − M is 4-choosable, and currently there is no other proof of this result. The second method for showing the coefficient of a certain monomial is nonzero is to calculate the coefficient through an interpolation formula. Uwe Schauz and Michal Laso´ n independently proved an Q interpolation formula that expresses the coefficient of a n monomial i=1 xtii of f (x) as a linear combination of the evaluations of f (x) on an arbitrary grid S1 ×S2 ×. . .×Sn with |Si | = ti +1. Such an interpolation formula is a generalization of CNS, because if f (x) has no nonzero point in a grid S1 ×QS2 × . . . × Sn with n |Si | = ti + 1, then certainly the coefficient of i=1 xtii is zero. The interpolation formula turns out to be quite useful in solving some graph colouring problems. In Chapter 4, we present a result of Schauz which asserts that if p is a prime, then the line graph of Kp+1 is p-choosable. Also it is shown that every r-edge colourable planar graph is r-edge choosable and the line graphs of cycle powers are chromatic choosable. The former result was first proved by Ellingham and Goddyn and the latter result was first proved by Woodall, and we present new proofs of these results by using the interpolation formula. Some other generalizations of CNS are also discussed in this chapter.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page xi — #7

Preface

xi

The third method for calculating the coefficient of a certain monomial is by calculating the permanents of certain matrices. This method is particularly applied to the edge-weighting and total-weighting of graphs. The celebrated 1-2-3 conjecture asserts that if G is a connected graph on at least 3 vertices, then there is a labeling f : E(G) so that for any two adjacent P → {1, 2, 3} P vertices u and v, f (e) = 6 e∈E(u) e∈E(v) f (e). The analogous 1-2 conjecture asserts that for any graph G, there is a labeling f : V (G) that for any two adjacent vertices u P ∪ E(G) → {1, 2} so P and v, e∈E(u) f (e) + f (u) 6= e∈E(v) f (e) + f (v). The list version of the 1-2-3 conjecture asserts that if L is a list assignment with |L(e)| = 3 for e ∈ E(G), then there is a required labeling f with f (e) ∈ L(e). Analogously, we have the list version of the 1-2 conjecture. CNS is used in most of the studies of the list version of these two conjectures. The coefficients of the corresponding polynomial turn out to be the permanents of matrices formed by columns of AG and BG , where AG , BG are matrices associated to a graph G. In Chapter 5, we present some results concerning the list version of the 1-2-3 conjecture and the 1-2 conjecture obtained by using this method. In particular, this method is used to show that complete graphs and complete bipartite graphs are (1, 3)-choosable, trees with an even number of edges are (1, 2)-choosable, and every graph is (2, 3)-choosable. CNS is now a widely used tool in the study of many combinatorial problems in a variety of areas in mathematics. This little book originates from the lecture notes of the first author given at a graduate student course and lecture notes the second author given at a workshop, with inputs from the first author. It is not intended to be a comprehensive survey of the application of this tool. Instead, it is intended to impress upon its readers of the usefulness of CNS and inspire them to further study and use this tool in their research.

i

i

“2020-12-10” — 2021/5/13 — 9:36 — page xiii — #9

Authors

Xuding Zhu is currently a Professor of Mathematics and also the Director of the Center for Discrete Mathematics at Zhejiang Normal University, China. His fields of interest are: Combinatorics and Graph Colouring. He published more than 260 research papers. He has served/is serving on the editorial board of SIAM Journal on Discrete Mathematics, Journal of Graph Theory, European Journal of Combinatorics, Electronic Journal of Combinatorics, Discrete Mathematics, Contribution to Discrete Mathematics, Discussiones Mathematicae. Graph Theory, Bulletin of Academia Sinica, Indian Journal of Discrete Mathematics and Taiwanese Journal of Mathematics. R. Balakrishnan is currently an Adjunct Professor of Mathematics at Bharathidasan University, Triuchirappalli, India. His fields of interests are: Algebraic Combinatorics and Graph Colouring. He is an author of three other books, one in Graph Theory and the other two in Discrete Mathematics. He is also one of the founders of the Ramanujan Mathematical Society. In addition, he is currently an Editor-in-Chief of the Indian Journal of Discrete Mathematics.

xiii

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page xv — #11

Acknowledgements

The first author (XZ) thanks Daniel Cranston, Uwe Schauz and Tsai-Lien Wong for their careful reading and valuable comments on the first draft of the manuscript. The second author (RB) thanks the authorities of Bharathidasan University, Tiruchirappalli for their support and his colleagues in the mathematics department for their encouragement which proved vital to this venture.

xv

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 1 — #13

Chapter 1 Some Definitions and Notations

In this chapter, we present some basic definitions and notations. Some other definitions are given at the appropriate places of the book. Definition 1.0.1 A graph G consists of a vertex set V (G) and edge set E(G). A graph G is a finite graph if both V (G) and E(G) are finite sets; otherwise, G is an infinite graph. The edges of a graph G are unordered pairs e = {u, v} of vertices in V . If e = {u, v} is an edge of G, then e is incident at u and v. Sometimes we write e = uv instead of e = {u, v}. When the graph G is clear from the context, we simply write V and E in place of V (G) and E(G), respectively. An edge of the form {v, v} is called a loop at vertex v. Two edges of G having the same end vertices are called parallel edges or multiple edges of G. A graph without loops and without multiple edges is called a simple graph. A graph H is a subgraph of G if V (H) ⊆ V (G) and E(H) ⊆ E(G). A subgraph H of G is called a spanning subgraph of G if V (H) = V (G). Definition 1.0.2 The number of edges of G incident at a vertex u of G is called the degree of u in G, and is denoted by dG (u) or simply by d(u) when the underlying graph G is clear from the context. A loop at u contributes 2 to the degree of u. Definition 1.0.3 A walk W in a graph G is an alternating sequence of vertices and edges of the form v1 e1 v2 e2 . . . vi ei vi+1 . . . vp−1 ep−1 vp , where ei is the edge vi vi+1 , 1 ≤ i ≤ p − 1. A trail in G is a walk in G in which no edge is repeated. A path is a trail in which no vertex is repeated. Definition 1.0.4 A walk (resp. trail, path) is closed if its initial and terminal vertices coincide. A closed path is called a cycle. Definition 1.0.5 A graph G is connected if there is a path between any two distinct vertices of G; otherwise, G is disconnected. 1

“2020-12-10” — 2021/5/8 — 18:39 — page 2 — #14

i

i

2

Combinatorial Nullstellensatz

A component of a graph G is a maximal connected subgraph of G. So if G is connected, it has just one component. Definition 1.0.6 Given a graph G, its line graph L(G) is defined as follows: Each vertex of L(G) corresponds to an edge of G, and two vertices of L(G) are adjacent in L(G) if, and only if, the corresponding edges have a common endvertex
Pin G. Therefore d(v) |V (L(G))| = |E(G)|. Moreover, |E(L(G))| = = v∈V (G) 2 P 1 2 (dG (v)) − |E(G)|. 2 v∈V (G)

Definition 1.0.7 A digraph D consists of a vertex set V (D) and an arc set A(D). The arcs of D, that is, the elements of A(D), are ordered pairs (u, v) of vertices of D. The definitions of a directed path, a directed cycle etc., in a digraph D are all analogous to the definitions in the undirected case. We presume that the reader is familiar with the basics of groups, rings, integral domains, fields (both infinite and finite) and vector spaces. For more details about graphs, digraphs and fields, the reader may consult the books [7], [63] and [8]. Definition 1.0.8 A graph G is planar if there exists a drawing of G in the plane in which no two edges intersect at a point other than a vertex of G, where each edge is a Jordan arc (that is, a simple arc in the plane). Such a drawing of a planar graph is called a plane representation of G. In this case, we also say that G has been embedded in the plane. A plane graph is a planar graph that has already been embedded in the plane. Definition 1.0.9 Let P be the set of points of the plane minus the points of the plane graph G (that is, points belonging to the edges of G). Define p ∼ p0 in P if and only if there exists a Jordan arc from p to p0 in P . This relation is an equivalence relation Π on P. The equivalence classes of Π are called the faces of the plane graph G. Let G be a connected plane graph with n vertices, m edges and f faces. There is a nice classical relation connecting n, m and f due to Euler. Theorem 1.0.10 (Euler’s Formula) For a connected plane graph G with n vertices, m edges and f faces. n − m + f = 2.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 3 — #15

Some Definitions and Notations

3

Let G be a graph and S = {1, 2, . . . , k}. A k-colouring of G is a map c from V (G) to S; it is proper, if adjacent vertices of G are mapped to distinct elements of S by c, that is, uv ∈ E(G) ⇒ c(u) 6= c(v). The elements of S are called colours. The set of vertices which receive the same colour q is an independent subset Vq of V (G) and is called a colour class. Definition 1.0.11 The chromatic number of a graph G, denoted by χ(G), is the least positive integer k such that G has a proper vertex colouring using k colours. Equivalently, the chromatic number χ(G) of G is the minimum size of a partition of the vertex set V (G) of G into independent subsets of G. Note that in considering proper vertex colourings of graphs, it is enough to restrict to simple graphs. Definition 1.0.12 A graph G is bipartite if its vertex set V (G) can be partitioned into two independent subsets. Definition 1.0.13 A list assignment of a graph G is a mapping L which assigns to each vertex v a set of permissible colours. An Lcolouring of G is a colouring c of G such that c(v) ∈ L(v) for each vertex v. We say that G is L-colourable if there exists a proper L-colouring of G. Definition 1.0.14 A k-list assignment of G is a list assignment L of G with |L(v)| = k for every vertex v of G. A graph G of is k-choosable if G is L-colourable for each k-list assignment L. The choice number ch(G) of G is the minimum k for which G is k-choosable. If ch(G) = k, then by considering the list assignment L with L(v) = {1, 2, . . . , k} for every vertex v, we conclude that G is kcolourable. So ch(G) ≥ χ(G). However, the difference ch(G)−χ(G) can be arbitrarily large as the following theorem shows:
Theorem 1.0.15 ((Erd˝ os–Rubin–Taylor)[23]) If m = 2k−1 , k then the complete bipartite graph Km,m is not k-choosable. Proof. Let (X, Y ) be the bipartition of the bipartite graph G = Km,m . Let L be the list assignment that assigns the distinct ksubsets of the set {1, 2, 3, . . . , 2k−1} to the m vertices of X and also to the m vertices of Y . We shall show that G is not L-colourable.

“2020-12-10” — 2021/5/8 — 18:39 — page 4 — #16

i

i

4

Combinatorial Nullstellensatz

Assume to the contrary that f is a proper L-colouring of G. Then f uses at least k colours for vertices in X; for otherwise, there is a k-set S ⊂ {1, 2, . . . , 2k − 1} not used for vertices in X. But then, there is a vertex v ∈ X with L(v) = S, a contradiction. Similarly, f uses at least k colours for vertices in Y . But there are 2k − 1 colours in total. So one colour is used for a vertex in X and a vertex in Y , and f is not a proper colouring. Notation 1 N = {1, 2, . . . } stands for the set of natural numbers. Furthermore, N0 = N ∪ {0} and Nn0 is the set of vectors (t1 , t2 , . . . , tn ) of length n, where ti ∈ N0 for each i ∈ {1, 2, . . . , n}. Notation 2 A grid of dimension n is the Cartesian product of n non-empty finite sets, written as: S = S1 × S2 × · · · × Sn , where S is given in bold. The size ||S|| of S is defined to be the integer vector (|S1 | − 1, |S2 | − 1, . . . , |Sn | − 1). Definition 1.0.16 Suppose that F is a field and f (x) = fQ(x1 , x2 , . . . , xn ) ∈ F [x1 , x2 , . . . , xn ] is a polynomial. Qn If fti(x) = n ti t x , then f (x) is called a monomial. We write i=1 i i=1 xi = x , where t = (t1 , t2 , . . . , tn ) ∈ Nn0 . By an expansion of a polynomial f (x), we mean the expression of f (x) as a linear combination of monomials. For t ∈ Nn , we denote P by cf,t the coefficient of xt in n the expansion of f (x) and let |t| = i=1 ti . Definition 1.0.17 Let f (x) = f (x1 , x2 , . . . , xn ) ∈ F [x1 , x2 , . . . , xn ] be a polynomial, and a = (a1 , a2 , . . . , an ) ∈ F n . If f (a) = 0, then a is called a zero point of f . If f (a) 6= 0, then a is called a non-zero point of f. Notation 3 Rn stands for the n-dimensional Euclidean space (over R). A canonical basis (or standard basis) for Rn over R is the set of n vectors {e1 , e2 , . . . , en }, where e1 = (1, 0, . . . , 0), e2 = (0, 1, . . . , 0), . . . , en = (0, 0, . . . , 0, 1). For sets A and B, the set of points of A not in B is denoted by ˙ A \ B. The disjoint union of A and B is denoted by A∪B. For a graph G and a subset M of E(G), we denote by G−M the subgraph obtained by removing edges in M from G. We use boldface lower case letters to denote vectors whose coordinates are denoted by the same (but not boldface) lower case letter; for example, a = (a1 , a2 , . . . , an ) and x = (x1 , x2 , . . . , xn ), where the dimension n is usually clear from the context.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 5 — #17

Some Definitions and Notations

5

Definition 1.0.18 Let D be a digraph. For a vertex x, denote − by AD (x) (respectively, A+ D (x), AD (x)) the set of arcs incident at x (respectively, the set of out-going arcs or incoming arcs at x), − + − d+ D (x) = |AD (x)| (respectively, dD (x) = |AD (x)|) is the out-degree + (respectively, the in-degree) of x. Let ∆D = max{d+ D (x) : x ∈ − V (G)} (respectively, ∆− = max{d (x) : x ∈ V (G)}). We say that D D − D is Eulerian if d+ (x) = d (x) for each vertex x of D. D D Definition 1.0.19 Let D be a digraph. An Eulerian subdigraph of D is an Eulerian digraph H with V (H) = V (D) and A(H) ⊆ A(D). An Eulerian subdigraph H of D is even (respectively, odd) if the number of arcs of H is even (respectively, odd). Definition 1.0.20 We denote by E(D) the set of Eulerian subdigraphs of D. Let Ee (D) Eo (D)

:= {H ∈ E(D) : H is even}, := {H ∈ E(D) : H is odd}.

The difference of D is defined as diff(D) = |Ee (D)| − |Eo (D)|. Note that diff(D) can be negative. A digraph D is called Alon–Tarsi (shortened as AT) if diff(D) 6= 0, i.e., |Ee (D)| 6= |Eo (D)|. Note that an Eulerian subdigraph H of D has the same vertex set as D; i.e., H is a spanning subdigraph of D. Also an Eulerian subdigraph H of D need not be connected (in contrast to the usual definition of Eulerian graphs). In particular, the empty subdigraph, which is the subdigraph having no arcs, of D is an even Eulerian subdigraph of D. So each digraph D has at least one even Eulerian subdigraph. For arcs a1 , . . . , ap , b1 , . . . , bq of D, we denote by E(D, a1 , . . . , ap , b1 , . . . , bq ) the set of Eulerian subdigraphs H ∈ E(D) whose arc set contains {a1 , . . . , ap } and is disjoint from {b1 , . . . , bq }. For E 0 ⊆ E(D),let Ee0 = {H ∈ E 0 : |A(H)| is even}, Eo0 = {H ∈ E 0 : |A(H)| is odd}, diff(E 0 ) = |Ee0 | − |Eo0 |.

“2020-12-10” — 2021/5/8 — 18:39 — page 7 — #19

i

i

Chapter 2 Combinatorial Nullstellensatz

2.1

Introduction

Combinatorial Nullstellensatz (abbreviated as CNS herein) is a landmark theorem in algebraic combinatorics, which is now a widely used tool in tackling combinatorial problems in diverse areas of mathematics. CNS was first introduced by Noga Alon in [4] wherein he has shown its applications in solving a variety of problems-providing new proofs to some of the classical results and solving new problems. The German word ‘Nullstellensatz’ may be roughly translated as ‘zero location theorem’ where the word zero refers to the zeros of a single polynomial or a set of polynomials. Alon’s proof of CNS, as given in [4] is based on the classical theorem in Algebra: Hilbert’s Nullstellensatz (HNS) (See for instance the book [43] for a proof), which is basic in Algebraic Geometry. However, our proof of CNS is the one given by Michalek [47] based on mathematical induction. We start by stating HNS. Definition 2.1.1 Let F be a field and let f (x) = f (x1 , x2 , . . . , xn ) be a polynomial in F[x1 , x2 , . . . , xn ]. We say xt is a non-vanishing monomial of f if cf,t 6= 0. The degree of f is deg f = max{|t| : cf,t 6= 0}. Theorem 2.1.2 (Hilbert’s Nullstellensatz) Let k be a field, and K its algebraic closure. Let I be an ideal in the polynomial ring k[x1 , x2 , . . . , xn ], and let V (I) be the set of all points x = (x1 , x2 , . . . , xn ) ∈ K n such that f (x) = 0 for every f ∈ I. If p ∈ k[x1 , x2 , . . . , xn ] is any polynomial that vanishes on all of V (I) (i.e., p(x) = 0 for each x ∈ V (I)), then there exists a positive integer t such that pt ∈ I, that is, p belongs to the radical of I. Recall that for subsets S1 , S2 , . . . , Sn of a field F , the set S := S1 × S2 × · · · × Sn = {(a1 , a2 , . . . , an ) : ai ∈ Si } 7

“2020-12-10” — 2021/5/8 — 18:39 — page 8 — #20

i

i

8

Combinatorial Nullstellensatz

is an n-dimensional grid over F and the size ||S|| of S is the integer vector (|S1 | − 1, |S2 | − 1, . . . , |Sn | − 1). Theorem 2.1.3 (Combinatorial Nullstellensatz (CNS [4]) Let F be a field and let f (x) = f (x1 , x2 , . . . , xn ) be a non-zero polynomial in F[x1 , x2 , . . . , xn ]. Suppose that xt is a highest degree non-vanishing monomial of f . Then for any n-dimensional grid S of F with ||S|| = t, there exists s ∈ S for which f (s) 6= 0. In other words, if cf,||S|| 6= 0, then the grid S contains a non-zero point of f . NOTE: Equivalently, we can as well replace ||S|| = t by ||S|| ≥ t, i.e., |Si | − 1 ≥ ti for i = 1, 2, . . . , n. Proof. (M. Michalek [47]) By induction on the degree of f . Assume that deg f = 1 so that f (x) = a1 x1 + a2 x2 + · · · + an xn + an+1 , ai ∈ F with at least one ai 6= 0, say a1 6= 0. Assume that S1 , S2 , . . . , Sn ⊆ F with |S1 | = 2, and |Si | = 1 for i = 2, 3, . . . , n. Assume that S1 = {s1 , s01 } and Si = {si } for i = 2, 3, . . . , n. Let s = (s1 , s2 , . . . , sn ) and s0 = (s01 , s2 , . . . , sn ). Then f (s) − f (s0 ) = a1 (s1 − s01 ) 6= 0. Hence at least one of s, s0 is a non-zero point of f . Assume f (x) ∈ F[x1 , x2 , x3 , . . . , xn ], deg f = r ≥ 2 and the result holds for polynomials of degree ≤ r−1. Suppose xt is a highest degree non-vanishing monomial of f , where t = (t1 , t2 , . . . , tn ) and |t| = r. Thus ti ≥ 0 for each i and t1 + t2 + · · · + tn = r. Without loss of generality, assume that t1 > 0. Suppose f does not satisfy the statement in the theorem, S1 , S2 , · · · , Sn are subsets of F with |Si | = ti + 1 and f (s) = 0 for all s = (s1 , s2 , . . . , sn ) ∈ S1 × S2 × . . . × Sn . Fix a ∈ S1 and write f (x) = (x1 − a)g(x) + R(x),

(2.1)

(that is, divide f (x) by (x1 −a) getting g(x) as quotient and R(x) as remainder, where R(x) does not contain x1 ). Then x1t1 −1 xt22 . . . xtnn is a non-vanishing monomial of g(x) of highest degree. For any s ∈ {a} × S2 · · · × Sn , f (s) = 0 by our assumption, which implies that R(s) = 0. By virtue of (2.1), we get R(s) = 0,

for every s ∈ S2 × S3 × . . . × Sn .

Thus both f (x) and R(x) vanish on the set (S1 − {a}) × S2 · · · × Sn ,

“2020-12-10” — 2021/5/8 — 18:39 — page 9 — #21

i

i

Combinatorial Nullstellensatz

9

and hence so does g(x). But this contradicts our induction assumption on g(x). Corollary 2.1.4 Let F be a field and let f (x) = f (x1 , x2 , . . . , xn ) be a non-zero polynomial in F[x1 , x2 , . . . , xn ]. Assume S = S1 × Pn S2 × . . . × Sn and j=1 (|Sj | − 1) > deg(f ). Then S contains either no non-zero point or more than one non-zero point. Pn Proof. Assume to the contrary that j=1 (|Sj | − 1) > deg(f ) and S contains exactly one non-zero point s = (s1 , s2 , . . . , sn ) of f . Let −1  n n Y Y Y Y (x − b). (si − b) g(x) = f (s)  i=1 b∈Si −{si }

i=1 b∈Si −{si }

Note that g(x) = 0 for all x ∈ S − {s} and g(s) = f (s). Thus (f − g)(x) = 0 for all x ∈ S, and cf −g,||S|| = c−g,||S|| = Q −1 n Q −f (s) 6= 0. This is a contradiction i=1 b∈Si −{si } (si − b) to Theorem 2.1.3. Corollary 2.1.5 Assume that F is finite field with |F| = q and P1P (x), . . . , Pm (x) are polynomials in F[x1 , x2 , . . . , xn ] with (q − Pn m 1) j=1 deg(Pj (x)) < i=1 (|Si | − 1). Then S = S1 × . . . × Sn contains either no common zero point or more than one common zero point of the Pj ’s. Qm Proof. Let P (x) = j=1 (1−Pi (x)q−1 ). Then P (x) 6= 0 if and only if x is a common zero point of allPthe Pj ’s. The assumption of this n corollary implies that deg(P ) < i=1 (|Si | − 1). So the conclusion follows from Corollary 2.1.4. CNS has applications to diverse areas of mathematics. In the book, we shall concentrate on application to graph colouring. However, in the remainder of this chapter, we present some applications of CNS to number theory, graph structure and geometry (hyperplanes in Rn ).

2.2

An application of CNS to additive number theory

We now present a proof due to Alon [4] of one of the celebrated classical theorems in additive number theory due to Cauchy and

“2020-12-10” — 2021/5/8 — 18:39 — page 10 — #22

i

i

10

Combinatorial Nullstellensatz

Davenport as an application of CNS. This result was first proved by Cauchy and rediscovered by Davenport after a gap of 122 years! Theorem 2.2.1 (Cauchy and Davenport) Let p be a prime, and let A and B be two nonempty subsets of the field Zp . Let C = {x ∈ Zp : x = a + b for some a ∈ A and b ∈ B, a 6= b}. Then |C| ≥ min{p, |A| + |B| − 3}. Proof. The case p = 2 is trivial. So assume that p ≥ 3. Case 1 p ≤ |A| + |B| − 3. Take any g ∈ Zp . Then |g − B| = |B| and |A| + |g − B| ≥ p + 3. Hence |A ∩ (g − B)| ≥ 3. Let a ∈ A ∩ (g − B) − {g/2}. (Note: g/2 exists in Zp as p(≥ 3) is an odd prime). Then g = a + b for some b ∈ B with a 6= b. So g ∈ C. As g is an arbitrary element of Zp ), C = Zp , and the theorem is true in this case. Case 2 |A| + |B| − 3 < p. Assume the contrary to the statement of the theorem so that |C| < min(p, |A| + |B| − 3) = |A| + |B| − 3. Then there exists D ⊆ F with C ⊆ D and |D| = |A| + |B| − 4. Let Y Q(x, y) = (x − y) (x + y − d). d∈D

As d varies over D, it varies over C as well. Taking x = a, y = b with a 6= b and d = a + b, we have x + y − d = 0. Hence Q(a, b) = 0 for all a ∈ A, b ∈ B. The degree of Q(x, y) is |D| + 1 and the coefficient cQ,(i,|D|+1−i) of xi y |D|+1−i in Q(x, y) arises in two ways: Q (i) when x in (x − y) is multiplied by xi−1 y |D|+1−i ∈ (x + d∈D

y − d), and (ii) when −y in (x−y) is multiplied by xi y |D|−i in

Q

(x+y −d).

d∈D

Hence  cQ,(i,|D|+1−i) =

      |D| − i + 1 |D| |D| |D| − = 1− . i−1 i i−1 i

This is zero if and only if i =

|D| + 1 in Zp . 2

“2020-12-10” — 2021/5/8 — 18:39 — page 11 — #23

i

i

Combinatorial Nullstellensatz

11

Therefore at least one of cQ,(|A|−1,|B|−2) and cQ,(|A|−2,|B|−1) is non-zero. By CNS, ∃α ∈ A and β ∈ B with Q(α, β) 6= 0. But this is a contradiction to the fact that Q(x, y) = 0 for every x ∈ A and y ∈ B. This contradiction establishes the theorem in Case 2. Remark The proof given above is again based on Michalek [47] but it is not substantially different from the original proof of Alon [4].

2.3

Application of CNS to geometry

We now present an application of CNS to geometry due to Alon and F¨ uredi [5]. Theorem 2.3.1 Let H1 , H2 , . . . , Hm be a family of affine hyperplanes in Rn that cover all corners of the unit cube Qn in Rn except one. Then m ≥ n, Proof. We can assume, without loss of generality, that the uncovered vertex is the origin (0, 0, . . . 0). Let ai1 x1 + ai2 x2 + · · · + ain xn = bi

(2.2)

be the equation of the affine hyperplane Hi . Set pi (x) = ai1 x1 + ai2 x2 + · · · + ain xn .

(2.3)

Here aij and bi are real numbers. By our hypothesis, none of the planes Hi contains the origin. Hence bi 6= 0 for each i. Suppose the assertion is not true so that m < n. Now consider the polynomial P (x) = (−1)n+m+1 (b1 b2 . . . bm )

n Y

(xi − 1) +

i=1

m Y

(pi (x) − bi ). (2.4)

i=1

In the RHS of the above equation, the degree of the first term is n while the degree of the second term is m. As m < n, the degree of P (x) is n and the coefficient of (x1 x2 . . . xn ) is (−1)n+m+1 b1 b2 . . . bm 6= 0 since none of the bi ’s is zero. Therefore by CNS (as the degree of each xi is 1 in the leading monomial), if we take Si to be the 2-element subset {0, 1} of R, there exists a point

“2020-12-10” — 2021/5/8 — 18:39 — page 12 — #24

i

i

12

Combinatorial Nullstellensatz

a = (a1 , a2 , . . . , an ) with ai ∈ {0, 1} for each i and with P (a) 6= 0. This point a is not the origin since P (0) = (−1)n+m+1 (b1 b2 . . . bm )(−1)n + (−b1 )(−b2 ) . . . (−bm ) = 0. Hence a must be some other corner (as ai = 0 or 1) of Qn . But in this case, pi (a) − bi = 0 for some i (as this corner is covered by some hyperplane Hi ). The first term of P (a) is zero (since some ai must be 1) and once again we have P (a) = 0. This contradiction proves that m ≥ n.

2.4

CNS and a subgraph problem

A well known conjecture of Berge and Sauer, proved by Tashkinov [59] and independently by Zhang [66], asserts that any simple 4-regular graph contains a 3-regular subgraph. This assertion is false for multi-graphs, i.e., graphs containing multiple edges. For instance, consider a 4-regular graph formed by duplicating the edges of an odd cycle. However, the existence of a 3-regular subgraph in any graph formed by adding a single edge (or increasing the multiplicity of an edge by 1) to any 4-regular multi-graph is a special case of the following more general result due to Alon, Friedland and Kalai [1], proved as an application of CNS. Theorem 2.4.1 For any prime p, and any loopless graph G, if the average degree of G is greater than 2p − 2 and maximum degree is at most 2p − 1, then G contains a p-reqular subgraph. Proof. Assume that G is of order n and size m. Let (av,e )v∈V,e∈E denote the n × m incidence matrix of G defined by ( 1 if v is incident to e ave = 0 otherwise. Assume that the edges of G are e1 , e2 , . . . , em and associate with

“2020-12-10” — 2021/5/8 — 18:39 — page 13 — #25

i

i

Combinatorial Nullstellensatz

13

each edge ei of G a variable xi , and consider the polynomial F (x)

= F (x1 , x2 , . . . , xm ) Y = (1 − (av,e1 x1 + av,e2 x2 + · · · + av,em xm )p−1 ) v∈V m Y

−

(1 − xi )

(2.5)

i=1

defined over the field GF (p). In (2.5), the first product has n terms while the second term has m terms. The degree of the second product is m while the degree of the first product is at most n(p−1). By hypothesis, 2m/n > 2p−2. So m > n(p − 1) and the second term in (2.5) is the dominant term with respect to the degree of F (x). Moreover, the coefficient of x1 x2 . . . xm in (2.5) is −(−1)m = (−1)m+1 6= 0. So by CNS, there are values si ∈ {0, 1} = Si for each i ∈ {1, 2, . . . , m} such that F (s1 , s2 , . . . , sm ) 6= 0. Note that not all sQ i are zero as this would m mean F (s1 , s2 , . . . , sm ) = 1−1 = 0. Hence i=1 (1−si ) = 0 in (2.5). This implies that for any v ∈ V, the term av,e1 s1 +· · ·+av,em sm = 0. If not, this term is a non-zero element of Zp and hence when raised to the power p−1 should give 1 in which case F (s1 , s2 , . . . , sm ) = 0. Let E 0 = {ei : si = 1} and G0 = (V, E 0 ). Then for each vertex v, the sum av,e1 s1 + · · · + av,em sm should be zero in GF (p) for each v ∈ V . This means that for each vertex v, dG0 (v) ≡ 0(mod p). But as the maximum degree of G is at most 2p − 1, this degree is either 0 or p. Therefore the non-isolated vertices of G0 induce a p-regular subgraph of G0 and hence a p-regular subgraph of G. By taking p = 3 in Theorem 2.4.1, we obtain the following corollary: Corollary 2.4.2 Let G be a graph obtained from a 4-regular graph (multiple edges are allowed) by adding one edge. Then G contains a 3-regular subgraph.

2.5

0–1 vectors in a hyperplane

A hyperplane in the n-dimensional vector space GF (q)n over the finite field GF (q) is defined by an equation a1 x1 + a2 x2 + . . . + an xn = 0, where a = (a1 , a2 , . . . , an ) is the normal vector of H. In

“2020-12-10” — 2021/5/8 — 18:39 — page 14 — #26

i

i

14

Combinatorial Nullstellensatz

other words, H = a⊥ is the set of n-vectors over GF (q) that are perpendicular to a. In this section, we present a result of Balandraud and Girard [9], which says that if p is a prime and n > p and all the aj ’s are non-zero, then the hyperplane H = a⊥ in GF (p)n is determined by the 0–1 vectors contained in H, that is, by the 0–1 vectors which are orthogonal to a. The proof presented below is given by Schauz and Honold [54]. Theorem 2.5.1 Assume that p is a prime and n > p and H is a hyperplane in GF (p)n that does not contain any of the coordinate axes ei , where e1 = (1, 0, . . . , 0) etc. Then H has a basis consisting of 0–1 vectors. Proof. We need to show that the 0–1 vectors in a⊥ span a subspace of dimension n − 1 (namely span the whole hyperplane a⊥ ). This is true if and only if for any other hyperplane b⊥ in GF (p)n , there is a 0–1 vector (x1 , x2 , . . . , xn ) ∈ a⊥ which is not contained in b⊥ . Assume that b = (b1 , b2 , . . . , bn ) and b⊥ 6= a⊥ , i.e., b is not a multiple of a. Let λi = abii . Since H does not contain any of the coordinate axes ei , ai 6= 0 for all i and λi is well-defined for all i. Let  ! n !p−1 n X X P (x) = λ i xi  xi − 1 , i=1

i=1

which is a polynomial of degree p in GF (p)[x1 , x2 , . . . , xn ]. Since b⊥ 6= a⊥ , we may assume that λ1 6= λ2 . Let t = (t1 , t2 , . . . , tn ), where tj = 1 for j ∈ {2, 3, . . . , p + 1}, and tj = 0 otherwise; and t0 = (t01 , t02 , . . . , t0n ), where t0j = 1 for j ∈ {1, 3, . . . , p + 1}, and t0j = 0 otherwise. As n > p, tj , t0j exist for j ≤ p + 1. Then for j ∈ {2, 3, . . . , p + 1},   !p−1 n X λ j xj  xi − 1 = λj (p − 1)!xt + other terms, i=1

and for j ∈ {1, 3, . . . , p + 1},   !p−1 n X 0 λj xj  xi − 1 = λj (p − 1)!xt + other terms. i=1

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 15 — #27

Combinatorial Nullstellensatz

15

Hence  cP,t

=

(p − 1)! 

p+1 X





λj  − λ1 , and

j=1

 cP,t0

=

(p − 1)! 

p+1 X





λj  − λ2 .

j=1

So cP,t − cP,t0 6= 0, and hence at least one of cP,t and cP,t0 is nonzero. By CNS, P has a non-zero point in {0, a1 } × {0, a2 } × . . . × {0, an }. In other words, there exists (x1 , x2 , . . . , xn ) ∈ {0, 1}n such that P (a1 x1 , a2 x2 , . . . , an xn ) 6= 0. This means that α = a1 x1 + a2 x2 + · · · + an xn = 0 (since otherwise, αp−1 = 1 in GF (p), and and so the second factor in P (a1 x1 , a2 x2 , · · · , an xn ) would be zero) and β = b1 x1 + b2 x2 + · · · bn xn 6= 0 (since otherwise the first factor in P (a1 x1 , · · · , an xn ) would be zero). Hence x ∈ a⊥ \b⊥ . In the examples shown in this chapter, in order to solve our problem using CNS, we had to judiciously choose a suitable polynomial and invoke CNS. Needless to stress that it may not always be possible. On the other hand, there are also many combinatorial problems where the associated polynomials follow naturally from the definitions. The remainder of this book concentrates on the applications of CNS to graph colouring and graph labelling problems, where the associated polynomials are easily constructed. We shall concentrate on methods in proving that a certain monomial has non-zero coefficient in the expansion of a polynomial.

“2020-12-10” — 2021/5/8 — 18:39 — page 17 — #29

i

i

Chapter 3 Alon–Tarsi Theorem and Its Applications

3.1

Alon–Tarsi theorem

To apply CNS to solve a problem, one need to construct a polynomial so that the problem is reduced to deciding if the polynomial has a non-zero point in a certain grid. We then show that the coefficient of a certain monomial in the expansion of the polynomial is non-zero. In this monograph, we concentrate on the calculation of the coefficients of monomials, which is usually the difficult step in the application of CNS. One of the most popular topics in graph theory is graph colouring. A widely studied variation of the conventional graph colouring, where one uses distinct colours for adjacent vertices, is the assignment of colours from out of a set of colours attached to each vertex of the graph. Such an assignment is known as list colouring. As mentioned by Alon [4], ‘This variant received a considerable amount of attention that led to several facinating conjectures and results and its study combines interesting combinatorial techniques with powerful algebraic and probabilistic ideas’. This study is called ‘choosability’ in graphs and was initiated by Vizing [61], and independently by Erd˝ os, Rubin and Taylor [23]. CNS is a useful tool in the study of choosability in graphs. Definition 3.1.1 Assume that G = (V, E) is a graph. A proper colouring of G is a mapping φ which assigns to each vertex v a colour φ(v) from a certain colour set such that for each edge uv of G, φ(u) 6= φ(v). A k-colouring of G is a proper colouring of G with colour set {1, 2, . . . , k}. We say that G is k-colourable if there is a proper k-colouring of G. The chromatic number χ(G) of G is defined as χ(G) = min{k : G is k-colourable}. 17

“2020-12-10” — 2021/5/8 — 18:39 — page 18 — #30

i

i

18

Combinatorial Nullstellensatz

Definition 3.1.2 Assume that G is a graph and f : V (G) → Z+ is a function which assigns a positive integer f (v) to each vertex v ∈ V (G). An f -list assignment of G is a mapping L which assigns to each vertex v of G a list L(v) of f (v) elements of a field F as permissible colours. If f (v) = k for all v ∈ V (G), then the f -list assignment of G is called a k-list assignment of G. An L-colouring of G is a proper colouring φ of G such that φ(v) ∈ L(v) for each vertex v. We say that G is L-colourable if there is an L-colouring of G, and we say that G is f -choosable (respectively, k-choosable) if G is L-colourable for each f -list assignment (respectively, k-list assignment) L of G. The choice number ch(G) of G is then defined as ch(G) = min{k : G is k-choosable}. It follows from the definition that for any graph G, χ(G) ≤ ch(G). On the other hand, the difference ch(G) − χ(G) can be arbitrarily large (cf. Theorem 1.0.15). Definition 3.1.3 Assume that G is an undirected graph whose vertices are linearly ordered and F is a field. We associate to each vertex v of G a variable xv . The graph polynomial ∈ F[x] of G is defined as Y fG (x) = (xu − xv ). uv∈E(G),u 2n, which is a contradiction. Thus after the qth step, all vertices are coloured and we obtain an L-colouring of G. This proves that ch(G) ≤ log2 n + 2. It is known [63] that for any positive integer n, ch(G) ≤ log2 n − ( 21 + o(1)) log2 log2 n. It was conjectured by Erd˝os and Lovasz (cf. [63]) that ch(G) = Ω(log2 n − (1 + o(1)) log2 log2 n) and the conjecture remains open.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 21 — #33

Alon–Tarsi Theorem and Its Applications

21

Definition 3.1.6 For f ∈ NG 0 and u ∈ V (G) for which f (u) > 0, define fˆu ∈ NG by setting 0 ( f (v) − 1, if v = u, ˆ fu (v) = f (v), otherwise. Lemma 3.1.7 If H is a subgraph of G, f ∈ NG 0 and G is f -AT, then H is f -AT. In particular, AT (H) ≤ AT (G). Proof. It suffices to prove that if H is obtained from G by deleting a single edge e = uv, then H is f -AT. In this case, fG = (xu − xv )fH = xu fH − xv fH . Assume that xg is a non-vanishing monomial in fG . Then at least one of xgˆu , xgˆv is a non-vanishing monomial in fH , for otherwise xg is a vanishing monomial in both xu fH and xv fH , and hence xg cannot be a non-vanishing monomial of fG . Therefore H is f -AT. Recall that a digraph D0 is called Eulerian if for every vertex − v, d+ D 0 (v) = dD 0 (v), E(D) the family of spanning Eulerian subdigraphs of D, and Ee (D) Eo (D) diff(D)

= {D0 ∈ E(D) : |A(D0 )| ≡ 0 = {D0 ∈ E(D) : |A(D0 )| ≡ 1 = |Ee (D)| − |Eo (D)|.

(mod 2)}, (mod 2)},

Definition 3.1.8 We say that D is Alon–Tarsi if diff(D) 6= 0. If an orientation D of G yields an Alon–Tarsi digraph, then we say D is an Alon–Tarsi orientation (or an AT-orientation, for short) of G. Alon and Tarsi [3] proved the following theorem. Its significance rests in the fact that it transforms the computation of the Alon– Tarsi number of a graph G from algebraic manipulations to the analysis of the structural properties of G. (See, for instance, Section 3.4). Theorem 3.1.9 (Alon–Tarsi Theorem) A graph G is f -AT if and only if G has an AT-orientation D with d+ D (v) ≤ f (v) − 1 for each vertex v of G. Therefore AT (G) = min{k : G has an AT-orientation D with ∆+ D = k − 1}.





“2020-12-10” — 2021/5/8 — 18:39 — page 22 — #34

22

Combinatorial Nullstellensatz  Proof. In calculating the expansion of fG (x) = uv∈E(G),u v, and let w(D) = e∈A(D) w(e). Then  fG (x) = w(D), (3.1) D

where the summation ranges over all the 2|E(G)| orientations D of G.

FIGURE 3.2: Example: an orientation D with |Ee (D)| = |Eo (D)| = 1. For instance, for  the oriented graph D of G of Figure 3.2, w(D) = w(e) = w(e1 )w(e2 )w(e3 )w(e4 ) = e∈A(D)

x1 (−x2 )(−x3 )(−x4 ) = −x1 x2 x3 x4 . Let m(D) be the number of arcs of D in ‘reverse’ directions, i.e., m(D) = |{(u, v) ∈ A(D) : u > v}|. It follows from the definition that

w(D) = (−1)m(D)



d+ (v)

xv D

.

v∈V (G)

We say D is even (or odd) if m(D) is even (or odd). For g ∈ NG 0, let E(G; g) = {D : D is an orientation of G for which d+ D (v) = g(v)for all v ∈ V (G)}

“2020-12-10” — 2021/5/8 — 18:39 — page 23 — #35

i

i

Alon–Tarsi Theorem and Its Applications

23

Let Ee (G; g) = {D ∈ E(G; g) : D is an even orientation of G}, Eo (G; g) = {D ∈ E(G; g) : D is an odd orientation of G}. Then ! fG (x) =

X

|Ee (G; g)| − |Eo (G; g)| xg .

g∈NG 0 0 Assume that g ∈ NG 0 and D ∈ E(G; g). For any D ∈ E(G; g), 0 let D ⊕ D be the set of arcs in D whose directions are reversed in D0 , i.e.,

D ⊕ D0 = {(u, v) ∈ A(D) : (v, u) ∈ A(D0 )}. + For each vertex v, since d+ D (v) = dD 0 (v), the number of outgoing arcs of D that got changed into incoming arcs of D0 equals the number of incoming arcs of D that have been changed into outgoing arcs of D0 . Hence D ⊕ D0 is an Eulerian subdigraph of D. i.e., D ⊕ D0 ∈ E(D). Conversely, if H ∈ E(D) and D0 is obtained from D by reversing the orientation of the arcs in H, then D0 and D have the same out-degree sequence. Moreover, D ⊕ D0 ∈ Ee (D) if D and D0 have the same parity (that is, either both D and D0 are even or both D and D0 are odd), and D ⊕ D0 ∈ Eo (D) if D and D0 have opposite parities. The mapping D0 → D ⊕ D0 is a bijection between De (G; g) ∪ Do (G; g) and the set of Eulerian subdigraphs of D. When D is even, it maps even orientations to even (Eulerian) subdigraphs and odd orientations to odd subdigraphs. Otherwise it maps even orientations to odd subdigraphs and odd orientations to even subdigraphs. In both the cases,

|De (G; g)| − |Do (G; g)| = ±diff(D). Qn So the coefficient of the monomial i=1 xdi i in the expansion of fG Qn is ±diff(D). In particular, i=1 xdi i is a non-vanishing monomial of fG if and only if D is an AT-orientation of G. Hence AT (G) = min{k : G has an AT-orientation D with ∆+ D = k − 1}. This completes the proof of Theorem 3.1.9.

“2020-12-10” — 2021/5/8 — 18:39 — page 24 — #36

i

i

24

Combinatorial Nullstellensatz

Example 3.1.10 The orientation in Figure 3.2 is the only orientation of the underlying graph G with maximum out-degree 1. As this orientation is not an AT-orientation, we conclude that AT (G) ≥ 3. On the other hand, by reversing the direction of one edge in the triangle, the resulting orientation is an AT-orientation. So AT (G) = 3. Lemma 3.1.11 Let d1 ≥ d2 ≥ · · · ≥ dn be the out-degree sequence of a digraph D, satisfying |Ee (D)| 6= |Eo (D)|. Then  for every k, n ≥ (n − k) k ≥ 0, G has an independent set of size at least . (dk+1 + 1) Proof. Relabel the vertices of G, if necessary, so that d+ (vi ) = di . By Theorem 3.1.9, there is a legal vertex colouring c : V → Z such that 1 ≤ c(vi ) ≤ di + 1 for each i, 1 ≤ i ≤ n. Let k + 1 ≤ j ≤ n. Then 1 ≤ c(vj ) ≤ dj + 1 ≤ dk+1 + 1. Hence to colour the n − k vertices vk+1 , vk+2 , . . . , vn , at most dk+1 +1 colours have been   used. (n − k) So at least one colour class has size at least . (dk+1 + 1) Assume that D is a digraph. For a subset A of arcs of D, denote by AR the set of arcs obtained by reversing the orientation of the arcs in A. For convenience, a digraph is usually viewed as a set of arcs. Thus (D − A) ∪ AR denotes the digraph obtained from D by reversing the direction of the arcs in A. Lemma 3.1.12 Assume that D is a digraph and C is a directed cycle in D. Let D0 = (D − C) ∪ C R . For an Eulerian subdigraph H of D, let φ(H) = (H − C) ∪ (C − H)R . Then φ is a one-to-one correspondence between E(D) and E(D0 ). In particular, |E(D)| = |E(D0 )|. Proof. First we show that φ(H) is an Eulerian subdigraph of D0 . It follows from the definition that φ(H) is a subdigraph of D0 . Assume that x ∈ V (D). We calculate the changes in the in-degree and out-degree of x as H is changed to φ(H). If x ∈ / V (C), then the in-degree and out-degree of x are not changed. Assume that x ∈ V (C). If both the arcs of C incident to x are contained in H, then both the in-degree and the out-degree of x decrease by 1. If both the arcs of C incident to x are not contained in H, then both the in-degree and the out-degree of x increase by 1. If exactly one arc incident to x is contained in H, then the in-degree and outdegree x are not changed. Thus φ(H) is an Eulerian subdigraph of D0 .

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 25 — #37

Alon–Tarsi Theorem and Its Applications

25

The same argument shows that for any Eulerian subdigraph H of D0 , ψ(H) = (H − C R ) ∪ (C R − H)R is an Eulerian subdigraph of D. Moreover, for any H ∈ E(D), ψ(φ(H)) = H. Thus φ is a one-to-one correspondence between E(D) and E(D0 ). Corollary 3.1.13 For any two orientations D1 , D2 of a graph G with the same out-degree sequence, |E(D1 )| = |E(D2 )|. Proof. The subdigraph D1 ⊕ D2 is an Eulerian subdigraph of D1 and hence it is the arc disjoint union of directed cycles. So D2 can be obtained from D1 by recursively reversing the directions of directed cycles, and hence |E(D1 )| = |E(D2 )|. Definition 3.1.14 A digraph D is strongly connected if for any two vertices u and v, there is a directed path from u to v. Definition 3.1.15 For a digraph D, a strongly connected component of D is a maximal subdigraph of D which is strongly connected. Observe that for any digraph D, there is a partial ordering ≤ of the strongly connected components of D such that for any two strongly connected components A and B of D, if A < B, then all the arcs between A and B (if any) are from A to B. ˙ 2. Lemma 3.1.16 Assume that D is a digraph and V (D) = X1 ∪X For i = 1, 2, let Di = D[Xi ] be the subdigraph of D induced by Xi . If all the arcs between X1 and X2 are from X1 to X2 , then D is Alon–Tarsi if and only if D1 , D2 are both Alon–Tarsi. Proof. Since any arc of an Eulerian digraph is contained in a directed cycle, each Eulerian subdigraph H of D is the arc disjoint union of H1 and H2 , where Hi is an Eulerian subdigraph of Di . Now H is even if and only if H1 , H2 have the same parity. Therefore |Ee (D)| = |Ee (D1 )| × |Ee (D2 )| + |Eo (D1 )| × |Eo (D2 )|, |Eo (D)| = |Ee (D1 )| × |Eo (D2 )| + |Eo (D1 )| × |Ee (D2 )|. So diff(D) = diff(D1 ) × diff(D2 ) and D is AT if and only if both D1 , D2 are AT. Corollary 3.1.17 A digraph is Alon–Tarsi if and only if each of its strongly connected components is Alon–Tarsi.

“2020-12-10” — 2021/5/8 — 18:39 — page 26 — #38

i

i

26

3.2

Combinatorial Nullstellensatz

Bipartite graphs and acyclic orientations

In general, it is difficult to determine if an orientation D of a graph G is AT. However, there are some cases where this problem is easy. Observe that every digraph D has at least one even Eulerian subdigraph, namely, the empty subdigraph, and every Eulerian digraph can be decomposed into directed cycles. So if D has no odd directed cycle, then D has no odd Eulerian subdigraph and hence D is AT. Thus we have the following result. Lemma 3.2.1 If G is a bipartite graph, then every orientation D of G is an AT-orientation. For any graph G, any acyclic orientation D of G is an AT-orientation of G. Corollary l 3.2.2m If G is a bipartite graph, then AT (G) = maxH⊂G |E(H)| |V (H) + 1. l m Proof. Assume that k = maxH⊂G |E(H)| |V (H) . For any orientation D of G and for any subgraph H lof G, the m restriction of D to H has maximum out-degree at least |E(H)| |V (H) . Hence AT (G) ≥ k + 1. We now show that there is an orientation D of G for which ∆+ D ≤ k. We construct a bipartite graph Q with partite sets E(G) and V (G) × [k], where [k] = {1, 2, . . . , k}, and (v, j) ∈ V (G) × [k] is adjacent to e ∈ E(G) if v is an end vertex of e. For any subset X of E(G), let Y be the set of vertices of G each of which is incident to at least one edge of X. Then NQ (X) = Y × [k]. Since |X| ≤ k|Y |, we conclude that |X| ≤ |NQ (X)|. By Hall’s Theorem, Q has a matching M that saturates every e ∈ E. We orient the edge e = uv ∈ E(G) as (u, v) if and only if (e, (u, j)) ∈ M for some j. Then the resulting orientation D has ∆+ D ≤ k. By Theorem 3.1.9, AT (G) ≤ k + 1. Hence AT (G) = k + 1. Corollary 3.2.3 Bipartite planar graphs G have AT (G) ≤ 3 and hence are 3-choosable. Proof. It follows from Euler’s formula (Theorem 1.0.10) that for any subgraph H of G, |E(H)|/|V (H)| ≤ 2. The colouring number of a graph G is defined as col(G) = 1 + max{δ(H) : H is a subgraph of G},





“2020-12-10” — 2021/5/8 — 18:39 — page 27 — #39

Alon–Tarsi Theorem and Its Applications

27

where δ(H) stands for the minimum degree of H. Alternately, col(G) is the minimum integer k such that there is a linear ordering < of the vertices of G so that each vertex v has at most k − 1 neighbours u with u < v. By orienting each edge uv as (u, v) if u > v, we obtain an acyclic orientation D of G with + = col(G) − 1. Hence we have the following corollary. δD Corollary 3.2.4 For any graph G, AT (G) ≤ col(G).

3.3

The Cartesian product of a path and an odd cycle

Another simple sufficient condition for an orientation D to be Alon–Tarsi is that |E(D)| is odd. For, if |Ee (D)| = |Eo (D)|, then |E(D)| = |Ee (D)| + |Eo (D)| is even. Hence if |E(D)| is odd, then  |Eo (D)| and D is Alon–Tarsi. In this section, this idea |Ee (D)| = is used to prove that the Cartesian product of a path and an odd cycle has Alon–Tarsi number 3. Definition 3.3.1 Assume that G = (V, E) and G = (V  , E  ) are graphs. The Cartesian product GG is the graph with vertex set V × V  , in which (x, x ) is adjacent to (y, y  ) if and only if either x = x and yy  ∈ E  or y = y  and xx ∈ E. Figure 3.3 is the graph P3 C5 .

FIGURE 3.3: The graph P3 C5 . Theorem 3.3.2 [34] If G is the Cartesian product C2k+1 Pn of an odd cycle and a path, then AT (G) = 3. Proof. Since G has chromatic number 3, we know that AT (G) ≥ 3.





“2020-12-10” — 2021/5/8 — 18:39 — page 28 — #40

28

Combinatorial Nullstellensatz

It remains to show that AT (G) ≤ 3. If n = 1, then the result is obvious. We assume that n ≥ 2. Assume that the vertices of G are {(vi , wj ) : 1 ≤ i ≤ 2k + 1, 1 ≤ j ≤ n}, where for each j, the set Xj = {(vi , wj ) : 1 ≤ i ≤ 2k + 1} induces an odd cycle of length 2k + 1, and for each i, the set Yi = {(vi , wj ) : 1 ≤ j ≤ n} induces a path on n vertices. Let D be the orientation of G in which the edges of G are oriented in such a way that for each j, G[Xj ] is a directed cycle, and for each i, G[Yi ] is a directed path from (vi , w1 ) to (vi , wn ) [See Figure 3.4]. Let D∗ be obtained from D by adding an arc e∗ from (v2 , wn ) to (v1 , w1 ). It is easy to check that D∗ has maximum out-degree 2. Now we show that D∗ has an odd number of Eulerian subdigraphs.

FIGURE 3.4: The digraphs D and D∗ for n = 3 and k = 2. Note that the only directed cycles in D are those directed cycles induced by Xj for j = 1, 2, . . . , n. All the directed cycles in D∗ not contained in D contain the arc e∗ . Let A be the family of Eulerian sub-digraphs Q of D ∗ such that for 1 ≤ j ≤ n, either all the arcs of D(Xj ) are contained in Q or disjoint from Q. That is, A = {Q ∈ E(D∗ ) : ∃j, E(D[Xj ]) ⊆ E(Q) or E(D[Xj ]) ∩ E(Q) = ∅}. For each Eulerian subdigraph Q in A, let Q be obtained from Q by deleting all the arcs of those directed cycles D[Xj ] contained in Q, and adding the arcs of those directed cycles D[Xj ] that are arc disjoint from Q. By definition, Q and Q are distinct members of A, and each member of A is contained in exactly one such pair. Therefore |A| is even. Let B be the set of all the other Eulerian subdigraphs of D∗ . Since each Eulerian subdigraph can be decomposed into an arcdisjoint union of directed cycles (and the only directed cycles in D are D[Xj ], for 1 ≤ j ≤ n), we conclude that each member of B is

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 29 — #41

Alon–Tarsi Theorem and Its Applications

29

a directed cycle containing e∗ and contains at least one arc and at most 2k arcs from each directed cycle D[Xj ]. To construct such a directed cycle C, we start from (v1 , w1 ), traverse at least one arc but at most 2k arcs from D[X1 ], then go to X2 and traverse at least one arc but at most 2k arcs from D[X2 ], and then continue in the same fashion to X3 , X4 , . . . , Xn . Such a directed cycle is uniquely determined by the positions in which it goes from Xi to Xi+1 for i = 1, 2, . . . , n − 1. Since the last arc of the directed cycle is from (v2 , wn ) to (v1 , w1 ), for the directed cycle to contain at least one and at most 2k arcs of D[Xn ], it is required that the arc from Xn−1 to Xn is not the arc joining (v2 , wn−1 ) to (v2 , wn ). Let Sn , Sn0 and Sn00 be sets of integer sequences defined as follows: Sn Sn0 Sn00

= {(i1 , i2 , . . . , in−1 ) : 1 ≤ ij ≤ 2k + 1, i1 6= 1, ij+1 6= ij }, = {(i1 , i2 , . . . , in−1 ) ∈ Sn : in−1 6= 2}, = {(i1 , i2 , . . . , in−1 ) ∈ Sn : in−1 = 2}.

By the argument above, |B| = |Sn0 |. To form a sequence (i1 , i2 , . . . , in−1 ) in Sn , each ij has 2k choices. So |Sn | = (2k)n−1 . It is obvious that |S20 | = 2k − 1 and |S200 | = 1 are odd. Each se00 quence in Sn−1 can be extended to a sequence in Sn0 in 2k different ways (in−1 can be any index distinct from in−2 ), and each sequence 0 in Sn−1 can be extended to a sequence in Sn0 in 2k − 1 different ways (in−1 can be any index distinct from both in−2 and 2). So 00 0 |Sn0 | = |Sn−1 |2k + |Sn−1 |(2k − 1). As |Sn00 | = |Sn | − |Sn0 |, Sn0 and Sn00 have the same parity. Thus by induction, |Sn0 | and |Sn00 | are both odd for all n. So |B| is odd, and hence D∗ has an odd number of Eulerian subdigraphs. Thus D∗ is an AT-orientation of G∗ = G + e, where e is an edge connecting (v1 , w2 ) to (v2 , wn ). Hence AT (G∗ ) ≤ 3, and by Lemma 3.1.7, AT (G) ≤ 3. Corollary 3.3.3 If G = C2k+1 Pn for some positive integers k, n, then χ(G) = ch(G) = AT (G) = 3. If G = C2k Pn and n ≥ 2, then G is a bipartite graph with 1 < |E(G)|/|V (G)| < 2. By Corollary 3.2.2, AT (G) = 3. So the Cartesian product of any cycle with a path with at least two vertices has Alon–Tarsi number 3. It is natural to ask as to what is the Alon–Tarsi number of the Cartesian product Cn Cm of two cycles. If both n and m are even,

“2020-12-10” — 2021/5/8 — 18:39 — page 30 — #42

i

i

30

Combinatorial Nullstellensatz

then Cn Cm is bipartite, and it follows from Corollary 3.2.2 that Cn Cm has Alon–Tarsi number 3. When at least one of n, m is odd, this question will be answered in Section 4.6.

3.4

A solution to a problem of Erd˝ os

Similar to the idea in the previous section, if |Ee (D)| = |Eo (D)| is even, then |E(D)| ≡ 0 (mod 4). Thus if |E(D)| ≡ 2 (mod 4) and Ee (D) is even, then D is an Alon–Tarsi orientation. In this section, this technique is used to prove a family of 4-regular graphs has Alon–Tarsi number 3. Assume that D is an Eulerian digraph. For any Eulerian subdigraph H of D, its complement D\H is also an Eulerian digraph. (Recall that as per our definition, an Eulerian digraph need not be connected). If D has an odd number of arcs, then H and D\H have opposite parity. Hence |Ee (D)| = |Eo (D)|. So D is not an ATorientation. On the other hand, if D has an even number of arcs, then H and D\H have the same parity. Thus |Ee (D)| and |Eo (D)| are even. Therefore, if |E(D)| ≡ 2 (mod 4), then |Ee (D)| = 6 |Eo (D)| and D is an Alon–Tarsi orientation. Assume that G is a 4-regular graph on 3n vertices whose edge set can be decomposed into a Hamiltonian cycle and n pairwise vertex-disjoint triangles. (See Figure 3.5 for an example of such a graph). Then an independent set of G contains at most one vertex from each triangle, and hence has size at most n. Du and Hsu conjectured in 1986 that G has independence number n. In 1990, Erd˝os asked whether G has chromatic number 3. This problem, often referred to as the ‘cycle plus triangles problem of Erd˝os’, was solved in the affirmative by Fleischner and Stiebitz in 1992 [25]. Indeed, Fleischner and Stiebitz proved the following stronger result. Theorem 3.4.1 Assume that G is a 4-regular graph on 3n vertices whose edge set can be decomposed into a Hamiltonian cycle and n pairwise vertex disjoint triangles. Then AT (G) = 3. To prove Theorem 3.4.1, we need to find an AT-orientaiton D of G in which each vertex has out-degree at most 2. Since P + + x∈V (D) dD (x) = |A(D)| = 2|V (D)|, we conclude that dD (x) =





“2020-12-10” — 2021/5/8 — 18:39 — page 31 — #43

Alon–Tarsi Theorem and Its Applications

31

FIGURE 3.5: An example of a graph G of Theorem 3.4.1. d− D (x) = 2 for each vertex x. So D is an Eulerian digraph. Theorem 3.4.1 follows from the next lemma. Lemma 3.4.2 Assume that D is an Eulerian digraph whose edge set can be decomposed into a directed Hamiltonian cycle C and n ≥ 0 pairwise vertex-disjoint directed triangles. Then |E(D)| ≡ 2 (mod 4). Proof. As the n triangles are vertex disjoint, |V (D)| ≥ 3n. We observe that the number of vertices of D can be greater than n. Thus some vertices of D may not be contained in any of the n triangles, and hence has in-degree and out-degree both equal to 1. We prove the lemma by induction on n. If n = 0, then D is a directed cycle and |E(D)| = 2 (the only Eulerian subdigraphs of D are the empty digraph and D itself). Assume that n ≥ 1 and the Lemma holds if there are at most n − 1 triangles. Let T be one of the triangles as indicated in Figure 3.6, where V (T ) = {x1 , x2 , x3 } and the arcs of T are a1 = (x3 , x2 ), a2 = (x1 , x3 ), a3 = (x2 , x1 ). Now an Eulerian subdigraph of D may contain one, two, all or none of the arcs a1 , a2 , a3 . Recall that for arcs z1 , z2 , . . . , zq , y1 , y2 , . . . , yp , E(D, z1 , z2 , . . . , zq , y 1 , y 2 , . . . , y p ) is the family of Eulerian subdigraphs H of D which contain z1 , z2 , . . . , zq but which do not contain any of y1 , y2 , . . . , yp . Hence E(D)

=

˙ ˙ a1 , a2 , a3 )∪E(D, a1 , a2 , a3 ), E ∗ (D)∪E(D,





“2020-12-10” — 2021/5/8 — 18:39 — page 32 — #44

32

Combinatorial Nullstellensatz

FIGURE 3.6: Digraph D. where E ∗ (D)

=

˙ ˙ E(D, a1 , a2 , a3 )∪E(D, a2 , a1 , a3 )∪E(D, a3 , a1 , a2 ) ˙ ˙ ˙ a2 , a1 , a3 )∪E(D, a3 , a1 , a2 ). ∪E(D, a1 , a2 , a3 )∪E(D,

Let D = D − A(T ), where A(T ) stands for the arc set of T . Then D is an Eulerian digraph whose arc set is decomposed into a Hamiltonian cycle and n − 1 pairwise vertex disjoint triangles. By induction hypothesis, |E(D )| ≡ 2 (mod 4). It is obvious that |E(D )| = |E(D, a1 , a2 , a3 )| = |E(D, a1 , a2 , a3 )|. So |E(D, a1 , a2 , a3 )| + |E(D, a1 , a2 , a3 )| ≡ 0 (mod 4). As |E(D)| = |E ∗ (D)| + |E(D, a1 , a2 , a3 )| + |E(D, a1 , a2 , a3 )|, we have |E(D)| ≡ |E ∗ (D)| (mod 4). So it remains to show that |E ∗ (D)| ≡ 2 (mod 4). For i = 1, 2, 3, let x− i be the predecessor of xi on C and let , x ). By Corollary 3.1.13, we may assume that the cycle bi = (x− i i C is directed as in Figure 3.6. For i = 1, 2, 3, let Ci be the unique directed cycle in C ∪ ai containing ai . For (i, j, k) ∈ {(1, 2, 3), (2, 3, 1), (3, 1, 2)}, let Di = (D − Ci ) ∪ CiR as in Figure 3.7 and let Di be obtained from Di by splitting    away arcs bj , ak , aj and bR k as in Figure 3.8. Let xi , xj , xk be the  new vertices that result from the splitting. Then Di is an Eulerian digraph which has a decomposition into a directed Hamiltonian cycle and (n − 1) pairwise vertex disjoint triangles.





“2020-12-10” — 2021/5/8 — 18:39 — page 33 — #45

Alon–Tarsi Theorem and Its Applications

33

FIGURE 3.7: Digraph Di .

FIGURE 3.8: Digraph Di . Since aj and ak share a degree 2 vertex, each Eulerian subdigraph H of Di either contains both aj and ak , or contains neither of aj and ak . Hence |E(Di )|

=

 R |E(Di , aR i , aj , ak )| + |E(Di , ai , aj , ak )|  R + |E(Di , aR i , aj , ak )| + |E(Di , ai , aj , ak )|.

“2020-12-10” — 2021/5/8 — 18:39 — page 34 — #46

i

i

34

Combinatorial Nullstellensatz

Claim 3.4.3 For (i, j, k) ∈ {(1, 2, 3), (2, 3, 1), (3, 1, 2)}, |E(Di0 , aR i , aj , ak )| = |E(D, ai , aj , ak )|, |E(Di0 , aR i , aj , ak )| = |E(D, ai , aj , ak )|, 0 R |E(Di , ai , aj , ak )| = |E(D0 , bj , bk )| = |E(D0 , bi , bj , bk )| +|E(D0 , bi , bj , bk )|, 0 0 |E(Di0 , aR i , aj , ak )| = |E(D , bj , bk )| = |E(D , bi , bj , bk )| 0 +|E(D , bi , bj , bk )|. Proof. For H ∈ E(D), let φ(H) = (H − Ci ) ∪ (Ci − H)R . Then φ is a one-to-one correspondence between E(D, ai , aj , ak ) and E(Di0 , aR i , aj , ak ), and a one-to-one correspondence between E(D, ai , aj , ak ) and E(Di0 , aR i , aj , ak ). For H ∈ E(Di0 ), ψ(H) = (H − CiR ) ∪ (CiR − H)R is a one0 to-one correspondence between E(Di0 , aR i , aj , ak ) and E(D , bj , bk ), 0 and ψ (H) = ψ(H) − {ai , aj , ak } is a one-to-one correspondence 0 between E(Di0 , aR i , aj , ak ) and E(D , bj , bk ). It follows from Claim 3.4.3 that |E(D10 )| + |E(D20 )| + |E(D30 )| = |E ∗ (D)| + m, where m =

X

|E(D0 , bi , bj , bk )| + |E(D0 , bi , bj , bk )|

(i,j,k)


+|E(D0 , bi , bj , bk )| + |E(D0 , bi , bj , bk )| , where the summation is over (i, j, k) ∈ {(1, 2, 3), (2, 3, 1), (3, 1, 2)}. By induction hypothesis, |E(Di0 )| ≡ 2 (mod 4) and hence |E(D10 )| + |E(D20 )| + |E(D30 )| ≡ 2

(mod 4).

To prove that |E ∗ (D) ≡ 2 (mod 4), it remains to show that m ≡ 0 (mod 4). By using the one-to-one correspondence φ : E(D0 ) → E(D0 ) defined as φ(H) = D\H, we conclude that |E(D0 , bi , bj , bk )| = |E(D0 , bi , bj , bk )|. It follows that |E(D0 , bi , bj , bk )| + |E(D0 , bi , bj , bk )| + |E(D0 , bi , bj , bk )| +|E(D0 , bi , bj , bk )| = 2(|E(D0 , bi , bj , bk )| + |E(D0 , bi , bj , bk )|).





“2020-12-10” — 2021/5/8 — 18:39 — page 35 — #47

Alon–Tarsi Theorem and Its Applications

35

Therefore m

= ≡

4(|E(D , b1 , b2 , b3 )| + |E(D , b1 , b2 , b3 )| + |E(D , b1 , b2 , b3 )|) 0 (mod 4).

This completes the proof of Lemma 3.4.2, and hence completes the proof of Theorem 3.4.1. Theorem 3.4.1 was extended in [24], where it was proved that if k ≥ 4, and G is a graph whose edge set can be decomposed into a Hamilton cycle and a family of vertex disjoint complete graphs of order at most k, then G is k-colourable. A different proof of Theorem 3.3.2 is given in [49].

3.5

Bound for AT (G) in terms of degree

It is obvious that for any graph G, col(G) ≤ ∆(G)+1. Moreover if G is connected, and G is not a regular graph, we have col(G) ≤ ∆(G). However if G is a regular graph, then col(G) = ∆(G) + 1. Theorem 3.5.3 below implies that even if G is a regular graph, we still have AT (G) ≤ ∆(G), provided that G is neither a complete graph nor an odd cycle. A graph G is called degree choosable if for any list assignment L for which |L(v)| = dG (v) for every vertex v, then G is L-colourable. We say that G is degree AT if G is f -AT for f (v) = dG (v) for every vertex v of G. By CNS, if a graph G is degree-AT, then it is degree choosable. A connected graph G is called a Gallai tree if every block of G is either a complete graph or an odd cycle.

FIGURE 3.9: A Gallai tree. Lemma 3.5.1 If G is a Gallai tree, then G is not degree choosable, and hence not degree-AT.

“2020-12-10” — 2021/5/8 — 18:39 — page 36 — #48

i

i

36

Combinatorial Nullstellensatz

Proof. Assume that G is a Gallai tree. The proof is by induction on the number of blocks of G. If G is a complete graph Kn , then each vertex has degree n − 1, and G is not (n − 1)-colourable, and hence not (n − 1)-choosable. If G is an odd cycle, then each vertex has degree 2 and G is not 2-colourable, and hence not 2-choosable. Assume that G has more than one block. Let B be a leaf block of G, i.e., B contains exactly one cut-vertex v. Let G0 = G − (V (B) − {v}). Then G0 is a Gallai tree and hence is not degree choosable. Let L0 be a list assignment of G0 with |L0 (x)| = dG0 (x) for each vertex x of G0 so that G0 is not L0 -colourable. Assume that B is k-regular and let C be a set of k colours disjoint from L0 (v). Let L(x) = L0 (x) for x ∈ V (G0 ) − {v}, L(v) = L0 (v) ∪ C and L(x) = C for x ∈ V (B) − {v}. Then |L(x)| = dG (x) for every vertex x. It remains to show that G is not L-colourable. Assume to the contrary that φ is an L-colouring of G. If φ(v) ∈ C, then the restriction of φ to B is a k-colouring of B, a contradiction. If φ(v) ∈ / C, then the restriction of φ to G0 is an L0 -colouring of G0 , again a contradiction. It was proved in [28] that the converse of Lemma 3.5.1 is also true. Before proving this result, we first prove the following lemma. Lemma 3.5.2 If G is a 2-connected graph which is neither complete nor an odd cycle, then G contains an even cycle which contains at most one chord. Proof. Let S be a minimum vertex cut of G. As G is 2-connected, |S| ≥ 2. Let u, v be two vertices of S. Let G1 , G2 be the two induced subgraphs of G with V (G1 ) ∩ V (G2 ) = S and V (G1 ) ∪ V (G2 ) = V (G). Let Pi be a shortest path in Gi −(S−{u, v})−E(S) connecting u and v. Then each Pi has length at least 2, and the union C = P1 ∪ P2 is a cycle of length at least 4. If C is an even cycle, then we are done, as C has at most one chord uv. So assume that C is an odd cycle. If e = uv is an edge, then either P1 ∪ e or P2 ∪ e is an induced even cycle and we are done. Assume therefore that uv is not an edge of G. So C is an induced odd cycle. As G is not an odd cycle, we may assume that G1 − C 6= ∅. If each vertex of G1 − C has at most one neighbour in C, then let P3 be a shortest path in G1 − E(C) connecting two vertices of C. Then the union P1 ∪ P2 ∪ P3 consists of three paths joining the two end vertices of P3 . Hence it contains an induced even cycle, and we are done. Thus we may assume that G1 − C has a vertex

“2020-12-10” — 2021/5/8 — 18:39 — page 37 — #49

i

i

Alon–Tarsi Theorem and Its Applications

37

w which has at least two neighbours in C. Let u1 , u2 , . . . , uk be the neighbours of w in C, in this order. For i = 1, 2, . . . , k, let Qi be the subpath of C connecting ui and ui+1 (with uk+1 = u1 ). If Qi has even length, then Qi ∪ {w} induces an even cycle and we are done. So assume that each Qi has odd length. If k = 3, |Q1 | = 3, |Q2 | = |Q3 | = 1, then Q2 ∪C3 ∪w is an even cycle with one chord. Otherwise, Q1 ∪ Q2 ∪ {w} induces an even cycle with one chord wu2 . Theorem 3.5.3 A connected graph G is degree-AT if and only if G is not a Gallai tree. Proof. By Lemma 3.5.1, it suffices to show that if G is not a Gallai tree, then G is degree-AT. Assume that G is a connected graph which is not a Gallai tree. By Lemma 3.5.2, G has an even cycle C which contains at most one chord. Let G0 be obtained from G by contracting C into a single vertex w. Order the vertices of G0 so that x precedes y if dG0 (y, w) > dG0 (x, w), and order x, y arbitrarily if dG0 (y, w) = dG0 (x, w). Then orient edge e = xy from x to y if x precedes y. Observe that each vertex other than w has at least one in-neighbour. Extend this orientation to G, so that C is oriented as a directed cycle, and if C has a chord, then the chord is oriented arbitrarily. Denote the resulting orientation by D. Then each vertex has at least one in-neighbour. So d+ D (x) < dG (x) for each vertex x. There are at least two even Eulerian subdigraphs in D: the empty subdigraph and the directed cycle C. On the other hand, there is at most one odd Eulerian subgraph in D, which is induced by the chord of C together with one path of C connecting the two ends of the chord. So |Ee (D)| − |Eo (D)| 6= 0. As |L(x)| = dG (x) > d+ D (x), we conclude that G is degree-AT.

3.6

Planar graphs

It was proved first by Voigt [62] that there are planar graphs that are not 4-choosable. The following example is modified from an example given by Choi and Kwon [17]. Theorem 3.6.1 There exists a non-4-choosable planar graph.





“2020-12-10” — 2021/5/8 — 18:39 — page 38 — #50

38

Combinatorial Nullstellensatz

Proof. Let H be the graph, and L, the list assignment of H as given in Figure 3.10.

FIGURE 3.10: The graph H. First we show that H is not L-colourable. Assume to the contrary that φ is an L-colouring of H. Since φ(u) = 1 and φ(v) = 2, we conclude that φ(x4 ) ∈ {4, 5}. If φ(x4 ) = 5, then each of the three vertices in the triangle {x1 , x2 , x3 } needs to be coloured by colours 3 and 4, a contradiction. If φ(x4 ) = 4, then each of the three vertices in the triangle {x5 , x6 , x7 } needs to be coloured by colours 3 and 5, again a contradiction. Now let G be the planar graph obtained from K4 by adding, for every ordered pair of vertices (x, y) of K4 , a copy of H, identifying u with x and identifying v with y. Let L(x) = {1, 2, 3, 4} for each vertex x of K4 , and for each copy of H and for each x ∈ V (H) − {u, v}, let L(x) be as defined above. Then L is a 4-list assignment of G. Assume that φ is an L-colouring of G. Then one ordered pair of vertices of K4 is coloured by 1 and 2. But such a colouring cannot be extended to an L-colouring of the copy of H associated with this pair of vertices. So G is not L-colourable. This non-4-choosable planar graph has 88 vertices. The smallest known non-4-choosable planar graph, due to Mirzakhani [48], has 63 vertices. Since G is not 4-choosable, we have AT (G) ≥ 5. The following theorem shows that AT (G) ≤ 5 for every planar graph G. Theorem 3.6.2 [67] If G is a planar graph, then AT (G) ≤ 5.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 39 — #51

Alon–Tarsi Theorem and Its Applications

39

Recall that ch(G) ≤ AT (G) so that Theorem 3.6.2 implies the following well-known result of Carsten Thomassen [60]: For every planar graph G, ch(G) ≤ 5. In other words, every planar graph is 5-choosable. Theorem 3.6.2 is a consequence of the following more technical result. Theorem 3.6.3 Assume that G is a plane graph and e = v1 v2 is a boundary edge of G. Let f (v1 ) = f (v2 ) = 1, f (v) = 3 for every other boundary vertex v and f (v) = 5 for every interior vertex v. Then G − e is f -AT. Proof. As per our assumption, G is a plane graph and e = v1 v2 is a boundary edge of G. Let f : V (G) → Z be defined as above. An AT-orientation D of G − e with d+ D (v) ≤ f (v) − 1 is called a nice orientation for (G, e). We shall prove that for any planar graph G and for any boundary edge e of G, (G, e) has a nice orientation. Assume that the theorem is not true and G is a minimum counterexample. If G has a cut vertex v, then let G1 , G2 be induced subgraphs of G with V (G1 ) ∩ V (G2 ) = {v} and V (G1 ) ∪ V (G2 ) = V (G). Assume that e ∈ E(G1 ). Let u be a boundary neighbour of v in G2 and e0 = vu. By the minimality of G, (G1 , e) has a nice orientation D1 , and (G2 , e0 ) has a nice orientation D2 . Let D be obtained from D1 ∪ D2 by adding the arc (u, v). Edges in D2 ∪ {(u, v)} incident to v are not contained in any directed cycles (since d+ D2 (v) = 0), and hence are not contained in any Eulerian subdigraph of D. Therefore |Ee (D)| = |Ee (D1 )| × |Ee (D2 )| + |Eo (D1 )| × |Eo (D2 )|, |Eo (D)| = |Ee (D1 )| × |Eo (D2 )| + |Eo (D1 )| × |Ee (D2 )|. So diff(D) = diff(D1 ) × diff(D2 ) 6= 0. Hence D is a nice orientation of (G, e). Thus we may assume that G is 2-connected and the boundary of G is a simple cycle BG = (v1 , v2 , . . . , vn ). If BG has a chord e0 = xy, then let G1 , G2 be the two e0 components with e ∈ G1 (i.e., G1 , G2 are induced subgraphs with V (G1 ) ∩ V (G2 ) = {x, y} and V (G1 ) ∪ V (G2 ) = V (G)). By the induction hypothesis, (G1 , e) has a nice orientation D1 , and (G2 , e0 ) has a nice orientation D2 . Let D = D1 ∪ D2 . The same argument as above shows that D is a nice orientation of (G, e).

“2020-12-10” — 2021/5/8 — 18:39 — page 40 — #52

i

i

40

Combinatorial Nullstellensatz

Assume that G has no chord. Let G0 = G − vn . Let v1 , u1 , u2 , . . . , uk , vn−1 be the neighbours of vn . By induction hypothesis, (G0 , e) has a nice orientation D0 . Let D be obtained from D0 by adding the arcs (vn , v1 ) and (vn , vn−1 ) and (ui , vn ) for i = 1, 2, . . . , k. We denote the arc (ui , vn ) by ai . Then D satisfies the out-degree condition for being a nice orientation of (G, e). So if D is an AT-orientation, then we are done. Assume that D is not an AT-orientation. For i = 1, 2, . . . , k, let E(D, ai ) denote the set of Eulerian subdigraphs H of D which contains the arc ai = (ui , vn ). Then Ee (D) = Ee (D0 )∪

k [

Ee (D, ai ), and Eo (D) = Eo (D0 )∪

k [

Eo (D, ai ).

i=1

i=1

Since vn has only one out arc, namely (vn , vn−1 ), which can be contained in a directed cycle, each Eulerian sub-digraph contains at most one in arc of vn . Thus for i 6= j, Ee (D, ai ) ∩ Ee (D, aj ) = ∅, Eo (D, ai ) ∩ Eo (D, aj ) = ∅, i.e., the unions above are disjoint unions. So 0

|Ee (D)| = |Ee (D )| +

k X

|Ee (D, ai )|, and |Eo (D)| = |Eo (D0 )|

i=1

+

k X

|Eo (D, ai )|.

i=1

Hence 0 = diff(D) = diff(D0 ) +

k X

diff(E(D, ai )).

i=1

As diff(D0 ) 6= 0, there exists i ∈ {1, 2, . . . , k} such that diff(E(D, ai )) 6= 0. Assume that diff(E(D, ai )) 6= 0, where i ∈ {1, 2, . . . , k}. Then D has a directed cycle C containing ai . Let D00 be obtained from (D−C)∪C R by reversing the arc aR i (thus the edge ui vn is oriented as (ui , vn ) in D00 ). It is straightforward to verify that D00 satisfies the out-degree conditions for being a nice orientation for (G, e). However, we shall show that D00 is an AT-orientation, and hence D00 is indeed a nice orientation of (G, e).

“2020-12-10” — 2021/5/8 — 18:39 — page 41 — #53

i

i

Alon–Tarsi Theorem and Its Applications

41

Observe that no arc incident to vn is contained in a directed cycle of D00 . Hence none of these arcs is contained in an Eulerian subdigraph of D00 . For each Eulerian subdigraph H of D00 , let φ(H) = (H − C R ) ∪ (C R − H)R . Then φ(H) is an Eulerian subdigraph of D containing ai . Conversely, if H is an Eulerian subdigraph of D containing arc ai , then ψ(H) = (H − C) ∪ (C − H)R is an Eulerian subdigraph of D00 . Further, ψ(φ(H)) = H. So φ is a one-to-one correspondence between E(D00 ) and E(D, ai ). Moreover, if C is an even cycle, then φ(H) and H have the same parity, and if C is an odd cycle, then φ(H) and H have opposite parities. Therefore diff(D00 ) = ±diff(E(D, ai )) 6= 0. Hence D00 is an AT-orientation and hence a nice orientation of (G, e). This completes the proof of Theorem 3.6.3.

3.7

Planar graph minus a matching

A d-defective colouring of a graph G is a colouring of the vertices of G such that each colour class induces a subgraph of maximum degree at most d. Thus, a 0-defective colouring of G is simply a proper colouring of G, while in a 1-defective colouring, a matching is allowed as a set of non-properly coloured edges. We say G is d-defective k-colourable if there is a d-defective colouring of G using a total number of k colours. Given a k-list assignment L of G, a d-defective L-colouring of G is a d-defective colouring c of G with c(v) ∈ L(v) for every vertex v of G. A graph G is d-defective k-choosable if for any klist assignment L of G, there exists a d-defective L-colouring of G. Clearly, every d-defective k-choosable graph is d-defective kcolourable. The converse is not true. It was proved by Cowen, Cowen and Woodall in [18] that every planar graph is 2-defective 3-colourable, and every outerplanar graph is 2-defective 2-colourable. Eaton and Hull [21], and ˇ Skrekovski [58] independently extended these results to list colouring, and showed that every planar graph is 2-defective 3-choosable, and every outerplanar graph is 2-defective 2-choosable. Cushing and Kierstead [19] proved that every planar graph is 1-defective ˇ 4-choosable. The result of Eaton and Hull and Skrekovski, and the result of Cushing and Kierstead can be formulated as follows:

“2020-12-10” — 2021/5/8 — 18:39 — page 42 — #54

i

i

42

Combinatorial Nullstellensatz

Assume that G is a planar graph. For any 3-list assignment L of G, there is a subgraph H of G with ∆(H) ≤ 2 such that G−E(H) is L-colourable; For any 4-list assignment L of G, there is a subgraph H of G with ∆(H) ≤ 1 such that G − E(H) is L-colourable. The subgraph H of G depends on the list assignment L. It is natural to ask whether one can choose a subgraph H of G independent of the list assignment L. In other words, we have the following questions: Question 3.7.1 Does every planar graph G have a subgraph H with ∆(H) ≤ 2 such that G\E(H) is 3-choosable? Question 3.7.2 Does every planar graph G have a subgraph H with ∆(H) ≤ 1 such that G\E(H) is 4-choosable? It turns out that the answer to Question 3.7.1 is negative, and the answer to Question 3.7.2 is positive. First we show that there is a planar graph G such that for any subgraph H of G with ∆(H) ≤ 3, G\E(H) is not 3-choosable. The following example was given in [35]. Let J1 and J2 be the two graphs depicted in Figure 3.11. For i = 1, 2, the edge ab in Ji is called the handle of Ji . a

a c

d

b

e

c

d

e

b J1

J2

FIGURE 3.11: The graphs J1 and J2 . Let J be the set of graphs obtained from the disjoint union of 6 copies of J1 or J2 by identifying the edges corresponding to ab from each copy. For each G ∈ J , let ci , di , ei be the vertices corresponding to c, d, e, respectively, for i ∈ [6], and the edge ab in G is called the handle of G. (See Figure 3.12.) Lemma 3.7.3 For i = 1, 2, there is a list assignment L of Ji such that |L(a)| = |L(b)| = 1, |L(c)| = |L(d)| = |L(e)| = 3 and Ji is not L-colourable.

“2020-12-10” — 2021/5/8 — 18:39 — page 43 — #55

i

i

Alon–Tarsi Theorem and Its Applications

43

a c1

d1

e1

c2

d2

e2

c6

d6

e6

b FIGURE 3.12: This illustrates a graph in J . For each i ∈ [6], the subgraph induced on {a, b, ci , di , ei } is isomorphic to J1 or J2 with isomorphism mapping a, b, ci , di , ei to a, b, c, d, e, respectively. That is, there exists exactly one edge between {a, b} and di . Proof. We consider the graph J1 . Let L(a) = {1}, L(b) = {2}, L(c) = {1, 2, 3}, L(d) = {1, 3, 4}, L(e) = {1, 2, 4}. It is straightforward to check that J1 is not L-colourable. Lemma 3.7.4 None of the graphs G ∈ J is 3-choosable. Proof. We construct a 3-list assignment of G as follows: L(a) = L(b) = {α, β, γ}. For each pair (x, y) of distinct colours from {α, β, γ}, there is a copy of J1 or J2 whose lists are given in such a way that there is no L-colouring φ of this copy of J1 or J2 with φ(a) = x and φ(b) = y. By Lemma 3.7.3, such a list assignment exists. Therefore G is not L-colourable. To be precise, the 3-list assignment L of G is defined as follows. Let (xi , yi , zi )i=1,...,6 be the six permutations of the colour set {α, β, γ}. • L(a) = L(b) = {α, β, γ}. • For each i ∈ [6], L(ci ) = {α, β, γ} and L(ei ) = {xi , yi , ω}. • For each i ∈ [6], L(di ) = {xi , zi , ω} if di is adjacent to a and L(di ) = {yi , zi , ω} if di is adjacent to b. Suppose there exists an L-colouring φ of G. We may assume that φ(a) = α and φ(b) = β. Without loss of generality, let x1 = α and y1 = β. Since L(c1 ) = {α, β, γ} and L(e1 ) = {x1 , y1 , ω} = {α, β, ω}, we have φ(c1 ) = γ and φ(e1 ) = ω. If d1 is adjacent to a, then L(d1 ) = {α, γ, ω} but φ(a) = α, φ(c1 ) = γ and φ(e1 ) = d, so there is no available colour for d1 . Similarly, if d1 is adjacent to b, then L(d1 ) = {β, γ, ω} but φ(b) = β, φ(c1 ) = γ and φ(e1 ) = ω, so there is no available colour for d1 . By the construction of

“2020-12-10” — 2021/5/8 — 18:39 — page 44 — #56

i

i

44

Combinatorial Nullstellensatz

G, d1 is adjacent to either a or b, therefore, there is no possible colour for d1 . This leads to a contradiction. Therefore G is not 3-choosable. a

h c

i d

j

e

f

g

k

b FIGURE 3.13: The graph J3 . Lemma 3.7.5 Assume that H is a subgraph of J3 (the graph depicted in Figure 3.13) with maximum degree at most three. If H does not contain any edge incident with a or b, then J3 − E(H) contains a copy of K4 , or a subgraph isomorphic to J1 or J2 with handle ab. Proof. Assume that H is a subgraph of J3 which does not contain any edge incident with a or b. If any of the edges cd, de, ef, f g is not contained in H, then J3 − E(H) contains a copy of K4 and we are done. Thus we assume that H contains {cd, de, ef, f g}. If H contains neither hd nor he, then the edge set {ab, ad, ah, ae, hd, he, bd, be} induces a copy of J1 with handle ab in J3 − E(H), and we are done. So we assume that H contains hd or he. Similarly, assume that H contains ie or if , jd or je, and ke or kf . Therefore |E(H) ∩ {hd, he, ie, if, jd, je, ke, kf }| ≥ 4. Since every edge in {hd, he, ie, if, jd, je, ke, kf } is incident with d, e or f , it follows from the pigeonhole principle that one of d, e and f is an end of at least two edges in E(H) ∩ {hd, he, ie, if, jd, je, ke, kf }. It implies that one of d, e and f has degree at least four in H because the edges cd, de, ef and f g are already contained in H, which yields a contradiction. This completes the proof. Let S be the graph obtained from nine copies of J3 by identifying the edges corresponding to ab from each copy. It is obvious

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 45 — #57

Alon–Tarsi Theorem and Its Applications

45

that S is a planar graph. The edge ab in S is called the handle of S. We obtain the following as a corollary of Lemma 3.7.5. Corollary 3.7.6 Assume that H is a subgraph of S with maximum degree at most three. If H does not contain any edge incident with a in S, then S − E(H) contains a copy of K4 or a member of J as a subgraph. Proof. Let S1 , S2 , . . . , S9 be the distinct subgraphs of S isomorphic to J3 with handle ab. Since H has maximum degree at most three, and H does not contain any edge incident with a in S, there are 1 ≤ i1 < . . . < i6 ≤ 9 such that every edge incident with b in Sij is not contained in H for j ∈ [6]. Without loss of generality, let ij = j for j ∈ [6]. Then, by Lemma 3.7.5, for each j ∈ [6], Sj − E(H) contains a copy of K4 or a subgraph isomorphic to J1 or J2 with handle ab. We are done if Sj − E(H) contains K4 , so, we may assume that Sj − E(H) has a subgraph Sj0 isomorphic to J1 or J2 with handle ab. Then, combining Sj0 for j ∈ [6], we obtain a subgraph of S − E(H) isomorphic to a member of J . This completes the proof. Now, we construct a planar graph G such that for every subgraph H of G with maximum degree at most 3, G − E(H) is not 3-choosable: Start with a star with four leaves v1 , v2 , v3 , v4 and centre c, and for each i ∈ [4], we add a copy of S with handle cvi to the star. That is, G consists of four edge-disjoint copies S1 , S2 , S3 , S4 of S where the handle of Si is cvi for i ∈ [4]. Theorem 3.7.7 For every subgraph H in G1 with ∆(H) ≤ 3, G1 − E(H) is not 3-choosable. Proof. Let H be a subgraph in G with ∆(H) ≤ 3. We claim that G − E(H) contains K4 or a member of J . Then, by the fact that K4 is not 3-colourable (so not 3-choosable) and Lemma 3.7.4, Theorem 3.7.7 follows. By adding isolated vertices to H, we consider H as a spanning subgraph of G with maximum degree at most three. Since c has degree at most three in H, there exists i ∈ [4] such that H does not contain any edge incident with c in Si . Then, by Corollary 3.7.6, Si − E(H) contains K4 or a member of J . This completes the proof. Next we show that Question 3.7.2 has a positive answer.

“2020-12-10” — 2021/5/8 — 18:39 — page 46 — #58

i

i

46

Combinatorial Nullstellensatz

Theorem 3.7.8 Every planar graph G contains a matching M such that AT (G\M ) ≤ 4. Theorem 3.7.8 was proved in [26], where the proof is a direct calculation of the coefficient of a certain monomial in the expansion of fG (x). Here we present a proof which is basically the same as the proof of Theorem 3.6.2, except that the technical statement is different. Definition 3.7.9 Assume that G is a plane graph, e = v1 v2 is a boundary edge of G, and M is a matching in G containing e. An orientation D of G\M is nice for (G, e, M ) if the following conditions hold: 1. D is an AT-orientation. + 2. d+ D (v1 ) = dD (v2 ) = 0.

3. d+ D (v) ≤ 2 − dM (v) for every other boundary vertex v. 4. d+ D (v) ≤ 3 for each interior vertex v. Notice that dM (v) = 1 if v is covered by M , and dM (v) = 0 otherwise. Theorem 3.7.8 follows from Theorem 3.7.10 given below. Theorem 3.7.10 Assume that G is a plane graph and e = v1 v2 is a boundary edge of G. Then G has a matching M containing e such that there exists a nice orientation D for (G, e, M ). Proof. Assume that the theorem is false and G is a minimum counterexample. The same argument as in the proof of Theorem 3.6.3 shows that G has no cut-vertex and the boundary cycle BG of G has no chord. Let G0 = G − vn . Let v1 , u1 , u2 , . . . , uk , vn−1 be the neighbours of vn . By induction hypothesis, G0 has a matching M 0 such that (G0 , e, M 0 ) has a nice orientation D0 . Let D be obtained from D0 by adding arcs (vn , v1 ) and (vn , vn−1 ) and arcs ai = (ui , vn ) for i = 1, 2, . . . , k. If D is an AT-orientation, then it is a nice orientation for (G, e, M 0 ) and we are done. Assume that D is not an AT-orientation. For i = 1, 2, . . . , k, E(D, ai ) is the set of Eulerian subdigraphs H of D which contains the arc ai = (ui , vn ). Then Ee (D) = Ee (D0 )∪

k [ i=1

Ee (D, ai ), and Eo (D) = Eo (D0 )∪

k [ i=1

Eo (D, ai ).

“2020-12-10” — 2021/5/8 — 18:39 — page 47 — #59

i

i

Alon–Tarsi Theorem and Its Applications

47

Moreover the unions are disjoint unions. So |Ee (D)| = |Ee (D0 )| +

k X

|Ee (D, ai )|, and |Eo (D)| = |Eo (D0 )|

i=1

+

k X

|Eo (D, ai )|.

i=1

Now |Ee (D)| = |Eo (D)| and |Ee (D0 )| = 6 |Eo (D0 )| imply that |Ee (D, ai )| = 6 |Eo (D, ai )| for some i ∈ {1, 2, . . . , k}. Assume that |Ee (D, ai )| 6= |Eo (D, ai )|, where i ∈ {1, 2, . . . , k}. Then D has a directed cycle C containing ai . If d+ D (ui ) ≤ 1, then let D00 be obtained from (D−C)∪C T by reversing the arc aR i (thus the edge ui vn is oriented as (ui , vn ) in D00 ). The same argument as in the proof of Theorem 3.6.2 shows that D00 is an AT-orientation. + 00 As d+ D 00 (ui ) = dD (ui )+2 ≤ 3, D satisfies the out-degree conditions of being a nice orientation for (G, e, M 0 ), and hence D00 is indeed a + nice orientation for (G, e, M 0 ). Otherwise d+ D (ui ) = 2. As dD (ui ) ≤ 2−dM 0 (ui ), we conclude that dM 0 (ui ) = 0, i.e., ui is not covered by M 0 . Let M = M 0 ∪ {ui vn }. Then M is a matching in G containing e. Let D00 be obtained from (D − C) ∪ C T by deleting the arc aR i . + 00 Then d+ (u ) = d (u ) + 1 = 3, and hence D satisfies the outi D 00 D i degree conditions of being a nice orientation for (G, e, M ). Again, the same argument as in the proof of Theorem 3.6.2 shows that D00 is an AT-orientation. Hence D00 is indeed a nice orientation for (G, e, M ).

3.8

Discharging method

The following lemma is a common tool used in proving that a graph G has an f -AT orientation D for some f : V (G) → N. Lemma 3.8.1 Assume that G is a graph, X is a subset of V (G) and f : V (G) → N is a mapping which assigns to each vertex v a positive integer f (v). For v ∈ X, let s(v) be the number of edges of G connecting v and V (G)\X and let g(v) = f (v) − s(v). If G − X has an f -AT orientation, and G[X] has a g-AT orientation, then G has an f -AT orientation.

“2020-12-10” — 2021/5/8 — 18:39 — page 48 — #60

i

i

48

Combinatorial Nullstellensatz

Proof. Let D1 be an f -AT orientation of G−X and D2 be a g-ATorientation of G[X]. Let D be obtained from D1 ∪ D2 by orienting all the edges between X and V (G)\X from X to V (G)\X. Then for each vertex v of G, we have d+ D (v) ≤ f (v)−1. Since no arcs between X and V (G)\X is contained in a directed cycle of D, and hence not contained in any Eulerian subdigraph of D, we conclude that diff(D) = diff(D1 ) × diff(D2 ) 6= 0. Hence D is an f -AT orientation of G. Discharging method is often used to find the existence of a subset X for which we can apply Lemma 3.8.1. This is explained in the proof of Theorem 3.8.2, which is given in [46]. For a positive integer l, we denote by P4,l the family of planar graphs with no cycles of length 4 and l. Theorem 3.8.2 Every graph G ∈ P4,5 has a matching M such that AT (G − M ) ≤ 3. For the purpose of using induction, instead of proving Theorem 3.8.2 directly, we shall prove a stronger and more technical result. Definition 3.8.3 Assume that G is a plane graph and v0 is a vertex on the boundary of G. A valid matching of (G, v0 ) is a matching M of G that does not cover v0 . Definition 3.8.4 Let G be a plane graph and v0 be a vertex on the boundary of G. An orientation D of G is good with respect to + v0 , if D is AT with ∆+ D (v) < 3 and dD (v0 ) = 0. We shall prove the following result, which implies Theorem 3.8.2. Theorem 3.8.5 Assume that G ∈ P4,5 , v0 is a vertex on the boundary of G. Then (G, v0 ) has a valid matching M such that there is a good orientation D of G − M with respect to v0 . Assume that Theorem 3.8.5 is not true and G is a counterexample with |V (G)| + |E(G)| minimum among all counterexamples. Let f0 denote the outer face of G. Lemma 3.8.6 G is 2-connected. Moreover, d(v) ≥ 3 for all v ∈ V (G)\{v0 }. Proof. Assume that G has a cut vertex. We may choose a block B of G that contains a unique cut vertex z ∗ and does not contain

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 49 — #61

Alon–Tarsi Theorem and Its Applications

49

v0 . Let G1 = G − (B − {z ∗ }). By the minimality of G, (G1 , v0 ) has a valid matching M1 and there is a good orientation D1 of G1 − M1 , and (B, z ∗ ) has a valid matching M2 and there is a good orientation D2 of B − M2 . Let M = M1 ∪ M2 and D = D1 ∪ D2 . Applying Lemma 3.8.1 (with X = V (B) − {z ∗ }), D is an ATorientation. So M is a valid matching of (G, v0 ), and G − M has a good orientation, a contradiction. If dG (v) ≤ 2, then G0 = G−v has a valid matching M such that 0 G − M has a good orientation D0 . Extending D0 to an orientation D of G in which v is a source vertex, it is obvious that D is a good orientation of G − M . Lemma 3.8.7 G does not contain two adjacent 3-vertices u and v, with u 6= v0 and v 6= v0 . Proof. Assume to the contrary that uv ∈ E(G) with d(u) = d(v) = 3 and u 6= v0 and v 6= v0 . Let G∗ = G − {u, v}. Then (G∗ , v0 ) has a valid matching M ∗ such that there exists a good orientation D∗ of G∗ − M ∗ . Let M = M ∗ ∪ {uv}. Then M is a valid matching of (G, v0 ). Extend D∗ to an orientation D of G − M in which u, v are sources. Then D is a good orientation of G − M . For positive integers a, b, c, an (a, b, c)-triangle is a triangle in G whose three vertices have degrees a, b, c, respectively. An a-vertex (respectively, an a+ -vertex or an a− -vertex) is a vertex in G of degree a (respectively, at least a or at most a). Definition 3.8.8 A 3-face f = [uvw] with v0 ∈ / {u, v, w} is called a minor triangle if f is a (3, 4, 4)-face. A 3-vertex v 6= v0 is called a minor 3-vertex if v is incident to a triangle. Definition 3.8.9 A triangle chain in G of length k is a subgraph of G − v0 consisting of vertices w1 , w2 , . . . , wk+1 , u1 , u2 , . . . , uk in which [wi wi+1 ui ] is a (4, 4, 4)-face for i = 1, 2, . . . , k. We denote by Ti the triangle [wi wi+1 ui ] and denote such a triangle chain by T1 T2 . . . Tk . Observe that a single [444]-triangle is also a triangle chain. We say that a triangle T intersects a triangle chain T1 T2 . . . Tk , if T has one common vertex with T1 . Lemma 3.8.10 If a minor triangle T0 intersects a triangle chain T1 T2 . . . Tk , then no vertex of Tk is adjacent to a 3-vertex, except possibly v0 ; and no vertex of a minor triangle is adjacent to a 3vertex, except possibly v0 .

“2020-12-10” — 2021/5/8 — 18:39 — page 50 — #62

i

i

50

Combinatorial Nullstellensatz u

D T

T

u

u

wk

wk

z

Tk

T

w

x

uk

u

T

w

Tk

w

u

T

uk

T

w

w

F

wk

u

T

T

w

wk

w

u

w

Tk

w

w

E

uk

u

w

wk

T

wk

y

x

FIGURE 3.14: (a) A triangle chain. (b) A triangle chain that intersects a minor triangle and adjacent to a 3-vertex. (c) A triangle chain that intersects a minor triangle and has distance 1 to another minor triangle. Proof. Assume to the contrary that T0 = [w0 w1 u0 ] is a minor triangle that intersects a triangle chain T1 T2 . . . Tk , with Ti = [wi wi+1 ui ] (1 ≤ i ≤ k), and wk+1 has a 3-neighbour x, as depicted in Figure 3.14(b). Assume that w0 is a 3-vertex. Let X = ∪ki=0 V (Ti ) ∪ {x} and G0 = G − X. By the minimality of G, (G0 , v0 ) contains a valid matching M 0 and there is a good orientation D0 of G0 − M 0 . Let M = M 0 ∪ {w0 u0 , w1 u1 , . . . , wk uk , wk+1 x}. Then M is a valid matching of (G, v0 ). Let D be an orientation of G − M obtained from D0 by adding arcs (wi , wi+1 ) and (wi+1 , ui ) for i = 0, 1, . . . , k, and all the edges between X and V \X are oriented from X to V \X. Since D[X] is acyclic, D[X] is AT. By Lemma + 3.8.1, D is AT. It is easy to see that ∆+ D (v) < 3 and dD (v0 ) = 0. Thus D is a good orientation of G − M . Lemma 3.8.11 If a triangle chain T1 T2 . . . Tk intersects a minor triangle T0 , then the distance between Tk and another minor triangle is at least 2, and the distance between two minor triangles is at least 2.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 51 — #63

Alon–Tarsi Theorem and Its Applications

51

Proof. Assume to the contrary that T1 T2 . . . Tk with Ti = [wi wi+1 ui ] (1 ≤ i ≤ k) is a triangle chain that intersects a minor triangle T0 = [w0 w1 u0 ], and the distance between Tk and another minor triangle T00 = [xyz] with d(x) = 3 is less than 2. Combining with Lemma 3.8.10, assume wk+1 y is a (4, 4)-edge connecting Tk and T00 , as in Figure 3.14(c). Let X = ∪ki=0 V (Ti ) ∪ V (T00 ) and G0 = G − X. Then (G0 , v0 ) has a valid matching M 0 and there is a good orientation D0 of G0 − M 0 . Let M = M 0 ∪ {w0 u0 , w1 u1 , . . . , wk uk , wk+1 y, xz}. Then M is a valid matching of (G, v0 ). Let D be an orientation of G − M obtained from D0 by adding arcs (x, y), (y, z), (wi , wi+1 ) and (wi+1 , ui ) for i = 0, 1, . . . , k, and all the edges between X and V \X are oriented from X to V \X. Obviously, D[X] is acyclic, so D[X] is AT. By Lemma 3.8.1, D is a good orientation of G − M , a contradiction. Lemma 3.8.12 Assume that f is a 6-face of G which is adjacent to five triangles, and none of the vertices in these triangles is v0 . If f has one 3-vertex, then there is at least one 5+ -vertex on the five triangles. Proof. Let f = [v1 v2 v3 v4 v5 v6 ], v1 be a 3-vertex and Ti = [vi vi+1 ui ] (i = 1, 2, . . . , 5) be the five triangles (see Figure 3.15(a)). Assume to the contrary that there is no 5+ -vertex on Ti for i = 1, 2, 3, 4, 5. By Lemma 3.8.10, we may assume all vi+1 and ui are 4-vertices for i = 1, 2, . . . , 5. Let X = ∪5i=1 V (Ti ) and G0 = G − X. Then (G0 , v0 ) has a valid matching M 0 and there is a good orientation D0 of G0 − M 0 . Let M = M 0 ∪{v1 u1 , v2 u2 , . . . , v5 u5 }. Then M is a valid matching of (G, v0 ). Let D be the orientation of G − M obtained from D0 by adding arcs (vi+1 , ui ), (vi , vi+1 ) for i = 1, . . . , 5 and (v1 , v6 ) and all the edges between X and V \X are oriented from X to V \X (see Figure 3.15(b)). Clearly, ∆+ D (v) < 3 and D is AT by Lemma 3.1.16, a contradiction. Now we use the discharging method to derive a contradiction. Let the initial charge ch be defined as ch(x) = d(x) − 4 for x ∈ V (G) ∪ F (G), of faces of G. ApplyP where F (G) is the set P ing the equalities d(v) = 2|E(G)| = d(f ) and Euler’s v∈V (G)

f ∈F (G)

formula |V (G)| − |E(G)| + |F (G)| = 2, we conclude that X ch(x) = −8. x∈V (G)∪F (G)

“2020-12-10” — 2021/5/8 — 18:39 — page 52 — #64

i

i

52

Combinatorial Nullstellensatz

We shall describe a set of rules that transfer charge from some vertices and faces to other vertices and faces. In the discharging procedure, ch(x → y) denotes the charge discharged from an element x to another element y, ch(x →) and ch(→ x) denote the charge totally discharged from or to x, respectively. The final charge ch∗ (x) of x ∈ V (G) ∪ F (G) is defined as ch∗ (x) = ch(x) − ch(x →) + ch(→ x). By applying appropriate discharging rules, we shall arrive at a final charge that ch∗ (x) ≥ 0 for all x ∈ V (G) ∪ F (G) \ {v0 , f0 }, and ch∗ (v0 ) + ch∗ (f0 ) > −8. As the total charge does not change in the discharging process, this would be a contradiction. u

u u

u

v

v

v

v

v

v u

v

v

u

v v

v

v

u

u

u

u u D

E

FIGURE 3.15: The configuration in Lemma 3.8.12. The discharging rules are as follows: R1 Assume that f 6= f0 is a 3-face. Then each face adjacent to f transfers 13 charge to f . R2 Assume that v 6= v0 is 3-vertex. If v is contained in a triangle, then each of the other two faces incident to v transfers 21 charge to v; otherwise each face incident to v transfers 13 charge to v. R3 Assume that u 6= v0 is a 5+ -vertex and f 6= f0 is a 6-face. (i) If f is incident to u and adjacent to two triangles that are incident to u, then u transfers 13 charge to f . (ii) If f is incident to u and adjacent to exactly one triangle which is incident to u, then u transfers 16 charge to f .

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 53 — #65

Alon–Tarsi Theorem and Its Applications

53

(iii) If f is not incident to u but adjacent to a triangle which is incident to u, then u transfers 16 charge to f . R4 f0 transfers 31 charge to each adjacent triangle, and 21 charge to each incident 3-vertex v 6= v0 . v0 transfers 31 charge to each 6-face f 6= f0 which is either incident to v0 , or is not incident to v0 but adjacent to a triangle T which is incident to v0 . Claim 3.8.13 If a 6-face f has three 3-vertices other than v0 and is adjacent to three triangles, then ch(→ f ) ≥ 12 . Proof. Assume that f = [v1 v2 v3 v4 v5 v6 ]. By Lemma 3.8.7, we may assume that v1 , v3 and v5 are the three 3-vertices other than v0 . Then each of v1 , v3 , v5 is incident to at most one triangle. Hence at most two of the three triangles adjacent to f intersect each other. Thus we may assume that the three triangles adjacent to f are either T1 , T2 , T4 , or T1 , T3 , T5 , where Ti = [vi vi+1 ui ]. Case 1 The three triangles incident to f are T1 , T2 , T4 . If v0 is a vertex of f or T1 , T2 or T4 , then v0 transfers 13 charge to f by R4. By Lemma 3.8.10, at least one of the three triangles has a 5+ -vertex v 6= v0 which sends at least 61 charge to f . So ch(→ f ) ≥ 13 + 16 = 12 . Assume that v0 is not a vertex of f, T1 , T2 or T4 . By Lemma 3.8.10, either v2 is a 5+ -vertex or both of u1 and u2 are 5+ -vertices. In both cases, f receives 31 charge in total from v2 , u1 and u2 . Moreover, by Lemma 3.8.10, either v4 or u4 is a 5+ -vertex, which transfers 16 charge to f . Hence, ch(→ f ) ≥ 13 + 61 = 21 . Case 2 The three triangles incident to f are T1 , T3 , T5 . By Lemma 3.8.10, each of these three triangles has a 5+ -vertex transferring 16 charge to f . Thus, ch(→ f ) ≥ 21 . Claim 3.8.14 If a 6-face f has two 3-vertices other than v0 and is adjacent to four triangles, then ch(→ f ) ≥ 31 . Proof. Assume that f = [v1 v2 v3 v4 v5 v6 ]. If edge vi vi+1 is contained in a triangle, then we denote the triangle by Ti = [vi vi+1 ui ]. If v0 is a vertex of f or Ti , then v0 transfers 13 charge to f by R4. Suppose v0 is not a vertex of f or Ti . By Lemma 3.8.7, we may assume that either v1 and v4 or v1 and v3 are the two 3-vertices. Analysing

“2020-12-10” — 2021/5/8 — 18:39 — page 54 — #66

i

i

54

Combinatorial Nullstellensatz

all the distribution of the four triangles, we need to consider the following five cases: Case 1 The four triangles incident to f are T1 , T2 , T3 , T5 while the two 3-vertices are v1 and v4 . If at least one of v2 and v3 is a 5+ -vertex, then by R3(i), ch(→ f ) ≥ 31 . Assume that both d(v2 ) and d(v3 ) are 4-vertices. By Lemma 3.8.10, at least two of u1 , u2 and u3 are 5+ -vertices each of which transfers 16 charge to f . That is, ch(→ f ) ≥ 13 . Case 2 The four triangles incident to f are T1 , T2 , T4 , T5 while the two 3-vertices are v1 and v4 . By Lemma 3.8.10 and R3, at least one of u1 , u2 , v2 and v3 is a 5+ -vertex transferring at least 61 charge to f . By symmetry, there is another such vertex in u4 , u5 , v5 and v6 . Thus, we are done. Case 3 The four triangles incident to f are T1 , T2 , T4 , T5 while the two 3-vertices are v1 and v3 . If v2 is a 5+ -vertex, then v2 transfers 31 charge to f by R3(i). Assume that v2 is a 4-vertex. From Lemma 3.8.10, both u1 and u2 are 5+ -vertices each of which transfers 16 charge to f . Case 4 The four triangles incident to f are T1 , T3 , T4 , T5 while the two 3-vertices are v1 and v3 . By Lemma 3.8.10, at least one of v2 and u1 is a 5+ -vertex transferring 61 charge to f . Moreover, using Lemma 3.8.10 again, at least one of v4 , v5 , v6 , u3 , u4 and u5 is a 5+ -vertex transferring at least 16 to f . Consequently, ch(→ f ) ≥ 13 . Case 5 The four triangles incident to f are T3 , T4 , T5 , T6 while the two 3-vertices are v1 and v3 . If one of v4 , v5 and v6 is a 5+ -vertex, then such a 5+ -vertex sends 31 charge to f by R3(i). Assume that all of v4 , v5 and v6 are 4-vertices. Then at least two of u3 , u4 , u5 and u6 are 5+ -vertices each sending 16 to f . Otherwise, it will contradict Lemma 3.8.10 or Lemma 3.8.11. Again ch(→ f ) ≥ 13 . Now we check the final charge of vertices v 6= v0 . If v is a 3-vertex, then by R2, v receives 1 from incident 6+ faces. So ch∗ (v) = ch(v) − ch(v →) + ch(→ v) = −1 − 0 + 1 = 0. If v is a 4-vertex, no charge is sent out or received. So ch∗ (v) = ch(v) = 0.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 55 — #67

Alon–Tarsi Theorem and Its Applications

55

If v is a 5+ -vertex, then by the rules, v only transfers charge to faces when it is incident with a triangle. Assume that v is incident with t triangles, then 0 < t ≤ b d(v) 2 c. Let T be a triangle incident with v, f1 be a 6-face not incident to v but adjacent to T , and f2 , f3 be 6-faces incident to v and adjacent to T . Note that, by R3, v transfers 21 charge to those 6-faces around T . This means that v sends out at most 12 t charge. So we have ch∗ (v) = ch(v) − ch(v →) ≥ d(v) − 4 − 12 t ≥ d(v) − 4 − 12 × b d(v) 2 c ≥ 0. Next we check the final charge of faces f 6= f0 . If f is a 3-face, then R1 guarantees ch∗ (f ) ≥ 0. Assume that f is a 6-face. Note that if f has no minor 3-vertex, then f sends out at most 31 × 6 = 2 to others and hence ch∗ (f ) ≥ 0. Assume that f has s minor 3-vertices for 1 ≤ s ≤ 3. When s = 3, f is adjacent to at most three triangles each of which transfers 13 charge from f . By Claim 3.8.13, we have ch∗ (f ) = d(f ) − 4 − ch(f →) + ch(→ f ) ≥ 2 − ( 12 × 3 + 13 × 3) + 12 = 0. When s = 2, we may assume that f has exactly two 3-vertices and is adjacent to four triangles. Otherwise ch(f →) ≤ 21 × 2 + 13 × 3 = 2. Thus, by Claim 3.8.14, we have ch∗ (f ) = d(f ) − 4 − ch(f →) + ch(→ f ) ≥ 2 − ( 12 × 2 + 1 1 3 × 4) + 3 = 0. Next we consider that s = 1. We may assume that f has exactly one 3-vertex and is adjacent to five triangles. Otherwise ch(f →) ≤ 12 + 31 × 4 = 11 6 < 2, and we are done. Thus, ch(f →) = 21 + 13 × 5 = 13 . On the other hand, by Lemma 3.8.12, 6 there is at least one 5+ -vertex transferring 16 charge to f on the 1 five triangles. This gives that ch∗ (f ) ≥ 2 − 13 6 + 6 = 0. + If f is a 7 -face with s 3-vertices, then f is adjacent to at most ) ∗ (d(f ) − s) triangles. Obviously, 0 ≤ s ≤ b d(f 2 c. Hence ch (f ) = 1 1 2 1 d(f ) − 4 − [ 2 × s + 3 × (d(f ) − s)] = 3 d(f ) − 6 s − 4 > 0. Finally we check the charges on f0 and v0 . By R4, v0 transfers at most (d(v0 ) − 1) × 13 charge to others. 7 That is, ch∗ (v0 ) ≥ d(v0 ) − 4 − (d(v0 ) − 1) × 13 = 32 d(v0 ) − 11 3 ≥ −3, as d(v0 ) ≥ 2. The face f0 is incident with at most b d(f20 ) c 3-vertices and each receiving 21 charge from f0 , and f0 is adjacent to at most d(f0 ) triangles and each receiving 13 charge from f0 . Therefore ch∗ (f0 ) ≥ 5 d(f0 ) − 4 − 12 b d(f20 ) c − 13 d(f0 ) ≥ 12 d(f0 ) − 4 ≥ − 11 4 . Consequently, the sum of the final charge of all vertices and faces is X x∈V ∪F

ch∗ (x) ≥ ch∗ (v0 ) + ch∗ (f0 ) ≥ −

61 , 12

“2020-12-10” — 2021/5/8 — 18:39 — page 56 — #68

i

i

56

Combinatorial Nullstellensatz P P contrary to the fact that x∈V ∪F ch∗ (x) = x∈V ∪F ch(x) = −8. This completes the proof of Theorem 3.8.5. Similar argument is used in [46] to show that for l ∈ {6, 7}, every graph G ∈ P4,l has a matching M such that G − M has Alon–Tarsi number at most 3.

3.9

Hypergraph colouring

Assume that k ≥ 2 and H = (V, E) is a k-uniform hypergraph with vertex set V = {v1 , v2 , . . . , vn }. Each hyperedge e ∈ E of H is a set of k vertices of H. Assume that e = {vi1 , vi2 , . . . , vik }, where i1 < i2 < . . . < ik . Then vij is the jth vertex of e. For convenience, we write a hyperedge e = {vi1 , vi2 , . . . , vik } as an injective mapping e : [k] → [n] defined as e(j) = i if the jth vertex of e is vi . A proper colouring of H is a mapping φ which assigns to each vertex vi a colour φ(vi ) such that no hyperedge is monochromatic, i.e., each hyperedge e has at least two vertices that are coloured by distinct colours. Ramamurti and West [52] extended the application of CNS to colouring of uniform hypergraphs. As for graphs, each vertex vi is associated with a variable xi . Let θ be a primitive k−th root of unity. For a hyperedge e, let he (x) = θxe(1) + θ2 xe(2) + . . . + θk xe(k) . Note that θk = θ0 = 1 and θ + θ2 + . . . + θk = 0. The hypergraph polynomial fH for H is defined as Y fH (x) = he (x). e∈E(H)

Assume that φ is a mapping from V to R. For a hyperedge e, if φ(ve(j) ) = c, a constant for j = 1, 2, . . . , k, then he (φ) = he (φ(x1 ), φ(x2 ), . . . , φ(xk )) = c(θ + θ2 + . . . + θk ) = 0. Note that this is a natural generalization of what had been defined for graphs for which k = 2 and θ = −1 = the primitive square root of unity. Hence if fH (φ) 6= 0, then each hyperedge uses at least two distinct colours and φ is a proper colouring of H. However, the converse is not true in general. For example, if k = 2q ≥ 4 is

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 57 — #69

Alon–Tarsi Theorem and Its Applications

57

even, then for e, let φ(ve(q) ) = φ(ve(k) ) = c 6= 0 and φ(ve(j) ) = 0 for j 6= k, q, then he (φ) = 0. However, if k is a prime, then the following holds ([6], p. 405): Lemma 3.9.1 If k is a prime number, and θ is P a primitive k-th k−1 root of unity and a1 , a2 , . . . , ak are rational, then i=0 ai θi = 0 if and only if a0 = a1 = . . . = ak−1 . Thus if k is a prime, then fH (φ) 6= 0 if and only if φ is a proper colouring of H. In the remainder of this section, we assume that k is a prime and H = (V, E) is a k-uniform hypergraph. Definition 3.9.2 An orientation of H is a mapping σ which assigns to each edge e a vertex of e as its source vertex. If v is the source vertex of e, then e is called an out-edge at v. Since each hyperedge e is an injective mapping from [k] to [n], an orientation of H can be encoded as a mapping σ : E → {1, 2, . . . , k}, where σ(e) = j means that the jth vertex of e is the source vertex of e. In other words, the source vertex of e under the orientation σ is ve(σ(e)) . We define the weight of e under the orientation σ as wσ (e) = θσ(e) xe(σ(e)) , and let w(σ) =

Y

wσ (e).

e∈E

For i = 1, 2, . . . , n, let Eσ (vi ) = {e ∈ E(v) : e(σ(e)) = i}, i.e., Eσ (vi ) is the set of out-edges at vi . The number di = |Eσ (vi )| is called the out-degree of vi , and the sequence (d1 , d2 , . . . , dn ) is ~ called the out-degree sequence of σ and is denoted by d(σ). Let X m(σ) = σ(e) (mod k). e∈E

~ It follows from the definition that if d(σ) = (d1 , d2 , . . . , dn ), then w(σ) = θm(σ)

n Y i=1

xdi i .

“2020-12-10” — 2021/5/8 — 18:39 — page 58 — #70

i

i

58

Combinatorial Nullstellensatz

Note that if k = 2, then H is a graph, and m(σ) = 1 or −1, according to whether σ has an even or odd number of reverse arcs. For an out-degree sequence d~ = (d1 , d2 , . . . , dn ), for i ∈ {0, 1, . . . , k − 1}, let ad,i ~ be the number of orientations σ with ~ ~ d(σ) = d and m(σ) = i. The same proof as Theorem 3.1.9 shows that   k−1 n X X Y j  fH (x) = ad,j xdi i . ~ θ Pn ~ d=(d 1 ,d2 ,...,dn ), i=1 di =|E|

j=0

i=1

Qn di Thus a monomial is vanishing if and only if i=1 xi Pk−1 Pk−1 j j ~ θ = 0. By Lemma 3.9.1, ~ θ = 0 if and only if j=0 ad,j j=0 ad,j there is a constant a such that ad,j ~ = a for j = 0, 1, . . . , k − 1. In particular, we have the following corollary. Corollary 3.9.3 If the total number of orientationsQσ of H with n ~ d(σ) = (d1 , d2 , . . . , dn ) is not a multiple of k, then i=1 xdi i is a non-vanishing monomial of fH . Assume that k = 2 and σ is an orientation of H with out~ Then all other orientations σ 0 of H with the degree sequence d. same out-degree sequence can be obtained from σ by reversing the orientation of the edges of an Eulerian subdigraph of σ. In some sense, the same holds for k-uniform hypergraphs with k ≥ 3. To describe the corresponding result for the k ≥ 3 case, we need a notion that corresponds to the operation of ‘reversing the orientation of edges in an Eulerian subdigraph of σ’. For the k = 2 case, using the notation introduced in this section, reversing the direction of an edge e of H is equivalent to increasing the value of σ(e) by 1 (modulo 2). To be precise, if E 0 is a subset E, and σ 0 is obtained from σ by reversing the orientations of edges in E 0 , then σ 0 is defined as ( σ(e) + 1 (mod 2), if e ∈ E 0 , 0 σ (e) = σ(e), otherwise. Alternatively, let τ : E → {0, 1} be defined as τ (e) = 1 if e ∈ E 0 and τ (e) = 0 otherwise. Then σ 0 = σ + τ , i.e., σ 0 (e) = σ(e) + τ (e) (mod 2). For k ≥ 3, the operation of ‘reversing the orientation of some edges’ can be defined in the same way. A shift function is a mapping

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 59 — #71

Alon–Tarsi Theorem and Its Applications

59

τ : E → {0, 1, . . . , k − 1}. The orientation σ 0 = σ + τ defined as σ 0 (e) = σ(e) + τ (e) (mod k) is said to be the τ -shift of σ. A τ shift of σ can be viewed as a modification of the orientation σ. We need to put restrictions on τ so that the τ -shift of σ has the same out-degree sequence as σ. In case k = 2, then σ 0 = σ +τ has the same out-degree sequence as σ if and only if the set {e : τ (e) = 1} is an Eulerian subdigraph of σ. Assume that e is an edge of G and v is a vertex of e. Let ie (v) be the position of v in e, i.e., v is the ie (v)-th vertex of e. Then σ(e) − ie (v) is the difference between the position of the source vertex of e and the position of v in e. In particular, σ(e) − ie (v) = 0 if and only if v is the source vertex of e. Assume that σ 0 = σ + τ . Let Eσ,τ (v) = {e ∈ E(v) : τ (e) = ie (v) − σ(e)}. Then σ 0 and σ have the same out-degree sequence if and only if for each vertex v, |Eσ (v)| = |Eσ,τ (v)|. Let Eτ∗ = {e ∈ E(v) : τ (e) = 0}. Then Eτ∗ consists of those edges e whose orientations are not changed. The set Eσ (v) − Eτ∗ (v) consists of those edges that are out-edges at v in σ, but not outedges at v in σ 0 ; and the set Eσ,τ (v) − Eτ∗ (v) consists of those edges that are out-edges at v in σ 0 , but not out-edges at v in σ. We call a shift function τ balanced for σ if for each vertex v, |Eσ (v) − Eτ∗ (v)| = |Eσ,τ (v) − Eτ∗ (v)|. Then balanced shift functions play the role of Eulerian subdigraphs in the k = 2 case, and every ~ 0 ) = d(σ) ~ orientation σ 0 with d(σ is obtained from σ by adding a balanced shift function, i.e., σ 0 = σ + τ P for some balanced shift function τ for σ. Moreover, let m(τ ) = e∈E τ (e). Then w(σ + ~ τ ) = θm(τ ) w(σ). Assume that Qn d(σ) = (d1 , d2 , . . . , dn ). Then the coefficient of the monomial i=1 xdi i in the expansion of fH is X θm(τ ) θm(σ) , τ

where the summation is over all balanced shift functions τ of σ. We say that a hypergraph H is f -Alon–Tarsi Qnchoosable, f -AT for short, if fH has a non-vanishing monomial i=1 xtii with ti < f (vi ). Hence we have the following corollary.

“2020-12-10” — 2021/5/8 — 18:39 — page 60 — #72

i

i

60

Combinatorial Nullstellensatz

Corollary 3.9.4 Assume that σ is an orientation of a k-uniform ~ hypergraph H with out-degree sequence d(σ) = (d1 , d2 , . . . , dn ), and L is a list assignment which assigns to each vertex vi a set L(vi ) of P di + 1 integers as permissible colours. If τ θm(τ ) over all balanced shift functions τ of σ is non-zero, then H is L-colourable. Corollary 3.9.5 Assume that σ is an orientation of H with out~ degree sequence d(σ) = (d1 , d2 , . . . , dn ). If the total number Qn of balanced shift functions for σ is not a multiple of k, then i=1 xdi i is a non-vanishing monomial of fH . Hence with f (vi ) = di + 1, H is f -AT. Observe that for any orientation σ, τ0 defined as τ0 (e) = 0 for each hyperedge e is a trivial balanced shift function for σ. Example 3.9.6 The Fano plane F as given in Figure 3.16 is a 3-uniform hypergraph with 7 vertices {1, 2, . . . , 7} and 7 edges {124, 235, 136, 157, 267, 347, 456}. Let σ be the orientation of F given as {124, 235, 136, 157, 267, 347, 456} (where the underlined vertex in each edge is the source vertex). Then there is only one non-trivial balanced shift function τ for σ defined as τ (124) = τ (235) = 1, τ (136) = 2 and τ (e) = 0 for all the other edges e. It is easy to verify that τ defined above is indeed a balanced shift function for σ. Indeed, σ + τ is the orientation {124, 235, 136, 157, 267, 347, 456}, which has the same out-degree sequence as σ. On the other hand, if τ is a balanced shift function for σ, then since the vertices 5, 6, 7 have out-degree 0, we have τ (e) 6= 1 for e ∈ {136, 157, 267, 456} and τ (e) 6= 2 for e ∈ {235, 157, 267, 347, 456}. Hence τ (456) = 0. This in turn implies that τ (347) 6= 1 and τ (124) 6= 2. Thus τ (e) = 1 only if e ∈ {124, 235} and τ (e) = 2 only if e = 136. It is straightforward to verify that τ (124) = 1 ⇒ τ (235) = 1 ⇒ τ (136) = 2 ⇒ τ (124) = 1. As the total number of balanced shift functions for σ is not a multiple of 3, and as each vertex of F has out-degree at most 2 in σ, we conclude that F is 3-choosable.





“2020-12-10” — 2021/5/8 — 18:39 — page 61 — #73

Alon–Tarsi Theorem and Its Applications

61

FIGURE 3.16: The Fano plane F .

3.10

Paintability of graphs

The Alon–Tarsi number AT (G) of G is not only an upper bound for the choice number of G, but it is also an upper bound for the painter number of G, which is an on-line version of the choice number. The painter number of a graph is defined through a twoperson game. Assume that G is a graph and f : V (G) → Z + is a function which assigns to each vertex a positive integer. Definition 3.10.1 [53] The f -painting game on G is a two-person game defined as follows: The two players are Lister and Painter. Initially no vertex of G is coloured and each vertex v is given f (v) tokens. In each round, Lister chooses a subset M of uncoloured vertices, and removes one token from each vertex in M . Painter chooses a subset I of M that is an independent set of G and colours vertices in I. If at the end of a certain round, there is an uncoloured vertex with no tokens left, then Lister wins the game. Otherwise, at the end of a certain round, all vertices of G are coloured, then Painter wins the game. Definition 3.10.2 We say that G is f -paintable if Painter has a winning strategy for this game. We say that G is k-paintable if G is f -paintable for the constant function f (v) = k for all v ∈ V (G). The painter number χP (G) is defined by χP (G) = min{k : G is k-paintable}.

“2020-12-10” — 2021/5/8 — 18:39 — page 62 — #74

i

i

62

Combinatorial Nullstellensatz

Consider the case when Lister plays with a trivial strategy: On each round, Lister chooses the set of all uncoloured vertices. If f (v) = k for all vertices v and Lister plays with this trivial strategy, then Painter can win the game if and only if G is k-colourable. With this trivial strategy of Lister, the game ends in at most k rounds: after k rounds, if there are still uncoloured vertices, then all the tokens of each uncoloured vertex are gone, and Lister wins the game. Otherwise all vertices are coloured and Painter wins. If G is k-colourable, then V (G) is covered by k independent sets I1 , I2 , . . . , Ik . Painter’s strategy is to colour Ij in the jth round. Conversely, if Painter wins the game, then let Ij be the set of vertices coloured in the jth round. Then I1 , I2 , . . . , Ik are k independent sets that form a partition of V (G). So G is k-colourable. This shows that χP (G) ≥ χ(G). Next we observe that the painter number of G is actually an upper bound for the choice number of G. Lemma 3.10.3 For any graph G, ch(G) ≤ χP (G). Proof. Assume that χP (G) = k and L is a k-list assignment of G. We shall Sprove that G is L-colourable. Let v∈V (G) L(v) = {1, 2, . . . , m}. We play the k-painting game on G. In the ith round, Lister chooses the set containing all uncoloured vertices from {v : i ∈ L(v)}. As G is k-paintable, Painter has a winning strategy, and hence after m rounds, as all the tokens are gone, all vertices are coloured. The result is an L-colouring of G. The graph shown in Figure 3.17 is the Θ2,2,4 graph. It is known that this graph is 2-choosable, however, it is not 2-paintable. Lister’s winning strategy is as follows: In the first move, Lister marks v1 , v2 . Painter can colour either v1 or v2 . Assume that Painter colours v1 . Lister will mark sets {v2 , v3 }, {v3 , v4 }, {v4 , u, w}, {v0 , u}, {v0 , w} in the next five moves. In each of these moves, Painter is forced to colour vertices v2 , v3 , v4 , u in the next four moves (otherwise there will be an uncoloured vertex with no tokens left and Painter loses), and in the fifth move, Painter loses. If Painter colours v2 in the first move, then Lister marks sets {v0 , v1 }, {v0 , u, w}, {u4 , u}, {u4 , w} in the next four moves and Painter will lose. It is known [20] that the difference χP (G) − ch(G) can be arbitrarily large.





“2020-12-10” — 2021/5/8 — 18:39 — page 63 — #75

Alon–Tarsi Theorem and Its Applications

63

FIGURE 3.17: The graph Θ2,2,4 . Nevertheless, many upper bounds for the choice number of graphs are also upper bounds for their painter number. For example, we say that a graph G is d-degenerate if its vertices can be ordered as v1 , v2 , . . . , vn so that each vi has at most d neighbours vj with j < i. Equivalently, G is d-degenerate if every subgraph of G contains a vertex of degree at most d. For example, every tree is 1-degenerate. It is well-known that ch(G) ≤ d + 1 for any d-degenerate graph G. This upper bound holds for painter number as well, i.e., for any d-degenerate graph G, χP (G) ≤ d + 1. Indeed, let v1 , v2 , . . . , vn be an ordering of the vertices of G for which each vi has at most d neighbours vj with j < i. In each round, Painter colours vertices greedily: traversing from v1 to vn , Painter colours vertex vi provided vi is marked in this round and no earlier neighbour of vi is coloured in this round. It is obvious that this is a winning strategy for Painter in the (d + 1)-painting game on G. It was proved by Schauz [53] that AT (G) is also an upper bound for χP (G). The remainder of this section is devoted to the proof of this result. S For f ∈ NG 0 and S ⊆ V (G), denote by OG (f ) the set of ori(v) = f (v) if v ∈ V (G)\S and entations D of G for which d+ D (v) ≥ f (v) if v ∈ S. Let < be an arbitrary fixed ordering of d+ D vertices of G, and the parity of an orientation D of G is the parity of the number of arcs (u, v) for which u > v. Let Oe,G (f S ) and Oo,G (f S ) be the sets of even and odd orientations in OG (f S ),

“2020-12-10” — 2021/5/8 — 18:39 — page 64 — #76

i

i

64

Combinatorial Nullstellensatz

respectively. Let diff G (f S ) = |Oe,G (f S )| − |Oo,G (f S )|. In case S = ∅, then OG (f S ), Oe,G (f S ), Oo,G (f S ), diff G (f S ) are denoted by OG (f ), Oe,G (f ), Oo,G (f ), diff G (f ), respectively. Note that OG (f ) is the set of orientations D with d+ D (v) = f (v) for all v. Lemma 3.10.4 Assume that f ∈ NG 0 and S ⊆ V (G) and diff G (f S ) 6= 0. Then the following hold: P • If v∈V (G) f (v) = |E(G)|, then for any subset S 0 of V (G), 0

diff G (f S ) 6= 0. G • There exists P g ∈ N0 such that g ≥ f , g(v) = f (v) for v ∈ V (G)\S, v∈V (G) g(v) = |E(G)| and diff G (g) 6= 0.

Proof. Assume that

P

v∈V (G)

f (v) = |E(G)|. Then for any sub0

0

set for any orientation D ∈ OG (f S ), |E(G)| = P S of V (G), P + + v∈V (G) f (v) ≤ v∈V (G) dD (v) = |E(G)|. Hence dD (v) = f (v) 0

0

for every vertex v. Thus OG (f S ) = OG (f S ) and diff G (f S ) = diff G (f S ) 6= 0. Let  F = g ∈ NG 0 : g ≥ f, g(v) = f (v)∀v ∈ V (G)\S,  X g(v) = |E(G)| . v∈V (G)

Then OG (f S ) is the disjoint union {OG (g) : g ∈ F } and 0 6= P |Oe,G (f S )|−|Oo,G (f S )| = g∈F |Oe,G (g)|−|Oo,G (g)|. Hence there exists g ∈ NG 0 such that g ≥ f , g(v) = f (v) for v ∈ V (G)\S, P = |E(G)| and diff G (g) 6= 0. v∈V (G) Lemma 3.10.5 If G has an edge e = uv such that f (u) = f (v) = 0 and u, v ∈ S, then diff G (f S ) = 0. Proof. For any D ∈ OG (f S ), let φ(D) be the orientation of G obtained from D by reversing the orientation of edge e. Then φ is a one-to-one correspondence from Oe,G (f S ) to Oo,G (f S ). So diff G (f S ) = 0.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 65 — #77

Alon–Tarsi Theorem and Its Applications

65

Lemma 3.10.6 If u ∈ S and f (u) > 0 and diff G (f S ) 6= 0, then diff G (fˆS ) 6= 0 or diff G (fˆS−u ) 6= 0. u

u

Proof. Assume that u ∈ S and f (u) > 0. For D ∈ OG (fˆuS ), + either d+ D (u) ≥ f (u) or dD (u) = f (u) − 1. In the former case, D ∈ OG (f S ). In the latter case, D ∈ OG (fˆuS−u ). It is obvious that OG (f S ) ∪ OG (fˆuS−u ) ⊆ OG (fˆuS ). Thus OG (fˆuS ) = OG (f S ) ∪ OG (fˆuS−u ). Thus if diff G (f S ) 6= 0, then diff G (fˆS ) 6= 0 or diff G (fˆS−u ) 6= 0. Lemma 3.10.7 Assume that f ∈ NG ⊆ V (G), and 0, S diff G (f S ) 6= 0. Then there exist R ⊆ S, g ∈ NG 0 for which the following hold: 1. g ≤ f and g(v) < f (v) for any v ∈ S\R. 2. diff G (g R ) 6= 0. 3. g(v) = 0 for every v ∈ R. P Proof. We prove the lemma by induction on v∈S f (v). If f (v) = 0 for all v ∈ S, then let R = S and g = f and we are done. So assume that f (u) > 0 for some u ∈ S. By Lemma 3.10.6, diff G (fˆuS ) 6= 0 or diff G (fˆuS−u ) 6= 0. Let f 0 = fˆu . In the first case, let S 0 = S and in the latter case, 0 0 let S 0 = S\{u}. P Then0 f andPS satisfy the condition of Lemma 3.10.7. Since v∈S 0 f (v) < v∈S f (v), by induction hypothesis, there exist R ⊆ S 0 and g for which the following hold: 1. g ≤ f 0 and g(v) < f 0 (v) for any v ∈ S 0 \R. 2. diff G (g R ) 6= 0. 3. g(v) = 0 for every v ∈ R. As g(u) < f (u) and f 0 ≤ f , this completes the proof of Lemma 3.10.7. By Lemma 3.10.5, g(v) = 0 for every v ∈ R and diff G (g R ) 6= 0 imply that R is an independent set of G. Combining Lemmas 3.10.4 and 3.10.7, we have the following corollary.

“2020-12-10” — 2021/5/8 — 18:39 — page 66 — #78

i

i

66

Combinatorial Nullstellensatz

Corollary 3.10.8 Assume that f ∈ NG 0 , S ⊆ V (G), and diff G (f S ) 6= 0. Then there exists an independent set R ⊆ S, g ∈ NG 0 for which the following hold: 1. g(v) ≤ f (v) for any v ∈ V (G)\S and g(v) < f (v) for any v ∈ S\R. 2. diff G (g) 6= 0. Lemma 3.10.9 Assume that G0 is obtained from G by deleting an edge e = uv. If diff G (g) 6= 0, then diff G0 (ˆ gu ) 6= 0 or diff G0 (ˆ gv ) 6= 0. Proof. We may assume that u < v. For orientation D ∈ OG (g), let φ(D) be the orientation of G0 obtained from D by deleting the edge uv. If e is oriented as (u, v) in D, then φ(D) ∈ OG0 (ˆ gu ) and hence D and φ(D) have the same parity. If e is oriented as (v, u) in D, then φ(D) ∈ OG0 (ˆ gv ) and D, φ(D) have opposite parities. Thus φ is a one-to-one correspondence from Oe,G (g) to Oe,G0 (ˆ gu ) ∪ Oo,G0 (ˆ gv ), and from Oo,G (g) to Oo,G0 (ˆ gu ) ∪ Oe,G0 (ˆ gv ). Note that the unions above are disjoint unions. So if diff G0 (ˆ gu ) = 0 and diff G0 (ˆ gv ) = 0, we would have diff G (g) = 0. Corollary 3.10.10 If G0 is obtained from G by deleting a set F of edges, and diff G (g) 6= 0, then there exists h ∈ N0G−F such that h ≤ g and diff G−F (h) 6= 0. Corollary 3.10.11 If R is an independent set and diff G (g) 6= 0, then there exists q ∈ N0G−R such that q(v) ≤ g(v) for v ∈ V (G)\R and diff G−R (q) 6= 0. Proof. Apply Corollary 3.10.10 to the set F of edges connecting vertices of R to V (G)\R to obtain a function h ∈ N0G−F such that h ≤ g and diff G−F (h) 6= 0. Since the vertices of R are isolated vertices in G−F , we have diff G−R (q) 6= 0, where q is the restriction of h to V (G − R). Now we are ready to prove the main theorem. Theorem 3.10.12 If G is f -AT, then G is f -paintable. Proof. We prove this theorem by induction on the number of vertices of G. If G has one vertex, then there is nothing to be proved. Assume that |V (G)| = n and the theorem holds for graphs with fewer vertices.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 67 — #79

Alon–Tarsi Theorem and Its Applications

67

Assume that G is f -AT. Then there is an orientation D of G such that d+ D (v) ≤ f (v) for each vertex v and |Ee (D)| 6= |Eo (D)|. Let g(v) = d+ D (v) for every vertex v. Then for any S ⊆ V (G), any D0 ∈ OG (g S ) has the same out-degree sequence as D, and hence is obtained from D by reversing the directions of arcs in an Eulerian subdigraph H of D. If H is an even Eulerian subdigraph of D, then D0 has the same parity as D while if H is an odd Eulerian subdigraph of D, then D0 and D have opposite parities. So diff G (g S ) = ±(|Ee (D)| − |Eo (D)|). Hence diff G (g S ) 6= 0. Assume that in the first move, the subset of uncoloured vertices chosen by Lister is S. Since diff G (g S ) 6= 0, by Corollary 3.10.8 and Corollary 3.10.11, there is an independent set R of G contained in S and a function q ∈ N0G−R such that diff G−R (q) 6= 0, and q(v) ≤ g(v) for all v ∈ V (G − R) and q(v) < g(v) for v ∈ S\R. Note that diff G−R (q) 6= 0 implies that there is an orientation D0 of G − R such that d+ D 0 (v) = q(v) for each vertex v and diff G−R (q) = ±(|Ee (D0 )| − |Eo (D0 )|) 6= 0. By induction hypothesis, Painter has a winning strategy for the q-painting game on G\R. After the first round, each vertex v of S\R has f (v) − 1 ≥ q(v) tokens, and every other vertex v has f (v) ≥ q(v) tokens. So by following Painter’s strategy for the q-painting game on G − R, Painter will win this game. The concept of d-defective k-choosability is naturally extended to d-defective k-paintability. A d-defective k-painting game on a graph G is played by two players: Lister and Painter. The rule is basically the same as the k-painting game on G, except that in each round, after Lister has chosen a subset M , the subset I of M of vertices coloured by Painter is not necessarily an independent set. Instead, it is required that the subgraph G[I] of G induced by I has maximum degree at most d. We say that G is d-defective k-paintable if Painter has a winning strategy in the d-defective kpainting game on G. So 0-defective k-painting game is the same as the k-painting game and 0-defective k-paintable is the same as k-paintable. By the same argument as in the proof of Lemma 3.10.3, we can show that d-defective k-paintability implies d-defective kchoosability. The converse is not true. It is proved in [27] that there are planar graphs that are not 2-defective 3-paintable. Also it is proved in [58] and [21] that every planar graph is 2-defective 3-choosable. Nevertheless, the following lemma holds.

“2020-12-10” — 2021/5/8 — 18:39 — page 68 — #80

i

i

68

Combinatorial Nullstellensatz

Lemma 3.10.13 If G has a subgraph H with maximum degree d and G − E(H) is k-paintable, then G is d-defective k-paintable. In particular, if G − E(H) has Alon–Tarsi number at most k, then G is d-defective k-paintable. Proof. Painter’s winning strategy for the k-painting game on G − E(H) is a winning strategy for the d-defective k-painting game on G. Corollary 3.10.14 Every planar graph is 1-defective 4-paintable.

“2020-12-10” — 2021/5/8 — 18:39 — page 69 — #81

i

i

Chapter 4 Generalizations of CNS and Applications

This chapter introduces some strenghtenings and generalizations of CNS and their applications.

4.1

Number of non-zero points

Assume that F is a field and f (x) = f (x1 , x2 , . . . , xn ) is a polynomial in F[x1 , x2 , . . . , xn ]. Assume that S = S1 × S2 × . . . × Sn is a finite grid in Fn . It follows from CNS that if x||S|| is a highest degree non-vanishing monomial in the expansion of f (x), then f (x) has a non-zero point in S. In many cases, f (x) has many non-zero points in S. For example, suppose that G is a planar graph and L is a 5-list assignment of G. By Thomassen’s Theorem, if we fix any two adjacent vertices u, v of G, then for any c ∈ L(u), c0 ∈ L(v) with c 6= c0 , G has an L-colouring φ for which φ(u) = c and φ(v) = c0 . As there are 5 choices of c and at least 4 choices of c0 , we conclude that G has at least 20 L-colourings. Intuitively, the number of L-colourings of G should be much more than 20, especially when G has many vertices. In this section, we obtain a lower bound on the number of non-zero points of a polynomial in a finite grid S. This result is proved in [13] and a generalization of this result can be found in [12]. The proof presented below is based on a note of Bosek, Grytczuk, Gutowski and Serra [13]. Theorem 4.1.1 Assume that F is a field and f (x) = f (x1 , x2 , . . . , xn ) is a polynomial in F[x1 , x2 , . . . , xn ]. Assume further that S1 , S2 , . . . , Sn are subsets of F and |Si | = si for eachPi. Let n d = deg f be the degree of f , t = maxni=1 si and s = i=1 si .

69

“2020-12-10” — 2021/5/8 — 18:39 — page 70 — #82

i

i

70

Combinatorial Nullstellensatz

If S := S1 × S2 × . . . × Sn contains a non-zero point of f , then S contains at least t(s−n−d)/(t−1) non-zero points of f . Proof. Assume that the theorem is not true and that f and S form a counterexample with (n, d, s) minimal in lexicographic order. It is obvious that (n, d, s) > (1, 1, 1). Without loss of generality, assume that |S1 | ≥ 2. Fix a ∈ S1 and write f (x) = (x1 − a)g(x) + R(x), (4.1) Note that x1 is absent in R(x). First we consider the case when for any x ∈ S2 × S3 × . . . × Sn , R(x) = 0. Then for x ∈ (S1 − {a}) × S2 × . . . × Sn , f (x) 6= 0 if and only if g(x) 6= 0. For g(x) and S1 − {a}, S2 , . . . , Sn , the corresponding parameters s0 , n0 , d0 , t0 satisfy the following: s0 = s − 1, n0 = n, d0 = d − 1, t0 ≤ t. By induction hypothesis, (S1 − 0 0 0 0 {a}) × S2 × . . . × Sn contains at least (t0 )(s −n −d )/(t −1) non-zero points of g(x). Observe that t1/(t−1) is a decreasing function. Hence 0 0 0 0 0 (t0 )1/(t −1) ≥ t1/(t−1) and (t0 )(s −n −d )/(t −1) ≥ t(s−n−d)/(t−1) , and we are done. Assume now that R(x) has a non-zero point in S2 × S3 × . . . × Sn . The corresponding parameters s0 , n0 , d0 , t0 for R(x) satisfy the following: s0 = s − s1 , n0 = n − 1, d0 ≤ d, t0 ≤ t. By 0 0 0 0 induction hypothesis, R(x) has at least (t0 )(s −n −d )/(t −1) non0 zero points in S2 × S3 × . . . × Sn . Since (t0 )1/(t −1) ≥ t1/(t−1) , 0 0 0 0 we have (t0 )(s −n −d )/(t −1) ≥ t{(s−n−d)/(t−1)}−{(s1 −1)/(t−1)} . If (a2 , a3 , . . . , an ) ∈ S2 × S3 × . . . × Sn is a non-zero point for R(x), then (a, a2 , a3 , . . . , an ) is a non-zero point for f (x). So f (x) has at least t{(s−n−d)/(t−1)}−{(s1 −1)/(t−1)} non-zero points in {a} × S2 × S3 × . . . × Sn . We apply the same argument for each element a of S1 . Then either for some a ∈ S1 , the corresponding R(x) has no nonzero point in S2 × S3 × . . . × Sn , and we are done. Otherwise, f (x) has at least t{(s−n−d)/(t−1)}−{s1 /(t−1)} non-zero points in {a} × S2 × S3 × . . . × Sn for each a ∈ S1 . Since s1 ≤ t, we 1/(s −1) have t1/(t−1) ≤ s1 1 . Hence t(s1 −1)/(t−1) ≤ s1 and f (x) has {(s−n−d)/(t−1)}−{(s1 −1)/(t−1)} at least s1 × (t ) ≥ t(s−n−d)/(t−1) nonzero points in S1 × S2 × . . . × Sn . Corollary 4.1.2 Assume that G is an n-vertex planar graph and L is a 5-list assignment of G. Then G has at least 5n/4 Lcolourings.

“2020-12-10” — 2021/5/8 — 18:39 — page 71 — #83

i

i

Generalizations of CNS and Applications

71

Proof. Apply Theorem 4.1.1 to the graph polynomial fG of G. Since s = 5n, d < 3n, t = 5, we conclude that G has at least 5n/4 L-colourings.

4.2

Multisets

Consider the polynomial f (x) = ak xk + ak−1 xk−1 + . . . + a1 x + a0 ∈ F[x] of degree k in the single variable x. Then f has at most k distinct roots in the field F. Thus if S is a subset of F of cardinality k + 1, then S contains a non-zero point of f (x). The CNS is basically an extension of this result to polynomials with n variables. Now if f (x) has repeated roots, then the number of distinct roots of f (x) is smaller than k. Hence if S is a subset of F of cardinality k, then S contains a non-zero point of f (x). This section discusses extension of this result to polynomials with more variables. Let F be a field. A multi-subset of F is a mapping ω : F → N, where ω(a) is the multiplicity of a. The support of ω is supp(ω) = P {a ∈ F : ω(a) ≥ 1}. The size of ω is |ω| = a∈F ω(a). ∂0 For f (x) ∈ F[x1 , x2 , . . . , xn ], for i = 1, 2, . . . , n, ∂x 0 f (x) = i f (x), and for k ≥ 1, ! ∂k ∂ ∂ k−1 f (x) = f (x) . ∂xi ∂xk−1 ∂xki i For u = (u1 , u2 , . . . , un ) ∈ Nn0 , ∂u ∂ u1 +u2 +...+un ∂ u1 f (x) = f (x) = u u u 1 2 n ∂xu ∂x1 ∂x2 . . . ∂xn ∂xu1 1



 ∂ u2 +...+un f (x) . ∂xu2 2 . . . ∂xunn

Lemma 4.2.1 Assume that g(x) ∈ F[x1 , x2 , . . . , xn ] and u1 ≥ 1. Then ∂ u1 ∂ u1 −1 ∂ u1 ((x − a)g(x)) = u g(x) + (x − a) g(x). 1 1 1 ∂xu1 1 ∂xu1 1 ∂xu1 1 −1

“2020-12-10” — 2021/5/8 — 18:39 — page 72 — #84

i

i

72

Combinatorial Nullstellensatz

Proof. By induction on u1 . If u1 = 1, then this is just the product rule. Assume that u1 ≥ 2 and the claim holds for u1 − 1. Then  ∂ u1 ∂ ∂ u1 −2 (u1 − 1) u1 −2 g(x) u1 ((x1 − a)g(x)) = ∂x1 ∂x1 ∂x1  ∂ u1 −1 +(x1 − a) u1 −1 g(x) ∂x1 ∂ u1 −1 ∂ u1 = (u1 − 1) u1 −1 g(x) + (x1 − a) u1 g(x) ∂x1 ∂x1 ∂ u1 −1 g(x) ∂xu1 1 −1 ∂ u1 ∂ u1 −1 = u1 u1 −1 g(x) + (x1 − a) u1 g(x). ∂x1 ∂x1 +

The following result was proved by K´ os and R´ongyai [40]. The proof given below is new and it is adapted from the proof of Theorem 2.1.3 of Michalek [47]. Theorem 4.2.2 [Combinatorial Nullstellensatz for multisets (CNSM)] Let F be a field and let f (x) = f (x1 , x2 , . . . , xn ) be a nonzero polynomial in F[x1 , x2 , . . . , xn ]. Suppose that xt11 xt22 . . . xtnn is a highest degree non-vanishing monomial of f . Then for any multisubsets ω1 , ω2 , . . . , ωn of F with |ωi | = ti + 1 for each i, there exist s ∈ Fn and u ∈ Nn0 with ui < ωi (si ) for i = 1, 2, . . . , n such that ∂u f (s) 6= 0. ∂xu Proof. Observe that for each i, 0 ≤ ui < ωi (si ). So ωi (si ) ≥ 1, i.e., si ∈ supp(ωi ). The proof is by induction on the degree of f . Assume that deg f = 1 so that f (x) = a1 x1 + a2 x2 + · · · + an xn + an+1 , ai ∈ F with at least one ai ∈ {a1 , . . . , an }, say a1 6= 0. Assume that ω1 , ω2 , . . . , ωn are multi-subsets of F with |ωi | ≥ 1 for i = 2, 3, . . . , n and |ω1 | ≥ 2. For i = 2, 3, . . . , n, assign any value si to xi for which ωi (si ) ≥ 1. This gives the element a2 s2 + · · · + an sn + an+1 of F. If |supp(ω1 )| = 2, then there exists s1 ∈ supp(ω1 ) so that a1 s1 + a2 s2 + · · · + an sn + an+1 6= 0.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 73 — #85

Generalizations of CNS and Applications

73

Otherwise, ω(s1 ) = 2 for some s1 ∈ F, and ∂ f (s1 , s2 , . . . , sn ) = a1 6= 0. ∂x1 Hence the result is true when deg f = 1. Now assume that f (x) ∈ F[x1 , x2 , x3 , . . . , xn ] is a polynomial of degree deg f = r for some r ≥ 2, and Theorem 4.2.2 holds for polynomials of degree ≤ r − 1. Suppose that xt11 xt22 . . . xtnn is a highest degree non-vanishing monomial of f . Without loss of generality, assume that t1 > 0. Suppose that f does not satisfy the statement in the theorem and there exist multi-subsets ω1 , ω2 , . . . , ωn of F with |ωi | = ti + 1 such that for any u ∈ Nn0 and any s ∈ Fn for which ui < ωi (si ) for i = 1, 2, . . . , n, ∂u f (s) = 0. ∂xu Fix a ∈ supp(ω1 ) and write f (x) = (x1 − a)g(x) + R(x),

(4.2)

(4.3)

As t1 > 0, g has a non-vanishing monomial x1t1 −1 xt22 . . . xtnn , and = (t1 − 1) + t2 + · · · + tn = deg f − 1.

deg g

Let ω10 be the multi-subset defined as ω10 (a) = 1 and ω10 (s) = 0 for s ∈ F − {a}, and for i = 2, 3, . . . , n, ωi0 = ωi . Assume that u0 ∈ Nn0 , s ∈ Fn , and u0i < ωi0 (si ) for i = 1, 2, . . . , n. Since ω10 (a) = 1 and ω10 (s) = 0 for s ∈ F − {a}, we have u01 = 0 and s1 = a. Hence 0

∂u ((x1 − a)g(x))|x=s = 0. ∂xu0 By assumption, 0

∂u f (s) = 0. ∂xu0 Therefore

0

∂u R(s) = 0. ∂xu0

“2020-12-10” — 2021/5/8 — 18:39 — page 74 — #86

i

i

74

Combinatorial Nullstellensatz

As R(x) does not contain x1 , it follows that for any s ∈ Fn and u ∈ Nn0 with ui < ωi (si ), ∂u R(s) = 0. ∂xu Plugging this and (4.2) into (4.3), we have ∂u ((x1 − a)g(x)) |x=s = 0. ∂xu

(4.4)

Let ω100 (a) = ω1 (a) − 1 and for b ∈ F − {a}, ω100 (b) = ω1 (b). So |ω100 | = |ω1 | − 1 > t1 − 1. By induction hypothesis, there exist u00 ∈ Nn0 and s0 ∈ Fn such that u001 < ω100 (s01 ) and u00i < ωi (s0i ) and 00

∂u g(s0 ) 6= 0. ∂xu00

(4.5)

We choose a minimal u00 so that (4.5) holds, i.e., for any u∗ < 00

u ,

∗

∂u g(s0 ) = 0. ∂xu∗ If s01 = a, then let u = (u001 + 1, u002 , . . . , u00n ). Otherwise let u = u00 . Then ui < ωi (s0i ) for i = 1, 2, . . . , n. If s01 = a, then u1 = u001 + 1 and by Lemma 4.2.1, 00

∂u ∂u 0 = u1 ((x − a)g(x)) | g(s0 ) 6= 0, 1 x=s ∂xu ∂xu00 contrary to (4.4). Assume that s01 6= a. Thus u = u00 . If u1 = 0, then 

   ∂u ∂u ((x1 − a)g(s)) |x=s0 = (x1 − a) u g(s) |x=s0 6= 0, ∂xu x

contrary to (4.4). Assume that u1 = u001 ≥ 1. Let u∗1 = u1 − 1 and u∗i = ui for i = 2, 3, . . . , n. Then by Lemma 4.2.1, ∗

∂u ∂u ∂u 0 0 ((x − a)g(s)) = u g(s ) + (s − a) g(s0 ). 1 1 ∗ 1 ∂xu ∂xu ∂xu Note that since u∗ < u00 , by our choice of u00 , ∗

∂u g(s0 ) = 0. ∂xu∗

“2020-12-10” — 2021/5/8 — 18:39 — page 75 — #87

i

i

Generalizations of CNS and Applications

75

Hence

∂u ∂u 0 ((x − a)g(s)) = (s g(s0 ) 6= 0, − a) 1 1 ∂xu ∂xu contrary to (4.4). This contradiction proves the theorem. Some further generalizations of CNS to multisets are discussed in [11, 39].

4.3

Coefficient of a highest degree monomial

The coefficient of a highest degree monomial in the expansion of a polynomial f (x1 , x2 , . . . , xn ), which is crucial when we invoke CNS, can be expressed by means of an interpolation of f at appropriate sets of points. This chapter explores applications of such interpolation formulas. Assume that f (x) ∈ F[x1 , x2 , . . . , xn ] is a polynomial over F, with degree deg f ≤ d1 + d2 + . . . + dn . Uwe Schauz [55] and independently Michal Laso´ n [44] proved a strengthening of CNS, which Qn shows that the coefficient of the monomial i=1 xdi i in the expansion of f can be written as an interpolation of its values on any grid of size (d1 , d2 , . . . , dn ). Let S := S1 × S2 × · · · × Sn , where each Si is a subset of F. For a := (a1 , a2 , . . . , an ) ∈ S, let NS (a) =

n Y

Y

(ai − b).

i=1 b∈Si −{ai }

We may think of NS (a) as a normalizing factor w.r.t. a. Let χa (x) = NS (a)−1

n Y

Y

(xi − b).

i=1 b∈Si −{ai }

Then χa vanishes at all points of the grid S, except that χa (a) = 1. Theorem 4.3.1 Assume that f (x) ∈ F[x1 , x2 , . . . , xn ] is a polynomial over F, S := S1 × S2 × · · · × Sn is an n-dimensional grid n P over F, and deg f ≤ s = |Si |. Then the coefficient cf,||S|| of the i=1 Qn |S |−1 monomial i=1 xi i in the expansion of f (x) is given by X cf,||S|| = NS (a)−1 f (a). a∈S

“2020-12-10” — 2021/5/8 — 18:39 — page 76 — #88

i

i

76

Combinatorial Nullstellensatz

Proof. For i ∈ {1, 2, . . . , n}, let di = |Si | − 1. Assume first that for each i ∈ {1, 2, . . . , n}, the highest degree of xi in f is at most di . Let X p(x) = χa (x)f (a). a∈S

By definition, p(x) is a polynomial in which the highest degree of xi is at most di . Since χa (x) = 1 if x = a and χa (x) = 0 if x ∈ S − {a}, we have p(a) = f (a) for a ∈ S. Thus f − p is a polynomial in which the highest degree of xi is at most di , and (f − p)(a) = 0 for all a ∈ S. If there is only one variable, then f − p is a polynomial of degree d1 in x1 and with d1 + 1 roots. So f − p = 0 is the zero polynomial. If there are more than one variable, then it is easy to prove by induction on the number of variables that f − p = 0 is the zero polynomial, i.e., f = p. (We can also apply Theorem 2.1.3 to Q conclude that f − p = 0 is the zero n di polynomial.) The coefficient of i=1 xi in p(x) is easily seen to P −1 be a∈S NS (a) f (a). So the theorem holds if the highest degree of xi in f is at most di for each i ∈ {1, 2, . . . , n}. So assume now that there is an index Qn j ∈ {1, 2, . . . , n} such that f has a non-vanishing monomial i=1 xsi i with sj ≥ dj + 1. Let Y d +1 gj (xj ) = (xj − a), and hj (xj ) = xj j − gj (xj ). a∈Sj

Then hj is a polynomial of degree dj . For any a ∈ Sj , since gj (a) = 0, we have adj +1 = hj (a). In the expansion of f , for each non-vanishing monomial Qn sj sj −dj −1 si hj (xj ). i=1 xi for which sj > dj , we replace xj by xj Denote the resulting polynomial by f˜. Then the following hold: (1) For any a ∈ S, f (a) = f˜(a). Qn (2) The coefficients of i=1 xdi i in the expansions of f and f˜ are the same. d +1

(1) holds because for a ∈ Sj , adj +1 = hj (a). So replacing xj j by hj (xj ) does not change the value of f (a) Qn for sai ∈ S. (2) holds because each monomial xi that we have Qi=1 n di changed is different from the monomial i=1 xi , and the polyQn si nomial that we used to replace i=1 xi is not of highest degree in Qn f . So this change does not affect the coefficient of i=1 xdi i .

“2020-12-10” — 2021/5/8 — 18:39 — page 77 — #89

i

i

Generalizations of CNS and Applications

77

Repeat this process whenever there is an index j for Qnwhich the current polynomial f˜ has a non-vanishing monomial i=1 xsi i for which sj > dj . Eventually, we arrive at a polynomial f˜ in which xj has degree at most dj , and the following hold: • For any a ∈ S, f (a) = f˜(a). Qn • The coefficient of i=1 xdi i in the expansions of f and f˜ are the same. By the coefficient of the monomial Qn thedifirst half of this proof, ˜ i=1 xi in the expansion of f is X X cf˜,||s|| = NS (a)−1 f˜(a) = NS (a)−1 f (a) = cf,||s|| . a∈S

a∈S

Qn

As the coefficients of i=1 xdi i in the expansion of f and f˜ are the same, we are done. Note that Theorem 2.1.3 (CNS) is a corollary of Theorem 4.3.1. Assume that f (x) ∈ F[x1 , x2 , . . . , xn ] P is a polynomial over F, and n di ≥ 0 are integers such that deg f ≤ i=1 di . Suppose that S := S1 ×S2 ×· · ·×Sn is an n-dimensional grid over F with |Si | = di +1. If f (a) = 0 for allQa ∈ S, then it follows from Theorem 4.3.1 that n the coefficient of i=1 xdi i is 0. Also, it follows from Theorem 4.3.1 that if cf,||S|| = 0, then either S contains no non-zero point of f or contains at least two non-zero points of f .

4.4

Calculation of NS (a)

To apply Theorem 4.3.1, we need to calculate NS (a) for all a ∈ S. This seems to be a nontrivial task. However, in some cases, we have simple formulas for NS (a). Lemma 4.4.1 below, proved in [55],Qshows that for some special sets S, there is a simple formula for b∈S−{a} (a − b). Q For subset S of F, let fS (x) = b∈S (x − b). Then Qn NS (a) = i=1 fSi −{ai } (ai ). The following lemma gives formulas for fS−{a} (a) for some special sets S.

“2020-12-10” — 2021/5/8 — 18:39 — page 78 — #90

i

i

78

Combinatorial Nullstellensatz

Lemma 4.4.1 Assume that F is a field. For a positive integer k, let Rk = {c ∈ F : ck = 1} be the set of kth roots of unity in the field F. Assume that |Rk | = k. 1. If S = Rk , then for a ∈ S, fS−{a} (a) = ka−1 . In particular, if Rk ∪ {0} is a finite subfield of F, then fS−{a} (a) = −a−1 . 2. If S = Rk ∪ {0}, then for a ∈ S − {0}, fS−{a} (a) = k1, and fS−{0} (0) = −1. 3. If S is a subfield of F, then for a ∈ S, fS−{a} (a) = −1. = {0, 1, . . . , d}, then for a ∈ S, fS−{a} (a) =
−1 (−1) d! ad .

4. If S

d+a

Proof. (1) Since Y fRk (x) = (x − b) = xk − 1 = (x − 1)(xk−1 + xk−2 + . . . + x0 ), b∈Rk

we conclude that fRk −{1} (x) = fRk (x)/(x − 1) = xk−1 + xk−2 + . . . + x0 and fRk −{1} (1) = k. For a ∈ Rk , aRk = {ba : b ∈ Rk } = Rk and Q a(Rk − {1}) = Rk − {a}. Hence fRk −{a} (a) = fa(Rk −{1}) (a) = b∈Rk −{1} (a − ab) = ak−1 fRk −{1} (1) = kak−1 = ka−1 . If Rk ∪ {0} is a finite subfield Fpt of F, then |Rk | = k = pt − 1 = −1 (mod pt ), and hence fRk −{a} (a) = −a−1 . This proves (1). (2) If S = Rk ∪ {0}, then for a ∈ S − {0}, fS−{a} (a) = afSk −{a} (a). By (1) fSk −{a} (a) = ka−1 . Hence fS−{a} (a) = k. As fS−{0} (x) = fRk (x) = xk − 1, we have fS−{0} (0) = −1. (3) If S is a finite subfield Fpt of F, then S = Rk ∪ {0} for k = pt − 1 = −1 and the conclusion follows from (2). (4) Plug in the values in the definition of fS−{a} (a), we have  fS−{a} (a)

= 

 Y

0≤b j is (−(i − j)2 )q . This contribution is independent of τ . So fG (στ ) Qn is a constant independent of τ . Hence i=1 xki is a non-vanishing monomial of fG and AT (G) ≤ k + 1. This completes the proof of Case 1.

“2020-12-10” — 2021/5/8 — 18:39 — page 84 — #96

i

i

84

Combinatorial Nullstellensatz

Case 2 dp/qe = 1. k+1 In this case, G is a subgraph of Cq(k+2) . Indeed, let the mapping φ : {1, 2, . . . , q(k + 1) + p} → {1, 2, . . . , q(k + 2)} be defined as ( i + b ki c, if i < (q − p)k φ(i) = i + (q − p), otherwise. k+1 k Then φ is an embedding of G = Cq(k+1)+p into Cq(k+2) . By Case k+1 1, AT (Cq(k+2) ) ≤ k + 2. By Lemma 3.1.7, AT (G) ≤ k + 2.

Case 3 dp/qe ≥ 2. In this case, n = q(k + 1) + p = q(k + 2) + (p − q), where 1 ≤ p − q ≤ k − q. Hence χ(Cnk+1 ) = (k + 2) + d(p − q)/qe = χ(G). By induction hypothesis AT (Cnk+1 ) = χ(Cnk+1 ). Therefore AT (G) = χ(G).

4.6

Alon–Tarsi numbers of toroidal grids

Let n, m ≥ 3 be integers. The Cartesian product Cn Cm of two cycles is called a toroidal grid, and we denote it by Tn,m . Note that Tn,m is a 4-regular graph. If both n, m are even, then Tn,m is a bipartite graph. It follows from Corollary 3.2.2 that AT (Tn,m ) = 3. On the other hand, by Theorem 3.5.3, for any integers n, m, AT (Tn,m ) ≤ 4. Of course, AT (Tn,m ) ≥ 3 for any integers n, m ≥ 3 (any orientation of Tn,m has maximum out-degree at least 2). The problem is to determine whether AT (Tn,m ) is 3 or 4, when at least one of n, m is odd. This problem is solved in [45]. Theorem 4.6.1 If both n, m are odd, then AT (Tn,m ) = 4. If one of n, m is odd and the other is even, then AT (Tn,m ) = 3. Observe that Pn Cm is a subgraph of Tn0 ,m for any n0 ≥ n. Hence AT (Pn Cm ) ≤ AT (T2n,m ) = 3 for n ≥ 2, which was proved in Section 3.3. The proof of Theorem 4.6.1 uses the concept of anti-Hermitian matrix and basic properties of complex numbers and matrices.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 85 — #97

Generalizations of CNS and Applications

85

Definition 4.6.2 The conjugation z of a complex number z = a + bi is defined as z = a − bi, where a, b are real numbers. The conjugate transpose M of a complex matrix M is defined as M i,j = Mj,i . (Here M i,j is the (i, j)th entry of the matrix M , and Mj,i is the conjugate of the complex number Mj,i .) A square complex matrix M is Hermitian if M = M and M is anti-Hermitian (or skew Hermitian) if M = −M . The following lemma lists some basic properties of conjugate transposes. These properties can be easily verified from the definitions. Lemma 4.6.3 Let A, B be complex matrices, x a complex vector and z, z 0 are complex numbers. Then the following hold: 1. A = A, z = z. 2. (A + B) = A + B, (z + z 0 ) = z + z 0 . 3. (zA) = zA, (1/z) = 1/z. 4. (AB) = B A, (zz 0 ) = z z 0 . Note that when we write the sum of two matrices, they are assumed to be of the same sizes; when we write the product AB of two matrices, the number of columns in A is assumed to be equal to the number of rows in B. We view an n-dimensional complex vector x as an n × 1 matrix. So x is a 1 × n matrix. Definition 4.6.4 Assume that u, v are n-dimensional complex vectors. Then uv is called the inner product √ of u and v. √ The norm of a complex number z = a + bi is |z|√= a2 + b2 = zz and the norm of a complex vector u is |u| = uu. Note that the inner product of two complex vectors is a complex number. Lemma 4.6.5 Assume that d is a positive integer and that H is a 2d-regular graph with vertex set {v1 , v2 , . . . , vn }. Let G = HCk and set N = nk. Fix a field F and a subset A ⊆ F of size |A| = d+2. Let U be the set of all proper colourings of H using colours from A. Each element of U is a vector u = (u1 , u2 , . . . , un ), where ui ∈ A is the colour assigned to vertex vi . For u, u0 ∈ U, let Mu,u0 = fH (u1 , . . . , un ) ·

n Y

ui − u0i . b∈A−{ui } (ui − b)

Q i=1

“2020-12-10” — 2021/5/8 — 18:39 — page 86 — #98

i

i

86

Combinatorial Nullstellensatz

Let M = [Mu,u0 ] be the matrix whose rows and columns are indexed by elements of U. Then cG,d+1 = trM k . Proof. By the coefficient formula in Theorem 4.3.1, cG,d+1

X

=

(a1 ,...,aN )∈AN

fG (a1 , . . . , aN ) , QN Q i=1 b∈A−{ai } (ai − b)

(4.6)

where cG,d+1 = cG,kSk with |S1 | = . . . = |Sn | = d + 2. We may restrict the summation to those (a1 , . . . , aN ) ∈ AN that are proper colourings of G (for otherwise fG (a1 , . . . , aN ) = 0). Each proper colouring (a1 , . . . , aN ) of G corresponds to a sequence u1 , u2 , . . . , uk of proper colourings of H, where ui is the colouring of the ith copy of H, i.e., colouring of those vertices of G × Ck whose second coordinate is vi (the ith vertex of Ck ). Let ui = (ui,1 , ui,2 , . . . , ui,n ). We denote by fH (ui ) = fH (ui,1 , ui,2 , . . . , ui,n ). Then fG (a1 , . . . , aN ) =

k Y i=1

fH (ui ) ·

k Y n Y

(ui+1,j − ui,j ).

i=1 j=1

In the summation above, uk+1,j = u1,j . Therefore fG (a1 , . . . , aN ) = Mu1 ,u2 · Mu2 ,u3 · . . . · Muk ,u1 . QN Q i=1 b∈A−{ai } (ai − b) The summation of all these products is exactly the trace of the matrix M k . The matrix M depends on A and H. Assume that H = C2s+1 . Let A = {1, w, w2 }, where w = e2πi/3 is a cube root of 1. We shall show that in this case, the trace of M k is non-zero if k is even, and trace of M k is zero if k is odd. Lemma 4.6.6 The matrix M is an anti-Hermitian matrix. Proof. In this proof, the indices are modulo 2s + 1. Let u = (u1 , u2 , . . . , u2s+1 ) be a proper 3-colouring of C2s+1 with colour set A = {1, w, w2 }. Let u∗i be the unique element in A − {ui , ui−1 }. Then ui − ui−1 1 Q = . (u − b) u − u∗i i b∈A−{ui } i

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 87 — #99

Generalizations of CNS and Applications Q2s+1 As fH (u1 , . . . , u2s+1 ) = i=1 (ui − ui−1 ), we conclude that = fH (u1 , . . . , u2s+1 ) ·

Mu,u0

2s+1 Y

ui − u0i b∈A−{ui } (ui − b)

Q i=1

=

2s+1 Y i=1

=

2s+1 Y i=1

87

(ui − u0i )(ui − ui−1 ) Q b∈A−{ui } (ui − b) ui − u0i . ui − u∗i

We shall now show that Mu,u0

= −Mu0 ,u .

(4.7)

u0i

If ui = for some i, then Mu,u0 = 0. So (4.7) holds. Assume therefore that ui 6= u0i for all i. Then (4.7) is equivalent to the following: 2s+1 Y i=1

(ui − u0i )(ui − u∗i ) |ui − u∗i |2

= −

2s+1 Y i=1

(u0i − ui )(u0i − u0∗ i ) . (4.8) 2 |u0i − u0∗ | i

2 ∗ 2 Since |u0i − u0∗ i | = |ui − ui | = 3, and z = (4.8) is equivalent to 2s+1 Y

(ui − u0i )

i=1

u∗i − ui ui u∗i

= −

1 z

for z ∈ {ui , u0i , u∗i },

2s+1 Y

(u0i − u0∗ i )

i=1

ui − u0i . ui u0i

(4.9)

Now (4.9) is equivalent to 2s+1 Y i=1

Let i =

ui ui−1

u∗i − ui u∗i

and δi =

u0i u0i−1 .

= −

2s+1 Y i=1

u0i − u0∗ i . u0i

Then

0 i , δi ∈ {w, w2 }, u∗i = ui i , u0∗ i = ui δi .

Hence

ui u0∗ = i , i0 = δi . ∗ ui ui

We re-write (4.10) as 2s+1 Y

2s+1 Y

i=1

i=1

(1 − i ) = −

(1 − δi ).

(4.10)

“2020-12-10” — 2021/5/8 — 18:39 — page 88 — #100

i

i

88

Combinatorial Nullstellensatz

As

2s+1 Y

i =

2s+1 Y

i=1

δi = 1,

i=1

(4.10) is equivalent to 2s+1 Y

(1 − δi )

i=1

= −

2s+1 Y i=1

i (1 − i ) =

2s+1 Y

(1 − i ).

(4.11)

i=1

√ √ √ Note that w = − 21 + 23 i and w2 = − 12 − 23 i. So 1 − w = 3iw2 √ √ and 1 − w2 = − 3iw. So for  ∈ {w, w2 }, 1 −  = ±i 32 and the Q2s+1 Q2s+1 signs are different for w and w2 . As i=1 2i = i=1 δi2 = 1, we conclude that (4.11) holds if and only if

|{i : i = w}| ≡ |{i : δi = w}| (mod 2).

(4.12)

Colour i with white if ui = u0i w, and colour i with black if ui = u0i w2 . Then i = δi if i and i − 1 have the same colour, and i 6= δi if i and i − 1 have distinct colours. As there is an even number of indices i for which i and i − 1 are of distinct colours, so there is an even number of indices i for which i 6= δi . Therefore (4.12) holds and hence M is an anti-Hermitian matrix. Recall that a complex number λ is pure imaginary if its real part is zero, which is equivalent to λ = −λ. Lemma 4.6.7 If M is an anti-Hermitian matrix, then all its eigenvalues are pure imaginary. Proof. Assume λ is an eigenvalue of M and x is the corresponding eigenvector. Then xM x = λ|x|2 . Hence λ|x|2 = xM x = xM x = −xM x = −λ|x|2 . Thus λ = −λ and hence λ is pure imaginary. Note that the same proof shows that if M is a Hermitian matrix, then all its eigenvalues are real. Now we are ready to complete the proof of Theorem 4.6.1. Assume that G = C2s+1 Ck . Let M be the matrix defined as above. Note that the matrix M is a non-zero matrix. If k is even, then since the eigenvalues of M are all imaginary, the eigenvalues of M k are all real numbers, and if k ≡ 0 (mod 4), then all the eigenvalues are non-negative; if k ≡ 2 (mod 4), then all the eigenvalues are non-positive. Since M is non-zero, M k has at least one nonzero eigenvalue. Hence trM k 6= 0 (recall that the trace of a square

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 89 — #101

Generalizations of CNS and Applications

89

matrix is the summation of all its eigenvalues). By Lemma 4.6.7, cG,2 6= 0 and hence AT (G) = 3. If k is odd, then the eigenvalues of M k are also all pure imaginary. Thus trM k is a pure imaginary number. On the other hand, trM k = cG,2 is a real number. Hence trM k = cG,2 = 0. So AT (G) ≥ 4. As observed above, AT (G) ≤ 4. So equality AT (G) = 4 holds. Corollary 4.6.8 Assume that n, m ≥ 2 are integers. If nm is even, then ch(Tn,m ) = 3. If nm is odd, then 3 ≤ ch(Tn,m ) ≤ 4. It was conjectured in [14] that ch(Tn,m ) = 3 for all n, m ≥ 3. It remains open whether the conjecture holds when n, m are odd.

4.7

List colouring of line graphs

Although there are graphs whose choice numbers are much larger than their chromatic numbers (cf: Theorem 1.0.15), there are many graphs whose choice numbers equal their chromatic numbers. Such graphs are called chromatic choosable. The results in the previous section show that K2?n and cycle powers Cnk are chromatic choosable. The problem as to which graphs are chromatic choosable has been studied a lot in the literature. One of the most wellknown problems is the so-called list colouring conjecture, which was suggested independently by many researchers, including Vizing, Gupta, Allbertson, Collins and Tucker. Conjecture 4.7.1 (List colouring Conjecture) For every graph G (parallel edges are allowed), the list edge-chromatic number χ0l (G) equals its edge chromatic number χ0 (G). Conjecture 4.7.1 is equivalent to the following conjecture, which is the restriction of Conjetcure 4.7.1 to r-edge colourable r-regular graphs. Conjecture 4.7.2 (List colouring Conjecture) Every r-edge colourable r-regular graph G is r-edge choosable. Lemma 4.7.3 Conjecture 4.7.1 and Conjecture 4.7.2 are equivalent.

“2020-12-10” — 2021/5/8 — 18:39 — page 90 — #102

i

i

90

Combinatorial Nullstellensatz

Proof. It is obvious that Conjecture 4.7.1 implies Conjecture 4.7.2. To see the converse implication, we assume that Conjecture 4.7.2 holds and G is an arbitrary r-edge colourable graph. Let H be obtained from two vertex disjoint copies of G, say G and G0 , where the vertex set of G0 is V 0 = {v 0 : v ∈ V (G)}, by adding r − dG (v) parallel edges between v and v 0 for each vertex v of G. Then H is r-regular and r-edge colourable. Since Conjecture 4.7.2 is assumed to be true for all r-edge colourable r-regular graphs, H is r-edge choosable. Hence G is also r-edge choosable. As G is an arbitrary r-edge colourable graph, Conjecture 4.7.1 is true. Let G be an n-vertex r-regular r-edge colourable graph. Let L(G) be the line graph of G. We are interested in the Alon–Tarsi number of L(G). In particular, we are interested in the problem whether L(G) has Alon–Tarsi number r. As a strengthening of Conjecture 4.7.2, one may wonder if it is true that for any r-edge colourable r-regular graph G, its line graph L(G) has Alon–Tarsi number r. However, this is not true. For example, if G = K3,3 , then G is 3-edge colourable and 3-regular, but its line graph L(G) has Alon–Tarsi number 4. Nevertheless, for some classes of r-edge colourable r-regular graphs G, L(G) does have Alon–Tarsi number r. In the next two sections, we shall show that the following hold: (1) If G is an r-edge colourable r-regular planar graph, then L(G) has Alon–Tarsi number r. (2) If G = Kp+1 and p is an odd prime, then L(G) has Alon– Tarsi number p. Let D be an orientation of L(G) and let Y fD (x) = (xe − xe0 ) (e,e0 )∈E(D)

be the graph polynomial of L(G) associated with D. Note that for a different orientation D0 of L(G), fD (x) = ±fD0 (x). We are only interested in whether a certain monomial in fD is non-vanishing or not. So the choice of the orientation D is irrelevant. However, for some of the proofs, we may choose appropriate orientations D to make the calculations easier. The line graph L(G) is 2(r − 1)-regular (if e, e0 are parallel edges in G, then e and e0 are connected by two parallel edges in L(G)). Thus the number of variables xe is |E(G)| = nr/2, and the degree of fD is |E(L(G))| = nr(r − 1)/2. ThusQthe Alon–Tarsi number of L(G) is r if and only if the monomial e∈E(G) xr−1 is e non-vanishing.

i

i

“2020-12-10” — 2021/5/8 — 18:39 — page 91 — #103

Generalizations of CNS and Applications By Corollary 4.4.3, the coefficient of the monomial in fD (x) is equal to X −nr/2 ((r − 1)!)

91 Q

e∈E(G)

xr−1 e

φ:E(G)→{0,1,...,r−1}

 ×

Y

e∈E(G)

   r−1  (−1)(r−1+φ(e)) φ(e)

×fD (φ). If φ : E(G) → {0, 1, . . . , r − 1} is not a proper edge colouring of G, then fD (φ) = 0. Thus the summation above can be restricted to proper edge colourings φ of G. On the other hand, if φ 1, . . . , r − 1}
is a proper edge colouring of G, then Q: E(G) → {0, (r−1+φ(e)) r−1 e∈E(G) (−1) φ(e) is a constant, as there are exactly n/2 edges e for which φ(e) = i for each i ∈ {0, 1, . . .Q , r − 1}. Therefore, to prove that the coefficient of the monomial e∈E(G) xr−1 in e P fD (x) is non-zero, it is equivalent to showing that φ∈C fD (φ) 6= 0, where C is the set of mappings φ : E(G) → {0, 1, . . . , r − 1} that are proper edge colourings of G. For a vertex v of G, let E(v) be the set of edges of G incident to v, and denote by Qv the r-clique of L(G) induced by E(v). Let Y fD,v (x) = (xe − xe0 ). (e,e0 )∈E(D),e,e0 ∈E(Qv )

Q Then fD (x) = v∈V (G) fD,v (x). If φ is a proper edge colouring of G, then for any vertex v of G, Y

|fD,v (φ)| =

(j − i), and hence |fD (φ)|

0≤i