Planar graphs without cycles of length 4 or 5 are (11:3)-colorable

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Abstract

A graph G is (a:b)-colorable if there exists an assignment of b-element subsets of {1,,a} to vertices of G such that sets assigned to adjacent vertices are disjoint. We show that every planar graph without cycles of length 4 or 5 is (11:3)-colorable, a weakening of recently disproved Steinberg’s conjecture. In particular, each such graph with n vertices has an independent set of size at least 311n.

Introduction

A function that assigns sets to all vertices of a graph is a set coloring if the sets assigned to adjacent vertices are disjoint. For positive integers a and ba, an (a:b)-coloring of a graph G is a set coloring with range {1,,a}b, i.e., a set coloring that to each vertex assigns a b-element subset of {1,,a}. The concept of (a:b)-coloring is a generalization of the conventional vertex coloring. In fact, an (a:1)-coloring is exactly an ordinary proper a-coloring.

The fractional chromatic number of G, denoted by χf(G), is the infimum of the fractions ab such that G admits an (a:b)-coloring. Note that χf(G)χ(G) for any graph G, where χ(G) is the chromatic number of G. The fractional coloring was first introduced in 1973 [10] to seek for a proof of the Four Color Problem. Since then, it has been the focus of many intensive research efforts, see [12]. In particular, fractional coloring of planar graphs without cycles of certain lengths is widely studied. Pirnazar and Ullman [11] showed that the fractional chromatic number of a planar graph with girth at least 8k4 is at most 2+1k. Dvořák et al. [7] showed that every planar graph of odd-girth at least 9 is (5:2)-colorable. Recently, Dvořák et al. [6] showed that every planar triangle-free graph on n vertices is (9n:3n+1)-colorable, and thus it has fractional chromatic number at most 333n+1.

Well-known Steinberg’s Conjecture asserts that every planar graph without cycles of length 4 or 5 is 3-colorable. Recently, Steinberg’s conjecture was disproved [4]. This conjecture, though disproved, had motivated a lot of research, see [3]. Since χf(G)χ(G) for any graph G, it is natural to ask whether there exists a constant c<4 such that xf(G)c for all planar graphs without cycles of length 4 or 5. In this paper, we confirm this is the case for c=113. In fact, we prove the following stronger theorem.

Theorem 1.1

Every planar graph without cycles of length 4 or 5 is(11:3)-colorable, and thus its fractional chromatic number is at most 113.

The independence number α(G) of a graph G is the size of a largest independent set in G. The independence ratio of G is the quantity α(G)|V(G)|. The famous Four Color Theorem [2] implies that every planar graph has independence ratio at least 14. In 1976, Albertson [1] proved a weaker result that every planar graph has independence ratio at least 29 without using the Four Color Theorem. In 2016, Cranston and Rabern [5] improved this constant to 313. If G is a triangle-free planar graph, a classical theorem of Grőtzsch [9] says that G is 3-colorable, and thus G has independence ratio at least 13. This bound can be slightly improved—Steinberg and Tovey [13] proved that the independence ratio is at least 13+13|V(G)|, and gave an infinite family of planar triangle-free graphs for that this bound is tight. Steinberg’s Conjecture would imply that every planar graph without cycles of length 4 or 5 has independence ratio at least 13, and it is not known whether this weaker statement holds or not. Since α(G)|V(G)|χf(G) for any graph G, we have the following corollary by Theorem 1.1.

Corollary 1.2

Every planar graph without cycles of length 4 or 5 has independence ratio at least311.

It is not clear whether the constant 113 from Theorem 1.1 is the best possible, and we suspect this is not the case. Hence, the following question is of interest.

Problem 1.3

What is the infimum of fractional chromatic numbers of planar graphs without cycles of length 4 or 5?

Let us remark that the counterexample to Steinberg’s conjecture constructed in [4] is (6:2)-colorable, and thus we cannot even exclude the possibility that the answer is 3.

The proof of Theorem 1.1 naturally proceeds in list coloring setting. A list assignment for a graph G is a function L that to each vertex v of G assigns a set L(v) of colors. A set coloring φ of G is an L-set coloring if φ(v)L(v) for all vV(G). For a positive integer b, we say that φ is an (L:b)-coloring of G if φ is an L-set coloring and |φ(v)|=b for all vV(G). If such an (L:b)-coloring exists, we say that G is (L:b)-colorable. For an integer ab, we say that G is (a:b)-choosable if G is (L:b)-colorable from any assignment L of lists of size a. We actually prove the following strengthening of Theorem 1.1.

Theorem 1.4

Every planar graph without cycles of length 4 or 5 is(11:3)-choosable.

Section snippets

Colorability of small graphs

Let us start with some technical results on list-colorability of small graphs, especially paths and cycles. In the proofs, it is convenient to work with a non-uniform version of set coloring. Let f:V(G)Z0+ be an arbitrary function. An (L:f)-coloring of a graph G is an L-set coloring φ such that |φ(v)|=f(v) for all vV(G). If such an (L:f)-coloring exists, we say that G is (L:f)-colorable. We repeatedly use the following simple observation.

Lemma 2.1

Let L be an assignment of lists to vertices of a graph G

Properties of a minimal counterexample

We are going to prove a mild strengthening of Theorem 1.4 where a clique (one vertex, two adjacent vertices, or a triangle) is precolored. A (hypothetical) counterexample (to this strengthening) is a triple (G,L,Z), where G is a plane graph without 4- or 5-cycles, Z is the vertex set of a clique of G, and L is an assignment of lists of size 11 to vertices of V(G)Z and pairwise disjoint lists of size 3 to vertices Z, such that G is not (L:3)-colorable. The order of the counterexample is the

Notation

Consider a minimal counterexample (G,L,Z). We say that the faces of G of length at least 6 are 6+-faces. Since G is 2-connected by Lemma 3.1, every face of G is bounded by a cycle, and in particular, every face of G is either a 3-face or a 6+-face. A vertex vV(G) is a k-vertex if v is internal and deg(v)=k. We say that v is a k+-vertex if either vZ or deg(v)k.

Let v1vv2 be a part of the cycle bounding a 6+-face f of G, and for i{1,2}, let fif be the face incident with the edge vvi. If both f

(11:3)-colorability of planar graphs

We are now ready to prove our main result.

Proof of Theorem 1.4

Suppose for a contradiction that there exists a plane graph G0 without 4- or 5-cycles and an assignment L0 of lists of size 11 to vertices of G0 such that G0 is not (L0:3)-colorable. Let z be any vertex of G0, let L0(z) be any 3-element subset of L0(z), and let L0(v)=L0(v) for all vV(G){z}. Then G0 is not (L0:3)-colorable, and thus (G0,L0,{z}) is a counterexample.

Therefore, there exists a minimal counterexample (G,L,Z). Let ch be the assignment of

Acknowledgments

The research leading to these results has received funding from the European Research Council under the European Union’s Seventh Framework Programme (FP/2007–2013)/ERC Grant Agreement n.616787. Xiaolan Hu is partially supported by NSFC, China under Grant Number 11601176 and NSF of Hubei Province, China under Grant Number 2016CFB146.

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