On choosability with separation of planar graphs with lists of different sizes

# On choosability with separation of planar graphs with lists of different sizes

H. A. Kierstead School of Mathematical and Statistical Sciences, Arizona State University, Tempe, AZ 85287, USA. E-mail: hal.kierstead@me.com. Research of this author is supported in part by NSA grant H98230-12-1-0212.    Bernard Lidický Department of Mathematics, University of Illinois. E-mail: lidicky@illinois.edu
August 28, 2019
###### Abstract

A -list assignment of a graph is a mapping that assigns to each vertex a list of at least colors and for any adjacent pair , the lists and share at most colors. A graph is -choosable if there exists an -coloring of for every -list assignment . This concept is also known as choosability with separation.

It is known that planar graphs are -choosable but it is not known if planar graphs are -choosable. We strengthen the result that planar graphs are -choosable by allowing an independent set of vertices to have lists of size 3 instead of 4.

## 1 Introduction

Given a graph , a list assignment is a mapping assigning to each vertex a list of colors . An -coloring is a vertex coloring such that for each vertex and for each edge . A graph is said to be -choosable if there is an -coloring for each list assignment where for each vertex . The minimum such is known as the choosability of , denoted . A graph is said to be -choosable if there is an -coloring for each list assignment where for each vertex and for each edge .

This concept is called choosability with separation, since the second parameter may force the lists of adjacent vertices to be somewhat separated. If is -choosable, then is also -choosable for all and . A graph is -choosable if and only if it is -choosable. Clearly, all graphs are -choosable for . Thus, for a graph and each , there is some threshold such that is -choosable but not -choosable.

The concept of choosability with separation was introduced by Kratochvíl, Tuza, and Voigt [4]. They used the following, more general definition. A graph is -choosable, if for every list assignment with for each and whenever are adjacent vertices, is -tuple -colorable. Since we consider only in this paper, we use a simpler notation. They investigate this concept for both complete graphs and sparse graphs. The study of dense graphs were extended to complete bipartite graphs and multipartite graphs by Füredi, Kostochka, and Kumbhat [2, 3].

Thomassen [5] proved that planar graphs are 5-choosable, and hence they are -choosable for all . Voigt [7] constructed a non--choosable planar graph, and there are also examples of non--choosable planar graphs. Kratochvíl, Tuza, and Voigt [4] showed that all planar graphs are -choosable and asked:

###### Question 1 ([4]).

Are all planar graphs -choosable?

Voigt [6] also constructed a non--choosable triangle-free planar graph. Škrekovski [8] observed that there are examples of triangle-free planar graphs that are not -choosable, and posed:

###### Question 2 ([8]).

Are all planar graphs -choosable?

Kratochvíl, Tuza and Voigt [4] proved a partial case of Question 2 by showing that every triangle-free planar graph is -choosable.

Choi et. al [1] proved that every planar graph without -cycles is -choosable and that every planar graph without -cycles and -cycles is -choosable.

In this paper we give a strengthening of the result that every planar graph is -choosable by allowing some vertices to have lists of size three. In a -list assignment on , for every holds that . In a -list assignment , for every holds that . An intermediate step is to investigate the case where for every holds that .

A -list assignment is a list assignment where and for every pair of adjacent vertices .

The main result of this paper is the following theorem.

###### Theorem 3.

Let be a planar graph and be an independent set. If is a -list assignment such that for every and for every then has an -coloring.

The following theorem shows it is not possible to strengthen Theorem 3 by allowing for every vertex and requiring that for every .

###### Theorem 4.

For every there exists a planar graph and a -list assignment such that for every , for every , and is not -colorable.

We first give some notation. In the next section, we prove Theorem 3 using Thomassen’s precoloring extension method. In the last section we show a construction proving Theorem 4.

### 1.1 Notation

Given a graph and a cycle , an edge of is a chord of if , but is not an edge of . If is a plane graph, then let be the subgraph of consisting of the vertices and edges drawn inside the closed disc bounded by , and let be the subgraph of obtained by removing all vertices and edges drawn inside the open disc bounded by . In particular, . Finally, denote the characteristic function of a set by . So if ; else .

## 2 Main theorem

In this section, we prove Theorem 3 by proving a slightly stronger theorem that is more amenable to induction. Observe that any list assignment satisfying the assumptions of Theorem 3 also satisfies the conditions of the following theorem.

###### Theorem 5.

Let be a plane graph with outer face and let be a subpath of containing at most two vertices. Let be an independent set. If is a -list assignment satisfying the following conditions:

1. for ,

2. is -colorable,

3. for every there is at most one with ,

then is -colorable.

###### Proof.

Let and be a counterexample where is as small as possible. Moreover, assume that the sum of the sizes of the lists is also as small as possible subject to the previous condition. Define if ; else . Since is minimal, we have:

###### Claim 1.

For all edges

1. ;

2. ; and

3. implies for every triangle .

###### Proof.

For (1), note that , and if then it suffices to -color , which is possible by minimality. For (2), the definitions imply , and if then -coloring , and then coloring with yields an -coloring of . Finally consider (3). By (1), there exists a color with . Thus , so by definition and (1), . ∎

###### Claim 2.

is 2-connected. In particular, is a cycle.

###### Proof.

Suppose not. Then there exists and two induced connected subgraphs and of where and . Moreover, both and have at least two vertices. By symmetry assume that . By the minimality of , there exists an -coloring of . Let be a list assignment on such that if , and otherwise. Since and satisfy the assumptions of Theorem 5, there exists an -coloring of . Colorings and coincide on ; hence is an -coloring of , a contradiction. ∎

###### Claim 3.

(1) for all , and (2) .

###### Proof.

The minimality of and (iii) imply (1). Using Claim 2, is a path. Since is independent, if then , contradicting (1). ∎

###### Claim 4.

does not contain a separating triangle with a vertex in .

###### Proof.

Let be a separating triangle in and let . Assume that and . By the minimality of , there exists an -coloring of .

Let , and . Define a list assignment on vertices in the following way:

 L′(u)=⎧⎪⎨⎪⎩φ(u) if u∈{x,y},L(u)−φ(z) if uz∈E(G−P′),L(u)otherwise.

Since , no neighbor of is in . Thus condition (iii) of Theorem 5 is satisfied for and . Condition (ii) is witnessed by . Since each vertex is on the outer face of , but not , it is straightforward to check that (i) is satisfied. Hence has an -coloring . The coloring is an -coloring of , a contradiction. ∎

###### Claim 5.

If is a chord of then neither nor is in , and there exists such that , , and .

###### Proof.

Suppose is a chord of . Let and be subgraphs of where and . Since is a chord, both and have at least three vertices. By symmetry assume that .

First suppose contains exactly three vertices, say . Using Claim 1, . So and . By condition (i), implies . Thus , since and are the only possible neighbors of . Finally, since , Claim 1.3 implies . Thus , and so .

Now suppose for a contradiction that has at least four vertices. Define in the following way. If there exists a vertex such that is adjacent to both and then is obtained from by adding a new vertex adjacent to and to the outer face of . Moreover, let and let be an extension of by defining . Notice that is unique if it exists, since Claim 4 implies has no separating triangles that contain a vertex of . If no such exists, let , , and . If contains , neither nor is in . Hence is indeed an independent set. Using that is on the outer face satisfies conditions (i,ii,iii). By the minimality of , there exists an -coloring of which gives an -coloring of .

Define a list assignment on by if , else . We wish to use as . Conditions (i,ii) of Theorem 5 hold since satisfies them. For (iii), consider a vertex with . As remarked above, . Since , and is an -coloring of , there exists with . Then implies , and (iii) holds. By the minimality of , there exists an -coloring of . Colorings restricted to and coincide on ; hence is an -coloring of , a contradiction. ∎

By the minimality of the sum of the sizes of the lists, we can assume . Let , where , identifying index with index . Choose with minimum index . Such an index exists by Claim 3. Claim 5 implies is not a chord, and condition (i) implies .

Select a set and an -coloring of by the following rules:

1. If is not a chord then set and pick .

2. Else, if there is , then set and .

3. Else set . Pick:

1. ;

2. , if ; else ;

3. .

See Figure 1 for an illustration of these rules. Observe that exactly one of (X1), (X2), or (X3) applies and is well defined. Also, in cases (X2) and (X3), Claim 5 implies , and so . Thus the sizes of their lists are as claimed in Figure 1. In (X3) either or . Also, by Claims 1(3) and 5, . Hence is also well defined. Moreover, , and so .

Let , , and be the list assignment on defined by

 L′(v)=L(v)∖{φ(x):x∈N(v)∩X}.

It suffices to show that , , and satisfy the assumptions of Theorem 5. Then by the minimality of , there is an -coloring of , and by the choice of , the function is an -coloring of , a contradiction.

Now we verify that , , and satisfy the assumptions of Theorem 5. Since is an independent set, so is . Let . Clearly condition (i) holds for vertices in . By Claim 5, all chords have the form . Thus was chosen so that . Hence the condition (i) is satisfied for . Condition (ii) holds since did not change. Since , Claim 3(1) implies condition (iii).

It remains to show that every satisfies condition (i). Let be the outer face of . Since each vertex of has a neighbor in , . Thus it suffices to show that . If then . Otherwise . Then is handled by rule (X3). So and . If then Claim 1(3) implies . Anyway, , and we are done. ∎

## 3 Lists of size 3 are necessary

In this section we give a proof of Theorem 4. The construction is analogous to the construction that bipartite graphs are not 2-choosable.

###### Proof of Theorem 4.

Let be given. Let be a complete bipartite graph with part of size and another part of size 2 formed by vertices and . Let be a list assignment assigning to a list of colors and to a list of colors . To the other vertices, assigns distinct lists of form where . There are such lists which is exactly the size of . Notice that for every edge . See Figure 2 for a sketch of and .

Suppose that there is an -coloring of . It assigns colors to and to for some . However, there is a vertex with list , a contradiction. Hence is not -colorable. ∎

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