Planar graphs without cycles of lengths 4 and 5 and close triangles are DP-3-colorable

# Planar graphs without cycles of lengths 4 and 5 and close triangles are DP-3-colorable

Yuxue Yin and Gexin Yu Department of Mathematics, Central China Normal University, Wuhan, Hubei, 430079 China.
Department of Mathematics, The College of William and Mary, Williamsburg, VA, 23185, USA.
###### Abstract.

Montassier, Raspaud, and Wang (2006) asked to find the smallest positive integers and such that planar graphs without -cycles and are -choosable and planar graphs without -cycles and are -choosable, where is the smallest distance between triangles. They showed that and . In this paper, we show that the following planar graphs are DP-3-colorable: (1) planar graphs without -cycles and are DP--colorable, and (2) planar graphs without -cycles and are DP--colorable. DP-coloring is a generalization of list-coloring, thus as a corollary, and . We actually prove stronger statements that each pre-coloring on some cycles can be extended to the whole graph.

The research of the last author was supported in part by the Natural Science Foundation of China (11728102) and the NSA grant H98230-16-1-0316.

## 1. Introduction

Coloring of planar graphs has a long history. The famous Four Color Theorem states that every planar graph is properly -colorable, where a graph is properly -colorable if there is a function that assigns an element to each so that adjacent vertices receive distinct colors.

Grötzsch [17] showed every planar graph without 3-cycles is 3-colorable. But it is NP-complete to decide whether a planar graph is -colorable. There were heavy research on sufficient conditions for a planar graph to be -colorable. Three typical conditions are the following:

• One is in the spirit of the Steinberg’s conjecture (recently disproved) or Erdős’s problem that forbids cycles of certain lengths. Borodin, Glebov, Raspaud, and Salavatipour [11] showed that planar graphs without -cycles are -colorable, and it remains open to know if one can allow -cycle.

• Havel [16] proposed to make large enough, where is the smallest distance between triangles. Dvor̂ák, Kral, and Thomas[14] showed that suffices.

• The Bordaux approach [12] combines the two kinds of conditions. Borodin and Glebov [10] showed that planar graphs without -cycles and are -colorable. It is conjectured [12] that suffices.

Vizing [27], and independently Erdős, Rubin, and Taylor [15] introduced list coloring as a generalization of proper coloring. A list assignment gives each vertex a list of available colors. A graph is -colorable if there is a proper coloring of such that for each . A graph is -choosable if is -colorable for each with . Clearly, a proper -coloring is an -coloring when for all .

While list coloring provides a powerful tool to study coloring problems, some important techniques used in coloring (for example, identification of vertices) are not feasible in list coloring. Therefore, it is often the case that a condition that suffices for coloring is not enough for the corresponding list-coloring. Thomassen [25, 26] showed that every planar graph is -choosable and every planar graph without -cycles is 3-choosable, but Voigt [28, 29] gave non--choosable planar graphs and non--choosable triangle-free planar graphs.

Sometimes we do not know if a stronger condition would help. For example, Borodin ([8], 1996) conjectured that planar graphs without cycles of lengths from to are -choosable.

In the spirit of Bordeaux conditions, Montassier, Raspaud, and Wang [24] gave the following conditions for a planar graph to be -choosable:

###### Theorem 1.1 (Montassier, Raspaud, and Wang [24]).

A planar graph is -choosable if

• contains no cycles of lengths and and , or

• contains no cycles of lengths from to and .

There exist planar graphs without -, -cycles and that are not -choosable.

They asked for the optimal conditions on for the same conclusions.

Very recently, Dvor̂ák and Postle [13] introduced DP-coloring (under the name correspondence coloring), which helped them confirm the conjecture by Borodin mentioned above. DP-coloring is a generalization of list-coloring, but it allows identification of vertices in some situations.

###### Definition 1.1.

Let be a simple graph with vertices, and be a list assignment of . For each vertex , let . For each edge in , let be a matching (maybe empty) between the sets and and let , called the matching assignment. Let be the graph that satisfies the following conditions

• .

• for all , the set forms a clique.

• if , then the edges between and are those of

• if , then there are no edges between and

If contains an independent set of size , then has an -coloring. The graph is DP--colorable if, for each -list assignment and each matching assignment over , it has an -coloring. The minimum such that is DP--colorable is the DP-chromatic number of , denoted by .

As in list coloring, we refer to the elements of as colors and call the element chosen in the independent set of an -coloring as the color of .

We should note that DP-coloring and list coloring can be quite different. For example, Bernshteyn [2] showed that the DP-chromatic number of every graph with average degree is , while Alon [1] proved that and the bound is sharp.

Much attention was drawn on this new coloring, see for example, [2, 3, 4, 5, 6, 7, 18, 19, 20, 23, 22]. We are interested in DP-coloring of planar graphs. Dvořák and Postle [13] noted that Thomassen’s proofs [25] for choosability can be used to show if is a planar graph, and if is a planar graph with no 3-cycles and 4-cycles. Some sufficient conditions were given in [18, 19, 23] for a planar graph to be DP--colorable. Sufficient conditions for a planar graph to be DP--colorable are obtained in [21] and [22]. In particular,

###### Theorem 1.2.

([21, 22]) A planar graph is DP-3-colorable if it has no cycles of length , where .

In this paper, we use DP-coloring to improve the results in Theorem 1.1. To state our results, we have to introduce extendability. Let be a graph and be a subgraph of . Then is DP-3-colorable if every DP-3-coloring of can be extended to .

A -cycle is bad if it is the outer -cycle in a subgraph isomorphic to the graphs in Figure 1. A -cycle is good if it is not a bad -cycle.

###### Theorem 1.3.

Let be a planar graph that contains no -cycles and . Let be a -, -, -, -cycle or a good -cycle in . Then each DP--coloring of can be extended to .

###### Theorem 1.4.

Let be a planar graph that contains no -cycles and . Let be a cycle of length or in . Then each DP--coloring of can be extended to .

The proofs of Theorem 1.3 and  1.4 use identification of vertices. We shall note that the planar graphs in the following corollary was not known to be 3-choosable.

###### Corollary 1.5.

The following planar graphs are DP--colorable (thus also -choosable):

• no -cycles and , or

• no -cycles and .

###### Proof.

Let be a planar graph under consideration. Note that is DP-3-colorable if contains no -cycle. So we may assume that contains a -cycle. Then by Theorem 1.3, is DP-3-colorable when . So we let and assume that contains no -cycles. By Theorem 1.2, we may assume that contains a cycle of length in . Now by Theorem 1.4, is DP-3-colorable. ∎

We use discharging method to prove the results. One part of the proofs is to show some structures to be reducible, that is, a coloring outside of the structure can be extended to the whole graph. The following lemma from [21] provides a powerful tool to prove the reducibility.

###### Lemma 1.6.

[21] Let and be a subgraph of . If the vertices of can be ordered as such that the following hold

• , and has no neighbor outside of ,

• and has at least one neighbor in ,

• for each , has at most neighbors in ,

then a DP--coloring of can be extended to a DP--coloring of .

We end the introduction with some notations used in the paper. All graphs mentioned in this paper are simple. A -vertex (-vertex, -vertex) is a vertex of degree (at least , at most ). The same notation will be applied to faces and cycles. We use and to denote the set of vertices and faces in , respectively. An -edge is an edge with . An -face is a -face with , respectively. Recall that two faces are adjacent if they share a common edge, and are intersecting if they share a common vertex. A vertex is incident to a face if it is on the face, and is adjacent to a face if it is not on the face but adjacent to a vertex on the face. A vertex in is light if it is incident to a -face. If is a cycle in an embedding of , we use and to denote the sets of vertices located inside and outside a cycle , respectively. The cycle is called a separating cycle if . An edge is straight if every satisfies . We note that if all edges in a subgraph are straight, then a DP--coloring on the subgraph is the same as a proper -coloring.

## 2. Proof of Theorem 1.3

Let be a counterexample to Theorem  1.3 with minimum number of vertices, where is a -,-,-,-cycle or a good -cycle. Below we let be a plane graph. The following was shown in [21] for every non-DP-3-colorable graphs.

###### Lemma 2.1.

For each , and for each , .

###### Lemma 2.2.

There exist no separating -cycles or good -cycle.

###### Proof.

First of all, we note that cannot be a separating cycle. For otherwise, we may extend the coloring of to both inside and outside , respectively, then combine them to get a coloring of . So we may assume that is the outer face of the embedding of .

Let be a separating -cycle or good -cycle in . By the minimality of , the coloring of can be extended to . Now that is colored, thus by the minimality of again, the coloring of can be extended to . Combine inside and outside of , we have a coloring of , which is extended from the coloring of , a contradiction. ∎

By Lemma 2.2, if is a bad -cycle, then the subgraph in Figure 1 that contains must be induced. From now on, we will let be the outer face of . Likewise, if contains a chord, then by Lemma 2.2, contains no other vertices, so the coloring on is also a coloring of . Therefore, we may assume that is chordless as well. A vertex is internal if it is not on and a face is internal if it contains no vertex of .

For convenience, a -face in is bad if and is adjacent to a -face, otherwise, it is good. Let be a (3,3,3,3,3,3)-face adjacent to a -face . We call the vertex on but not on the roof of , and the base of .

###### Lemma 2.3.

Let be an internal -face in and be an internal -face adjacent to . Then each of the followings holds:

1. cannot contain vertices of another -face;

2. If is a -face such that and share a common -edge, then the other -edge of cannot be on another internal -face.

3. If is a -face, then cannot be adjacent to an internal -face. This means a -vertex on an internal -face cannot be a roof.

###### Proof.

(a) follows from the condition on the distance of triangles. To show (b) and (c), let so that is the common edge of and and . Let be the -face adjacent to .

(b) We have and for . Order the vertices on and as

 y, v4, v3, v2, v1, z, x, u4, u3, u2, u1.

Let be the set of vertices in the list. By Lemma  1.6, a DP--coloring of can be extended to , a contradiction.

(c) We have and for , and . Order the vertices on and as

 x, z, v1, v2, v3, y, u1, u2, u3, u4.

Let be the set of vertices in the list. By Lemma  1.6, a DP--coloring of can be extended to , a contradiction. ∎

###### Lemma 2.4.

Let be an internal -face that is adjacent to an internal -face , then or .

###### Proof.

We assume that , and use to denote the neighbor of other than . First we may rename the lists of vertices in so that each edge in is straight.

Consider the graph obtained from by identifying and . We claim that no new cycles of length from to or multiple edges are created, for otherwise, there is a path of length or from to in , which together with forms a separating -cycle or good -cycle, a contradiction to Lemma 2.2. Clearly, . Finally, we claim that no new chord in is formed in , for otherwise, and is adjacent to a vertex on , then there is a path between and on with length at most four, which with forms a separating -cycle or good -cycle.

By minimality of , the --coloring of can be extended to a --coloring of . Now keep the colors of all vertices in and color and with the color of the identified vertex. Now properly color , and then color with the color of , which we can do it because the edges are straight and the color of is different from the one on and . Now color properly in the order, we obtain a coloring of , a contradiction. ∎

###### Lemma 2.5.

Let be a path in and be an internal -face so that . If , then . (And similarly, if , then .)

###### Proof.

Assume that . Since there is no separating -cycles by Lemma 2.2, the -cycles and are both -faces. Then by Lemma 2.4, . Let be the fourth neighbor of . We may rename the lists of vertices in so that the edges are straight.

Consider the graph obtained from by identifying and . Since , and cannot be on triangles. We claim that no new cycles of length from to are created, for otherwise, there is a path of length or from to in , which together with forms a separating -cycle or good -cycle, a contradiction to Lemma 2.2. Clearly, . Finally, we claim that no new chord in is formed in , for otherwise, and is adjacent to a vertex on , then there is a path between and on with length at most four, which again forms a good separating cycle with of forbidden length.

By minimality of , the --coloring of can be extended to a --coloring of . Now keep the colors of all vertices in and color and with the color of the identified vertex. For , let and . Then , and . So we can extend to a --coloring of by properly coloring and coloring with the color of , and coloring in order, a contradiction. ∎

We use to denote the initial charge of a vertex or face in and to denote the final charge after the discharging procedure. We use for each vertex , for each face , and . Then by Euler formula, To lead to a contradiction, we shall prove that for all and is positive. For shortness, let .

We use the following discharging rules:

1. Each internal -vertex gives to its incident -face, and to its base or incident -face. Each internal -vertex gives to its adjacent -face and to its incident -faces that are not adjacent to its adjacent -face, and each internal -vertex gives to its adjacent -face and to its incident -faces that are not adjacent to its adjacent -face.

2. Each -face or non-internal -face other than gives to each of its adjacent internal -faces and the rest to the outer face. Each internal -face gives to its adjacent internal -face when it shares an -edge with the -face, or contains a -vertex that is not adjacent to a -face, or it is a -face.

3. The outer face gets from each , gives to each intersecting -face and to each adjacent bad -face with an internal -face.

We first check the final charge of vertices in . By (R3), each vertex on has final charge . So let be an internal vertex of . Then by Lemma 2.1, . Note the if .

Let . If is on a -face, then it is not adjacent to other -faces, so by (R1), it gives to the -face, to each other incident face and possibly to its base (at most one by definition), so . If is adjacent to a -face, then it is not on or adjacent to other -faces, so by (R1), it gives at most to the -face, and to each other incident -faces that are not adjacent to the -face, hence . If is not on or adjacent to -faces, then by (R1), its final charge is .

Let . Let for be the incident face of in clockwise order. First assume that is on a -face. By Lemma 2.3 (b) and (c), cannot be a roof and on a -face at the same time, so by (R1), gives out at most to -faces and to the -face, thus . Now assume that is adjacent to a -face. Then cannot be adjacent other -faces. By (R1), gives at most to the -face and to each of the other -faces that are not adjacent to the -face, and . Finally assume that is not on or adjacent to any -face. Then by (R1), .

Now we check the final charge of faces. Let . If contains vertices of , then by (R3), . So we assume that is internal. If is incident with at least two -vertices, then gets from each of the incident -vertices by (R1), thus If is incident with exactly one -vertex, then gets from the incident -vertex by (R1) and gets from each of the incident -face by (R2).

Now we assume that is an internal -face. Let and let be the three adjacent faces of so that contains and contains . If is adjacent to three - or non-internal -faces, then it gets from each by (R2) and its final charge is at least . So we may assume that it is adjacent to an internal -face, say . By Lemma 2.4, is adjacent to at least one internal -vertex (say ) which is on . If is adjacent to three internal -faces, then by Lemma 2.4, one of and is a -vertex, and by Lemma 2.5, either one of is a -vertex, in which case by (R1), , or they are all -vertices, in which case by (R1), , or one of them (say ) is a -vertex and other two are -vertices, in which case by Lemma 2.5, and both contain -vertices that are not adjacent to so by (R1) and (R2), gets from the two -vertices and from and . Likewise, if and are both - or non-internal -faces, then by (R1) and (R2), . So we may assume that one of or is an internal -face and the other is a - or non-internal -face. If is an internal -face, then by Lemma 2.4, or is a -vertex, thus by (R1) gets from the two adjacent -vertices and by (R2) gets from . So we may assume that is an internal -face and is a - or non-internal -face, and furthermore assume that are -vertices and . Now by Lemma 2.5, and both contain -vertices that are not adjacent to , so by (R2), gets from and , from , and by (R1), from , and we have .

Since contains no - or -cycles, we only need to check the -faces. If , then is adjacent to at most -faces, so after (R1), .

Let . If is good or contains vertices of , then . Now we assume that is an internal bad -face that is adjacent to an internal -face on edge with .

• If , then gives nothing to . So .

• If and , then gets from and gives to . Thus .

• If , then by (R2), gives to only when contains a -vertex that is not adjacent to the -face, in which case, gets from the -vertex by (R1). So we always have .

• Let and . If is an internal -face, then it gets from , or else contains another -vertex, from which gets . Thus .

We call a bad -face in special if is adjacent to one internal -face.

###### Lemma 2.6.

The final charge of is positive.

###### Proof.

Assume that . Let be the set of edges between and . Let be the number of edges in that is not on a -face and be the number of charges receives by (R3). Let and be the number of special -faces. By (R3) and (R4), the final charge of is

 μ∗(C0) =d(C0)+6+∑v∈C0(2d(v)−6)−3f3−f6+x =d(C0)+6+∑v∈C02(d(v)−2)−2d(C0)−3f3−f6+x =6−d(C0)+2|E(C0,G−C0)|−3f3−f6+x ≥6−d(C0)+f3+2e′−f6+x,

where the last equality follows from that each -face in contains two edges in .

Note that for each special -face , no edge in is on -faces. Then . When , is adjacent to at least three -faces, so , and it follows that and and , in which case, we have a bad -cycle as in the second graph in Figure 1. So we may assume that . Thus

 μ∗(C0)≥6−d(C0)+f3+2e′−f6+x≥6−d(C0)+f3+x+f6+2.

Since , . So if , then and . Now that the -face shares at most four vertices with , is adjacent to a -face that contains at least five consecutive -vertices on , thus by (R3), , a contradiction.

Therefore, we may assume that , and . So .

Let . It follows that .

• Let . Then and . Since is not a bad -cycle, is adjacent to a -face and is adjacent to the -face, so by (R3), gives at least to , that is, , a contradiction.

• Let . Then and . Note that is adjacent to a -face that contains at least consecutive -vertices, thus by (R3), gives at least to , a contradiction to .

Finally let . Then , and each edge in is on a -face. Note that we may assume that , for otherwise . Now follow the boundaries of the -faces adjacent to , each of the triangles is encountered twice, thus the -faces do not give charge to at least triangles, so . It follows and . In this case, is adjacent to a -face that contains at least consecutive -vertices on . Then by (R3), gives at least to , a contradiction to . ∎

## 3. Proof of Theorem 1.4

Let be a counterexample to Theorem  1.4 with minimum number of vertices, where is a -, -, - or -cycle. Let be a plane graph.

For each , .

###### Proof.

Let be a vertex with . Any -coloring of can be extended to since has at most elements of forbidden by the colors selected for the neighbors of , while . ∎

###### Lemma 3.2.

The graph has no separating cycles of length or .

###### Proof.

First of all, we note that cannot be a separating cycle. For otherwise, we may extend the coloring of to both inside and outside , respectively, then combine them to get a coloring of . So we may assume that is the outer face of the embedding of .

Let be a separating cycle of length or in . By the minimality of , the coloring of can be extended to . Now that is colored, thus by the minimality of again, the coloring of can be extended to . Combine inside and outside of , we have a coloring of , which is extended from the coloring of , a contradiction. ∎

So we may assume that is the outer face of the embedding of in the rest of this paper. Like in the previous section, we may assume that is chordless. A face is internal if none of its vertices is on , and a vertex is internal if it is not on .

###### Lemma 3.3.

Let be an internal -face that is adjacent to an internal -face and is incident with at least six -vertices. Then none of the followings occur

• contains a -edge that is on an internal -face.

• contains seven -vertices and is adjacent to an internal -face.

• is adjacent to another internal -face.

###### Proof.

Let , and be the -edge that is on an internal -face . Since , by symmetry we may assume that is on a -face .

(a) or (b): If and , then let be the set of vertices listed as:

 v2,v3,v4,v45,v5,v6,v7,v1,v12.

If , then let be the set of vertices listed as:

 v1,v7,v6,v5,v45,v4,v3,v2,v12.

By Lemma  1.6, a DP--coloring of can be extended to , a contradiction.

(c) Suppose otherwise that the -face is an internal -face. Let be the neighbor of not on . Since is incident with at least six -vertices, by symmetry we may assume that . We can rename the lists of vertices in so that each edge in is straight.

Consider the graph obtained from by identifying and . We claim that no new cycles of length from to are created, for otherwise, there is a path of length or from to in , which together with forms a separating cycle of length or , a contradiction to Lemma 3.2. Since none of and is on a triangle, . Finally, we claim that no new chord in is formed in , for otherwise, and is adjacent to a vertex on , then there is a path between and on with length at most four, which again forms a separating cycle with of forbidden length.

By minimality of , the --coloring of can be extended to a --coloring of . Now keep the colors of all other vertices in and color and with the color of the identifying vertex. For , let