Planar graphs without 7-cycles and butterflies are DP-4-colorable
Seog-Jin Kim, Runrun Liu, Gexin Yu

TL;DR
This paper proves that planar graphs lacking 7-cycles and butterflies are DP-4-colorable, extending known results for graphs without smaller cycles and clusters of triangles, using a proof adaptable to other forbidden structures.
Contribution
It establishes a new sufficient condition for DP-4-colorability of planar graphs by forbidding 7-cycles and butterflies, broadening the class of graphs known to be DP-4-colorable.
Findings
Planar graphs without 7-cycles and butterflies are DP-4-colorable.
The proof method can be adapted to other forbidden cluster conditions.
Extends previous results on DP-colorability for graphs without smaller cycles.
Abstract
DP-coloring (also known as correspondence coloring) is a generalization of list coloring, introduced by Dvo\v{r}\'ak and Postle in 2017. It is well-known that there are non-4-choosable planar graphs. Much attention has recently been put on sufficient conditions for planar graphs to be DP--colorable. In particular, for each , every planar graph without -cycles is DP--colorable. In this paper, we prove that every planar graph without -cycles and butterflies is DP--colorable. Our proof can be easily modified to prove other sufficient conditions that forbid clusters formed by many triangles.
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Taxonomy
TopicsAdvanced Graph Theory Research
Planar graphs without 7-cycles and butterflies are DP-4-colorable
Seog-Jin Kim1, Runrun Liu2, Gexin Yu2,3
1Department of Mathematics Education, Konkuk University, Seoul, 05029, Korea.
2Department of Mathematics, Central China Normal University, Wuhan, Hubei, China.
3Department of Mathematics, The College of William and Mary, Williamsburg, VA, 23185, USA.
[email protected] (S.-J. Kim), [email protected] (R. Liu), [email protected] (G. Yu)
Abstract.
DP-coloring (also known as correspondence coloring) is a generalization of list coloring, introduced by Dvořák and Postle in 2017. It is well-known that there are non-4-choosable planar graphs. Much attention has recently been put on sufficient conditions for planar graphs to be DP--colorable. In particular, for each , every planar graph without -cycles is DP--colorable. In this paper, we prove that every planar graph without -cycles and butterflies is DP--colorable. Our proof can be easily modified to prove other sufficient conditions that forbid clusters formed by many triangles.
1. Introduction
We only consider simple graphs in this article. A list assignment of a graph is a mapping that assigns each vertex of a set of colors . An -coloring of is a proper coloring of such that for any . A list assignment is called a -list assignment if for any . A graph is -choosable if admits an -coloring for each -list assignment . The list-chromatic number (or the choice number) of , denoted by , is the minimum integer such that is -choosable.
Thomassen [12] showed that every planar graph is 5-choosable, and Voigt [13] showed that there are planar graphs that are not 4-choosable. This makes it an interesting question to determine which planar graphs are -choosable.
A graph is said to be -degenerate if each of its subgraph contains a vertex of degree at most . Let be the cycle with vertices. For each , it is shown that a planar graph without is 3-degenerate, thus -choosable, see [5, 8, 9, 15]. It is further shown in [4, 9] that for each , planar graphs without -cycles is -choosable. The proof for the case of -cycle is quite involved.
Some powerful tools (for example, vertex identification) in coloring are not feasible for list coloring. In an effort to overcome this, Dvor̂ák and Postle [3] introduced the notion of -coloring, which is a generalization of list coloring. By using this notion, they solved a long-standing conjecture of Borodin [1] on list coloring of planar graphs.
Definition 1.1**.**
Let be a graph with vertices and let be a list assignment for . For each edge in , let be a matching between the sets and and let (called a matching assignment). Let be the graph that satisfies the following conditions
- •
each corresponds to a set of vertices in ;
- •
for all , the set forms a clique in ;
- •
if , then the edges between and are those of ; and
- •
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 any matching assignment in which for each , it has an -coloring. The minimum value of such that is DP--colorable is the DP-chromatic number of , denoted by .
Let be a matching assignment for . An edge is straight if every satisfies . If all the edges under are straight, then an -coloring is exactly a list-coloring. So DP-coloring is a generalization of list-coloring. Since more vertices in will only make it easier to find an independent set of size , we may assume that each of has size , namely . The elements in sometimes are still called colors of . We may also assume that for each is a perfect matching.
Dvor̂ák and Postle [3] observed that Thomassen’s proof also implies that planar graphs are DP--colorable. As a planar graph without -cycles with is -degenerate, it is also DP--colorable. Kim and Ozeki [6] showed that planar graphs without -cycles are DP-4-colorable. More sufficient conditions for a planar graph to be -4-colorable have been found in [2, 6, 7, 10, 11], and we summarize them below.
Theorem 1.1**.**
([2, 6, 7, 10, 11]) The following planar graphs are DP-4-colorable
- •
without -cycles, where , or
- •
without -cycles adjacent to -cycles, where , and two cycles are adjacent if they share an edge.
Note that it is conjectured independently in [8, 16] that every plane graph without adjacent triangles is 4-choosable and it remains open. We would like to conjecture that every plane graph without adjacent -cycles is -choosable, or even DP--colorable.
A cluster in a plane graph is a subgraph of that consists of a minimal set of 3-faces such that no other 3-face is adjacent to 3-faces in the set. It is called a -cluster if it contains -faces. Below is the set of possible clusters with distinct vertices in a plane graph without -cycles (in [4], all 23 clusters in such plane graphs are given).
It turns out that all known results on DP--coloring of planar graphs use certain kinds of condition to forbid one or more of the given clusters in Figure 1, especially the clusters with more -faces. What makes the proof of list-4-coloring of planar graphs without -cycles difficult is that one has to take care of all the clusters in the figure.
In this article, we give a sufficient condition for a planar graph to be DP--colorable. This condition allows the existence of all clusters in Figure 1.
Theorem 1.2**.**
Every planar graph without 7-cycles and butterflies is DP-4-colorable, where a butterfly is a graph isomorphic to the configuration depicted in Figure 2.
Without much effort, our proof can be modified to provide more sufficient conditions for a planar graph to be DP-4-colorable. Here is a potential list: planar graphs without -cycles adjacent to -cycles, where .
Let be the set of planar graphs without 7-cycles and butterflies. A -cycle in a plane graph is bad if and interior of form a -cluster. So in Figure 1 (11), is a bad -cycle. A -cycle is called good if it is not bad. We actually prove the following stronger result.
Theorem 1.3**.**
Any DP-4-coloring of a good 3-cycle in plane graph can be extended to a DP-4-coloring of .
By [6], every planar graph without triangles is DP-4-colorable. So we may assume that contains a triangle. In particular, we can always find a good triangle in . By Theorem 1.3, we can get a DP-4-coloring of by extending a DP-4-coloring of a triangle in .
We use a discharging argument in our proof. This involves the proof of some reducible configurations. The proof of reducibility of some - and -clusters in Lemma 2.7 and 2.8 involves careful consideration of matching assignments, thus is essentially different from that of the structures in list-coloring. Also, some -clusters are reducible in list--coloring but not reducible in DP--coloring, which makes the addition of forbidding butterflies necessary for our proof, see the Final Remarks section. On the other hand, by strengthening Theorem 1.2 to Theorem 1.3, instead of 23 clusters in [4], we only need to discuss 11 clusters (see Figure 1), so our proof can be modified to give a simplified proof of 4-choosability of planar graphs without -cycles.
The paper is organized as follows. In Section 2, we show the reducible structures useful in our proof. In Section 3, we show the discharging process to complete the proof. In Section 4, we give some examples to show the necessary of adding the butterflies to be forbidden structure.
2. Reducible configurations
The following are some notions used in the paper. A -vertex (-vertex, -vertex, respectively) is a vertex of degree (at least , at most , respectively). The same notation will be applied to faces and cycles. An -face is a -face with for . Let be a cycle of a plane graph . We use (resp. ) to denote the sets of vertices located inside (resp. outside) the cycle . The cycle is called separating if both and are nonempty. The next lemma follows immediately from ([3], Lemma 7).
Lemma 2.1**.**
Let be a graph with a matching assignment . Let be a subgraph of which is a tree. Then we may rename for to obtain a matching assignment for such that all edges of are straight in .
Let be a minimal counterexample to Theorem 1.3, where is a good -cycle in that has a DP-4-coloring . If is a separating cycle, then any precoloring of can be extend to and , respectively. Then we get a DP-4-coloring of , a contradiction. So we let be a plane graph so that is the boundary of the outer face in the rest of this paper. We still denote the outer face by . Call a vertex in internal if , and a subgraph in internal if . Assume that is an independent set in with that extends . For each with , define Intuitively, contains the available colors for after a partial coloring .
Lemma 2.2**.**
Every internal vertex in has degree at least 4.
Proof.
Suppose otherwise that there exists an internal -vertex in . By the minimality of , has a DP-4-coloring that extends . Thus there is an independent set in with . Since and is a -vertex, we have . So we can pick a vertex such that is an independent set of with vertices, a contradiction. ∎
Lemma 2.3**.**
G contains no separating good 3-cycles.
Proof.
Let be a separating good -cycle in . By the minimality of , can be extended to . After that, is precolored, then again the coloring of can be extended to . Thus, we get a DP-4-coloring of , a contradiction. ∎
Lemma 2.4**.**
Two internal -faces cannot share exactly one common edge unless they form a .
Proof.
Suppose for a contradiction that and are two internal -faces so that . Let . By the minimality of , the graph has a DP-4-coloring that extends . Thus there is an independent set in with . Then So we can select a vertex in for such that has at least two available colors. Color in order, we can find an independent set with . So is an independent set of with , a contradiction. ∎
We call a cluster special if it is one of Figure 1 (7)(9)(10)(11) with three internal -vertices . For an internal -vertex in a cluster , we shall call -type to if is incident with exactly edges in ; furthermore, we call good when is special, and call bad otherwise. For example, vertex in Figure 1 (9) is good when is special and 3-type to , and vertex in Figure 1 (8) is bad and 3-type to .
Lemma 2.5**.**
Every internal -vertex in cannot be on two special clusters.
Proof.
Let be an internal -vertex with . Suppose otherwise that is on two special clusters . By symmetry assume that in and in are internal -faces. By the minimality of , the graph has a DP-4-coloring extends . Thus there is an independent set in with . Note that . So we can select a vertex such that has no neighbors in , and a vertex such that has no neighbors in . Color in order, we can find an independent set with . So is an independent set of with , a contradiction. ∎
We call an internal -vertex special if is good -type to an internal -cluster and good -type to a -cluster with and , see Figure 3 for an illustration.
Lemma 2.6**.**
Let be a special -vertex on a -cluster and a -cluster . Let be a -face in such that . Let be a DP-coloring of that extends . Then among the four colors in , one can precolor with all but at most one color so that can be colored.
Proof.
We may assume that for some color in , cannot be all colored if we color with . Then each of has exactly two available colors that induce two triangles in . But then we can precolor with any other color in such that can be colored in order. ∎
Lemma 2.7**.**
Let be an internal special -cluster isomorphic to Figure 4(a). If , then cannot be a -vertex or a special -vertex.
Proof.
By Lemma 2.4, . Suppose that is a -vertex or a special -vertex. By the minimality of , has a DP-4-coloring that extends . Thus there is an independent set in with . Note that . If , then . When is a special -vertex, by Lemma 2.6, we may assume that has at least three available colors to use when coloring .
By Lemma 2.1, we can assume that the edges are straight. If a color in and a color in have a common neighbor in , then we select , and can be colored in order, a contradiction. So we assume that none of the two colors from and have common neighbors in . For each , if we can choose such that and forbidden at most one color at both and , then can be colored in order, a contradiction. So we may assume that and forbidden two colors at or for each . By symmetry we have two cases, see Figure 4(b)(c). Choose at Case (b) and choose at Case (c). If , then we select , then can be colored in order. If for , then we select , then can be colored in order, a contradiction. ∎
Lemma 2.8**.**
Let be an internal -cluster, see Figure 5(a). If , then is incident to at most one -vertex or special -vertex.
Proof.
Suppose otherwise, by symmetry let be -vertices or special -vertices. By the minimality of , the graph has a DP-4-coloring that extends . Thus there is an independent set in with . Note that , and . By Lemma 2.6, each of has at least three available colors to use when coloring .
By Lemma 2.1, we can assume that the edges are straight. Let and by straightness, for , let . We further assume that . Let and .
We may assume that . For otherwise, we extend to include . Then . We pick a color in such that still has two available colors. Then can be colored in order, a contradiction.
For each , if , then we select , then can be colored in order. So . Similarly, .
We observe that must have a neighbor in . Suppose by symmetry that has no neighbors in . Then we select and can be properly colored in order, a contradiction. Thus either or . Note that if , then we have . Therefore, we have either or . By symmetry of colors, we may assume that . Then either or .
Consider the case . If , then we select . So . Then can be colored in order, a contradiction. So . Similarly, we have . Therefore we have since we showed that for , which gives us Figure 5(b).
Consider the case of . If , we select , then can be colored in order, a contradiction. Thus . Next, if , we select , then can be colored in order, a contradiction. Thus . So we have , which gives us Figure 5(c).
Now, we show how to color Figure 5 (b) and (c).
Case 1: . In this case, we select in both (b) and (c). Then , and can be colored in order, a contradiction.
Case 2: for some . In this case, we select in both (b) and (c). Then , and can be colored in order, a contradiction.
Case 3: . In this case, for (b), we select and can be colored in order. For (c), either for some , or for some . In the former case, we select and color in order; in the latter case, we select and color in order.
This completes the proof of Lemma 2.8. ∎
3. Discharging procedure
Proof of Theorem 1.3. We are now ready to complete the proof of Theorem 1.3 by a discharging procedure. Let has an initial charge of , and . By Euler’s Formula, . Let be the charge of after the discharge procedure. To obtain a contradiction, we shall prove that for all and .
Observation 3.1**.**
Every good -vertex is on at most one special cluster.
Proof.
If a good -vertex is on two special clusters, then has four neighbors such that and are both on -faces. This contradicts Lemma 2.5. ∎
The discharging rules:
- (R1)
Each -face gives to each adjacent -face, and moreover, if is not on a -face, then gives to each of and . Each internal -vertex having at leasts three incident edges in a cluster gives the charge it obtained from incident -faces to the cluster. 2. (R2)
Let be a -cluster with . Then gets from each incident -type -vertex that is either good or on a -face adjacent to , from each -type -vertex that is bad to or incident -type -vertex, from each good -type -vertex, and from each incident -type -vertex. Note that -clusters cannot be adjacent to -faces. 3. (R3)
A -cluster gets from each incident bad -type -vertex, from each incident good -type -vertex, from each incident -type -vertex or -type -vertex, from each incident good -type -vertex if contains two -type -vertices. 4. (R4)
A -cluster gets from each incident -vertex or special -vertex, from each incident non-special -vertex, and from each incident -vertex. 5. (R5)
The outer-face gets from each incident vertex and gives to each non-internal -face.
Remark: (a) Let be a -vertex that incident with four consecutive faces in order. If and , then by (R1)(R2) gives to the cluster containing , when .
(b) Let be a cluster and be an internal -vertex. By (R2)-(R4) gives at most to if is -type to (note that -clusters contain no 2-type vertices), at most 1 to if is -type -vertex to , to if is a -type -vertex or -type -vertex of , at most to if is a -type -vertex of , at most to if is a -type -vertex of .
First we check the final charge of vertices in . Let be a vertex in . If , then by (R5). So we may assume that . By Lemma 2.2 . If , then by (R1), .
Suppose . Then is on at most two clusters. If is on at most one cluster, then by (R1)-(R3), gives out more than to the cluster only if is a -type vertex to the cluster. But in this case, each face adjacent to the cluster is a -face since contains no -cycles. So is incident with two consecutive -faces, thus obtains by (R1). So . Now let be on two clusters. Then is -type to one cluster and -type to the other cluster, and none of the cluster can be a -cluster. By (R2)(R3), gives more than to one cluster when is a good -type vertex. In this case, is a -type -vertex to the other cluster, say . It follows that cannot be on a -face, and by Observation 3.1, is not good to . So by (R2)(R3), gives no charge to . Therefore, .
Now let . Assume that gives to clusters, to clusters, to clusters, and to clusters. Since contains no butterfly, . Then by (R2)-(R4), and . It follows that
[TABLE]
So if , then only if , and . In this case, is on a -cluster and -type to a -cluster, so by (R2)(R4), .
For , . If , then is -type to its clusters so is on at most two clusters. Note that is on at most one -cluster since has no butterfly. If is on a -cluster, then by (R2)(R3), either is -type to a -cluster, thus , or is -type to a -cluster, which gets at most from since contains no butterflies. So . If is not on any -cluster, then by (R2), gives at most to each of the two clusters. So . Now let , then only if since . It implies . By (R2)(R4), .
Let . By (R2)-(R4), . So . We may assume that . Assume first that . Then is -type to at most two clusters, and if is -type to a cluster, then is only on one cluster. Thus by (R2)(R3), . So let . Then is on at most one -cluster, and in particular, . Thus only if and . So is not on a -face, and by (R2)-(R4), is good -type to one -cluster and good -type to one -cluster which contains two -type -vertices, contrary to Lemma 2.7.
Now we check the final charge of faces in . Let be a face in . Since contains no -cycles, a -face is adjacent to at most one -face and a -face is not adjacent to -faces. By (R1), when is a -face. So we only need to consider the final charge of -faces. Let . Since contains no -cycles or separating -cycles, must be in a -cluster for some , which are depicted in Figure 1. Let , where are -faces in , and define . We shall show that for each , which would imply that all -faces can get enough charge to have nonnegative final charges.
Case 1. -cluster . Since contains no -cycles, shares at least two edges with -faces. So either gets from by (R5) or from adjacent -faces by (R1). Thus, .
Case 2. is a -cluster isomorphic to Figure 1 (2). Since contains no -cycles, shares edges only with -faces or -faces; furthermore, shares at least two edges with -faces, which gives at least to by (R1). So we only need 1 more charge to make . We may assume that , for otherwise, by (R5) gets at least from . If shares at least three edges with -faces, then by symmetry say are on -faces. So gets from by Remark (a) and extra from adjacent -faces by (R1). If shares exactly two edges with -faces, then by symmetry are on -faces or are on -faces. In the former case, and thus gives to by (R2), and by Remark (a), gives at least to as well. Consider the latter case. If both are -vertices, then each of them transfers they got from -faces to , and by Lemma 2.4, or is a -vertex, which by (R2), gives at least to , therefore . By symmetry, assume that . Now by (R2), gives to , and each of gives (if it is a -vertex) or at least (if it is a -vertex) to . Therefore, .
Let be a -cluster with . All faces adjacent to are -faces since has no -cycles.
Case 3. is a 3-cluster isomorphic to Figure 1 (3). By (R1), gets from adjacent -faces. In addition, either gets at least from by (R5) or from each of and by (R2) (see Remark (a)). So .
Case 4. is a -cluster isomorphic to one in Figure 1 (4)-(7).
Case 4.1. is isomorphic to one of Figure 1 (4)-(6). By (R1), gets from adjacent -faces in (4),(5),(6), respectively. If any vertex in is on , then by (R5), gets at least from in (4),(5),(6) respectively, so . If is internal, then by (R2), in (4) it gets from each of , in (5) it gets from each of and , and in (6), it gets from each of . In any case, .
Case 4.2. is isomorphic to Figure 1 (7). By (R1), gets from adjacent -faces. We need to find more charge to make . We may assume that is internal, for otherwise, gets at least from by (R5). If one of is a -vertex, say , then gets from by (R2). Otherwise, by Lemma 2.4, . By (R2), each of gives to .
Case 5. is a -cluster isomorphic to Figure 1 (8) or (9).
Case 5.1. is isomorphic to Figure 1 (8). By (R1)(R2), gets from and from adjacent -faces. If is not internal, then it gets at least from by (R5), thus . If is internal, then gets at least from each of by (R2), thus .
Case 5.2. is isomorphic to Figure 1 (9). By (R1), gets from adjacent -faces. We may assume that both and are internal, for otherwise, gets from by (R5) and . First assume that both and are internal -vertices. Then by Lemma 2.4, are -vertices or on . Note that if are internal -vertices, then they are good -type -vertices to . By (R2)(R5), gets at least through and at least from by (R2), and . Now by symmetry, let . By (R2), gets from . Now, whether and are -vertices, -vertices, or on , gets at least from each of them, so .
Case 6. is a -cluster isomorphic to Figure 1 (10). By (R1), gets from adjacent -faces. So we need to find another to make . Since is a good -cycle, . If , then gets at least from by (R5). If , then by symmetry either or . In the case of , gets from by (R5). By Lemma 2.4, at least one of is a -vertex, by symmetry say . Then by (R3), gets from . If , then by Lemma 2.4 , so gives at least to by (R3); If , then gives to by (R3); So gets at least . Now consider the case . By (R5), gets from . By Lemma 2.4 at least one of is a -vertex, which gives at least to by (R3). So we assume that . By Lemma 2.4, at least one of and is a -vertex. If both and are -vertices, then by (R3), gets from each of and , and from each of and . So gets . If one of and , by symmetry say is -vertex, then by Lemma 2.4, and by Lemma 2.7, one of has degree more than . If , then by (R3) gets from and from each of and ; and , then by (R3), gets at least from each of and and at least from ; and , then by (R3), gets at least from each of and and at least from . In any case, gets at least .
Case 7. is a -cluster isomorphic to Figure 1 (11). By (R1), gets from adjacent -faces. So we need to find another to make . By Lemma 2.4 each of is either on or an internal -vertex; by (R3), gets at least from each of when it is internal. Since is a good -cycle, . If , then by (R5), gets at least from . So gets at least . So we assume that is internal. If one of is a -vertex, then by (R4), . So we assume that . By Lemma 2.8, contains at most one -vertex or special -vertex, so by (R4), .
To finish the proof, we show that the outer face has positive final charge. Let be the number of non-internal -faces. Let be the set of edges between and and let be its size. Note that . Then by (R5),
[TABLE]
4. Final Remarks
In Figure 6, we illustrate two matching assignments in a -cluster and a -cluster that prevent us from finding an independent set of order . The examples show that Lemma 2.7 and Lemma 2.8 cannot be improved. The examples also show that it would be difficult to improve our results by removing the requirement of forbidding butterflies. For example, in our proof, -vertices have no burden to afford the charges, but if we allow butterflies, even a -vertex may not be able to afford to give charges when it is on two -clusters.
Acknowledgement. The work is done while the second author is studying at the College of William and Mary as a visiting student, supported by the Chinese Scholarship Council. Seog-Jin Kim’s work was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT) (NRF-2018R1A2B6003412). Gexin Yu’s work was supported in part by the Natural Science Foundation of China (11728102) and the NSA grant H98230-16-1-0316. The authors would like to thank the referees for their valuable comments.
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