1 Introduction
A list assignment is a function which assigns a list of colors to each vertex . An coloring of is a proper coloring of such that for all . A list assignment is a list assignment with for each vertex . A graph is choosable if it admits an coloring whenever is a list assignment. The list chromatic number is the least integer such that is choosable. List coloring is a generalization of ordinary proper coloring, thus we have that . The Four Color Theorem states that every planar graph is colorable. It was proved that not every planar graph is choosable, see [16, 18]. Some sufficient conditions for planar graphs to be 4choosable are well studied, see [5, 21, 1, 7, 19, 4].
Correspondence coloring is introduced by Dvořák and Postle [3] as a generalization of list coloring, now it is also called DPcoloring.
Definition 1.
Let be a simple graph and be a listassignment for . For each vertex , let ; for each edge , let be a matching between the sets and , and let , call it the matching assignment. The matching assignment is called a matching assignment if for each . A cover of is a graph (simply write ) satisfying the following two conditions:

[label =(C0)]

the vertex set of is the disjoint union of for all ; and

the edge set of is the matching assignment .
Note that the matching is not required to be a perfect matching between the sets and , and possibly it is empty. It is easy to see that the induced subgraph is an independent set for each vertex .
Definition 2.
Let be a simple graph and be a cover of . An coloring of is an independent set in such that for each vertex . The graph is DPcolorable if for any list assignment and any matching assignment , it has an coloring. The DPchromatic number of is the least integer such that is DPcolorable.
For DP3colorable graphs, we refer the reader to [12, 11, 17, 22]; for "weakly" DP3colorable graphs, we refer the reader to [3, 9, 13]. Dvořák and Postle [3] observed that every planar graph is DP5colorable. In this paper, we concentrate on DP4coloring of planar graphs.
Two cycles (or faces) are adjacent if they have at least one common edge. Two cycles are intersecting if they shares at least one vertex. Liu and Li [10] showed that every planar graph without 4cycles adjacent to two triangles is DP4colorable. Kim and Ozeki [6] showed that for each integer , every planar graph without cycle is DP4colorable. Chen et al. [2] showed that every planar graph without 4cycles adjacent to 6cycles is DP4colorable. Zhang and Li [23] showed that every planar graph without 5cycles adjacent to 6cycles is DP4colorable. Lv, Zhang and Li [15] showed that every planar graph without intersecting 5cycles is DP4colorable.
Li and Wang [8] considered planar graphs without some mutually adjacent 3, 4, and 5cycles.
Theorem 1.1 (Li and Wang [8]).
Every planar graph without subgraphs isomorphic to the configurations in Fig. 1 is DP4colorable.
An edge in is straight in a matching assignment if satisfies . An edge in is full in a matching assignment if is a perfect matching. Let with be a closed walk of length in , a matching assignment is inconsistent on , if there exists for such that is an edge in for and . Otherwise, the matching assignment is consistent on .
Lemma 1.1 (Dvořák and Postle [3]).
Let be a graph with a matching assignment , and let be a subgraph of . If for every cycle in , the assignment is consistent on and all edges of are full, then we may rename for to obtain a matching assignment for such that all edges of are straight in .
A vertex, vertex or vertex is a vertex of degree , at least or at most respectively. Similar definitions can be applied to faces and cycles. Liu and Li [10] showed that every planar graph without a cycle adjacent to two triangles is DPcolorable. Actually, they showed the following stronger result.
Theorem 1.2 (Liu and Li [10]).
Let be a planar graph without a cycle adjacent to two triangles, and let be a matching assignment for . If consists of exactly one vertex or all the vertices on a face of , then every coloring of can be extended to an coloring of .
The first result of this paper is the following theorem which generalizes Theorem 1.2.
Theorem 1.3.
Let be a plane graph without subgraphs isomorphic to the configurations in Fig. 2, and let be a matching assignment for . If consists of exactly one vertex or all the vertices on a cycle of , then every coloring of can be extended to an coloring of .
In the above theorem, adjacent triangles are forbidden in the graph. In the second result of this paper, adjacent triangles are allowed but two triangles sharing exactly one vertex is forbidden.
Theorem 1.4.
Let be a plane graph without subgraphs isomorphic to the configurations in Fig. 3, and let be a matching assignment for . If is the vertices of a cycle of , then every coloring of can be extended to an coloring of .
Wang and Lih [20] showed that every planar graph without intersecting triangles is choosable. Luo [14] improved it to toroidal graph, showing that every toroidal graph without intersecting triangles is choosable. As a corollary of Theorem 1.4, the following corollary generalizes Wang and Lih’s result.
Corollary 1.5.
Every planar graph without intersecting triangles is DPcolorable.
Li and Wang showed the following result as a corollary in [8].
Theorem 1.6.
Every toroidal graph without subgraphs isomorphic to the configurations in Fig. 4 is DPcolorable.
For planar graphs, we give the following result which is stronger than Theorem 1.6.
Theorem 1.7.
Let be a plane graph without subgraphs isomorphic to the configurations in Fig. 4, and let be a matching assignment for . If is the vertices of a cycle of , then every coloring of can be extended to an coloring of .
We need some definitions for plane graphs in the following section. The unbounded face is called the outer face, and the other faces are called inner faces. An internal vertex is a vertex that is not incident with outer face. An internal face is a face having no common vertices with the outer cycle. Let be a cycle of a plane graph , the cycle divides the plane into two regions, the subgraph induced by all the vertices in the unbounded region is denoted by , and the subgraph induced by all the vertices in the other region is denoted by . If both and contain at least one vertex, then we call the cycle a separating cycle of . The subgraph induced by the complement of is denoted by , and the subgraph induced by the complement of is denoted by .
Let be a plane graph and the outer face be bounded by a cycle. A vertex is a vertex if it is incident with exactly triangularfaces. An inner face is a face if it has exactly common vertices with the outer cycle. A face is a special face if it is an internal face incident with a vertex. A special edge is an edge having a common vertex with the outer cycle but it is not an edge of the outer cycle. Let be the set of inner faces having at least one common vertex with the outer face.
2 Proof of Theorem 1.3
See 1.3
Suppose that is a minimal counterexample to Theorem 1.3. That is, there exists an coloring of that can not be extended to an coloring of such that
(1) 
Subject to (1),
(2) 
The following structural results Lemma 2.11–6 are almost the same with that in [10, 13], so the proofs are omitted here. Let be an internal face adjacent to five 3faces. If is incident with one of the five 3faces but not on , then we call a related source of and a sink of .
Lemma 2.1.

[label = ()]

;

is connected, and thus the boundary of every face is a cycle;

each vertex not in has degree at least four;

either or is an induced cycle of ;

there is no separating cycle for ;

is an induced cycle of , so we may assume that is the outer cycle;

if and are two nonconsecutive vertices on , then they have no common neighbor in ;

if is a sink in , then at most one of its source is on the outer cycle ;

there is no vertex incident with a face. ∎
7 Suppose that and are two nonconsecutive vertices on , and they have a common neighbor . The two vertices and divide the cycle into two paths and . It is observed that and are two cycles with length at most , thus these two cycles are not separating cycles by Lemma 2.15, and thus must be an internal vertex, but this contradicts Lemma 2.13.
8 Let be an internal 5face adjacent to five 3face for . It is observed that and are five distinct vertices. Suppose that and are on the outer cycle , where the subscripts are taken module . By Lemma 2.17, and are consecutive on , thus is a triangle, but this contradicts that there is no adjacent triangles.
Suppose that and are on the outer cycle , where the subscripts are taken module . By the above arguments, and are internal vertices. The outer cycle and the path form two cycles containing . It is observed that these two cycles are separating cycles, thus the lengths are at least 7 by Lemma 2.15, and thus , a contradiction.
The next two structural results focuses on the internal vertices.
Lemma 2.2.
Let be a vertex with four neighbors in a cyclic order. If and are all internal vertices, then at most one of and is a vertex.
Suppose to the contrary that both and are vertices. By Lemma 1.1, we may assume that all the four edges incident with are straight. Let , let be obtained from by identifying and , and let be the restriction of on . Let be a shortest path from to in . If the length of is at most four, then is a separating cycle with length at most six, a contradiction. Therefore, the distance of and in is at least five. Note that the precolored cycle has length at most six and the length of is at least five in , thus and is also an coloring of . Moreover, is a simple plane graph without subgraphs as in Fig. 2. By the minimality, can be extended to an coloring of . There are at most three forbidden colors for each of and , so we can extend the coloring to and . Since and have the same color and are straight, we can further extend the coloring to , a contradiction. ∎
Lemma 2.3.
Let be an internal vertex with four neighbors in a cyclic order. If is incident with a face, then cannot be an internal vertex.
Suppose to the contrary that is an internal vertex. By Lemma 1.1, we may assume that all the four edges incident with are straight. Let , let be obtained from by identifying and , and be the restriction of on . Suppose that is a path from to in with length at most three. Thus, is a cycle of length at most five. If is not on , then is a separating cycle, a contradiction. Otherwise is on , thus Lemma 2.15 implies that is incident with a face and another face, but this contradicts the fact that the configurations in (a) and (b) are forbidden in , a contradiction. Therefore, the distance of and in is at least four, and at least one of and is an internal vertex.
Suppose that the identification creates a chord of , where is the new vertex. The outer cycle is divided into two paths and by and , see Fig. 5. Recall that the distance of and is at least four in , thus and have lengths at least three. Since is a cycle of length at most six, and have lengths exactly three, which implies that is a separating cycle of length six, a contradiction. Therefore, the identification does not create any chord of , and is an coloring of . Since the distance of and in is at least four, has no loops and no multiple edges. It is easy to check that there is no subgraphs isomorphic to the configurations in Fig. 2. By the minimality, can be extended to an coloring of . There are at most three forbidden colors for , so we can extend the coloring to . Since and have the same color and are straight, we can further extend the coloring to , a contradiction. ∎
We give the initial charge for any , for the outer face , and for any face other than . By Euler’s formula, the sum of the initial charges is zero. That is,
We design discharging rules to redistribute the charges, preserving the sum, such that every element in has a nonnegative final charge . Moreover, there exists a face having a positive final charge, which leads to a contradiction.

[label = R0]

Let be an internal vertex and be the number of incident faces.

[label = ]

If , then sends to each incident face.

If and it is incident with a face , then sends to the incident face, sends to the incident face not adjacent to , and sends to each incident face that shares an edge with .

If and it is not incident with a face, then sends to each incident face.

If , then sends to each incident face.


Every internal vertex sends to each incident special face, sends to each of the other incident face, and sends to each incident face.

Every vertex on the outer cycle sends its initial charge to the outer face , and sends to each special edge, and each special edge immediately sends to each incident face.

Every internal vertex sends via incident 3face to the sink.
Remark 1.
Note that each special edge receives from the outer face and sends to each incident face, thus each special edge has final charge zero and each face in receives at least from the outer face via special edges.
Each vertex on has final charge zero by 3, thus it suffices to consider all the faces and internal vertices. By Lemma 2.13, every internal vertex has degree at least four. Since there is no adjacent faces, every vertex is incident with at most triangularfaces. In particular, every vertex is incident with at most two faces.
Let be an internal vertex. If is an internal vertex, then by 11a. If is an internal vertex and it is incident with a face, then by 11b. If is an internal vertex but it is not incident with any face, then by 11c. If is an internal vertex, then by 11d.
Let be an internal vertex. Every internal vertex sends at most to/via each incident face, and sends to each incident face by 2 and 4. Thus,
Let be an inner face. It is observed that there is no faces. If is a face, then it receives from the incident internal vertex, and then by 1, 2 and Remark 1. If is a face, then it receives at least from each incident internal vertex, and then by 1, 2 and Remark 1. If is an internal face but not a special face, then by 11a and 2. Assume that is a special face. By Lemma 2.3, is a face. Thus, receives from the incident vertex and from each incident vertex by 11c and 2, which implies that .
Let be an inner face. If is a face in , then by Remark 1. If is an internal face, then no vertex on can be a vertex by Lemma 2.19, which implies that . If is an internal face, then . So we may assume that is an internal face in the remaining of this paragraph. If is incident with at least two vertices, then by 2. So we may further assume that is a face. If is adjacent to a face via , then neither nor is a vertex, and then each of and sends to , which implies that . Hence, is an internal face adjacent to five 3faces. By Lemma 2.18, at least four sources are internal vertices. By Lemma 2.2, every internal source is a vertex, and it sends to the sink by 4, which implies that .
By 3, the final charge of the outer face is
Since is connected and , there are at least two special edges. Since there is no adjacent faces, there exists a face incident with a special edge. If is a face, then by Remark 1. So we may assume that is a face incident with an internal vertex . Since is in , it receives from . If is a vertex, then it sends to by 2; while is a vertex, it is a vertex or vertex, so it sends at least to by 1. In these two cases, . Therefore, there must exist a face incident with a special edge such that its final charge is positive. This completes the proof. ∎
3 Proof of Theorem 1.4
See 1.4
Suppose that is a minimal counterexample to Theorem 1.4. That is, there exists an coloring of that can not be extended to an coloring of such that
(3) 
Subject to (3),
(4) 
The following properties are very similar to Lemma 2.1, so we omit the proofs here.
Lemma 3.1.

[label = ()]

and is connected;

is an induced cycle of ;

there is no separating cycle for , so we may assume that is the outer cycle;

each internal vertex has degree at least four.

every vertex is incident with at most two faces;

there is no vertex incident with a face. ∎
The following result and its proof are almost the same with Lemma 2.3, so we only present the result but without giving the proof. Note that is incident with exactly one face in Lemma 3.2.
Lemma 3.2.
Let be an internal vertex with four neighbors in a cyclic order. If is incident with a face, then cannot be an internal vertex.
We give the initial charge for any , for the outer face , and for any face other than . By Euler’s formula, the sum of the initial charges is zero. That is,
Next, we give the discharging rules to redistribute the charges, preserving the sum, such that every element in has a nonnegative final charge . Moreover, there exists a face having a positive final charge, which leads to a contradiction.

[label = R0]

Let be an internal vertex and be the number of incident faces.

[label = ]

If , then sends to each incident face.

If , then sends to each incident face, and sends to each incident internal face, internal face and each of the other incident face, and sends to each incident face in .


Every internal vertex sends to each incident face, and sends to each incident face.

Each vertex on the outer cycle sends its initial charge to the outer face . The outer face sends to each special edge, and each special edge immediately sends to each incident face.
Remark 2.
Note that each special edge receives from the outer face and sends to each incident face, thus each special edge has final charge zero and each face in receives at least from the outer face via special edges.
Each vertex on has final charge zero by 3, thus it suffices to consider all the faces and internal vertices. By Lemma 3.14, every internal vertex has degree at least four. By Lemma 3.15, every vertex is incident with at most two faces.
Let be an internal vertex. If is an internal vertex, then by 11a. If is an internal vertex and it is incident with a face, then by 11b. If is an internal vertex but it is not incident with any face, then by 11b. If is an internal vertex, then by 11b.
Let be an internal vertex. By 2, sends to each incident face, and sends to each incident face. Thus, .
Let be an inner face. It is observed that there is no faces. If is a face, then it receives from the incident internal vertex, and then by 1, 2 and Remark 2. If is a face, then it receives at least from each incident internal vertex, and then by 1, 2 and Remark 2. If is an internal face but not a special face, then by 11a and 2. Assume that is a special face. By Lemma 3.2, is a face. Thus, receives from the incident vertex and from each incident vertex by 11b and 2, which implies that .
Let be an inner face. If is a face in , then by Remark 2. If is an internal face, then no vertex on can be a vertex by Lemma 3.16, which implies that . If is an internal face, then . Let be an internal face. Since there is no subgraph isomorphic to the configuration in (a), is adjacent to at most two faces. If is adjacent to a face with on , then at most one of and is a vertex. Hence, is incident with at most two vertices. By the discharging rules, if is a vertex on but it is not a vertex, then it sends to . This implies that .
By 3, the final charge of the outer face is
Lemma 3.3.
There is a face incident with a special edge.
Since is connected and , there is at least one special edge. Suppose that is a path on the outer cycle and it is incident with a special edge . If every inner face incident with is a face, then must has degree three and it is incident with two faces, say and , for otherwise there is a subgraph isomorphic to the configuration in (a). Since every cycle bounds a face and every internal vertex has degree at least four, and are nonadjacent. This implies that is incident with a face, for otherwise there is a subgraph isomorphic to the configuration in (a). ∎
By Lemma 3.3, there is a face incident with a special edge and is an internal vertex. If is a face, then by Remark 2. So we may assume that is a face. By Lemma 3.16, there is no vertex incident with face . If is a vertex, then it is incident with at most one face, and sends at least to by 11b, thus . If is a vertex, then by 2 and Remark 2. This completes the proof. ∎
4 Proof of Theorem 1.7
See 1.7
Suppose that is a minimal counterexample to Theorem 1.7. That is, there exists an coloring of that can not be extended to an coloring of such that
(5) 
Subject to (5),
(6) 
Similarly, we only give the following structural results but without giving the proofs.
Lemma 4.1.

[label = ()]

and is connected;

is an induced cycle of ;

there is no separating cycle for , so we may assume that is the outer cycle of ;

each internal vertex has degree at least four.

if and on are nonadjacent, then they have no common neighbor not on ;

every face is incident with four face. ∎
We give the initial charge for any , for any inner face , and for the outer face . By Euler’s formula, the sum of the initial charges is zero. That is,
Next, we give the discharging rules to redistribute the charges, preserving the sum, such that every element in other than has a nonnegative charge after applying 1, 2 and 3. Moreover, the outer face has a positive final charge after applying 4, which leads to a contradiction.

[label = R0]

Let be an internal vertex and be the number of incident faces.

[label = ]

If , then sends to each incident face.

If and it is incident with a face , then sends to the face , and sends to the incident face not adjacent to , and sends to each incident face that shares an edge with .

If , then sends to each incident face.


Every internal vertex sends to each incident face, and sends to each incident face.

Each vertex on the outer cycle sends its initial charge to the outer face. The outer face sends to each special edge, and each special edge immediately sends to each incident face.

Every inner face transfers its surplus charge to .
Remark 3.
Note that each special edge receives from the outer face and sends to each incident face, thus each special edge has final charge zero and each face in receives at least from the outer face via special edges.
Each vertex on has final charge zero by 3, thus it suffices to consider all the faces and internal vertices. By Lemma 4.14, every internal vertex has degree at least four.
By the absence of (a), every vertex is incident with at most triangularfaces. In particular, every vertex is incident with at most two faces. If is an internal vertex, then by 11a. If is an internal vertex, then by 11b. If is an internal vertex, then by 11c.
Let be an internal vertex. By 2, sends to each incident face, sends to each incident face. Thus,
Let be a face in . If is a face in , then it receives from and receives from each incident internal vertex, so . If is a face in , then by Remark 3. If is a face in , then by Remark 3.
If is an internal face, then it receives from each incident vertex, and then . If is an internal face, then it is adjacent to four faces, and it receives from each incident vertex, which implies that . If is an internal face, then .
Let be an internal face. Since (b) is forbidden in , is adjacent to at most two faces. By symmetry, we may assume that neither nor is adjacent to a face. If is adjacent to a face with on , then at most one of and is a vertex, for otherwise (a) or (b) will appear in . Hence, at most two of are vertices. By the discharging rules, if but it is not a vertex, then it sends at least to . Moreover, whenever is a ,  or vertex, it sends to . This implies that by 11b, 11c and 2.
Finally, we show that has a positive final charge . By 3 and 4, the final charge of the outer face is
where is the total charge sent to by 4. So we may assume that is a  or face, otherwise we are done.
Lemma 4.2.
Every face in sends at least to by 4.
After applying 1, 2 and 3, every face in has charge at least , thus it sends at least to by 4. Suppose that is a face in . Since is an induced cycle and , there are at least one internal vertex incident with . Moreover, there are at least two internal vertices incident with , for otherwise it contradicts Lemma 4.15. Since is a face, it is adjacent to four faces. Therefore, each internal vertex on sends to , which implies that sends at least to by 4. ∎
If there is a face in , then sends at least to by 4, which implies that . So every face in is a face. By Lemma 4.2, if there are at least two faces in , then . So we may further assume that there is at most one face in . It follows that is adjacent to a face , where is an internal vertex. Suppose that each of and is adjacent to exactly one face. Thus, each of and is adjacent to a face in , this contradicts that there is at most one face in . Suppose that is adjacent to another face . If is on , then each of and is adjacent to a face in , but this is impossible. If is an internal vertex, then each of and is adjacent to a face in , but this is also impossible. This completes the proof. ∎
Acknowledgments. This work was supported by the Natural Science Foundation of China and partially supported by the Fundamental Research Funds for Universities in Henan (YQPY20140051).
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