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Abstract
A total coloring of a graph is an assignment of colors to its vertices and edges such that no two adjacent or incident elements receive the same color. The Total Coloring Conjecture (TCC) states that every simple graph has a total ()coloring, where is the maximum degree of . This conjecture has been confirmed for planar graphs with maximum degree at least 7 or at most 5, i.e., the only open case of TCC is that of maximum degree 6. It is known that every planar graph of or with some restrictions has a total coloring. In particular, in [Shen and Wang, “On the 7 total colorability of planar graphs with maximum degree 6 and without 4cycles”, Graphs and Combinatorics, 25: 401407, 2009], the authors proved that every planar graph with maximum degree 6 and without 4cycles has a total 7coloring. In this paper, we improve this result by showing that every diamondfree and housefree planar graph of maximum degree 6 is totally 7colorable if every 6vertex is not incident with two adjacent 4cycles or not incident with three cycles of size for some .
On a Sufficient Condition for Planar Graphs of Maximum Degree 6 to be Totally 7Colorable]On a Sufficient Condition for Planar Graphs of Maximum Degree 6 to be Totally 7Colorable

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Enqiang Zhu and Chanjuan Liu and Yongsheng Rao^{†}^{†}thanks: Corresponding author: ysrao2018@163.com
Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
School of Computer Science and Technology, Dalian University of Technology, Dalian, China
1 Introduction
Throughout the paper, we consider only simple, finite and undirected planar graphs, and follow Bondy and Murty (2008) for terminologies and notations not defined here. Given a graph , we use and to denote the vertex set and the edge set of , respectively. For a vertex , we denote by the degree of in and let . A vertex, vertex or vertex is a vertex of degree , at most or at least . For a planar graph , we always assume that is embedded in the plane, and denote by the set of faces of . The degree of a face , denoted by , is the number of edges incident with , where each cutedge is counted twice. A face of degree , at least or at most is called a face, face, or face. A face with consecutive vertices along its boundary in some direction is often said to be a . Two faces are called adjacent if they are incident with a common edge.
A total coloring of a graph is a coloring from to such that no two adjacent or incident elements have the same color. A graph is said to be totally colorable if it admits a total coloring. The total chromatic number of , denoted by , is the smallest integer such that is totally colorable. The Total Coloring Conjecture (TCC) states that every simple graph is totally ()colorable Behzad (1965); Vizing (1968), where is the maximum degree of . This conjecture has been proved for graphs with in Kostochka (1996). For planar graphs, the only open case of TCC is that of maximum degree 6; see Borodin (1989); Jensen and Toft (1995); Sanders and Zhao (1999). More precisely, if is a planar graph with , then . For planar graphs with maximum degree 7 or 8, some related results can be found in Chang et al. (2011, 2013); Du et al. (2009); Hou et al. (2011, 2008); Liu et al. (2009); Shen and Wang (2009b); Wang and Wu (2011, 2012); Xu and Wu (2014); Wang et al. (2014). Moreover, for planar graphs of maximum degree 6, it is proved that if does not contain 5cycles Hou et al. (2011) or 4cycles Shen and Wang (2009a). In this paper, we show that every planar graph with has a total 7coloring if contains no some forbidden 4cycles, which improves the result of Shen and Wang (2009a).
Theorem 1.1
Suppose that is a planar graph with . If does not contain a subgraph isomorphic to a diamond or a house, as shown in Figure 1, and every 6vertex in is not incident with two adjacent 4cycles or three cycles with sizes for some , then .
2 Reducible configurations
Let be a minimal counterexample to Theorem 1.1, in the sense that the quantity is minimum. That is, satisfies the following properties:
(1) is a planar graph of maximum degree 6.
(2) contains no subgraphs isomorphic to a diamond or a house.
(3) Every 6vertex of is incident with neither two adjacent 4cycles, nor three cycles with sizes for some .
(4) is not totally 7colorable such that is minimum subject to (1),(2),(3).
Notice that every planar graph with maximum degree 5 is totally 7colorable Kostochka (1996). Additionally, it is easy to check that every subgraph of also possesses (2) and (3). Therefore, every proper subgraph of has a total 7coloring using the color set =. For a vertex , we use to denote the set of colors appearing on and its incident edges, and =. This section is devoted to investigating some structural information, which shows that certain configurations are reducible, i.e. they can not occur in .
Lemma 2.1
Let be an edge of such that or . Then .
The subgraph induced by all edges, whose two ends are 2vertex and 6vertex respectively in is a forest.
For any component of the forest stated in Lemma 2.1 (2), we can see that all leaves (i.e. 1vertices) of are 6vertices. Therefore, has a maximum matching that saturates every 2vertex in . For each 2vertex in , we refer to the neighbor of that is saturated by as the master of , see Borodin et al. (1997). Clearly, for a given , each 6vertex can be the master of at most one 2vertex, and each 2vertex has exactly one master.
The following result follows from Lemma 2.1 directly.
Lemma 2.2
Every 4face in is incident with at most one 2vertex.
Lemma 2.3
Let be a 3face incident with a 2vertex. Then every 6vertex incident with has only one neighbor of degree 2.
Proof: Let be the 2vertex incident with , and be the two 6vertices incident with . We first show that the result holds for , and then holds for analogously. Assume to the contrary that has another neighbor of degree 2, say . Let be a total 7coloring of by the minimality of . Erase the colors on and . Without loss of generality, we assume =. If , then can be properly colored with 7. Hence, has a total 7coloring by coloring properly (Since are 2vertices, there are at least three available colors for each of them), and a contradiction. So we assume . Let . When , we can color with and recolor with 7. When , let and . We first exchange the colors of and , and then color with and recolor with . Therefore, we can obtain a 7totalcoloring of by coloring with two available colors. This contradicts the assumption of .
Lemma 2.4
has no (4,4,4)face.
Proof: Suppose that has a (4,4,4)face with three incident vertices , and . By the minimality of , has a total 7coloring. Erase the colors on for . Clearly, each element in has at least three available colors. Since every 3cycle is totally 3choosable, it follows that has a total 7coloring, and a contradiction.
Lemma 2.5
has no (3,5,3,5)face.
Proof: Assume to the contrary that has a (3,5,3,5)face , where , . By the minimality of , has a total 7coloring . Erase the colors on . Clearly, each edge of has at least two available colors. Since even cycles are 2edgechoosable, we can properly color edges and . Additionally, since , are 3vertices and they are not adjacent (Since contains no subgraph isomorphic to a diamond), we can properly color and with two available colors. Hence, we obtain a total 7coloring of , and a contradiction.
Lemma 2.6
Every 6vertex incident with a 2vertex in is adjacent to at most five vertices.
Proof: Let be a 6vertex incident with a 2vertex in . Assume to the contrary that contains six vertices. Let , where for . By the minimality of , has a total 7coloring . Without loss of generality, we assume . Erase the colors on for . If 7 does not appear on the edges incident with , then we can properly color with 7. Otherwise, we can properly color with by recoloring with 7. Additionally, since are vertices, there is at least one available color for each of them and by Lemma 2.1 (1) for any , . Hence, we can obtain a 7totalcoloring of , and a contradiction.
Lemma 2.7
contains no configurations depicted in Figure 2, where the vertices marked by have no other neighbors in .
Proof: For configuration (1), by the minimality of , has a 7totalcoloring . Without loss of generality, we assume that . If (or ), then we can properly color with 7 (or with by recoloring with 7). If , then we can properly color with by recoloring with 7. So, we assume . Let . Obviously, . If , then we can recolor with 7, with , with , and then properly color with 7. If , then . Therefore, we can safely interchange the colors of and , recolor with 7, with , with , and then properly color with 7. Thus, we obtain a 7totalcoloring of , and a contradiction.
For configuration (2), let be a 7totalcoloring of . Assume that , where for , and . By a similar argument as in (1), we assume . Let and . Obviously, . First, if , then we can recolor with 7, with , with , and then properly color with 7. Second, if , then . When , we can safely interchange the colors of and , recolor with 7, with , with , and then properly color with 7. When , we can safely interchange the colors of and , recolor with 7, with , with , and then properly color with 7. Third, if , then . When , we can recolor and with 7, recolor , and with , and then properly color with 7. When , we can safely interchange the colors of and , recolor and with 7, recolor , and with , and then properly color with 7. Hence, we obtain a 7totalcoloring of , and a contradiction.
For configuration (3), let be a 7totalcoloring of . Assume that , where for , and . By a similar argument as in (1), assume that . Let and . Obviously, and . If , then we can recolor with 7, with , with , and then properly color with 7. If , then . Therefore, we can recolor with 7, with , with , with , and then properly color with 7. So, we obtain a 7totalcoloring of , and a contradiction.
3 Discharging
In this section, to complete the proof of Theorem 1.1, we will use discharging method to derive a contradiction. For a vertex , denote by and (or simply by and ) respectively the number of 3faces and 4faces incident with . For a face , denote by and (or simply by and ) respectively the number of 2vertices and 3vertices incident with .
According to Euler’s formula , we have
Now, we define to be the initial charge of . Let for each . Obviously, . Then, we apply the following rules to reassign the initial charge that leads to a new charge . If we can show that for each , then we obtain a contradiction, and complete the proof.
 (R1)

From each vertex to each of its incident face , transfer
, if , and is a ()face;
, if , and is a ()face or ()face;
, if , and is incident with a 2vertex or 3vertex.
 (R2)

From each 6vertex to each of its incident 4face , transfer
, if is a ()face;
, if is a ()face;
, if is a ()face or ()face
, if is a ()face;
, if is a ()face;
, if is a ()face.
 (R3)

From each 6vertex to each of its adjacent 2vertex , transfer
, if is incident with a 3face;
, if is not incident with a 3face and is a master of ;
, if is not incident with a 3face and is not a master of .
 (R4)

From each 4face to each of its adjacent vertex , transfer
, if and is not incident with a 3face;
, if and is not incident with a 3face.
 (R5)

From each face to each of its adjacent vertex , transfer
, if and is incident with a 3face;
, if and is not incident with a 3face;
, if and is incident with a 3face;
, if and is not incident with a 3face.
 (R6)

Each face transfer to its adjacent ()face.
 (R7)

Every face with positive charge after R1 to R6 transfers its remaining charges evenly among its incident 6vertices.
The rest of this article is to check that for every .
4 Final charge of faces
Let be a face of . Suppose that is a 3face. By Lemma 2.1(1) and Lemma 2.4, it follows that is incident with at most two vertices. If is incident with at most one vertex, then by (R1) or . If is incident with two vertices, then by (R1) and (R6), .
Suppose that is a 4face. Clearly, is not adjacent to a 3face, since does not contain any subgraph isomorphic to a house. If is incident with neither a 2vertex nor a 3vertex, then ; If is incident with a 2vertex, then is a (2,6,,6)face by Lemma 2.1 (1) and Lemma 2.2, and the 2vertex is not incident with any 3face (Since contains no subgraph isomorphic to a diamond). So, by (R2), (R4) and (R7), (When is incident with a 3vertex), or (When is not incident with any 3vertex); If is not incident with a 2vertex but is incident with a 3vertex, then is either a (3,,,)face, or a (3,,,)face by Lemma 2.1 (1) and Lemma 2.5. For the former case, after (R1), (R2) and (R4), has at least (When is (3,)face), or (When is ()face), or (When is (3,6,4,6)face). Therefore, by (R7). For the latter case, by (R1), (R2), (R4) and (R7) (When is (3,)face), or by (R2), (R4) and (R7) (When is (3,)face).
Suppose that is a 5face. Since does not contain any subgraph isomorphic to a house, it has that every 2vertex incident with it is not incident with a 3face. Obviously, . If , then has at least after to (R6), and hence by . If , then is not adjacent to any ()face. Hence, has at least after to (R6), and .
Suppose that is a 6face. Then, at most one 2vertex incident with is incident with a 3face. Otherwise, contains a subgraph isomorphic to a house. By Lemma 2.1 (1) and (2), it is easy to see that and . When , the number of faces adjacent to is at most . Therefore, has at least after to (R6), and hence by . When , it follows that is not adjacent to any faces. Therefore, has at least after to (R6), and .
For the convenience of proving for every , we first introduce the following Lemma, which indicates that every face has positive charges.
Lemma 4.1
Let be a 6vertex. Then receives at least from each of its incident face by .
Proof: Let be a ()face incident with . Clearly, the number ()faces adjacent to is at most .
Suppose . Then has at least charges after (R5) to (R6). Since by Lemma 2.1 (2) and by Lemma 2.1 (1). Therefore, receives at least (when ) from .
Suppose . Clearly, . Particularly, in the case of , is not adjacent to any ()face. First, when , it has that is incident with at most two 2vertices that are incident with a 3face (Otherwise, there is a subgraph isomorphic to a house, and a contradiction). Therefore, has at least charges after (R5) to (R6). Second, when , it follows that . If , then is adjacent to at most three ()faces (Note that when is adjacent to a ()face, has to be incident with a 4vertex. So, is incident with at most four 6vertices in this case). Therefore, has at least charges after (R5) to (R6). If , then is not adjacent to any ()face. Therefore, has at least charges after (R5) to (R6). Third, when , it has that . In this case, we can see that has at least charges after (R5) to (R6). All of the above show that sends at least by (R7).
By Lemma 4.1, we can see that for every face .
4.1 Final charge of vertices
We start with an observation and a lemma.
Observation. Let be a vertex of . Since has no subgraph isomorphic to a diamond, we have . Moreover, if is a 6vertex, then by the condition of Theorem 1.1, and .
Lemma 4.2
Suppose that is a 6vertex incident with three consecutive faces of size 4, 6 and 4, respectively, where the 6face is denoted by ; See Figure 3 (a). Then by , gives at least
, if is incident with at most two vertices;
, if is incident with three vertices and (See Figure 3 (b));
, if is incident with three vertices and (See Figure 3 (c));
, if is incident with three vertices and .
Proof: Since contains no subgraph isomorphic to a diamond or a house, and are not incident with a 3face if and , and is incident with at most one 2vertex that is incident with a 3face.
For (1), if and , then is adjacent to at most two ()faces. Therefore, has at least charges after (R5) to (R6), and receives at least from .
If exact one of and is a vertex, say , then we consider two cases. First, . In this case, is adjacent to at most one ()face. Particular, if is a 2vertex, denoted by =, then is not adjacent to another 2vertex that is incident with a 3face by Lemma 2.3. This implies that when is incident with a 2vertex that is incident with a 3face, the 2vertex is a neighbor of and so is not incident with ant ()face. Consequently, has at least ( is a 2vertex), ( is a 3vertex)}= charges after (R5) to (R6), and receives at least from . Second, , i.e. is incident with only one vertex . Then, is adjacent to at most three ()faces. Therefore, has at least charges after (R5) to (R6), and receives at least from .
If and , then . When , one can readily check that is not adjacent to any ()face. Therefore, has at least charges after (R5) to (R6), and receives at least from . When , it has that is adjacent to at most two ()faces. Therefore, has at least charges after (R5) to (R6), and receives at least from . When , it has that has at least charges after (R5) to (R6), and receives more than from .
For (2) and (3), it follows that is not adjacent to any ()face. If , then the other vertex incident with is a 3vertex. Therefore, has at least charges after (R5) to (R6), and receives at least from . If , then has at least charges after (R5) to (R6), and receives at least from .
For (4), it is clear that and are vertices. Without loss of generality, we assume and ; See Figure 3 (d), where is another vertex incident with . If , then is not incident with a 3face by Lemma 2.3. Therefore, has at least charges after (R5) to (R6), and receives at least from .
In the following, we turn to the proof of for every . Let be a vertex of . By Lemma 2.1 (1), we have .
Suppose that is a 2vertex. Then has two neighbors with degree 6 by Lemma 2.1 (1). If is incident with a 3face, then is incident with a face. So, receives from each of its neighbors by (R3), and receives 1 from its incident face. Hence, . If is not incident with a 3face, then by (R3) receives from its master and from its other neighbor of degree 6, and receives from each of its adjacent face by (R4) and (R5). Therefore, .
Suppose that is a 3vertex. If is incident with a 3face, then is incident with two faces since does not contain any subgraph isomorphic to a house. So, by . If is not incident with any 3face, then is incident with three faces. Hence, by (R4) and (R5), .
Suppose that is a 4vertex. By the discharging rules (R1) to (R7), we have .
Suppose that is a 5vertex. By the observation, we have . If , then is incident with at most five 4faces. So, by (R1), =0. If , then by the condition of Theorem 1.1. So, by (R1), =. If =2, then by the same reason. So, by (R1) =.
Suppose that . By the observation we have +. Denote by the number of 2vertices adjacent to . Then by Lemma 2.6. When , it is clear that by (R1) and (R2). When , denote by the unique 2vertex adjacent to . First, . Then, by (R1), (R2) and (R3). Second, . In this case, by the condition of Theorem 1.1, either or . For the former, we have by (R1),(R2) and (R3). For the latter case, if is incident with a face, then is incident with at most one (3,5,3,6)face by Lemma 2.6. Therefore, by (R2) and (R3); If is not incident with a face, then is incident with at most two (3,5,3,6)faces. Therefore, by (R2) and (R3). In what follows, we assume , and then by Lemma 2.3, we have that every 2vertex is not incident with a 3face. Thus, when +, by (R3). Now, we further consider the following three cases.