#### Abstract

In this paper, by applying the discharging method, we show that if is a planar graph with a maximum degree of that does not contain any adjacent 8-cycles, then is of class 1.

#### 1. Introduction

Graph coloring is a very important problem in graph theory. Since the four-color problem was first proposed, many other forms of coloring problems have been put forward and extensively studied. In this paper, we study the edge coloring of a planar graph. It is assumed that all graphs are finite, without parallel edges, and all edges are undirected. A planar graph is a graph that can be embedded in the plane satisfying the condition that no two edges intersect geometrically except at a vertex to which they are both incident. Table 1 presents some symbols and definitions used in this paper.

A graph is said to be -edge-colorable if the edges of can be colored with colors such that adjacent edges have different colors. In all edge colorings of , the edge chromatic number of a graph is the minimum number of colors used. In 1964, Vizing proved that, for any graph of maximum degree , or . This result divides all graphs into two classes: graphs satisfying are in class 1, whereas graphs satisfying are in class 2. For a connected graph , if it is of class 2 and for , then is critical. A -critical graph is a critical graph of maximum degree .

For planar graphs, Vizing [1] proved that any planar graph with is of class 1 and illustrated that there are class 2 planar graphs with . Therefore, Vizing conjectured that any planar graph with is of class 1. Zhang [2] and Sanders and Zhao [3] proved the case of independently. The case of has not yet been solved, but some positive results with restrictive conditions have been obtained, as in Theorem 1.

Theorem 1. If is one of the following planar graphs, then is of class 1:(1) and , or and , or and ([1])(2) without intersecting 3-cycles ([4])(3) without 4- or 5-cycles ([5])(4) without 5- or 6-cycles with chords [6](5) without 5-cycles ([7]) or 6-cycles ([8]) with two chords(6) without 7-cycles ([9]) or 6-cycles ([10]) with three chords(7) and any vertex is incident with at most three triangles ([11])(8) and without adjacent -cycles, where ([12])(9) and any two 7-cycles are not adjacent ([13]).

In this paper, two cycles sharing at least a common edge are said to be adjacent. We present a result concerning the edge chromatic number of planar graphs with which do not contain any adjacent 8-cycles.

#### 2. Main Result and Its Proof

To prove our main result, we first present several lemmas.

Lemma 1 (see [2, 3]). Suppose that is a planar graph with . Then, .

Lemma 2 (Vizingâ€™s Adjacency Lemma [1]). Let be a -critical graph, and let with .(1)If , then has at least -neighbors(2)If , then has at least two -neighbors

Lemma 3 (see [2]). Let be a -critical graph, and let with . Then,(1)Every vertex of is a -vertex(2)Every vertex of has a degree of at least (3)If , then every vertex of is a -vertex

Lemma 4 (see [3]). No -critical graph has distinct vertices , , and such that is adjacent to and , , and is in at least triangles not containing .

We use to denote a 3-face with three vertices satisfying . A face is denoted by a bad 4-face in the event that and form a 3-face. Let denote the number of bad 4-faces incident with such that is incident with a bad 4-face with . Let , so that .

Lemma 5. Let be a 6-critical plane graph with no two adjacent 8-cycles. Let .(1)If , and , then the remaining face is a -face. Moreover, if , then there is at least one 5-neighbor of on the -face and .(2)If and , then the remaining face is an -face or a 4-face (as in case 1.1 of Figure 1). Moreover,â€‰(2.1) if the remaining face is an -face and any 3-neighbors and 5-neighbors are not adjacent, then for each 5-neighbor of â€‰(2.2) if the remaining face is a 4-face and , then for each - neighbor of â€‰(2.3) if the remaining face is a 4-face and , then there are two cases for each 6-neighbor of â€‰(2.3.1) if , then , and â€‰(2.3.2) if , then and (3)If and , then the remaining face is an -face other than cases 1.2â€“1.7 in Figure 1.â€‰(3.1) and in Figure 1 (cases 1.2, 1.4, and 1.5)â€‰(3.2) , and in Figure 1 (case 1.3)â€‰(3.3) and in Figure 1 (cases 1.6 and 1.7)(4)If and , then , , or (as in Figure 1, case 1.8). Furthermore, if , then there are three cases for each 6-neighbor of â€‰(4.1) if , then , and â€‰(4.2) if , then and â€‰(4.3) if , then , and (5)If is on a bad 4-face satisfying and the remaining face is a 4-face or 5-face (as in Figure 1, cases 1.9â€“1.25), then there are four cases for vertex :â€‰(5.1) , and in Figure 1, cases 1.9â€“1.15â€‰(5.2) , and in Figure 1, cases 1.16â€“1.21â€‰(5.3) and in Figure 1, cases 1.22â€“1.24â€‰(5.4) , and in Figure 1, case 1.25(6)If , and is on a 3-face satisfying and another face incident with is a 5-face (as in Figure 1, cases 1.26â€“1.30), then there are three cases for vertex :â€‰(6.1) , and in Figure 1, case 1.26â€‰(6.2) , and in Figure 1, case 1.27â€‰(6.3) , and in Figure 1, cases 1.28â€“1.30

Theorem 2. Suppose that is a planar graph without any adjacent 8-cycles. If , then .

Proof. Suppose that is a counterexample to this theorem with the minimum number of edges and that is embedded in the plane. Then, is a 6-critical graph by Lemma 1, and it is 2-connected by Lemma 2. By Eulerâ€™s formula , we have thatDefine to be the initial charge and . Suppose that . Therefore, . For , we allocate a new charge in accordance with our discharging rules (see below). Because our rules only transfer charge between vertices and faces, they have no effect on the sum. In the event that for each , we obtain an apparent contradiction: . This completes the proof.
The discharging rules are as follows:â€‰R1: let in . Then, each vertex on obtains from .â€‰R2: let and . If is on a bad 4-face such that form a 3-face, then obtains from and from each neighbor. If is on two 4-faces, then obtains 1 from each neighbor. If is on a -face, then obtains from , and at least from each neighbor.â€‰R3: let . Each 3-vertex obtains from each neighbor.â€‰R4: let and . If , and satisfies (1) , (2) and , or (3) , and , then sends to .â€‰R5: let be a 3-face satisfying . If and , then obtains from and from . If and , then obtains 1 from . If , then obtains from each of .â€‰R6: let .â€‰R6.1: suppose that and . If , then obtains 0 from . Otherwise, if is on a -face satisfying and , then obtains from . Otherwise, obtains from .â€‰R6.2: if , , and , then , and or . If , then obtains from ; if , then obtains from .â€‰R7: let .â€‰R7.1: suppose that and is on a 3-face satisfying and . Then, obtains from .â€‰R7.2: suppose that , and is incident with a 3-face such that . If another face that is incident with is an -face () (i.e., cases 1.2â€“1.7 in Figure 1), then obtains from . If another face that is incident with is an -face, then obtains from .â€‰R7.3: suppose that , and . If is a 6-neighbor of , then by Lemma 5 (2.3) and obtains from .â€‰R7.4: suppose that , and (i.e., case 1.1 in Figure 1). If is a 6-neighbor of , then by Lemma 7 (4) and obtains from .â€‰R7.5: if is on a bad 4-face satisfying , and (i.e., cases 1.9â€“1.25 in Figure 1), then obtains from .â€‰R7.6: if , and is on a 3-face satisfying and another face that is incident with is a 5-face (i.e., cases 1.26â€“1.30 in Figure 1), then obtains from .In the following, we start to verify for all . Suppose that , and obviously, . In the event of , by R1. In the event of , . Suppose that . As , must be a -face, -face, -face, or -face by Lemma 2. Therefore, by R5.
Let . Obviously, . In the event of , has two 6-neighbors by Lemma 2. Therefore, by R2. In the event of , by Lemma 2. Therefore, by R3. In the event of , or by R1 and R4.
There are only two cases left to test.

Case 1. Suppose that . By Lemma 2, and .
Suppose that . If , then by R5 and R6.1; otherwise, , and then by R5 and R6.1.
Suppose that (denote one 5-neighbor of as ). Then, by Lemma 2. There are three situations. First, suppose that . If , then or because there are no two adjacent 8-cycles in . By R6.2, sends or to . If , then by R1, R5, and R6.1. If , then by R1, R5, and R6.1. Otherwise, , and sends no charge to . Then, by R5 and R6. Second, if , then by R5 and R6.1. Third, suppose that . If , then by R5 and R6.1; otherwise, , and then or because there are no two adjacent 8-cycles in . Therefore, sends or to by R6.2 and or by R5 and R6.1.
Suppose that . By Lemma 2, , and, by Lemma 4, is on at most two -faces. There are several subcases. First, and . Thus, . If , then by R5 and R6.1; otherwise, , and then by R5 and R6.1. Second, and (mark the 5-vertex as ). If , then , and or because no two adjacent 8-cycles appear in . By R6.2, sends or to . If , then by R1, R5, and R6.1. If , then by R1, R5, and R6.1. Otherwise, , and sends no charge to . If , then by R5 and R6.1. If , then or because contains no adjacent 8-cycles. Thus, sends or to by R6.2, and or by R5 and R6.1. Third, . Suppose that . If , then by R5 and R6.1; otherwise, , and then by R5 and R6.1. If , then by R5 and R6.1.
Suppose that (mark the 3-vertex as ). Then, by Lemma 2. Suppose that . If , then by R3, R5, and R6.1; otherwise, , and then by R3, R5, and R6.1. If , then by R5 and R6.1.

Case 2. Let . By Lemma 2,