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 (Vizing’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, (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 (2.2) if the remaining face is a 4-face and , then for each - neighbor of (2.3) if the remaining face is a 4-face and , then there are two cases for each 6-neighbor of (2.3.1) if , then , and (2.3.2) if , then and (3)If and , then the remaining face is an -face other than cases 1.2–1.7 in Figure 1. (3.1) and in Figure 1 (cases 1.2, 1.4, and 1.5) (3.2) , and in Figure 1 (case 1.3) (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 (4.1) if , then , and (4.2) if , then and (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 1.9–1.25), then there are four cases for vertex : (5.1) , and in Figure 1, cases 1.9–1.15 (5.2) , and in Figure 1, cases 1.16–1.21 (5.3) and in Figure 1, cases 1.22–1.24 (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 1.26–1.30), then there are three cases for vertex : (6.1) , and in Figure 1, case 1.26 (6.2) , and in Figure 1, case 1.27 (6.3) , and in Figure 1, cases 1.28–1.30
Proof. Let be arranged in a counterclockwise sequence. Let be the face incident with , , and (). Note that all the indices in this lemma are modulo 6 if it is larger than 6. Obviously, by Lemma 3.(1)If , then by Lemma 3. If , then by Lemma 3. Obviously, there are adjacent 8-cycles. Suppose that and is a bad 4-face. In that case, , and . Three cases are discussed below from the perspective of symmetry. First, suppose that . In the event of , there are at least two adjacent 8-cycles in . Thus, . Moreover, if , then is on the -face. Let be the faces that are incident with in an anticlockwise sequence. If , then there are at least two adjacent 8-cycles in . Thus, and . Second, suppose that . In the event of , there are at least two adjacent 8-cycles in . Therefore, . Moreover, if , then and are on the -face. Without loss of generality (WLOG), we consider the vertex . Let be the faces that are incident with in an anticlockwise sequence. In the event of , there are at least two adjacent 8-cycles in . Hence, and . Third, suppose that . In the event of , there are at least two adjacent 8-cycles in . Therefore, . Moreover, if , then and are on the -face. WLOG, we consider the vertex . Let be the faces that are incident with in an anticlockwise sequence. In the event of , there are at least two adjacent 8-cycles in . Therefore, and .(2)Suppose that are 3-faces. If , then there is only one case, as in Figure 1 (case 1.1). If , then there are at least two adjacent 8-cycles in . Therefore, . (2.1) Suppose that the remaining face is an -face and any 3-neighbors and any 5-neighbors of are not adjacent. WLOG, we suppose that , , and is not adjacent to . Let be the faces that are incident with in an anticlockwise sequence. If , then and ; otherwise, there are at least two adjacent 8-cycles in . Thus, . If , then ; otherwise, . Thus, . If and is not adjacent to , let be the faces that are incident with in an anticlockwise sequence. If , then ; otherwise, there are at least two adjacent 8-cycles in . Thus, and . The proofs of cases are similar, so we only give the proof for here. (2.2) Suppose that the remaining face is a 4-face satisfying . Suppose that , another face incident with edge is , and another face incident with edge is . If , then there are at least two adjacent 8-cycles in . Therefore, . If , then there are at least two adjacent 8-cycles in . Thus, . The proofs of cases are similar, so we omit them here. (2.3) Suppose that the remaining face is a 4-face and . (2.3.1) Suppose that and are the faces that are incident with in an anticlockwise sequence. If , then there are at least two adjacent 8-cycles in . Therefore, . In the event of , (otherwise, there are at least two adjacent 8-cycles in ); otherwise, . Thus, , and . The proofs of cases and are similar, so we omit them here. (2.3.2) Suppose that and are the faces incident with in an anticlockwise sequence. If or , then ; otherwise, because no two adjacent 8-cycles appear in . If or , then because no two adjacent 8-cycles appear in ; otherwise . The proofs of cases and are similar, so we omit them here.(3)Suppose that , and are 3-faces. In the event of , only three cases 1.2–1.7 in Figure 1 exist. If , then there are at least two adjacent 8-cycles in . Thus, . Let be the faces that are incident with in an anticlockwise sequence. (3.1) First, if , then and . Second, if , then there are at least two adjacent 8-cycles in . Therefore, . If , then ; otherwise, there are at least two adjacent 8-cycles in . If , then there are at least two adjacent 8-cycles in . Thus, . Therefore, and in Figure 1 (case 1.2). Cases 1.4 and 1.5 in Figure 1 are similar, so we omit them here. (3.2) If , then there are at least two adjacent 8-cycles in . Thus, . If , then there are at least two adjacent 8-cycles in . Therefore, . Hence, , and in Figure 1, case 1.3. (3.3) If , then there are at least two adjacent 8-cycles in . Therefore, . If , then there are at least two adjacent 8-cycles in . Therefore, . Hence, and in Figure 1 (case 1.6). Case 1.7 in Figure 1 is similar, so we omit it here.(4)If , then . WLOG, suppose that are 3-faces and is a bad 4-face. If , then there are at least two adjacent 8-cycles in . Therefore, and . Hence, . If , then . Two cases are discussed below from the perspective of symmetry. First, suppose that are bad 4-faces. If and are 3-faces, then and ; otherwise, there are at least two adjacent 8-cycles in . If and are 3-faces, then and ; otherwise, there are at least two adjacent 8-cycles in . If and are 3-faces, then and ; otherwise, there are at least two adjacent 8-cycles in . If and are 3-faces, then and ; otherwise, there are at least two adjacent 8-cycles in . Second, suppose that and are bad 4-faces. If and are 3-faces, then and ; otherwise, there are at least two adjacent 8-cycles in . If and are 3-faces, then and ; otherwise, there are at least two adjacent 8-cycles in . Thus, . Suppose that and is a bad 4-face. Then, . Five cases are discussed below from the perspective of symmetry: , , , , or is a 3-face. As the discussion of each case is similar, we only present the proof for the case in which is a 3-face. If , then ; otherwise, and . Thus, or . Suppose that and . Three cases are discussed below from the perspective of symmetry. First, suppose that . If , then ; if , then . Therefore, or . Second, suppose that . If , then ; if , then . Therefore, or . Third, suppose that . If , there are two cases: , as in case 1.8 of Figure 1, or . If , then . Thus, , , or . Suppose that . Symmetry means that there are three distinct cases: , , and . As the discussion of each case is similar, we only present a proof of here. Let be the faces that are incident with in an anticlockwise sequence. If , then there are at least two adjacent 8-cycles in . Thus, . If , then there are at least two adjacent 8-cycles in . Therefore, . Hence, , and .(5)Suppose that is on a bad 4-face satisfying . In this case, the remaining face is a 4-face or a 5-face, as in cases 1.9–1.25 in Figure 1. For vertex , there are four cases. As the discussion of each case is similar, we only present the proof for case 1.9 in Figure 1. Let be the faces that are incident with in an anticlockwise sequence. If , then there are at least two adjacent 8-cycles in . Thus, . If , then there are at least two adjacent 8-cycles in . Thus, . Hence, , and in case 1.9 of Figure 1.(6)Suppose that , and is on a 3-face satisfying . In this case, another face that is incident with is a 5-face (as in Figure 1, cases 1.26–1.30). For vertex , there are three cases. As the discussion of each case is similar, we only present the proof of case 1.26 in Figure 1.Let be the faces that are incident with in an anticlockwise sequence, besides the known faces. Face cannot be a 3-face; otherwise, there would be a multiedge. If , then there are at least two adjacent 8-cycles in . Therefore, . Hence, , and in case 1.26 of Figure 1.
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 Euler’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: R1: let in . Then, each vertex on obtains from . 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. R3: let . Each 3-vertex obtains from each neighbor. R4: let and . If , and satisfies (1) , (2) and , or (3) , and , then sends to . 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 . R6: let . R6.1: suppose that and . If , then obtains 0 from . Otherwise, if is on a -face satisfying and , then obtains from . Otherwise, obtains from . R6.2: if , , and , then , and or . If , then obtains from ; if , then obtains from . R7: let . R7.1: suppose that and is on a 3-face satisfying and . Then, obtains from . 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 1.2–1.7 in Figure 1), then obtains from . If another face that is incident with is an -face, then obtains from . R7.3: suppose that , and . If is a 6-neighbor of , then by Lemma 5 (2.3) and obtains from . 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 . R7.5: if is on a bad 4-face satisfying , and (i.e., cases 1.9–1.25 in Figure 1), then obtains from . R7.6: if , and is on a 3-face satisfying and another face that is incident with is a 5-face (i.e., cases 1.26–1.30 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, , , and .
In the process of discharging, there are some special cases. That is, some 6-vertices discharge to other 6-vertices. Therefore, we now individually verify these special cases. In the following, we need to check the two following subcases.
Subcase 1. gives some charge to some 6-vertex (see R7).
Suppose that is on a 3-face satisfying and . If , then , and because no two adjacent 8-cycles appear in . As , we have that by Lemma 2, by Lemma 3, and by Lemma 4. Thus, by R1, R3, R5, R6.1, and R7.1. Otherwise, , and then and . Because , we have that by Lemma 2 and by Lemma 4. Therefore, by R1, R3, R5, R6.1, and R7.1.
Suppose that is on a 3-face satisfying . If a 4-face is incident with , then and by Lemma 7 (3). By R1, R2, R5, and R7.2, . If a 5-face is incident with , then there are two cases: (1) , and . In this case, by R1, R2, R5, and R7.2. (2) and . By R1, R2, R5, and R7.2, . If a 6-face is incident with , then there are two cases: (1) and . In this case, by R1, R2, R5, and R7.2. (2) and . In this case, by R1, R2, R5, and R7.2. If an -face is incident with , then , and because no two adjacent 8-cycles appear in . Therefore, by R1, R2, R5, and R7.2.
Suppose that and such that , and (i.e., case 1.1 in Figure 1). Because , we know that by Lemma 3. First, if , then , and . By R7.3, sends to . If , by R1, R2, and R5. If , by R1, R2, and R5. If , by R1, R2, and R5. Second, if , then and . By R7.3, sends to . If , then by R1, R2, and R5. Assume that . If is not on any -face, then, by R1, R2, and R5, . Therefore, is on a -face, and by R1, R2, and R5. If , by R1, R2, and R5.
Suppose that and such that (i.e., case 1.8 in Figure 1). First, if , then , and . By R7.4, sends to . If , then by R1, R2, and R5. If , then by R1, R2, and R5. If , then by R1, R2, and R5. If , then by R1, R2, and R5. Second, if , then and . By R7.4, sends to . If , then by R1, R2, and R5. If , then by R1, R2, and R5. Assume that . If is not on any -face, then by R1, R2, and R5. Thus, is on a -face. By R1, R2, and R5, . If , then by R1, R2, and R5. Third, if , then and . By R7.4, sends to . If , then by R1, R2, and R5. If , then by R1, R2, and R5. If , then by R1, R2, and R5. If , then by R1, R2, and R5.
Suppose that is on a bad 4-face satisfying with satisfying , . By Lemma 7 (5), there are four cases: (1) and . Then, by R1, R2, R5, and R7.5. (2) , and . Then, by R1, R2, R5, and R7.5. (3) and . Then, by R1, R2, R5, and R7.5. (4) and . Then, by R1, R2, R5, and R7.5.
Suppose that is on a 3-face satisfying and another face incident with is a 5-face, with satisfying . By Lemma 7 (6), there are three cases: (1) , and . Therefore, by R1, R2, R5, and R7.6. (2) , and . Then, by R1, R2, R5, and R7.6. (3) , and . Then, by R1, R2, R5, and R7.6.
Subcase 2. gives no charge to other 6-vertices.
Let be neighbors of in an anticlockwise order and let be the faces that are incident with such that is incident with and for all , where .
Subcase 3. .
In the event of , by Lemma 2. Therefore, by R5 and R6.1.
Suppose that . By Lemma 2, . Suppose that is not on any -face. In the event of , by R5 and R6.1; otherwise, , and so by R5 and R6.1. In light of this, is on a -face, and so by R5.
Suppose that (mark this vertex as ). By Lemma 2, . If (mark this vertex as ), then and are not on any 3-face by Lemma 4. Thus, by R2 and R3. In the event of (mark this vertex as ), is on at most one 3-face and is not on any 3-face by Lemma 4. Therefore, by R3 and R5. If (mark this vertex as ), then gives at most to by R6.1. Hence, by R3 and R5.
Suppose that (mark this vertex as ). By Lemma 3, and . In the event of , by R2 and R5; otherwise, . If is on a bad 4-face, then by R2 and R5. If is on a 3-face and a -face, then by R1, R2, and R5. If is on two -faces, then by R2 and R5.
Subcase 4. .
Subcase 5. .
By Lemma 7 (4), , , or .
Subcase 6. or .
Suppose that , so that by Lemma 2. Thus, or by R1, R5, and R6.1.
Suppose that . Then, by Lemma 2. Suppose that is not on any -face. If , then or by R1, R5, and R6.1; otherwise, , and so or by R1, R5, and R6.1. Therefore, is on one -face. Then, or by R1, R5, and R6.1.
Suppose that (mark this vertex as ). Then, by Lemma 2 and by Lemma 5. If (mark this vertex as ), then and are not on any 3-face by Lemma 4 and . This is impossible. If (mark this vertex as ), then is on at most one 3-face and is not on any 3-face by Lemma 4. Therefore, by R3 and R5. If (mark this vertex as ), then gives at most to by R6.1. Then, or by R1, R3, R5, and R6.1.
Subcase 7. .
In the event of , by R5.
Suppose that . By Lemma 2, . Then, by R5, R6.1, R6.3, and R7.4.
Suppose that . Then, by Lemma 2. Suppose that is not on any -face. If , then . If , then by R5, R6.1, R6.3, and R7.4. Therefore, is on one -face. Then, by R5, R6.1, R6.3, and R7.4.
Suppose that . At this point, by Lemma 2 and by Lemma 4. Thus, by R5, R6.1, R6.3, and R7.4.
Subcase 8. .
Mark as the 2-neighbor of and let be another neighbor of . There are five scenarios to discuss: (1) If is on a 3-face and an -face, then the -face gives at least to by R1. Therefore, by R2 and R5. (2) If is on a 3-face and a 7-face, then by R1, R2, and R5. (3) If is on one 3-face and one 6-face, then by R1, R2, and R5. (4) If is incident with one 3-face and one 5-face, then the remaining face of is a -face or a 4-face. If the remaining face of is a -face, then by R1, R2, and R5; otherwise, the remaining face of is a 4-face, and then gives to by R7.6, so by R1, R2, and R5. (5) is on a bad 4-face satisfying . If the remaining face of is a 4-face or a 5-face, then gives to by R7.5, so by R1, R2, and R5. If the remaining face of is a -face, then by R1, R2, and R5.
Subcase 9. .
Subcase 10. .
By the proof of Lemma 7 (1), we know that and . By Lemmas 7 (1) and Lemma 3, the remaining face is a -face and the neighbors of are -vertices. If , then by R1 and R5. Suppose that . By Lemma 2, . If , then by R1, R5, and R6.1; otherwise, . Then, there is at least one 5-vertex on the -face, and this 5-vertex is on another -face by Lemma 7 (1). Hence, by R6.1, sends no charge to the 5-vertex and by R1 and R5.
Subcase 11. .
WLOG, we assume that are 3-faces. In the following, we need to consider two situations.
Subcase 12. .
The remaining face is an -face or a 4-face by Lemma 7 (2).
Subcase 13. The remaining face is an -face.
If , then by R1 and R5.
Suppose that . By Lemma 2, . By Lemma 7 (2.1), each 5-neighbor of is on at least two -faces. Thus, by R6.1, sends no charge to the 5-neighbor and .
Suppose that . Then, by Lemma 2. Suppose that is not on any -face. If , then by R1 and R5; otherwise, . If , then by Lemma 7 (2.1). Therefore, by R1, R5, and R6.1. The cases are the same as , so we omit them. Therefore, is on one -face. If or is a -face, then by R1 and R5. If is a -face, then or because no two adjacent 8-cycles appear in . By R4, or sends at least to . Thus, by R1 and R5. If is a -face, it is similar to . If is a -face, then or because no two adjacent 8-cycles appear in . By R4, or sends at least to . Hence, by R1 and R5.
Suppose that . Therefore, by Lemma 4 and by Lemma 2. In the event of , we find that by R1, R3, and R5. Suppose that . If the 3-neighbor of is adjacent to a 5-neighbor, then sends no charge to the 5-neighbor by R6.1 and by R1, R3, and R5. Thus, the 3-neighbor of is not adjacent to the 5-neighbor. WLOG, we assume that , and so , , or . Now, we can assume that , in which case by Lemma 7 (2.1). Thus, by R1, R3, and R5. If (), then this is similar to the case , and so we omit it here.
Subcase 14. The remaining face is a 4-face.
If , then by R5.
Suppose that . Then, by Lemma 2 and every 5-neighbor of is on at least two -faces by Lemma 7 (2.2). Therefore, sends no charge to its 5-neighbors and by R5.
Suppose that . Then, by Lemma 2 and every 4-neighbor of is on at least two -faces by Lemma 7 (2.2). Thus, the 4-neighbor gives at least to by R4 and by R5.
Suppose that . Then, by Lemma 4 and by Lemma 2. Each 6-neighbor sends to by R7.3. Thus, by R5.
Subcase 15. .
The remaining face is an -face or cases 1.2–1.7 in Figure 1.
If the remaining face is a 4-face, then sends to by R7.2. Thus, by R2 and R5. If the remaining face is a 5-face, then sends to by R7.2. Then, by R2 and R5. If the remaining face is a 6-face, then sends to by R7.2. Thus, by R2 and R5. If the remaining face is an -face, then sends to by R7.2. Then, by R2 and R5.
Subcase 16. .
Obviously, . If , then by R5.
Suppose that . Then, by Lemma 2. WLOG, we suppose that and the other three faces of are denoted as in anticlockwise order. As no two adjacent 8-cycles appear in , we have that , and sends no charge to its 5-neighbors by R6.1. Thus, by R5.
Suppose that . Then, by Lemma 2. Suppose that is not on any -face. WLOG, we suppose that and the other two faces of are denoted as in anticlockwise order. As no two adjacent 8-cycles appear in , we have that , and sends to by R4. If is adjacent to a 5-neighbor, then the 5-neighbor is incident with at least two -faces because no two adjacent 8-cycles appear in . Thus, sends no charge to its 5-neighbors. If , then by R5. If , then by R5. If , then by R5. Thus, is incident with one -face. WLOG, we suppose that is a -face and that and are on two -faces because no two adjacent 8-cycles appear in . Thus, and send at least to by R4. Therefore, by R5.
Suppose that . Therefore, by Lemma 4 and by Lemma 2. There are two subcases. First, we suppose that and (WLOG), and then . By R7.1, and each send to . Thus, by R3 and R5. Second, we suppose that . (1) Suppose that the 3-neighbor of is adjacent to a 5-neighbor. WLOG, we suppose that and . Then, sends to by R7.1. Thus, by R5. (2) Suppose that the 3-neighbor of is not adjacent to a 5-neighbor. WLOG, we assume that , and then and . Now, , , or . Obviously, the 5-neighbor is on at least two -faces because contains no adjacent 8-cycles. Thus, sends no charge to the 5-neighbor. By R7.1, and each send to . Thus, by R5.
Data Availability
No data were used to support this study.
Conflicts of Interest
The author declares that there are no conflicts of interest.
Acknowledgments
This work was partially supported by a research grant from the NSFC (no. 12001332) and the Shandong Provincial Natural Science Foundation, China (no. ZR2017BA009).