List Edge Colorings of Planar Graphs without Adjacent 7-Cycles
In this paper, we get that is edge--choosable () for planar graph without adjacent 7-cycles.
Edge coloring and list edge coloring of graphs are very old fashioned problems in graph theory, and the research on such problems has a long history. Denote as the set of the integers. Now, we only consider the list edge coloring problem of finite simple undirected graphs. Before describing the concept of list edge coloring in detail, we have to revisit fundamental conception of normal edge coloring. A graph is -edge-colorable if all edges of can be colored by colors such that no two adjacent edges get the same colors. Denoted as the edge chromatic number of a graph , which is the smallest satisfying is -edge-colorable. For each edge of graph , if we can assign a list of colors to it, then is an edge assignment. Graph is edge--colorable if has a proper edge-coloring such that for each edge of , and is an edge--coloring. Graph is edge--choosable if for every satisfying for each edge . Denote as the list-edge chromatic number of which is the smallest in such that is edge--choosable.
We will study the list edge colorings of planar graphs. Planar graph is a kind of graph broadly studied in graph coloring theory. The so-called plane graph is actually a special drawing method of planar graph, which can be embedded in the plane satisfying no two edges intersect geometrically except at a vertex to which they are both incident. Let us introduce some definitions and symbols needed. Given a plane graph , we use , and to indicate its vertex set, edge set, face set, maximum degree, and minimum degree, respectively. For a vertex , let or be the set of edges which are incident with . We use or to indicate the degree of in , which is the number of edges in . We use or to indicate the set of the vertices which are adjacent to in . Denoted -vertex, -vertex, or -vertex as a vertex of degree , at most or at least , respectively. A (or )-neighbor of is a (or )-vertex which is adjacent to a vertex . A -cycle is a cycle of length . Two cycles are adjacent, that is, the two cycles share at least a common edge. A 2-alternating cycle is an even cycle in which the 2-vertices appear alternately. For , we use to indicate the degree of a face, which is the number of edges incident with where each cut edge is counted twice. Denote -, -face as a face of degree , at least . For a -face of , we called it -face if the vertices incident with it are of degrees , respectively. We use to indicate the number of -faces which are incident with , the number of -vertices which are incident with , and the number of -vertices which are incident with .
Conjecture 1. For every multigraph , .
So far, Conjecture 1 has only been proved to be true for a few graphs, including bipartite multigraphs , complete graphs of odd order , multicircuits , graphs embedded in a surface of nonnegative characteristic and , and outer planar graphs . For planar graphs, the readers can see [8–12].
Wang and Wu  proved that is edge--choosable for a planar graph without 6-cycles with three chords. Now, we will study the planar graphs without adjacent 7-cycles and obtain:
Theorem 1. Suppose thatis a planar graph which contains no adjacent 7-cycles. Then, is edge--choosable, where .
From Theorem 1, we can obtain the following corollary.
Corollary 1. Suppose thatis a planar graph which contains no adjacent 7-cycles, andfor.
2. The Proofs of Theorem
Proof. Let be a minimal graph satisfying the number of as little as possible; then, the graph has the following properties.
Lemma 1. (see ). Let be a planar graph, by the minimality hypothesis of graph, and we have(1)and is a connected graph(2) does not contain edgesatisfyingand(3) does not contain 2-alternating cycleDenote as the induced subgraph of by all 2-vertices of , where . By Lemma 1(2) and (3), contains no odd cycle and even cycle. Therefore, must be a forest. Thereby, there must be a matching in and all 2-vertices in are saturated. If and , then is named the 2-master of and is the dependent of . Obviously, each 2-vertex has a 2-master and each -vertex may be the 2-master of no more than one 2-vertex.
Lemma 3. Suppose that is a planar graph which contains no adjacent 7-cycles and . Then,(1)If and one of its edges is not incident with any 3-face (as in Figure 1,), then (2)If , then
Proof. Let , where all the neighbors of are in an anticlockwise order. We use to indicate the face which is incident with , , and .(1)Now, we prove (1–10) in Figure 1. Suppose that and are 3-faces. If is a 4-face or 5-face or 6-face, then there will be adjacent 7-cycles in . It must be and . The proof process of (1–11) and (1–13) is similar, so we will not repeat it.(2)Its proof process is similar to (1), so we will not repeat it.Similarly, we can get the following two lemmas.
Lemma 4. Suppose that is a planar graph which contains no adjacent 7-cycles and . Then,(1)If and four of its edges are not on any 3-face (as shown in Figure 2), then (2)If (as shown in figure ), then (3)If (as shown in figure ), then (4)If (as shown in figure ), then
Lemma 5. Suppose that is a planar graph which contains no adjacent 7-cycles and . Then,(1)If and one of its edges is not on any 3-face (as shown in Figure 3), then (2)If (as in figure ), then Now, we complete the proof by Euler’s formula. In , the authors proved for every planar graph with , so we only assume that in our proof. Suppose that is already embedded in the plane. We can obtainby practical Euler’s formula .
Firstly, denote as the original charge. For each , let . Therefore, . Secondly, we formulate some rules to redistribute the original charge and each will get a new charge . Note that the rules we formulated only move between the vertices and faces of the graph and have no effect on the total charge. Thirdly, we will show that (). If we can do, then we will obtain an apparent contradiction (). The proof of Theorem 1 is completed.
The discharging rules are formulated as in R1–R5. We use to indicate the charge from to : R1. Let , be a 3-master vertex and be 2-master vertex of . Firstly, . Secondly, if is on a face with , then and , otherwise . R2. Let and and be two 3-masters of , then and . R3. Every 5-vertex receives from each of its -neighbors. R4. Let be a 3-face with . R4.1 If or 4, then and . R4.2 If , then for . R5. Let and . Each of 2-vertices on receives from and other vertices remaining on receives from .Now, let us start to test and verify is greater than or equal to 0 for all vertices and faces. It is easy to verify faces, so let us verify the new charge of every face firstly. Obviously, . If , then by R4; if , then ; otherwise , by R5 .
Let us verify the new charge for every vertex. If for , then by R1, by R2, and , respectively. If , then must be adjacent to -vertices by Lemma 1(2). So, by R3 and R4, . If , then must be adjacent to -vertices by Lemma 1(2). So, by R4 . If , then , and by Lemma 1(2), must be adjacent to -vertices. So, by R3 and R4, .
Suppose that be a 8-vertex. Then, and must be adjacent to -vertices by Lemma 1(2). If , then by R3-R4 . Otherwise, . Since all the neighbors of 5-vertex must be -vertices by Lemma 1(2), so every 3-face which is incident with 8-vertex has no more than a 5-vertex, that is, . So, by R3-R4, .
Suppose that be a 9-vertex. Then, and must be adjacent to -vertices by Lemma 1(2), and it may be the 3-master of no more than two 3-vertices. In the following, we will test it from three cases. Firstly, it is not the 3-master of any 3-vertex, and then, by R3 and R4.
Secondly, is the 3-master of only one 3-vertex. If , then by R3, R4, and R5. Suppose (as shown in Figure 1, (1–10)–(1–15)). Since all the neighbors of 5-vertex should be -vertices by Lemma 1(2), so every 3-face incident with 9-vertex has no more than one 5-vertex, that is, . So, by R3–R5. Suppose (as shown in figure, (1–16)–(1–17)). If has a 5-neighbor , then the 3-face which is incident with and must be a -face and by R4 this 3-face receives from . So, by R3, R4, and R5.
Thirdly, is the 3-master of two 3-vertices. If , then by R3, R4, and R5. Suppose (as shown in Figure 1, (1-1)–(1–9)). Then, by R3, R4, and R5. Suppose (as shown in Figure 1: (1–10)–(1–15)). In Figure 1, (1–12), (1–14), (1–15), by R3, R4, and R5. Now, we consider Figure 1: (1–10), (1–11), and (1–13). By Lemma 3, . By R5, the -face gives no less than to . So, by R3, R4, and R5. Suppose (as shown in Figure 1, (1–16), (1–17)), then, by Lemma 3, . By R5, each -face gives no less than to . So, by R3, R4, and R5.
Suppose that be a 10-vertex. Obviously, and it could be the 3-master of no more than two 3-vertices and the 2-master of no more than one 2-vertex. If , then by R3–R5, . Let us discuss it in four ways.
Firstly, (as shown in Figure 2, (2-1)–(2-15)). From the figure, (2-2)–(2-15), and obviously, by R3, R4, and R5. By Lemma 4, in figure (2-1). So, by R5, -face gives no less than to . Therefore, by R3–R5 .
Suppose that be an 11-vertex. It follows that , and it could be the 3-master of no more than two 3-vertices and the 2-master of no more than one 2-vertex. If , then, by R3–R5, . Let us discuss it in four ways.
Firstly, (as shown in Figure 3, (3-1)–(3-25)). Obviously, by R3–R5, .
Fourthly, (as shown in the figure, (3-47)), then, by Lemma 5, . Hence, by R3–R5.
No data were used to support this study.
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.
This work was supported by the National Natural Science Foundation of China (NSFC) under Grant 12001332.
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