Total colorings of planar graphs with small maximum degree


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1 Total colorings of planar graphs with small maximum degree Bing Wang 1,, JianLiang Wu, SiFeng Tian 1 Department of Mathematics, Zaozhuang University, Shandong, 77160, China School of Mathematics, Shandong University, Jinan, 50100, P.R. China Zaozhuang No. middle school, Shandong, 77100, P.R. China Abstract Let G be a planar graph of maximum degree Δ and girth g, and there is an integer t(> g) such that G has no cycles of length from g + 1 to t. Then the total chromatic number of G is Δ + 1 if (Δ, g, t) {(5, 4, 6), (4, 4, 17)}; or Δ = and (g, t) {(5, 1), (6, 11), (7, 11), (8, 10), (9, 10)}, where each vertex is incident with at most one gcycle. Keywords: Total coloring; Planar graph; Cycle; Girth MSC: 05C15 1 Introduction All graphs considered in this paper are simple, finite and undirected, and we follow [6] for the terminologies and notations not defined here. Let G be a graph. We use V (G), E(G), Δ(G) and δ(g) (or simply V, E, Δ and δ) to denote the vertex set, the edge set, the maximum degree and the minimum degree of G, respectively. For a vertex v V, let N(v) denote the set of vertices adjacent to v, and let d(v) = N(v) denote the degree of v. A kvertex, a k + vertex or a k vertex is a vertex of degree k, at least k or at most k respectively. A kcycle is a cycle of length k, and a cycle is usually called a triangle. A totalkcoloring of a graph G is a coloring of V E using k colors such that no two adjacent or incident elements receive the same color. The total chromatic number χ (G) of G is the smallest integer k such that G has a totalkcoloring. Clearly, χ (G) Δ + 1. This work is supported by a research grant NSFC ((109711) of China. Corresponding author. addresses: 1
2 Behzad [1] posed independently the following famous conjecture, which is known as the Total Coloring Conjecture (TCC). Conjecture.For any graph G, Δ + 1 χ (G) Δ +. This conjecture was confirmed for a general graph with Δ 5. But for planar graph, the only open case is Δ = 6 (see [9],[]). Interestingly, planar graphs with high maximum degree allow a stronger assertion,that is, every planar graph with high maximum degree Δ is (Δ + 1)totallycolorable. This result was first established in [] for Δ 14, which was extended to Δ 9 (see [10]). For 4 Δ 8, it is not known whether that the assertion still holds true. But there are many related results by adding girth restrictions, see [7, 8, 11, 1, 14, 15, 16]. We present our new results in this paper. Theorem 1. Let G be a planar graph of maximum degree Δ and girth g, and there is an integer t(> g) such that G has no cycles of length from g + 1 to t. Then the total chromatic number of G is Δ + 1 if (a) (Δ, g, t) = (5, 4, 6) or (b)(δ, g, t) = (4, 4, 17). Borodin et al. [5] obtained that if a planar graph G of maximum degree three contains no cycles of length from to 9, then χ (G) = Δ+1. In the following, we make further efforts on the totalcolorability of planar graph on the condition that G contains some kcycle,where k (5,, 9). We get the following result. Theorem. Let G be a planar graph of maximum degree and girth g, each vertex is incident with at most one gcycle and there is an integer t(> g) such that G has no cycles of length from g + 1 to t. Then χ (G) = Δ + 1 if one of the following conditions holds. (a) g = 5 and t 1, (b) g = 6 and t 11, (c) g = 7 and t 11; (d) g = 8 and t 10, (e) g = 9 and t 10. We will introduce some more notations and definitions here for convenience. Let G = (V, E, F ) be a planar graph, where F is the face set of G. The degree of a face f, denoted by, is the number of edges incident with it, where each cutedge is counted twice. A kface or a k + face is a face of degree k or at least k, respectively. Let n k (v) be the number of kvertices adjacent to v and n k (f) be the number of kvertices incident with f. Proof of Theorem 1 Let G be a minimal counterexample to Theorem 1 in terms of the number of vertices and edges. Then every proper subgraph of G is (Δ + 1)totallycolorable. Firstly, we investi
3 gate some structural properties of G,which will be used to derive the desired contradiction completing our proof. Lemma. G is connected and hence, it has no vertices of degree 1 and the boundary b(f) of each face f in G is exactly a cycle (i.e. b(f) cannot pass through a vertex v more than once). Lemma 4. [4] G contains no edge uv with min{d(u), d(v)} Δ and d(u) + d(v) Δ + 1. Lemma 5. [] The subgraph of G induced by all edges joining vertices to Δ vertices is a forest. Lemma 6. [5] If Δ 5, then no vertex is adjacent two vertices. Let G be the subgraph induced by all edges incident with vertices of G. Then G is a forest by Lemma 5. We root it at a 5vertex. In this case, every vertex has exactly one parent and exactly one child, which are 5vertices. Since G is a planar graph, by Euler s formula, we have v V (d(v) 4) + f F ( 4) = 8 < 0. Now we define the initial charge function ch(x) of x V F to be ch(v) = d(v) 4 if v V and ch(f) = 4 if f F. It follows that x V F ch(x) < 0. Note that any discharging procedure preserves the total charge of G. If we can define suitable discharging rules to change the initial charge function ch to the final charge function ch on V F, such that ch (x) 0 for all x V F, then we get an obvious contradiction. Now we design appropriate discharging rules and redistribute weights accordingly. For a vertex v, we define f k (v) or f + k (v) to be the number of kfaces or k+ faces incident with v, respectively. To prove (a), our discharging rules are defined as follows. R11. Each vertex receives from its child. R. Each vertex v receives 1 f + 7 (v) from each of its incident 7+ faces. R1. Each 5 + vertex receives 1 from each of its incident 7+ faces. Next,we will check ch (x) 0 for all x V F. Let f F (G). If = 4, then ch (f) = ch(f) = 0. Suppose = 7. Then n (f) 4 by Lemma 6. Moreover, every 7 + face sends at most 1 to its incident vertices by R and 1 to its incident 5vertices by R1. So we
4 have ch (f) ch(f) = 0. Suppose 8. Then n (v) by Lemma 4 and Lemma 6. Thus, ch (f) ch(f) ( 1 ) ( ) by R and R1. Let v V (G). If d(v) =, then ch (v) = ch(v)+ = 0. If d(v) =, then f + 7 (v) and it follows from R that ch (v) = ch(v)+f 7 + (v) 1 = 0. If d(v) = 4, then f + 7 (v) ch (v) = ch(v) = 0. If d(v) = 5, then f 7 + (v). Moreover, it may be the child of at most one vertex. Thus ch (v) ch(v) + 1 = 0 by R1. Suppose d(v) 6. Then v is incident with at most d(v) 4faces and it may be the the parent of at most one vertex. So ch (v) ch(v) + (d(v) d(v) ) 1 = 7d(v) 6 > 0. 6 Note that (a) implies that (b) is true if Δ 5. So it suffice to prove (b) by assuming Δ = 4. Since G is a planar graph, by Euler s formula, we have ( 6) = < 0. v V (d(v) 6) + f F Now we define the initial charge function ch(x) of x V F to be ch(v) = d(v) 6 if v V and ch(f) = 6 if f F. It follows that x V F ch(x) < 0. To prove (b), we construct the new charge ch (x) on G as follows. R1 Each ( 18)face gives 1 6 to its incident vertices. R Each vertex gets from its child and 1 from its parent. R Let f be a 4face. If f is incident with a vertex, then it gets from each of its incident + vertices. If f is incident with no vertices, then it gets 1 from each of its incident vertices. The rest of this paper is devoted to checking ch (x) 0 for all x V F. Let f F (G). If = 4, then ch (f) = ch(f) + max{, 1 4} = 0. If 18, then ch (f) = ch(f) r (1 6 ) = 0 by R1. r Let v V (G). If d(v) =, then ch (v) = ch(v) = 0 by R. If d(v) =, then f 18(v) + and f 4 (v) 1,and it follows from R1 and R that ch (v) = ch(v)+ > 0. Suppose that d(v) = 4. Then ch(v) = 4 6 =. If n (v) 1, then v sends at most n (v)+ to all its adjacent vertices by R. If n (v) 4, then f 4 (v) 1 by Lemma 5, and it follows that ch (v) ch(v) n (v)+ + = 14 n (v) > 0 by R1 and R. If 6 1 n (v), then f 4 (v), and it follows that ch (v) ch(v) n (v)+ + = n (v) 0. If n (v) = 0, we have f 4 (v). Moreover, each 4face incident with v contains no vertices. By R, we have ch (v) ch(v) + 1 > 0. Now we complete the proof of Theorem 1. 4
5 Proof of Theorem A (k) vertex is a vertex adjacent to exactly k vertices. Let G be a minimal counterexample to Theorem in terms of the number of vertices and edges. By minimality of G, it has the following result. Lemma 7. [5] (a) no vertex is adjacent to two vertices; (b) no vertex is adjacent to a vertex and a ()vertex ; (c) no vertex is adjacent to three vertices. Let G be the bipartite subgraph of G comprising V and all edges of G that join a  vertex to a vertex. Then G has no isolated vertices by Lemma 7(a), and the maximum degree is at most by Lemma 7(c), and any component of G is a path with more than one edges must end in two vertices by Lemma 7(b). It follows that n n. So we can find a matching M in G saturating all vertices. If uv M and d(u) =,v is called the master of u. Each vertex has one master and each vertex of degree Δ can be the master of at most one vertex. Since G is a planar graph, by Euler s formula, we have v V (d(v) 6) + f F ( 6) = < 0. Now we define the initial charge function ch(x) of x V F to be ch(v) = d(v) 6 if v V and ch(f) = 6 if f F. It follows that x V F ch(x) < 0. Note that any discharging procedure preserves the total charge of G. If we can define suitable discharging rules to change the initial charge function ch to the final charge function ch on V F, such that ch (x) 0 for all x V F, then we get an obvious contradiction. Now we design appropriate discharging rules and redistribute weights accordingly. R1 Each ( 5)face gives 6 to its incident vertices. R Each vertex receives 6 from its master. t+1 g Let ch (x) be the new charge obtained by the above rules for all x V F. If f F (G), then ch (f) = ch(f) 6 = 0 by R1. Let v V (G). Suppose d(v) =. Then v can be the master of at most one vertex, and v sends at most 6 to vertex t+1 g by R. In addition, If v is incident with a gface, then the other faces incident with v are two (t + 1) + faces, for G has no cycles of length from g + 1 to t. Thus, v receives ( 6 g ) from its incident gface and ( 6 ) from each of its incident (t + t+1 1)+ face by R1. So ch (v) ch(v) + ( 6 ) + ( 6 ) ( 6 ) = 0 for all g and t. Otherwise, v t+1 g t+1 g 5
6 is incident with three (t + 1) + faces, then ch (v) ch(v) + ( 6 ) ( 6) = t+1 t+1 g 6 6 > 0, for t + 1 > g. Suppose d(v) =. Then v receives at most 6 g t+1 t+1 g from its master by R1. If v is incident with a gface, since G has no cycles of length from g + 1 to t, then the other face incident with v is a (t + 1) + face, and it follows that ch (v) ch(v)+( 6 )+( 6 )+( 6 ) = 0 for all g and t. Otherwise, v is incident t+1 g t+1 g with two (t + 1) + faces, then ch (v) ch(v) + ( 6 ) + ( 6) = 4 6 > 0. t+1 t+1 g t+1 g From the above, we can see that ch (f) = ch(f) 6 Suppose d(v) =.So ch (v) ch(v) + ( 6 ) + ( 6 ) ( 6 t+1 g t+1 g When v is incident with three (t+1) + faces, then ch (v) ch(v)+( g = 0 for all f F (G). ) = 0 for all g and t. ) ( 6) = t+1 t+1 g > 0, for t+1 > g. Suppose d(v) =. If v is incident with a gface and a t+1 (t+1)+ face, then ch (v) ch(v) + ( 6 t+1 ) + ( 6 g ) + ( t+1 6 g ) = 0 for all g and t.when v is incident with two (t + 1) + faces, then ch (v) ch(v) + ( 6 ) + ( 6) = 4 6. So t+1 t+1 g t+1 g when g = 5, then t 1;when g = 6, then t 11;when g = 7, then t 11; when g = 8, then t 10; when g = 9, then t 10, and it follows that ch (v) 0. Our proof of Theorem is now complete. References [1] M. Behzad, Graphs and their chromatic numbers, Ph. D. Thesis, Michigan State University, [] O. V. Borodin, On the total coloring of planar graphs, J. Reine Angew. Math. 94 (1989) [] O.V. Borodin, A.V. Kostochka and D.R. Woodall, List edge and list total colourings of multigraphs, J. Combin. Theory Ser. B 71 (1997) [4] O. V. Borodin, A.V. Kostochka and D.R. Woodall, Total colorings of planar graphs with large maximum degree, J. Graph Theory 6 (1997) [5] O. V. Borodin, A.V. Kostochka and D.R. Woodall, Total colourings of planar graphs with large girth, Europ. J. Combin. 19 (1998) [6] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, MacMillan, London,
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