Graphs with Bounded Maximum Average Degree and Their Neighbor Sum Distinguishing Total-Choice Numbers

Abstract

Let be a graph and be a -total coloring. Let denote the sum of color on a vertex and colors assigned to edges incident to . If whenever , then is called a neighbor sum distinguishing total coloring. The smallest integer such that has a neighbor sum distinguishing -total coloring is denoted by . In 2014, Dong and Wang obtained the results about depending on the value of maximum average degree. A -assignment of is a list assignment of integers to vertices and edges with for each vertex and for each edge . A total--coloring is a total coloring of such that whenever and whenever . We state that has a neighbor sum distinguishing total--coloring if has a total--coloring such that for all . The smallest integer such that has a neighbor sum distinguishing total--coloring for every -assignment is denoted by . In this paper, we strengthen results by Dong and Wang by giving analogous results for .

1. Introduction

Let be a simple, finite, and undirected graph. We use , , and to denote the vertex set, edge set, and maximum degree of a graph , respectively. A vertex is called a -vertex if . The length of a shortest cycle in is called the girth of a graph , denoted by . The maximum average degree of is defined by . The well-known observation for a planar graph is mad. Let be a -total coloring. We denote the sum (set, resp.) of colors assigned to edges incident to and the color on the vertex by (, resp.); that is, and . The total coloring of is a neighbor sum distinguishing (neighbor distinguishing, resp.) total coloring if (, resp.) for each edge . The smallest integer such that has a neighbor sum distinguishing (neighbor distinguishing, resp.) total coloring is called the neighbor sum distinguishing total chromatic number (neighbor distinguishing total chromatic number, resp.), denoted by (, resp.). In 2005, a neighbor distinguishing total coloring of graphs was introduced by Zhang et al. [1]. They obtained for many basic graphs and brought forward the following conjecture.

Conjecture 1 (see [1]). If is a graph with order at least two, then .

Conjecture 1 has been confirmed for subcubic graphs, -minor free graphs, and planar graphs with large maximum degree [2–4].

In 2015, Pilśniak and Woźniak [5] obtained for cycles, cubic graphs, bipartite graphs, and complete graphs. Moreover, they posed the following conjecture.

Conjecture 2 (see [5]). If is a graph with at least two vertices, then .

Li et al. verified this conjecture for -minor free graphs [6] and planar graphs with the large maximum degree [7]. Wang et al. [8] confirmed this conjecture by using the famous Combinatorial Nullstellensatz that holds for any triangle free planar graph with maximum degree of at least 7. Several results about for planar graphs can be found in [9–11].

In 2014, Dong and Wang [12] proved the following results.

Theorem 3. If is a graph with , then .

Corollary 4. If is a graph with and , then .

Corollary 5. Let be a planar graph. If and , then ; and if and only if has two adjacent vertices of maximum degree.

The concept of list coloring was introduced by Vizing [13] and by Erdös et al. [14]. A -assignment of is a list assignment of integers to vertices and edges with for each vertex and for each edge . A total--coloring is a total coloring of such that whenever and whenever . We state that has a neighbor sum distinguishing total--coloring if has a total--coloring such that for all . The smallest integer such that has a neighbor sum distinguishing total--coloring for every -assignment , denoted by , is called the neighbor sum distinguishing total-choice number.

Qu et al. [15] proved that for any planar graph with . Yao et al. [16] studied of -degenerate graphs. Later, Wang et al. [17] confirmed Conjecture 2 true for planar graphs without 4-cycles. For , we let denote a list restricted to any proper subgraph of . In this paper, we strengthen Theorem 3 by giving analogous results for .

2. Main Results

The following lemma is obvious, so we omit the proof.

Lemma 6. Let and . Then .

Proof. We proceed by induction on .
If , then ; then Lemma 6 holds. Assume that . Suppose that Lemma 6 holds for . Let . Without loss of generality, let . Let be such that and for . By induction hypothesis, we have . Thus , where for and for . So . Let with for and for and . Thus and . Therefore, we obtain .

Lemma 7 (see [12]). Let be two sets and let . If and , then .

Theorem 8. If is a graph with , then , where .

Proof. The proof is proceeded by contradiction. Assume that is a minimum counterexample. Let for each vertex and for each edge in . For any proper subgraph of , we always assume that there is a neighbor sum distinguishing total--coloring of by minimality of . For convenience, we use a total--coloring of to denote a neighbor sum distinguishing total--coloring of and we use for and for .

Let be the graph obtained by removing all leaves of . Then is a connected graph with mad. The properties of the graph are collected in the following claims.

Claim 1. Each vertex in has degree of at least 2.

Proof. Suppose to the contrary that contains a vertex with . If , then is the star and ; then we obtain a total--coloring of , a contradiction to the choice of . Assume that . Let and be the neighbors of where and . Let . First, we uncolor where . Then we color with a color in . Next, we color with a color in for ; then we obtain a total--coloring of , a contradiction to the choice of .

Claim 2. If , then .

Proof. Suppose to the contrary that . Let be the neighbors of and be all neighbors of which are leaves in for .
Case 1 (). Let and . We color with a color in and color with a color in . Thus we obtain a total--coloring of , which is a contradiction to the choice of .
Case 2 (). Let , where . Let , where . Then , where . By Lemma 6, we have at least color sets available for the edge set to guarantee for . Since at most two color sets may cause or , we have at least one color set available for the edge set . Finally, we color with the color in for ; then we obtain a total--coloring of , which is a contradiction to the choice of .

Claim 3. A 2-vertex is not adjacent to a 3-vertex.

Proof. Suppose to the contrary that is adjacent to a 3-vertex in . Let , be the neighbors of and be the other neighbor of .
Case 1 (). Let . First, we uncolor . Next, we color with a color in . Later, we color with a color in ; then we obtain a total--coloring of , which is a contradiction to the choice of .
Case 2 (). Let be the other neighbors of such that for all . Let . First, we uncolor all vertices and , . Consider and . We can see that and . By Lemma 7, we can choose and such that and . Next, we color with a color in and color with a color in for ; then we obtain a total--coloring of , which is a contradiction to the choice of .

Claim 4. A 4-vertex is adjacent to at most two 2-vertices.

Proof. Suppose to the contrary that is adjacent to three 2-vertices and the other vertex . Let be the neighbor of for .
Case 1 (). Let and . First, we uncolor all vertices . Next, we color with a color in and color with a color in for . Thus we obtain a total--coloring of , which is a contradiction to the choice of .
Case 2 (). Let be the neighbors of such that for all . Let . First, we uncolor vertices and where , . Next, we choose . After that, we color with a color in for and color with a color in for . Thus we obtain a total--coloring of , which is a contradiction to the choice of .

Claim 5. A 5-vertex is adjacent to at most four 2-vertices.

Proof. Suppose to the contrary that is adjacent to five 2-vertices . Let be the other neighbors of (if they exist) such that for all and be the neighbor of for . Let and and . First, we uncolor vertices and . Next, we color with a color in . After that, we color with a color in . Finally, we color with a color in . Thus we obtain a total--coloring of , which is a contradiction to the choice of .

By Claim 1, we have .

Suppose that . By Claims 1 and 2, is a cycle. One can obtain that , a contradiction to the choice of .

Suppose that . By Claim 3, is a 3-regular graph. Thus we have mad, which is a contradiction.

Suppose that . We complete the proof by using the discharging method. Define an initial charge function for every . Next, rearrange the weights according to the designed rule. When the discharging is finished, we have a new charge . However, the sum of all charges is kept fixed. Finally, we want to show that for all . This leads to the following contradiction:

Let . Assume that and . Then vertex gives charge to .

Consider a vertex . By Claim 1, we have .

If , then is adjacent to at least two 4-vertices by Claim 3. Hence .

If , then .

If , then is adjacent to at most two 2-vertices by Claim 4. Hence .

If , then is adjacent to at most four 2-vertices by Claim 5. Hence .

If , then .

From the above discussion, we have , which is a contradiction. This completes the proof of Theorem 8.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The first author is supported by University of Phayao, Thailand. In addition, the authors would like to thank Dr. Keaitsuda Nakprasit for her helpful comments.