Neighbor Sum Distinguishing Total Colorings of Triangle Free Planar Graphs

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1 Acta Mathematica Sinica, English Series Feb., 2015, Vol. 31, No. 2, pp Published online: January 15, 2015 DOI: /s y Acta Mathematica Sinica, English Series Springer-Verlag Berlin Heidelberg & The Editorial Office of AMS 2015 Neighbor Sum Distinguishing Total Colorings of Triangle Free Planar Graphs Ji Hui WANG Qiao Ling MA Xue HAN School of Mathematical Sciences, University of Ji nan, Ji nan , P. R. China wangjh@ujn.edu.cn ss maql@ujn.edu.cn ss hanx@ujn.edu.cn Abstract A total k-coloring c of a graph G is a proper total coloring c of G using colors of the set [k] ={1, 2,...,k}. Let f(u) denote the sum of the color on a vertex u and colors on all the edges incident to u. Ak-neighbor sum distinguishing total coloring of G is a total k-coloring of G such that for each edge uv E(G), f(u) f(v). By χ nsd(g), we denote the smallest value k in such a coloring of G. Pilśniak and Woźniak conjectured that χ nsd(g) Δ(G) + 3 for any simple graph with maximum degree Δ(G). In this paper, by using the famous Combinatorial Nullstellensatz, we prove that the conjecture holds for any triangle free planar graph with maximum degree at least 7. Keywords Neighbor sum distinguishing total coloring, combinatorial Nullstellensatz, triangle free planar graph MR(2010) Subject Classification 05C15 1 Introduction The terminology and notation used but undefined in this paper can be found in [2]. Graphs considered in this paper are finite, simple and undirected. Let G =(V,E) be a graph. We use V (G), E(G), Δ(G) andδ(g) to denote the vertex set, edge set, maximum degree and minimum degree of G, respectively. Let d G (v) orsimplyd(v) denote the degree of a vertex v in G. Given a graph G =(V,E) and a positive integer k, a total k-coloring of G is a mapping φ : V E {1, 2,...,k} such that (a) φ(u) φ(v) for every pair u, v of adjacent vertices; (b) φ(v) φ(e) for every vertex v and every edge e incident to v; (c) φ(e) φ(e ) for every pair e, e of adjacent edges. Given a total k-coloring φ of G, letc φ (v) denote the set of colors of the edges incident to v and the color of v. A total k-coloring is called adjacent vertex distinguishing if for each edge uv, C φ (u) is different from C φ (v). The smallest k admitting such coloring is called the adjacent vertex distinguishing total chromatic number, denoted by χ nd (G). Zhang [23] put forward the following conjecture. Conjecture 1.1 ([23]) For any graph G with at least two vertices, χ nd (G) Δ(G)+3. Received February 21, 2014, accepted July 2, 2014 Supported by National Natural Science Foundation of China (Grant No ) and the Scientific Research Foundation of University of Ji nan (Grant No. XKY1120)

2 Neighbor Sum Distinguishing Total Colorings of Triangle Free Planar Graphs 217 Conjecture 1.1 has been confirmed for subcubic graphs, K 4 -minor free graphs and planar graphs with large maximum degree, see [3, 7, 18, 19]. Recently, colorings and labellings related to sums of the colors have received much attention. The family of such problems includes, e.g., vertex-coloring k-edge-weightings [20], total weight choosability [16, 21], magic and antimagic labellings [8, 22] and the irregulaity strength [13, 14]. Among them there are the Conjecture due to Karoński et al. [9] and 1-2 Conjecture due to Przyby lo and Woźniak [15]. For more information, see the survey paper [17]. In a total k-coloring of G, letf(v) denote the total sum of colors of the edges incident to v and of v itself. If for each edge uv, f(u) f(v), we call such total k-coloring a k-neighbor sum distinguishing total coloring. The smallest number k admitting such coloring is called the neighbor sum distinguishing total chromatic number, denoted by χ nsd (G). For neighbor sum distinguishing total colorings, we have the following conjecture due to Pilśniak and Woźniak [12]. Conjecture 1.2 ([12]) For any graph G with at least two vertices, χ nsd (G) Δ(G)+3. Clearly, Conjecture 1.2 implies Conjecture 1.1 since it is easy to check that χ nd (G) χ nsd (G). Pilśniak and Woźniak [12] proved that Conjecture 1.2 holds for complete graphs, cycles, bipartite graphs and subcubic graphs. Dong and Wang [6] showed that Conjecture 1.2 holds for sparse graphs. Li et al. [11] confirmed this conjecture for K 4 -minor free graphs. By using the famous Combinatorial Nullstellensatz, Ding et al. [4] proved that χ nsd (G) 2Δ(G)+col(G) 1, where col(g) is the coloring number of G. Later Ding et al. [5] improved this bound to Δ(G)+2col(G) 2. Moreover they proved this assertion in its list version. In [10], Li et al. proved that Conjecture 1.2 holds for planar graphs with maximum degree at least 13. In this paper, we prove that Conjecture 1.2 holds for triangle free planar graphs with maximum degree at least 7. We have the following result. Theorem 1.3 Let G be a triangle free planar graphs with maximum degree Δ(G) 7. Then χ nsd (G) Δ(G)+3. 2 Preliminaries To prove our main result, we need to introduce some notations. A vertex v is called an l-vertex if d(v) =l, similarly, we call it an l + -vertex or an l -vertex if d(v) l or d(v) l. Suppose that H is a subgraph of a given planar graph G. Forx V (H), let d H (x) denote the degree of x in H. WeuseNl H(x) todenotethesetofl-vertices adjacent to x in H and set dh l (x) = N l H(x). Similarly, we can define d H l (x) andd H + l (x). If there is no confusion in the context, we usually write d G l (x), dg l (x), d G + l (x) asd l (x), d l +(x), d l (x), respectively. In order to prove the main result, we need some lemmas. The famous Combinatorial Nullstellensatz [1] presented below will be used frequently in our proofs. Lemma 2.1 ([1]) Let F be an arbitrary field, and let P = P (x 1,...,x n ) be a polynomial in F[x 1,...,x n ].Supposethedegreedeg(P ) of P equals n k i,whereeachk i is a non-negative k1 k integer, and suppose the coefficient of x 1 x n n in P is non-zero. Then if S 1,...,S n are subsets of F with S i >k i, there are s 1 S 1,...,s n S n so that P (s 1,...,s n ) 0. Lemma 2.2 ([10]) Suppose S is a set of integers and S = n. LetT = { m x i x i S, x i x j (i j)}, wherem n. Then T mn m 2 +1.

3 218 Wang J. H., et al. 3 Proof of Theorem Unavoidable Configuration Let k = Δ(G) + 3. For simplicity, we use k-nsd-total coloring to denote k-neighbor sum distinguishing total coloring. Suppose that φ is a k-nsd-total coloring of triangle free planar graph G with Δ(G) 7. Assume that v V (G) withd(v) 3. It is easy to see that v has at most 3 adjacent vertices and 3 incident edges, so v has at most 9 forbidden colors. Since k 10, we may erase the color of v and recolor it at the end. In other words, we will omit the coloring for all 3 -vertices in the following discussion. Our proof proceeds by reduction and absurdum. Assume that G is a counterexample to Theorem 1.3 such that V (G) + E(G) is as small as possible. Obviously, G is connected. In the following, we give some properties of G. Claim 1 There is no edge uv E(G) such that d(u) 4andd(v) 3. Proof It is enough to show that there is no 4-vertex adjacent to any 3 -vertex. Suppose to the contrary that G contains a vertex u of degree 4 which is adjacent to a 3 -vertex v. Let H = G uv. By the minimality of G, thereisak-nsd-total coloring φ of H. We shall now extend the coloring φ to G. Since d(v) 3, we can erase the color of v and recolor it at the end. Assume that N(u) ={v, u 1,u 2,u 3 }. Now, we erase the color of u, and color the edge uv and the vertex u with x 1 and x 2 (x 1,x 2 [k]), respectively. To guarantee that the coloring is a nsd-total coloring, we can color the edge uv with colors which are different from the colors of the edges incident with u and v. Let S 1 denote the feasible color set which uv canuse. Thenwehave S 1 Δ Similarly, we cannot use the colors of uu i and u i (i =1, 2, 3) to color the vertex u. Hence, let S 2 denote the feasible color set which u can be used. Then S 2 Δ Let w i denote the total sum of colors assigned to u i and these edges incident to u i in H for i =1, 2, 3, and let w 0 denote the contemporary sum at u (not counting the color of uv and u). In order to guarantee the sum obtained at u must be distinct from the other already fixed sums of the neighbors of u. It is then enough to find appropriate substitutions for the variables x 1 and x 2 such that the value of the following polynomial is non-zero: P (x 1,x 2 )=(x 1 x 2 )(x 1 + x 2 + w 0 w 1 )(x 1 + x 2 + w 0 w 2 )(x 1 + x 2 + w 0 w 3 ). Since deg(p ) = 4, by Lemma 2.1, it is enough to show that the coefficient of the monomial x 3 1 x 2 in P is non-zero. However, this coefficient is the same as the coefficient of the same monomial in Q(x 1,x 2 )=(x 1 x 2 )(x 1 + x 2 ) 3. Since = 12, then we know that the 4 Q x 13 x 2 coefficient of the monomial x 3 1 x 2 in P is 2. Since S 1 5 > 3, S 2 4 > 1, by Combinatorial Nullstellensatz, we can find s 1 S 1 and s 2 S 2 such that P (s 1,s 2 ) 0. Thus we know that G has a k-nsd-total coloring, a contradiction. Claim 2 If u is a 5-vertex of G, thend 1 (u) =d 2 (u) =0andd 3 (u) 1. Proof First, we prove that d 1 (u) =d 2 (u) =0. Suppose to the contrary that G contains a vertex u of degree 5 which is adjacent to a 2 - vertex v. LetH = G uv. ThenH has a k-nsd-total coloring. Let N(u) ={v, u 1,u 2,u 3,u 4 }. Similarly to the proof of Claim 1, we erase the color of u, and use colors x 1,x 2 to color the

4 Neighbor Sum Distinguishing Total Colorings of Triangle Free Planar Graphs 219 edge uv and the vertex u, respectively. Let w i denote the total sum of colors assigned to u i and those edges incident to u i in H for i =1, 2, 3, 4, and let w 0 denote the contemporary sum at u (not counting the color of uv and u). Then we let P (x 1,x 2 )=(x 1 x 2 )(x 1 +x 2 +w 0 w 1 )(x 1 +x 2 +w 0 w 2 )(x 1 +x 2 +w 0 w 3 )(x 1 +x 2 +w 0 w 4 ). Obviously, deg(p )=5. LetQ(x 1,x 2 )=(x 1 x 2 )(x 1 + x 2 ) 4.Then 5 Q x 14 x 2 = 72, and we know that the coefficient of the monomial x 1 4 x 2 in Q is 3. This coefficient is the same as the coefficient of the same monomial in P (x 1,x 2 ). Let S i denote the feasible color set which x i can be used for i =1, 2. Since d(v) 2, then we know that S 1 Δ > 4, S 2 Δ > 1. By Lemma 2.1, we can find s 1 S 1 and s 2 S 2 such that P (s 1,s 2 ) 0. Thus we know that G has a k-nsd-total coloring, a contradiction. Now, we prove that d 3 (u) 1. Suppose to the contrary that G contains a vertex u of degree 5 which is adjacent to two 3-vertexs v 1,v 2. Let N(u) ={v 1,v 2,u 1,u 2,u 3 } and let H = G uv 1 uv 2. Then H has a k-nsd-total coloring. Similarly, we erase the color of u, and use colors x 1,x 2,x 3 to color the edges uv 1,uv 2 and vertex u, respectively. Let w i denote the total sum of colors assigned to u i and the edges incident to u i in H for i =1, 2, 3, and let w 0 denote the contemporary sum at u (not counting the colors of uv 1,uv 2 and u). Let P (x 1,x 2,x 3 )= 3 ( 3 ) (x i x j ) x j + w 0 w i and 1 i<j 3 Q(x 1,x 2,x 3 )= 1 i<j 3 j=1 ( 3 ) 3 (x i x j ) x i. Since 6 Q x 3 1 x 2 2 x 3 = 12, then we know that the coefficient of the monomial x 3 1 x 2 2 x 3 in P (x 1,x 2,x 3 )is1. LetS i denote the feasible color set which x i can be used for i =1, 2, 3. Since d(v 1 )=d(v 2 )=3,then S 1 = S 2 Δ+3 5 5, S 3 Δ By Lemma 2.1, we can find s 1 S 1, s 2 S 2 and s 3 S 3 such that P (s 1,s 2,s 3 ) 0. ThenG has a k-nsd-total coloring, a contradiction. Claim 3 Let u be a 6-vertex of G. Thend 2 (u) 1andifd 2 (u) =1,thend 3 (u) =0. Proof In order to prove the claim, we may only to prove that u cannot be adjacent to a 2 -vertex and a 3 -vertex. Suppose to the contrary that G contains a vertex u of degree 6 which is adjacent to a 2 - vertex v 1 and a 3 -vertex v 2. Let N(u) ={v 1,v 2,u 1,u 2,u 3,u 4 } and let H = G uv 1 uv 2. Then H has a k-nsd-total coloring. Similarly, we erase the colors of u, v 1,v 2, and use colors x 1,x 2,x 3 to color the edges uv 1,uv 2 and vertex u, respectively. Let w i denote the total sum of colors assigned to u i and those edges incident to u i in H for i =1, 2, 3, 4, and let w 0 denote the contemporary sum at u (not counting the colors of uv 1,uv 2 and u). Let P (x 1,x 2,x 3 )= 4 ( 3 ) (x i x j ) x j + w 0 w i 1 i<j 3 j=1

5 220 Wang J. H., et al. and Q(x 1,x 2,x 3 )= 1 i<j 3 ( 3 ) 4 (x i x j ) x i. Since 7 Q x 4 1 x 2 2 x 3 = 144, then we know that the coefficient of the monomial x 4 1 x 2 2 x 3 in P (x 1,x 2,x 3 )is3. LetS i denote the feasible color set which x i can be used for i =1, 2, 3. Since d(v 1 ) 2, d(v 2 ) 3, then S 1 Δ > 4, S 2 Δ > 2, S 3 Δ > 1. By Lemma 2.1, we can find s 1 S 1, s 2 S 2 and s 3 S 3 such that P (s 1,s 2,s 3 ) 0. ThenG has a k-nsd-total coloring, a contradiction. Claim 4 Let u be a 7-vertex of G. Thend 2 (u) 2and (1) If d 2 (u) =1,thend 3 (u) 1. (2) If d 2 (u) =2,thend 3 (u) =0. Proof In order to prove the claim, we may only to prove that u cannot be adjacent to a 2 -vertex and two 3 -vertexs. Suppose to the contrary that G contains a vertex u of degree 7 which is adjacent to a 2 -vertex v 1 and two 3 -vertexs v 2,v 3. Let N(u) ={v 1,v 2,v 3,u 1,u 2,u 3,u 4 } and let H = G uv 1 uv 2 uv 3. Then H has a k-nsd-total coloring. Similarly, we erase the colors of u, v 1,v 2,v 3, and use colors x 1,x 2,x 3,x 4 to color the edges uv 1,uv 2,uv 3 and the vertex u, respectively. Let w i denote the total sum of colors assigned to u i and those edges incident to u i in H for i =1, 2, 3, 4, and let w 0 denote the contemporary sum at u (not counting the color of uv 1,uv 2,uv 3 and u). Let P (x 1,x 2,x 3,x 4 )= 4 ( 4 ) (x i x j ) x j + w 0 w i 1 i<j 4 j=1 and Q(x 1,x 2,x 3,x 4 )= 1 i<j 4 ( 4 ) 4 (x i x j ) x i. Since 10 Q x 4 1 x 3 2 x 2 3 x 4 = 288, then we know that the coefficient of the monomial x 4 1 x 3 2 x 2 3 x 4 in P (x 1,x 2,x 3,x 4 )is1.lets i denote the feasible color set which x i canbeusedfori =1, 2, 3, 4. Since d(v 1 ) 2, d(v 2 ) d(v 3 ) 3, then S 1 Δ > 4, S 2 Δ > 3, S 3 Δ > 2, S 4 Δ > 1. By Lemma 2.1, we can find s 1 S 1, s 2 S 2, s 3 S 3 and s 4 S 4 such that P (s 1,s 2,s 3,s 4 ) 0. ThenG has a k-nsd-total coloring, a contradiction. Claim 5 Let u be an l-vertex of G with l 8. If d 2 (u) 0,thend 3 (u) l 2. Proof Let t = d 3 (u), s = l 2. Suppose to the contrary that t>s,letu 1,u 2,...,u l be the neighbors of u with d(u 1 ) d(u 2 ) d(u l ), and d(u 1 ) 2. Without loss of generality, we may assume that d(u 1 ) = 2, and d(u 2 )=d(u 3 )= = d(u t )=3. LetH = G uu 1.ThenH has a total k-nsd-coloring φ. We erase all the colors of u i and uu i for 1 i t and denote this coloring by φ.letc = {1, 2,...,k}, C φ (v) denote the set of colors assigned to a vertex v and those edges incident to v in H and φ(uv) denote the color that assigned to a edge uv in H.

6 Neighbor Sum Distinguishing Total Colorings of Triangle Free Planar Graphs 221 First suppose that there exists an integer i, 2 i t, such that φ(u 1 v 1 ) C φ (u i ), without loss of generality, we may assume that φ(u 1 v 1 ) C φ (u 2 ). Since Δ d(u) = l, C\(C φ (u) {φ(u 1 v 1 )}) Δ+3 (l t +2) t +1. Thus we can choose t colors from C\(C φ (u) {φ(u 1 v 1 )}) (at least different t + 1 colors) to color uu 1,...,uu t. Since t>s, by Lemma 2.2, we can find at least t +1 different ways to color uu 1,...,uu t which result in at least t + 1 different sums of the vertex u. Hence, we can find at least one color set C C\(C φ (u) {φ(u 1 v 1 )}) to color uu 1,...,uu t, such that the sums of u does not conflict with its neighbors. In the following we will give a k-nsd-coloring φ of G, which will yield a contradiction. We color uu t,...,uu 1 one-by-one with colors in C as follows: Since d(u i )=3 for 2 i t, we color uu i with color a i in (C \{a t,...,a i+1 })\C φ (u i ). Since d(u 2 )=3, φ(u 1 v 1 ) / C, but φ(u 1 v 1 ) C φ (u 2 ), we color uu 2 with color a 2 in (C \{a t,...,a 3 })\C φ (u 2 ). Since d(u 1 )=2andφ(u 1 v 1 ) / C, we color uu 1 with color in C \{a t,...,a 2 }. Now we can assume that for every integer i, 2 i t, φ(u 1 v 1 ) / C φ (u i ). Without loss of generality, let a C φ (u 2 ). Since Δ d(u) =l, C\(C φ (u) {a}) Δ+3 (l +2 t) t +1. Thus we can choose t colors from at least t + 1 colors to color uu 1,...,uu t. Since t>s,by Lemma 2.2, we can find at least t +1 different ways to color uu 1,...,uu t,whichgett +1 different sums of vertex u. Hence, we can find one color set C C\(C φ (u) {a}) to color uu 1,...,uu t, such that f(u) does not conflict with its conflicting vertices. In the following, we will properly color uu 1,...,uu t with colors in C to get a total k-nsd-coloring φ of G, which is a contradiction. Case 1 φ(u 1 v 1 ) / C. Since d(u i )=3for3 i t, we color uu i with color a i in (C \{a t,...,a i+1 })\C φ (u i ). Since d(u 2 )=3,a / C and a C φ (u 2 ), we color uu 2 with color a 2 (C \{a t,...,a 3 })\C φ (u 2 ). Since d(u 1 )=2andφ(u 1 v 1 ) / C, we color uu 1 with color in C \{a t,...,a 2 }. We get a total k-nsd-coloring of G. Case 2 φ(u 1 v 1 ) C. We use φ(u 1 v 1 )=a t to color uu t.sinced(u i )=3for3 i t 1, we color uu i with color a i in (C \{a t,...,a i+1 })\C φ (u i ). Since d(u 2 )=3, a / C and a C φ (u 2 ), we choose one color a 2 in (C \{a t,...,a 3 })\C φ (u 2 ) to color uu 2.Sinced(u 1 )=2andφ(u 1 v 1 )=a t, we color uu 1 with color in C \{a t,...,a 2 }. Then we get a total k-nsd-coloring of G. 3.2 Discharging Process Let H be the graph obtained by removing all leaves of G. Then H is a connected planar graph with δ(h) 2. By Claims 1 5, we display the relation between d(v) andd H (v) in Table 1. d(v) d H (v) = d(v) Table 1 The relation between d(v) andd H (v) Claim 6 Let v be a vertex of H. Ifd H (v) =2ord H (v) = 3, then the neighbors of v must be 5 + -vertices in H.

7 222 Wang J. H., et al. Proof If there is a vertex u with d(u) =8andd H (u) = 4. By Claim 5, u must have 4 leaves in G and have no neighbor of 2-vertex and 3-vertex. If d(u) 8, it is trivial by Table 1 and Claim 1. Using Euler s formula V (H) E(H) + F (H) =2, we have (d H (v) 4) + (d H (f) 4) = 8. v V (H) f F (H) In order to complete the proof, we use the discharging method. First, we give an initial charge function w(v) = d H (v) 4 for every v V (H) andw(f) = d H (f) 4 for every f F (H). Next, we design a discharging rule and redistribute weights accordingly. Let w be the new charge after the discharging. We will show that w (x) 0 for all x V (H) F (H). This leads to the following obvious contradiction: 0 w (x) = w(x) = 8 < 0, x V (H) F (H) x V (H) F (H) hence demonstrates that no such a counterexample can exist. The discharging rules are defined as follows: (R) For every 5 + -vertex u, it gives 1 to each adjacent 2-vertex and gives 1 3 to each adjacent 3-vertex. By the rule (R), we have the following results: 1. For every 5-vertex u in G, by Claim 2, d H (u) =5andu has at most one neighbor of 3-vertex in H. Sow (u) > For every 6-vertex u in G, by Claim 3, d 2 (u) 1, and we have (1) If d 2 (u) =0,thend H (u) = 6, and the neighbors of u must be all 3 + -vertex in H. So w (u) =0. (2) If d 1 (u) =1,thend 2 (u) =d 3 (u) =0. Thend H (u) =5andtheneighborofu must be all 4 + -vertex in H. Sow (u) =5 4 > 0. (3) If d 2 (u) =1,thend 1 (u) =d 3 (u) =0. Thend H (u) =6andw (u) =6 4 1 > For every 7-vertex u in G, by Claim 4, d 2 (u) 2 and we have (1) If d 2 (u) =0,thend H (u) = 7, and the neighbor of u must be all 3 + -vertex in H. So w (u) > 0. (2) If d 1 (u) =1,thend 2 (u) =0,d 3 (u) 1. Then d H (u) =6andw (u) > 0. (3) If d 1 (u) =0,thend 2 (u) =1,d 3 (u) 1. Then d H (u) =7andw (u) > 0. (4) If d 1 (u) =2,thend 2 (u) =0,d 3 (u) =0. Thend H (u) =5andw (u) =5 4 > 0. (5) If d 1 (u) =0,thend 2 (u) =2,d 3 (u) =0. Thend H (u) =7andw (u) =7 4 2 > 0. (6) If d 1 (u) =1,thend 2 (u) =1,d 3 (u) =0. Thend H (u) =6andw (u) =6 4 1 > For every l-vertex(l 8) u in G, by Claim 5, we have (1) If d 2 (u) =0,thend H (u) =l, and the neighbors of u must be all 3 + -vertex in H. Since l 8, we have w (u) l l>0. (2) If d 2 (u) 0,thend 3 (u) l 2.Letd 1(u) =x, d 2 (u) =y. Thend 3 (u) l 2 x y.

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