△=3的图的邻和可区别全可选性（英文）  

Neighbor Sum Distinguishing Total Choosability of Graphs With △=3

作　　者：姚京京 邵泽玲[1] 徐常青[1] 

机构地区：[1]河北工业大学理学院,天津300401

出　　处：《数学进展》2016年第3期343-348,共6页Advances in Mathematics（China）

基　　金：Supported by NSFC(No.11301134,No.11301135);HUSTP(No.ZD2015106);HNSF(No.A2015202301,No.A2012202067)

摘　　要：设图G=(V,E),φ:V∪E→{1,2,…,k}为图G的一个正常全染色.令f(v)表示点v及所有与其关联的边的颜色的加和.若对任意uv∈E(G),有f(u)≠f(v),则称φ是图G的邻和可区别全染色.Pilsniak和Wozniak最早研究了邻和可区别全染色,并猜想对于任意图G,若k≥△(G)+3,则其存在邻和可区别全染色.图G的最大平均度,记为mad(G),是G的所有非空子图的平均度的最大值.本文运用组合零点定理与权转移方法证明了:若图G满足△(G)=3且mad(G)<(44)/(15),则ch_Σ″(G)≤6(其中ch_Σ″(G)为图G的邻和可区别全可选性).Let G =（V,E） be a graph and φ：V U E → {1,2,…,k} be a proper total coloring of G.Let f（v） denote the sum of the color on vertex v and the colors on the edges incident with v.We say that the proper total coloring φ is neighbor sum distinguishing if for each edge uv ∈ E（G）,f（u） ≠ f（v）.Pilsniak and Wozniak first introduced this coloring and conjectured that such coloring exists for any graph G if k ≥ △（G） ＋ 3.The maximum average degree of G is the maximum of the average degree of its non-empty subgraphs,which is denoted by mad（G）.In this paper,by using the Combinatorial Nullstellensatz and the discharging method,we prove that the conjecture holds for some graphs in their list versions.More precisely,we prove that if G is a graph with △（G） = 3 and mad（G） （44）/（15）,（G） ≤6（where ch_∑″（G） is the neighbor sum distinguishing total choosability of G）.

关 键 词：邻和可区别全可选性 最大平均度 组合零点定理 

分 类 号：O157.5[理学—数学]