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作 者:车向凯[1]
机构地区:[1]东北大学理学院
出 处:《东北大学学报(自然科学版)》1999年第3期333-336,共4页Journal of Northeastern University(Natural Science)
基 金:国家自然科学基金
摘 要:设G为n阶3连通无爪图·δ=min{d(x)|x∈V(G)},δ=min{max(d(x),d(y))|x,y∈V(G),d(x,y)=2},则C(G)≥min{n,3δ+δ,6δ}·采用反证法,将图G分为若干情形·在每一种情形中,利用图G的3连通性和无爪性,构造若图G的最长圈不满足已给条件的矛盾·It is proved that if G is a 3 connected claw free graph on n vertices with the minimum degree δ and δ *=min{max( d(x),d(y))|x,y∈V(G),d(x,y)=2} ,then the circumference of the graph G is at least min{ n,3δ *+δ,6δ} .The graph G was sorted into several types and the method of reduction to absurdity was used to get the result. The properties of 3 connected claw free were used to structure the contradiction if the longest cycle of graph G does not satisfy a given condition in each type to prove the above conclusion.
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