设▽(G)表示最少的点数,这些点去掉后图中无圈(即森林).称这个数▽(G)为图G的消圈数.通常,确定图的消圈数是NP完全的.Bau和Beineke曾提出以下问题:哪些阶数为n的3正则图G的消圈数满足▽(G)=[(n+2)/4]?本文回答了这个问题:阶数为n的3正则图G的消圈数满足▽(G)=[(n+2)/4]当且仅当G是上嵌入的(即以最多两个面嵌入在可定向曲面上).其次,对于一般3正则图,得出其消圈数的计算公式为▽(G)=γ_M(G)+ζ(G),这里γ_M(G)表示图的最大亏格,ζ(G)表示图G的Betti亏数.由此可知,3正则图的最大亏格的计算的多项式算法是存在的,所以3正则图的消圈数的计算也是多项式可解的.
Let ▽(G) denote the minimum number of vertices whose removal results in an acyclic graph(i.e.,a forest).Such a number ▽(G) is called the decycling number of graph G.In general,deciding the decycling number of a graph is NP-complete.Bau and Beineke proposed the following question:Which cubic graphs G of order n satisfy ▽(G) =[(n+2)/4]? In this paper we solve this problem and show that a cubic graph G of order n has decycling number▽(G) =[(n+2)/4]if and only if G is upper-embeddable(i.e.,may be embedded in an orientable surface having at most two faces).Furthermore,we find that the decycling number of a cubic graph G is ▽(G) = γ_m(G) + ζ(G),where γ_m(G) and ζ(G) are,respectively,the maximum genus of G and the Betti deficiency of G.Since there exists a polynomial time algorithm to find the maximum genus of cubic a graph,there also exists a polynomial time algorithm to find the decycling number of a cubic graph.