Maximizing the number of spanning trees in a graph with n nodes and m edges

D. Shier
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引用次数: 33

Abstract

Consider the class of undirected graphs hav ing 11. nodes and m edges. The proble m to be addressed here is that of finding specific configurations of III edges on the given n nodes so that the resulting graph will contain the largest number of spanning trees. In particular, an explicit solution to this problem will be exhibited for graphs which have "enough" edges. To be specific, let the set E of k edges be deleted from the comp lete graph KII on 11. nodes: KII has an edge between every pair of distinct nodes and thus contains 11.(11.-1)/2 edges. For th e case when k ~ 11./2, we will demonstrate that the number of spanning trees T(n, E) in the res ulting graph is maximized by choosing the k deleted edges to be mutually nonadjacent. The (apparently more complicated) cases with k > 11./2 await resolution. Let Pic denote a set of k nonadjacent ("parallel") edges in K II , where k ~ n/2. We will show that lEI = k implies T(n, E) ~ T(n, Pic). First the case when the edge set E is disconnected will be di sposed of. This will be done in the context of the inductive hypothesis that if i < k and 151 = i, then T(n, S) ~ T(n, Pi) whether or not 5 is connected. Suppose lEI = k and that E can be decomposed into connected edge sets C I , C 2, •.• , C p with p 3 2. Certainly, when k = 2 the only disconnected set E possible consists of two nonadjacent edges, so that the inductive hypothesis holds. More generally, if lEI = k then IC,I < k and n 3 2k > 21C 11, whence 0 ~ T(n, C d ~ T(n, Pa ), (X= IC d, by the inductive hypothesis. Similarly, we have O~T(n, C,)~T(n, P(3), {3=IC ,I, where CI = C2 U ... U Co in the decomposition E=C 1 U C2 U ... U C/!. From [2, p. 106] it is known that ( 2)i T(n,P j )=nn2 1-;; ,
在有n个节点和m条边的图中最大化生成树的数量
考虑一类有11个向量的无向图。节点和m条边。这里要解决的问题是在给定的n个节点上找到III条边的特定配置,从而使生成的图包含最多数量的生成树。特别是,对于具有“足够”边的图,将展示该问题的显式解决方案。具体地说,设k条边的集合E从11上的完整图KII上删除。节点:KII在每对不同的节点之间都有一条边,因此包含11.(11.-1)/2条边。对于k ~ 11的情况。/2,我们将证明,通过选择k个被删除的边相互不相邻,生成树的数量T(n, E)在结果图中是最大化的。(显然更复杂的)k bbbb11的情况。/2等待决议。设Pic表示k II中k条不相邻(“平行”)边的集合,其中k ~ n/2。我们将证明lEI = k意味着T(n, E) ~ T(n, Pic)。首先处理边集E断开的情况。这将在归纳假设的背景下完成,如果i < k且151 = i,那么T(n, S) ~ T(n, Pi)无论5是否连通。设lEI = k,且E可以分解为连通边集c1, c2,•。•,C p与p 32。当然,当k = 2时,唯一可能的分离集E由两条不相邻的边组成,因此归纳假设成立。更一般地说,如果lEI = k,则IC,I < k, n 3 2k > 21C 11,则由归纳假设,0 ~ T(n, C d ~ T(n, Pa), (X= IC d)。同样地,我们有O~T(n, C,)~T(n, P(3), {3=IC,I,其中CI = C2 U…U Co在分解E= c1 U C2 U…U C / !由[2,p. 106]可知(2)i T(n, p. j)=nn2 1-;;,
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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