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On the number of spanning trees of some irregular line graphs

机译:关于一些不规则线图的生成树数

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摘要

Let G be a graph with n vertices and m edges and δ and δ the maximum degree and minimum degree of G. Suppose ~(G ') is the graph obtained from G by attaching δ-dG(v) pendent edges to each vertex v of G. It is well known that if G is regular (i.e., δ = δ, G = ~(G ')), then the line graph of G, denoted by L(G), has ~(2m -n +1δm -n -1)(G) spanning trees, where t(G) is the number of spanning trees of G. In this paper, we prove that if G is irregular (i.e., δ ≠ δ), then t(L(~(G '))) = ~(2m -n +1δm +s -n -1)t(G), where s is the number of vertices of degree one in G '.
机译:令G为具有n个顶点和m个边的图,而δ和δ为G的最大程度和最小程度。假设〜(G')是通过将δ-dG(v)垂边附加到每个顶点v而从G获得的图众所周知,如果G是规则的(即δ=δ,G =〜(G')),则G的线形图由L(G)表示为〜(2m -n +1δm)。 -n -1)(G)生成树,其中t(G)是G的生成树数。本文证明,如果G是不规则的(即δ≠δ),则t(L(〜 (G')))=〜(2m -n +1δm+ s -n -1)t(G),其中s是G'中一阶顶点的数目。

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