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arXiv 1012.5710math.CO
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完全二部图的广义连通度

The generalized connectivity of complete bipartite graphs

Shasha Li, Wei Li, Xueliang Li

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18 pages
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摘要：

设$G$是一个阶为$n$的非平凡连通图，$k$是一个具有$2\leq k\leq n$的整数.对于$G$的$k$顶点的集合$S$，令$κ(S)$表示$G$中的边不相交树$T_1,T_2,...,T_\ell$的最大数目$\ell$，使得对于具有$1\leq i,j\leq \ell$的每个不同整数对$i,j$，$V(T_i)\cap V(T_j)=S$。Chartrand等人将连通性的概念概括如下：$G$的$k$-$connectivity$，用$κ_k(G)$表示，由$κ_k(G)=$ min $\{κ(S)\}$定义，其中最小值取$V(G)$的所有$k$ -子集$S$。因此，$κ_2(G)=κ(G)$，其中$κ(G)$是$G$的连通性。此外，$κ_{n}(G)$是$G$的边不相交生成树的最大数目。本文主要研究了完全二部图PDMATHBNF的PDMATHBMO-连通性.首先，我们得到了$K_{a,b}$的边不交生成树的个数，即$\lfloor\frac{ab}{a+b-1}\rfloor$，并具体给出了$\lfloor\frac{ab}{a+b-1}\rfloor$边不交生成树。在此基础上，我们得到了所有$2\leq k \leq a+b$的$K_{a,b}$的$k$连通性.即，如果$k>b-a+2$和$a-b+k$是奇数，则$κ_{k}(K_{a,b})=\frac{a+b-k+1}{2}+\lfloor\frac{(a-b+k-1)(b-a+k-1)}{4(k-1)}\rfloor,$;如果$k>b-a+2$和$a-b+k$是偶数，则$κ_{k}(K_{a,b})=\frac{a+b-k}{2}+\lfloor\frac{(a-b+k)(b-a+k)}{4(k-1)}\rfloor,$;如果$k\leq b-a+2$，则$κ_{k}(K_{a,b})=a. $

英文摘要：

Let $G$ be a nontrivial connected graph of order $n$, and $k$ an integer with $2\leq k\leq n$. For a set $S$ of $k$ vertices of $G$, let $κ(S)$ denote the maximum number $\ell$ of edge-disjoint trees $T_1,T_2,...,T_\ell$ in $G$ such that $V(T_i)\cap V(T_j)=S$ for every pair $i,j$ of distinct integers with $1\leq i,j\leq \ell$. Chartrand et al. generalized the concept of connectivity as follows: The $k$-$connectivity$, denoted by $κ_k(G)$, of $G$ is defined by $κ_k(G)=$min$\{κ(S)\}$, where the minimum is taken over all $k$-subsets $S$ of $V(G)$. Thus $κ_2(G)=κ(G)$, where $κ(G)$ is the connectivity of $G$. Moreover, $κ_{n}(G)$ is the maximum number of edge-disjoint spanning trees of $G$. This paper mainly focus on the $k$-connectivity of complete bipartite graphs $K_{a,b}$. First, we obtain the number of edge-disjoint spanning trees of $K_{a,b}$, which is $\lfloor\frac{ab}{a+b-1}\rfloor$, and specifically give the $\lfloor\frac{ab}{a+b-1}\rfloor$ edge-disjoint spanning trees. Then based on this result, we get the $k$-connectivity of $K_{a,b}$ for all $2\leq k \leq a+b$. Namely, if $k>b-a+2$ and $a-b+k$ is odd then $κ_{k}(K_{a,b})=\frac{a+b-k+1}{2}+\lfloor\frac{(a-b+k-1)(b-a+k-1)}{4(k-1)}\rfloor,$ if $k>b-a+2$ and $a-b+k$ is even then $κ_{k}(K_{a,b})=\frac{a+b-k}{2}+\lfloor\frac{(a-b+k)(b-a+k)}{4(k-1)}\rfloor,$ and if $k\leq b-a+2$ then $κ_{k}(K_{a,b})=a. $

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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