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arXiv 2610.09483math.CO

带匹配桥的笛卡尔积的图灵敏度

Graph Sensitivity of Cartesian Products with Matched Bridges

Zhen-Mu Hong, Zi-Yi Wu, Zheng-Jiang Xia

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中文总结 AI 辅助

该论文将 Huang 定理从超立方体推广到带匹配桥的二分图笛卡尔积,证明树积的灵敏度等于根号 k 的上取整,并确定路径积的灵敏度,解决偶数路径情形,同时给出超立方体 f_2 的精确值。

中文摘要 AI 辅助

对于图 $G$,令 $f_t(G)$ 表示具有 $\alpha(G)+t$ 个顶点的诱导子图的最大度的最小值,其中 $\alpha(G)$ 是独立数,并记 $f(G)=f_1(G)$。Huang 定理给出 $f(Q_k)\ge\lceil\sqrt{k}\rceil$ 对于 $k$ 维超立方体 $Q_k$。我们将此下界推广到具有完美匹配的 $k$ 个二分图的笛卡尔积,并证明当因子连通且每个因子具有匹配桥时等式成立。特别地,我们证明当每个 $T_i$ 是具有完美匹配的树时,$f(T_1\Box\cdots\Box T_k)=\lceil\sqrt{k}\rceil$。我们确定了每个路径笛卡尔积的灵敏度,解决了 Zeng 和 Hou [J. Graph Theory 107 (2024), 169--180] 留下的偶数路径情形。对于这些树积,令 $D=\lceil\sqrt{k}\rceil$,我们还证明当 $1\le t\le 2^{D-\lceil\log_2D\rceil-1}$ 时,$f_t(T_1\Box\cdots\Box T_k)=D$。当 $t=2$ 时,此等式对所有 $k\ge 2$ 成立,前提是至少一个因子不是 $K_2$。匹配割给出偶数阶路径积的额外精确范围。最后,我们证明对于每个 $k\ge 2$ 且 $k\not\in \{4,9\}$,$f_2(Q_k)=\lceil\sqrt{k}\rceil$,而 $f_2(Q_4)=3$ 且 $3\le f_2(Q_9)\le 4$。

英文摘要

For a graph $G$, let $f_t(G)$ denote the minimum of the maximum degree of an induced subgraph with $α(G)+t$ vertices, where $α(G)$ is the independence number, and write $f(G)=f_1(G)$. Huang's theorem gives $f(Q_k)\ge\lceil\sqrt{k}\rceil$ for the $k$-dimensional hypercube $Q_k$. We extend this lower bound to Cartesian products of $k$ bipartite graphs with perfect matchings and prove that equality holds when the factors are connected and each has a matched bridge. In particular, we prove that $f(T_1\Box\cdots\Box T_k)=\lceil\sqrt{k}\rceil$ whenever each $T_i$ is a tree with a perfect matching. We determine the sensitivity of every Cartesian product of paths, settling the even-path case left open by Zeng and Hou [J. Graph Theory 107 (2024), 169--180]. For these tree products, with $D=\lceil\sqrt{k}\rceil$, we also prove that $f_t(T_1\Box\cdots\Box T_k)=D$ whenever $1\le t\le 2^{D-\lceil\log_2D\rceil-1}$. When $t=2$, this equality holds for all $k\ge 2$, provided that at least one factor is not $K_2$. Matching cuts give an additional exact range for products of even-order paths. Finally, we prove that $f_2(Q_k)=\lceil\sqrt{k}\rceil$ for every $k\ge 2$ with $k\not\in \{4,9\}$, whereas $f_2(Q_4)=3$ and $3\le f_2(Q_9)\le 4$.

发表机构

  • Anhui University of Finance & Economics(安徽财经大学)

机构由 AI 辅助整理,请以论文原文为准。

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