AI 中文总结
本文针对$(K_{t_1}, \ldots, K_{t_k})$-共临界图，在最小度猜想成立的前提下，利用$q$-邻居自举渗流闭包方法渐近证明了Hanson-Toft关于其边数下界的猜想。
AI 中文摘要
给定整数$k\ge2$和$t_1, \ldots, t_k\ge2$，若图$G$的每条$k$-边着色中，都存在某个$i\in\{1, \ldots, k\}$对应的单色$K_{t_i}$副本，则记为$G \rightarrow (K_{t_1}, \ldots, K_{t_k})$。非完全图$G$是$(K_{t_1}, \ldots, K_{t_k})$-共临界图，当且仅当$G \nrightarrow (K_{t_1}, \ldots, K_{t_k})$，但对每条不在$G$边集$E(G)$中的边$e$，都有$G+e \rightarrow (K_{t_1}, \ldots, K_{t_k})$。令$r=R(K_{t_1}, \ldots, K_{t_k})$表示拉姆齐数。1987年，Hanson和Toft猜想，每个顶点数$n\ge r$的$(K_{t_1}, \ldots, K_{t_k})$-共临界图$G$满足边数$|E(G)|\ge (r-2)n-\binom{r-1}{2}$，该界对所有$n\ge r$都是最优的。更近一些，本文作者猜想这类图的最小度至少为$r-2$。利用$q$-邻居自举渗流闭包方法，本文证明：若最小度猜想成立，则Hanson-Toft猜想渐近成立；更准确地说，若每个$(K_{t_1}, \ldots, K_{t_k})$-共临界图的最小度至少为$r-2$，则存在仅依赖$r,k$的常数$C$，使得每个顶点数$n\ge r$的$(K_{t_1}, \ldots, K_{t_k})$-共临界图$G$满足$|E(G)|\ge (r-2)n-C$。
英文摘要
Given integers $k\ge2$ and $t_1, \ldots, t_k\ge2$, we write \emph{$G \rightarrow (K_{t_1}, \ldots, K_{t_k})$} if every $k$-coloring of the edges of a graph $G$ contains a monochromatic copy of $K_{t_i}$ in color $i$ for some $i\in\{1, \ldots, k\}$. A non-complete graph $G$ is \emph{$(K_{t_1}, \ldots, K_{t_k})$-co-critical} if $G \nrightarrow (K_{t_1}, \ldots, K_{t_k})$, but $G+e\rightarrow (K_{t_1}, \ldots, K_{t_k})$ for every edge $e\notin E(G)$. Let $r=R(K_{t_1}, \ldots, K_{t_k})$ denote the Ramsey number. In 1987, Hanson and Toft conjectured that every $(K_{t_1}, \ldots, K_{t_k})$-co-critical graph $G$ on $n\ge r$ vertices satisfies \[|E(G)|\ge (r-2)n- \binom{r- 1}{2}.\] This bound is best possible for every $n\ge r$. More recently, the present author conjectured that every such graph has minimum degree at least $r-2$. Using the $q$-neighbor bootstrap percolation closure method, here we prove that the Hanson-Toft Conjecture holds asymptotically, provided that the minimum-degree conjecture is true; more precisely, If every $(K_{t_1},\ldots,K_{t_k})$-co-critical graph has minimum degree at least $r-2$, then there is a constant $C=C(r,k)$ such that every $(K_{t_1},\ldots,K_{t_k})$-co-critical graph $G$ on $n\ge r$ vertices satisfies $|E(G)|\ge (r-2)n-C$.
