最小化$\boldsymbol{\bigcup_{[4,6]}}$-饱和图的边数
Minimizing the number of edges in $\mathcal{C}_{[4,6]}$-saturated graphs
AI总结：
本文研究$\boldsymbol{\bigcup_{[4,r]}}$-饱和图的最小边数问题，证明了$\boldsymbol{r\boldsymbol{\boldsymbol{6}}}}$时Ma的猜想不成立，并确定了$\boldsymbol{r=6}$时的饱和数精确值。
AI中文摘要：
令$\boldsymbol{\bigcup_{[4,r]}}$为圈族$\boldsymbol{\bigcup_{4},\bigcup_{5},\boldsymbol{\bigcup_{r}}}$。若图$\boldsymbol{G}$不含$\boldsymbol{4\boldsymbol{\boldsymbol{i}\boldsymbol{\boldsymbol{r}}}}$的圈副本，但添加任意一条不属于$\boldsymbol{G}$的边$\boldsymbol{e}$都会产生至少一个$\boldsymbol{4\boldsymbol{\boldsymbol{i}\boldsymbol{\boldsymbol{r}}}}$的圈副本，则称$\boldsymbol{G}$为$\boldsymbol{\bigcup_{[4,r]}}$-饱和图。饱和数$\boldsymbol{sat(n, \boldsymbol{\bigcup_{[4,r]}})}$是$\boldsymbol{n}$个顶点的$\boldsymbol{\bigcup_{[4,r]}}$-饱和图的最小边数。2025年，Ma确定$\boldsymbol{sat(n, \boldsymbol{\bigcup_{[4,5]}})=\boldsymbol{\boldsymbol{\frac{5n}{4} - \frac{3}{2}}}}$，并猜想对任意$\boldsymbol{r\boldsymbol{\boldsymbol{5}}}$，当$\boldsymbol{n}$足够大时$\boldsymbol{sat(n, \boldsymbol{\bigcup_{[4,r]}})=\boldsymbol{\boldsymbol{\frac{5n}{4} - \frac{3}{2}}}}$成立。本文证明，当$\boldsymbol{n\boldsymbol{\boldsymbol{r}}+1}$时，$\boldsymbol{sat(n, \boldsymbol{\bigcup_{[4,r]}})\boldsymbol{\boldsymbol{\frac{5n}{4} - \frac{r+1}{4}}}}$，这推翻了$\boldsymbol{r\boldsymbol{\boldsymbol{6}}}}$时Ma的猜想；当$\boldsymbol{r=6}$时，确定$\boldsymbol{sat(n, \boldsymbol{\bigcup_{[4,6]}})=\boldsymbol{\boldsymbol{\frac{5n}{4} - \frac{7}{4}}}}$。
英文摘要：
Let $\mathcal{C}_{[4,r]}$ be the family of cycles $\{C_4, \dots, C_r\}$. A graph $G$ is said to be $\mathcal{C}_{[4,r]}$-saturated if $G$ does not contain a copy of cycle $C_i$ for $4\le i\le r$, but the addition of any edge $e\notin E(G)$ creates at least one copy of $C_i$ for $4\le i\le r.$ The saturation number $sat(n, \mathcal{C}_{[4,r]})$ is the minimum number of edges in an $n$-vertex $\mathcal{C}_{[4,r]}$-saturated graph. In 2025, Ma determined that $sat(n, \mathcal{C}_{[4,5]})=\lceil \frac{5n}{4} - \frac{3}{2} \rceil$, and conjectured that for any $r \ge 5$, $sat(n, \mathcal{C}_{[4,r]}) = \lceil\frac{5n}{4} - \frac{3}{2} \rceil$ holds for large $n$. In this paper we prove that $sat(n, \mathcal{C}_{[4,r]}) \le \lceil\frac{5n}{4} - \frac{r+1}{4}\rceil$ for $n \ge r+1$, which disproves Ma's conjecture for $r\ge 6.$ For $r=6,$ we determine that $sat(n, \mathcal{C}_{[4,6]})=\lceil\frac{5n}{4}-\frac{7}{4}\rceil.$ {\bf Keywords}: Saturation graphs; Saturation number; Cycles; Edge minimization
