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arXiv 2009.02503math.CO
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3-连通可平面图圈谱中的紧间隙

Tight gaps in the cycle spectrum of 3-connected planar graphs

Qing Cui, On-Hei Solomon Lo

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摘要：

对于任何正整数$k$，定义$f(k)$（分别为$f_3(k)$）为最小整数$\ge k$，使得周长$\ge k$的每个3-连通平面图$G$（分别为3-连通立方平面图$G$）具有长度在区间$[k, f(k)]$（分别为$[k, f_3(k)]$）中的圈。Merker表明，对于任何$k \ge 2$，都有$f_3(k) \le 2k + 9$，对于任何偶数$k \ge 4$，都有$f_3(k) \ge 2k + 2$。他确认$f_3(k) \le 2k + 2$适用于任何$k \ge 2$。这个猜想被Zamfirescu否定了，他给了一个无穷族的反例给每个偶$k \ge 6$，它的图在$[k, 2k + 2]$中没有圈长，即$f_3(k) \ge 2k + 3$给任何偶$k \ge 6$。然而，$f_3(k)$的确切值仅为$k \le 4$所知，并且对于$k \ge 5$，确定$f_3(k)$是开放的。本文改进了Merker的上界，并对每个$k \ge 5$给出了$f_3(k)$的精确值.我们表明，$f_3(5) = 10$，$f_3(7) = 15$，$f_3(9) = 20$，和$f_3(k) = 2k + 3$的任何$k = 6, 8$或$\ge 10$。对于一般的3连通可平面图，Merker证明了存在某个正整数$c$，使得对任意正整数$k$，有$f(k) \le 2k + c$.我们对这个猜想给出了完全肯定的回答。我们证明了$f(k) = 5$对于任何$k \le 3$，$f(4) = 10$，和$f(k) = 2k + 3$对于任何$k \ge 5$。

英文摘要：

For any positive integer $k$, define $f(k)$ (respectively, $f_3(k)$) to be the minimal integer $\ge k$ such that every 3-connected planar graph $G$ (respectively, 3-connected cubic planar graph $G$) of circumference $\ge k$ has a cycle whose length is in the interval $[k, f(k)]$ (respectively, $[k, f_3(k)]$). Merker showed that $f_3(k) \le 2k + 9$ for any $k \ge 2$, and $f_3(k) \ge 2k + 2$ for any even $k \ge 4$. He conjectured that $f_3(k) \le 2k + 2$ for any $k \ge 2$. This conjecture was disproved by Zamfirescu, who gave an infinite family of counterexamples for every even $k \ge 6$ whose graphs have no cycle length in $[k, 2k + 2]$, i.e. $f_3(k) \ge 2k + 3$ for any even $k \ge 6$. However, the exact value of $f_3(k)$ was only known for $k \le 4$, and it was left open to determine $f_3(k)$ for $k \ge 5$. In this paper we improve Merker's upper bound, and give the exact value of $f_3(k)$ for every $k \ge 5$. We show that $f_3(5) = 10$, $f_3(7) = 15$, $f_3(9) = 20$, and $f_3(k) = 2k + 3$ for any $k = 6, 8$ or $\ge 10$. For general 3-connected planar graphs, Merker conjectured that there exists some positive integer $c$ such that $f(k) \le 2k + c$ for any positive integer $k$. We give a complete positive answer to this conjecture. We prove that $f(k) = 5$ for any $k \le 3$, $f(4) = 10$, and $f(k) = 2k + 3$ for any $k \ge 5$.

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