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

Abstract

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 $k \ge 5$.