We study the positional game where two players, Maker and Breaker, alternately select respectively 1 and $b$ previously unclaimed edges of $K_n$. Maker wins if she succeeds in claiming all edges of some odd cycle in $K_n$ and Breaker wins otherwise. Improving on a result of Bednarska and Pikhurko, we show that Maker wins the odd cycle game if $b\le \left((4-\sqrt{6})∕5+o\left(1\right)\right)n$. We furthermore introduce “connected rules” and study the odd cycle game under them, both in the Maker–Breaker as well as in the Client–Waiter variant.

中文翻译：

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关于奇数周期游戏和相关规则

我们研究了位置游戏，其中两个玩家Maker和Breaker分别选择1和 $b$ 以前无人认领的边缘 ${ķ}_{ñ}$。如果制造商成功声明了某个奇数周期的所有边缘，则她将获胜。${ķ}_{ñ}$否则断路器获胜。根据Bednarska和Pikhurko的结果进行改进，我们表明Maker赢得了奇周期游戏$b\le （（4-\sqrt{6}）∕5+Ø（\mathrm{1个}））ñ$。我们还将介绍“关联规则”，并研究根据规则进行的奇数循环博弈，无论是在Maker-Breaker还是Client-Waiter变体中。