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arxiv: 1210.7618 · v1 · pith:JCSGYCJ6new · submitted 2012-10-29 · 🧮 math.CO

Biased Games On Random Boards

classification 🧮 math.CO
keywords gamesgamebiascriticalfracmakermaker-breakeravoider-enforcer
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In this paper we analyze biased Maker-Breaker games and Avoider-Enforcer games, both played on the edge set of a random board $G\sim \gnp$. In Maker-Breaker games there are two players, denoted by Maker and Breaker. In each round, Maker claims one previously unclaimed edge of $G$ and Breaker responds by claiming $b$ previously unclaimed edges. We consider the Hamiltonicity game, the perfect matching game and the $k$-vertex-connectivity game, where Maker's goal is to build a graph which possesses the relevant property. Avoider-Enforcer games are the reverse analogue of Maker-Breaker games with a slight modification, where the two players claim at least 1 and at least $b$ previously unclaimed edges per move, respectively, and Avoider aims to avoid building a graph which possesses the relevant property. Maker-Breaker games are known to be "bias-monotone", that is, if Maker wins the $(1,b)$ game, he also wins the $(1,b-1)$ game. Therefore, it makes sense to define the critical bias of a game, $b^*$, to be the "breaking point" of the game. That is, Maker wins the $(1,b)$ game whenever $b\leq b^*$ and loses otherwise. An analogous definition of the critical bias exists for Avoider-Enforcer games: here, the critical bias of a game $b^*$ is such that Avoider wins the $(1,b)$ game for every $b > b^*$, and loses otherwise. We prove that, for every $p=\omega(\frac{\ln n}{n})$, $G\sim\gnp$ is typically such that the critical bias for all the aforementioned Maker-Breaker games is asymptotically $b^*=\frac{np}{\ln n}$. We also prove that in the case $p=\Theta(\frac{\ln n}{n})$, the critical bias is $b^*=\Theta(\frac{np}{\ln n})$. These results settle a conjecture of Stojakovi\'c and Szab\'o. For Avoider-Enforcer games, we prove that for $p=\Omega(\frac{\ln n}{n})$, the critical bias for all the aforementioned games is $b^*=\Theta(\frac{np}{\ln n})$.

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