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Asymptotically Optimal Threshold Bias for the $(a : b)$ Maker-Breaker Minimum Degree, Connectivity and Hamiltonicity Games

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arxiv 2406.11051 v1 pith:VYAJV3K7 submitted 2024-06-16 math.CO

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keywords gamefracmakersubgraphbiasboundbreakerdegree
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abstract

We study the $(a:b)$ Maker-Breaker subgraph game played on the edges of the complete graph $K_n$ on $n$ vertices, $n,a,b \in \mathbb{N}$ where the goal of Maker is to build a copy of a specific fixed subgraph $H$. In our work this is a spanning graph with minimum degree $k=k(n)$, a connected spanning subgraph or a Hamiltonian subgraph. In the $(a:b)$ game in each round Maker chooses $a$ unclaimed edges of $K_n$ and Breaker chooses $b$ unclaimed edges. Maker wins, if he succeeds to build a copy of the subgraph under consideration, otherwise Breaker wins. For the $k$-minimum-degree, we present a winning strategy for Maker leading to a bound that generalizes a bound of Gebauer and Szab{\'o} for the $(1:b)$ case. Moreover, we give an explicit strategy for Breaker for $b >(1+o(1)) \frac{an}{a+\ln(n)}$ in case of $a=o\left(\sqrt{\frac{n}{\ln(n)}}\right)$ and $k=o(\ln(n))$. Note that this bound is the same as the Maker bound presented by Hefetz et al. (2012) for the $(a:b)$ connectivity game, which implies that the asymptotic optimal bias for this game is $\frac{an}{a+\ln(n)}$. This resolves the open problem stated by these authors. We also study the $(a:b)$ Hamiltonicity game in which Maker's goal is to create a Hamiltonian subgraph. For the $(1:b)$ variant Krivelevich proved that $\left(1+o(1) \right)\frac{n}{\ln n}$ is the exact threshold bias. Controlling Breaker's vertex degree in the $(a:b)$ Maker-Breaker minimum degree game enables us to the asymptotic optimal generalized threshold bias for the $(a:b)$-game, both for $a=o\left(\sqrt{\frac{n}{\ln n}} \right)$ and $a=\Omega\left(\sqrt{\frac{n}{\ln n}} \right)$.

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  1. Maker playing against an invisible Breaker

    math.CO 2025-07 accept novelty 7.0 of 10

    In a hidden-move variant of Maker-Breaker games on the complete graph, Maker almost surely wins exactly when her bias is at least half the Breaker's bias for matching, connectivity and minimum-degree-one, and at least...

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