arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 2109.03083math.CO
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追踪3- AP博弈的阈值偏差

Chasing the Threshold Bias of the 3-AP Game

Albert Cao, Felix Christian Clemen, Sean English, Xiaojian Li, Tatum Schmidt, Leeann Xoubi, Weian Yin

摘要:

在制造者-破坏者游戏中,有两个玩家,制造者和破坏者,如果他们创造了一个指定的结构,制造者获胜,而破坏者如果他们阻止制造者无限期地获胜,则获胜。$3$ -term算术数列,或$3$-AP,是三个不同整数$a, b, c$的序列,使得$b-a = c-b$。$3$ -AP游戏是在$[n]$上进行的有偏见的制造者-破坏者游戏,其中每一轮破坏者为每个制造者的一个整数选择$q$未声明的整数。Maker试图选择点,使其具有$3$ -AP,而Breaker试图阻止此操作。主要问题是确定阈值偏差$q^*(n)$,即Breaker具有获胜策略的最小值$q=q(n)$。Kusch、Rué、Spiegel和Szabó最初提出了这个问题,并证明了$\sqrt{n/12-1/6}\leq q^*(n)\leq \sqrt{3n}$。我们发现新的策略,使制造商和断路器,改善现有的界限,以\[ (1+o(1))\sqrt{\frac{n}{5.6}} \leq q^*(n) \leq \sqrt{2n} +O(1). \]

英文摘要:

In a Maker-Breaker game there are two players, Maker and Breaker, where Maker wins if they create a specified structure while Breaker wins if they prevent Maker from winning indefinitely. A $3$-term arithmetic progression, or $3$-AP, is a sequence of three distinct integers $a, b, c$ such that $b-a = c-b$. The $3$-AP game is a biased Maker-Breaker game played on $[n]$ where every round Breaker selects $q$ unclaimed integers for every Maker's one integer. Maker is trying to select points such that they have a $3$-AP and Breaker is trying to prevent this. The main question of interest is determining the threshold bias $q^*(n)$, that is the minimum value of $q=q(n)$ for which Breaker has a winning strategy. Kusch, Rué, Spiegel and Szabó initially asked this question and proved $\sqrt{n/12-1/6}\leq q^*(n)\leq \sqrt{3n}$. We find new strategies for both Maker and Breaker which improve the existing bounds to \[ (1+o(1))\sqrt{\frac{n}{5.6}} \leq q^*(n) \leq \sqrt{2n} +O(1). \]

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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