用有限自动机进行奇偶博弈
Playing odds and evens with finite automata
摘要：
本文研究了由有限自动机表示的重复博弈的渐近行为。证明了对于每一个$n$，都有一个具有$2^n \cdot \mathrm{poly}(n)$状态的自动机击败每一个$n$状态自动机，在这个意义上，它赢得了所有的回合，除了1000回合。此外，每个这样的自动机至少有$2^n \cdot (1 - o(1))$状态，这意味着上限紧到多项式因子。这是一个显着的改进，在特殊情况下的“赔率和偶数”的经典结果的Ben-Porath。此外，我猜想，该方法可以推广到任意零和游戏。
英文摘要：
This paper is concerned with asymptotic behaviour of a repeated game of "odds and evens", with strategies of both players represented by finite automata. It is proved that, for every $n$, there is an automaton with $2^n \cdot \mathrm{poly}(n)$ states which defeats every $n$-state automaton, in the sense that it wins all rounds except for finitely many. Moreover, every such automaton has at least $2^n \cdot (1 - o(1))$ states, meaning that the upper bound is tight up to polynomial factors. This is a significant improvement over a classic result of Ben-Porath in the special case of "odds and evens". Moreover, I conjecture that the approach can be generalised to arbitrary zero-sum games.
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。
