镜像博弈中的一个随机策略
A randomized strategy in the mirror game
摘要：
Alice和Bob轮流（Alice先玩）宣布集合$[1,2N]$中的数字。如果一个玩家宣布了一个之前宣布的数字，那么这个玩家输了，另一个玩家赢了。如果所有号码都不重复，结果是平局。如果两个玩家都有无限的记忆和发挥最佳，那么游戏将是平局。Garg和Schneider [ITCS 2019]表明，如果Alice具有无界内存，则Bob可以使用$\log N$内存确保平局，而如果Bob具有无界内存，则Alice需要$N$中的内存线性才能确保平局。Garg和Schneider还考虑了一个{ \em辅助匹配}模型，在这个模型中，Alice得到了一个额外的输入，一个随机匹配的$M$，它与数字$[1,2N]$匹配，存储这个输入不计入Alice使用的内存。他们表明，在这个模型中，爱丽丝有一种策略，它与概率至少$1 - \frac{1}{N}$有关，并且只使用$O(\sqrt{N} (\log N)^2)$记忆。我们将展示如何修改Alice的策略，使其只使用$O((\log N)^3)$空间。
英文摘要：
Alice and Bob take turns (with Alice playing first) in declaring numbers from the set $[1,2N]$. If a player declares a number that was previously declared, that player looses and the other player wins. If all numbers are declared without repetition, the outcome is a tie. If both players have unbounded memory and play optimally, then the game will be tied. Garg and Schneider [ITCS 2019] showed that if Alice has unbounded memory, then Bob can secure a tie with $\log N$ memory, whereas if Bob has unbounded memory, then Alice needs memory linear in $N$ in order to secure a tie. Garg and Schneider also considered an {\em auxiliary matching} model in which Alice gets as an additional input a random matching $M$ over the numbers $[1,2N]$, and storing this input does not count towards the memory used by Alice. They showed that is this model there is a strategy for Alice that ties with probability at least $1 - \frac{1}{N}$, and uses only $O(\sqrt{N} (\log N)^2)$ memory. We show how to modify Alice's strategy so that it uses only $O((\log N)^3)$ space.
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。
