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一个空储物柜与两次检查:一种精确的最优团队策略

One Empty Locker and Two Inspections: An Exact Optimal Team Strategy

Igor Kleiner, David Perry

arXiv 2608.13074首次发表:更新:

AI 中文总结

该研究针对含N个储物柜的合作搜索游戏,证明所有玩家找到分配物品的最大概率为I_N/N!,提出基于虚拟符号N和指针的最优策略,通过混合整数验证了N=2至5的情况。

AI 中文摘要

我们研究一个包含N个储物柜、N-1个带标签物品、一个空储物柜和N-1个玩家的合作搜索游戏。玩家i寻找物品i,最多可检查两个储物柜,第二次检查可基于第一次检查的内容决定;玩家可预先协调,但游戏开始后无法获取其他玩家搜索的信息。我们证明所有玩家找到各自分配物品的最大概率为I_N/N!,其中I_N是N个元素的对合数。最优策略用通用虚拟符号N表示空位,遵循指针规则:玩家i先打开储物柜i,再打开观察到编号的储物柜。该上界对所有确定性或随机自适应策略均成立,由删除引理和两个递推式推导得出。我们给出明确示例,将三门问题对应到同构的“Monty Hall归来”游戏,并报告N=2、3、4、5的可复现混合整数验证,该计算独立于证明且非证明必需。

英文摘要

We study a cooperative search game with $N$ lockers, $N-1$ labelled objects, one empty locker, and $N-1$ players. Player $i$ seeks object $i$ and may inspect at most two lockers; the second inspection may depend on the content of the first. The players may coordinate beforehand but receive no information about the searches of other players after play begins. We prove that the maximum probability that every player finds the assigned object is $I_N/N!$, where $I_N$ is the number of involutions of $N$ elements. An optimal strategy represents the blank by the common fictitious symbol $N$ and follows pointers: player $i$ first opens locker $i$, then opens the locker whose number was observed. The upper bound holds for every deterministic or randomized adaptive strategy and follows from a deletion lemma and two recurrences. We give explicit examples, identify the three-door case with an isomorphic "Return of Monty Hall" game, and report a reproducible mixed-integer verification for $N=2,3,4,5$. The computation is independent of, and not needed for, the proof.

CommentsWithdrawn at the request of the co-author due to an unresolved authorship issue. The withdrawal is not related to an error in the mathematical results

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