We examine a new variant of the classic prisoners and lightswitches puzzle: A warden leads his $n$ prisoners in and out of $r$ rooms, one at a time, in some order, with each prisoner eventually visiting every room an arbitrarily large number of times. The rooms are indistinguishable, except that each one has $s$ lightswitches; the prisoners win their freedom if at some point a prisoner can correctly declare that each prisoner has been in every room at least once. What is the minimum number of switches per room, $s$, such that the prisoners can manage this? We show that if the prisoners do not know the switches' starting configuration, then they have no chance of escape -- but if the prisoners do know the starting configuration, then the minimum sufficient $s$ is surprisingly small. The analysis gives rise to a number of puzzling open questions, as well.

中文翻译：

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囚犯、房间和电灯开关

我们研究了经典囚犯和电灯开关谜题的新变体：监狱长带领他的 $n$ 个囚犯进出 $r$ 个房间，一次一个，以某种顺序，每个囚犯最终访问每个房间的数量是任意多的次。除了每个房间都有 $s$ 的电灯开关外，房间没有区别。如果囚犯在某个时候能够正确地声明每个囚犯至少在每个房间里去过一次，那么囚犯就会赢得自由。每个房间最少需要多少个开关，$s$，这样囚犯才能做到这一点？我们表明，如果囚犯不知道开关的起始配置，那么他们就没有机会逃脱——但如果囚犯知道起始配置，那么最小的足够 $s$ 是惊人的小。