# The Prisoners' Switch Problems

### Version 2

This can be reduced to version 1's solution
(almost) if we take switch `A` to be
the “signal” switch
let prisoners use switch `B` if they can't send a signal
(because `A` is already on or because they have already sent their signal.

The problem is, we don't know the initial state of `A`.
If switch `A` is initially on, Fred will receive a false signal.

The solution to this problem is to
go through the process twice. Every prisoner is told to flip
switch `A` *twice*, the first two times they
enter the room and see the signal switch off. In the end,
Fred will necessarily see 44 signals (22 other prisoners times
2 signals each). After 44, Fred knows she received 2 signals from
everybody, or possibly 2 from 21 of the corresponding prisoners and
one from another, but, either way, every prisoner has entered the
room. Fred can now say with certainty that everybody has been in
the room at least once.

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