[sfenu]

Here's the best solution I've heard for this:

The prisoners meet together and designate a leader. The leader will maintain a tally. They also designate one of the switches to not matter. The prisoners then use the following strategy:

When a prisoner is lead to the room, if the designated switch is off and it is the prisoner's first time flipping the relevant switch, it is turned on. Else, flip the irrelevant switch. If the prisoner is the designated leader, and the relevant switch is on, he adds a count to his tally and turns the relevant switch off. Otherwise he just flips the irrelevant switch. This way, once the leader has seen the light be on a sufficient number of times he can definitively say that they have all been in the room.

[callahad]

> once the leader has seen the light be on a sufficient number of times he can definitively say that they have all been in the room.

Ah, but what's a sufficient number of times? The unknown initial state is what tripped me for a bit. :)

[gberger]

n-1, where n is the number of prisoners. -1 to account for the leader.

[evanb]

It's actually 2(n-1), because the designated switch can start in either state.

[linux_devil]

"In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the on or the off position. I am not telling you their present positions. The switches are not connected to anything.

The switches are not connected to anything, correct me if I am wrong but how will you find out light is on sufficient number of times

[jaltekruse]

The person tallying keeps a count in their head.

[rokhayakebe]

Good luck if the leader is the 22nd person to go in.