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The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison there 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. BOTH SWITCHES ARE IN THEIR OFF POSITIONS NOW.* The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."

"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."

"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

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This sol'n may not be the fastest or the most elegant but it should work:

make a rule that you can only switch the left switch 'on' if you have never switched it 'on' before and never switch it 'off' except for the 1 guy in charge of counting. So the right switch is the dummy/default switch that is used in all other cases. so every time the assigned guy sees the left switch 'on' he switches it back 'off' and counts that as a new prisoner and when he counts up to 22 (no need to include himself) then he declares everyone has visited and everyone is free

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