Elect a SpokesmanDate: 07/02/2002 at 06:42:52 From: Vaibhav Kamthan Subject: Puzzle Ponder This Challenge from IBM (select July 2002): http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/ This puzzle has been making the rounds of Hungarian mathematicians' parties. The warden meets with the 23 prisoners when they arrive. He tells them: You may meet together today and plan a strategy, but after today you will be in isolated cells and have no communication with one another. There is in this prison a "switch room" which contains two light switches, labelled "A" and "B", each of which can be in the "on" or "off" position. I am not telling you their present positions. The switches are not connected to any appliance. 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", and this prisoner will select one of the two switches and reverse its position (e.g. if it was "on", he will turn it "off"); the prisoner will then be led back to his cell. Nobody else will ever enter the "switch room". Each prisoner will visit the switch room arbitrarily often. That is, for any N it is true that eventually each of you will visit the switch room at least N times.) At any time, any of you may declare to me: "We have all visited the switch room." If it is true (each of the 23 prisoners has visited the switch room at least once), then you will all be set free. If it is false (someone has not yet visited the switch room), you will all remain here forever, with no chance of parole. Devise for the prisoners a strategy which will guarantee their release. Date: 07/21/2002 at 22:46:09 From: Doctor Shawn Subject: Solution to this puzzle One prisoner is designated as the "caller." He and only he will make the determination of when everyone's been in the room. Whenever he sees that the left-hand switch is "down," he will move it "up" and add 1 to his count; otherwise, he will flip the right-hand switch and not think anything of it. When the other prisoners enter the room, they have a different strategy. If the left-hand switch is "up," they will switch it "down," but they will only do this the first two times they encounter the left-hand switch "up." If they've moved the left switch twice already, or if the left-hand switch is "down," they will flip the right-hand switch. When the caller's count is 44, each prisoner must have been in the room at least once, regardless of the initial configuration of the switches. Neat problem. For IBM's explanation, see (select July2002.html): http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/solutions/ -Doctor Shawn http://mathforum.org/dr.math/ |
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