Three wise men are locked in three rooms. Each day there there is, or there is no light in each room.During a certain finite initial period there is no rule about how the rooms are illuminated, and after one of two cases are possible:1. Each day exactly one room is illuminated2. Each day exactly two rooms are illuminatedThis lasts infinitely.(Let's assume that wise men can't die during this process, for exemple each next day is twice shorter than previous one)I other words wise men are given three binary sequences a,b,c, and there exists n such as for each m>nAfter all this they are independently asked what was the constant (how many rooms were lighted each day and which case was actually realised). If two of them give a right answer they stay alive if not, they are killed.Is there is a strategy for them to stay alive?

just a quick guess:If there is no light in the room, the men should say 1, otherwise they should say 2.There are 6 possible cases:Room1 Room2 Room3-----------------------------------------Dark Dark Light1 1 2 -> SurvivalDark Light Dark1 2 1 -> SurvivalLight Dark Dark 2 1 1 -> SurvivalLight Light Dark2 2 1 -> SurvivalLight Dark Light 2 1 2 -> SurvivalDark Light Light1 2 2 -> SurvivalDid I guess right?

Re reading the problem, I am not sure I fully understood the question.My solution should work if they ask everyday, but it should not be true after few days.

Let A be the fraction of times for which a_m = 1, i.e. and define B and C similarly. Then A, B, and C are each between 0 and 1 (inclusive) and they sum to const, which is either 1 or 2. If the constant is 1, there can't be two of A, B, and C which are both larger than 1/2, whereas if the constant is 2, there can't be two of A, B, and C which are smaller than 1/2. So if a wise man's fraction is < 1/2, he says 1, but if it is > 1/2, he says 2. Some care must be taken when it equals 1/2 though, since we could have either 1/2, 1/2, 0 (if the constant is 1) or 1/2, 1/2, 1 (if the constant is 2). The problem only arises when 2 wise men both have a ratio of 1/2, in which case we want to guarantee that one says 1 and the other says 2 (thus letting the third wise man be the tie breaker). I'm confident there is a clever way to do this, but for now it evades me. I'll think about it more later.

QuoteOriginally posted by: LapinRe reading the problem, I am not sure I fully understood the question.My solution should work if they ask everyday, but it should not be true after few days.They are asked only at the end, when they already dispose an infinite sequence of 0s and 1s.

QuoteOriginally posted by: mhughesLet A be the fraction of times for which a_m = 1, i.e. and define B and C similarly. .These limits are not necessarily exist.Imagine the sequence 0 11 0000 11111111 ....

Ah, true enough... Back to square 1.

The 3 wise men will give their answers on the basis of the last day they spent in the room. If in a room the light was OFF on the last day, he should say 1, otherwise he should say 2. In this case exactly 2 men will give correct answer.