Deadly | Strategy |

A countably infinite number of prisoners, each with an unknown and randomly assigned red or blue hat line up single file line. Each prisoner faces away from the beginning of the line, and each prisoner can see all the hats in front of him, and none of the hats behind. Starting from the beginning of the line, each prisoner must correctly identify the color of his hat or he is killed on the spot. As before, the prisoners have a chance to meet beforehand, but unlike before, once in line, no prisoner can hear what the other prisoners say. The question is, is there a way to ensure that only finitely many prisoners are killed?

Solution

It uses Axiom of choice in a special way. Before night of execution, prisoners will create some equivalent classes among the set of all possible infinitely long binary strings. Let us say two infinite strings s1 and s2 are related, if they differ at finitely many places, and are same otherwise. This creates the equivalence relation. Thus they create all equivalence class. From each class, we mark one representative string.

Now, a person pi looks ahead, and instantly knows his equivalent class & recalls the representative string S. When his turn comes, he speaks the ith digit of S as his guess. Since S and actual (current) string only differ upto a finite (say N) number of position. Thus only atmost N people may die using this technique.

Now, first person will count number of places S differs from current string, and answer it in modulo 2. Based on answer of first person, second person can deduce his hat color. now the third person can deduce his color and so on. Except possibly for the first person, everyone will guess correctly.

Now, a person pi looks ahead, and instantly knows his equivalent class & recalls the representative string S. When his turn comes, he speaks the ith digit of S as his guess. Since S and actual (current) string only differ upto a finite (say N) number of position. Thus only atmost N people may die using this technique.

Now, first person will count number of places S differs from current string, and answer it in modulo 2. Based on answer of first person, second person can deduce his hat color. now the third person can deduce his color and so on. Except possibly for the first person, everyone will guess correctly.

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