Man, I meant to blog one more time before the go congress, but blogging is so hard. Anyway, solution time. Puzzle from last time was another hat one. The solution is rather similar to puzzles I have posted here before, but it still took me a little while to get it.
As always, it is easier to start with the two person, two colour case. Suppose the two colours are white and black, and we have two rooms. Label the two rooms white and black and go into the room that corresponds to the hat colour of the other person. If you both have the same colour, you will go to the same room, but if you have different hat colours you will go to different rooms.
Naturally, a strategy for the more general case must just be a function that maps from the hats you see to the room you go to. Lets try two hat colours and N people next.
We have two rooms again, call one room even and one odd. If you see an odd number of black hats, go to the odd room, if you see an even number of black hats, go to the even room. Then, if the total number of black hats is even, white-hatted people will go to the even room and black-hatted people will go to the odd room. Naturally it works the other way if the total number of black hats is odd. One could also arrive at this strategy by assigning the rooms numbers 0 and 1 and call white 0 and black 1. Then add the hat colours you see mod 2 and go to that room.
This generalizes to the strategy for N people and K colours. First number the hat colours 0 through K-1 and number the rooms 0 through K-1. Then add the hats colours you see together mod K and go to that room. We can see this strategy works by considering the sum of all the hats (call that S), then a person with hat colour x will go to the room S-x mod K. If two people with hat colours x and y go to the same room, then S-x=S-y mod K and so x=y mod K and so x=y, two people go to the same room means they have the same hat colour, so this solution works.
Generalizing this to the case of countably infinite hat colours is trivial, you can just use the same solution and forget the mod K stuff. Each hat is a natural number, and add together the hats you see and go to that room.
With infinite logicians it is a bit more tricky. In this case there are infinitely many of them to add together, so you can't just go to the 'infinity room' and be done with it. Since we have been handed the Axiom of Choice, we might as well make use of it.
What can one do with the Axiom of Choice? In most cases, the most efficient thing to do is the same old equivalence class thing. Let us construct an equivalence class of hat sequences. Two sequences are said to be the same if they differ in only finitely many places. Now we can use the axiom of choice to construct a set S which contains exactly one element from each equivalence class. Since each logician can see all but their own hat, they can all determine what equivalence class we are in, and thus can select an element x in S that differs from the real sequence of hats in finitely many places. We will add together those finitely many hats to construct our solution.
Alright, so everybody sees most of the full sequence of hats yi and they all have agreed on a sequence xi that differs from the sequence yi in finitely many places, they must now choose a room to go to. Let us suppose that all the hats are correct. Then it is simple, you just go to the room that xi says you go and it will all work (the hats and the rooms have both been numbered as naturals). Now, let us suppose that person j is wearing the wrong hat colour, that is xj ≠ yj but all the other xi=yi. Now he is going to go to the room suggested by the hat xj, which is xj-yj too many, so everybody else must also go xj-yj too many also. It is possible for this number to be negative, so first we must number the rooms as integers instead of naturals. So, the strategy is that when you are person n and you see hat j wrong, you go to room xn+xj-yj. Naturally, if you see more than one wrong hat, just add up xi-yi of all the hats that are wrong (there are only finitely many) and add that to your own xn and go to that room.
Tanya Khovanova has a complaint with this solution that even though the Axiom of Choice allows you to select the set S that we will use for the solution, it does not guarantee that there is a method of distributing the set S to a collection of people. If they cannot distribute the set S, they cannot guarantee that they all have the same one (and I would suppose they almost certainly do not, but that is a bit of an odd statement to make). I am not sure if I agree with her on this point, but I generally am against the Axiom of Choice in the first place, so it is a bit of a non-issue for me. At any rate, I suppose one could distinguish another axiom where you have the Axiom of Choice, and you may actually describe the set it gives you, then this puzzle simply needs that more powerful axiom, to solve.
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