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**Spoiler (click to open)**]Although it's easier to solve this for 2 prisoners/colors than for 10, solving for 10 is actually easier than solving for 9.

I'll probably post the answer early next week.

- Colored Hat Puzzle - Last Call
- No one has yet messaged me with a correct answer to the Colored Hat Puzzle, so here's Hint #3.

[**Spoiler (click to open)**]Although it's easier to solve this for 2 prisoners/colors than for 10, solving for 10 is actually easier than solving for 9.

I'll probably post the answer early next week.

corwyn_apkgbooklogOkay, now to figure out how to expand it to more than 2 prisoners.

kgbooklogSo far:[

Spoiler (click to open)]1 goes free if all hats are different, 2 goes free if 1&2 are same and others different, 3 goes free if 1&3 are same and others different, 4 goes free if 1&4 are same and others different. I also found a way for 1 to go free half the time when 2&3 are same and others different, but can't think of how to make 4 go free the other half of the time.Any hints on how to simplify things by combining cases? And is it necessary to number the colors, or am I missing something easier?

Edited at 2013-01-13 06:53 pm (UTC)alexx_kay[

Spoiler (click to open)]Numbering the colors is one route to combining cases to simplify. It isn't *strictly* necessary to do so, but for N=10, I can't think of any other way that doesn't require that the prisoners effectively all have eidetic memory. (or they make huge complex cheat-sheets, which probably take more than 3 hours to prepare.)kgbooklog[

Spoiler (click to open)]Number the prisoners 0 to N-1, and number the colors 0 to N-1. Each prisoner calculates what color his hat needs to be in order that the sum of all hats (including his own) modulo N equals his prisoner number.I may have had an easier time reaching this point if you hadn't given me hint #3 (I wouldn't have been so willing to give up on N=3).