![]() If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. So: : If there are two blue-eyed people on the island, they will each leave the 2nd night. I must have blue eyes." And each leaves the second night. And if he's the only person, by THEOREM 1 he will leave tonight." They each wait and see, and when neither of them leave the first night, each realizes "My HYPOTHESIS 1 was incorrect. ![]() ![]() They will each realize that " if I don't have blue eyes, then If there are two blue-eyed people, they will each look at the other. So: If there is oneīlue-eyed person, he leaves the first night. He looksĪround and sees no one else, and knows he should leave. Knows he's the only one the Guru could be talking about. You can show that he obviously leaves the first night, because he If you consider the case of just one blue-eyed person on the island, If you're really confused by something, let me know. It's correct, but the explanation/wording might not be the best. Note - while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It's pretty convoluted logic and it took me a while toīelieve the solution, but here's a rough guide to how to get there. The answer is that on the 100th day, all 100 blue-eyed people If you like formal logic, graph theory, sappy romance, bitter sarcasm, puns, or landscape art, check out my webcomic, xkcd. Blue Eyes - The Hardest Logic Puzzle in the World - Solution
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