1wingangel > RE: The Official Riddle Thread (7/15/2009 0:57:54)

The dreaded... BLUE EYES LOGIC PUZZLE quote:
From xkcd.com/blue_eyes.html A group of people with assorted eye colors live on an island. They are all perfect logicians  if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph. On this island there are 100 blueeyed people, 100 browneyed people, and the Guru (she happens to have green eyes). So any given blueeyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes. The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following: "I can see someone who has blue eyes." Who leaves the island, and on what night? There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blueeyed person on this island who isn't me." And lastly, the answer is not "no one leaves." I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know. A word of warning: The answer is not simple. This is an exercise in serious logic, not a lateral thinking riddle. There is not a quickandeasy answer, and really understanding it takes some effort. HINTS spoiler:
1. What is the quantified piece of information that the Guru provides that each person did not already have? 2. Each person knows, from the beginning, that there are no less than 99 blueeyed people on the island. How, then, is considering the 1 and 2person cases relevant, if they can all rule them out immediately as possibilities? 3. Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know? ANSWER spoiler:
The answer is that on the 100th day, all 100 blueeyed people will leave. It's pretty convoluted logic and it took me a while to believe the solution, but here's a rough guide to how to get there. 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 throwntogether. It's correct, but the explanation/wording might not be the best. If you're really confused by something, let me know. If you consider the case of just one blueeyed person on the island, you can show that he obviously leaves the first night, because he knows he's the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: [THEOREM 1] If there is one blueeyed person, he leaves the first night. If there are two blueeyed people, they will each look at the other. They will each realize that "if I don't have blue eyes [HYPOTHESIS 1], then that guy is the only blueeyed person. 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. I must have blue eyes." And each leaves the second night. So: [THEOREM 2]: If there are two blueeyed people on the island, they will each leave the 2nd night. If there are three blueeyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities  "I have blue eyes" or "I don't have blue eyes." He will know that if he doesn't have blue eyes, there are only two blueeyed people on the island  the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes  THEOREM 2 says that if he didn't, the other guys would have left. When he sees that they didn't, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave. This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they'll each wait 99 days, see that the rest of the group hasn't gone anywhere, and on the 100th night, they all leave. From xkcd.com/solution



