There seems to me to be a contradiction that everyone can observe something (and everyone know it can be observed) and still believing that somehow someone may not be able to observe it. I, however, am failing to grasp why the fact that everyone knows that blue-eyes are directly observable does not provide any information to anybody. They are a very good explanation of the logic to arrive to the general solution. Great answers by everyone who have commented or answered. In summary, the state of "What A knows B knows C knows D sees there are 0 blue-eyed people" is not possible given that "ABCD knows ABCD knows there are >0 blue-eyed people" and only possible states should be taken into consideration. If you see 3 blue-eyed people, you can prove that everyone can prove that everyone can prove there are >0 blue-eyed people If you see 2 blue-eyed people, you can prove that everyone can prove there are >0 blue-eyed people If you see 1 blue-eyed person, you can prove that there are >0 blue-eyed people If it can be proven that everybody knows that no one thinks there are no blue-eyed people then the oracle does not provide extra information and no one leaves the island
Therefore, I propose the following proof: This combined induction covers all purely sequential inductions past k=4 blue eyed peoples. Which when combined with the sequential induction collapses toĪ knows ABCD knows ABCD knows >0 blue eyes The generally accepted solution states that all the blue-eyed people leave on the 100th day by the inductive reasoning of
Simply saying "I count at least one blue-eyed person on this islandĪnd lastly, the answer is not "no one leaves." The Guru is not making eye contact with anyone in particular she's Wording or anyone lying or guessing, and it doesn't involve peopleĭoing something silly like creating a sign language or doing genetics. Trick question, and the answer is logical. There are no mirrors or reflecting surfaces, nothing dumb. Who leaves the island, and on what night? The Guru is allowed to speak once (let's say at noon), on one day inĪll their endless years on the island. Or 100īrown, 99 blue, and he could have red eyes. (and one with green), but that does not tell him his own eye color asįar as he knows the totals could be 101 brown and 99 blue. Person can see 100 people with brown eyes and 99 people with blue eyes On this island there are 100 blue-eyed people, 100 brown-eyed people,Īnd the Guru (she happens to have green eyes). Island knows all the rules in this paragraph. Themselves), but they cannot otherwise communicate. Everyone can see everyone else at all times and keeps aĬount of the number of people they see with each eye color (excluding Any islanders who haveįigured out the color of their own eyes then leave the island, and the Night at midnight, a ferry stops at the island. They areĪll perfect logicians - if a conclusion can be logically deduced,
The Blue-eyes Riddle is commonly expressed asĪ group of people with assorted eye colors live on an island. I would like help understanding my flawed logic in my following reasoning:
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