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Originally posted by Ramo
A bar. The person who told me it seemed to imply the same.
A bar. The person who told me it seemed to imply the same.
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Were you and/or the person who told it to you drinking heavily enough that an important additional detail was lost somewhere between then and now?THEY!!111 OMG WTF LOL LET DA NOMADS AND TEH S3D3NTARY PEOPLA BOTH MAEK BITER AXP3REINCES
AND TEH GRAAT SINS OF THERE [DOCTRINAL] INOVATIONS BQU3ATH3D SMAL
AND!!1!11!!! LOL JUST IN CAES A DISPUTANT CALS U 2 DISPUT3 ABOUT THEYRE CLAMES
DO NOT THAN DISPUT3 ON THEM 3XCAPT BY WAY OF AN 3XTARNAL DISPUTA!!!!11!! WTF -
I was drinking heavily. But to demonstrate that I'm not full of ****, I'll show you a solution for N = 2:
Person 1 guesses Person 2's color
Person 2 guesses the opposite of Person 1's color"Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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That much is obvious. I'm considering a sum and modulo or something like that... hm.Comment
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You're on the right track (or at least one of them)..."Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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one person looks at the other ones hat and will say his color. then when its that guys turn... HE SAYS THE SAME THINGThe Wizard of AAHZComment
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The function has to take into account the order. Consider the cases that all but one are red and the last blue, and all be one are red and the last green. Any function of just "what is everyone else wearing" would have them all make the same guess (and not telling the last guy which color he was), so it must also be a function of their respective positions.Comment
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Is the person allowed to look at all of the other hats so that he knows the distribution of the colours? If each individual person can do this, each person simply notes which colour is the most common and everyone says that. After the hats have been given out the people don;t have to communicate.You just wasted six ... no, seven ... seconds of your life reading this sentence.Comment
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i see what you mean Kuciwalker...The Wizard of AAHZComment
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Though that doesn't work if number of people = number of colours, but in reality you're screwed if that happened anyway.You just wasted six ... no, seven ... seconds of your life reading this sentence.Comment
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You have no information about the distribution.Originally posted by Krill
Is the person allowed to look at all of the other hats so that he knows the distribution of the colours? If each individual person can do this, each person simply notes which colour is the most common and everyone says that. After the hats have been given out the people don;t have to communicate.Comment
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so they cant hear eachother when its guessing time? damn...The Wizard of AAHZComment
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if they can see eachothers hat, then they can read lips... and one will look at the guys hat next to him and lip- RED very distinctly.. then when its the next guys turn... he simply says RED.
edit- this is assuming the hat is red... it could be green, etc...The Wizard of AAHZComment
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Spoiler:Comment
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theoretically my way COULD work... i played DnD for 15 years... im just not good with numbers...
The Wizard of AAHZComment
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"Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-BokononComment
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