Timeline for N logicians wearing hats of N colors
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jun 4, 2015 at 0:50 | history | edited | Mike Earnest |
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| Apr 21, 2015 at 10:03 | answer | added | LogicianWithAHat | timeline score: 0 | |
| Feb 1, 2015 at 17:52 | history | edited | Gamow |
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| Jan 8, 2015 at 13:49 | comment | added | Gilles 'SO- stop being evil' | @ghosts_in_the_code They call out simultaneously, i.e. what one logician says cannot depend on what another one has said. | |
| Jan 8, 2015 at 13:40 | comment | added | ghosts_in_the_code | Are you sure they call out simultaneously, not 1 at a time? | |
| Nov 16, 2014 at 10:54 | review | Close votes | |||
| Nov 17, 2014 at 7:12 | |||||
| Jun 13, 2014 at 19:26 | history | edited | Gilles 'SO- stop being evil' | CC BY-SA 3.0 |
reverted mathjax delimiter
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| Jun 2, 2014 at 21:59 | vote | accept | Gilles 'SO- stop being evil' | ||
| Jun 2, 2014 at 18:34 | answer | added | kaine | timeline score: 9 | |
| May 24, 2014 at 23:14 | history | edited | Gilles 'SO- stop being evil' | CC BY-SA 3.0 |
add the mathematical difficulty in spoiler text
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| May 24, 2014 at 9:02 | answer | added | klm123 | timeline score: 7 | |
| May 21, 2014 at 19:20 | answer | added | kaine | timeline score: 8 | |
| May 16, 2014 at 13:20 | comment | added | Jaydles | @gilles, +1 - I thought I remembered this one from my past, but put a good hour of fun into solving it for cases where N>2 last night. | |
| May 16, 2014 at 2:38 | comment | added | Ross Millikan | I don't think the amount of modular arithmetic that is required for the linked solution is very tough. It is beyond basic addition, but not far. I don't know a solution that involves less math. | |
| May 16, 2014 at 2:21 | answer | added | kBisla | timeline score: -1 | |
| May 16, 2014 at 0:03 | comment | added | user88 | Anyway, for N=2, the solution seems to be that one person will always say the same colour as the one he sees, and one will always say the opposite colour. | |
| May 16, 2014 at 0:02 | comment | added | Gilles 'SO- stop being evil' | It doesn't matter if one of them gets it wrong: they win as long as at least one color is called correctly. | |
| May 16, 2014 at 0:01 | comment | added | user88 | I just wanted to make sure it's not one of those where if one of them gets it wrong, they also all get killed, and some of them have to choose not to say anything. | |
| May 15, 2014 at 23:59 | comment | added | Gilles 'SO- stop being evil' | @JoeZ. I'm not interested in probabilistic solutions, only in a guaranteed solution (which exists). If the logicians lose, they'll be killed. And none of them are suicidal. | |
| May 15, 2014 at 23:59 | history | edited | Gilles 'SO- stop being evil' | CC BY-SA 3.0 |
added 161 characters in body
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| May 15, 2014 at 23:53 | comment | added | user88 | Well, as a preliminary solution, it seems as if every logician just calls out randomly, they have about a 63.2% chance of winning anyway. You're looking for one that works 100%, though, right? | |
| May 15, 2014 at 2:43 | history | asked | Gilles 'SO- stop being evil' | CC BY-SA 3.0 |