EPTMath Puzzle - 100 Guards - Page 2
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Bill Murray
United States9292 Posts
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FatChicksUnited
Canada214 Posts
+ Show Spoiler +
+ Show Spoiler + First post! | ||
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Ftrunkz
Australia2474 Posts
On October 08 2010 18:52 FatChicksUnited wrote: Potential Hard Solution: + Show Spoiler +
+ Show Spoiler + First post! whoa, this is pretty mathematical but i understand the idea behind what you've done here, that's pretty cool and i would've never figured this out, gj (asuming you're correct :D) | ||
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Sadir
Vatican City State1176 Posts
Potential Hard Solution: + Show Spoiler +
+ Show Spoiler + First post! pretty damn impressive really, I got to the situation with the P1-G1 and P2-G2 (while G1 possible as well), but the information handling....puh...I couldnt think of how to solve this pretty good  chapeau EDIT: I have to rethink that, I am not sure any more if I get it =) EDIT2: ok I definitely got it now hehe | ||
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garbanzo
United States4046 Posts
If they can differentiate are they able to agree upon a unique way to number them in the strategy phase? I mean in a way that once they are imprisoned then they will know when they are guarded by guard #1 or guard #38. | ||
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FatChicksUnited
Canada214 Posts
+ Show Spoiler + I can't take credit actually. A few months ago Slithe or someone else put another puzzle of similar nature up, and the solution also was to encode information using modular arithmetic. Reading about it a few months ago totally blew my mind my as well  . | ||
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Bill Murray
United States9292 Posts
prisoner A could dismiss one, and depending on the sign, wait 1-12 days to dismiss the last guard. They would know when this was occuring, because the guards would have a gap in between when they guarded the prisoners. They eliminate down to 2 guards. At that point it would be 1, 2, 1, 2, 1, 2, 1, 2 for them both when the prisoner dismisses the last guard, it would go from 1, 0 , 1, 0, to 1, 1, 1, 1, and he could correspond it with days of the gap relating to sign, because he would go back to 1, 2, 1, 2. The other prisoner wouldn't have a guard to guess, and would therefore "not get it wrong" When he dismisses one, where the guards switch, the other guard will be stuck at his cell everyday, though he doesn't want to be stuck there whatsoever. when he waits the corresponding amount of days to dismiss him, the other prisoner will actually gain a guard. He would know said guard's sign based upon silent communication of days, and the other dude wouldn't even have a guard to guess and get wrong. So when he dismisses the guard, they will both be locked on the prisoner he waits 1 day for ares, 12 days for pisces on the 13th day, prisoner 2 would know said guard was pisces, for instance, because he would be back in rotation at his cell with the other guard whose sign he doesn't know | ||
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Suspense
United States500 Posts
+ Show Spoiler + If there are 100 guards and they must answer what their sign is the first time then that is a positive way of knowing one guards sign. Also they are not to guard the same prisoners cell after that, unless no one else is available. The prisoners would have to agree on a sign that when they heard it they would start the count. This being said if on the first day one prisoners would ask the guards what their signs are and it was Leo.. Well heres and example on the first day to make things easier. In the event that Leo is answered the day count starts at one. Now with 100 guards both guarding each cell when possible it will give you 54 days without a guard being repeated on that prisoner. On the 55th day the prisoner who did not ask the question on the first day would be guarded by the guard that guarded the other prisoner on the first day. Making the other prisoner positive that Leo is his sign. Perhaps this only works if the sign is hit on the first day however.. but that would still give the prisoners a 1 and 6 shot I believe. I'm not sure if the rotation I have in this solution is how it has to work, but i don't see it stated otherwise, so in this scenario it would Either way I'm way too tired to try and work this out.. This was just what I initial thought I might read some of the other spoilers to see what other people have come up with Bah nvm mine.. it relies on them speaking to each other | ||
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Bill Murray
United States9292 Posts
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FiBsTeR
United States415 Posts
On October 08 2010 19:19 Bill Murray wrote: They eliminate down to 2 guards. At that point it would be 1, 2, 1, 2, 1, 2, 1, 2 for them both You can't guarantee that the same two guards are left for each prisoner. P1 could be left with G1 and G2 and P2 left with G3 and G4. | ||
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Slithe
United States985 Posts
On October 08 2010 18:52 FatChicksUnited wrote: Potential Hard Solution: + Show Spoiler +
+ Show Spoiler + First post! Congratulations, you have found the answer. It's also pretty much the fastest answer that you can have in terms of days. I actually put a little bit of spread between this problem and the 7 hats problem because they use the similar trick of modular arithmetic, and I was trying to make sure people's brains would have to rediscover this neat idea. | ||
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Yukidasu
Australia125 Posts
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Bill Murray
United States9292 Posts
On October 08 2010 23:21 FiBsTeR wrote: You can't guarantee that the same two guards are left for each prisoner. P1 could be left with G1 and G2 and P2 left with G3 and G4. easy. short, bald, brown eyes, etc distinguishable you can pre-game strategy | ||
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Slithe
United States985 Posts
On October 09 2010 15:08 Bill Murray wrote: easy. short, bald, brown eyes, etc distinguishable you can pre-game strategy This is considered beyond the abilities of the prisoners. To make the problem more clear, I will say that all guards look the same, but they are arbitrarily numbered so that a prisoner can differentiate between them. However, they wear different numbers for each prisoner, so they cannot use number to jointly identify a guard. | ||
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