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The prisoners are back for more puzzles! And this time there's a light bulb in the mix.
There are 100 prisoners, each in their own isolated rooms. There's another special room with a light bulb and a switch, and nothing else. The light bulb is initially off.
One at a time, a prisoner is chosen at random to go into the special room. The prisoner can choose to either flip the switch or not. The prisoner will then leave the room and a guard will ask him whether he he thinks all 100 prisoners have been in the room at least once. The prisoner can either answer "yes" or "don't know".
If he answers yes and he's correct, all the prisoners go free.
If he answers yes and he's wrong, all the prisoners are killed.
If he answers don't know, then the prisoner goes back to his room and the process continues.
Before the prisoners are put in their rooms, they get to decide beforehand on a strategy. What strategy should the prisoners use to assure the survival of the prisoners?
Key Point: The prisoners can go into the room multiple times, and the order is random. For example, prisoner 1 could potentially go into the room 10 times before prisoner 5 goes in even once.
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+ Show Spoiler +If it is your first time in the room, flip the switch. Wait until you have been brought in for the hundred thousanth time and all you have seen is an off light-bulb. On the hundred thousanth time, declare that all the prisoners have been there. The laws of chance say that it's just a little bit unlikely that somebody would be chosen a hundred thousand times before somebody else was chose once.
Better make that a million.
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Thats laws of odds though, still a small chance you could get shot in the face.
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+ Show Spoiler +One solution I managed to come up with...it's safe too.
Before 100 days, leave the switch off if it's your first time in. Then if someone comes in two times, they turn it on. The next person to come in flicks the switch off. He now knows that the number of days before him is the number of unique people that have been in the room. After that, you use a new rule. Leave the light on if you've been in the room. Turn it off if it's your first time. Now, whoever has been in the room two times knows that he has the remainder to count. So he counts off each time he goes in the room if the light is off, until it's been 100 times.
GG? Takes too long I think. Should be another solution.
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Isn't this very similar to the other prisoner problem with the thingy they turned up and down, but then it was more complicated, because they had that stupid king that could mess with them?
The same solution would work, right?
I dont know what to say about the recuring apearance of prisoners and light bulbs either. :o
edit: leejas:
+ Show Spoiler +So the "yes" condition would be that you had seen the light being switched 100 times? But if two prisoners is in the room between two of your visits, you will not see their switches, and you will never get to 100. You want a solution that always works. And I dont think "but this will take a lot of time" is a flaw. These prisoners in general have A LOT of time on their hands... 
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On March 12 2008 17:12 EmeraldSparks wrote:+ Show Spoiler +If it is your first time in the room, flip the switch. Wait until you have been brought in for the hundred thousanth time and all you have seen is an off light-bulb. On the hundred thousanth time, declare that all the prisoners have been there. The laws of chance say that it's just a little bit unlikely that somebody would be chosen a hundred thousand times before somebody else was chose once.
Better make that a million.
+ Show Spoiler +The riddle is not correct. The prisoners may communicate and discuss a plan beforehand. Your solution is not possible, because you will just have 1 prisoner / day which means you would have to wait 100000 days with your solution or even 1000000 which is equal to 273 years (100000) or 2739 years (1000000). Given the fact that they may discuss beforehand, one person is chosen as the counter. If the light switch is on, no prisoner will touch it. If it is off, and a prisoner who has never flicked the switch enters, then he flicks the switch on. It stays like that until the counter returns. The counter never flicks it up, only flicks it down. Once he notices that it's been flicked up 99 times, he says that 100 prisoners have visited, and they go free. This is the solution.. ;o NOTE: It could still take years.. or never happen if its really random 
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The problem, lee, is that you don't know that this is going to be once a day. Slithe didn't specify an interval. You can only break it into two phases like that with either a set interval between iterations or if every prisoner knows what iteration they are on.
Edit: Grammar.
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Sorry, it's 5am here and I'm studying for an exam :D.
Anyway, you're gonna have to do it the other way then.
Leave the light on all the time. If someone new comes in, they turn it off. If you go in, the light is off, someone new was there, thus, it's one note. After that, you have to count off how many times you've been in the room to 100. After that, everyone's been there.
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On March 12 2008 17:43 G.s)NarutO wrote:Show nested quote +On March 12 2008 17:12 EmeraldSparks wrote:+ Show Spoiler +If it is your first time in the room, flip the switch. Wait until you have been brought in for the hundred thousanth time and all you have seen is an off light-bulb. On the hundred thousanth time, declare that all the prisoners have been there. The laws of chance say that it's just a little bit unlikely that somebody would be chosen a hundred thousand times before somebody else was chose once.
Better make that a million. + Show Spoiler +The riddle is not correct. The prisoners may communicate and discuss a plan beforehand. Your solution is not possible, because you will just have 1 prisoner / day which means you would have to wait 100000 days with your solution or even 1000000 which is equal to 273 years (100000) or 2739 years (1000000). Given the fact that they may discuss beforehand, one person is chosen as the counter. If the light switch is on, no prisoner will touch it. If it is off, and a prisoner who has never flicked the switch enters, then he flicks the switch on. It stays like that until the counter returns. The counter never flicks it up, only flicks it down. Once he notices that it's been flicked up 99 times, he says that 100 prisoners have visited, and they go free. This is the solution.. ;o NOTE: It could still take years.. or never happen if its really random
nice one
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Man, saw this on the william wu riddle website years ago
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It feels like it's the same riddle all the time.
Prisoners are presented with a a set of rules where ultimately their behavior leads to death or freedom. They get to discuss a strategy to beat the system for their freedom. Thereafter, they are instructed not to communicate directly. The system often has a certain set of artifacts such as light switches, light bulbs and rooms which they have to interact with. Any other interaction is not allowed. Strangest of all, the system they have to beat is always beatable. But in such a situation, can the guards be trusted? Why would they make a game like this that could guarantee the prisoners freedom? Can the rules be trusted? Can the prisoners be trusted to follow the strategy? Can the prisoners be trusted not to cheat? These are all reasons why people should not be used in riddle, unless the solution is about thinking outside the box.
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On March 12 2008 20:00 stenole wrote:
It feels like it's the same riddle all the time.
Prisoners are presented with a a set of rules where ultimately their behavior leads to death or freedom. They get to discuss a strategy to beat the system for their freedom. Thereafter, they are instructed not to communicate directly. The system often has a certain set of artifacts such as light switches, light bulbs and rooms which they have to interact with. Any other interaction is not allowed. Strangest of all, the system they have to beat is always beatable. But in such a situation, can the guards be trusted? Why would they make a game like this that could guarantee the prisoners freedom? Can the rules be trusted? Can the prisoners be trusted to follow the strategy? Can the prisoners be trusted not to cheat? These are all reasons why people should not be used in riddle, unless the solution is about thinking outside the box.
It's just a riddle not real life situation ffs.
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I've not read all the posts but this does it:
+ Show Spoiler +
Pick one man, DAVE,
if you are not DAVE:
if it's your first time in the room: flick the on if it's off, leave it on if it's on
if you've been there before: leave it
if you're DAVE:
if it's on count:=count +1 and switch it off,
if it's off leave it
when you count to 100 (or 99 really) you are done
edit: G.s)NarutO got it first, gg
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Congratulations to Naruto, I think he got it first.
@betaben: That puzzle is a pretty good one. I've heard a very similar version with the key difference that there is no time limit, and the question is "How many days does it take for the diseased ones to kill themselves off". I dunno if I would call it "The hardest logic puzzle" though.
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the strategy is this: if you flipped the switch and the gaurd asks you if you flipped it, say yes. if + Show Spoiler +you didn't flip it say don't know -_- how easy xp
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+ Show Spoiler +the strategy is this: if you flipped the switch and the gaurd asks you if you flipped it, say yes. if you didn't flip it, don't say that you flipped the switch! -_- easy xp
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On March 12 2008 20:00 stenole wrote:
It feels like it's the same riddle all the time.
Prisoners are presented with a a set of rules where ultimately their behavior leads to death or freedom. They get to discuss a strategy to beat the system for their freedom. Thereafter, they are instructed not to communicate directly. The system often has a certain set of artifacts such as light switches, light bulbs and rooms which they have to interact with. Any other interaction is not allowed. Strangest of all, the system they have to beat is always beatable. But in such a situation, can the guards be trusted? Why would they make a game like this that could guarantee the prisoners freedom? Can the rules be trusted? Can the prisoners be trusted to follow the strategy? Can the prisoners be trusted not to cheat? These are all reasons why people should not be used in riddle, unless the solution is about thinking outside the box.
idiot, it's a math riddle. the setup is similarbecause you don't want to say "x y to the squre root of 2 lbahlbahlbha, solve it." assume! assume the rules acn be trusted assume th eprisoners can follow the same strategy assume they can be trusted not to cheat geez. it's not thinking about outside the box. even my though answers always are, im just kidding. u couldn't even figure out the question itself. -_-
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On March 12 2008 23:33 drift0ut wrote:I've not read all the posts but this does it: + Show Spoiler +
Pick one man, DAVE,
if you are not DAVE:
if it's your first time in the room: flick the on if it's off, leave it on if it's on
if you've been there before: leave it
if you're DAVE:
if it's on count:=count +1 and switch it off,
if it's off leave it
when you count to 100 (or 99 really) you are done
edit: G.s)NarutO got it first, gg
+ Show Spoiler +
A minor correction is that if you are not DAVE you leave it not if you have been there before, but if you have switched the light before. If you leave the rule as it is and SAM's first time in the room is when the light is on, he will never flip the switch.
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I didn't think this puzzle was the hardest puzzle ever
.. this is my solution (sorry if I donn't explain it very well)
+ Show Spoiler +
The answer, as far as I can tell, is to leave as soon as you are sure you are infected.
So, if a single monk is infected, on the first evening he will see all of his fellow monks and see that none of them are infected. Knowing that at least one person is infected, he will assume it is himself and then leave.
If two monks are infected, they will each see one other person infected on the first night. On the second night, upon seeing that the other infected person has not left, they must assume that that person also saw one infected person (themself.) And so on the second night, both of these monks would leave.
If three monks are infected they will each see two infected people, and on the third night, when no one has left, they will realise that each of the other two people can also see two people (hence making themself the third infected person), and so these three would leave.
And so on...
(i.e. if you see x infected people, and it reaches night x+1 and they are still all there, you must be infected and so you leave.)
It takes 14 nights to work out how many monks are infected, so there must have been 14 infected people.
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On March 13 2008 06:39 LiAlH4 wrote:I didn't think this puzzle was the hardest puzzle ever .. this is my solution (sorry if I donn't explain it very well) + Show Spoiler +
The answer, as far as I can tell, is to leave as soon as you are sure you are infected.
So, if a single monk is infected, on the first evening he will see all of his fellow monks and see that none of them are infected. Knowing that at least one person is infected, he will assume it is himself and then leave.
If two monks are infected, they will each see one other person infected on the first night. On the second night, upon seeing that the other infected person has not left, they must assume that that person also saw one infected person (themself.) And so on the second night, both of these monks would leave.
If three monks are infected they will each see two infected people, and on the third night, when no one has left, they will realise that each of the other two people can also see two people (hence making themself the third infected person), and so these three would leave.
And so on...
(i.e. if you see x infected people, and it reaches night x+1 and they are still all there, you must be infected and so you leave.)
It takes 14 nights to work out how many monks are infected, so there must have been 14 infected people.
+ Show Spoiler + But how would that single monk know in the first place that he is infected?
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On March 13 2008 12:31 omfghi2u2 wrote:Show nested quote +On March 13 2008 06:39 LiAlH4 wrote:I didn't think this puzzle was the hardest puzzle ever .. this is my solution (sorry if I donn't explain it very well) + Show Spoiler +
The answer, as far as I can tell, is to leave as soon as you are sure you are infected.
So, if a single monk is infected, on the first evening he will see all of his fellow monks and see that none of them are infected. Knowing that at least one person is infected, he will assume it is himself and then leave.
If two monks are infected, they will each see one other person infected on the first night. On the second night, upon seeing that the other infected person has not left, they must assume that that person also saw one infected person (themself.) And so on the second night, both of these monks would leave.
If three monks are infected they will each see two infected people, and on the third night, when no one has left, they will realise that each of the other two people can also see two people (hence making themself the third infected person), and so these three would leave.
And so on...
(i.e. if you see x infected people, and it reaches night x+1 and they are still all there, you must be infected and so you leave.)
It takes 14 nights to work out how many monks are infected, so there must have been 14 infected people.
+ Show Spoiler + But how would that single monk know in the first place that he is infected?
+ Show Spoiler +because somebody is, that's a given. if he looks at ev eryone else and sees nobody, it's gotta be him.
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rule 8. this is your first night at light club, you HAVE TO flip
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Haha.
If there are 100 of them, alot of them must have been breaking the first rule
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EPT
