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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
