Fri Nov 21, 2014 3:05 pm by JanPil |
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It seems to me that for 3 prisoner the leader must flip the first switch down four times. Because if the switch was up in beginning he switch it down and the same other prisoner can switch it up two times before the third prisoner have been in the switch room. Only if it was down in beginning three is enough
It also seems to me that a general solution for n prisoners is 2(n-1)
which fit the solution given for n=23
Best regards
Jan |
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Thu Jul 21, 2016 2:50 pm by bathfilms |
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Could someone please post the question?
Ta |
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Thu Aug 25, 2016 6:08 am by trident |
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The prisoners will mark the first person to go inside as a flag/leader(say A).
A will flip the first switch(switch 1) down if already up otherwise leave it as it is and start the counter with 1.. Now when a new prisoner enters the room he will flip that switch up which can be down only by A. If a person enters the room for the second or subsequent time he doesn't need to touch switch 1, he need to play with switch 2 only.
Prisoner who enters the room for the first time can only flip the switch 1 up. When A enters the room next time he will again flip the switch 1 down & increase the counter by 1 and same process goes on till the count reaches to 23.
This should work, correct me if I'm wrong. |
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