# 5 Star Logic Problem (100 prisoners and a light bulb)

Once a prisoner is called, he goes in and he can either turn the light on or off, it doesn't matter. Once he's done this, he'll kill himself. Since he can't be called again, the warden must call another person. That person will go in, and then kill himself. This will continue, for about 99 days, and on the 100th day the last person will enter and say, "all 100 have entered if you include me." Pretty gruesome, I know, but it works (I think).

Since he can't be called again, the warden must call another person.
I'm not sure that works-- I don't see anything in the OP that restricts the warden into choosing living prisoners:

Everyday, the warden picks a prisoner at random, and that prisoner goes to the central living room.
I suppose it would work if one assumed that a dead person was incapable of being a prisoner and was instead a "former prisoner".

DaveE

My Solution:

1. First prisoner who goes to central room acts as counter and he leaves the switch as it is i.e off
2. Different prisoner goes to central room turns the light on if it is switched off and he/she
should not turn it on/off from then, even he/she gets chance to enter central room again(goes to infinity)..
3. Different prisoner goes to central room leaves the light as it is, if it is switched on
4. So whenever the first prisoner gets chance to go to central room again,he will count 1 if
the light is switched on and make the switch off.
5. when the first prisoner who is counting gets 99,then he will tell that all 100 prisoners
have been to the central room

This method takes a long long time but will work. please tell me if there are any mistakes
There is a common mistake in many of these plans, namely the assumption that a prisoner will know that he is "first" (or WHO is first).

Doesn't need to know if he's first, once he's been in the living room, he kills himself. This goes on for 100 days. The only person left at the last 100th day is the only one who will walk free.

I'm not sure that works-- I don't see anything in the OP that restricts the warden into choosing living prisoners:

I suppose it would work if one assumed that a dead person was incapable of being a prisoner and was instead a "former prisoner".

DaveE
If he's dead, he's got to buried you know. It's not like you let him rot in his cell.

If he's dead, he's got to buried you know. It's not like you let him rot in his cell.
I figure they'll just cremate him and take his urn into the room.

DaveE

Begin with Phlegmy's #71 post of even/odd days for (say) 100 day intervals. The first repeat prisoner will reverse the order. The SECOND repeat prisoner becomes counter and begins "regular" (post #2) strategy, starting at the next 100 day (100 is probably not optimal) interval.
(There MUST be a second repeat prisoner by day 100).
This plan may need some work, but I like the odd/even idea......

To clarify, strategy is two parts: days 1-100 use even/odd method and designate Counter (second repeater);
days 101 till end is "regular" (post#2) strategy.
They won't know who Counter is, but they'll know who it ain't (them).
When the first second repeater leaves room, he resets to original even/odd days, so that any future second repeaters know they're not Counter.
Now if I could only figure out the average time till second repeater and.....

English is not my first language so sorry for some mistakes

something isn't accurate with the problem. read this problem and there is some similarity between these 2 problems

The riddle: 100 prisoners. tomorrow will be stood by a line: the 100th one can see other 99, 99th other 98, 2nd-only first one and 1st can't see anybody. each of them will wear a hat either black or white, and nobody can know his color. the count of each color is unknown it can be 50bl 50wh or 100bl 0wh or 73bl 27wh etc., they have to say the color of their hat, using words "black" or "white" no other words and can't prompt others with the intonation. who says wrong will be killed. 1st will begin the 100th one then 99-98-97. . .-1
Q: Knowing all this what plan can they make at night for the tomorrow's issue in order to survive AT LEAST 99 prisoners???

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.
I know this is an old thread, but it struck me that 'optimal' could be reached by combining more than one strategy. I saw good and other solutions in this thread, but hardly any cooperation between users to reach a more optimal solution.

I suggest a different strategy that happens for only the first 100 days: the first to come into the room switches on the light. The second person leaves the light on unless it is the same person who came in the room the first day. The 3rd person also leaves the light on unless he is the person who was there day one or day two. Same for the 4th, 5th, 6th, etc. In fact they all leave the light on UNTIL someone comes in for the second time while the light is on within those first 100 days. This person switches the light off and becomes the counter.
Once the light is off, the group has to wait for day 101 to indicate to people on later days that the first strategy has stopped working. From day 101 on I use the above AKG strategy. Anyone who was in the room and switched on the light during this first cycle does not switch the light on again, ever. This extra strategy gives the counter a head start. (It even gives the group the possibility, albeit a small one I agree, to get out in 100 days.)

Example: the first 30 people to go in the room leave the light switch on, because they all went in the room only once. Person 31 is in fact the person who went in day 8 so he switches off the light, so next persons know the chain is broken. Everyone now leaves the light off until day 101. Then they start the above AKG strategy with person 31 being the counter, having already 30 on his tally. Only 70 more to go. Without doing the math I think it's quicker to use this extra strategy.