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I'm curious if there is a solution,however ineffective, for the puzzle when the prisoners do not know whether initially the bulb is on or off.

Second, has several bulbs modification of the problem been studied? Where, if yes? There may be for instance two bulbs and prisoners are either randomly or regularly put into one of the two rooms each time, this fact being known to them in advance. Does this modification increase or decrease the average time? I believe that in the case of regularly, then the average time for being them free will decrease, since they can encode more information into two bulbs.

user122424
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    Sorry, you'll have to tell us what the puzzle is before we can consider variations of it. – Mark Bennet Jan 26 '14 at 20:25
  • Could you provide a statement of the problem? – Lost Jan 26 '14 at 20:25
  • Sorry but I was not allowed to add comment to the existing link in stackexchange. Here it is http://math.stackexchange.com/questions/116340/100-prisoners-and-a-lightbulb – user122424 Jan 26 '14 at 20:28
  • Your two-bulb problem is underspecified. Do they know in which of the rooms they are brought, or are both rooms identical, and they have no idea which of them they are in? And what is the condition the prisoner has to correctly state for getting free? Is it that each prisoner has been in at least one of the rooms? That each prisoner has been in the same room? That each prisoner has been in both rooms? – celtschk Jan 26 '14 at 23:01
  • I'd say that a natural setting is this.They know which room they are brought(we want to allow them to exploit the power of two rooms),they will get free if everyone was at at least one room(we don't want them to be punished by being in the wrong room). Certainly other specifications are also interesting (I'd even say more difficult to handle) but let's consider the one above.I'd like to get an idea how much another room helps. To have another room exactly corresponds to the above specification. – user122424 Jan 31 '14 at 18:56

1 Answers1

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I'm curious if there is a solution,however ineffective, for the puzzle when the prisoners do not know whether initially the bulb is on or off.

Assuming that it is known on which day the first prisoner is sent into the room, it is indeed possible: The first prisoner to be sent into the room is chosen as the counter (everyone knows if he was in the room on day one), and if at day one he finds the light on, he switches it off without counting, because he knows that nobody has yet been in the room. After that, the protocol goes on the same way as in the original problem.

If the starting day is not known, the following variation works: Each of the $99$ non-counters is asked to switch the light on two times. If only $98$ of the non-counters have switched the light on twice, the total count is at most $197$ ($196$ switch-ons from the $98$ prisoners, and $1$ count from the initial-on lamp). Thus if the count reaches $198$, then either the lamp was initially off and all non-counters have switched on the light twice, or the lamp was initially on, at least $98$ prisoners have switched the light on twice and the $99$th at least once. In any case, all prisoners have been in the room.

celtschk
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  • Of course, the question now is what about without this assumption.I suspect very much that some sophisticated method exists to avoid any further assumptions to achieve prisoners' goal without knowing the initial state of the bulb. – user122424 Jan 26 '14 at 21:44
  • Of course, your assumption is equivalent to the knowledge WHO is sent first to the room with the bulb. – user122424 Jan 26 '14 at 22:10
  • Not exactly equivalent, because everyone only knows if it was him who was first into the room. So 99 of 100 prisoners don't know who was first, except that it was not them. – celtschk Jan 26 '14 at 22:27
  • I mean know at their meeting before they start being sent to the room?That is they are all told before that you are the counter and you will be sent first to the room about one particular prisoner. – user122424 Jan 26 '14 at 22:31
  • Well, they then know who is the counter. For the standard protocol it doesn't make a difference; however I don't know if there are possible protocols which may make use of that information (that is, protocols where more than one person is given a special role, and it is an advantage to know the one who is sent in first). – celtschk Jan 26 '14 at 22:40
  • Is there an optimal protocol for the puzzle with 100prizoners?I think there must be one,which cannot be beaten, but it may be impossible to find it and if found it may be impossible to prove its optimality.Is this correct? – user122424 Jan 27 '14 at 20:49
  • I think the problem is still solvable using the standard protocol of picking a counter beforehand. If the prisoners can keep track of the days, whoever enters the room on the first day turns off the light bulb if it is on. If the prisoners do not know the days, @celtschk's answer above works. Neither protocol requires the counter to be the first person who enters the room. – Quanquan Liu Dec 25 '14 at 21:20