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Topic: 100 prisoners & a light bulb (Read 166633 times) |
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Mattis
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Re: 100 prisoners & a light bulb
« Reply #300 on: Dec 20th, 2004, 4:11am » |
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on Dec 11th, 2004, 7:12pm, Rezyk wrote:I suggest this modification to the problem:
The warden will listen in on all the planning and, if possible, pick an ordering that will result in a false assertion. Otherwise he uses random selection. |
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I think the rule above gives a fun twist to it if you still allow solutions that don't have a 100% chance of success. In that case you must make an elaborate plan that the warden will listen to. You then follow the plan for say three years, and then you claim your freedom (none of the other prisoners will know about your plan though).
The scary thing here is that every other prisoner may reason the same way: "I make up an elaborate plan and play along for X years/months/weeks, and then I take a chance". So, if any of the other prisoners is a stone-cold gambler you might all be free or dead after say 200 days.
I liked the thinking about parity-checking suggested long ago, where each prisoner toggled the lightswitch the first time around. After say two years, if the ligh is toggled off, you can be pretty darn sure that everyone has been there, since the odds that there are two prisoners would have never been in the room is really small.
Maybe, if you were to be the one to lay out the plan you could suggest a plan that sounds like a reasonable and 100% bulletproof plan, but in secret the plan would in fact give you something like parity-checking or something else that gives you higher chance of successfully claiming freedom when you visit the room after say three years or so.
(I know the discussion has come to discuss the real failsafe solutions with only 1 bit of information passed each day, but I thought that lying to the warden and your fellow prisoners was fun enough to motivate a post of my own).
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Patashu
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Re: 100 prisoners & a light bulb
« Reply #301 on: Dec 21st, 2004, 1:46am » |
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I've been wondering something. What if the light bulb had three states? How much quicker could you guarantee the escape of all prisoners with a three-state bulb? How about four? Five? Six?
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Mattis
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Re: 100 prisoners & a light bulb
« Reply #302 on: Dec 21st, 2004, 11:06am » |
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Actually, the more I think about it, the better the solution where you take a chance after three years is. If everyone has been to the mainroom, then fine, you are all free.
But, if there are prisoners who haven't, there are two possibilities:
Either you have had a stroke of bad luck. You should have used the badges and crown-system maybe.
But more likely, if there are prisoners who in three years have never been picked, the warden is being a bitch. Maybe he will never take some of the prisoners to the light-room. Maybe you are all just part of his twisted little game.
If that is the case, no badges or crowns in the world will save you, you are doomed to rot in this wicked place.
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Reege Lafai
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Re: 100 prisoners & a light bulb
« Reply #303 on: Jan 3rd, 2005, 2:23pm » |
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Hi, this is my first time to the forums and I am impressed! So many smart people!
I have an idea. Perhaps this solution will work, perhaps not. In any case, I thought that the people here could build off of this and see what you come up with.
The average lifespan of a lightbulb is 1500 hours. Divide that by 100 people and you get 150 hours per person.
Each time a prisoneer is in the room, he/she turns the light on, regardless of whether or not they have been in the main room before.
Each prisoneer keeps track of the amount of time the lightbulb is left on in their pressence. Once a prisoneer has left the lightbulb on for a total of 150 hours, they turn it off and don't turn it on again, even if they are shown to the main room after that.
In this way, once every prisoneer has used up their 150 hours, (thereby also having visited the main room,) the lightbulb will burn out and the prisoneer on whose "shift" the lightbulb burns out will know that everyone has been in the main room.
Holes in this solution:
1. How would the prisoneers keep track of exactly 150 hours? The solution requires that the prisoneers know how long they are in the main room.
2. What if the lightbulb is slightly more or less than average in its lifespan? That could mess things up a bit. Here is a modified solution if the lightbulb is special (not average):
To account for the fact that the lightbulb may be more than average, that is lasting for longer than 1500 hours, each prisoneer should leave the lightbulb on for 151 hours. This would add 100 hours to the initial time the lightbulb is planned to be left on, and the 100 hours would eat into the time of the last prisoneer's time to leave the lightbulb on. However, it would still go off sometime during his "shift" so it would still work.
To account for the fact that the lightbulb may be less than average and burn out earlier than 1500 hours, the prisoneer on whose shift the lightbulb burns out should not assert that all the prisoneers have been in the main room. The prisoneer after him/her should, upon realizing that the lightbulb has burned out, wait until just before he/she gets put back into their cell and then assert that every prisoneer has been in the main room. This part of my solution I have not thought through as thoroughly. Does anyone see any problems with it that I've overlooked?
The minimum amount of time it would take for this solution to work is slightly more than two years. Of course, the maximum amount of time can never really be computed because of the randomness factor: there could always be one prisoneer left out of the random cycle, sitting in his/her cell waiting to use their 150 hours. I think this randomness factor is what has been messing up all the other solutions from achieving an optimal answer. Even if my solution doesn't work, it would be mighty helpful if someone were to figure out a way to get around this factor.
If this solution works, many thanks to my two brothers who discussed this with me and helped me divise a solution. =)
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mattian
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Re: 100 prisoners & a light bulb
« Reply #304 on: Jan 3rd, 2005, 2:37pm » |
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What if the light bulb is on when the first prisoner enters the room on the first day? He might wonder how long it has been on.
Let me put it this way - I would not want to be a prisoner relying on this solution. But an interesting idea, nevertheless.
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| « Last Edit: Jan 3rd, 2005, 2:38pm by mattian » |
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Icarus
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Re: 100 prisoners & a light bulb
« Reply #305 on: Jan 3rd, 2005, 4:58pm » |
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You're not the first to suggest this approach, though you have taken it farther than before. The problem of the light bulb not being nice and giving up at exactly 1500 hours (each prisoner should leave it on for 15 hours, by the way, not 150) is much larger than you figure, though. In addition to mattian's point that the bulb may not be at the beginning of its life, the variance in light bulbs is almost certainly too great for this to work with any reliability. Even with 150 hours, so each person could have it on an extra hour, there is a good chance the light bulb would still not burn out, and then everyone is left sitting in a dark room, without ever having the chance to deduce everyone has been there.
on Jan 3rd, 2005, 2:23pm, Reege Lafai wrote:| To account for the fact that the lightbulb may be less than average and burn out earlier than 1500 hours, the prisoneer on whose shift the lightbulb burns out should not assert that all the prisoneers have been in the main room. The prisoneer after him/her should, upon realizing that the lightbulb has burned out, wait until just before he/she gets put back into their cell and then assert that every prisoneer has been in the main room. This part of my solution I have not thought through as thoroughly. Does anyone see any problems with it that I've overlooked? |
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How would this help at all? What is the point of sitting around in the dark room for a while before making the announcement? The same people will have visited the room at the beginning of this visit as at the end, so why wait? And the only advantage of having the first person to enter the room with a burnt-out light announce, rather than the one it burnt out on, is the very slim possibility that the one extra guy happens to be the sole person who hasn't visited before.
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Grimbal
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Re: 100 prisoners & a light bulb
« Reply #306 on: Jan 3rd, 2005, 5:05pm » |
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Another problem is that it is not only the burning, but also the turning on and off that uses up the bulb. And to check wether the bulb is burned out, you have to try it. So you are using up the bulb even if one prisonner is unlucky enough to remain in his cell.
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Reege Lafai's brother
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Re: 100 prisoners & a light bulb
« Reply #307 on: Jan 3rd, 2005, 6:02pm » |
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on Dec 21st, 2004, 1:46am, Patashu wrote:| I've been wondering something. What if the light bulb had three states? How much quicker could you guarantee the escape of all prisoners with a three-state bulb? How about four? Five? Six? |
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Actually, the bulb has 4 states (unless it is too high to reach): On, off, Screwed in, and Unscrewed.
This led me to thinking, if the bulb can be reached, what about each prisoner unscrewing it 1/100 of the way out the first time (and only the first time) they get into the room? The prisoner that finally unscrews the light bulb demands their freedom.
In response to the problem with our (My sister's and mine) solution about the light bulb not being at full capacity on day 1, the first prisoner could turn (and leave) the light on. No one touches the light switch until the light bulb burns out. When they replace it (and this they must do), you are left with a brand new light bulb with which to start your strategy.
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mattian
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Re: 100 prisoners & a light bulb
« Reply #308 on: Jan 4th, 2005, 7:11am » |
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There is only one logical interface which has only two states - on and off. As indicated in the red text at the top of this puzzle, no interpretation other than these two states may be obtained from the bulb/switch combination in the room, thus excluding the option of screwing and unscrewing the bulb, making marks on the wall or filling the room with sand and making snow angels. Just a switch - on or off. That's all. After all, this is a logical puzzle and there aren't really any prisoners out there waiting for the solution. As long as we're hypothesising, therefore, we may as well pretend the prisoners are unable to perform any function other than walking, toggling the switch and pondering.
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| « Last Edit: Jan 4th, 2005, 7:13am by mattian » |
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mattian
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Re: 100 prisoners & a light bulb
« Reply #309 on: Jan 4th, 2005, 7:18am » |
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on Jan 3rd, 2005, 6:02pm, Reege Lafai's brother wrote:
Actually, the bulb has 4 states (unless it is too high to reach): On, off, Screwed in, and Unscrewed. |
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Not true. Even if we allowed physical states of the bulb (as opposed to logical states), these would only be three.
On, off and unscrewed. Screwed in is a superset of on and off unless by "screwed in" you mean without making electrical contact.
But if we're unscrewing the bulb anyway, why not place it in significant positions about the room and leave in a 150 days?
This is precisely why we limit the puzzle to purely logical solutions including ideal bulbs and switches.
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| « Last Edit: Jan 4th, 2005, 7:20am by mattian » |
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rmsgrey
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Re: 100 prisoners & a light bulb
« Reply #310 on: Jan 4th, 2005, 8:13am » |
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on Jan 3rd, 2005, 6:02pm, Reege Lafai's brother wrote:| In response to the problem with our (My sister's and mine) solution about the light bulb not being at full capacity on day 1, the first prisoner could turn (and leave) the light on. No one touches the light switch until the light bulb burns out. When they replace it (and this they must do), you are left with a brand new light bulb with which to start your strategy. |
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The problem with this fix is that you only know the bulb's been burned out and replaced when it burns out while a prisoner is in the room (to be able to use this method, the bulb has to be left on while burning the first one, and left off between visits while burning the second)
Besides, what stops the warden from replacing the bulb with the one he's been using in the guard room, and using the new bulb himself? Or, for that matter, changing the bulb from time to time himself...
The only guarantee we have is that the switch will be in the same on/off state when the next prisoner enters as when the previous prisoner left - for all we know, the entire room is dismantled and replaced overnight...
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Icarus
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Re: 100 prisoners & a light bulb
« Reply #311 on: Jan 4th, 2005, 5:45pm » |
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Besides which, only one prisoner can know that the old bulb has burnt out, and so must have been replaced. The other 99 would not know it was time to put this plan in action.
The idea of working off bulb life, while interesting, is critically dependent on the prisoners knowing to a very high degree of accuracy the lifetime of the bulb at the start. It is very unlikely they would have this knowledge.
I think, if I were to restate the problem, I would despose of the whole room & light bulb, and just have a guard visit a prisoner every day, deliver a message of "yes" or "no" from the previous prisoner, and take down this prisoner's own "yes" or "no" answer, to give to the next guy. This would preclude anything other than the 1-bit transfer (assuming an unbribable guard).
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Mattis
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Re: 100 prisoners & a light bulb
« Reply #312 on: Jan 5th, 2005, 4:47am » |
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Even if lightbulbs had a very reliable lifespan, there is another problem with the lightbulb-lifespan-strategy:
Sure, the warden, and the whole prison-system seems wicked and evil. But that don't mean that they don't take their environmental responsibility.
It would be a nasty surprise to agree on the lifespan-strategy only to find on the first day that the prison uses low-energy long-life-lightbulbs that lasts 10 times longer than ordinary bulbs. 
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rmsgrey
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Re: 100 prisoners & a light bulb
« Reply #313 on: Jan 5th, 2005, 6:41am » |
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on Jan 4th, 2005, 5:45pm, Icarus wrote:| Besides which, only one prisoner can know that the old bulb has burnt out, and so must have been replaced. The other 99 would not know it was time to put this plan in action. |
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I started out assuming this, then realised that, in order to control the burn time, the individual prisoners would have to leave the bulb off between visits once the strategy started, while, during the initial burn, it makes some sense to leave the light on permanently (or spend your time in the room flipping the switch), which allows a simple signal to be sent (if the light's on we haven't started yet...)
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Reege Lafai
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Re: 100 prisoners & a light bulb
« Reply #314 on: Jan 5th, 2005, 1:32pm » |
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Yeah, you guys had good responses. I knew there were flaws in the special-lightbulb theory.
But like I said before, I don't think this will ever be solved because of the randomness factor.
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Icarus
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Re: 100 prisoners & a light bulb
« Reply #315 on: Jan 5th, 2005, 5:19pm » |
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Several solutions have already been discovered. None of the solutions guarantee the prisoners with release because this cannot be guaranteed: it is possible that the warden's random picks will never happen to pick one of the prisoners. What is desired is a scheme such that, if it does lead one of the prisoners to make a declaration, that declaration will be sucessful. The second desire is to find a scheme that will on average produce a positive result in the least amount of time.
Plenty of solutions meeting the first criteria have been proposed. The simplest is the "Leader" proposal. The prisoners pick one guy to be leader at their meeting. Only the leader is allowed to turn off the light. Everyone else turns it on the first time only that they enter the room with it off, but otherwise never touches it. The 99th time the leader turns off the light, he can safely declare that everyone has been to the room. The problem with this method is that it takes a considerable amount of time to work (I don't remember how long - a variation that speeds things up significantly, and which was William's answer when he first posted the problem, takes 27 years on average).
Several other solutions have been mentioned. Most of them variations on the general "leader" concept. The fastest make use of a concept of "tokens" collected by subleaders and past on to the leader. Currently the best scheme has a average run-time of about 9 years.
So the basic problem is solved. It is the optimization problem that remains open. And you are correct that it will remain that way, though not because of randomness. Rather, because it is extremely difficult to prove that no better scheme exists.
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mattian
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Re: 100 prisoners & a light bulb
« Reply #316 on: Jan 6th, 2005, 7:19am » |
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I've been thinking about this problem again and I am beginning to take issue with the manner in which optimum strategies are determined. I don't think average escape time is a sufficient measure of the success of a given strategy.
For example:
Suppose two strategies are run 10 times and the results are:
Strategy 1:
6345; 6443; 5987; 6302; 1995; 6744; 6229; 6221; 5883; 2013
AVG: 5416.2
Strategy 2:
5437; 5498; 5972; 5164; 5375; 5248; 5335; 5657; 5634; 5043
AVG: 5436.3
Strategy 1 has the better average but Strategy 2 is more stable.
I encountered this problem in my implementation of the token solution which produced extremely low numbers (under 2000) about 1 in 5 iterations, but higher numbers (above 6000) for the others. Clearly, the strategy favours certain patterns in the sequence which relies on probability. So there is a 1 in 5 chance that this strategy, for example, will flatten the other strategies, but a 4 in 5 chance that it will not do as well.
So how should the success of a given strategy be measured?
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| « Last Edit: Jan 6th, 2005, 7:20am by mattian » |
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rmsgrey
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Re: 100 prisoners & a light bulb
« Reply #317 on: Jan 6th, 2005, 7:55am » |
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The expected time to release isn't a bad measure - if a million prisons all ran this experiment, and all the prisoners used the same strategy (or the 100 prisoners had to earn their escape a million times in a row) then they should use the strategy with the least expected time to release.
If you are prepared to gamble on a relatively slim chance of an early release, then minimising expected log release time would probably be better.
In general, you probably want to minimise the expected value of some function of release-time.
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Grimbal
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Re: 100 prisoners & a light bulb
« Reply #318 on: Jan 6th, 2005, 8:35am » |
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That makes sense.
Doing one more year of prison is more painful when you did just one year than when you did 10. So there should be some decrease in the penalty of an additional day of prison when you average.
Some people might like taking risks. They'd prefer to have a slight chance for a very early escape, even if it means a longer average time. Some might be overcautious. They'd prefer to minimize the chances of a long stay, even if it means no chance for an early liberation.
Then there might be reasons they want to be free at some specific date and not at others. They might want to spend christmas outside, but if they miss christmas, they'd rather spend the winter in prison, where they have shelter.
All this amounts to what rmsgrey said: "In general, you probably want to minimise the expected value of some function of release-time." the function being the dissatisfaction for some liberation date.
Now, just to complicate the problem, each prisonner might have a different opinions about what is the objective. It should be fun to see all the prisonners arguing and negociating to settle to a solution that suits everybody enough so that nobody will want to sabotage it.
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Leo Broukhis
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Re: 100 prisoners & a light bulb
« Reply #319 on: Jan 6th, 2005, 5:48pm » |
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on Jan 6th, 2005, 7:19am, mattian wrote:
So how should the success of a given strategy be measured?
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It does not make sense to average the release times because the prisoners only need to escape once and they do not care about the repeat performance of a strategy. The median or some other agreed upon percentile is a better measure in this case: what really matters is odds of escaping after so many days.
To give a better example of misappication of average:
A patient asks the doctor: What are my options?
Doctor: The average lifespan without the operation is 1.5 years, with the operation - 4 years and counting.
Patient: Great, I'll have the operation! By the way, what are the median lifespans?
Doctor: Well, without the operation it's slightly less than 1.5 years, and among those 10 people who had the operation the only person who survived it 40 years ago still lives, so the median must be 0.
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| « Last Edit: Jan 6th, 2005, 6:03pm by Leo Broukhis » |
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towr
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Re: 100 prisoners & a light bulb
« Reply #320 on: Jan 7th, 2005, 12:52am » |
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It's a matter of opinion whether that's a misapplication of statistic measures.
A utilitarian might simply go for the best expected outcome (average). Whereas an overly risk averse person might go for the best worst-case.
What the best measure is for a strategy for this problem depends on how the prisoners feel about it. If they feel 9 years is only twice as bad as 3, and 16 twice as bad as 4, then a logarithmic scale of 'cost' might work best.
But they might also just feel very strongly about getting out in the weeks before christmas, and care much less for the rest of the year.
Or maybe they just want to be free before they die.
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| « Last Edit: Jan 7th, 2005, 12:55am by towr » |
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riadov
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Re: 100 prisoners & a light bulb
« Reply #321 on: Feb 17th, 2005, 5:11pm » |
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hi i am new here
after hours of reading 22 pages i think i know what the problem's are(your's i mean)
1)u didnt read well the riddle 
2)u have concentrated on meaningless words in the riddle
3)u see numbers and u start calculating
some of u thought of having the meeting in the central living room ,well of course its wrong :THE PRISONERS CANT CHOOSE.
well let us say there is a tunel X that gides from A to B(and it's the only way),if i want to get from A to B the only way is thru X  i am not joking here
let :
-X be the central living room
-A the cells
-B the meeting place or home sweet home
soooooo to be in the cell: THE PRISONERS HAVE TO PASS THRU THE CENTRAL LIVING ROOM and to get out from the cell to go to the gym toilet or wherever the meeting is THEY ALSO HAVE TO PASS THRU THE CENTRAL LIVING ROOM (its not the Disney Land out there)
so the first prisonner to be chosen will claim that all the prisoners were in the central living room :the day they entered their cells or the day they went to the meeting
P.S: if i am wrong ,don't be rude to me
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mattian
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Re: 100 prisoners & a light bulb
« Reply #322 on: Feb 18th, 2005, 5:26am » |
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CONGRATULATIONS!!! Hey everyone, Riadov has solved the riddle - why didn't we think of this? There we were, silly us, trying to do calculations *spit* while the answer was staring us in the face. In fact, why do we bother to think at all? I mean the whole point in life is apparently to find short cuts rather than solve problems (whether it is the correct means or not). Apparently the people who subscribe to this forum are interested only in thinking and problem solving - how very short-sighted of them - instead of just knowing the answer or finding a spelling error that makes the question null and void. I didn't realise until now how much easier it is to solve riddles by technicality - or even better, by introducing factors which help solve it. For example, the solution to this particular riddle is actually "one day", because all the prisoners actually had a key to their cells and their cell doors opened up directly into their living rooms at home, so the solution is this: they turn the key, open the door and walk into their living rooms. The solution is so simple - I battle to figure out why all of you waist so much time thinking! In fact, why not extend this principle into all things - in a game of chess, for example, my pawn needs to be diagonally adjacent to a piece belonging to the opposition in order to take it - however, with this incredibly ingenious new way of thinking, I've just realised, that there is no need for this at all; the opponent's piece can simply be taken at any time - my pawn (or any piece for that matter) need not even be there, since the act of picking up the opponent's piece and removing it such a trivial task requiring very little energy and absolutely no thought. From now on, when I play chess, I will simply take my opponent's pieces willy-nilly whenever I feel like it. Or move my pawns backwards, or throw the board in the air while making gurgling noises. This concept of protocol and good spirit is becoming an annoyance and I feel liberated now that I've come to realise that it is so unnecessary - I will make up my own rules, stand on my colleague's heads to get to the top, take the shortest path to every outcome because my name is SOCIETY_IN_THE_21ST_CENTURY!
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| « Last Edit: Feb 18th, 2005, 8:25am by mattian » |
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rmsgrey
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Re: 100 prisoners & a light bulb
« Reply #323 on: Feb 18th, 2005, 8:14am » |
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on Feb 18th, 2005, 5:26am, mattian wrote:| my name is SOCIETY_IN_THE_21ST_CENTURY! |
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My name is "good-old-days"
What? this isn't the Sandman Duel thread?
Well, the prisoners is obvious - it must be 100 days because the warden has to pick a different prisoner each day, so 100 days will have them all visiting.
What? What happens on day 101? They're all at home by then...
Besides, the tally marks on the wall will solve it!
Or they remove the bulb and use the socket to electrocute themselves! Whoever's left alive on day 100 knows they're the last prisoner!
Or stage a revolt! Kill the warden and take over the prison!
******************************
As for tuning for reaching a given probability of escape as early as possible, my understanding is that existing multi-stage methods already do something similar - the end of each stage of the first full cycle is designed as a trade off between taking longer to get to the release stage, and risking having to do a second full cycle. Depending on the threshhold probability, they're probably already fairly close to optimum evaluated that way too.
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Icarus
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Re: 100 prisoners & a light bulb
« Reply #324 on: Feb 18th, 2005, 3:26pm » |
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Now, play nice! riadov has misunderstood the point of this puzzle, though I do wonder why he didn't pick it up after reading all the previous posts, since it has been explained before.
riadov - finding a way of interpreting the puzzle so that the prisoners can get out easily is NOT the point here. Those sorts of puzzles are found in the "What Happened?" forum. In the Hard forum are, preferably, puzzles that do not depend on tricky wording, but rather deep and imaginative thought to come up with a solution. (Maintaining this standard is hard, because people tend to assume that if they don't know the answer, it must be hard - so a lot of people keep posting trivial puzzles that they are hung up on.)
The point of this puzzle is to figure out the best scheme the prisoners can use when:
1) the only time anyone gets to enter the room is during their visit.
2) the only way a prisoner has to communicate any message to another prisoner is by leaving the light on, or leaving it off: one bit only, and only readable by the randomly selected prisoner of the next day.
This is the real challenge, and as an intriquing challenge it beats the pants off of coming up with different interpretations.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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