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Title: Forcefield Detainment Post by Jeremy on Sep 29th, 2002, 12:49pm A group of prisoners are trapped in a forcefield. These prisoners are perfectly brave, meaning that they would attempt an escape on any positive probability of success. The prisoners are monitored by a guard who has only one bullet in his gun, but who also has perfect marksmanship skills (he never misses). A maintenance technician needs to tune up the forcefield generator, and so for one second, the forcefield is released. How can the guard still keep all the prisoners detained? HINT: Those pesky students in the logic class may have an idea of how to solve this. |
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Title: Re: Forcefield Detainment Post by Pietro K.C. on Sep 29th, 2002, 1:43pm I came up with two different solutions... but I guess they're kind of analogous. First one: the force field is presumably a 2-surface in 3-space, and contains a finite number of prisoners. So the guard could just stand in front of them and say, 'I will shoot the first one who comes out, and if more than one comes out at the same time, I will shoot the rightmost one nearest to me.' Obviously, no one will go out alone, and if two go out at the same time the guard's criterion is enough to deterministically single one of them out. So that guy's probability of escaping is zero, and he will know it beforehand - so he will stay put, ruining the others' escape. And here's what I think the "correct" one is; it's really REALLY similar, but perhaps simpler: say there are n prisoners, p1, ... , pn and that the guard numbers them out loud, so they can hear it. The guard could then say: 'if any subset of you tries to leave, I will shoot the highest-numbered one in the subset.' That would prevent pn from ever participating in an escape attempt, since his probability of survival would be zero. This knowledge in turn would prevent pn-1 of ever trying to make a break for it; and so on until p1, which, if he attempted to escape by himself, would be shot. Hence they all stay perfectly put, the fools. They must also be supposed to be perfectly honest, otherwise the n-1 smallest-numbered ones would just gather around pn and throw him outside of the cell. :) Now that's just ironic, because they're prisoners. What could they have done? |
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Title: Re: Forcefield Detainment Post by DarkBronzePlant on Nov 1st, 2002, 12:25pm That's better than my approach, which was to shoot the maintenance technician before he could shut off the force field. ;) |
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Title: Re: Forcefield Detainment Post by Chronos on Nov 4th, 2002, 10:57pm I think that Pietro's second solution will work, but not his first one. If the guard just threatens to kill the first one out, then the prisoners can draw straws to see who goes first. You can't give the prisoners any control over who's the priority kill. |
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Title: Re: Forcefield Detainment Post by Pietro K.C. on Nov 5th, 2002, 2:38am Ah, but would the one who drew the shortest straw go out after he drew it? I view the problem statement as outlawing such a possibility. And, if you suppose the others throw him out by force, then I don't see the point of drawing straws at all. Just throw the ugliest one out, or whatever. :) Notice that the brute-force approach of N-1 prisoners also breaks my second solution. |
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Title: Re: Forcefield Detainment Post by towr on Nov 5th, 2002, 8:49am The guard shouldn't make the fact he has only one bullet common knowledge.. Perhaps even lie and say he has more.. Also one second isn't a lot of time to escape, so the prisoners would have to get close to the forcefield so they can jump out the one second it's down.. Just say you'll shoot the closest, then they'll all be at the back of the cell and won't have time to jump out (without getting sliced by the reactivating forcefield). The guard could also pick several prisoners and order them to keep the others in, and threaten to shoot the first one that fails and let's anyone break through.. |
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Title: Re: Forcefield Detainment Post by Icarus on Nov 5th, 2002, 6:54pm Chronos is correct that Pietro's first solution is flawed, even if the senario provided doesn't prove it. The solution still allows the prisoners an opportunity to pick a sacrificial victim by random chance. They just need to find a way for the victim to not have an opportunity to back out. For instance, they could clasp hands and start spinning. When they see the forcefield go down, they throw themselves out en masse. The guard would probably be too shocked at seeing his prisoners play Ring-Around-The-Rosie to remember to shoot! (Which of course is good, since the prisoners will be too dizzy to do more than just lie there. ;)) The second method prevents this, since one guy knows he's dead not matter what they do, so he won't go. And so the next won't go, and so on. As for preventing the brute force approach, the guard should threaten to kill the highest numbered one to move, not to leave. (This assumes he has a clear shot at everyone.) |
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Title: Re: Forcefield Detainment Post by jordan on Oct 19th, 2003, 4:21pm What if, with this problem, all the prisoners must leave AT THE SAME TIME? So that there is no FIRST or LAST prisoner? And each prisoner must be convinced that, if they run, they have a 100% chance of dying--and they are convinced of this by what the guard says? And the GUARD must be telling the 100% truth?? This is how the puzzle was presented to me.....can anyone solve it? |
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Title: Re: Forcefield Detainment Post by Icarus on Oct 19th, 2003, 8:14pm Pietro's second solution still works, since it does not depend on who leaves first or last. And the problem of the other prisoners throwing the highest numbered guy out can be overcome by the guard threatening to kill on any movement, not just escape. |
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Title: Re: Forcefield Detainment Post by william wu on Oct 19th, 2003, 9:50pm on 09/29/02 at 12:49:49, Jeremy wrote:
A year later, I think I might know what Jeremy was referring to ... the Unexpected Hanging paradox, otherwise known on this site as the Pop Quiz problem. If you recall the paradox, the professor says there is a quiz next week, although he does not specify the day. The students then conclude there can be no quiz on Friday because that could be deduced after there is no quiz by Thursday. And thus the quiz cannot be on Thursday either; et cetera. Yet a pop quiz can happen. So maybe Jeremy was thinking that despite what ordering-based solution you use, the domino effect of chickening out will not occur, and someone really will get shot. However I don't think this reduction holds. I find Chronos's objection troubling, in that I'm not sure I buy it. On one hand, if the prisoners agree a priori that the drawer of the shortest straw must go forth, then any individual prisoner has probability (n-1)/n > 0 of succeeding, so they will all agree to take part in the draw. However, after a prisoner draws the shortest straw, will he keep his word and be first to exit the forcefield? Only if he's an extreme moralist -- a quality not stipulated in the problem statement. I think Icarus's objection to Pietro's first solution is handled by the fact that the guard "shoots the rightmost one nearest to me" (Pietro's words) in the event that prisoners come out at the same time. Imposing a left-right order is not much different from imposing a numerical order. So this will single out a particular prisoner X, who must stop. And assuming the prisoner to the right of X also knows X's policy, that prisoner will also stop. By induction, they will all chicken out. One might argue that the prisoners in Icarus's ring-around-the-rosie contraption will spiral outward with such momentum that there's no way for anyone to back out, but I think we must assume it is possible for the prisoners to make instantaneous decisions and halts of motion, and for the guard to have no such emotions as bewilderment. As I hinted at in my previous paragraph, I think the problem statement is flawed. It does not say that 1) the prisoners know that the guard has perfect marksmanship, nor 2) the prisoners know that every other prisoner holds the same policy (this is needed for the domino effect), nor 3) the guard only has one bullet. Thus the prisoners could just all run out at once, because the expected accuracy of all marksmen in the universe is not 0. Or, they can blindfold themselves and each generate a random number k from 1 to N, and then start walking toward the exit after k seconds have passed. (Random number generators are standard issue in most modern day prisons.) Finally, even if we fix these aforementioned problems with total information awareness, the problem is still flawed! If I'm in the forcefield and I know the guard has one bullet, and that he will shoot the highest number to come out, all I have to do is shove some poor sap through the forcefield. After he gets capped, I'll be free! Thus the prisoners will engage in a violent brawl, and at the end of the day, perhaps the physically weakest prisoner in the bunch will be dead. So we must change the scenario; I offer two suggestions: 1) Each prisoner is locked in his own isolated cell, although the prisoners are allowed to communicate with each other, perhaps via teleconferencing. The event of the forcefield turning off is then analogous to the opening of each cell's door. Or ... 2) (I like this better) All the prisoners are infinitely skillful sumo wrestlers which cannot be pushed out of the forcefield, even when being rushed by arbitrarily large gangs of equally skullful sumo wrestlers. And all the prisoners are aware of everyone else's mad 1337 sumo skills. ... Ok back to work ... good god dude, I have a problem set due soon and I'm sitting here constructing E.Honda collages ... lol |
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Title: Re: Forcefield Detainment Post by Icarus on Oct 20th, 2003, 4:05pm on 10/19/03 at 21:50:02, william wu wrote:
But I LIKED the idea of the guard standing there gaping at hardened criminals playing child games! The point is, the Pietro's first solution allows for the possibility of some sort of randomized "force out a sacrificial victim" strategy. You set up matters so that this one is no longer possible, but that still does not mean another scheme might not be made to work. The second strategy does not have this weakness. And sumo wrestlers or separate cells are not needed if the threat is to shoot the highest numbered of those who move at all, not just escape. Then there is no physical way that they can force out a victim (I am assuming no "psychic powers", or hypnotic suggestions) without being shot in the process. I don't see the need to change the problem to insure that the first solution will actually work when there is an alternative that works just fine. |
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Title: Re: Forcefield Detainment Post by aero_guy on Oct 21st, 2003, 11:39am There is a problem though. If the guard is going to shoot the highest numbered person that moves at all, they just have to beat the highest numbered prisoner (HP) to a pulp before the field goes down, and then second highest numbered guy (2HP) tosses him out first. You could say that 2HP is the one who would get shot, but not if he has HP as a human shield. Hell, he doesn't even need to toss HP out, he can just sit at the back of the cell with HP as a shield and wiggle while everyone else makes it out and beats the guard up. He can follow last, or if he doesn't have the time, just make the deal with the others to let him out once the guard is taken out. I think this makes the seperate cell idea a better puzzle. |
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Title: Re: Forcefield Detainment Post by william wu on Oct 21st, 2003, 1:02pm on 10/20/03 at 16:05:15, Icarus wrote:
It's not clear to me how one could devise any such strategy without resorting to mechanical contraptions that both assume the presence of technological resources, and also make the problem look silly. Such a strategy would require that the randomly chosen martyr have no opportunity of turning back. Since we can't be assured that any prisoner always keeps his word, this comes down to physical force. Maybe they could construct an n-spooned catapult into which prisoners can synchronously self-lock themselves. A central computer then generates a random number between 1 and n, and fires that spoon. But this is ridiculous. If you have a reasonable scheme that guarantees randomized force out, I'd like to hear it. Quote:
"Shooting the highest numbered of those who move at all" looks like it will work under the current problem statement, assuming the prisoners lack telekinetic abilities (in which case they will try to move others using their minds, a la the mutant Jean Grey from X-Men :)). However, I think it's interesting to note that you're threatening to do something your prison supervisor probably wouldn't let you do anyways. If I was a prisoner and you capped me just for scratching my chin, you'd probably be in the forcefield the next day. If we modify the problem statement such that you are only allowed to kill prisoners who cross the forcefield (a reasonable clause I believe), and if we assume the prisoners know your restraint, then they will freely engage in The Pushing Orgy, because they know you are unable to execute your policy (assuming you care about your future as an unimprisoned citizen). Separate cells, or the power of sumo, is necessary to avoid solutions that reduce to physical violence. You guys seem to have ignored what I think is the more poignant rewording suggestion, which is that of total information awareness. The prisoners must know that 1) that the guard shoots with 100% accuracy, 2) every other prisoner holds the same desperado policy (attempting escape on any positive success probability), and 3) the guard has only one bullet. If 1) does not hold, then they'll all attempt an escape, because the average marksmanship accuracy among humans -- and perhaps even robots (exponentially distributed electronic failure) -- is less than 100%, and there's always a possibility that the guard accidentally left his bullets at home that day. If 2) does not hold, then 2HP doesn't know if HP will attempt an escape anyways. So 2HP might assume that there is some nonzero probability that HP will make an attempt; likewise for 3HP, 4HP ... nHP. Of course, this produces ill-defined questions such as "Can escapes be attempted in 0 time?" and "Can 2HP deduce whether HP is attempting an escape solely from the behavior of HP?" Basically, it is unclear whether the domino effect of the desired elegant solution (imposing an order) will execute properly. Also note that 2HP cannot simply ask HP whether he is attempting an escape. HP could lie to 2HP, and say he is attempting one. Then they will run out together, but at distance dx from the forcefield boundary, HP could stop and trick 2HP into being the first to cross the boundary. Or something like that. I realize that these probability assertions assume various worldly facts and may seem frivolous, but I think they are reasonable considerations to make when studying such extremist statements as "any positive probability of success." |
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Title: Re: Forcefield Detainment Post by rmsgrey on Apr 5th, 2005, 7:47am on 10/21/03 at 13:02:57, william wu wrote:
That seems wrong - the desperado prisoners will always attempt an escape if there's a non-zero probability of the guard having no bullets, rather than being intimidated by the possibility of everyone dying. |
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Title: Re: Forcefield Detainment Post by william wu on Apr 5th, 2005, 9:22am You're right ... dunno what I was thinking there. |
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Title: Re: Forcefield Detainment Post by The Knife on Apr 10th, 2005, 6:59am This always happens - I find a very simple solution and then I see people writing long and excessively complex answers which make me doubt whether I'm being a naive simpleton who hasn't really put enough thought into his answer. This time I'm having the courage of my convictions though, I am right :o Someone mentioned it before - the guard threatens to shoot the technician if anyone tries to escape. This way no matter how many of the prisoners charge the forcefield and try to escape, none of them will make it because the moment they move, the technician will be shot and the chance of escape will disappear. Knowing this, the prisoners should all stay put during the maintenance since they know there is a zero probability of them escaping and the process will pass off just hunky-dorey. The only scenario I can think that poses a problem for the above is if the prisoners were a sadistic bunch who liked seeing technicians get shot... in this case they would charge the forcefield every time maintenance was taking place, not in attempt to escape (since this would still have zero probability if the guard sticks to his plan of shooting the technician) but to stop the maintenance successfully taking place... if they did this for long enough then I suppose it is possible that the forcefield would eventually fail due to lack of maintenance! You guys probably all realised that at once though and were just having fun making up answers involving sumo wrestling and ring o' roses :) |
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Title: Re: Forcefield Detainment Post by rmsgrey on Apr 10th, 2005, 10:04am on 04/10/05 at 06:59:08, The Knife wrote:
So what happens if the technician gets shot while the forcefield is open? Who re-establishes the containment field? |
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Title: Re: Forcefield Detainment Post by alien on Apr 10th, 2005, 1:13pm The guard can keep all the prisoners detained only if they are all asleep. |
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Title: Re: Forcefield Detainment Post by Alex on Jun 7th, 2005, 3:48am ok here is a thought it plays on what the nazis did in wwII if you redifine the perameters of the force feild to narrow the containment feild to the width of a single man and told the prisoners to stand shortest to tallest so that you could see all the men in a line and hold the gun to the head of the shortest man then if anyone moves you pull the trigger and they all die no fuss no muss and since the feild will only be down for a second then who will dare to move for the second that it is down. it takes the human body longer than a second to process and react to any chage anyway so if it a computer controlled shutdown for exactly one second then there shouldnt be any time to move. anyway that is my thought ;D |
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Title: Re: Forcefield Detainment Post by andrewc32569 on Jun 14th, 2005, 1:27pm Here is something that is bugging me though, If they are perfectly brave, wouldnt that mean they would die for a chance at getting out? I mean, if they all jumped out at the same time, most of them are bound to escape, even if the guard did have more then one bullet. Also, Why the heck is only ONE guard with ONE bullet in there? |
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Title: Re: Forcefield Detainment Post by towr on Jun 14th, 2005, 3:00pm on 06/14/05 at 13:27:34, andrewc32569 wrote:
Quote:
Quote:
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Title: Re: Forcefield Detainment Post by bobsagot on Jun 24th, 2005, 5:04am on 06/14/05 at 13:27:34, andrewc32569 wrote:
Asking for logic in a logic riddle is like asking for freshwater out in the sea! |
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Title: Re: Forcefield Detainment Post by Ardea on Sep 27th, 2005, 7:37am Hello to all, my first post, so be gentle: What about this solution: Assuming that none of the prisoners are of exactly equal height and the guard can always tell who is taller in any set of two prisoners, it would suffice to say he will shoot the tallest of any set of fleeing prisoners. So the tallest of all prisoners will say: Oh, if I flee, IŽll definitely get shot and so he stays put. The secondmost tallest will think: Oh, the tallest knows he will be shot, so he wont move, then I will be the one shot and so on down to the smallest. What do you think? |
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Title: Re: Forcefield Detainment Post by Neelesh on Sep 27th, 2005, 7:56am Yeah, this will work. This is very similar to Pietro's second solution, the numbering is based on height in your case. |
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Title: Re: Forcefield Detainment Post by BaH on Dec 28th, 2005, 10:45pm Same solution, different method, make everyone lay down, tell then the first to stand gets shot, and that if more than 1 stands/moves, the closest will be shot. Alternately, have them sit in a row clasping the neck of the person in front of them, and inform them that if someone stands/breaks free, you will shoot the one who let go or if 2 let go, the rearmost. Lastly, hogtie them with their sheets. |
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Title: Re: Forcefield Detainment Post by dbaker84 on Feb 23rd, 2006, 2:53am If we assume two conditions: (1) The guard monitors the prisoners 24/7 and can hear what they are saying to each other, and can order them to do whatever he desires. (2) The guard knows that they will try to escape and needs to thwart this plan. There is a relatively simple solution. First, the guard should order the prisoners not to communicate or else they will be shot/punished/whatever. The guard can then call each prisoner to his observation point individually and talk to them for a short time about any random subject, it doesn't matter. Shortly after, he will call them each back individually again and tell them that another prisoner ratted the subject prisoner out as ring-leader of the escape, and therefore he will be shot first when they try to escape. He can also tell them that if they try any communication with any other prisoners "to change the escape plan" (to keep them from reconciling their talks with the guard), they will also be shot. Now every prisoner will think that they have been ratted and tagged as the mastermind of the escape, and will be shot first if they attempt to escape or even talk to the other prisoners. Because they have a perceived 0% chance of escape, each will fail to try and escape when the field opens. |
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Title: Re: Forcefield Detainment Post by moooo on Mar 28th, 2006, 4:16am I think there is a solution to the original problem and that it isn't flawed nor does it need any additional assumptions about their sumo skills. It was mentioned before by "The Knife". The guard threatens to shoot the technician if any of the prisoners dares to move. If the technician is shot, there is no chance of any of the prisoners escaping, so none of them is going to attempt to escape. on 04/10/05 at 10:04:45, rmsgrey wrote:
The guard shouldn't shoot the technician if the forcefield is down. This solution assumes that none of the prisoners is located close enough to the edge of the forcefield that it would take them less than 1 second to run out of it. That takes care of the possibility that the prisoners make their escape as soon as the technician shuts off the forcefield. ...or maybe there is something I've missed completely after reading the previous posts. btw this is an awesome forum. Wish I had discovered it earlier! |
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Title: Re: Forcefield Detainment Post by Icarus on Mar 28th, 2006, 4:03pm The flaw in that is that there must be some reason for the technician to need to shut off the field in the first place. That reason is not going to go away just because you shot him. Possibly, the field will enventually fail if the tech is not allowed to fix the problem. In this case, it is to the prisoners' advantage to crowd the field. If you follow through with your threat, or the tech, realizing that you are psycho, flees for his life, then the field will eventually fail on its own, and everyone escapes. If you don't follow through, then they can escape while it is down. |
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Title: Re: Forcefield Detainment Post by flamingdragon on Nov 19th, 2006, 5:21pm I disagree, the only way to be 100% certain that no one will escape is to shoot the technician. The ringleader/tallest/number solutions will not work b/c one might sacrifice himself for the others to escape. And even if he did say he'd shoot them first, there's still a probability of escape, such as the guard decides not to shoot them for some reason, such as he's defecting or something, furthermore they probably don't know he never misses or wont believe him if he says that. Shooting the technician is the only way to gurantee the prisoners dont have a probability of escaping and guranteeing u can detain them all. |
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Title: Re: Forcefield Detainment Post by towr on Nov 20th, 2006, 12:59am on 11/19/06 at 17:21:41, flamingdragon wrote:
Quote:
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Title: Re: Forcefield Detainment Post by flamingdragon on Nov 20th, 2006, 8:22pm Ok, they maybe they wont sacrifice themselves, but still they would think there is some probability of escape since they dont know he has 100% aim and him not shooting them for some reason also works. If the guard never misses, he could shoot the technician in the head and he wouldnt have a dying breath. ::) And u dont know why the technician is updating the forcefield, it could just be that hes upgrading it while it doesn't need to be. And i'm sure the riddle doesn't mean to keep them secure indefinately, just for the moment. If it meant indefinately then it would be impossible as the forcefield would eventually run out of power or the prisoners and the guard would eventually die. |
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Title: Re: Forcefield Detainment Post by towr on Nov 21st, 2006, 1:02am I don't think the forcefield is battery operated, it's probably on a powergrid with a few backup generator. Guards and prisoners inevitably change. But no one will service the forefield because of the risk of being shot by the guard, so in the end it will fail. :P |
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Title: Re: Forcefield Detainment Post by flamingdragon on Nov 21st, 2006, 11:49am They would probably die b4 the forcefield failed, and their empire or whatever would fall eventually when their sun goes supernova, or some other disaster. They could not be detained forever. That has to be the correct answer, its the ONLY way to gurantee none will escape. |
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Title: Re: Forcefield Detainment Post by towr on Nov 21st, 2006, 12:39pm on 11/21/06 at 11:49:02, flamingdragon wrote:
Or kill them with radiation or something (such that you don't have to drop the forecfield to do it); being dead usually stops people running off (except when they turn into zombies). |
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Title: Re: Forcefield Detainment Post by flamingdragon on Nov 21st, 2006, 2:06pm I think keeping them detained might mean keeping them alive as well, but I could be wrong about that. There probably isn't enough time to do any of that b4 the forcefield goes down though. |
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Title: Re: Forcefield Detainment Post by Icarus on Nov 21st, 2006, 7:21pm on 11/21/06 at 11:49:02, flamingdragon wrote:
The intent of this riddle is that the guard can and will - with complete certainty - kill whichever one prisoner he chooses. The trick to the riddle is the inductive effect that allows you to conclude none of the prisoners will move as each knows that the ones who would be shot before him will not move; so if he moves, he will be shot. The inductive solution is the intended one for the riddle. |
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Title: Re: Forcefield Detainment Post by flamingdragon on Nov 25th, 2006, 6:34pm on 11/19/06 at 17:21:41, flamingdragon wrote:
Hmm, somehow I don't see this being disproven... |
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Title: Re: Forcefield Detainment Post by rmsgrey on Nov 26th, 2006, 9:43am on 11/25/06 at 18:34:50, flamingdragon wrote:
Well, if you want to discuss low probability events, its possible that the prisoners would all spontaneously fall through the back walls of the cells and escape that way (macroscopic quantum tunneling's incredibly low probability but still (I believe) theoretically possible...) |
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Title: Re: Forcefield Detainment Post by flamingdragon on Nov 26th, 2006, 1:58pm LOL, very well then. They would try it and fail and still all be detained. However it says on ANY chance of escape. It is most likely the guard is not defecting, however they would still take it as it is possible, one would then most likely get shot and the others would escape, same thing with them not believing he has perfect aim. |
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Title: Re: Forcefield Detainment Post by Zatanna on Aug 15th, 2007, 3:26pm I agree with towr. There's not enough time for a mob to escape; the threat of a re-activated force-field would insure that EVERYONE would want to be the first out, causing a jam at the exit. Appointing a "trustee" or two could stop the escape attempt cold, it's worked for Southern chaingangs for decades. |
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Title: Re: Forcefield Detainment Post by srn347 on Aug 26th, 2007, 9:32pm Either shoot the technician or threaten to shoot the first to move at all. |
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Title: Re: Forcefield Detainment Post by mikedagr8 on Aug 26th, 2007, 9:56pm Get n guards for n prisoners each with n bullets who have perfect marksmanship. Now the forcefield can be down forever. ::) |
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Title: Re: Forcefield Detainment Post by srn347 on Aug 27th, 2007, 10:45am You can't do that. How about numbering them and threatening to shoot the highest number that tries escaping(gets a certain distance of the forcefield). |
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Title: Re: Forcefield Detainment Post by mikedagr8 on Aug 27th, 2007, 2:11pm ::) |
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Title: Re: Forcefield Detainment Post by Tucan_Sam on Feb 7th, 2009, 4:57pm how about you offer a reward to the strongest prisoner if he breaks the legs of all the other prisoners. If he does not complete this mission. You offer the next strongest prisoner a reward if he finishes the mission... so on and so on until only one prisoner remains without broken legs. You have him drag all the others to the middle of the force field. When the force field is lowered there is only one person who can escape and that person is aware he will be shot and wont attempt to escape. |
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