The New Adventures of Mr Stephen Fry » Topic: Three Prisoners

The New Adventures of Mr Stephen Fry » Topic: Three Prisoners http://www.stephenfry.com/forum/topic/three-prisoners Just another bbPress community en-US Thu, 20 Jun 2013 10:23:35 +0000 http://bbpress.org/?v=1.0.2 <![CDATA[Search]]> q http://www.stephenfry.com/forum/search.php The Oracle on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners/page/2#post-291343 Sun, 27 Jan 2013 20:47:38 +0000 The Oracle 291343@http://www.stephenfry.com/forum/ Can someone explain it again please. Drokles on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284819 Fri, 08 Jun 2012 20:30:30 +0000 Drokles 284819@http://www.stephenfry.com/forum/ He knows he is not the first prisoner because he knows that it is no longer the first day of their sentence when he enters the room. Therefore the first prisoner must have already been there :). With a hundred prisoners there are a few different strategies. An article was written about the different solutions and the expected jail times :D. It can be found here: <a href="http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf" rel="nofollow">http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf</a> Thanks for playing guys <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_biggrin.gif" title=":D" class="bb_smilies" /> Stefi on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284793 Thu, 07 Jun 2012 18:12:53 +0000 Stefi 284793@http://www.stephenfry.com/forum/ Great riddle! Although, how does the third prisoner know that he is not, in fact, the first to be called in? And how would it work with 100 prisoners? Desideria on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284785 Thu, 07 Jun 2012 10:53:08 +0000 Desideria 284785@http://www.stephenfry.com/forum/ But only with your help! I had been thinking in the wrong direction again but your summary got me back on track. Thank you for the riddle, I enjoyed trying to solve it. <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_smile.gif" title=":-)" class="bb_smilies" /> Drokles on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284779 Wed, 06 Jun 2012 21:15:34 +0000 Drokles 284779@http://www.stephenfry.com/forum/ A winner is you! Desideria on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284777 Wed, 06 Jun 2012 21:02:18 +0000 Desideria 284777@http://www.stephenfry.com/forum/ Ah! So the third one who enters the room for the first time will see that the light is switched off. He therefore knows that he can't be the second one to enter the room because if that were the case the light would still be switched on. However, the light is switched off when he enters the room, therefore he knows that the second prisoner has also already been there. This means that he can now announce that all three prisoners, himself included, have been in the interrogation room. This protocol, however, does only work, when those who have been in the room several times do nothing and just leave the light switched off. Drokles on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284771 Wed, 06 Jun 2012 16:36:26 +0000 Drokles 284771@http://www.stephenfry.com/forum/ You've actually almost solved it now. So to sum up your protocol thus far: You've got the first prisoner coming in there to turn on the light, and the light will stay on as long as it is this same guy who is called in. When a new prisoner then is called in, this new prisoner will see the light is on and think 'one of the others has been here before me. I'm the second prisoner to enter the room.' The new prisoner then turns off the light. This can also be formulated as a set of instructions to the prisoners: If you enter the room and it is the first day, turn on the light, and if you reenter the room on subsequent days, do nothing. If you enter the room for the first time and it is not the first day and the light bulb is on, turn off the light. So. You now have three distinct prisoners at this point. One who came in there on the first day and turned on the light. One who came in there as the second one to enter the room for the first time who turned off the light. One who hasn't been in the room yet. Desideria on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284767 Wed, 06 Jun 2012 16:22:08 +0000 Desideria 284767@http://www.stephenfry.com/forum/ I'm on the right track? <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_biggrin.gif" title=":D" class="bb_smilies" /> Never thought that I could actually be on the right track with something that involves logical thinking. Thank you for the hints! Quite helpful, although I don't think I will find the correct solution. However, this is what I can come up with at this point: I think that the position of the toggle switch can be used to indicate if someone has been in the interrogation room more than once. Therefore, they will need a counting system of some sorts and they will have to take notice of the status of the light bulb (on/off). They can perhaps decide that the first person who goes into the room on the first day switches the light on and the light has to stay switched on for as long as the person who came into the interrogation room on the first day goes in there for a second time. Then this person will have to switch the light off as a signal to the others. And then the counting starts anew with another prisoner ... Or not. Now I'm lost. <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_lol.gif" title=":lol:" class="bb_smilies" /> Drokles on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284757 Tue, 05 Jun 2012 20:02:01 +0000 Drokles 284757@http://www.stephenfry.com/forum/ In a fair and just world that would certainly work. Unfortunately the prisoners do not have lawyers to represent them, and the warden values game theory over justice. Stefi on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284755 Tue, 05 Jun 2012 19:38:21 +0000 Stefi 284755@http://www.stephenfry.com/forum/ Hmm ok with logical thinking I'd say they should demand a lawyer and try to sue the prison for this frankly very weird way of determining guilt or innocence...? Maybe it's a trick question and that's the solution. It's worth a try. Drokles on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284753 Tue, 05 Jun 2012 19:36:08 +0000 Drokles 284753@http://www.stephenfry.com/forum/ That's not quite it, although I can see the appeal of having four independent indicators of who has been there (the four walls) instead of just one (the light bulb). It is possible though, communicating only through the light bulb :). <blockquote><cite>Desideria <a href=""//www.stephenfry.com/forum/topic/three-prisoners#post-284737""">said</a>:</cite> It's a fascinating riddle but I'm really bad at logical thinking. Perhaps they could agree that Prisoner 1 always switches the light on, Prisoner 2 switches it off and that Prisoner 3 never touches the toggle switch. However, I have no idea what would happen if, say, Prisoner 2 would be placed in the interrogation room on two consecutive days because that would mean that he wouldn't be allowed to touch the toggle switch, an action which is only allowed to Prisoner 3. Hm. However, what I am very sure of is that they must try to communicate via the toggle switch. There doesn't seem to be a way around that. A hint, perhaps? <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_biggrin.gif" title=":D" class="bb_smilies" /> </blockquote> You are on the right track :D. Here's a hint: [spoiler] Think about how you plan to assign the roles of prisoner #1, #2 and #3. Do they decide who gets to be who from the beginning? Or are roles assigned dynamically? Remember that when a prisoner enters the room, he knows <ul> What the current state of the light bulb is Whether he himself has been there already or not (and how many times if any) How many days have passed </ul> The actions he takes inside the room should depend on this information, and possibly his role as either prisoner #1, #2 or #3. I hope I haven't just made it confusing now :D. [/spoiler] Edit: Oh, spoiler tags don't work? Oh well. Popeye on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284747 Tue, 05 Jun 2012 16:45:45 +0000 Popeye 284747@http://www.stephenfry.com/forum/ Assuming the room in which the light bulb is has just four walls, they could agree to each move the toggle in the direction of a different wall. Once one of them had seen it in the other two positions to their own then they would know everyong had visited the room. Is this in any way close or am I talking gobbledegook!?! <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_confused.gif" title=":?" class="bb_smilies" /> Desideria on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284737 Tue, 05 Jun 2012 16:25:01 +0000 Desideria 284737@http://www.stephenfry.com/forum/ It's a fascinating riddle but I'm really bad at logical thinking. Perhaps they could agree that Prisoner 1 always switches the light on, Prisoner 2 switches it off and that Prisoner 3 never touches the toggle switch. However, I have no idea what would happen if, say, Prisoner 2 would be placed in the interrogation room on two consecutive days because that would mean that he wouldn't be allowed to touch the toggle switch, an action which is only allowed to Prisoner 3. Hm. However, what I am very sure of is that they must try to communicate via the toggle switch. There doesn't seem to be a way around that. A hint, perhaps? <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_biggrin.gif" title=":D" class="bb_smilies" /> Drokles on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284625 Fri, 01 Jun 2012 16:16:57 +0000 Drokles 284625@http://www.stephenfry.com/forum/ Sadly, it is not that easy ;). But nice idea, I wouldn't have thought of it. Some of my friends came up with the idea that you could smash the light bulb at some point, giving you an extra degree of freedom. Of course, you could do that, but there's no reason to. It is a riddle that encourages logical thinking - there are no tricks. Stefi on "Three Prisoners" http://www.stephenfry.com/forum/topic/three-prisoners#post-284331 Wed, 16 May 2012 20:07:55 +0000 Stefi 284331@http://www.stephenfry.com/forum/ Hm I would probably do something like let every prisoner bind a small thread on the toggle switch when they are in the room, and when a prisoner sees two threads on there he knows that both have been there <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_smile.gif" title=":)" class="bb_smilies" /> But I don't think that's really the answer... more of an idea <img src="http://www.stephenfry.com/bb-content/plugins//bb-smilies/default/icon_wink.gif" title=":wink:" class="bb_smilies" />