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Title: BROWN EYES AND RED EYES Post by Axela on Jul 24th, 2002, 2:09am I can think of only 3 cases: 1) All the monks have brown eyes: mass suicide 2) Only 1 monk has red eyes: he kills himself 3) There at least 2 monks with red eyes: nothing happen. I'm missing something? |
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Title: Re: BROWN EYES AND RED EYES Post by SFoster on Jul 24th, 2002, 3:22am If 1 person has red eyes they commit suicide on the first night, if 2 people they both do the deed on the second night, in general if n people have red eyes all n commit suicide on the nth night. |
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Title: Re: BROWN EYES AND RED EYES Post by David Elstob on Jul 25th, 2002, 1:51am Yeh but why??? That explantion might be a reason but I can't see how you followed it through to get that answer. Please explain this one hurts my head. |
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Title: Re: BROWN EYES AND RED EYES Post by David Elstob on Jul 25th, 2002, 2:06am Oh I get it now. Cool. 8) |
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Title: Re: BROWN EYES AND RED EYES Post by Ad on Jul 25th, 2002, 3:34am I thought they'd just use the reflection of the water surrounding the 'island monastary'. ;) |
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Title: Re: BROWN EYES AND RED EYES Post by Eli on Jul 25th, 2002, 5:23am Ok, explain it to me then, because I am still not getting it. As long as two monks have red eyes, noone should kill themselves. As any 'red-eyed monk', can always see another red-eyed monk, and then not be sure that it is themself that is red-eyed, and therefore, noone kills themself. So explain to me this 'n days' theory. |
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Title: Re: BROWN EYES AND RED EYES Post by SFoster on Jul 25th, 2002, 7:48am Say there's two monks with red eyes. They each look about on the first day and see one other guy, so they expect him to commit suicide that night. When that doesn't happen he figures out he must have red eyes too, so both guys commit suicide on the second night. When there's 3 monks with red eyes they commit suidice on the third night, and so on. |
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Title: Re: BROWN EYES AND RED EYES Post by blahedo on Jul 25th, 2002, 8:04am That's what I thought at first too, until I read that solution. Here's the reasoning. Say I'm the only red-eyed monk. When the tourist says that, knowing that everyone else has brown eyes, I commit suicide. Now say that I'm one of two red-eyed monks. When the tourist says that, I think, "I might be ok, that other guy has red eyes." But when he *doesn't* commit suicide, I know that he must be seeing at least one *other* red-eyed monk, since otherwise he would have committed suicide the night the tourist said that (due to the reasoning in the previous paragraph). Now I know that there are at least two red-eyed monks, and I can only see one, so I'm the other one, so I commit suicide---on the second night. If I'm one of three red-eyed monks, when the first night goes by with no suicide I realise that the two red-eyed monks I can see are going through the reasoning in the previous paragraph, and will commit suicide that night (the second night). But when they don't, I know that they must each see at least two others, and thus I know that one of them must be me. So I commit suicide, on the third night. The number of brown-eyed monks is actually totally irrelevant. |
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Title: Re: BROWN EYES AND RED EYES (solution) Post by ThatTallGuy on Jul 25th, 2002, 8:32am I think I got it... you have to assume that every monk knows every other monk and knows how many _other_ monks on the island have red eyes. Under those assumptions, look at the case where one monk has red eyes. He knows, immediately after the visitor arrives, that _he_ is the one because he can see that everybody else has brown eyes. Honorable hara-kiri follows that night. If two monks on the island have red eyes, nothing happens the first night because each of the red-eyes sees that somebody _else_ has red eyes and the visitor might only have been referring to them. But each of the red-eyes wakes up the next morning and realizes that there is _more_than_one_ set of red eyes on the island, because the one he can see didn't kill himself. Since red-eyed monk A knows that there is only one _other_ red-eyed monk (B), since he can eliminate all the rest of the inhabitants (he knows their eyes are all brown), then _he_ (A) must be the one who has kept the other cursed fellow (B) alive -- the existence of A let B believe that the visitor might have only been talking about A. So A and B now know they both have red eyes. Everybody else knows that there were two red-eyed monks and so does nothing. If there are three monks on the island, same thing happens, except it takes three days for the realization to strike. Etc., etc. Cool. |
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Title: Re: BROWN EYES AND RED EYES Post by bl@ke on Jul 25th, 2002, 7:01pm Taking a different view - the tourist is color blind! They all have brown eyes but the color blind tourist tells them that "at least one of them has red eyes". Every monk then looks to every other monk to see that they actually have brown eyes. By the process of elimiation every monk thinks that they are the one with red eyes and so every monk commits suicide. |
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Title: Re: BROWN EYES AND RED EYES Post by jkl on Jul 25th, 2002, 10:12pm what if there are 4 monks with red eyes? then if i am one of them.. and i see the three other monks not killing themselves.. what am i supposed to think?? |
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Title: Re: BROWN EYES AND RED EYES Post by Rhaokarr on Jul 26th, 2002, 1:07am And on the N+1 night, all the remaining monks breath easily... |
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Title: Re: BROWN EYES AND RED EYES Post by KNOWITALL on Jul 26th, 2002, 10:14am Answer to Red Eyes/Brown Eyes We know the following: 1. No monk had killed themselves to date. 2. A tourist has introduces new information that causes a definitive action. Hence the tourist's information must have been unknown to at least one monk. The tourist's information is that at least one monk has red eyes. Therefore, at least one monk must have been unaware that there were any monks with red eyes. However, there could not have been more than one such monk, If there were two monks with red eyes, they would certainly have seen each other and already knew the tourist's information. If they both already knew the tourist's information, there would have been no 'definitive' action as a result. Likewise with three, etc. Therefore, there was exactly one monk with red eyes. After the tourist revelation, he knew that he must be the monk with red eyes (he can see that one else has red eyes), and he properly kills himself at midnight. Answer: Exactly one red-eyed monk kills himself at midnight, leaving no more red-eyed monks. |
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Title: Re: BROWN EYES AND RED EYES Post by Alex on Jul 26th, 2002, 10:30am Respectfully, taking your scenario where there are two monks with red eyes: Until the tourist introduced the information, each monk would be uncertain whether he had red or brown eyes, and would be uncertain if the other monk was aware that he had red eyes. From each monk's point of view, the other monk may simply be "unaware of his eye color." When the tourist announced that there is at least one with red eyes, the resolution then becomes unavoidable, as both become aware that the other is aware that at least one monk has red eyes. The same scenario being true as well with more than two. The information in these scenarios that changes is not what the monks know of themselves, but what they know of the knowledge of other monks. |
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Title: Re: BROWN EYES AND RED EYES Post by tot on Jul 26th, 2002, 11:51am on 07/24/02 at 02:09:14, Axela wrote:
In the case 3, mass suicide on the second night? After the first night, when nobody kills himself, everybody thinks that they must have red eyes too. |
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Title: Re: BROWN EYES AND RED EYES Post by tot on Jul 26th, 2002, 11:52am on 07/24/02 at 02:09:14, Axela wrote:
In the case 3, mass suicide on the second night? After the first night, when nobody kills himself, everybody thinks that they must have red eyes too. |
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Title: Re: BROWN EYES AND RED EYES Post by srowen on Jul 26th, 2002, 12:25pm The answer given by others above is correct: if there are N monks with red eyes, then they will all kill themselves on the Nth night. They have been told that at least one person has red eyes. If there is exactly one, the one will easily deduce his eye color and kill himself on the first night. If nobody kills themselves the first night, then all can deduce that there must be at least two red-eyed monks in total. If there are exactly two, then each of the two will see only one other red-eyed monk and easily deduce that his own eyes must be red, and both will commit suicide on night 2. Same goes for 3, 4, ... |
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Title: Re: BROWN EYES AND RED EYES Post by I.M._Smarter_Enyu on Jul 26th, 2002, 12:53pm I think you've got the logical part down... but I don't think that's what really happens, though! See the other post about the "Unexpected Quiz." |
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Title: Re: BROWN EYES AND RED EYES Post by srowen on Jul 26th, 2002, 1:33pm Really, then what do you think happens? I don't think is the same sort of problem as the "unexpected quiz" puzzle you posted. |
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Title: Re: BROWN EYES AND RED EYES Post by Rhaokarr on Jul 26th, 2002, 8:53pm All that the tourists information does is set time to 'zero', acting as a marker point in time. There is also the assumption that every monk hears the tourist's statement, and also knows that every monk knows that the others heard as well. |
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Title: Re: BROWN EYES AND RED EYES Post by Frost on Jul 28th, 2002, 4:52am A long time ago, before the tourist visited, the first red-eyed monk came to live at an all-brown-eyed monestrary. Nothing happened. When a second red-eye arrived, all monks knew there was at least one red-eyed monk, but nothing happened yet. When a third red-eye came to live with the monks, all monks knew all others monks knew at least one of them had red eyes, and all red-eyes died three days later. Therefore, no more than two red-eyed monks can exists at the monestary. |
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Title: Re: BROWN EYES AND RED EYES Post by srowen on Jul 28th, 2002, 8:04am Interesting, but it's equally possible that all the red-eyed monks were there before any brown-eyed monks arrived. Surely this is not a material aspect of the problem though. |
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Title: Re: BROWN EYES AND RED EYES Post by Frost on Jul 29th, 2002, 1:16am If, at any time, the monestary has >2 red-eyed monks, each monk can see at least two red-eyed monks. From this information they know each other monk knows at least one of them has red eyes. This results a Nth-day suicide. There are only four stable situations in which this monestary can exists: 1) no red-eyed monks 2) one red-eyed monk 3) two red-eyed monks 4) a new red-eyed monk is added each day (hardly stable) When the visitor reveals his additional information, this will happen in each of the above cases: 1) all monks kill themselves tonight 2) he kills himself tonight 3) they kill themselves tomorrownight 4) nothing happens |
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Title: Re: BROWN EYES AND RED EYES Post by Thott on Jul 29th, 2002, 2:49am I've done a similar riddle before, and I expect this one is intended to be a copy of it. What is being assumed in previous posts, but is lacking in this riddle description, is the phrase 'every monk sees every other monk every day'. Without that piece of information, it's impossible to solve. One can't even assume that every monk has ever seen every other monk, much less it happening every day. --Thott |
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Title: Re: BROWN EYES AND RED EYES Post by pa0pa0 on Jul 29th, 2002, 5:57am There's lots of interesting stuff to discuss about this riddle. For example, did you know that the tourist is actually giving out more information than he needs to? (Well, to be more precise, he's actually increasing the monks' knowledge more than he needs to.) (Yes, and I'm aware that alot of people here think he isn't increasing the monks' knowledge at all.) But before we get on to that, could we clear the ground by convincing Frost that his arguments don't hold water? I'm not sure how to do that, but let's try this: Frost, as soon as there are three red-eyed monks, BUT no tourist, it's true, as you say, that Everyone knows that everyone knows that there's at least one red-eyed monk. It is also true that Everyone knows that there are at least two red-eyed monks (BUT not everyone knows that everyone knows that there are at least two - in particular only the brown-eyed monks know that everyone knows this) It is also true that There are at least three red-eyed monks (BUT not everyone knows this - in particular, only the brown-eyed monks know this.) Can you see that without the tourist, nothing will happen? In particular, nothing will change as a result of no one dying overnight on the first night. However, IF it were ever true that Everyone knows that everyone knows that everyone knows that there's at least one red-eyed monk THEN on the second day it would become true (if no one died) that Everyone knows that everyone knows that there are at least two red-eyes THEN on the third day it would become true (if no one died) that Everyone knows that there are at least three red-eyes AS A COROLLARY OF WHICH the red-eyed monks would know their own eye colour. |
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Title: Re: BROWN EYES AND RED EYES Post by srowen on Jul 29th, 2002, 5:58am Not true - say there are three red-eyed monks. They each see two others. It's quite possible, to each, that the other two are the only one red-eyed ones. The other two would not be able to figure out whether they both had red eyes or not, as you argue. None can deduce anything from this - this is stable. 0 red-eyed monks is stable, but not part of this problem. 1 is stable, until the visitor arrives! n is stable as long as n-1 is, along the same lines as above. So indeed the whole thing falls down when the visitor arrives, and monks can start making conclusions about the state of things. And yes, I think it is assumed that they have all seen each other. They have to eat and do monk-chores every day! I don't think they have to see each other daily, but it doesn't matter... this is not important to the problem. on 07/29/02 at 01:16:44, Frost wrote:
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Title: Re: BROWN EYES AND RED EYES Post by Frost on Jul 29th, 2002, 10:06am I agree. Any system before the tourist's magic phrase is stable. |
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Title: Re: BROWN EYES AND RED EYES Post by pa0pa0 on Jul 30th, 2002, 8:53am on 07/29/02 at 05:58:33, srowen wrote:
Hmm, that's a fascinating question. I *think* that the following is the case: (1) IF death-yells are guaranteed to resound throughout the monastery, AND they are willing to accept the risk of the logic going haywire should all the red-eyed monks die in the interim of natural causes, THEN no one needs to see each other, or have any communication whatsoever, other than listening out for and if necessary making death-yells, once the tourist has departed. (They do all have to be together when the tourist makes his statement, if they are to start with common knowledge of his statement.) Admittedly, the situation is quite bizarre. Suppose there are 100 monks, of which 42 have red eyes. When the tourist departs, everyone can go to their separate cells, and stay there. Absolutely nothing appears to happen, until midnight on the 42nd night, when unearthly yells start resounding through the monastery. (The yells don't need to be synchronized, nor do the remaining monks need to try counting the yells; everyone already knew that if any yell occurs on the 42nd night, then there will be 42 yells.) The monks' algorithm is even more simple than the above might suggest: actually, each monk only has to listen out for death-yells on one particular night. His algorithm is: if at the tourist's gathering (on day 1) I saw R red-eyed monks, and there are no death-yells on night R, then I die on night R+1. One might ask, how can that possibly be? How can it take 42 days for something to happen, when nothing appears to happen in the interim to provide any relevant lead-up? It's not as though *absence* of yells on the 17th night (for example) is needed in order to increase anyone's knowledge on day 18. Everyone already knew that there wouldn't be any yells on the 17th night. Bizarre, yes. But anyway, when one thinks about it a bit more, the situation isn't really any more bizarre than (and is in fact identical to) the normal scenario where the monks see each other (but don't say anything) every day. (2) If death-yells don't work, then I think the situation is much more complicated. Let us suppose, for definiteness, that the only time when the monks can be sure of seeing each other (if they are still alive) is at breakfast. And again, let us have 42 red-eyed monks (Rs) and 58 brown-eyed monks (Bs). Now, before the tourist arrives, each R knows that there are either 41 or 42 Rs, and each B knows that there are either 42 or 43 Rs. That knowledge stays the same after the tourist's statement on day 1, but they now additionally know the following: Each R knows that: Either at the end of day 41, 41 monks (not including himself) will die; or at the end of day 42, 42 monks (including himself) will die. And each B knows that: Either at the end of day 42, 42 monks (not including himself) will die; or at the end of day 43, 43 monks (including himself) will die. Thus everyone knows that no one will die (unless of natural causes) until the end of day 41 at the earliest. So does anyone actually need to see each other before that time? The natural inclination of monks being to fast, couldn't they all just skip breakfast until day 41? Or even day 42? After all, each individual knows that although breakfast on day 41 might be interesting to someone else, it won't be interesting to himself. Not only that, but perhaps the Bs don't even need to turn up until day 43? (The fact that they don't know yet that they are Bs is not relevant, at least not in any straightforward way. And if they don't turn up on day 42, that will certainly tell the Rs something, but it is something that those Rs are going to discover anyway.) I'm still thinking about that. Does anyone have the answer? I have a strong feeling that fasting is illegal - not because it disturbs the logic of the riddle, but because it violates the rule that the monks mustn't reveal anything about each others' eye colour. Except that if it does violate that rule, it could cause an R to die too early, which *would* disturb the logic of the riddle. Hmm... |
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Title: Re: BROWN EYES AND RED EYES Post by pa0pa0 on Jul 30th, 2002, 9:17am P.S. to my note a moment ago: upon reading it through now that it's all nicely formatted by the forum software, I've suddenly realized how the absence of any yells on night 17 *does* increase everyone's knowledge on day 18, even though everyone already *knew* there wouldn't be any yells on night 17. Consider propositions of the form "everyone knows that everyone knows that ... (J times) ... that there won't be any yells on night 17". Before night 17, those propositions were true for values of J only up to some critical value K (round about 16, but if I try to work it out exactly, I'll get it wrong). On day 18, those propositions are true for *all* values of J, since the absence of yells on night 17 has now become common knowledge. (In fact, I think the logic of the riddle only requires that the proposition become true for the single extra value of K+1, so the absence of those yells is telling the monks more than they need to know!) |
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Title: Re: BROWN EYES AND RED EYES Post by cnmne on Jul 30th, 2002, 4:19pm Assuming that suicide is a sin if you do not have red eyes, then the only way a monk would kill himself after that statement is if he was the only one with red eyes. Otherwise he would see someone else with red eyes, assume it was not him, and not suicide. If the tourist was looking at a group photo, and saw red eye, not red eyes, then they would all suicide because they would see no one else with red eyes and have to assume it was themselves. If more than one had red eyes, both would assume the other was alone and neither would suicide. |
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Title: Re: BROWN EYES AND RED EYES Post by srowen on Jul 30th, 2002, 4:56pm See the preceding discussion - we've been over this. Imagine you see one other red-eyed monk. What do you think to yourself when, the following morning, you see him again in the courtyard? What must he see, and what does it mean about you? Extend from there. |
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Title: Re: BROWN EYES AND RED EYES Post by mook on Aug 2nd, 2002, 9:30am While I don't dispute the fact assuming that all of the monks have seen each others eyes and all of the monks heard the tourist, what we must also assume is that these monks must never gesture or write someone a note, or hand them a knife or a rope, letting them know that they have red eyes. I'm willing to assume that this red-eye curse thing is a tenent of their religion, that as monks(I'm assuming they would be devout followers of said religion) they would be morally obligated to somehow signal the red-eyed monks what their true eye-color is. With that in mind, I see only two possiblities: 1) the monks are devout followers of their religion, the tourist lied and they commit mass suicide. 2) no one commits suicide. the monks with red eyes have been living there for this long with all of the other monks knowing that their eyes were red not making a big deal about it, the tourist didn't know anything about the eye curse, and with the monk's all having taken a vow of silence, who's going to say anything? |
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Title: Re: BROWN EYES AND RED EYES Post by Tony Mountifield on Aug 13th, 2002, 8:22am But surely: If there were 2 or more monks with red eyes, and they could all see each other, they ALL already knew that "at least one of them has red eyes". So the visitor did not add any new information. He could not then have kickstarted a countdown of days. The only way the visitor increased the information known by any of the monks was if there was only one with red eyes. The red-eyed monk would then know that it was him, since he could only see brown eyes, and commit suicide on the first night. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 13th, 2002, 8:51am Good point: the key is that the visitor's comment tells them more than that: all the monks now know for sure that all the monks must know that at least one has red eyes. In the case of two monks with red eyes, agreed - both already know that at least one has red eyes. We, the riddlers that posit this situation, "know that." But crucially, each of those two monks doesn't know that the other knows that, because each is not sure of his own eye color. Monk A might see that B has red eyes, but doesn't know if he himself has red eyes. Therefore he isn't sure if B knows that "at least one monk has red eyes." Once the visitor makes his comment, things change, not because they now know "at least one monk has red eyes" - they each knew that - but because they are sure the other knows it. Then the chain of reasoning starts to get somewhere. This holds for any number of red-eyed monks. |
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Title: Re: BROWN EYES AND RED EYES Post by tot on Aug 13th, 2002, 1:35pm If the monks see each other every day, they already knew for sure that there is at least one monk with red eyes. The visitor did not give any new information to the monks. What he really did was a reference point in time that is known to everybody. They needed day 0 to make the Nth day mass suicide. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 13th, 2002, 1:44pm on 08/13/02 at 13:35:24, tot wrote:
Yes, the visitor did give them new information. See above - after that they all know that all monks know that at least one has red eyes. This is what allows them to start deducing things about themselves. This "marking a reference point in time" changes nothing about their deductions or information, so that alone cannot be the new thing that makes some of them deduce that they have red eyes. |
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Title: Re: BROWN EYES AND RED EYES Post by Tony Mountifield on Aug 13th, 2002, 1:48pm OK, I don't get the bit about a time reference, but I can now see S Owen's reasoning for 2 red-eyed monks. Each of them could only see one red-eyed monk. But they didn't know whether they were also red-eyed. So A would reason: "if I have brown eyes, then B can't see a red-eyed monk - now he knows there is one, he has to conclude it is himself and commit suicide tonight. If he doesn't, then I must have red eyes too." B reasons similarly, and the following day both A and B see the other alive and realise they each have red eyes. I'm still trying to fathom how it works with 3 red-eyes, because the ALL the monks already could see more red-eyes than the visitor mentioned, AND knew that all the others could already see at least one. :-/ |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 13th, 2002, 1:53pm Well imagine there are three red-eyed monks. Each sees two others. Each is not sure whether the others are the only two or not. You see how it works for one and two red-eyes monks, yes? One kills himself the first night; or both kill themselves on the second night, respectively. So if on the third morning, you still see those same two red-eyed monks, then you know they must be seeing another set of red eyes - yours! The reasoning in this situation still relies indirectly on knowing that everyone knows there is at least one red-eyed monk. Until you're sure everyone knows that, it doesn't mean anything that those two other red-eyed monks haven't killed themselves. Once everyone knows that, you can infer things from who is and isn't killing themselves. |
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Title: Re: BROWN EYES AND RED EYES Post by Tony Mountifield on Aug 13th, 2002, 2:02pm on 08/13/02 at 13:48:25, Tony Mountifield wrote:
OK, I get it now. For 3 red-eyes: A reasons "if B and C are the only two, they will know that tomorrow morning, and will die tomorrow night." When B and C survive another night, A knows he has red eyes too. Same for B and C, by symmetry. For 4 red-eyes, he can see the other three. When they survive a day longer than expected, he and they all know. Yes? |
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Title: Re: BROWN EYES AND RED EYES Post by Tony Mountifield on Aug 13th, 2002, 2:04pm (still waiting for my password) Yes, I'm glad to say I worked it out and didn't see S Owen's last post (thanks anyway!) until I'd posted mine! Right onto the next..... |
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Title: Re: BROWN EYES AND RED EYES Post by tot on Aug 14th, 2002, 4:46pm on 08/13/02 at 13:48:25, Tony Mountifield wrote:
What I mean by time reference, is that every monk needs to have a day 0 to start the logic that S. Owen has already defined. It really does not matter exactly what the visitor says, as long as it is a clear reference to all the monks start to wonder why the red-eyed monks have not yet committed suicide, even they should have done it already. The riddle just says that the current situation is what it is and that there are red-eyed monks in the island. All that is needed is some way to start the counting of nights to find out how many red-eyed monks everyone else sees. Of course, if the red-eyed monks do not follow this logic but the brown-eyed ones do, it could be that brown-eyed ones commit suicide instead on Nth + 1 night, or just part of the red-eyed do, or that everybody commits suicide next night thinking they are the missing red-eyed one. But I believe that the intetion of the riddle is the Nnth night sucide. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 15th, 2002, 7:35am on 08/14/02 at 16:46:15, tot wrote:
No, a point from which to count nights is decidedly not all that they need. If the visitor had simply said, "Hey monks! Tonight is night zero, if you know what I mean!" then nothing would have happened. No monk would then be able to deduce anything more about his own eye color. It does matter what the visitor says - the crucial fact that he communicates to all the monks is that "all monks now know that at least one monk has red eyes." This new information, not the simple fact that anything at all was said, allows a chain of reasoning to begin. |
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Title: Re: BROWN EYES AND RED EYES Post by pa0pa0 on Aug 15th, 2002, 1:06pm S.Owen, I don't think you've quite got the nub of it yet. on 08/15/02 at 07:35:36, S. Owen wrote:
Well, that was not new information if there are at least 2 red-eyed monks. You, 13th Aug >Yes, the visitor did give them new information. See above - > after that they all know that all monks know that at least >one has red eyes. This is what allows them to start >deducing things about themselves. Better, but no cigar. That was not new information if there were at least 3 red-eyed monks. If you had said "The visitor did give them new information: after his statement they all know that all monks know that all monks know that at least one has red eyes" - still no cigar. That was not new information if there were at least 4 red-eyed monks. It is precisely this difficulty which is causing people in this thread to keep going back to square 1 and ask what new information did the tourist really give. Please think carefully about the answer to this. Does anyone have any views on whether the monks are allowed to fast? (ref my earlier note, July 30th) |
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Title: Re: BROWN EYES AND RED EYES Post by Jonathan_the_Red on Aug 15th, 2002, 1:32pm Yes, thinking about the problem this way really highlights just how delicate the induction is. Suppose there are N monks with red eyes. Pick a random red-eyed monk (REM). He knows there are either N or N-1 REMs. He also knows: If there are N-1 REMs, they know there are either N-1 or N-2 REMs, and they also know: If there are N-2 REMs, they know there are either N-2 or N-3 REMs, and they also know: If there are N-3 REMs, they know there are either N-3 or N-4 REMs, and they also know: ... If there are 2 REMs, they know there are either 2 or 1 REMs, and they know also know: If there is 1 REM, he knows there are either 1 or 0 REMs. The tourist's statement settles the indecision of this hypothetical lone REM at the very bottom of this long chain of "if"s. One by one, day by day, the premises are shattered, until the chain reaches the top and all the REMs realize that there are in fact N REMs, not N-1, and they all kill themselves. Or... the monks, being smart, realize what's going to happen, all draw straws, with the one who picks the short straw selecting a REM at random and telling him: "You have red eyes." That one guy kills himself, the induction is broken, and the rest of the monks live happily ever after, mourning their lost brother but celebrating his sacrifice for the good of all. A month later, they get sensitivity training, get over their REMophobia, and learn that it's okay to have red eyes and nothing to kill oneself over. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 15th, 2002, 2:23pm on 08/15/02 at 13:06:27, pa0pa0 wrote:
OK, and I ask that you think carefully about this! I agree that my argument needs a more complete statement, but I maintain that it is correct. I understand and agree with your point about new information in the case of more than 2 red-eyed monks. For more than 2 red-eyed monks, I agree that the "new information" does not directly affect the reasoning of each monk - it does however affect it indirectly, if you will. Take 4 red-eyed monks. Each of them doesn't know whether there are a total of 3 or 4 red-eyed monks. Each is wondering whether the other 3 are the only red-eyed monks. And thus, they wonder whether those 3 are themselves wondering whether there are 2 or 3 red-eyed monks. And thus, whether those 3 monks are wondering whether the 2 others are wondering whether there are 2 or 1. And thus... whether each of those 2 is wondering whether the other is the only one or not. But that's where it all stops, at this "base case" of 2 monks wondering about each other. They can't figure anything out, and so won't do anything. Knowing this, the supposed 3 monks wondering about the other 2 won't do anything. Knowing this, the 4 monks wondering about the other 3 won't do anything. Therefore, the fact that nobody is committing suicide does not provide any information. But when they learn from the visitor that "all monks know that at least one monk has red eyes" (and I stress that this is all they need), then the base case is no longer indeterminate. Now two monks can figure it out. Now it does mean something when they do or do not kill themselves. Thus it is precisely the new fact that "all monks know that at least one monk has red eyes" that drives the chain of reasoning. Or to put it another way, what precisely are you arguing? That there is not a mass suicide on the Nth night? Or that there is, but that it's somehow the "marking a point in time" that makes it possible? |
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Title: Re: BROWN EYES AND RED EYES Post by pa0pa0 on Aug 16th, 2002, 7:23am on 08/15/02 at 14:23:21, S. Owen wrote:
Hmm. I invite you to reconsider your "I stress that this is all they need". In particular, I argue that to make the riddle work, it is not sufficient that they "learn from the visitor that all monks know that at least one monk has red eyes", as you put it. To try to make this clear, let me introduce a sequence of scenarios. To make these scenarios work, it has to be the case that there is no question about the trustworthiness of the tourist (that also has to be the case in the original riddle). Scenario 1: the tourist goes round to each monk, whispering to each "at least one monk has red eyes" and then leaves. Scenaro 2: as scenario 1, but before leaving the tourist goes around again, whispering to each "in my first round I whispered the same thing to each monk, namely that at least one monk has red eyes". Scenario 3: as scenario 2, but before leaving the tourist goes around again, whispering to each "in my second round I whispered the same thing to each monk, namely that in my first round ..." I hope you will agree with the following: - that in scenario 1, each monk learns from the visitor that at least one monk has red eyes; - that in scenario 2, each monk learns from the visitor that all monks know that at least one monk has red eyes - which according to you is sufficient to make the riddle work. I hope you will also agree - with scenario 1, if there is more than 1 red-eyed monk, then nothing will happen; - with scenario 2, if there are more than 2 red-eyed monks, then nothing will happen; - with scenario 3, if there are more than 3 red-eyed monks, nothing will happen. (In thinking this through, you have to keep bearing in mind that each monk learns the contents of all whispers in non-final rounds, but knows the content of only one whisper in the final round. Suppose that there are 3 red-eyed monks A, B and C. The problem is that in scenario 2, C never knows whether B knows that A knows that there is at least one red-eyed monk, because he is not privy to the tourist's final whisper to B.) Implication: the monks must be learning something more from the tourist than you have nailed down. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 16th, 2002, 8:19am I agree with your analysis of your scenarios, but aren't they slightly different than the riddle's scenario? In the original riddle, the visitor tells them all at once that "at least one monk has red eyes", presumably in a group. Your scenarios make a different assumption. That is, in the riddle they all have all the first-order, second-order, etc. knowledge at the outset, instead of in progressive rounds. |
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Title: Re: BROWN EYES AND RED EYES Post by AlexH on Aug 16th, 2002, 11:45am Jonathan has the right logic. The new information is that all monks know that "If there were 1 REM, he would know it now". If there are >2 REMs then they will all know that this information doesn't immediately apply to them, but it is still new information. Each morning they learn new information about a hypothetical which they know isn't realized. Note that they will all be able to predict what they will learn up until the next to last morning since they will know the hypothetical for that morning is false, but thats not the same as actually learning it because the statement wouldn't be true until that morning. The morning prior to the suicides, the hypothetical they learn about will be one which the REMs will know is realized. |
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Title: Re: BROWN EYES AND RED EYES Post by pa0pa0 on Aug 16th, 2002, 11:49am on 08/16/02 at 08:19:39, S. Owen wrote:
Yes, that's exactly it, and the point of my scenarios was to illustrate how the essential effect of the tourist's visit in the original riddle is to give them this nth-order etc. knowledge without which the riddle doesn't work. (Incidentally, for any particular number of red-eyed monks, there is a number of rounds in my scenario which would make the riddle work.) |
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Title: Re: BROWN EYES AND RED EYES Post by AlexH on Aug 16th, 2002, 12:01pm Thats true. The "if there were 1 he would know it now" has to be known to the nth order otherwise we stop learning about hypotheticals before we reach the one that applies. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Aug 16th, 2002, 12:10pm OK, sure. I guess I was just saying that they don't need to be explicitly told that all the nth-order propositions are true. I thought that was what you were saying. Rather, they know those just from the fact that the visitors tells them all in a group that there's at least one REM. That's why I'm saying that first statement is all they need, need to be given that is... |
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Title: Re: BROWN EYES AND RED EYES Post by loucura on Aug 22nd, 2002, 1:47pm The monks kill the tourist. Reasoning: If there are N+1 Red-Eyed Monks (REMS), then they should already have deduced that they are the final REM, and committed suicide LONG before the tourist arrives. Since they haven't, they are either slow, or busy, and have more important things to do with their time, like killing tourists. I don't think that the tourist "resetting" the day to zero can really hold water for N+1 REMs, because they've seen each other and -should- have deduced that the other REM sees another REM. So, either there is either only 1 REM, but since the hint says to generalise for N Red-Eyed Monks, (which is really N+1, since the hypothetical 'self-REM' doesn't know it's an REM), the only answer we can assume is that the monks kill the tourist. |
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Title: Re: BROWN EYES AND RED EYES Post by AlexH on Aug 22nd, 2002, 7:23pm Actually no, the monks are perfectly stable as is until the tourist speaks. This is one of the weird things about this riddle. The tourist says something which everyone already knows but saying it changes things at a much deeper order of "I know that you know that...". Even if new monks come and go there will never be a suicide until the tourists statement because no one ever knows something of the form "If there were k REMs they would know it now". |
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Title: Re: BROWN EYES AND RED EYE Post by James on Sep 4th, 2002, 5:56pm (Sorry for the long post, but the question has a long history - I hope that placing my comments in the middle of your notes is okay in this forum) on 08/15/02 at 14:23:21, S. Owen wrote:
What you seem to be saying here is that if there were (say) 100 monks with red-eyes then they eventually imagine that there might be monks that could only see 1 REM? This appears to be a subtle flaw in the argument (indeed, the person who first put this question to me many years ago noted that this was the case). Put simply, you have a chain of implications A->B, B->C, C->D, D->E etc leading to a final point which can only be resolved by the arrival of the tourist. However, the problem with this argument is that B true ->A is not true which brings into question the whole chain of implications. on 08/15/02 at 14:23:21, S. Owen wrote:
One further point to note - in this long chain of comments on the puzzle, you have put forward two slightly different inductive arguments. The first begins with N=1 and builds up (if nothing happens on night N then this means...) and the second (mostly here) starts with N=number of REMs and works back downwards. The chains of reasoning in each case appears inductive but probably isn't - becase inductive arguments build on the chain N assumed true -> N+1 must be true, but N+1 true has no automatic implication for N. However, in this case that is not exactly true. As an example, consider the standard example of induction - the sum of the first N integers is 1/2*N*(N+1). If N is 6 the sum of the first N integers is 21 - in our example one the 6th night, 6 REMs kill themselves (lets assume this is proved). However, if N is 6, the sum of the first five integers is still 15, but if N is 6, on the fifth night, 5 REMs don't kill themselves. So the question remains, is your answer an actual inductive proof, or is it subtly different. ;) |
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Title: Re: BROWN EYES AND RED EYE Post by S. Owen on Sep 4th, 2002, 9:46pm on 09/04/02 at 17:56:08, James wrote:
No, agreed, that wouldn't make sense. Each of the 100 wonders whether there are 100 or 99 REMs. And thus, whether the other 99 are the only ones and they are wondering whether there are 99 or 98, etc. So it's wondering about wondering... on 09/04/02 at 17:56:08, James wrote:
Yeah there is a sort of induction here. Showing what happens in the case of N REMs does depend on showing what happens in the case of N-1 REMs. Indeed, the whole thing hinges on the monks figuring out this induction too - maybe that makes this somewhat different. However, I don't see that the reasoning is invalid. I'm not sure if I'd call the sub-argument about first/second/third nights an induction... well it kind of looks like one I suppose. Again, whether or not it exactly follows the form of induction you've cited doesn't make it right or wrong... do you think this argument is flawed? |
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Title: Re: BROWN EYES AND RED EYE Post by James on Sep 5th, 2002, 12:06am on 09/04/02 at 21:46:34, S. Owen wrote:
However, while it is wondering about wondering, eventually you are saying that it is logical to continue to wonder based on conditions that you yourself know to be false. If you can see 99 REMs you know that everyone else can see at least 98, so is it logical to base your answer on a line of reasoning that assumes at some point that there are people who can see fewer than 98 REMs? on 09/04/02 at 21:46:34, S. Owen wrote:
Well, here's the thing... The question is actually leading to an important philosophical point - namely, because a progression of statements can be linked logically from one point to the next, does this imply that the whole chain, viewed as a whole, is logical. That is, the reasoning makes perfect sense to go from N=1 to N=2 and from N=2 to N=3 and so on, but, does this actually mean that the whole chain up to N=k is proved. There are two further points: 1) Are you sure that you are not making additional assumptions at each step. Expressed in formulae, you might think you are constructing a chain, F(n) true -> F(n+1) and then going on to say F(n+1) true -> F(n+2) true etc which would imply F true for all n, when you are actually proving F(n) true -> G(n+1) true and then saying F(n+1) true -> G(n+2) true when F and G seem to be the same, but are actually very slightly different. 2) The problem with my questions is this - at what point does the reasoning break down. It seems obvious that the answer is true for n=1, and very little thought would seem to show it is correct for n=2. Most people will accept 1 and 2 as correct. A few have trouble with n=3, but it does seem to make sense (I couldn't tell you why it was wrong - but the arguments above - if true - would imply the answer is wrong for n=3). At n=4 you get the problems really kicking in - the general problem is as expressed earlier in the forum "What new information is given". Remember however, the lack of a proof that your answer is wrong is not a proof of its correctness. Maybe there is actually no answer? Maybe logic isn't applicable in this case? :-/ |
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Title: Re: BROWN EYES AND RED EYE Post by S. Owen on Sep 5th, 2002, 7:42am on 09/05/02 at 00:06:17, James wrote:
But the reasoning is predicated on what the other 99 know, not what I know, so I don't believe this points out an inconsistency. I see 99 REMs; I don't know whether they see 99 or 98. They might be seeing 98. In which case they are wondering the same thing, minus one... but that's all reasoning about what they know; I know there are at least 99 REMs. I'd like to separate the reasoning that can occur before and after the tourist. Before the tourist, I see 99 REMs, and don't know whether I am the 100th or not. The other 99 might see only 98, and be wondering a similar thing, I'm not sure. But that's it, there is nothing further. Maybe that is a good way to answer this concern? No real reasoning happens until after the tourist. But then, it's not really a chain of reasoning that we consider, but a sequence of events (tourist, then suicide or no suicide each night) that reveals progressively more information to the monks about what the other monks know. Again, I'm not sure I am really tackling your objection regarding the long chain of reasoning effectively... on 09/05/02 at 00:06:17, James wrote:
Yeah I think we all appreciate that this is a subtle problem and that it's easy to reason incorrectly here. But similarly, that alone doesn't make the given reasoning wrong; I think it's been well considered by some smart folks here, all of whom grasp the nuances of proofs and reasoning, thank you. So can you spot a flaw in the reasoning along the lines you have proposed? I believe that's the useful question. |
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Title: Re: BROWN EYES AND RED EYE Post by James on Sep 5th, 2002, 4:57pm on 09/05/02 at 07:42:06, S. Owen wrote:
But the whole situation is also grounded in reality, a reality in which there are at least the 99 REMs that you can see, so can you base your logic on something inconsistent with that - so you know that everybody else can see at least 98. Can you justify that there might be people that can see just 3 REMs - which of the other monks could possibly be in that group? Your answer the predicated on the assumed reasoning of people you know don't exist. So, an interesting side-questions are "can non-existent people reason?", "is their reasoning the same as ours?", "are the non-existent people they imagine more or less imaginary than the non-existent people that we (the existent) imagine?" Quote:
Except that there is still a straight rush to accept a "each step is logical, whole process is logical" answer in this case. Quote:
Yes and no. Each step in the deduction is correct, but the whole answer is lacking. The best way I can describe it is that at some point, reality and logic part company - but I am not sure at which point. For a simpler example, consider the case of the card with two sides A and B. Side A says "Side B is wrong", Side B says "Side A is right" - is Side A right or wrong. We could assume A is right, so logically B is wrong. If B is wrong then logically A is wrong. Both steps themselves are totally logical, but the whole chain of reasoning comes unstuck. Could you tell me where? You might answer "the assumption that A was right" which was the only assumption we made. Then do the same with assuming A is wrong and reach the same point. Therefore the only thing we could have done wrong is assume that A is right or wrong - ie that there is an answer. This question is similar. Maybe your initial assumption "this question has an answer" is in fact wrong. However, for an alternative "answer" to the REM question, try this one for size. Assume there are N REMs (N>0). If N<k then on the Nth night, the N REMs kill themselves If N>=k then on the Nth night two things happen 1) the N REMs kill themselves; and 2) the N REMs dont kill themselves. The value of k is unable to be determined as yet, but would seem to be bigger than 2, or 3, or maybe 4. |
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Title: Re: BROWN EYES AND RED EYE Post by S. Owen on Sep 5th, 2002, 7:38pm on 09/05/02 at 16:57:18, James wrote:
No, again, this is not the reasoning. As one of 100 REMs, I see 99 REMs. I don't know whether those are the only 99 REMs or not. I could also then wonder whether those 99 see 99 or 98 REMs. But that's as far as their reasoning can go, and this is why nobody kills themselves until some new information comes along. Nothing less than 98 REMs figures into the reasoning. There is no induction in the pre-tourist reasoning, so I don't see where anything less than 98 REMs would fit into their reasoning. on 09/05/02 at 16:57:18, James wrote:
This sounds like portentous hand-waving... I appreciate your skepticism but do not share it on this particular problem. I do not see that there is a paradox here like in your card example, and I guess you don't either necessarily, but if you can argue it, by all means. on 09/05/02 at 16:57:18, James wrote:
Explain this one? |
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Title: Re: BROWN EYES AND RED EYE Post by James on Sep 5th, 2002, 7:55pm on 09/05/02 at 19:38:23, S. Owen wrote:
Fair enough, I must admit, this is the counter-argument to the "top-down" justification of the answer. Here is another possible "flaw" in the "bottom-up" answer. To prove the answer is correct for the case of N REMs we assume two things. 1) The answer is true for the case of N-1 REMs 2) There are N REMs - if there aren't then we know nothing happens. Given these two assumptions, the answer is certainly true for N REMs. However, the two assumptions are mutually contradictory. Because (1) can only be proved by assuming 2 things. 1a) The answer is true for the case of N-2 REMs and 1b) There are N-1 REMs. Therefore, our list of assumptions has now expanded to 3. 1) The answer is true for N-2 REMs 2) There are exactly N REMs (no more, no fewer) and 3) There are exactly N-1 REMs. This is the subtle flaw. |
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Title: Re: BROWN EYES AND RED EYES Post by AlexH on Sep 5th, 2002, 8:26pm It may be subtle, but its not a flaw. We can talk about and prove things about hypotheticals without knowing that they match reality. You are confusing proving "If there are n REMs, then they suicide on the nth night" with simply proving "they will suicide on the nth night". Obviously you can't prove the second statement inductively, because if they've already suicided on night n, they can't on night n+1. To put it a different way, your assumption #2 "There are n REMs" is never made. |
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Title: Re: BROWN EYES AND RED EYES Post by James on Sep 5th, 2002, 8:43pm on 09/05/02 at 20:26:42, AlexH wrote:
No, I disagree entirely. As currently stated, the answer is "If there are N REMs they all commit suicide on the Nth night". Can you rephrase answer to the question without reference to the number of monks? |
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Title: Re: BROWN EYES AND RED EYES Post by James on Sep 5th, 2002, 8:49pm on 09/05/02 at 20:26:42, AlexH wrote:
But, can we talk about and prove things about hypotheticals knowing that they don't match reality? |
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Title: Re: BROWN EYES AND RED EYES Post by AlexH on Sep 6th, 2002, 12:01am Perhaps I wasn't sufficiently clear. I fully agree that the statement we prove is of the form "If there are N REMs then they suicide on night N", and this is exactly why your statement about the assumptions is false. Quote:
We never make assumption 2. We assume "If there are N-1 REMs then they will suicide on night N-1" and use that to prove "If there are N REMs then they will suicide on night N". |
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Title: Re: BROWN EYES AND RED EYES Post by Jonathan_the_Red on Sep 6th, 2002, 1:18pm I honestly don't see why this is causing so much problems. Proof by induction works as follows: We want to prove that some proposition P(x) is true for all x>=N. To do so, we must prove two things: 1. P(N) is true. 2. If P(y) is true (y>N), then P(y+1) is true. Let P(x) be the proposition: if there are x REMs, they commit suicide on the xth day. Let N=1. The base case then becomes: if there is only one REM, he commits suicide on the first day. This is obviously true. The inductive case is almost as easy. We need to prove that if y REMs would kill themselves on day y, y+1 REMs would kill themselves on day y+1. We need not prove that y REMs would kill themselves on day y, we need only show that the second part of the statement (y+1 REMs die on day y+1) follows from assuming that they do. And it does: if there are y+1 REMs, they all know that there are either y REMs or y+1 REMs. When day y comes and goes and nobody dies, they know that there are not y REMs, therefore they all deduce that there are y+1 REMs and they all die the next day. The inductive step follows. Therefore, the proposition (for x=>1, x REMs will die on the xth day) is proven. |
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Title: Re: BROWN EYES AND RED EYES Post by James on Sep 9th, 2002, 3:30pm on 09/06/02 at 13:18:34, Jonathan_the_Red wrote:
So it doesn't concern you that you are basically saying the monks kill themselves twice. For P(y) to be true, you need to have dead monks. If you don't have dead monks, P(y) is no longer proven. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Sep 9th, 2002, 3:52pm No, again, P(x) is the proposition that "if there are x REMs then they kill themselves on night x," not "they kill themselves on night x." Yeah, Jonathan's statement of the inductive step might be more properly stated as follows: "If, when there are y REMs then they kill themselves on night y, then when there are y+1 REMs they will kill themselves on night y+1." ...but the reasoning he gives remains sound. |
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Title: One assumption may be incorrect Post by DeeK on Sep 9th, 2002, 8:15pm I propose that the answer to the riddle, is that the monk population is completely wiped out, for every case other than REM=1 Everyone is making a base assumption that is probably not true .... that every REM is a logic master. Assuming that at least one REM is not a riddle guru, then they will not commit suicide on the required night, causing every other monk with logical abilities to assume that they have red eyes. This leaves all the less cluey monks left. Suddenly deprived of all the monks who managed the monastry, the place falls into disrepair, and the rest of the monks eventually starve. For REM=1, even a simpleton should be able to deduce that they have red eyes. |
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Title: Re: BROWN EYES AND RED EYES Post by Jonathan_the_Red on Sep 10th, 2002, 11:55am on 09/09/02 at 15:30:57, James wrote:
No, I'm not saying that. As S. Owen said, P(y) is the proposition that if there are y REMs, they will kill themselves on day y. "If A then B" is true for all values of A and B except A true, B false. For example, if I say, "If it doesn't rain today, I will go to the movies" and it rains, I can either go to the movies or stay home and either way, my statement will be true. The only way my statement will be false is if it doesn't rain and I don't go to the movies. Nobody need to kill himself for the statement "if there are Y REMs, they kill themselves on day Y" to be true. In particular, if there are NOT Y REMs, the statement is true whether there are suicides or not. Perhaps it will be clearer to you if P(y) is expressed in the contrapositive: If nobody kills himself on day y, there are not exactly y REMs. This is logically equivalent to the first proposition, but doesn't involve the conditional suicide. The inductive step then becomes. The induction procedes the same way regardless. |
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Title: Re: One assumption may be incorrect Post by Carl_Cox on Sep 16th, 2002, 4:45pm on 09/09/02 at 20:15:02, DeeK wrote:
I do believe this is over complicating the problem. I would argue that not all simpletons would realize they are the REM, even if they know there is at least one. A fool proof system doen't take into account the ingenuity of fools. However, given that this is a riddle, one has to work with basic assumptions of human ability, in this caes, assuming that they are all "logic masters." Otherwise it's a question of psycology and not logic. Besides; these people are monks. What else are they going to do, besides figure out who's going to die when? It's gotta be the social event of the Nth days for them! |
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Title: Re: BROWN EYES AND RED EYES Post by James on Oct 7th, 2002, 10:25pm on 09/10/02 at 11:55:18, Jonathan_the_Red wrote:
Now, this is certainly not true given the "you only die once" criteria. Of course, you could take that away but I would be surprised if you are happy with the answer. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Oct 8th, 2002, 7:33am No, Jonathan is right... this is basic logic. The statement "If A then B" is false only when A is true but B is false. If A is false, then the statement is true. Another way to say it is that it's logically equivalent to "not A, or B". So, let Kn be the statement: "There are n REMs and they kill themselves on night n." Let In be the statement: "If there are n REMs, then they kill themselves on night n." You are right that it is not possible for Kn to be true for more than one value of n. But it is quite possible for In to be true for all n, in fact it is! |
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Title: Re: BROWN EYES AND RED EYES Post by James on Oct 9th, 2002, 4:57pm Sorry if this pops up more than once, first time got timed out Looks like we're going to have to walk you through this one. Here is the initial claim. on 09/10/02 at 11:55:18, Jonathan_the_Red wrote:
Let's break this one down. Quote:
Okay, let's say there are Y+3 REMs Quote:
Okay, let's say there are suicides. So, on night Y, Y+3 REMs suicide. Quote:
So apparently, on night Y+3, the Y+3 REMs suicide - AGAIN. These guys have a serious death wish problem. Can you see the minor problem with the argument yet. The induction proof is ignoring the fact that nights and REMs are two independent variables - something that standard induction is not built to handle. Nowhere does the current proof say why the Y+1 (or Y+n) REMs don't suicide on night Y. In fact, the argument goes something like this. On night Y, Y REMs suicide. Quick, change everything to Y+1 and hope no-one notices that we haven't said anything about why only Y REMs can suicide at this point. Now, on night Y+1, the Y+1 REMs realise that there must be more than Y REMs and hence kill themselves. This is the currently missing piece of the puzzle. Don't assume anything - it's always the unwritten assumptions that stab you in the back |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Oct 9th, 2002, 6:03pm on 10/09/02 at 16:57:05, James wrote:
You're still not understanding the proposition. It is: "If there are exactly n REMs, then those n REMs commit suicide on night n." Call this proposition In. I think we've been leaving out the word "exactly" - does that help? Yeah, that's important. The claim is that In is true for all n, and that is what is shown inductively. If there are Y+3 REMs, then In is definitely true for all n != Y+3, "by default," because there aren't exactly Y or Y+1 or Y-1 REMs. That is to say, they don't apply, but are still considered true. That's all. You cannot connect these statements to show that both "Y+3 REMs commit suicide on night Y" and "Y+3 REMs commit suicide on night Y+3" are true. That's good, since that would indeed be a contradiction. But you have failed to show that such a contradiction arises from the propositions. on 10/09/02 at 16:57:05, James wrote:
No no no no no! See above. The argument is that In holds for all n, which is shown inductively. This is crucial - you must see the difference between the following: "If there are exactly n REMs, then those n REMs will commit suicide on night n." "n REMs will commit suicide on night n." The first is true for all values of n (>= 1 of course). The second is true for exactly one value of n. I also do not understand your objection to the inductive step... the proposition In has just one variable in it... n! And the inductive step shows that "if In, then In+1." It doesn't get much more classic than that. Not to ratchet up the level of condescension more than necessary... but have you ever taken a class on logic, or read anything about it? |
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Title: Re: BROWN EYES AND RED EYES Post by James on Oct 9th, 2002, 7:07pm PART I Read VERY SLOWLY AND CAREFULLY - this has nothing to do with "The proposition" as you call it, it is to do with this partcular quote Quote:
Note - You assume (as you assert in "the proposition") that there ARE Y REMS, so how does that reflect on any point made about where there are "NOT Y REMs"? To wit.. [quote author=S. Owen] "If there are exactly n REMs..." [/quote] The points made in my response related to a proposition made when there are "NOT Y REMs". [quote author=S. Owen] have you ever taken a class on logic [/quote] Not one where they taught me that "exactly equal" meant the same thing as "not equal" |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Oct 10th, 2002, 6:01am OK, I will show you why you are not really getting a contradiction from this quote. Jonathan says... "In particular, if there are NOT Y REMs, the statement is true whether there are suicides are not." The statement is the proposition that I repeated above - substitute in Y. You say, consider what happens when there are really Y+3 REMs, which is fine. You say, assume there are suicides on night Y - that is the problem. Well, you can assume whatever you like, but "there are suicides on night Y" does not follow logically from anything. You have shown that if there are Y+3 REMs and suicides on night Y, we have a contradiction, and I agree. That's fine, but doesn't apply to the original argument, or Jonathan's statement, since you tossed in suicides on night Y. If anything you have proved that there can't be suicides on night Y. You are welcome to have the last word on this... I don't think I have anything else useful to add. |
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Title: Re: BROWN EYES AND RED EYES Post by Jonathan_the_Red on Oct 10th, 2002, 11:29am I gotta jump back in here. Let's try doing this completely manually. Suppose there is one and only one REM. We're agreed that he'll kill himself on the first day, right? He'll look around, he'll see that there are no other REMs, deduce his eye color and go off and kill himself. Is there any disagreement on this point? If not, then we'll consider the following proposition proven: I1: If there is exactly 1 REM, he will kill himself on day 1. Suppose there are two REMs. Both of those REMs look around and see one other REM. Both of them deduce the following (true) statement: P1: If I myself am not a REM, there is exactly one REM. Since you've taken a logic class, you understand the concept of a contrapositive. The contrapositive of P1 is: C1: If there is not exactly one REM, I myself am a REM. Nobody is able to deduce his own eye color, so an entire day goes by and nobody kills himself. Both of the REMs now follow this chain of logic:
Both REMs now know that they are REMS, so they kill themselves on day 2. We have now proven the following: I2: If there are exactly two REMs, they kill themselves on day 2. Now, suppose there are three REMs. They look around, they see two other REMs, and deduce the following: P2: If I myself am not a REM, there are exactly two REMs. C2: If there are not exactly two REMs, I myself am a REM. Day two comes and goes, nobody kills themselves. The three REMs reason as follows:
All three of them now know that they're REMs, they kill themselves, and we've now proven: I3: If there are exactly three REMs, they kill themselves on day 3. I can keep going with this for as long as you'd like. Or you can accept that the induction is valid. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 10th, 2002, 9:07pm The Induction is valid. But I say the conclusion is wrong: Any REMs smart enough to figure this out on the 100th midnight, should also be smart enough to figure out it's time for a change of profession on day 99! ;D |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 11th, 2002, 9:58pm On a more serious note: While I agree with the analysis, that every REM should suicide on night N, where N is the number of REMS, there are some curious facts that have been missed, or at least not brought out clearly (Okay - at least that I missed seeing mentioned!) Assume 100 REMS. Each REM knows that there are either 99 or 100 REMS. Each BEM knows that there are either 100 or 101 REMS. And each REM knows that every other REM can see at least 98 REMS. Thus every monk knows, and knows that every other monk knows, that there are at least 98 REMS, and so, every monk knows, and knows that every other monk knows, that nothing will happen for 97 nights. They also all know nothing will happen the 98th night, but the REMS do not know that all monks know this. (Still with me?) What this means, is that neither the tourist's remark, nor the first 98 nights afterward, introduce any new information to the system. It is not until night 99 after the tourist's remark that the REMS learn something that they didn't know before: because they see that no-one is commiting suicide on day 99, they learn that no-one was expecting suicide on day 98, as they would have if there were only 99 REMS. It is this delayed effect of tourist's remark - that for 98 days, it makes no change to the system, but still has an effect on night 99, that I find intriguing. The question that boggles me as I think about it, is "If they all know that nothing is going to change for 98 days, why does it still require 99 days to have an effect?" |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 11th, 2002, 11:12pm I seriously doubt that anything I write here will make anyone say “I see the light, I was wrong”… I think this post has been going on way too long for anyone to readily change their position. But what the hell I’m going to try anyway. In a small number of REM (1) you have the REM who is ready to die on this night (in this case night 1), and you have BEM who are ready to die on night 2 (N+1). If there are 2 REM, then the REM are prepared to die on night 2 (like the BEM of last round), and the BEM are prepared to die on night 3. Up another level. The REM become the BEM of last round, and are now prepared to die on night 3. The BEM are now prepared to die on night 4. The only thing different between the BEM and REM is that the REM see one REM less than the BEM, and are then prepared to die one night earlier then the BEM… and since they do this prevents the BEM from having to kill themselves. But actually, I think this will just make sense to people who believe the answer to this riddle is “All the REM will die on the Nth night (where N is equal to the number of REM”… the key to coming to this conclusion is realizing the tourists remark made sure everyone knows that everyone knows that everyone knows … etc… that there is at least one REM. And the introduction of this information started the countdown so to speak. I really think mathematical induction is the best way to go with this. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 12th, 2002, 8:39am Quote:
My point is: There is NO information introduced by the tourist's remark, or for the first 97 nights (given 100 REMS) afterward. The tourist does introduce information, but not until the 98th night after his remark! (This is a slight change from my previous argument, I've realized that something is indeed learned on night 98. What the monks learn is: "if any monks thought there might be only 98 REMS, they now know there are at least 99". This does not apply to night 97 or earlier because every monk knows that every monk knows that there are at least 98 REMS.) The tourist does not introduce any "everybody knows that everybody knows that ..." information when he makes his statement. Before the tourist, it was already the case that everybody knew that everybody knew that there are at least 98 REM. (You can tack as many "everybody knew that"s as you feel like on the front of this statement too, The statement will still be true.) Quote:
My comments are a consequence of the induction, not an argument against it, or an attempt to go a different way. To borrow a term from physics, the tourist's remark introduces a state-change in the monastary. This state-change does not include the introduction of any new information. However, it does allow for new information to be deduced 98 nights later. If you disagree, consider again the 3 REM case. Tell me exactly what information our good REMs and BEMs had immediately after the tourist's remark that was not available to them before. I argue that it is not until midnight that night that anything is learned. |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 12th, 2002, 9:58am Ok, with 3 REM, everyone knows there is at least one REM. But each REM looks at the other 2 and thinks "those 2 are looking at each and might be thinking "I know there is at least one REM, and that one REM might be thinking "I wonder if there are any REM"... if you ask any individual REM if there is at least one REM then he will of course say yes (provided there is more than one REM), but when asked what these other REM know, the first REM will say "well they know there are at LEAST*The number i said minus 1*. And when you ask the first REM what this new group of monks know he will say, "well that group knows there are at LEAST *the number i said minus 1* REM." this will continue all the way down. Granted no one learns "oh i guess there really is at least one REM"... once this is said everyone knows that everyone knows that ... etc.. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 12th, 2002, 11:00am So what you are saying is: before the tourist, given 100 REMs (sometimes a nice solid number helps, if only to make clearer statements), each of the 100 REMs is thinking (or at least able to think): If I am a BEM, then there are 99 REMs, each of which is thinking: If I am a BEM, then there are 98 REMs, each of which is thinking: If I am a BEM, then there are 97 REMs, each of which is thinking: . . . If I am a BEM, then there is 1 REM, who is thinking If I am a BEM, then there are no REMs. Immediately after the tourist's remark, this whole tower is the same except the last line becomes: I must be the REM! As each night goes by, 1 more "if" line is removed from the bottom of the reasoning chain, and the last line changes to: I must be one of the N REMs. With N increasing by 1 each night. Finally, after the 99th night, all the "if" lines are gone, and each REM is thinking of themselves that they are 1 of the 100 REMS, so they all commit suicide on the 100th night. OUCH!! These REMs are smarter than me!, but mea culpa! :-[ You are right. The tourist does indeed immediately impart information! Thanks for an interesting discussion. |
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Title: Re: BROWN EYES AND RED EYES Post by Jonathan_the_Red on Oct 12th, 2002, 1:48pm I'm not one to toot my own horn or anything (well, okay, actually I am) but I said basically the exact same thing as the above back on page 2. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 12th, 2002, 2:17pm Mea culpa2 :-[ :-[ Yes, you did! |
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Title: Variant: The repentent tourist Post by Chronos on Oct 14th, 2002, 5:59pm So, we've all seen what happens when the tourist introduces this tiny little fragment of information. So let's look at a variation. Warmup problem: Suppose that instead of saying "There is at least one red-eyed monk", the tourist instead makes the statement "Brother Bob has red eyes". What happens? Of course, Brother Bob goes and kills himself, and everyone else is safe. But the statement "Brother Bob has red eyes" implies the statement "There is at least one red-eyed monk". So the tourist is actually giving more information in this example, and yet most of the monks (all but the unfortunate Br. Bob) gain less information! How is this possible? OK, if you've stewed over that enough, variant number 2: The tourist announces at breakfast that there is at least one REM. But later in the day, he learns of the peculiar customs of these monks, and realizes what he's set in motion. Guilt-stricken, he addresses the monks again at supper that night, and tells them all "Monks, I realize now the significance of what I told you this morning. To reduce the number of inevitable deaths, I now tell you that Brother Bob has red eyes". What happens? Explain. Variant number 3: The tourist doesn't learn about the monks' habits (pun not intended) until day M, where M < N. What should the tourist say? Variant number 4, which I still can't quite get my brain around: What happens if, the evening of the first day, the tourist says "I was lying this morning"? Alternately, what if he says "I'm unreliable"? |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 14th, 2002, 8:33pm well for most of these cases there are a couple ways to interpret the information... so if we assume the monks believe it's unnacceptable for BEM to kill themselves (eliminating the "better safe than sorry" approach) i think they will go back to stable systems in a day. Variation 1: every learns there is at least one REM... ok lets look at a 2 REM island. You got Bob and Joe. Bob says "well f*ck, i'm done", and joe says "well, i don't know what my color eyes are, but Bob is going to kill himself tonight, so it won't tell me anything... so i can't really do anything" Variation 2: either there is just one REM, or more than one, but by saying there is at least one, and then pointing to Bob and saying it's him, no one knows if there was JUST one... so after bob throws himself out awindow, everything will return to normal... well as normal as this island ever gets. Variation 3: i'll get back to that Variation 4: again, everyone is just unsure... so nothing can really happen. back to 3. i'm not quite sure. if there are N REM monks, the REM will know there won't be any deaths until at least N-1 days, and BEM know there won't be any deaths until at least N days. ok... i'm thinking all the tourist has to do to save the day is name off one REM. before N-1 days happen. any REM looks at all the other REM and figures "ok, it'll take them N-1 days to figure out THEY are all REM monks... and if they don't kill themselves after that day, i know i'm a REM." But if a REM is named before N-1 days any given monk thinks "well i don't know if i'm a REM yet... since N-1 days has not passed... and every other monk is thinking exactly whta I am..." however, lets consider 3 REM Bob, Joe, and Jim. Mr. Tourist says there is at least 1 REM. one day passes. then the tourist says "hey guys.. bob has red eyes." So that night bob kills himself. both joe and jim are thinking "well i don't know if joe/jim was looking at me, and was planning on killing himself tonight" so they are both ok. ok yeah that's my answer. as long as the tourist names one REM name before N-1 days, the other REM can live, because they are just uncertain. |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 14th, 2002, 8:41pm to encourage some more thinking (that i don't want to do) what if the tourist says : "There are 10 Brown Eyed Monks" "There are at lesat two Red Eyed Monks" "There is an odd number of Red Eyed Monks" "There is an even number of Red Eyed Monks" "There is more than one Red Eyed Monk" |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Oct 14th, 2002, 8:53pm These are great... I wonder if Mr. Wu can make these an addition to the original problem? Or maybe we can at least start a new thread. I agree that the answer to #1 is clear, though I'm searching for the most satisfying way of digesting how this is different than the original. They are given more information initially... too much. Because Bob is explicitly told that he's an REM, his death signals nothing else to the other monks. So it's really that the monks aren't able to make stronger conclusions about themselves each night that passes without suicide - that's why the end up knowing less. #2, yeah, if the tourist only means to imply that the one REM is Bob, then it's just case #1. If he means there is at least one REM besides Bob, we have the original problem starting the next day. If he doesn't specify, then everyone besides Bob is back to where they were before the tourist arrived, and nothing else happens. The rest I am still working on... |
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Title: Re: BROWN EYES AND RED EYES Post by william wu on Oct 15th, 2002, 1:23am on 10/14/02 at 20:53:50, S. Owen wrote:
done http://www.ocf.berkeley.edu/~wwu/riddles/medium.shtml#brownEyesRedEyes |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 15th, 2002, 7:49am I just realized something. My example doesn't work. N-1 days has passed (2) and if Joe or Jim doesn't kill himself that night, it will tell the other they were not planning on it. Which will tell them both they are REM. Since conclusions are made on the N-1 day, to make everyone unsure a monk has to be pointed out before then. |
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Title: Re: BROWN EYES AND RED EYES Post by Jonathan_the_Red on Oct 15th, 2002, 3:40pm You know, I don't recall the initial statement of this problem. Do you kill yourself if you deduce your eye color, or do you kill yourself if you deduce that you have red eyes? For this post, I'll assume the former. on 10/14/02 at 20:41:26, Jeremy wrote:
Everyone deduces his eye color right away and dies that very day. Quote:
This gives the induction a "jump start." All the REMs die on day N-1, instead of day N. Quote:
Everyone deduces his eye color right away and dies on day 1. See "Kabouters, communication between" and "Prisoners, hatted, execution of." |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 15th, 2002, 6:20pm Good points, Jonathan_the_Red, (I hope that's not short for "the red-eyed monk" ;)) It is only the REMs who kill themselves. Otherwise all the BEMs would kill themselves the night after the REMs and another perfectly good bizarre religion would come to an uphappy end! Your remarks still hold of course, except only the REMs die. My comments on Chronos' excellent questions: 1) The tourist actually gives more information only to Bob. Everyone else already knew Bob was a REM. To everyone else the only change is that they can exclude Bob from the reasoning chain - it still applies to the remaining REMs unchanged, including the uncertainty of the "lone REM" at the bottom as to his REM status. So only Bob dies. 2) If the tourist says exactly what Chronos quoted, the suicides are a go on night N-1 (the number of REMs less Bob). This is because the way it is stated makes it clear that there are more REMs (otherwise fingering Bob would not cut down on the suicides). This is equivalent to the tourist stating right at the start, "Bob is a REM, and there is another". It allows the remaining REMs to establish a reasoning excluding Bob, and ending with the "lone REM" at the bottom realizing he must be the other REM. If the tourist wants to stop the killing, he should finger Bob (who he obviously doesn't like) without saying why, or admit to being a liar. 3) I have to disagree with Jeremy. By day 2, the information is now apparent at the bottom of reasoning chain that there are 2 REMs. Naming Bob alone on day 2 is equivalent to the tourist saying nothing on day 1, and saying "Bob is a REM, and there is another" on day 2. Bob offs himself that night, the remaining REMs on night N. To stop a complete suicide, on day M the tourist will need to finger M REMs. 4) Now I agree with Jeremy, though the situation is not as simple as he states it (the reasoning is sound, there's just more to it than he actually said). After all, except in the one REM case, all the monks know what the tourist said is true, regardless of any later statements. But it is not what they know that makes the difference. It is what that hypothetical "lone REM" at the bottom of the reasoning chain knows that determines whether they live or die. Since a lone REM would see no evidence to support the tourist's first remark, the unreliability established by the second remark would rule the day (and the Nth night after). |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 16th, 2002, 11:38am I don't think so (about naming M monks on night M to save the rest). if there are 100 monks, and 10 nights have passed, the case is NOT "10 monks know they are REM, and 90 are not sure yet"... the case is "Nothing surprising has happened yet, and everyone need to wait till night 99 before anything is actually learned". But actually i'm not sure. Can you explain yourself any better? I'm starting to think that naming one monk on a random day will cut it. if there are 2 monks, and on the second day, one of those 2 are named, they will both kill themselves that night. if there are 3 REM, each will think it will take 2 days for the other 2 to figure out they are REM and kill themselves, however if a monk is named on the second day, and kills himself that night... the other 2 monks will see the other REM, and know he did not suicide because there are more than 2 REM. So that kills 3 REM on day 3 (regardless of naming a REM). with 4 REM, hmm... yeah naming a single REM doesn't help at all. So what are you saying? the tourist has to name M REM on night M? so with 3 REM, on the second day he names 2 REM. with 3 monks, they 3rd monk is unsure if they were going to kill themselves anyway on that night. and then it just works it's way up the system... crap, you're right. Alright, try this... lets say the tourist belongs to a different religion, and he wants to kill off ALL the monks (REM and BEM). What statement can he make that will eventually have all the monks commit suicide? Assume the monks will believe the tourist, unless they have a reason not to... in that case they will not believe him (this means saying "you're all REM" will not work because some REM will be able to see some BEM, and they will not trust the tourist. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Oct 16th, 2002, 7:40pm That one might have been harder had I not already noticed one of the tourist statements above will have exactly this effect, provided it is false. I had been trying to find ways for the repentant tourist to undo his damage without taking it out on poor Bob! (And without proclaiming himself a liar.) I thought I had found one with this lie which would convince all the REMs that they were BEMs. But then I realized it would also convince all the BEMs that they were REMs, leading to their deaths. When the REMs woke up the next morning, it would not take them long to figure out what happened, which would also let them know that they are REMs. Personally, I believe the REMs will foreswear their suicides until they have hunted down the dastardly tourist and exacted a ghastly revenge! Or else, one will have an epiphany that since the BEMs all died, it must have been them that were cursed all along! |
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Title: Re: BROWN EYES AND RED EYES Post by Chronos on Oct 17th, 2002, 6:36pm Quoth Icarus: Quote:
I see now, by the way, why it works for the tourist to say that he's lying. All of the real monks know he wasn't, but there's a multimetahypothetical monk way down the line who doesn't know, and that multimetahypothetical monk would be forced to conclude that the tourist would be unreliable and therefore ignore him. On to Jeremy's variations: Is there any reason that there should be a difference between "There are at least two REMs" and "There is more than one REM"? And if the tourist tells them that there are exactly ten brown-eyed monks, I see how that results in instant mass suicide (if they know exactly how many browns there are, each can easily deduce his own color). But if the tourist only says that there are at least ten BEMs (or, in general, at least n), I don't think that that has any significant effect, since a monk isn't obligated to signal in any way, should he determine that he's brown. |
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Title: Re: BROWN EYES AND RED EYES Post by Jeremy on Oct 17th, 2002, 9:21pm Actually, i just kinda named off some things the tourist could have said that might be a little more difficult to solve for. But i think all of them are pretty much "nothing changes, or all the REM's die". Here's a couple that have a little more thought: "There is a prime number of REM monks" Each day the tourist says "There are at least 2N REMs" where N equals the number of days that have passed. (e.g. on day 1 : "There are at least 2 REMs", day 2 : "There are at least 4 REMs" etc) until the tourist can not make the statement and still be telling the truth. "There is at least one colorblind REM" enjoy... or if you think i should just stop bothering with this riddle ignore... and have fun with the last one. :P |
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Title: Re: BROWN EYES AND RED EYES Post by Chronos on Oct 21st, 2002, 10:27pm Assuming that the tourist is truthful? For the prime one: If there are five or more REMs, then they all instantly kill themselves. See the answers to the parity version above. If there are two, each of them knows that there is either one or two, and one isn't prime, so they also know that they're red and kill themselves. If there are three REMs, then the first night, they all wait, since after all, there might be only two. But then when nobody dies the first night, all three off themselves the second night. For the 2N version, you're obviously just speeding things up, though I'm not sure by exactly how much (is there a closed-form expression?). Now, for the colorblind one, the answer will depend on how many of each sort are colorblind, and whether the monks know who's colorblind. I'll assume that the monks don't know, but leave the other open. It's obvious that a colorblind monk, of either eye color, will never kill himself: There's no way he can get enough information, without seeing the other monks. He can never make a conclusion that a brown can't make. So at least one (and possibly more) of the REMs is safe. If there's exactly one REM, then, nothing happens. But let's consider the case where all monks but one are colorblind. Then none of them ever commit suicide, and the one monk with color vision never gets any information from suicides of others (since there aren't any). So he never kills himself, either. But here's the key: For all any individual monk knows, he's the only monk with color vision. So he can't assume anything from a lack of deaths, so he doesn't kill himself. This applies to all color-vision monks, and we've already shown that colorblind monks don't suicide, either, so nobody suicides, ever. |
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Title: Re: BROWN EYES AND RED EYES Post by SgtAcid on Dec 8th, 2002, 4:30pm I think i may have found an alternate solution to this riddle. What would you say if I said the answer was a mass suicide? You would probably ask Mr. Wu to terminate my membership, because we can't have anyone that stupid associating with us. But what S. Owen, Jonathan the Red, Icarus, and Jeremy said all makes sense to me and I thought they were right. But when I read the riddle again, the last sentence says something "DRAMATIC" happens at the island. Which made me think, if the REM killed themselves thats not really that dramatic, because thats what they were supposed to do. Which made me think of an alternate solution. Consider this: We have already established that if there is 1 REM, he would wake up the next day and notice that no one has commited suicide so it must be him, so that night he would kill himself accordingly, and that would be the end of the story. But the tourist says "at LEAST one of you has red eyes". Which means we can't determine how many people have red eyes, and neither could the monks. So the monks are thinking what if there are two of us with red eyes. When S. Owen, Jonathan the Red, Icarus, and Jeremy gave an answer they supposed that there are 3 monks with red eyes, or 99 monks with red eyes, or whatever and they tried to solve the riddle basing it on how many monks had red eyes. But since the Monks didn't know how many of them were REM, neither should we. To show you exactly what i mean here is an example. A BEM sees 3 REM but how does he know he isn't the 4th, since we don't know how many there are in total? and the rest of the BEM are thinking the same exact thing, that they may be the 4th and thats when the mass suicide comes in, on the second night, because everyone else thinks they might be a REM. If the tourist said there were 10 REM the BEM would count 10 REM and know it wasn't them, and now its up to the REM to figure it out. But since there wasn't a definite number given they all thought it might be them and let tourist be the only person left alive on the island, and that is my definition of "dramatic". I know this is a different answer, but I tried to find a flaw in it, and i couldn't. So I would really appreciate your thoughts about mine, and respond to this. |
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Title: Re: BROWN EYES AND RED EYES Post by Chronos on Dec 8th, 2002, 7:40pm SgtAcid, you're using the same reasoning that was used to get the "all REM die", but you're not carrying it through completely. We're not assuming that we know how many REMs there are; we're saying "If there is one, this happens. If there are two, this happens", etc. Specifically, what happens is that for any number n of red-eye monks, they all kill themselves on night n. So, suppose there's one: He looks around, sees that there are no others, and realizes he's red, so he kills himself. We agree on that. Now suppose that there are two. The first night, each of them looks around and says "Well, maybe it's just that guy, so maybe I'm safe". Neither one dies the first night. But then, the second night, they both know that there are at least two, so both know and suicide. Now suppose that there are three. Indeed, as you mention, nothing happens the first night, and nobody expects anything to happen the first night. The browns are all unsure as to whether there are 3 or 4 reds, and the reds are all unsure as to whether there are 2 or 3 reds. So they're all unsure, and nothing happens... Yet. But if they just wait a little bit longer, then they'll know for certain. Unless we assume that the monks are all impatient, and so pious that they can't stand even risking the possibility of being red, the browns will stay alive and the reds will at least wait until day N to die. |
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Title: Re: BROWN EYES AND RED EYES Post by SgtAcid on Dec 8th, 2002, 7:49pm I get it now |
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Title: Re: BROWN EYES AND RED EYES Post by datarocks on Dec 9th, 2002, 8:39pm guys, help me here... there was a very similar riddle to this one involving three men at a table each wearing one of two colored hats.... but i forget how it goes... coudl someone please post it if they know it? thanks |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Dec 10th, 2002, 1:33am But how can you know they use days to pace their counting? Obviously they should use something, because if they don't and there are for instance two REM, then one may think faster than the other that if the toher was the only one he would have killed himself, so I must be one, so I must kill myself and then does so, before the other has thought that much and consequently considers, "well some REM killed himself, and that was the only one I saw, so they're all dead now." But they can't really discus how to pace their counting, yet, if they're smart, they know there has to be some pacing. So it should be forced on them from the outside by someone repeating 'there is (still) at least one REM', counting at each utterance. So unless there is any pacing most monks that think too fast, and don't consider the pacing-paradox, will kill themselves. And any others would keep on living.. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Dec 10th, 2002, 5:50am The riddle says that red-eyed monks must kill themselves at midnight. So, that condition defines the discrete "rounds" in which this process takes place. |
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Title: Re: BROWN EYES AND RED EYES Post by S. Owen on Dec 10th, 2002, 6:04am on 12/09/02 at 20:39:30, datarocks wrote:
There is a riddle under "Easy" called "Three Hats" that might be what you're looking for. I remember that some guy actually played out a riddle like what you're talking about on the Johnny Carson show a long time ago.. it involved white and black dots on each of three persons' foreheads, and guessing one's own color dot. Heh, maybe that will lead you to some info on the internet about it. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Dec 10th, 2002, 8:37am on 12/10/02 at 05:50:32, S. Owen wrote:
I guess I missed that.. [mental_note]Must learn to read the problems better..[/mental_note] |
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Title: Re: BROWN EYES AND RED EYES Post by UofLis#1 on Mar 10th, 2003, 8:59pm :o Wow icant believe i finally have the answer. A friend told me this riddle once and it took me forever i just couldnt figure it out. With the hints i did but it was because of this forum i have finally the answer and i know what. i didnt get the answer from him because he wouldnt tell me it was because of the hints on this forum that i got it. Thanks all!!!!!!!!!!!!!!! ;D GO CARDS!!!!!!!!!!! -Ford |
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Title: Re: BROWN EYES AND RED EYES Post by Lusus on Apr 26th, 2003, 1:13am Keep in mind, the curse and the suicide are two different things. After all, if Red Eyed monks existed on the Island for years, then every monk would've realized that suicide is not a part of the curse of the Red Eyes. If there was one Red Eyed Monk, he may kill himself that night to avoid the curse. But if there were two REM's then as one of them who wakes up the next day and sees that the other hasn't killed himself yet I have two things to consider: 1) either there is another REM that he sees (logic leads to this other person being me), or... 2) This person bravely chose to endure the curse. Even before the tourist arrived, I don't know that he doesn't already know he's red eyed. He may have chosen to endure the curse long ago. It has already been seen a monk can live for years with the curse, why not live a few more? This alternative view remains in situations where there are more than two REM's. |
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Title: Re: BROWN EYES AND RED EYES Post by Cage Hiu on May 14th, 2003, 9:55pm Alternate sloution Lets say you are a monk (doesn't matter what eye color) , you can draw attention to yourself by stamping your feet or what not.... As the other monks are watching you, you remove one of your EYES. Therefore, the other monks will understand the logic and follow suit... you see your eye color with your other eye. So therefore the next day, you are left with a temple full of one brown-eyed MONKS. Also all monks uphold their vow of slience Yes I am well aware the sloution is not optimal to the Nth death on the Nth day logic but the premises are also fullfilled by THIS means... it also aviods the chance of MASS SUCIDE if the monks are tricked by the tourist. -C.H |
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Title: Re: BROWN EYES AND RED EYES Post by Mike V on Aug 5th, 2003, 8:08pm I know it's been a long time, but the concept of colorblind monks (CBM) intrigues me. A CBREM (colorblind redeyed monk) could indeed commit suicide. Although he never can see the other colors, if all monks assume all other monks can see color. If there is only one CBREM he will commit suicide on the n+1th day, unless n = 1, in which case he kills himself on the 3rd day (after everyone else kills themselves on the 2nd day). The reason is that since no one knows about his colorblindness, except him, they'll follow the normal logic. If he doesn't kill himself on day one, they all assume they must be red eyed and kill themselves on day two. He then realizes what had happened and that he has red eyes. If there are more redeyes than just him, they follow normal logic and kill themselves on day n. Seeing that n-1 monks killed themselves on day n, our CBREM realizes that he must have red eyes and kills himself on day n+1 (leaving the other monks confused, but alive). This gets more complicated with more CBMs (red or non), but follows similar logic. A) One interesting case is what happens when all monks are colorblind (but each assumes he is the only one). B) Or, as mentioned before, all but one are colorblind (and each still assumes others see just fine). Finally, when you take away the assumption about the others seeing ok (perhaps by having the tourist say, 'also, at least one of you is colorblind') I find that there generally is not enough information for all the red eyes to kill themselves (by generally I mean I've tried several different amounts of info they get, such as an exact number of CBMs or an at least, and similarly for REM). Ok, sorry, but I just found this site and I love this riddle, finally finally, I want to thank Jonathan_the_Red if he happens to still be around. While I understood how it worked without them, he gave the best explanations (3 unique ones at that!) I've ever heard for this phenomenon. I've been inspired to take some logic classes. |
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Title: Monks and the stock market Post by S. Owen on Aug 5th, 2003, 9:41pm Also on the subject of color-blind monks -- I read an interesting article about investor behavior that bears some relation to this aspect of the riddle, so I thought I'd share: I think one of the original points regarding color-blind monks was that there is a big assumption in the canonical solution to the riddle -- that all monks are very smart and arrive at this canonical solution. There is a certain circularity to it. Some monk could be color-blind, or reason differently -- things other monks cannot know -- and then the canonical solution is no longer valid. The statement of the riddle kind of waves this all aside, but it is interesting to think about -- what assumptions really underpin the canonical solution? But, that's not quite the point of this post. Consider a game played by a group of very smart people, in which each person chooses a number between 0 and 100 independently and in secret. After this, all numbers are revealed and averaged. The person whose choice is closest to 80% of that average wins. And everybody wants to win of course. What happens? Well, a normal person might consider that the average will probably be about 50, and guess 40. But of course everyone is thinking that. So, 32? Same thing. Following the reasoning in this riddle (sort of), the solution is that all participants guess 0. I take it this is a classic problem. The article I read presented this as a simplified version of the reasoning that underpins the stock market -- the goal is not really to guess which companies will do best, but rather to guess which companies most other investors think will do best. While the solution to what happens in the game above is 0, of cousre, we all know that if we took 20 of our friends and tried this, that nobody would guess 0. It's not a matter of intelligence -- even though you know the solution, you would not be wise to guess 0! Your optimal guess depends on your assessment of other participants' guesses, which in turn depends on their assessments and you don't know those. This, the article says, is what keeps the stock market moving. So the color-blind monk question is, I think, an important one. Is the canonical solution really a solution at all? I haven't worked out whether the "very smart monks" assumption is sufficent to resolve this, and would like to hear from people on this question. In any event, in the stock market, it is no solution at all. |
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Title: Re: BROWN EYES AND RED EYES Post by BaH on Dec 28th, 2005, 1:45pm lolol, there is one rem, and he tries to commit suicide on day one and fails, on day 2 he is sad and grieving because of his failed suicide attempt, and the other monks think he is just grieving becasue he has realized he is the 'other' rem. night 2 everyone dies except the rem, who is incarcerated for mass murder. day 3, everyone who has waded through this 5 pages (so far) of posts join him in the sanitarium. seriously tho, there comes a point where the chain logically gets broken, a rem will be colorblind, or too daft to follow, or wimpy to carry through. however, if you are walking around the compound and see people walking around pointing at the rems, and apparently counting, and they point at you in the process :P |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 18th, 2006, 11:39pm Hi, hope I am not dragging up an old sore with this one. ;) This logic problem bugs the tar out of me. It's my belief that this problem breaks down at larger numbers. I do agree the dieing on n days is the correct formula, but I donot believe it would be triggered by the statement made to a large population of REMs. Note I am talking the original problem here not the additional what if's (colorblind etc..) The number of BEM is inconsequential in the solving, so I will only address the number of REMs. We will say there are 10 REM's at the island. Each BEM thinks the following: There are Either 10 REM (and I am BEM) or there are 11 REM (and I am one of them). Each REM thinks the following: There are Either 9 REM (I am BEM) or there are 10 (and I am REM) The REM's want to be BEM, so they are each HOPING there are only 9 REM's so they want the Other REM's to be considering the possiblity of either 9 or 8 REM's. (this is the start of the induction and sequencing giving as Nth night solution) However the problem is None of the other REMs are thinking these possabilities. Making the jump to continue the cycle downward after noone is considering the possible outcomes in my opinion makes the statement of "I see red eyed monks" mean nothing to the monks. (they knew this already) There is nothing that says REMs should start dying after the statement. The fact giving of confirming 0 REM's (which breaks the bottom tier If statement and theoretically starting the cycle) was already known by every single Monk. We have proven beyond a doubt for small numbers the formula is obvious. But what I am failing to see is why everyone is granting that the cycle would start with the tourists statement. However if the problem had been along the lines of: "Monks who live with vows of silence, and no mirrors, have either Brown or Red eyes, none of them know their own eye color. One day their god decrees that the red eye monks are cursed and if you are a red eye monk you should kill yourelf at midnight." I would agree the cycle would be started. The difference is not that they learn there are REMs, (they already knew that, though they donot know for sure if they are one or not) the difference is they learn they should kill themselves. Now they are watching the other monks to see if they kill themselves or not. The change of action starts the cycle not "new" knowledge. (they all already knew there were REMs. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Mar 19th, 2006, 8:19am OK, let's try this variation. There are 6 monasteries, numbered 0,1,2,3,4,5, each with that many REMs. The monks all know this, but don't know which monastery is which (and aren't allowed to visit to find out either) When the inconsiderate tourist visits all 6 monasteries in one day, at 0 he says nothing, but at each other, he starts the process going. At 1, the lone REM realises he's not at 0 after all, and duly suicides at midnight At 2, each REM is hoping they're living at 1, so hopes to find the other dead in the morning. When they meet, they realise the awful truth and suicide the second night. At 3, each REM hopes they're at 2, realises on the second morning that they're not, and suicides on the third night At 4, the REMs hope they're at 3 (where there will be bodies on the third morning) and duly suicide on the fourth night. Or taking things in the other direction: At 5, all 5 REMs (call them A,B,C,D,E) are hoping they're living at 4. In particular, E hopes that A,B,C and D are the only REMs. So E hopes that D is hoping he lives at 3, in which case D would hope that C hopes he's living at 2, in which case C would be hoping that B hopes he's at 1 where A has just given up hope that he's living at 0. So, before the tourist visits, E hopes that they're at 4 and that D hopes that they're at 3 and that C hopes that they're at 2 and that B hopes that they're at 1 and that A hopes that he's living at 0. When the tourist makes his announcement, it's clear that no-one could possibly expect A to believe he could be at 0, so now E hopes that they're at 4 and that D hopes that they're at 3 and that C hopes that they're at 2 and that B hopes that they're at 1 and that A knows it. The next day, it's clear that no-one could possibly believe that A's the only REM, so E hopes that they're at 4 and that D hopes that they're at 3 and that C hopes that thy're at 2 and that B and A know it. The following day, E hopes that they're at 4 and that D hopes that they're at 3 and that C, B and A know it. Next day E hopes that they're at 4 and that D, C, B and A all know it. On the final day, E, D, C, B and A all realise they're at 5, much to F's relief the following morning. It's not important that everyone knows there's at least one REM, or even that they know everyone knows everyone knows everyone knows there's at least one REM. The point is that the 5 REMs don't know that everyone knows everyone knows everyone knows everyone knows there's at least one REM until the tourist speaks... The argument is much simpler to follow if you start at the bottom and work up - if you accept that the REMs at monastery k will suicide on the k'th night, then the REMs at monastery k+1, who are smart enough to also have realised this, will realise they can't possibly be at k when they wake up after the k'th night and there hasn't been a massacre. The only thing keeping REMs alive before the tourist visits is the fact that none of the monks at any monastery could know for sure which monastery they're at, so no monastery would do anything different from the others to enable it to be distinguished. The tourist's comment distinguishes 0 from 1. The events of the first night distinguish 1 from 2. The events of the second night distinguish 2 from 3. The events of the third night distinguish 3 from 4, and so on. Any monks unsure whether they're at k or k+1 will find out when they discover what happens on the k'th night. If not, then what's the minimum number of REMs required for no-one to suicide? |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 19th, 2006, 8:25pm Ok let me state it simplier.. I agree if the process starts that for any N number of REMs they will kill themselves on the Nth Night. (and I said this above as well) I donot agree that telling the monks there are REM's when they already know this would start the process given N is a large number of REMs. There is absolutely no change in the monks perceptions or knowledge level or proper etiquette. (It apprears to me the critical number is around 5 from below) My postulation was that for larger numbers the chain would not start, with this information. For small numbers it obviously does. Using a single monestary with 5 REMs (A,B,C,D,E) A is hoping there are 4 REMs, thus He Hopes B is seeing only 3. Further hoping B hopes C sees only 2. However A knows C cannot be seeing just 2 Because you know B is a REM. C can only hope there are just 3 REMs, because, A and C both see BDE, so A knows C cannot think there are only 2 REMs. If everyone knows that no-one can hope there are just 2 REMs then In My Opinion the chain would not start when given the information as presented. The monks would not jump to the Theoretical case of only 2 monks as they know NONE of the other monks can even HOPE that is a possability. Thats where the chain breaks in a single monestary problem. Having multiple monestaries in your case would occur as forecast by N Rems killing themselves on the Nth night at each monestary. (only of course if they communicated that monks had died at other monestaries on each night) |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 20th, 2006, 2:04am on 03/19/06 at 20:25:37, Thallion wrote:
The tourist's statement delivers more. He says that there's at least one, but to the monks it means that now everyone knows that everyone knows that everyone knows ..<arbitrarily many 'that everyone knows'>.. that there is at least one REM. It's now common knowledge, whereas before it was not. Sure everyone knew there were a certain minimum, but it wasn't common knowledge. It's like the puzzle of the three armies, one large one in the middle, and two smaller allies on opposite sides. The two allies have to agree when to attack, because they can't win on their own. But any messenger they send might be caught. So you send one messenger over, but you don't know if he arrived untill they send one back. But they don't know if their messenger arrived untill you send another one back. So you can never be sure the last messenger arrived and will always be slightly unsure if the plan goes through or not. |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 20th, 2006, 5:00am Ah, but that is not true. With 5 REM's all of them know that all of the others know there is more than 1 REM. A believes he is not a REM. He knows the other 4 are REM's so he knows all of the other monks know that there are at least 3 REMs. This is fact. Regardless of his own Eye color. He Knows that all of the other monks see at least 3 REMs, as he sees the same monks. Everyone already knew there is at least 1 REM. It was common knowledge that every monk saw more than 1 REM. Wether or not the fact is stated outloud is irregardless of the monks perception of the number of REMs (for a large enough N) This is why I think 5 is the critical Number(though 4 is on the edge). If none of the monks can consider the 2 monk case, then I do not think the chain would start. And with 5 REMs every monk knows the other monks see at least 3 REMs. Hence my current position that given a large number of REMs (5 seems enough to me) the given information will not start the chain reaction. To put it another way: Given N number of REMs. Everyone knows that everyone else knows that there are AT LEAST N-2 REMs. Derived from assuming they are not a REM, they know the other REMS see at least the same monks that they see in common. Namely not themselves and not the other viewer. Hence N-2. If N-2 > 2 Then the cycle cannot start by the information given. As they all knew 1 was not a possability, hence nothing changes by seeing no-one commit suicide, as that is the same event that has happened every night for the last years and years. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 20th, 2006, 5:28am on 03/20/06 at 05:00:17, Thallion wrote:
Suppose we have a total of 3 REMs. Each of them obviously knows that every other monks must see at least one, regardless of wether he is one himself. However, A doesn't know whether B knows if C can see a monk with red eyes, even though he himself does know that C sees one. Because if A doesn't have red eyes (as he hopes), and A thinks that B thinks he doesn't have red eyes (as B would hope) than C wouldn't see any red eyes. And thus wouldn't kill himself unless someone came buy to tell them that at least one of them does in fact have red eyes. And that works for all N > 0 You need all levels of knowledge, so the monks can reason about eachother's reasoning. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Mar 20th, 2006, 3:25pm The change in information that the tourist provides is in the statement "If there was only one REM, he would know he was an REM." Before the tourist's unfortunate remark, everyone knows this statement to be false. After the remark, everyone knows that it is true. It is from the truth of this remark that the monks can derive the statement "For all K, if there were exactly K REMs, they will commit suicide on the Kth night". This derivation has nothing to do with how many REMs really exist. It is strictly hypothetical. Only after this statement is derived does the real situation comes into play. If there are actually N REMs, then the REMs know that there are either N-1 or N REMS, while the BEMs know that there are either N or N+1 REMs. So they all wait for N-1 nights (the BEMs are actually waiting for night N). When nothing happens on night N-1, the REMs all realize there is one more REM than they can see. It does not matter that everyone knows there are at least N-1 REMs. The information change and deduction are not based on what the actual monks know. It is based on what hypothetical numbers of REMs would know if the hypothetical situation were real. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Mar 20th, 2006, 6:34pm Quote:
A knows that C sees at least 3 REMs (BDE), and B knows that C sees at least 3 REMs (ADE), but A, who's hoping the only REMs are the ones he knows about (BCDE) hopes that B only knows about C seeing at least 2 REMs (DE). If A's hope were true, then B (only seeing CDE) would hope that C only saw DE. A knows C cannot think that there are only 2 REMs, but doesn't know that B also knows this, so hopes that B hopes that C sees only 2 (DE). Similarly, D sees 4 REMs (ABCE), and C knows D sees at least 3 REMs (ABE) but B only knows that C knows that D sees at least 2 REMs (AE) and A only knows about B knowing about C knowing about D knowing about the 1 REM (E) because A doesn't know there's anything to know about himself being one. Finally, while B knows that C knows that D knows that E knows about 1 REM (A), A doesn't know that, so hopes B hopes that C hopes that D hopes that E doesn't know there are any REMs. Once the tourist speaks, A knows that B knows that C knows that D knows that E knows that at least one of ABCDE is an REM. Once the first night passes, A knows that B knows that C knows that D knows that E isn't the only REM, so A knows that B knows that C knows that D knows that at least one of ABCD is an REM. After the second night, A knows that B knows that C knows that D saw a second REM as well as E so A knows that B knows that C knows that at least one of ABC is an REM After the third night, A knows that B knows that C saw a third REM as well as DE, so A knows that B knows that at least one of AB is an REM After the fourth night, A knows that B saw a fourth REM as well as CDE, so A knows that A is an REM. On the fifth night, A kills himself (so do BCDE having followed the same logic with the letters assigned differently) |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 20th, 2006, 9:37pm on 03/20/06 at 05:28:05, towr wrote:
In a case of N-2 > 2 they DO know that they know that they know. 3 is not large enough for n-2 >2 to be true, so in your example case I agree the chain starts. on 03/20/06 at 18:34:38, rmsgrey wrote:
Agree - A knows BCDE, A Wants B to only know CDE(even though wrong), A wants B to want C to only see DE. A knows that is not possible. Thus A knows (as BCDE all have the same knowledge) That Each other person already knows there are at least 3 REMs. The best case hope is that each of the others are hoping that the others are only seeing 2. But A knows the others know that everyone sees at least 2. Hence N-2 > 2 became my theory of when the process breaks. on 03/20/06 at 15:25:38, Icarus wrote:
This is a very good explaination, One of these best I have seen. But this is not what the tourist says. Though I see where you can make that conclusion of his statement. I can't really contradict that the process would start if that had been the statement made by the tourist. If though is the key word. Which is also the key to this whole problem. IF brings in the theoretical cases, and indeed if you worked the problem UP and you said "I see Red Eyes, if there were only one person he would know he had them." Indeed the process would start. However It's my ascertation that people would not jump to the hypothetical if the statement was only "i see red eyes" as it is in the problem. When the Tourist says, "I see Red Eyes" Each monk (given N of REMs with N-2 > 2 is true) would simply say to themselves, "DUH so do I" Recursive thought is not a natural tendancy, and this problem is recursive logic when done from the monks perspective. Thats why given a large enough N the knowledge that there are REMs wouldn't change anything. |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 20th, 2006, 11:32pm Most people when I propose this solution get hung up on the chain of events, so let me be clear. This is a chain reaction problem. Given that the chain starts the chain Will follow the pattern N REMs will die on the Nth Night. No arguements from me. The question becomes "what is truely required to start the chain?" Chains start because of a state change. Adding more vinegar to a beaker wont do anything it's only when you add in baking soda does anything start to happen. If everyone knows that everyone else knows there is at least 1 REM, then the statement there is at least 1 REM, does not constitute a change in state. Hence my postulation of if N is sufficiently large the information given is not enough. In this case I use N-2>2 though you could simply say n>4. But I like the n-2 >2 Because it indicates the level of knowledge that each person would have, as well as the cut off point, where that knowledge is important. The first side is each person knows that every other person knows there are at least n-2 REMs as both the viewer and every other person will see that same n-2 set of people. The second side of the equation is the number that indicates when the information there is at least 1 REM becomes valueable. At only 2 REMs they hope the other doesnt see any and hence would provide that other person with the knowledge that they are the only one. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 21st, 2006, 2:21am on 03/20/06 at 21:37:58, Thallion wrote:
if there are N REMs they consider that the other N-1 consider there may be only N-2 that consider there are N-2 ... etc untill you come to one REM that considers there is hopefull no REM. And every REM has such a chain leading to zero. You can model any N REM as an N dimensional hypercube where each monk considers all worlds along one direction equally possible (the ones where he is and the ones where he isn't REM himself) starting at any of those world (where a subset of N has red eyes) you can move along the edges to a world where none of the monks has red eyes. As long as any monk can hope that some monk hopes that ... etc that a monk hopes there is no REM, then they all live on happily doing what they do. The tourist destroys this hope by make a public announcement that there is at least one. So the no-REM world is no longer possible, and no path can lead there. And so the hypercube unravels step by step untill monks start killing themselves. (Each Kth steps all world with K-1 REMs are eliminated) Quote:
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 21st, 2006, 2:38am on 03/20/06 at 23:32:34, Thallion wrote:
Let's look at the case of 4 monks A,B,C,D using some mathematical abbreviation (and for any permutation of A,B,C,D) we have at the start: KA(>=3) {A knows that there are >= 3 REM) KAKB(>=2) {A knows that B knows ...) KAKC(>=2) KAKD(>=2) KAKBKC(>=1) KAKBKD(>=1) KAKCKB(>=1) KAKCKD(>=1) KAKDKB(>=1) KAKDKC(>=1) KAKBKCKD(>=0) KAKBKDKC(>=0) KAKCKBKD(>=0) KAKCKDKB(>=0) KAKDKBKC(>=0) KAKDKCKB(>=0) Now the publis statement by the tourist adds the relevant knowledge update K*(>=1) K*K*(>=1) K*K*K*(>=1) K*K*K*K*(>=1) where * is a wildcard for any A,B,C,D (so all 4,16,64,256 combinations respectively). So the relvant change in knowledge is that the last 6 lines become KAKBKCKD(>=1) KAKBKDKC(>=1) KAKCKBKD(>=1) KAKCKDKB(>=1) KAKDKBKC(>=1) KAKDKCKB(>=1) Because of this change people start dying. And it works for any N, it'd just be more work to write out. (not to mention I've already left out a lot) |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 21st, 2006, 4:52am To illustrate the hypercube model, here's a picture (for 4 REMs): http://www.ai.rug.nl/~towr/4REM.png magenta connects worlds indistinguishable to the first REM green connects worlds indistinguishable to the second REM blue connects worlds indistinguishable to the third REM red connects worlds indistinguishable to the fourth REM (All worlds are also reflexive for all monks, of course, but it's cluttered enough as it is) For any actual world you start in, you can get to 0000 in 4 steps or less along different colored edges. Meaning that some monk believes it's possible that some other monk believes it's possible that a third monks believes it's possible that the fourth believes there are possibly no REMs and won't kill himself because of that.. [edit]oops, there's a typo in the picture, there's two worlds named 1101, the left one should be 1100[/edit] |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Mar 21st, 2006, 5:23am Thallion, I'm guessing you're still unconvinced since towr still used only 4 monks in his example rather than five (your threshold value), so let me try. :) I. From the bottom up (1) If there were only one lone REM (and any number of BEMs), that lone REM would know there were either zero or one REMs and hope there were none; each BEM would know there were one or two REMs and hope there was only one. The tourist's statement would give no new information to the BEMs, but would dash the hopes of the lone REM, who would commit suicide that first night. (2) If there were two REMs (and any number of BEMs), each REM would know there were one or two REMs and hope there was only one--that is, they would each hope they were living in situation (1). However, after no monks committed suicide the first night the REMs' hopes would be dashed--they could no longer hope to be living in situation (1)--and they would each commit suicide the second night. (3) If there were three REMs (and any number of BEMs), each REM would know there were two or three REMs and hope there were only two--that is, they would each hope they were living in situation (2). However, after no monks committed suicide the second night the REMs' hopes would be dashed--they could no longer hope to be living in situation (2)--and they would each commit suicide the third night. (4) If there were four REMs (and any number of BEMs), each REM would know there were three or four REMs and hope there were only three--that is, they would each hope they were living in situation (3). However, after no monks committed suicide the third night the REMs' hopes would be dashed--they could no longer hope to be living in situation (3)--and they would each commit suicide the fourth night. (5) If there were five REMs (and any number of BEMs), each REM would know there were four or five REMs and hope there were only four--that is, they would each hope they were living in situation (4). However, after no monks committed suicide the fourth night the REMs' hopes would be dashed--they could no longer hope to be living in situation (4)--and they would each commit suicide the fifth night. (6...N) and so on. N REMs (with any number of BEMs) would each hope to be living in situation (N-1), but would have their hopes dashed after night N-1 when no monks committed suicide. II. From the top down There are five REMs, call them A, B, C, D and E. Without loss of generality, consider monk A. Monk A knows there are either four or five REM, and hopes there are only four. Specifically, A hopes that > w/out l. of g., B knows that A is a BEM and so there are either three or four REMs, and hopes there are only three REMs. Now here's where I think confusion may be coming in; A knows that B's hopes are false--that there are in fact four or five REMs, but A is still hoping that B is hoping there are only three. That is, A hopes that > B hopes that, > > WLOG, C knows that A and B are BEMs and so there are either two or three REMs, and hopes there are only two REMs. That is, A hopes that > B hopes that > > C hopes that, > > > WLOG, D knows that A, B, and C are BEMs and so there are either one or two REMS, and hopes there is only one REM. That is, A hopes that > B hopes that > > C hopes that > > > D hopes that > > > > E knows that A, B, C, and D are BEMs and so there are either zero or one REMs, and hopes there aren't any REMs. That is, A hopes that B hopes that C hopes that D hopes that E hopes that there are no REMs. But since at each step we chose an arbitrary representative monk, without loss of generality, it holde for all combinations of monks; that is, if -> represents "hopes that": A->B->C->D->E->there are no REMs B->A->C->D->E->there are no REMs C->B->A->D->E->there are no REMs D->C->B->A->E->there are no REMs E->D->C->B->A->there are no REMs etc. But after the tourist's statement, A can only hope that B hopes that C hopes that D hopes that E now knows he is the only REM. But after the first night, when no monks commit suicide, A can only hope that B hopes that C hopes that D ane E now know they are the only two REMs. But after the second night, when no monks commit suicide, A can only hope that B hopes that C, D and E now know that they are the only three REMs. But after the third night, when no monks commit suicide, A can only hope that B, C, D and E now know that they are the only four REMs. But after the fourth night, when no monks commit suicide, A now knows that all five of them are REMs, and so sommits suicide on the fifth night. B, C, D and E, reasoning likewise, do the same. Whew, a bit long, but I hope it's clear enough. --SMQ |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 21st, 2006, 5:46am on 03/21/06 at 05:23:43, SMQ wrote:
Like that isn't enough work already :P Oh well, it's not like I believed it the first time my professor tried to convince me. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 21st, 2006, 6:09am here's the entire proces modeled for 5 REMs http://www.ai.rug.nl/~towr/5REM.png Ever more worlds get eliminated as long as no one dies. Because if that possible world was the actual one, people would have killed themselves (Due to not being able to believe another world is possible, i.e. there not being a connection to another world with their color) |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Mar 21st, 2006, 8:28am on 03/20/06 at 21:37:58, Thallion wrote:
A doesn't know any such thing. As far as A is concerned, it's perfectly possible for B to want C to only see DE... A knows that B would be hoping for something that isn't actually the case, but we know that A is hoping for something that isn't the case when he hopes he doesn't have red eyes... Anyway, you seem to agree that, if the tourist makes his statement and there are only 4 REMs, all four commit suicide on the 4th night. If there are 5 REMs and the tourist makes his statement, then each REM knows that there are either 4 or 5 REMs, and knows that if there are 4 REMs, they will suicide on the 4th night. What conclusions will the 5 REMs draw when they wake up after the 4th night and no-one's suicided? You don't need to look at what people know about what people know about what people know (etc.) only at what everyone knows would happen to a monastery with a given number of REMs on a given night. Suppose, each afternoon, each of ten monks at a monastery is asked, in secret, by an outsider, the following questions: 0) How many REMs could there be at this monastery? 1) What would have happened last night at a monastery with 1 REM? 2) What would have happened last night at a monastery with 2 REMs? 3) What would have happened last night at a monastery with 3 REMs? 4) What would have happened last night at a monastery with 4 REMs? 5) What would have happened last night at a monastery with 5 REMs? 6) What would have happened last night at a monastery with 6 REMs? 7) What would have happened last night at a monastery with 7 REMs? 8) What would have happened last night at a monastery with 8 REMs? 9) What would have happened last night at a monastery with 9 REMs? 10) What would have happened last night at a monastery with 10 REMs? Until the tourist shows up, each REM will give the same answers every time, and so will each BEM, and they'll agree with each other on everything except question 0 - nothing special would have happened. Once the Tourist makes his statement, the answers to the questions may start to change. |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 21st, 2006, 9:56pm Wow, I just hadn't had a chance to come check recently. goodness. hypercubes.. :o you all are going all out. If the monks say, "well if there had only been one REM he would kill himself tonight" then I guess the process starts. If it is taken 1 question at a time forward then there is no reason it wouldn't start. Mind you I think that is a pretty big IF. I do have a problem with the fact that if A knows B knows that everyone else sees more than 1 REM, that the information means anything to him. Regardlses of what A knows B knows that C knows that D knows. What B knows C knows about D is irrelevant to A because A relies on his knowledge of the others to make his conclusion, not someone else's thoughts about 2 other people furhter away from him. However having said that, if you follow the whole a knows b knows c knows.... long enough then again you could come to the 0 case and 'start' the chain. I also have a problem with the lack of any proof being sufficient evidence that the event was true/false. Can the lack of an action prove a binary question is true or false? What if you already knew the answer, then does it prove anything? But then again I guess that is part of the problem with inductive logic it does deal with probability that an outcome is correct. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 22nd, 2006, 12:38am on 03/21/06 at 21:56:09, Thallion wrote:
Quote:
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If there are N REMs they must kill themselves on the Nth night. If no REMs kill themselves on the Nth night, then this proves there must be more than N. Quote:
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 22nd, 2006, 3:19am on 03/22/06 at 00:38:07, towr wrote:
I think you are misreading me there. I said the knowledge that everyone knows >=1 is already known, thus that bit of information means nothing to them. Noone learns anything until N-1 nights. Everything Before N-2 nights is all hypothetical. (Note BEMs dont learn anything until N Nights) On N-1 Night REMs learns the other REMs see the same number of REMs as themselves. On N Night REMs kill themselves and BEMs learn that they were not a REM. Your statement is describing this interchange on night N-1, it has nothign to do with what was "learned" when the tourist opened his big mouth. You can hope that the other REMs learn something on an earlier night (which would thus move you into the other category), but until N-1 Night You ( and the other REMS) donot learn anything. The premise comes down to this(commonly accepted answer): For any N of REMs, when the statement of "I see REMs" is made that every REM will think "Hmm that means that if there was only 1 he would now know who he is, which follows that if there were 2 they would find out tonight, that 3 would...." When in reality given a sufficiently large number of REMs they say "DUH I knew that" Thats really what this boils down to. What is the probability that of 100 REMs they all think exactly the same thing that is required to start the chain? I like my N-2 > 2 postulation mainly out of the facts that, People may care about someone knows about someone else knowing (a knows B knows C knows) IE 2 hops from N) so N-2, and the case when the answer of NOt 0 is relevant is the 2 REM case so > 2. But I fully agree IF the trail/regression is followed by all N, and everyone acknowledges night 1. Then the formula works. I just think that is a pretty big IF. |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 22nd, 2006, 3:59am I think you all can agree with me on this.. All N must acknowledge the hypothetical case of N being 1. AND All N must acknowledge that night one occurs. If either of these are false the process cannot begin. Can we agree with this? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 22nd, 2006, 4:15am on 03/22/06 at 03:19:39, Thallion wrote:
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Until the final night when there is just one hypothesis (possible world) left, which then becomes the actual working theory of them committing suicide. Quote:
on 03/22/06 at 03:59:02, Thallion wrote:
Every single one of the N REMs knows there are at least N-1. They can't consider the hypothesis that it's less. But they can consider that other monks might consider that so. Quote:
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 22nd, 2006, 5:18am Wow, apparently no matter what I say you are going to disagree with me. For if you had said Yes I agree then you would have proved the process works and that the Nth night is true. Now I don't know what you are doing, unless you are just trying to be "know it all" because that is how you are treating this discourse. You say No to my statement: " All N must acknowledge the hypothetical case of N being 1." Regardless of the monks perceptions of what everyone knows, the monks have to be able to say "If there was 1 monk that monk would now know he was the only one." Thus they must Hypothesize that if there was only one REM that that monk would kill himself the first night. Are you really telling me you disagree with this. Do you want to consider your answer again. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 22nd, 2006, 5:28am on 03/22/06 at 05:18:49, Thallion wrote:
They don't have to consider that "there might be only be one monk". But that "if there was one monk, he'd kill himself tonight" (and everyone knows this) |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 22nd, 2006, 6:00am The statement: All N must acknowledge the hypothetical case of N being 1. Is the same as: The Monks must acknowledge that if there was 1 monk that monk would now know he was the only one. The second more wordy certainly, The second certainly spells out the hypothesis and thus makes it easier to draw the conculsion. But they do say the same thing. A hypothetical case is a case where a hypothesis is drawn (webster's dictionary) So I am not sure how you can say these are different statements. When the second part of the second line is simply the hypothesis referenced in the first statement - > "if there was 1 monk that monk would now know he was the only one" This is especially relavent in this riddle For everyone has said over and over in this thread that the statement of " I see red eyes" is equivalent to saying "I see red eyes, So if there was only 1 REM that monk would now know he was the only one." And then running from there. That is all semantics, though so I will reword the question to make it more clear. Can we all agree that: All Monks must acknowledge that if there was only 1 monk that he would now know he was the only one. AND All Monks must agree that night 1 has occured. Note I do see that actually if the first statement becomes true at the same time for every monk that the second statement becomes a given. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 22nd, 2006, 6:19am on 03/22/06 at 06:00:13, Thallion wrote:
So that lead to a misunderstanding.. Quote:
Before the tourist speaks the one monk may still have hope there aren't any. And if the statement isn't public (to all monks at once where they all know they all hear it), not all other monks may know that the one REM knows he should kill himself that night. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Mar 22nd, 2006, 6:54am OK, the standard solution does rely on certain assumptions which are commonly accepted in solving logic puzzles involving reasoning beings. 1) The monks are "smart enough" to work out (by the time it becomes relevant) anything that we can work out, and will do so (and this is common knowledge). 2) The conditions of the problem are common knowledge among the monks. 3) The monks don't cheat. No clever murders disguised as suicides, no refusal to suicide, etc... The key assumption is the first one - it means that the monks will be aware that, when N=4, all REMs will kill themselves on the 4th night, so if N=5, then all 5 will know where they stand after that night... As far as nothing happening proving something, there's the notorious curious incident of the dog in the night-time. The answer to the binary question "Did any monks suicide last night?" is definitely given by the lack of events of the night before... |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Mar 22nd, 2006, 8:29am Thanks, Thallion, for forcing us to be completely clear about our reasoning. :) (seriously; it's good mental exercise!) You seem to agree that the "bottom up" argument at least looks reasonable, so let's reformulate the "top down" argument in a more "bottom up" way: First, let's look at the state of things before the tourist makes his unfortunate observation. Consider an abbey, First Abbey, with only one red eyed monk, Br. Albert. Br. Albert looks around and sees only brown eyes, so fervently hopes his own eyes are brown. We, as outside observers (or as brown-eyed monks in First Abbey), know that Br. Albert is hoping for a false thing--that his own eyes are really red--but Br. Albert doesn't know that, and so he goes on fervently hoping despite our superior knowledge. :) Next, consider Second Abbey, with two REMs, Brs. Albert and Bennie. Br. Bennie looks around and sees all brown eyes except for Br. Albert, so fervently hopes his own eyes are brown and that Br. Albert is the only REM. But in hoping his own eyes are brown, Br. Bennie hopes that he is an observer in First Abbey where Br. Albert is a lone REM; that is, Br. Bennie hopes Br. Albert hopes there are no REMs, and, being an expert logician, Br. Benie realizes he is hoping this about Br. Albert's hopes. In so doing, Br. Bennie places himself in our position in First Abbey, reasoning about Br. Albert's hoped-for hopes even though he knows that reality is other than Br. Albert hopes for. Now, let's move on to Third Abbey where there are three REMs, Brs. Albert, Bennie and Charles. Br. Charles looks around and sees that Br. Bennie's and Br. Albert's eyes are red, so fervently hopes his own eyes are brown and Brs. Albert and Bennie are the only REMs. But in so hoping, Br. Charles hopes Br. Bennie considers himself to be living in Second Abbey. And just as in Second Abbey, the Br. Bennie of Br. Charles' hopes himself hopes Br. Albert considers himself to be living in First Abbey where there might not be any REMs at all. That is, Br. Charles hopes Br. Bennie hopes Br. Albert hopes there are no REMs, and, being an expert logician, Br. Charles puts himself in our place in Second Abbey and realises he is hoping all this. Likewise, in Fourth Abbey, Br. David is hoping Fourth Abbey's Br. Charles considers himself to be living in Third Abbey, etc., and, of course, in Fifth Abbey Br. Ernie is fervently hoping that Br. David is hoping that Br. Charlie is hoping that Br. Bennie is hoping that Br. Albert is hoping that there aren't any REMs. It should be clear that the names don't matter; that every REM in Fifth Abbey puts himself in the place of Br. Ernie and hopes things about the hopes of the hopes of the hopes of the hopes of all his Brothers, and, being himself an expert logician, realizes all the things he is hoping. In general, every REM hopes that his own eyes are brown, and in so hoping, implicitly hopes things about his red-eyed Brother's hopes--specifically that each of them is hoping exactly what they would be hoping if they each saw one fewer REM than he himself does--and, being an expert logician, is able to reason out all the myriad things he is hoping, not only about his Brothers and about their immediate hopes, but about the hopes of the hopes of the hopes of ... of all his Brothers, right down to that completely hypothetical, many-times-removed lone monk who everyone knows doesn't exist in reality but everyone is hoping that someone is hoping that someone else is hoping that ... he hopes there are no REMs. Now along comes the unsuspecting tourist and his unfortunate remark. On the surface, it appears nothing changes--everyone already knew that there was at least one REM, of course--but let's look at what it does to Br. Ernie's hopes. Still hoping he himself has brown eyes, but now realizing that no monk, however hypothetical could reasonably believe that there are no REMs, Br. Ernie has begun to hope that Br. David hopes that Br. Charles hopes that Br. Bennie hopes that Br. Albert now realises he has red eyes. That is, in First Abbey, Br. Albert would now have had his hopes dashed. He could no longer hope that there were no REMs, and, seeing that all his Brothers had brown eyes, would know that he must be the lone REM. In Second Abbey, Br. Bennie could no longer hope that Br. Albert hoped there were no REMs; he (Br. Bennie) could only hope that Br. Albert now knows himself to have red eyes. We the observers know that this still isn't true; Br. Albert is, in reality, hoping the same things about Br. Bennie, but Br. Bennie doesn't know that, so he goes on hoping. Likewise in Third Abbey Br. Charles puts himself in the place of an observer at Second Abbey and hopes Br. Bennie hopes Br. Albert now knows he has red eyes; and so on up to Br. Ernie's hopes in Fifth Abbey. But after the first night passes, Br. Enrie's hopes are changed again. Now, in a hypothetical First Abbey, Br. Albert would have committed suicide. In Second Abbey, Br. Bennie, noticing that Br. Albert didn't actually commit suicide, can no longer hope that he is an observer in First Abbey; he must accept the reality that he's been living in Second Abbey all along and that his own eyes are red. In Third Abbey, Br. Charles, still hoping he is an observer in Second Abbey, now hopes that Brs. Bennie and Ablert both now realize they have red eyes. And so on, right up to Br. Ernie in Fifth Abbey who hopes he is an observer in Fourth Abbey where Br. David hopes he is an observer in Third Abbey where Br. Charles hopes he is an observer in Second abbey where Brs. Bernie and Albert now realize they must have red eyes. Likewise, after the second night Br. Ernies hopes must change yet again, now hoping that Br. David hopes that Brs. Charles, Bennie, and Albert now realize they are all REMs. And after the third night the best Br. Ernie can hope is that Brs. David, Charles, Bennie and Ablert all now realize they must have red eyes. (And, of course, in reality, Brs. David, Charles, Bennie and Albert are all hoping exactly the same thing about all the monks except themselves.) Finally, after the fourth night, when still no one in Fifth Abbey commits suicide, Br. Ernie is forced to abnadon his last remaining hope and accept that he must have red eyes himself. |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Mar 22nd, 2006, 8:32am In summary then, to begin with, every red-eyed monk in Fifth Abbey hopes he is living as a brown-eyed monk in Fourth Abbey where each of four red-eyed monks hopes he is living as a brown-eyed monk in Third Abbey where each of three red-eyed monks hopes he is living as a brown-eyed monk in Second Abbey where each of two red-eyed monks hopes he is living as a brown-eyed monk in First Abbey where a lone red-eyed monk hopes his own eyes are brown. On the day the tourist makes his unfortunate remark this delicate skien of hopes begins to unravel. Now every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where every REM hopes to be a BEM in Third Abbey where every REM hopes to be a BEM in Second Abbey where every REM hopes to be a BEM in First Abbey where a lone REM just found out his own eye color. On the day after the remark, every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where every REM hopes to be a BEM in Third Abbey where every REM hopes to be a BEM in Second Abbey where two REMs just found out their own eye color. On the third day, every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where every REM hopes to be a BEM in Third Abbey where three REMs just found out their own eye color. And the fourth day, every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where four REMs just founf out their own eye color. And finally, the fifth day, every REM in Fifth Abbey finds out his own eye color. Whew! :) --SMQ |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 22nd, 2006, 8:22pm It was a bit of a trick question.. The tendancy is that once you believe someone doesn't agree with you, you tend to try to refute everything that person says, closing off your mind to the fact that that person could be right about something. The trick is, that if the two statements made above are true the process starts wether you use top down or bottom up logic. So the trick is if you try to refute either of the above statements then you end up trying to say the riddle breaks down. Work with me on this I am being devils advocate now... The riddle states that the monks have a vow of silence. The riddle states the tourist makes the statement, "I see people with Red Eyes". Situation 1: The riddle does not say that all of the monks heard the comment. And we know they have a vow of silence so they cannot tell the others what they heard. Do you agree this situation could break the problem? Situation 2:All of the monks are eating breakfast when the tourist makes the statement, and the statement is made in the same room as all of the monks. Do all of the monks know for certain that every other monk heard the statement? Does the process start? Situation 3: What if one of the REMs is deaf? What if one of the BEM is deaf? Situation 4: What if the tourist tells each monk individually that he sees monks with red eyes? Situation 5: What if the tourist is at the monestary for two days. The tourist tells every monk he has told every other monk that he sees REMs. What happens? A few other misc comments. RMSGrey's statement #1. My issue with this is, I would imagine we are all pretty smart people. Smarter than your average joe certainly. Anytime you put people in a situation you cannot count on anything. but if this is a 'tenant' of logic problems, then forgive me, as it has been 13 years since i was in my logic class in college. ;) |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 22nd, 2006, 10:43pm I have another one. Situation 6: What if when the tourist makes the fateful comment, all of the REM's hear the comment, but not all of the BEM hear it? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 23rd, 2006, 1:10am Or what if a REM when he find out decides not to kill himself. Or what if BEM decide to kill every REM they see? You can almost always sidestep the intended puzzle and make something else of it. That's usually covered in the first pages of a thread ;) S1/2/6- As long as all REM hear the statement at the same time and know they all hear it, the process will start. The BEM don't matter in this case. If not all REM hear it, or think they don't all hear it the process doesn't start. If some monks think all hear it, but in fact some don't, awfull things may start to happen. S3- If X of the REMs is deaf, then the others if they know this will kill themselves on the N-Xth night, or if they don't know this will kill them on the Nth night. The deaf ones survive (not having a clue that anything went on) Of course after that mass uicide, they may in the latter case deduce there are X deaf REMs and seeing only X-1 conclude they are the last one. But that's assuming monks don't commit mass suicide for other reasons (otherwise they may not be sure that the red eyes thing was the cause). S4- If each monk is told individually, then there is no common knowledge and the process doesn't start, because now they really don't learn anything new from his statement. S5- second level knowledge is not enough to start the process. The tourist must also tell every REM that he told every REM that he told every REM... etc up to at least the level of the number of REM. If there is just one REM, telling him there is at least one is enough to bring him to kill himself. If there are two, he must also tell both that he told the other one, so that they know the other one would kill himself if he were the only one. Etc. |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on Mar 24th, 2006, 3:31am On the second day after the tourist departed, Br. Zoe leaves his room and wonders: Where is everybody? It is breakfast time but nobody is there. This place has become crazy! The brothers have acted strangely lately. In fact, since that guy came to visit. He said something about red-eyed monks. I don't understand what it was about. There are no red-eyed monks on this island. Hello? Anybody there? |
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Title: Re: BROWN EYES AND RED EYES Post by Thallion on Mar 24th, 2006, 8:16pm Ok, I agree with your assessment of all of the cases except 5. in 5, I have broken the agreement on night one conditional. Since the tourist was at the monestary two days, monks could have found out on day 1 or day two, that uncertainty means the monks can't pick a night 1, thus they can't start eliminating numbers. Also I disagree that he would have to tell, that he told, that he told. I believe if the tourist told each monk exactly this it would be sufficient, "I have told every monk this same bit of information, I see red eyed monks" Thus every monk would have the knowledge that every other monk also knew it. Here is an interesting side effect of this. If the tourist tells the monks the above statement (all in one day), but he doesnt tell them all, he got confused on who he told, they all look the same in those robes anyway. Well.. all of the REMs who he has told will kill themselves on nigth N regardless of how many REM he doesn't tell. all the REMs he hasn't told will live on. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Mar 25th, 2006, 4:52am on 03/24/06 at 20:16:30, Thallion wrote:
And it seems once again I interpreted somethign differently from you. The statement "I have told every monk this same bit of information, I see red eyed monks" is different from "I see red eyes, and I told every monk I see red eyes" Which is how I read "The tourist tells every monk he has told every other monk that he sees REMs" So in that case you have insufficient depth of knowledge. Interpreted your (intended) way, the statement seems to be recursive and has infinite depth. It's a big problem in such word problem to be precise and unambiguous. Which I suppose is why logicians invented formal logic notations. Quote:
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Mar 25th, 2006, 5:20am on 03/22/06 at 20:22:42, Thallion wrote:
The problem with throwing "real people" into a problem is that everyone has a different idea of how "real people" behave... It's a sensible default assumption that people in logic problems, having grown up in a world built around logic, are going to be competent logicians... @Towr: "I have told every monk this same bit of information, I see red eyed monks" is less likely to be believed than "I am telling every monk this and that I have seen red eyes here" |
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Title: Re: BROWN EYES AND RED EYES Post by fireashwinter on May 21st, 2006, 9:23pm "There is at least one Colorblind REM." Cronos (I think) was correct to say that this information would not lead to suicides. A CB monk knows that all the other monks might be CBREM. Normal monks know that all the REMs they see might be CBREM. So, no monk will act, as uncertainty remains. This makes me wonder. Assume that the tourist simply said to the monks all at once: "At least one of you is a REM." Then, the tourist realizes at some day M<N the suicide rule. As Jeremy asked, what might be a very best solution to prevent suicides? What if the tourist says: "You all know there is at least one REM." or "All of you have for some time known that there is at least one REM." Once we have opened up the possibility that the monks are not perfectly perceptive in sight, those variations may have meaning. As another question, consider what would happen if the tourist says to all the monks at once: "You are each aware of the physical state of every other monk." |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 22nd, 2006, 9:31am on 05/21/06 at 21:23:22, fireashwinter wrote:
Sounds good. Quote:
I don't think there's any way to prevent at least some suicides from happening as long as the monks are aware of whether other monks have suicided (unless some later information comes along to change things) Quote:
Certainly the second is equivalent to saying "there are at least 2 REMs and no-one is colour-blind" - the first might count as "there is at least one" Quote:
The statement will do nothing on its own - it's one of the assumptions of the original... |
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Title: Re: BROWN EYES AND RED EYES Post by fireashwinter on May 22nd, 2006, 11:17am rmsgrey, Perhaps I was unclear. I understand CBREM is outside of the normal possibilities, so the original suggestion that we finger monks is the solution to minimizing suicides is correct, assuming we can't claim to have lied, etc. I was just wondering-- once we say CBREM is possible, might a solution be to tell the monks "I had neglected to tell you that at least one of you is colorblind"? And my subsequent 2 were along the same lines-- how would the tourist have to alter the CBREM solution if he had slightly different original statements. "You all know that there is at least one REM" is a statement that can be a cause of its content. Therefore, it does not necessitate prior knowledge, and does not necessitate the conclusion that it is the same as "at least 2 REM." So, the tourist's solution might be safe. NOTE: I am sorry if this seems hair-splitting. "All of you have for some time known..." is different, because it does necessitate prior knowledge, and does mean, as you say, that there are at least two REM. The last suggested tourist statement: "You are each aware of the physical state of every other monk." You say that this statement is part of the original assumptions, and so not new. I believe that you are confusing what we know with what the monks know. It is an original assumption that every monk is so aware, but it is not an original assumption that every monk knows that of every other monk. Thus the statement is new information, and may in certain circumstances have an effect. I will continue this in another post. |
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Title: Re: BROWN EYES AND RED EYES Post by fireashwinter on May 22nd, 2006, 1:38pm Sorry, Although my other points were valid, I've been thinking through the last one. I think I may have been mistaking a few of my assumptions about the problem as excluding what I suppose are standard assumptions (I have no background in such) in these problems. I have to think about this a bit more. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 23rd, 2006, 11:21am on 05/22/06 at 13:38:13, fireashwinter wrote:
One standard assumption for any logic problem that requires the agents to reason about each other's knowledge is that the details of the scenario are "common knowledge" - everyone knows it and everyone knows that (and that everyone knows that everyone knows that everyone knows, and so on) except where stated otherwise. If you remove the assumption that the scenario is common knowledge then there's no reason to conclude that the monks would expect each other to suicide - they each know that they would suicide were they ever to be convinced they were an REM, but they also know that they don't know anyone else would, so the entire process fails (OK, if there's just the one REM, then he'll still suicide, but the others won't be expecting it) If there are monks who don't know everyone else's eye-colour, then they also disrupt the logic - even the possibility that there are REMs with incomplete information is enough to kill the process - "either I have red eyes, or Brother Bob (who does have red eyes) doesn't realise that he's the only REM so hasn't suicided after all" - good news if you want to save everyone, but not so good if you want an interesting situation... On day one (the day before the hypothetical lone REM would suicide) or day two, saying "there is exactly one CBREM" would stop the process. On any day before the suicides, saying "there may be at least one CBREM" will also stop the process (each (hypothetical) REM that would suicide that night thinks "either I'm an REM so no-one suicided last night, or all the REMs I see are colour-blind, so didn't realise they should suicide") By day 3, it's too late for "there is exactly one CBREM" - a hypothetical pair of REMs would have one with normal vision, who would have suicided, expecting the CBREM to suicide too. |
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Title: Re: BROWN EYES AND RED EYES Post by Aravis on Jul 13th, 2006, 8:38am I was reading the posts, and never saw the answer to the question "how can the tourist kill everybody with one statement?" Icarus said he found one, but I was unsure of which one he was referring to, so I came up with one myself. Note: I have tried to use as few pronouns as possible to reduce confusion, so the grammar may not be quite right, but the ideas are there. Given R red eyes and B brown eyes, the tourist must say "There are exactly B-1 brown eyed monks". The red eyed monks will look around and see that there are actually B brown eyed monks, and thus will realize the tourist was lying. However, they have taken a vow of silence, and thus cannot say anything to this effect. WLOG, a hypothetical brown eyed monks Bx will look around and indeed see B-1 brown eyed monks, thus eveything Bx sees will be consistent with what the tourist said, and thus Bx will deduce that Bx himself must have red eyes, and thus must kill himself. Extend this to all brown eyed monks. Thus after one night, all the brown eyed monks will have killed themselves. WLOG, take red eyed monk Rx. Rx, only seeing monks with red eyes, will immediately realize that everybody Rx knew who had brown eyes killed themselves due to what the tourist said. Rx will then deduce that the brown eyed monks could not deduce that the tourist had been lying, thus there were originally B brown eyed monks. Therefore the Rx will realize that since he saw B brown eyed monks to begin with, Rx must have red eyes himself. Therefore at this point Rx will deduce correctly that since all the brown eyed monks killed themselves, and since these monks are infinitely logical, Rx must have red eyes, otherwise hewould have killed himself last night. Thus Rx commits suicide the second night. Extend this to all red eyed monks. Therefore, with one statement, the tourist has killed everybody in just two nights. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Jul 15th, 2006, 7:13pm That will work. However, there is a version that works even if the Monks are not vowed to silence - just to not talking about eye color. I.e., one where the REMs are just as in the dark as the BEMs are about the accuracy of the tourist's statement - at least, until the BEMs all kill themselves. But by then, it is too late. I'll let you think on it a while longer. If you want a hint: [hide]Consider Jonathan's questions[/hide]. |
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Title: Re: BROWN EYES AND RED EYES Post by Aravis on Jul 15th, 2006, 11:10pm Of course. It works the same way as my first idea, but gives less specific data that cannot be proven wrong or right. BTW, I guess this would kill off everybody in the single file hat execution too, huh. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Jul 16th, 2006, 10:28am on 07/13/06 at 08:38:35, Aravis wrote:
Corner case: with no BEMs, the tourist looks pretty silly saying "There are exactly -1 brown-eyed monks" and none of the monks are likely to act on the information in any way... Of course, saying "there are no BEMs" would work in that situation. |
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Title: Re: BROWN EYES AND RED EYES Post by JA on Jul 27th, 2006, 5:24am Hmmm. I've read the whole list of stuff and have a bit of a problem. I have found the "solution" that is stated a couple of times. But, sadly, it has a big (and unproven) assumption stuck in the middle of it. It may be a little thing, but then, hidden assumptions in logic proofs often tend to have unfortunate implications. Well, actually there are two big assumptions. The first (or the second as it happens) is that a simple induction proof suffices. The other assumption ... well. Have at it. JA /I have checked through to see if this has been addressed, but couldn't find it on two readings through. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jul 27th, 2006, 6:13am on 07/27/06 at 05:24:36, JA wrote:
Quote:
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Title: Re: BROWN EYES AND RED EYES Post by JA on Jul 27th, 2006, 6:41am on 07/27/06 at 06:13:05, towr wrote:
When a number of conditions are satisfied, yes. If they are not, then no (obviously). It's all a question of dimension JA |
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Title: Re: BROWN EYES AND RED EYES Post by BNC on Jul 27th, 2006, 6:49am on 07/27/06 at 06:41:20, JA wrote:
Like you say: obviously. But what condition do you think is not satisfied? And what do you mean by "dimension"? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jul 27th, 2006, 7:07am on 07/27/06 at 06:41:20, JA wrote:
I don't see what more should be to it. |
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Title: Re: BROWN EYES AND RED EYES Post by Aravis on Jul 27th, 2006, 8:44am on 07/27/06 at 05:24:36, JA wrote:
It would be a heck of a lot easier to discuss this issue if you'd actually say what the problem is. This deep into the thread, its not like there are going to be people who don't want to know the answer. Post the problem, and then somebody will either have a solution for that issue, or will discuss how the problem would have to be modified to take the issue into account. However, until the exact question is posted, nobody can really give an educated answer. |
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Title: Re: BROWN EYES AND RED EYES Post by JA on Jul 30th, 2006, 7:28am on 07/27/06 at 07:07:39, towr wrote:
Yeah, that sounds like induction doesn't it. But is that what we actually have? Take this proof (on page one) Quote:
Base case, got it. Quote:
This is one of two things. At best, it is an inductive proof that if n=2 we are okay. I suspect, however, that it is merely the proof of the case when n=2 with no true inductive basis. I could be convinced otherwise. What it definitely is not is a proof that if n is true, then n+1 is true. Quote:
This is just handwaving. There is another version of this proof on page one, and another that covers the specific case when n=3. Solving n=1, n=2 and n=3 leaves us a fair way short of infinity however. So, I haven't seen anything that actually covers the required inductive step - therefore I submit that the accepted answer remains to be actually proven. JA |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Jul 30th, 2006, 12:02pm on 07/30/06 at 07:28:22, JA wrote:
No, it isn't. It is a statement that if you are willing to think it out yourself, the same argument applies to any n, rather than just to n=2. (And if you are not willing to think it out yourself, then these forums are not for you!) Since you need it spelled out: Suppose that if there are exactly n REMs, then on night n, they will all kill themselves. Now examine the case of n+1 REMs. Each REM can see that either: (i) he is a BEM, and there are exactly n REMs, or (ii) he is one of exactly n+1 REMs. By the supposition, he knows that if (i) is true, then the REMs will all kill themselves on night n. So he will wait until the morning after night n. On that morning he will see that nobody has killed themselves, and hence there are not n REMs. He now knows that case (ii) is true, and he must be a REM. Therefore on night n+1, he will kill himself. Since this is true for each REM, on night n+1 all n+1 REMs will kill themselves, which is what was to be proven. |
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Title: Re: BROWN EYES AND RED EYES Post by JA on Jul 30th, 2006, 10:59pm on 07/30/06 at 12:02:12, Icarus wrote:
Cute dig. I suppose I'm not an "uberpuzzler" so I'm not worthy to even talk to you - but I wonder if you've actually thought this one out for yourself, or iif you're merely parrotting stuff other people tell you. Quote:
Why didn't he kill himself the night before? JA |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jul 31st, 2006, 1:46am on 07/30/06 at 22:59:42, JA wrote:
Quote:
base case: A single REM will kill himself on the first night. induction hypothesis (for M < N): if there are exactly M REMs they will kill themselves on the Mth night Or in contraposition: if no one kills themselves on the Mth night, there are not exactly M REMs. induction step: If there are N REMs, then any of those will know after the N-1th night, that there are not in fact just N-1 REMs, and seeing no others, they now know they complete the set. And thus kill themselves the next night, night N. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Jul 31st, 2006, 4:13pm As towr has said, it wasn't a dig. It was a point. You complained that no one had proven the induction case. Yet the example you gave clearly outlines how to prove the induction case. It does it for a particular case, but the author indicates that the same argument applies for all values, with appropriate modification. I.e., he is giving an example of how to prove the induction case rather than bothering to "dot the i's and cross the t's". He does this because most people are not mathematicians, and examples like this are what they understand best, not abstraction. Possibly this is the case for the author himself (I haven't bothered to look up who you are quoting). If he were wrong about the argument applying to the induction case, and you could show a failure of it to apply more generally, then you would have a point to your complaint. But it isn't hard to see that it does apply more generally, and you haven't pointed out any any failure to do so. Admittedly, there is a basic assumption here that the monks will not commit suicide unless they discover that they are definitely REMs. This is one of many basic assumptions needed to solve the puzzle. While the puzzle does not state this explicitly, it is obvious that this is the intent. It is common for puzzles like this to leave a number of things implicit rather than spelling everything out. Indeed, if most puzzles spelled out every condition, they would be insufferably long. Note that prior to the morning after the nth night, the monk still has the two cases I described, either one of which could be true according the induction hypothesis. In one of those cases, he is a BEM. So until the n+1st night, he has no cause to kill himself. It is only after the nth night that the induction hypothesis supplies information contradicting case (i), causing him to learn that he is a REM. |
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Title: Re: BROWN EYES AND RED EYES Post by mypalmike on Dec 7th, 2007, 6:05pm Any monk who sees 1 or more REMs already knew the information relayed in the tourist's comment, so it was not news to himself. Any monk who sees 2 or more REMs can deduce that the tourist's comment was not news to any monk (he knows that the REMs he can see can see other REMs). Any monk who sees 3 or more REMs can deduce that the tourist's comment was neither news to any monk, nor would any monk have reason to believe it was news to any other monk. So, we consider the case where there are 4 REMs or more. Each monk would see 3 or more REMs. In this situation, all monks know that the tourist's comment was not news to any monk, and that no monk would have reason to suspect it was news to any monk. They would, logically, disregard the tourist's comment entirely as information previously known to all, and not take any action. This is obviously inconsistent with what I believe to be the perfectly valid induction argument (where N REMs kill themselves after N days). Hence, it appears to me that the problem statement itself comprises an inconsistent logic system (e.g. "The next sentence is true. The previous sentence was false. How many of the previous two sentences is true?") However, depending on how you read the question, the system may be considered consistent if you rigorously interpret the phrase "Having acquired this new information", which, as has been mentioned previously, implies that there was exactly one REM, since in all other cases, the information was not new to any of the monks. There are many logic puzzles that can only be solved using implied constraints such as this. |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Dec 8th, 2007, 6:40am on 12/07/07 at 18:05:39, mypalmike wrote:
I believe your analysis is one level too shallow. Consider the case where there are two monks, each with red eyes. While each of them indeed already knew the tourist's information, neither of them knew for sure that the other monk already knew the tourists information. --SMQ |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on Dec 8th, 2007, 4:49pm Everybody knows the information, but everybody doesn't know that everybody knows it. |
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Title: Re: BROWN EYES AND RED EYES Post by mypalmike on Dec 9th, 2007, 10:48pm SMQ and Grimbal, please actually go back read my post, not just the first sentence. It's annoying that you both obviously responded without doing so. I specifically addressed the issue of "who knows who knows what". Indeed, it forms the entire core of my argument. If I need to restate in shorter terms that you'll actually read: If there are 4 REMs or more, all monks see at least 3 REMs. Any monk who sees 3 or more REMS when hearing the tourist's message can logically deduce that a) he himself already knew it, b) every other monk knew it, and c) every other monk knew that every other monk knew it. Being logical, they each realize that the tourist's comment can not possibly be new information in any sense to any monk. It's a non-signal, and triggers no events on the island. |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on Dec 10th, 2007, 1:36am You are right, I only read the first sentence. But the argument doesn't stop at 4 REMs. The effect of the tourist telling about a REM in front of everybody is that - Everybody knows that there is a REM. - Everybody knows that everybody knows that there is a REM. - Everybody knows that everybody knows that everybody knows that there is a REM. - etc. So every monk learns something he didn't know before. With 1 REM, the REM learns that there is a REM. With 2 REMs, the REMs learns that everybody knows that there is a REM. With 3 REMs, the REMs learns that everybody knows that everybody knows that there is a REM. With 4 REMs, the REMs learns that everybody knows that everybody knows that everybody knows that there is a REM. etc. The non-REMs also learn something. With 1 REM, the non-REMs learn that the REM knows that there is a REM. With 2 REM, the non-REMs learn that the REMs know that everybody knows that there is a REM. etc. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Dec 10th, 2007, 1:47am on 12/09/07 at 22:48:22, mypalmike wrote:
The tourist brings the common knowledge that the monks cannot have developed on their own; because they can't distinguish things at the deepest level. You can reduce the case of N monks back to N-1 hypothetical monks, and for N=1 there is a problem; increasing N doesn't help. |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Dec 10th, 2007, 5:34am on 12/09/07 at 22:48:22, mypalmike wrote:
Excuse me, but I did read your entire post, and after considering it decided to only quote the first line because that's where I saw (what I perceive as) the problem with your analysis starting: all your statements are one level too shallow -- they need to consider one further step of who knows who knows what. on 12/07/07 at 18:05:39, mypalmike wrote:
But he doesn't know who it may have been news to -- at least not right away. After the first night, he knows it wasn't news to anyone (specifically the one REM he sees) and so knows that he, too, must have red eyes. --SMQ |
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Title: Re: BROWN EYES AND RED EYES Post by StallionMang on Jan 2nd, 2008, 4:09pm Here's a little twist on the original that I thought up: Immediately after making his statement, the tourist murders one of the monks in front of all the others. How does this affect the result? Does the eye color of the murdered monk matter? |
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Title: Re: BROWN EYES AND RED EYES Post by StallionMang on Jan 2nd, 2008, 4:34pm Oh, and one other thing, I don't know if anyone posted this yet because I haven't read through all 8 pages of this thread, but here are my answers to the follow-up quesitons asked in the original riddle, as posted on this site. What happens if we change the tourist's statement to each of the following? "There are 10 Brown Eyed Monks" This will result in all of the REMs killing themselves on the first night, as they will all count 10 brown eyes and know that they must not have brown eyes. "There are at lesat two Red Eyed Monks" All REMs will kill themselves on the N-1 night. The reason why it is no longer the N night is because 2 REMs now becomes the base case, rather than 1 REM. "There is an odd number of Red Eyed Monks" All REMs will kill themselves on the first night, because each of them will count an even number of red eyes, and will therefore know that they must be the additional one that makes the number odd. "There is an even number of Red Eyed Monks" Assuming the number of REMs is greater than 0, the result is the same as for odd. "There is more than one Red Eyed Monk" This is the same as him telling them that there are "at least two", which is addressed above. Now keep in mind, I am assuming in my responses that the explorer is always telling the truth when he makes these statements. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jan 3rd, 2008, 1:35am on 01/02/08 at 16:09:17, StallionMang wrote:
What if the tourist says "the number of REMs is a a multiple of 3 or a multiple of 4"? |
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Title: Re: BROWN EYES AND RED EYES Post by StallionMang on Jan 3rd, 2008, 6:52am towr, you got it! In response to your question, hmmmm. 3 REMs : All 3 die on first night 4 REMs : All 4 die on second night 6 REMs : All 6 die on first night 8 REMs : All 8 die on first night 9 REMs : All 9 die on second night Okay, I think I have got it. Where N is a number of REM's that is a multiple of 3 or 4, then: If N-1 is a multiple of 3 or 4, then all N monks will die on second night. If N-1 is not a multiple of 3 or 4, then all N monks will die on first night. |
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Title: Re: BROWN EYES AND RED EYES Post by StallionMang on Jan 3rd, 2008, 6:57am One other thing I forgot to mention. The only thing that could cause a different result than the ones I mentioned above would be if N, N-1, and N-2 were all multiples of 3 or 4. But I don't think there exist 3 consecutive numbers that fit that bill. But I am not a big math guy, so I could be wrong. |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on Jan 3rd, 2008, 7:25am No, you are right. Along the same line, what now if the visitor just says: "The number of REMs is not a prime number". |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jan 3rd, 2008, 7:35am on 01/03/08 at 07:25:00, Grimbal wrote:
I think the number of nights till the suicides will be [hide]N minus the largest prime number smaller than N. If they see a prime number of REMS, they must be one themselves. If they see one more than a prime number, then they consider that if they aren't a REM, the other REMs see a prime number and would kill themselves on day one, and if they don't he must thus be one; etc.[/hide] |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jan 3rd, 2008, 7:48am We can say a similar thing in general. If a set S of numbers is excluded (in the original case {0}), then the REMs kill themselves in night [hide]N - max{x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif S | x < N}[/hide] |
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Title: Re: BROWN EYES AND RED EYES Post by StallionMang on Jan 3rd, 2008, 8:20am Cool. Okay, how about this? The visitor gets greedy and decides that he would prefer to have the entire island to himself as quickly as possible. Think of a statement he can make to get all of the monks on the island to kill themselves as expeditiously as possible. Note: One assumption to be made is that monks kill themselves in the privacy of their own rooms. Either that, or they form a big gathering but in a dark enough environment that they can't see each other's eyes at the time of the suicide... :) |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Jan 3rd, 2008, 8:36am on 01/03/08 at 08:20:19, StallionMang wrote:
If the population consists entriely of REMs he can simply state the true number of REMs. Each will see one fewer REM than stated and immediately conclude he is a REM himself. If the population consists of both REMs and BEMs he can state that there is one more REMs than there truly is. Each BEM will see one fewer REM than stated and immediately conclude (erroniously) he is a REM himself. Each REM will initially know the statement was a lie, but after seeing the BEMs' actions will deduce the true number of REMs. --SMQ |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on Jan 3rd, 2008, 8:45am That was exactly what I was after. I thought it is an enlightening way to look at the problem. |
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Title: Re: BROWN EYES AND RED EYES Post by StallionMang on Jan 3rd, 2008, 8:56am OK. Cool. Now....if the visitor lies, and any of the monks are able to immediately detect that he is lying, they will kill him before he can get off the island. Think of a statement he can make that will allow him the opportunity to get off the island alive before the suicides commence. |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Jan 3rd, 2008, 8:11pm He can state that there are an even number of REMs, if the actual number is odd, or an odd number of REMs, if the actual number is even. No one will be able to tell that the statement is false until they see who didn't make it through the night. |
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Title: Re: BROWN EYES AND RED EYES Post by ima1trkpny on Jan 3rd, 2008, 8:23pm on 01/03/08 at 20:11:23, Icarus wrote:
Icarus is back! ;D (You had us all worried!) ::) |
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Title: Re: BROWN EYES AND RED EYES Post by Aryabhatta on Jan 3rd, 2008, 9:40pm on 01/03/08 at 20:23:22, ima1trkpny wrote:
Welcome back Icky! You did have us worried there... |
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Title: Re: BROWN EYES AND RED EYES Post by JiNbOtAk on Jan 4th, 2008, 2:20am on 01/03/08 at 21:40:35, Aryabhatta wrote:
And guess what ? Your nick name stuck !! :P |
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Title: Re: BROWN EYES AND RED EYES Post by Icarus on Jan 4th, 2008, 5:34pm >:( Maybe I should go away again... I've been under a considerable amount of stress at work this last year, and the time I was spending here wasn't helping, so I decided to stay away for awhile. But things have loosened up a bit, so here I am. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jan 5th, 2008, 11:12am Good to see you back, now we can get all those answers that eluded us the last year ;D |
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Title: Re: BROWN EYES AND RED EYES Post by Ghost Sniper on Jan 5th, 2008, 1:09pm Icarus is back, ;D but what about everyone else that went AWOL, such as Raven, BNC, etc.? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Jan 5th, 2008, 2:00pm on 01/05/08 at 13:09:30, Ghost Sniper wrote:
(* not that people need to get leave to be away from here in the first place ;)) |
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Title: Re: BROWN EYES AND RED EYES Post by Shadow on Oct 24th, 2008, 1:50pm I have just read this entire thread and I think a flawed mental model occurs for several posters in the middle of this thread. I agree that the N REM will commit suicide on the Nth night if a change in the knowledge of all monks occur. I do not agree that a tourist saying "at least one monk has red eyes" is the catalyst that sets this in motion for a monastery that has 3 or more REMs. The premise is that these are logical monks, therefore all monks will realize this statement has added nothing new to their knowledge of how many monks have red eyes. Take monks A,B, and C. A hopes he has brown eyes He knows B and C have red eyes He knows both B and C know at least one monk has red eyes. The same argument holds for B. B hopes he has brown eyes He knows A and C have red eyes He knows both A and C know at least one monk has red eyes. The same argument holds for C. C hopes he has brown eyes He knows A and B have red eyes He knows both A and B know at least one monk has red eyes. ALL BEM know their are at least three REM. Therefore no new is information is given to any of the monks and life goes on as usual. The backward argument is made that this statement will cause all monks to notice how many others have red eyes and commit suicide on the Nth night. However, no change is made in the what they knew. All monks already knew that at least one monk has red eyes AND they also knew that all monks already knew that at least one monk has red eyes. A rewording that would cause all REM monks to commit suicide on the Nth day is for god to declare on day 0 that all REM must commit suicide at midnight. This would set into motion N REM commit suicide on the Nth day. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Oct 24th, 2008, 2:13pm on 10/24/08 at 13:50:20, Shadow wrote:
Quote:
Even if you had only two REMs; obviously they both know there is at least one REM. But after the public announcement, they now know that the other one knows. It told them nothing about how many REMs there are, it told them something different. Quote:
Just take a course in multi-agent modal logic, if you haven't already, and then work it out. (I'd suggest using a model, rather than working it out syntactically, but both approaches are equivalent.) |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Oct 24th, 2008, 2:21pm It seems some of the images I posted where unavailable (due to our faculty's sysadmin screwing around with the server over the past few years). I've modified the posts to make the available again. On the off chance it brings the point across, they're here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?action=display;board=riddles_medium;num=1027806383;start=124#124) and here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?action=display;board=riddles_medium;num=1027806383;start=127#127). |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Oct 24th, 2008, 3:36pm on 10/24/08 at 13:50:20, Shadow wrote:
A knows that B hopes he has brown eyes. A knows that B knows that C has red eyes. Before the announcement: A knows that B knows that C knows that at least zero monks have red eyes. After the announcement: A knows that B knows that C knows that at least one monk has red eyes. A knows that, if A has brown eyes, then B knows that, if B has brown eyes, then C knows that C has red eyes and will suicide overnight. The next day: A knows that, if A has brown eyes, then B knows that B doesn't have brown eyes, so must have red eyes, and will suicide overnight. The next day: A knows that A doesn't have brown eyes, so must have red eyes, and will suicide overnight. The next day, A is dead. Looking at it another way: A knows he's living in one of two possible universes. In one, A, B and C are REMs; in the other, only B and C are. A, being good at logic, knows that, in the universe with only two REMs, they will both suicide on the second night. Why? Because in that universe, B would know that he were living in either a universe where C was the only REM, or one where both of them were, and, being good at logic, would know that, if C were the only REM, he would suicide on the first night. When C doesn't suicide, in this 2-REM universe, B knows that he's the second REM, so B (and C) would suicide on the second night if the three monks were in the universe where A was not an REM. When they don't, A knows that he's living in the universe where all three of them are REMs |
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Title: Re: BROWN EYES AND RED EYES Post by Shadow on Oct 24th, 2008, 9:21pm "Before the announcement: A knows that B knows that C knows that at least zero monks have red eyes. " This is one monastery. A knows that both B and C KNOW that at least one monk has red eyes.( these are not separate monasterys) Therefore the statement above makes no sense to the logical monk who KNOWS that it is not possible that "C knows that at least zero monks have red eyes". Previous posts show that C already knows that at least one (not zero) monks have red eyes. "After the announcement: A knows that B knows that C knows that at least one monk has red eyes. " This last quote is from your post, but this was known before the tourist arrived. What caused anything to change? Aside: TWOR - I agree that that I have ignored the all BEM / REMsituation. Aside: TOWR is very upset that I have questioned his/her opinion. You have not addressed the opinions expressed by me, but have only attacked my right to express them. You are an Uberpuzzler and a moderator and I respect these credentials as a newbie. I contend that you have a mental model that prevents you from seeing any other option than the one your mental model agrees with. I respect your opinion, and would also appreciate if you would respect mine. Look at this from a new beginning and explain where my comments are in error. I am more than willing to admit I am missing something. |
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Title: Re: BROWN EYES AND RED EYES Post by Shadow on Oct 24th, 2008, 11:34pm TOWR This is your quote (Being a newbie I am not sure how quoting from others should be formatted) "Yeah, but the point is that it adds to their knowledge about the knowledge^N of other monks. That's a point so often repeated in this thread it's hard to believe you missed it. Even if you had only two REMs; obviously they both know there is at least one REM. But after the public announcement, they now know that the other one knows. It told them nothing about how many REMs there are, it told them something different." The point is that this holds true only if their are 2 REMs(or less). After there is more than 2 monks with red eyes, this premise becomes invalid. See previous posts. You're comment makes it seem like I am a complete idiot for not seeing things " your way". You keep going back to the N-1 theory but completely ignore that this theory does not hold true when there are more than 3 REMs. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Oct 25th, 2008, 4:03am on 10/24/08 at 21:21:19, Shadow wrote:
You have to follow the chain of knowledge down to at least N levels. Quote:
At some point, especially late at night, you start to wonder if it's all worth it anymore. But you're right that I shouldn't vent that frustration at you. Quote:
All monks know that there is at least one REM, that doesn't change. But the knowledge about monks having knowledge about monks having knowledge about ... etc .. does change. And the simplest way to imagine how that works is to take one REM and have him think "Suppose I have brown eyes, how then would the other monks behave". Which places him in a hypothetical world that after N-1 nights gets disproven, leaving only the other hypothesis, that he does have red eyes. Quote:
Quote:
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Title: Re: BROWN EYES AND RED EYES Post by towr on Oct 25th, 2008, 4:41am on 10/24/08 at 23:34:17, Shadow wrote:
Quote:
If you have two or more monks than the statement "there is at least one REM" does not add tell any monk anything with regard to how many monks with red eyes there are. They all already knew there was at least one. If two monks nonetheless kill themselves, then this is not a reason you can use for objecting larger amounts of monks. Quote:
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Let's take monks A, B and C. They all have red eyes. A can see two possibilities: world A0 where he has brown eyes, and world A1 where he has red eyes. In world A0, monk B (to the best of A's knowledge) may consider himself in world A0B0, where he and A both have brown eyes, or world A0B1, where he has red eyes and A doesn't. In world A0B0 (to the best of A's knowledge of B's knowledge), monk C might hold the hypothesis that he's in world A0B0C0, where none of the monks have brown eyes. This option is eliminated were it to become common knowledge there is at least one monk with red eyes; but at the moment it is possible that A thinks that B think that C thinks this is possible. There is also the hypothesis A0B0C1, where C is the only monk with red eyes. If it were to become common knowledge that there is at least one REM, then to the best of A's knowledge about B's knowledge C would kill himself after one night. Should this not happen, then the B's hypothesis A0B0 has failed, and only A0B1 is left. So, A considers A0B1. In this case, C might once again consider (according to A's knowledge of B's knowledge) that either he's in world A0B1C0, where only B has red eyes. Which means that if it is made common knowledge someone has red eyes, C would think (to the best fo A's knowledge of B's knowledge) that B should kill himself that night, because A thinks that B thinks that C thinks that B can't see any other REM. If this doesn't happen this hypothesis is falsified. The other world A believes it is possible for B to believe that C holds possible is that we're in A0B1C1. So on the second night (A0B1C0 and AB0C1 having been eliminated), because to the best of A's knowledge both B and C would see A has brown yes, they would kill themselves. If this doesn't happen, A can no longer hold this hypothesis, and A0 is eliminated. Which means A can only believe he has red eyes. Now, we have 4 more cases to go; and I'm sick of it already. (And you can get a further 40 cases by changing the orders of A,B,C.) If A sees N-1 REMs, then he believes it possible that B sees N-2, and A believes it is possible that B believes it is possible that C sees N-3, and A believes it is possible that B believes it is possible that C believes it is possible that D sees N-4, etc. Down to the last monk in the chain which is believed to believe there might be none. This last monks is eliminated from the graph of hypotheses when it is made common knowledge one monk has red eyes. |
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Title: Re: BROWN EYES AND RED EYES Post by Shadow on Oct 25th, 2008, 9:51am Quote: The point is that this holds true only if their are 2 REMs(or less). Why? You gave a reason why the process wouldn't apply to 3 or more monks; and that exact same reason also applies to two monks. Yet you concede two monks would work out that they have to commit suicide, so obviously the given reason cannot be correct. If you have two or more monks than the statement "there is at least one REM" does not add tell any monk anything with regard to how many monks with red eyes there are. They all already knew there was at least one. If two monks nonetheless kill themselves, then this is not a reason you can use for objecting larger amounts of monks. " Still can't get the grasp of the quote in top right corner. I highlight the text I want to quote and then hit this Quote link and it takes your entire response into a reply box(not just the the text I had highlighted, but the entire response) The reason it works for 2 monks is that A might (and hopes he has) brown eyes, so he doesn't know that all other monks know that at least 1 monk has red eyes(he hopes monk B sees him with brown eyes). As soon as there are 3 monks with red eyes, all 3 monks with red eyes know that the other 2 with red eyes already know that at least one monk has red eyes. A tourist telling this already known fact should not start the sequence described since this "fact" was already known by all monks. I sense the frustration in your responses. I just do not understand the cascading principal you keep describing. My intent is not to keep you responding as above. Your responses are very intricate and detailed, but rely on the fact that all monks do not realize the fact all other monks do not know that all other monks already know that at least 1 monk has red eyes. My observation is that this is already known by all monks when there are 3 or more. Perhaps we should just agree to disagree? |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on Oct 25th, 2008, 10:44am on 10/25/08 at 09:51:18, Shadow wrote:
Once you have the quoted text included, you can delete the parts you don't want just as you would any other text in your post. Quote:
With two REMs, B knows that A has red eyes, and A knows that B has red eyes, but B doesn't know that A knows B has red eyes. With three REMs, C knows that B and A have red eyes, B knows that C and A have red eyes, and A knows that C and B have red eyes, so C knows that B knows that A has red eyes, and C knows that A knows that B has red eyes, but C also knows that B doesn't know that A knows that B has red eyes, and doesn't know that B knows that A knows that C has red eyes, so, while B knows that A knows there's at least one REM (C), and C knows that A knows that there's at least one REM (B), C doesn't know that B knows that A knows that there's at least one REM |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Oct 25th, 2008, 11:18am on 10/25/08 at 09:51:18, Shadow wrote:
Quote:
All three know there is at least one; all three know that all three know there is at least one; but it's not the case that all three know that all three know that all three know. And for every extra red eyed monk, we'd get another level. |
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Title: Re: BROWN EYES AND RED EYES Post by ooa4oo on Jan 28th, 2009, 2:16am soo.. 0 reds = evil tourist. mass suicide on 1st night (all think they are the red eye monk) 1 red = red eye monk thinks "i don't see any monks w/ red eyes.." suicide on 1st night 2 reds = all red eye monks think "i see one but he didn't kill himself last night. uh oh.." deaths on 2nd night 3 reds = all red eye monks think "i see two, but they didn't kill themselves last night. uh oh.." deaths on 3rd night. etc... effects of tourist's statement: -brings the red eye issue to the front of all monks' minds. -sets a reference in time for all monks to count and figure out how man red eyes there are. -if there is only one red eye monk, he is introducing new info to that one monk. brown eye monks always count one more red eye monk than red eye monks count(except in the case of zero red eye monks), so they wait one day longer before thinking about killing themselves. when the suicides occur the day before.. all the brown eyes can relax. |
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Title: Re: BROWN EYES AND RED EYES Post by SMQ on Jan 28th, 2009, 5:43am That just about covers it, yep. :) --SMQ |
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Title: Re: BROWN EYES AND RED EYES Post by Shadow on Apr 13th, 2009, 2:41pm I think you got through to my thick skull. Still a bit confusing, but I finally do see the light. ( I think) |
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Title: Re: BROWN EYES AND RED EYES Post by noodles on May 6th, 2009, 10:11pm OK the logic is clear. The riddle states that something drastic happens. The most drastic result is the mass suicide. Is it safe to draw a conclusion that the best answer for this riddle is a lying tourist starting a tragic mass suicide? |
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Title: Re: BROWN EYES AND RED EYES Post by Ankitt_sk on Oct 28th, 2009, 1:37am where's the problem statement .. i don't see it in the first page ? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on Oct 28th, 2009, 1:56am on 10/28/09 at 01:37:20, Ankitt_sk wrote:
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 13th, 2013, 2:38pm Cool riddle Got me puzzled for some time, but now im pretty sure that COMMON SOLUTION IS WRONG (meaning n monks die on n-th day) Proper solution in my opinion is 1, 2, 3 red eyed Monks - they die 4 or more red eyed monks - noone dies Hint: [hide]Why 4 red eyed monks expect 3 other to die on 3rd night?[/hide] |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 13th, 2013, 10:14pm Hint: [hide]For the reason why you think three monks would die on the third night if there were three.[/hide] |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 14th, 2013, 8:46am This reason doesn't hold for 4 monks I guess noone will even try to think about this riddle again so i'll just put an explanation here --- EXPLANATION: Something will happen only if any monk might think that some monk sees just one red eyed monk and will expect him to react on new knowledge Otherwise no monk would ever think that any monk might think that someone will suicide on the first night. --- 3 and 4 CASES from point of view of red eyed: 3 case - I see 2 red eyed monks and if he have brown eyes, they would die on 2nd night. WE DON'T STOP OUR REASONING HERE. Why I think these 2 would die? Because each of them would see just 1 in that case. Therefore each of them would expect this other guy to die. We can deduce from this point. 4 case - i see 3 red eyed monks, if i have brown eyes they would die on third day. AGAIN WE DON'T STOP HERE. Why I think these 3 would die? Because each of them would see just 2 in that case and expects them to die on 2nd day. AGAIN WHY? Because each of these 2 would see just 1 in that case. WE STOP HERE. That's impossible. I know they all see at least 2. Noone will ever die. --- IN OTHER WORDS AGAIN With 4 red eyed monks it's not possible to make assumption that anyone might think that anyone might think that there is just one red eyed monk. We can paste this phrase more times it doesn't change the fact. With 4 red eyed monks it's not possible to make assumption that anyone might think that anyone might think that anyone might think that anyone might think that anyone might think that anyone might think that there is just one red eyed monk. Every monk knows that every monk knows that every monk sees at least 2 red eyed monks. Every reasoning based on the assumption that anyone might think that anyone might think that someone will die on first night is not valid. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 14th, 2013, 9:04am The thing you fail to understand is that for the three monks that the four monks hypothesize, two monks can be hypothesized that hypothesize there is only one. And each day one of these hypotheses fails. If I'm the fourth monk, than I imagine that if there were only three, they would kill themselves on the third day; because if there were only three monks, and I was the third, I'd imagine that if there were only two monks, they'd kill themselves on the second day and so if they didn't I'd kill myself along with them on the third. So when no one kills themselves on the third day, I know I must be a fourth red-eyed-monk, because I can only see the three that would have killed themselves if there wasn't a fourth. My master's thesis was on modal logic, trust me, the canon solution is correct. Just be sure to hypothesize a separate world with n-1 REMs to figure out what the nth REM would conclude. I think that somewhere in this thread there's an illustration for up to 5 REMs. (Or in a similar thread; since this puzzle comes in many guises.) |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 14th, 2013, 9:12am Another consideration: If there are 4 REMs every non-REM knows everyone can see at least 3 REMs every REM knows only everyone can see at least 2 REM every non-REM knows that everyone knows that everyone can see at least 2 REMs every REM knows only that everyone knows that everyone can see at least 1 REM every non-REM knows that everyone knows that everyone knows that everyone can see at least 1 REM But no REM knows that, so it is NOT common knowledge. No non-REM knows that everyone knows that everyone knows that everyone knows that everyone can see at least 1 REM That's why it matters when it is announced, because that makes it common knowledge, in the sense that only after that moment everyone knows that everyone knows that everyone knows that everyone knows that everyone knows ... that there is at least 1 REM, which starts the inevitable fatal cascade. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 14th, 2013, 10:30am You know, the thing I find really weird, is that at the same time you say that if there are three red eyed monks, they'll kill themselves on the third night. But if I see three red-eyed monks and they're alive after the third night, you don't think it's a valid conclusion that I must be a fourth. That's not consistent. It really is this simple: * If I don't see any REM, even though I know there must be one, then it must be me, and I kill myself. >> Therefore if there is one REM, he kills himself on the first night. * If I see one REM, and he doesn't kill himself on the first night, even though I know that if there is only one REM he will kill himself on the first night, then I must be a second REM, and thus I kill myself on the second night (along with the other). >> Therefore if there are two REMs they all kill themselves on the second night. * If I see two REMs, and they don't kill themselves on the second night, even though I know that if there are only two REMs they will kill themselves on the second night, then I must be a third REM, and thus I kill myself on the third night (along with the others). >> Therefore if there are three REMs they all kill themselves on the third night. * If I see three REMs, and they don't kill themselves on the third night, even though I know that if there are only three REMs they will kill themselves on the third night, then I must be a fourth REM, and thus I kill myself on the fourth night (along with the others). >> Therefore if there are four REMs they all kill themselves on the fourth night. * If I see N-1 REMs, and they don't kill themselves on the N-1st night, even though I know that if there are only N-1 REMs they will kill themselves on the N-1th night, then I must be an Nth REM, and thus I kill myself on the Nth night (along with the others). >> Therefore if there are N REMs they all kill themselves on the Nth night. Simple induction is the easiest proof. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 15th, 2013, 6:58am on 05/14/13 at 08:46:09, riddler358 wrote:
But what about the 3 case from the brown-eyed viewpoint? Red eyes - I see 2 red eyed monks and if I have brown eyes, they would die on 2nd night. WE DON'T STOP OUR REASONING HERE. Why I think these 2 would die? Because each of them would see just 1 in that case. Therefore each of them would expect this other guy to die. We can deduce from this point. Brown eyes - i see 3 red eyed monks, if i have brown eyes they would die on third day. AGAIN WE DON'T STOP HERE. Why I think these 3 would die? Because each of them would see just 2 in that case and expects them to die on 2nd day. AGAIN WHY? Because each of these 2 would see just 1 in that case. WE STOP HERE. That's impossible. I know they all see at least 2. Noone will ever die. So in the 3 REM case, the brown-eyed monks have "proved" that no-one will die. That's a problem because we've just got done proving that the three REMs will die, so something's wrong - either the three REMs won't die because the BEMs are right, or the REMs are right and will die. Either way, the logic of the other group must be wrong, but we've assumed that the monks are all (and know each other to be) expert logicians, so any conclusions reached by any monk must be correct - the monks have different information to start with, but that just means that their correct conclusions don't have to be identical - they each have to include the true outcome, so they still have to overlap. In the 3 case, each REM concludes: Either those two die on the second night, or those two and myself die on the third night. In the usual solution, each BEM concludes: Either those three die on the third night, or those three and myself die on the fourth night. In your solution, each BEM concludes: No-one dies. The usual solution provides an overlap between the two conclusions - that all three REMs die; your solution does not. That doesn't prove that the usual solution is correct, but it does prove that your solution is wrong. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 16th, 2013, 9:55am on 05/14/13 at 10:30:36, towr wrote:
It makes it obvious after reading this. Thanks. It's still a bit confusing for me because in cases 4 and more evey monk knows that every other monk knows that noone will die on first day. And still they use this possibility to contradict their nested hypothesis. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 16th, 2013, 1:21pm Well, look at it this way; if you see three red-eyed monks, you're in one of two possible worlds: either there are 4 red-eyed monks, and you're one of them, or there are three red-eyed monks and you're not one of them. For every red-eyed monk in the first possible world, the options they can consider are exactly the same; they would see what you see think what you think. But in the second case, the red-eyed monks are in a different situation; so not everything you know and see and think necessarily applies to them. They would consider either a world where they are one of three red-eyed monks, or one where they're not one of two red-eyed monks. Basically, the same knowledge does not automatically apply to the world you're in and the worlds you can imagine someone else might think to be in. The three REMs consider there might be only two REM, who imagine there might be only one REM, who might have imagined there wasn't one until some bloody tourist ruined his ignorance. Each shift in perspective and into a deeper nested possible-world can change what knowledge is available. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 18th, 2013, 4:53am I know it works like you said, still on 05/16/13 at 13:21:40, towr wrote:
for me that's the confusing part, because on the other hand everyone knows that everyone knows that this is only imaginary situation (for cases 4 and more) |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on May 18th, 2013, 1:01pm Sorry, I come in a bit late, but maybe here is a less confusing way to explain the solution. Imagine there are N islands, numbered from 1 to N and on island n there are n REMs. But nobody knows on which island they are, except that if they see k REMs, they can tell they are on island k or k+1. As before, the tourist goes to each island and tells there is at least one REM. The bastard. On the first night, the REM on island 1 commits suicide. Now everybody on island 2 knows they are not on island 1. On the second night, the 2 REMs on island 2, seeing only 1 REM and knowing they are not on island 1, conclude they are a REM and commit suicide. Now, everybody on island 3 knows they are not on island 2. On the third night, the REMs on island 3, seeing only 2 REMs and knowing they are not on island 2, conclude they are a REM and commit suicide. Now, everybody on island 4 knows they are not on island 3. Etc. This makes clear that the hypothetical case you reason is about another world that can have a number of monks that contradicts what you observe. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 19th, 2013, 4:04am Yeah but monks on island 4 all knew at the very beginning that everyone on island 4 knows that on island 1 and 2 noone will die, that's the confusing part my initial mistake was by this reasoning - why would 4 monks bother at all what would happen to 1 red eyed monk if they all know that everyone knows it's not possible there is only one and now the best explanation i got for ppl who think the same is - ok, 4 monks might bother only about 3 monks case, but these 3 monks (imagined by 4 monks) bother about 2 monks case, and these 2 monks (imagined by 3 monks that are imagined by 4 monks) they bother about 1 monk case, therefore 4 monks are indirectly bothered by 1 monk case. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 20th, 2013, 10:44am The "simplest" explanation is probably the superficial: "A monk who sees 3 REMs knows that either those 3 will die on the third night, or he's the fourth REM". Provided you can prove that, when there are only 3 REMs, they will die on the third night, the above explanation is mathematically rigorous (okay, you also have to assume that all the monks are at least as good at logic as you are, and that it's common knowledge among the monks that that is the case - but you need that assumption to get anywhere anyway - if the monks don't trust each other's ability to follow the logic, then with more than a couple of REMs, everyone lives anyway) That simple proof also generalises to an inductive proof that, when there are n REMs, they will all die on the nth night. You can also argue directly that 4 REMs each hope that the other 3 only see 2 REMs, so hope those 2 each only see 1 REM (despite the original REM knowing that that couldn't be true) and the hypothetical 2 monks would each hope that the other saw no REMs and would face suicide that night, only to be disappointed, and suicide the second night, but the 3 monks would be disappointed and suicide the third night, but the original REM is, himself, doomed to further disappointment when the other 3 don't suicide, so kills himself on the fourth night. On the other hand, any explanation like that is hard to follow, so people tend to find it unconvincing - which is fair enough - a long and confusing argument is more likely to slip in a subtle flaw without it being noticed, so, all else being equal, should be less convincing than one you can grasp in its entirety... |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on May 23rd, 2013, 4:56am There still is a problem: If there is only one REM, he will reason that if he doesn't commit suicide, then everybody else will commit suicide on the next night. So there will be nobody to blame him for his misconduct. With 2 REMs, after the first night, each REM will consider the possibility that the other one is bluffing. They will wait another day just to be sure. And so on... |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 24th, 2013, 3:41am on 05/23/13 at 04:56:43, Grimbal wrote:
Yeah, as soon as the monks no longer trust each other to follow the rules, the deductions based on other monks following the rules break down. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 6th, 2016, 11:30pm This one haunts me I keep thinking so called cannon solution breaks, I think it breaks at 3 REM case For 3 REM imagine his point of view, he expects 2 to die only because each of them expects 1 to die, and therefore if 2 don't die he concludes he have red eyes but that breaks his deduction because if he have red eyes his deduction was incorrect I came up with an example (maybe not the best) We wonder if biggest natural number N is odd or even. We assume it's odd. But we can make N+1 and it would be even. Then we can make N+2 and this would be odd. And so on. And we can go forever, but our base assuption is broken, N is no longer biggest natural number, so how can we base anything that follows on this? EDIT: another example (maybe better) Willy Wutang goes on a family road trip with their camper, he likes to be prepared and have 2 spare tires, his wife sometimes gets 1 more spare tire, sometimes not, so Willy have either 2 or 3 spare tires, during first stop their kid says - dad, we have at least one spare tire, should willy panic and buy one more spare tire? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 7th, 2016, 12:01am Try drawing out the graph of possible worlds. The place people always go wrong is looking at the various what-if scenarios from the wrong perspective. When reasoning about what what a monk in a 3-REM world thinks a monk in a 2-REM world would think, you _shouldn't_ use the 3-REM world monks perspective, but the 2-REM worlds monk, because that's whose thinking you're considering. The problem isn't with the canon solution. The solution is a mathematical fact; you can prove it from the axioms of modal logic. So the only problem is reconciling it with the way you think about the puzzle. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 7th, 2016, 12:36am i think i'm trying to do that, but 2 REM perspective makes it possible and proves it was all along 3 REM perspective where it's impossible EDIT: for me it's like this: 1 REM is 1D world, 2 REM is 2D world, 3 REM is 3D world, one might jump from 1D to 2D in both ways, one might jump from 2D to 3D both ways, but one jump from 1D to 3D seems impossible to me, sure you can make 2 jumps, but is that the question? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 7th, 2016, 2:39am on 05/07/16 at 00:36:59, riddler358 wrote:
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 7th, 2016, 5:43am on 05/07/16 at 02:39:00, towr wrote:
But if it turns out to be 3REM case then every one of them in the first place saw 2REM and knew each one of them saw at least 1REM, so no new information arises, and noone was to think 0REM case to begin with. I see it like this |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 7th, 2016, 9:20am on 05/07/16 at 05:43:22, riddler358 wrote:
So, Alex, Bob and Charlie are the 3 REMs: Charlie looks at Alex and Bob and thinks "I can see two REMs, so we're either in 3REM or 2REM world. If we are in 2REM world, then Bob would be looking at Alex and thinking 'I can see one REM, so we're either in 2REM world or 1REM world. If we are in 1REM world, then Alex has just found out that we aren't in 0REM world and will suicide tonight. If Alex is still around tomorrow, then we must be in 2REM world.' so Bob (and Alex) will suicide tomorrow night. If they don't, then we must be in 3REM world." The key is that in the 2REM world Charlie thinks is possible, Charlie does not have red eyes. So those versions of Alex and Bob both only see 1 REM so are deciding between 2REM and 1REM worlds. The 2REM Bob is not the same person as the actual 3REM Bob |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 7th, 2016, 12:47pm on 05/07/16 at 05:43:22, riddler358 wrote:
You're also reasoning like the people in the 2REM world have knowledge from the 3REM world, but they don't. Maybe the model is to simplistic. So, instead consider the following starting model (attached). Each world is exactly characterized by which of Alex, Bob and Charlie is a red-eyed-monk. And the links labeled A,B,C are what they respectively consider possible. Before the announcement all relations are symmetrical: they don't know whether they have red eyes or not, so in the world where they have red eyes, they consider it possible they don't and vice versa. But the eyes of their fellow monks are known to them, so those are the same in both worlds linked for a given monk. The announcement breaks that symmetry and unravels the graph like before. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 8th, 2016, 12:55am i read and tried to undarstand your point of view, but it's even less convincing for me now, you guys are most likely right, but i'll try to explain my point of view once again - let's agree that BEM have zero influence on the situation, only REM matter - from monk point of view there are 2 possible worlds in every situation: one is real and one is imaginary (we don't know which one is real) they differ only by 1 REM - we all agree that in 2 REM case they die - crucial case is 3 REM -> it splits into: -> imaginary 2 REM and 1 BEM which is exactly the same as 2 REM we can remove ourselves from this case they die we agreed on that -> real case 3 REM in which evey REM sees 2 REM he can imagine 2 REM case but he cannot hyphotesise 1 REM case, none of 3 REM can do that, it is not possible, so my conclusion here is => every monk sees 2 REM and he can't hyphotesise himself world with 1 REM, he might think such world exists in one of other monks head, but it doesn't, maybe that's the new information i'm trying to grasp, lack of imaginary world with 1 REM that reveals for him after 2 nights it's still very confusing, somehow i cannot imagine 100 red eyed monks sitting there, and each one of them thinking yeah in their heads they think there might be only 1 red eyed monk that will suicide this night EDIT: one other example that might be related i insult towr => he tells grimbal i insulted him => grimbal acts => i insult tower for telling grimbal => he tells grimbal i insulted him => grimbal acts ... and so on if grimbal tells me to apologize for each insult then we can roll this back i apologize => towr accepts => grimbal asks me to apologize for previous insult => i apologize ... and so on to the first one but if grimbal blocks my account we cannot roll this back unless i create new account so above implication is perfectly correct but if grimbal acts blocking my account it breaks the loop and in order to insult towr again i need to create new account (new universe), we cannot go back apologizing for each insult, because i have no access to account from which i insulted |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 8th, 2016, 1:18am one more take on 3 REM case I'm sitting there thinking: if i have brown eyes then those two guys can imagine 1 REM world then after 2 nights: whoaa, they didn't suicide, therefor i have red eyes, but this means, they couldn't ever imagined 1 REM world in the first place |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 8th, 2016, 6:11am on 05/08/16 at 01:18:16, riddler358 wrote:
That's right - and the reason they couldn't have imagined 1REM world is the same as the reason why they survived the second night - that you have red eyes. If it helps, you could think of the process as trying to figure out whether they think 1REM world is possible. Also, the same issue arises when there are only 2 REMs to start with - they each look at the other and wonder whether they think 0REM is possible. When the other guy survives the first night, that means they couldn't ever have imagined 0REM world in the first place, so the original announcement didn't tell them anything. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 8th, 2016, 11:50pm yet another take on this riddle I really like analogy to n-dimensional world, let us make another riddle based on that: - we all live in 3D world or in 2D world (virtual reality), we cannot tell which one - every person who will come to knowledge in which world he lives in needs to commit suicide and this should be announced in the morning to everyone - some alien comes and states -> your world have at least one dimension - what will happen? in 3 REM world we all know it's either 2 REM or 3 REM, and possibility of imagining 1 REM world doesn't change that it is 2 REM world or 3 REM world, why should one act upon impossible imaginary possibility? |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 9th, 2016, 7:45am on 05/08/16 at 23:50:31, riddler358 wrote:
If it were actually a 2REM world, then 1REM world would be possible, which is why events there get taken into account indirectly. The n-dimensional world analogy doesn't quite work because removing me from the equation doesn't turn the world into a 2D one that the inhabitants think could be 2D or 1D. Let's try another approach: Rather than starting by saying there are definitely a specific number of REMs present, which is information none of the monks has, let's start with what one of the monks does know: suppose you're a monk and you see exactly 2 REMs. What happens then? |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 9th, 2016, 10:35am Huh, I thought we were done already. How about a different approach all together. Base case (N=1): If there is 1 REM, then he will kill himself at the end of the first day. Induction hypothesis (what we want to prove for all N): If there are N REMs, these kill themselves at the end of the Nth day. Induction step (assuming it's true for all 1<=k<=N, then is it true for N+1?): If there are N+1 REMs, then by the induction hypothesis, after the Nth day we know there were not N or fewer REMs, because they would have killed themselves. Therefore, on the N+1th day, the N+1 REMs * know the number of REMs is not any number from 1 to N (inclusive), * know the number of REMs is not 0, because it was announced there's at least one and also they can see N REMs * will realize that because they see only N REMs, they must be the N+1th one and thus they will kill themselves at the end of the N+1th day. Therefore by induction, from the base case and the induction step, for all N >=1 if there are N REMs, they kill themselves at the end of the Nth day. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 9th, 2016, 12:16pm on 05/09/16 at 07:45:28, rmsgrey wrote:
as i mentioned before there are 2 cases: 2 REM case -> both die because each of them can imagine that other one is imagining 0 REM case 3 REM case -> noone can imagine that any of other 2 are imagining 0 REM case to begin with |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 9th, 2016, 12:58pm on 05/09/16 at 12:16:41, riddler358 wrote:
If I'm the third REM, then I'm looking at the other two and thinking, huh, I'm sure hoping they can imagine someone considering the 0 REM case, otherwise I'm screwed. Because that would be what's going on if it were the 2REM case. The only thing I know is that either of them can see at least one person with red eyes (and I hope it's only one). But if they only see one person with red eyes, then the person they see would be considering there might be no REMs. Even though I know that that's not the case, I don't know whether they know that. The only way they would definitely not consider that is if I have red eyes, but since I don't know whether I have red eyes, I don't know whether they can or can't imagine anyone considering a 0 REM world. (Until after two days they're still alive.) |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 9th, 2016, 1:12pm @towr i think induction is flawed in this case, because you go ahead and say (N=1) => (N=2) => (N=3) but if (N=3) then (N=1) is impossible i say i agree on (N=1) => (N=2) and i agree on (N=2) => (N=3) but connection between (N=1) and (N=3) is lost because your base case says "If there is 1 REM" i think i have great example for this one let's imagine a world where people live short lives and noone ever met their grandparents in such world my grandfather can talk to my father and my father can talk to me, but that doesn't mean my grandfather can talk to me, link is lost, but my father who knew both of us can imagine us talking with each other very vividly PS just a thought - how about proving by induction that sum 1 + 2 + 3 + ... is a positive number? |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 9th, 2016, 1:25pm on 05/09/16 at 12:58:25, towr wrote:
i think that the fact that you don't know if they can consider something doesn't mean they can consider it let's try different approach, convince me that in 100 REM case anyone thinks that anyone can imagine 0 REM world. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 9th, 2016, 10:28pm on 05/09/16 at 13:12:14, riddler358 wrote:
Sure, if there are 3 REMs, then there is not just 1 REM. 3 is not 1. We can all agree on that. And 3 is not 2 either. But that's not the issue. If there is one single REM, then he would kill himself on the first night. If N REMs kill themselves on the Nth night, then logically, N+1 REMs kill themselves on the N+1th night. Therefore by the rule for induction, it's true for all N. There's just no wiggle room. Quote:
What you're suggesting is more equivalent to saying you haven't got a (by now possibly deceased) great-grandfather because you can't image talking to him. But whether you can talk to him is irrelevant. Your father's father could talk to him, therefore once upon a time he existed. In this case each world with N+1 REMs can imagine (talk to) the N REM world. And therefore the 0 REM world has influence. Not because there is any direct link from N+1 REMs to 0 REMs, anymore than you can talk to your oldest ancestor. Quote:
But 1 + .. + N is a positive natural number for any positive natural number N * 1 is a positive natural number * if 1 + .. + N is a positive natural number, then 1 + .. + N + (N+1) is a positive natural number (because a positive natural number plus another positive natural number is always a positive natural number) * therefore by induction, for any positive natural number N: 1 + .. + N is a positive natural number (infinity is not a positive natural number, so it doesn't work for 1 + 2 + 3 + ... where ... means to go on forever) |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 9th, 2016, 10:34pm on 05/09/16 at 13:25:56, riddler358 wrote:
Quote:
No one needs to think that someone can imagine a 0 REM world, what they need to consider is whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider whether anyone might consider a 0 REM world. Which is an entirely different proposition. Because we're so deep into hypothetical worlds that the final person doesn't need to exist in the real world (or in fact the second), anymore than that your hundredth ancestor needs to still be alive for your 99th ancestor to have talked to him. |
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Title: Re: BROWN EYES AND RED EYES Post by Grimbal on May 10th, 2016, 3:37am If you buy something in a shop and say you forgot something and need to return in the same shop. You are still carrying your purchase in your hands. You migh be worried that when leaving again a shop employee could think you are stealing the item. You know you woud never ever steal anything. But you still consider yourself in the role of a potential thief. |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 10th, 2016, 8:16am on 05/10/16 at 03:37:52, Grimbal wrote:
but if someone else sees entire situation (let's say person A) he doesn't think that you are a thief, moreover person B sees person A and entire situation he doesn't think person A thinks you are a thief unless he thinks person A doesn't see entire situation what i fail to understand is why going every hypothetical step further we omit bit of information that is an obvious observational fact and go imagine universe that is not possible and base our conclusions on that. for me it's like this in implication terms p => q and for p we substitute obvious false sentence (if there are 0 REM) now for q we substitute whatever and it is always truth you can say N REM die on Nth night and i can say they don't from my point of view it's only possible for someone to think p is possible to imagine for someone else in 1 REM and 2 REM case tourist might have as well said: at least one of you have red eyes and thinks earth is flat. |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 10th, 2016, 10:25am on 05/10/16 at 08:16:15, riddler358 wrote:
It's an obvious observation fact that your tenth generation paternal ancestor cannot speak. Because he's dead. But all your ancestors up to 70 thousand years or so ago quite probably talked to the one before. We can imagine your 9th ancestor speaking to your tenth ancestor, even though we can't imagine either of them speaking now. Observations about the here and now don't necessarily apply to the situations we imagine. When you see 2 REMs, and you try to imagine what would happen if they were the only two, then you really need to commit to the idea of the 2 REM world, look at it entirely from their viewpoint, and not from your own. Suppose I can see that Alex and Bob have red eyes. If I were Bob, I would (in the 2REM case) only see that Alex had red eyes. If I were Bob imagining to be Alex, then if my imaginary Bob was considering the 1 REM case, he can imagine Alex seeing no REMs. Because I'm imagining that I don't have red eyes, and I'm imagining that Bob is imagining he doesn't have red eyes. Each step adds another red eyed monk imagining that he doesn't have red eyes. That goes on until there's none left. And then the announcement happens and some very highly hypothetical down-the-line imaginary red eyed monk's illusions get shattered. If I don't know that I have red eyes, and Bob doesn't know that he has red eyes, then why would I think that Bob must somehow know that Alex can see someone with red eyes? If I think that Bob knows for certain that Alex sees someone with red eyes, then Bob must be thinking that Alex sees either Bob or me as a REM. So either I think that Bob does know that he's a REM, or I know that I am. Which is a contradiction. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 10th, 2016, 12:33pm on 05/09/16 at 12:16:41, riddler358 wrote:
Can you ever figure out which case you're in? |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 10th, 2016, 1:30pm on 05/10/16 at 12:33:25, rmsgrey wrote:
you can't but you know for sure it's one of just 2 cases N (number of REM you see) or N+1, moreover everyone else is either in same situation as you or in situation N or N-1, for everyone N-2 is impossible and everyone knows that @towr i need some more time to understand your last post |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 10th, 2016, 10:52pm on 05/10/16 at 13:30:14, riddler358 wrote:
Quote:
Suppose Alex in secret tells Bob something, and Alex tells Charlie the same thing also in secret. Bob doesn't know that Alex told Charlie; Charlie doesn't know Alex told Bob. Everyone knows the secret, but Bob and Charlie do not know that they both know it. So everyone knowing X does not mean everyone knows that everyone knows X. And that's important to realize in this puzzle. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 11th, 2016, 7:32am on 05/10/16 at 13:30:14, riddler358 wrote:
If you don't see any red eyes, I assume you can figure out that you're in the 1REM world when the starting announcement is made. If you see 1 REM, what happens if you're in a 1REM world? If that doesn't happen, can you then conclude that you're in a 2REM world? If you know what would happen in a 2REM world, and that doesn't happen when you see 2 REMs, can you then conclude that you're in a 3REM world? |
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Title: Re: BROWN EYES AND RED EYES Post by riddler358 on May 11th, 2016, 2:19pm i found some paradox that might seem related http://plato.stanford.edu/entries/sorites-paradox/ it might be called eubilides' heap paraphrasing if someone is fat then guy that weights 1 kilo less is also fat, and if someone is thin then guy that weights 1 kilo more is also thin monks do official weighting before breakfast tourist comes and says at least one of you is fat imagine there are many monks and all possible weights of monks are present there in the monastery but another tourist comes and says at least one of you is thin by these assumptions every monk would be fat and thin at the same time we build up to something comparing one-to-one but don't bother looking at the bigger picture |
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Title: Re: BROWN EYES AND RED EYES Post by towr on May 11th, 2016, 10:24pm Having red eyes or not is a binary property, and not an ambiguous judgement like fat/slim. |
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Title: Re: BROWN EYES AND RED EYES Post by rmsgrey on May 12th, 2016, 10:09am on 05/11/16 at 22:24:21, towr wrote:
Yeah, the sorites paradox rests on trying to be precise about something vague - at the ends of the process, you have something that's definitely a heap, or definitely not a heap, but in between you have things that might or might not be heaps - even if you ask the same person about the same ?heap? they might give different answers depending on the context. With red eyes, either someone has red eyes or not - there are no awkward edge cases, no eyes where some people say they're red and others say they're blue - so there's no sorites-style ambiguity to grab on to. |
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