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Topic: BROWN EYES AND RED EYES (Read 74752 times) 

S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #50 on: Aug 16^{th}, 2002, 12:10pm » 
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OK, sure. I guess I was just saying that they don't need to be explicitly told that all the nthorder 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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loucura
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Re: BROWN EYES AND RED EYES
« Reply #51 on: Aug 22^{nd}, 2002, 1:47pm » 
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The monks kill the tourist. Reasoning: If there are N+1 RedEyed 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 RedEyed Monks, (which is really N+1, since the hypothetical 'selfREM' doesn't know it's an REM), the only answer we can assume is that the monks kill the tourist.


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AlexH
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Re: BROWN EYES AND RED EYES
« Reply #52 on: Aug 22^{nd}, 2002, 7:23pm » 
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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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James
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Re: BROWN EYES AND RED EYE
« Reply #53 on: Sep 4^{th}, 2002, 5:56pm » 
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(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 Aug 15^{th}, 2002, 2:23pm, S. Owen wrote: I understand and agree with your point about new information in the case of more than 2 redeyed monks. For more than 2 redeyed 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 redeyed monks. Each of them doesn't know whether there are a total of 3 or 4 redeyed monks. Each is wondering whether the other 3 are the only redeyed monks. And thus, they wonder whether those 3 are themselves wondering whether there are 2 or 3 redeyed 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. 
 What you seem to be saying here is that if there were (say) 100 monks with redeyes 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 Aug 15^{th}, 2002, 2:23pm, S. Owen wrote: 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? 
 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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S. Owen
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Re: BROWN EYES AND RED EYE
« Reply #54 on: Sep 4^{th}, 2002, 9:46pm » 
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on Sep 4^{th}, 2002, 5:56pm, James wrote:What you seem to be saying here is that if there were (say) 100 monks with redeyes then they eventually imagine that there might be monks that could only see 1 REM? 
 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 Sep 4^{th}, 2002, 5:56pm, James 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. 
 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 N1 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 subargument 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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James
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Re: BROWN EYES AND RED EYE
« Reply #55 on: Sep 5^{th}, 2002, 12:06am » 
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on Sep 4^{th}, 2002, 9:46pm, S. Owen 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... 
 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 Sep 4^{th}, 2002, 9:46pm, S. Owen wrote: ... do you think this argument is flawed? 
 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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S. Owen
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Re: BROWN EYES AND RED EYE
« Reply #56 on: Sep 5^{th}, 2002, 7:42am » 
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on Sep 5^{th}, 2002, 12:06am, James wrote: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? 
 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 Sep 5^{th}, 2002, 12:06am, James wrote: 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? 
 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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James
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Re: BROWN EYES AND RED EYE
« Reply #57 on: Sep 5^{th}, 2002, 4:57pm » 
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on Sep 5^{th}, 2002, 7:42am, S. Owen 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. 
 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 sidequestions are "can nonexistent people reason?", "is their reasoning the same as ours?", "are the nonexistent people they imagine more or less imaginary than the nonexistent people that we (the existent) imagine?" Quote: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. 
 Except that there is still a straight rush to accept a "each step is logical, whole process is logical" answer in this case. Quote:So can you spot a flaw in the reasoning along the lines you have proposed? I believe that's the useful question. 
 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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S. Owen
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Re: BROWN EYES AND RED EYE
« Reply #58 on: Sep 5^{th}, 2002, 7:38pm » 
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on Sep 5^{th}, 2002, 4:57pm, James 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? 
 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 pretourist reasoning, so I don't see where anything less than 98 REMs would fit into their reasoning. on Sep 5^{th}, 2002, 4:57pm, James wrote:Your answer the predicated on the assumed reasoning of people you know don't exist. So, an interesting sidequestions are "can nonexistent people reason?" ... 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. ... This question is similar. Maybe your initial assumption "this question has an answer" is in fact wrong. 
 This sounds like portentous handwaving... 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 Sep 5^{th}, 2002, 4:57pm, James wrote: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. 
 Explain this one?


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James
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Re: BROWN EYES AND RED EYE
« Reply #59 on: Sep 5^{th}, 2002, 7:55pm » 
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on Sep 5^{th}, 2002, 7:38pm, S. Owen wrote: No, again, this is not the reasoning. 
 Fair enough, I must admit, this is the counterargument to the "topdown" justification of the answer. Here is another possible "flaw" in the "bottomup" 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 N1 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 N2 REMs and 1b) There are N1 REMs. Therefore, our list of assumptions has now expanded to 3. 1) The answer is true for N2 REMs 2) There are exactly N REMs (no more, no fewer) and 3) There are exactly N1 REMs. This is the subtle flaw.


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AlexH
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Re: BROWN EYES AND RED EYES
« Reply #60 on: Sep 5^{th}, 2002, 8:26pm » 
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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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James
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Re: BROWN EYES AND RED EYES
« Reply #61 on: Sep 5^{th}, 2002, 8:43pm » 
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on Sep 5^{th}, 2002, 8:26pm, AlexH wrote: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. 
 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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James
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Re: BROWN EYES AND RED EYES
« Reply #62 on: Sep 5^{th}, 2002, 8:49pm » 
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on Sep 5^{th}, 2002, 8:26pm, AlexH wrote:It may be subtle, but its not a flaw. We can talk about and prove things about hypotheticals without knowing that they match reality. 
 But, can we talk about and prove things about hypotheticals knowing that they don't match reality?


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AlexH
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Re: BROWN EYES AND RED EYES
« Reply #63 on: Sep 6^{th}, 2002, 12:01am » 
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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: 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 N1 REMs 2) There are N REMs  if there aren't then we know nothing happens. 
 We never make assumption 2. We assume "If there are N1 REMs then they will suicide on night N1" and use that to prove "If there are N REMs then they will suicide on night N".


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Jonathan_the_Red
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Re: BROWN EYES AND RED EYES
« Reply #64 on: Sep 6^{th}, 2002, 1:18pm » 
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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.

« Last Edit: Sep 6^{th}, 2002, 1:19pm by Jonathan_the_Red » 
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James
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Re: BROWN EYES AND RED EYES
« Reply #65 on: Sep 9^{th}, 2002, 3:30pm » 
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on Sep 6^{th}, 2002, 1:18pm, Jonathan_the_Red wrote: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. 
 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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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #66 on: Sep 9^{th}, 2002, 3:52pm » 
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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.

« Last Edit: Sep 9^{th}, 2002, 3:56pm by S. Owen » 
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DeeK
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One assumption may be incorrect
« Reply #67 on: Sep 9^{th}, 2002, 8:15pm » 
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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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Jonathan_the_Red
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Re: BROWN EYES AND RED EYES
« Reply #68 on: Sep 10^{th}, 2002, 11:55am » 
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on Sep 9^{th}, 2002, 3:30pm, James 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. 
 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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Carl_Cox
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Re: One assumption may be incorrect
« Reply #69 on: Sep 16^{th}, 2002, 4:45pm » 
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on Sep 9^{th}, 2002, 8:15pm, DeeK wrote: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. 
 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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James
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Re: BROWN EYES AND RED EYES
« Reply #70 on: Oct 7^{th}, 2002, 10:25pm » 
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on Sep 10^{th}, 2002, 11:55am, Jonathan_the_Red wrote: 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. 
 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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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #71 on: Oct 8^{th}, 2002, 7:33am » 
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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 K_{n} be the statement: "There are n REMs and they kill themselves on night n." Let I_{n} be the statement: "If there are n REMs, then they kill themselves on night n." You are right that it is not possible for K_{n} to be true for more than one value of n. But it is quite possible for I_{n} to be true for all n, in fact it is!


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James
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Re: BROWN EYES AND RED EYES
« Reply #72 on: Oct 9^{th}, 2002, 4:57pm » 
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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 Sep 10^{th}, 2002, 11:55am, Jonathan_the_Red wrote: 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. 
 Let's break this one down. Quote: Okay, let's say there are Y+3 REMs Quote: whether there are suicides or not 
 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 noone 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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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #73 on: Oct 9^{th}, 2002, 6:03pm » 
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on Oct 9^{th}, 2002, 4:57pm, James wrote: Let's break this one down. Okay, let's say there are Y+3 REMs Okay, let's say there are suicides. So, on night Y, Y+3 REMs suicide. So apparently, on night Y+3, the Y+3 REMs suicide  AGAIN. These guys have a serious death wish problem. 
 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 I_{n}. I think we've been leaving out the word "exactly"  does that help? Yeah, that's important. The claim is that I_{n} is true for all n, and that is what is shown inductively. If there are Y+3 REMs, then I_{n} is definitely true for all n != Y+3, "by default," because there aren't exactly Y or Y+1 or Y1 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 Oct 9^{th}, 2002, 4:57pm, James wrote:In fact, the argument goes something like this. On night Y, Y REMs suicide. 
 No no no no no! See above. The argument is that I_{n} 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 I_{n} has just one variable in it... n! And the inductive step shows that "if I_{n}, then I_{n+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?

« Last Edit: Oct 9^{th}, 2002, 6:04pm by S. Owen » 
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James
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Re: BROWN EYES AND RED EYES
« Reply #74 on: Oct 9^{th}, 2002, 7:07pm » 
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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: In particular, if there are NOT Y REMs, the statement is true whether there are suicides are not. 
 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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