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Topic: BROWN EYES AND RED EYES (Read 80039 times) |
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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #75 on: Oct 10th, 2002, 6:01am » |
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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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Jonathan_the_Red
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Re: BROWN EYES AND RED EYES
« Reply #76 on: Oct 10th, 2002, 11:29am » |
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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:
- Nobody killed himself on day 1.
- Therefore, there is not exactly one REM (by I1)
- Therefore, I am a REM (by C1)
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:
- Nobody killed himself on day 2.
- Therefore, there are not exactly two REM (by I2)
- Therefore, I am a REM (by C2)
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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #77 on: Oct 10th, 2002, 9:07pm » |
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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! 
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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #78 on: Oct 11th, 2002, 9:58pm » |
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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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #79 on: Oct 11th, 2002, 11:12pm » |
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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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #80 on: Oct 12th, 2002, 8:39am » |
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Quote: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.
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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:I really think mathematical induction is the best way to go with this.
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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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #81 on: Oct 12th, 2002, 9:58am » |
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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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #82 on: Oct 12th, 2002, 11:00am » |
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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:
.
.
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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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Jonathan_the_Red
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Re: BROWN EYES AND RED EYES
« Reply #83 on: Oct 12th, 2002, 1:48pm » |
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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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #84 on: Oct 12th, 2002, 2:17pm » |
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Mea culpa2 Yes, you did!
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Chronos
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Variant: The repentent tourist
« Reply #85 on: Oct 14th, 2002, 5:59pm » |
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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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #86 on: Oct 14th, 2002, 8:33pm » |
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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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #87 on: Oct 14th, 2002, 8:41pm » |
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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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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #88 on: Oct 14th, 2002, 8:53pm » |
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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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #90 on: Oct 15th, 2002, 7:49am » |
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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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Jonathan_the_Red
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Re: BROWN EYES AND RED EYES
« Reply #91 on: Oct 15th, 2002, 3:40pm » |
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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 Oct 14th, 2002, 8:41pm, Jeremy wrote:"There are 10 Brown Eyed Monks"
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Everyone deduces his eye color right away and dies that very day.
Quote:
"There are at lesat two Red Eyed Monks"
"There is more than one Red Eyed Monk"
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This gives the induction a "jump start." All the REMs die on day N-1, instead of day N.
Quote:
"There is an odd number of Red Eyed Monks"
"There is an even number of Red Eyed Monks"
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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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #92 on: Oct 15th, 2002, 6:20pm » |
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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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #93 on: Oct 16th, 2002, 11:38am » |
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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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Re: BROWN EYES AND RED EYES
« Reply #94 on: Oct 16th, 2002, 7:40pm » |
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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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Chronos
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Re: BROWN EYES AND RED EYES
« Reply #95 on: Oct 17th, 2002, 6:36pm » |
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Quoth Icarus: Quote:| 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". |
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Yup, there's a reason I put that in quote marks, rather than paraphrasing. I rather wondered how long it would take you guys to see it . And i meant for it to be assumed that the tourist chose Bob randomly from among the red-eyed monks, not that he has anything particular against him.
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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Jeremy
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Re: BROWN EYES AND RED EYES
« Reply #96 on: Oct 17th, 2002, 9:21pm » |
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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. 
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Chronos
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Re: BROWN EYES AND RED EYES
« Reply #97 on: Oct 21st, 2002, 10:27pm » |
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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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PSesulka
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Re: BROWN EYES AND RED EYES
« Reply #98 on: Dec 8th, 2002, 4:30pm » |
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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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Chronos
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Re: BROWN EYES AND RED EYES
« Reply #99 on: Dec 8th, 2002, 7:40pm » |
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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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