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Topic: BROWN EYES AND RED EYES (Read 80027 times) |
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fireashwinter
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Re: BROWN EYES AND RED EYES
« Reply #150 on: May 22nd, 2006, 11:17am » |
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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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fireashwinter
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Re: BROWN EYES AND RED EYES
« Reply #151 on: May 22nd, 2006, 1:38pm » |
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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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rmsgrey
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Re: BROWN EYES AND RED EYES
« Reply #152 on: May 23rd, 2006, 11:21am » |
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on May 22nd, 2006, 1:38pm, fireashwinter wrote: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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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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Aravis
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Re: BROWN EYES AND RED EYES
« Reply #153 on: Jul 13th, 2006, 8:38am » |
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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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #154 on: Jul 15th, 2006, 7:13pm » |
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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: Consider Jonathan's questions.
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Aravis
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Re: BROWN EYES AND RED EYES
« Reply #155 on: Jul 15th, 2006, 11:10pm » |
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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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rmsgrey
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Re: BROWN EYES AND RED EYES
« Reply #156 on: Jul 16th, 2006, 10:28am » |
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on Jul 13th, 2006, 8:38am, Aravis wrote:
Given R red eyes and B brown eyes, the tourist must say "There are exactly B-1 brown eyed monks".
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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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JA
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Re: BROWN EYES AND RED EYES
« Reply #157 on: Jul 27th, 2006, 5:24am » |
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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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towr
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Re: BROWN EYES AND RED EYES
« Reply #158 on: Jul 27th, 2006, 6:13am » |
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on Jul 27th, 2006, 5:24am, JA wrote:| Well, actually there are two big assumptions. The first (or the second as it happens) is that a simple induction proof suffices. |
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Why wouldn't a simple induction proof suffice? Logical induction is a sound proving method.
Quote:| The other assumption ... well. |
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JA
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Re: BROWN EYES AND RED EYES
« Reply #159 on: Jul 27th, 2006, 6:41am » |
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on Jul 27th, 2006, 6:13am, towr wrote:
Logical induction is a sound proving method.
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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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BNC
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Re: BROWN EYES AND RED EYES
« Reply #160 on: Jul 27th, 2006, 6:49am » |
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on Jul 27th, 2006, 6:41am, JA wrote:
When a number of conditions are satisfied, yes. If they are not, then no (obviously).
It's all a question of dimension
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Like you say: obviously.
But what condition do you think is not satisfied? And what do you mean by "dimension"?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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towr
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Re: BROWN EYES AND RED EYES
« Reply #161 on: Jul 27th, 2006, 7:07am » |
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on Jul 27th, 2006, 6:41am, JA wrote:| When a number of conditions are satisfied, yes. If they are not, then no (obviously). |
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We have a base case for which it is true, and for every case N the same is implied for case N+1.
I don't see what more should be to it.
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Aravis
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Re: BROWN EYES AND RED EYES
« Reply #162 on: Jul 27th, 2006, 8:44am » |
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on Jul 27th, 2006, 5:24am, JA wrote: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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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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JA
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Re: BROWN EYES AND RED EYES
« Reply #163 on: Jul 30th, 2006, 7:28am » |
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on Jul 27th, 2006, 7:07am, towr wrote:
We have a base case for which it is true, and for every case N the same is implied for case N+1.
I don't see what more should be to it.
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Yeah, that sounds like induction doesn't it. But is that what we actually have?
Take this proof (on page one)
Quote:| If there is exactly one, the one will easily deduce his eye color and kill himself on the first night. |
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Base case, got it.
Quote: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. |
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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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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #164 on: Jul 30th, 2006, 12:02pm » |
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on Jul 30th, 2006, 7:28am, 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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JA
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Re: BROWN EYES AND RED EYES
« Reply #165 on: Jul 30th, 2006, 10:59pm » |
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on Jul 30th, 2006, 12:02pm, Icarus 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!) |
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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: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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Why didn't he kill himself the night before?
JA
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towr
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Re: BROWN EYES AND RED EYES
« Reply #166 on: Jul 31st, 2006, 1:46am » |
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on Jul 30th, 2006, 10:59pm, JA wrote:That wasn't a dig. It's simply that a lot of answers are left partly open, so that people can finish the thought themselves.
Quote:| Why didn't he kill himself the night before? |
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Because he didn't know he was a REM the night before. They only kill themselveswhen they know they're REM. Otherwise they might as well all kill themselves the first night and not take the chance.
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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| « Last Edit: Jul 31st, 2006, 1:55am by towr » |
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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #167 on: Jul 31st, 2006, 4:13pm » |
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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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| « Last Edit: Jul 31st, 2006, 4:37pm by Icarus » |
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mypalmike
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Re: BROWN EYES AND RED EYES
« Reply #168 on: Dec 7th, 2007, 6:05pm » |
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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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SMQ
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Re: BROWN EYES AND RED EYES
« Reply #169 on: Dec 8th, 2007, 6:40am » |
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on Dec 7th, 2007, 6:05pm, mypalmike wrote:| 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. |
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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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Grimbal
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Re: BROWN EYES AND RED EYES
« Reply #170 on: Dec 8th, 2007, 4:49pm » |
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Everybody knows the information, but everybody doesn't know that everybody knows it.
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mypalmike
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Re: BROWN EYES AND RED EYES
« Reply #171 on: Dec 9th, 2007, 10:48pm » |
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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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| « Last Edit: Dec 9th, 2007, 11:50pm by mypalmike » |
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Grimbal
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Re: BROWN EYES AND RED EYES
« Reply #172 on: Dec 10th, 2007, 1:36am » |
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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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| « Last Edit: Dec 10th, 2007, 4:18am by Grimbal » |
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towr
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Re: BROWN EYES AND RED EYES
« Reply #173 on: Dec 10th, 2007, 1:47am » |
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on Dec 9th, 2007, 10:48pm, mypalmike wrote:| 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. |
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But not every monk knew that every monk knew that every monk knew. And that's as far as you should go; that is what the tourist conveys.
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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| « Last Edit: Dec 10th, 2007, 1:48am by towr » |
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SMQ
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Re: BROWN EYES AND RED EYES
« Reply #174 on: Dec 10th, 2007, 5:34am » |
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on Dec 9th, 2007, 10:48pm, mypalmike wrote:| 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. |
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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 Dec 7th, 2007, 6:05pm, mypalmike wrote:| 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. |
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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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