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Re: BROWN EYES AND RED EYES  
« Reply #125 on: Mar 21st, 2006, 5:23am »
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Thallion, I'm guessing you're still unconvinced since towr still used only 4 monks in his example rather than five (your threshold value), so let me try. Smiley
 
 
I. From the bottom up
 
(1) If there were only one lone REM (and any number of BEMs), that lone REM would know there were either zero or one REMs and hope there were none; each BEM would know there were one or two REMs and hope there was only one.  The tourist's statement would give no new information to the BEMs, but would dash the hopes of the lone REM, who would commit suicide that first night.
 
(2) If there were two REMs (and any number of BEMs), each REM would know there were one or two REMs and hope there was only one--that is, they would each hope they were living in situation (1).  However, after no monks committed suicide the first night the REMs' hopes would be dashed--they could no longer hope to be living in situation (1)--and they would each commit suicide the second night.
 
(3) If there were three REMs (and any number of BEMs), each REM would know there were two or three REMs and hope there were only two--that is, they would each hope they were living in situation (2).  However, after no monks committed suicide the second night the REMs' hopes would be dashed--they could no longer hope to be living in situation (2)--and they would each commit suicide the third night.
 
(4) If there were four REMs (and any number of BEMs), each REM would know there were three or four REMs and hope there were only three--that is, they would each hope they were living in situation (3).  However, after no monks committed suicide the third night the REMs' hopes would be dashed--they could no longer hope to be living in situation (3)--and they would each commit suicide the fourth night.
 
(5) If there were five REMs (and any number of BEMs), each REM would know there were four or five REMs and hope there were only four--that is, they would each hope they were living in situation (4).  However, after no monks committed suicide the fourth night the REMs' hopes would be dashed--they could no longer hope to be living in situation (4)--and they would each commit suicide the fifth night.
 
(6...N) and so on.  N REMs (with any number of BEMs) would each hope to be living in situation (N-1), but would have their hopes dashed after night N-1 when no monks committed suicide.
 
 
II. From the top down
 
There are five REMs, call them A, B, C, D and E.  Without loss of generality, consider monk A.  Monk A knows there are either four or five REM, and hopes there are only four.
 
Specifically, A hopes that
> w/out l. of g., B knows that A is a BEM and so there are either three or four REMs, and hopes there are only three REMs.
 
Now here's where I think confusion may be coming in; A knows that B's hopes are false--that there are in fact four or five REMs, but A is still hoping that B is hoping there are only three.
 
That is, A hopes that
> B hopes that,  
> > WLOG, C knows that A and B are BEMs and so there are either two or three REMs, and hopes there are only two REMs.
 
That is, A hopes that
> B hopes that
> > C hopes that,
> > > WLOG, D knows that A, B, and C are BEMs and so there are either one or two REMS, and hopes there is only one REM.
 
That is, A hopes that
> B hopes that
> > C hopes that
> > > D hopes that
> > > > E knows that A, B, C, and D are BEMs and so there are either zero or one REMs, and hopes there aren't any REMs.
 
That is, A hopes that B hopes that C hopes that D hopes that E hopes that there are no REMs.
 
But since at each step we chose an arbitrary representative monk, without loss of generality, it holde for all combinations of monks; that is, if -> represents "hopes that":
 
A->B->C->D->E->there are no REMs
B->A->C->D->E->there are no REMs
C->B->A->D->E->there are no REMs
D->C->B->A->E->there are no REMs
E->D->C->B->A->there are no REMs
etc.
 
But after the tourist's statement, A can only hope that B hopes that C hopes that D hopes that E now knows he is the only REM.
 
But after the first night, when no monks commit suicide, A can only hope that B hopes that C hopes that D ane E now know they are the only two REMs.
 
But after the second night, when no monks commit suicide, A can only hope that B hopes that C, D and E now know that they are the only three REMs.
 
But after the third night, when no monks commit suicide, A can only hope that B, C, D and E now know that they are the only four REMs.
 
But after the fourth night, when no monks commit suicide, A now knows that all five of them are REMs, and so sommits suicide on the fifth night.  B, C, D and E, reasoning likewise, do the same.
 
 
Whew, a bit long, but I hope it's clear enough.
 
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Re: BROWN EYES AND RED EYES  
« Reply #126 on: Mar 21st, 2006, 5:46am »
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on Mar 21st, 2006, 5:23am, SMQ wrote:
Thallion, I'm guessing you're still unconvinced since towr still used only 4 monks in his example rather than five (your threshold value), so let me try. Smiley
Damn, I misread, I thought his minimum was 4..  
Like that isn't enough work already Tongue
 
Oh well, it's not like I believed it the first time my professor tried to convince me.
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Re: BROWN EYES AND RED EYES  
« Reply #127 on: Mar 21st, 2006, 6:09am »
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here's the entire proces modeled for 5 REMs

Ever more worlds get eliminated as long as no one dies. Because if that possible world was the actual one, people would have killed themselves (Due to not being able to believe another world is possible, i.e. there not being a connection to another world with their color)
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Re: BROWN EYES AND RED EYES  
« Reply #128 on: Mar 21st, 2006, 8:28am »
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on Mar 20th, 2006, 9:37pm, Thallion wrote:

Agree - A knows BCDE, A Wants B to only know CDE(even though wrong), A wants B to want C to only see DE.  A knows that is not possible.  

A doesn't know any such thing. As far as A is concerned, it's perfectly possible for B to want C to only see DE... A knows that B would be hoping for something that isn't actually the case, but we know that A is hoping for something that isn't the case when he hopes he doesn't have red eyes...
 
 
Anyway, you seem to agree that, if the tourist makes his statement and there are only 4 REMs, all four commit suicide on the 4th night.
 
If there are 5 REMs and the tourist makes his statement, then each REM knows that there are either 4 or 5 REMs, and knows that if there are 4 REMs, they will suicide on the 4th night.
 
What conclusions will the 5 REMs draw when they wake up after the 4th night and no-one's suicided?
 
You don't need to look at what people know about what people know about what people know (etc.) only at what everyone knows would happen to a monastery with a given number of REMs on a given night.
 
Suppose, each afternoon, each of ten monks at a monastery is asked, in secret, by an outsider, the following questions:
 
0) How many REMs could there be at this monastery?
1) What would have happened last night at a monastery with 1 REM?
2) What would have happened last night at a monastery with 2 REMs?
3) What would have happened last night at a monastery with 3 REMs?
4) What would have happened last night at a monastery with 4 REMs?
5) What would have happened last night at a monastery with 5 REMs?
6) What would have happened last night at a monastery with 6 REMs?
7) What would have happened last night at a monastery with 7 REMs?
Cool What would have happened last night at a monastery with 8 REMs?
9) What would have happened last night at a monastery with 9 REMs?
10) What would have happened last night at a monastery with 10 REMs?
 
Until the tourist shows up, each REM will give the same answers every time, and so will each BEM, and they'll agree with each other on everything except question 0 - nothing special would have happened.
 
Once the Tourist makes his statement, the answers to the questions may start to change.
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Re: BROWN EYES AND RED EYES  
« Reply #129 on: Mar 21st, 2006, 9:56pm »
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Wow, I just hadn't had a chance to come check recently.  goodness. hypercubes.. Shocked you all are going all out.  
 
If the monks say, "well if there had only been one REM he would kill himself tonight"  then I guess the process starts. If it is taken 1 question at a time forward then there is no reason it wouldn't start.
 
Mind you I think that is a pretty big IF.
 
I do have a problem with the fact that if A knows B knows that everyone else sees more than 1 REM, that the information means anything to him.  Regardlses of what A knows B knows that C knows that D knows.  What B knows C knows about D is irrelevant to A because A relies on his knowledge of the others to make his conclusion, not someone else's thoughts about 2 other people furhter away from him.  
 
However having said that, if you follow the whole a knows b knows c knows.... long enough then again you could come to the 0 case and 'start' the chain.
 
I also have a problem with the lack of any proof being sufficient evidence that the event was true/false.
 
Can the lack of an action prove a binary question is true or false?  What if you already knew the answer, then does it prove anything?
 
But then again I guess that is part of the problem with inductive logic it does deal with probability that an outcome is correct.
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Re: BROWN EYES AND RED EYES  
« Reply #130 on: Mar 22nd, 2006, 12:38am »
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on Mar 21st, 2006, 9:56pm, Thallion wrote:
If the monks say, "well if there had only been one REM he would kill himself tonight"  then I guess the process starts. If it is taken 1 question at a time forward then there is no reason it wouldn't start.
That's why it's important REMs should kill themself at midnight. this means each morning when they get together, the process goes one step forward.
 
Quote:
I do have a problem with the fact that if A knows B knows that everyone else sees more than 1 REM, that the information means anything to him.
Among other things B learns that A knows this about him. And A learns that B must have learned this.
 
Quote:
Can the lack of an action prove a binary question is true or false?
I don't see why not. If it's an action that would have to take place in one case, or would have to not take place in the other case.
If there are N REMs they must kill themselves on the Nth night. If no REMs kill themselves on the Nth night, then this proves there must be more than N.
   
Quote:
What if you already knew the answer, then does it prove anything?
It may prove something to others, and you know what it will prove to them.
 
Quote:
But then again I guess that is part of the problem with inductive logic it does deal with probability that an outcome is correct.
There is nothing probabilistic about this. You can cleanly prove what happens in epistemic (update) logic. Using a model does the same thing, except it's much simpler.
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Re: BROWN EYES AND RED EYES  
« Reply #131 on: Mar 22nd, 2006, 3:19am »
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on Mar 22nd, 2006, 12:38am, towr wrote:
Among other things B learns that A knows this about him. And A learns that B must have learned this.

 
I think you are misreading me there.  I said the knowledge that everyone knows >=1 is already known, thus that bit of information means nothing to them.
 
Noone learns anything until N-1 nights.  Everything Before N-2 nights is all hypothetical.  (Note BEMs dont learn anything until N Nights)  
 
On N-1 Night REMs learns the other REMs see the same number of REMs as themselves.  On N Night REMs kill themselves and BEMs learn that they were not a REM.  Your statement is describing this interchange on night N-1, it has nothign to do with what was "learned" when the tourist opened his big mouth.
 
You can hope that the other REMs learn something on an earlier night (which would thus move you into the other category), but until N-1 Night You ( and the other REMS) donot learn anything.
 
The premise comes down to this(commonly accepted answer):  For any N of REMs, when the statement of "I see REMs" is made that every REM will think "Hmm that means that if there was only 1 he would now know who he is, which follows that if there were 2 they would find out  tonight, that 3 would...."
 
When in reality given a sufficiently large number of REMs they say "DUH I knew that"
 
Thats really what this boils down to.  What is the probability that of 100 REMs they all think exactly the same thing that is required to start the chain?
 
I like my N-2 > 2 postulation mainly out of the facts that, People may care about someone knows about someone else knowing (a knows B knows C knows) IE 2 hops from N) so N-2, and the case when the answer of NOt 0 is relevant is the 2 REM case so > 2.  
 
But I fully agree IF the trail/regression is followed by all N, and everyone acknowledges night 1.  Then the formula works.  I just think that is a pretty big IF.
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Re: BROWN EYES AND RED EYES  
« Reply #132 on: Mar 22nd, 2006, 3:59am »
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I think you all can agree with me on this..
 
All N must acknowledge the hypothetical case of N being 1.
 
AND  
 
All N must acknowledge that night one occurs.
 
If either of these are false the process cannot begin.  
 
Can we agree with this?
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Re: BROWN EYES AND RED EYES  
« Reply #133 on: Mar 22nd, 2006, 4:15am »
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on Mar 22nd, 2006, 3:19am, Thallion wrote:
I think you are misreading me there.  I said the knowledge that everyone knows >=1 is already known, thus that bit of information means nothing to them.
But that isn't the only information that comes with the statement. That's the point.
 
Quote:
Noone learns anything until N-1 nights.  Everything Before N-2 nights is all hypothetical.  (Note BEMs dont learn anything until N Nights)  
They DO learn something. The hypotheses people hold change, that's what they learn. Step by step.
Until the final night when there is just one hypothesis (possible world) left, which then becomes the actual working theory of them committing suicide.
 
Quote:
Thats really what this boils down to.  What is the probability that of 100 REMs they all think exactly the same thing that is required to start the chain?
This isn't a probability riddle, this is an abstract logics puzzle. All monks are infinitely smarts and infinitely fast at reasoning and know that the other monks are in that respect like them. They're not actual real people.
 
on Mar 22nd, 2006, 3:59am, Thallion wrote:
I think you all can agree with me on this..
 
All N must acknowledge the hypothetical case of N being 1.
No, they just have to acknowledge that some monk may believe it's possible that some monk believes it's possible that.. etc that some monk believes he knows  (after the tourist's statement) N is 1 and he must be that one.
Every single one of the N REMs knows there are at least N-1. They can't consider the hypothesis that it's less. But they can consider that other monks might consider that so.
 
Quote:
All N must acknowledge that night one occurs.
That's a given since any monk that finds out he is a REM must kill himself the first night that comes along. And so the hypothetical monk from N=1 covers this.
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Re: BROWN EYES AND RED EYES  
« Reply #134 on: Mar 22nd, 2006, 5:18am »
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Wow, apparently no matter what I say you are going to disagree with me.  For if you had said Yes I agree then you would have proved the process works and that the Nth night is true.
 
Now I don't know what you are doing, unless you are just trying to be "know it all" because that is how you are treating this discourse.
 
You say No to my statement: " All N must acknowledge the hypothetical case of N being 1."  
 
Regardless of the monks perceptions of what everyone knows, the monks have to be able to say "If there was 1 monk that monk would now know he was the only one."
 
Thus they must Hypothesize that if there was only one REM that that monk would kill himself the first night.
 
Are you really telling me you disagree with this.  Do you want to consider your answer again.
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Re: BROWN EYES AND RED EYES  
« Reply #135 on: Mar 22nd, 2006, 5:28am »
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on Mar 22nd, 2006, 5:18am, Thallion wrote:
You say No to my statement: " All N must acknowledge the hypothetical case of N being 1."  
 
Regardless of the monks perceptions of what everyone knows, the monks have to be able to say "If there was 1 monk that monk would now know he was the only one."
 
Thus they must Hypothesize that if there was only one REM that that monk would kill himself the first night.
Yes, but that's something different from what you said before.
 
They don't have to consider that "there might be only be one monk". But that "if there was one monk, he'd kill himself tonight" (and everyone knows this)
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Re: BROWN EYES AND RED EYES  
« Reply #136 on: Mar 22nd, 2006, 6:00am »
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The statement:  
 
All N must acknowledge the hypothetical case of N being 1.
 
Is the same as:
 
The Monks must acknowledge that if there was 1 monk that monk would now know he was the only one.
 
The second more wordy certainly,  The second certainly spells out the hypothesis and thus makes it easier to draw the conculsion.  But they do say the same thing.  A hypothetical case is a case where a hypothesis is drawn (webster's dictionary) So I am not sure how you can say these are different statements.  When the second part of the second line is simply the hypothesis referenced in the first statement - > "if there was 1 monk that monk would now know he was the only one"
 
This is especially relavent in this riddle For everyone has said over and over in this thread that the statement of " I see red eyes" is equivalent to saying "I see red eyes, So if there was only 1 REM that monk would now know he was the only one."  And then running from there.  
 
That is all semantics, though so I will reword the question to make it more clear.
 
Can we all agree that:
All Monks must acknowledge that if there was only 1 monk that he would now know he was the only one.
 
AND
 
All Monks must agree that night 1 has occured.
 
Note I do see that actually if the first statement becomes true at the same time for every monk that the second statement becomes a given.
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Re: BROWN EYES AND RED EYES  
« Reply #137 on: Mar 22nd, 2006, 6:19am »
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on Mar 22nd, 2006, 6:00am, Thallion wrote:
The statement:  
 
All N must acknowledge the hypothetical case of N being 1.
 
Is the same as:
 
The Monks must acknowledge that if there was 1 monk that monk would now know he was the only one.
The way I read it, it meant "All N must acknowledge that N might be 1"
So that lead to a misunderstanding..
 
Quote:
Can we all agree that:
All Monks must acknowledge that if there was only 1 monk that he would now know he was the only one.
 
AND
 
All Monks must agree that night 1 has occured.
After the public statement by the tourist, and the first night following. Then yes. Otherwise no.
Before the tourist speaks the one monk may still have hope there aren't any. And if the statement isn't public (to all monks at once where they all know they all hear it), not all other monks may know that the one REM knows he should kill himself that night.
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Re: BROWN EYES AND RED EYES  
« Reply #138 on: Mar 22nd, 2006, 6:54am »
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OK, the standard solution does rely on certain assumptions which are commonly accepted in solving logic puzzles involving reasoning beings.
 
1) The monks are "smart enough" to work out (by the time it becomes relevant) anything that we can work out, and will do so (and this is common knowledge).
 
2) The conditions of the problem are common knowledge among the monks.
 
3) The monks don't cheat. No clever murders disguised as suicides, no refusal to suicide, etc...
 
The key assumption is the first one - it means that the monks will be aware that, when N=4, all REMs will kill themselves on the 4th night, so if N=5, then all 5 will know where they stand after that night...
 
As far as nothing happening proving something, there's the notorious curious incident of the dog in the night-time. The answer to the binary question "Did any monks suicide last night?" is definitely given by the lack of events of the night before...
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Re: BROWN EYES AND RED EYES  
« Reply #139 on: Mar 22nd, 2006, 8:29am »
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Thanks, Thallion, for forcing us to be completely clear about our reasoning. Smiley (seriously; it's good mental exercise!)
 
 
 
You seem to agree that the "bottom up" argument at least looks reasonable, so let's reformulate the "top down" argument in a more "bottom up" way: First, let's look at the state of things before the tourist makes his unfortunate observation.
 
Consider an abbey, First Abbey, with only one red eyed monk, Br. Albert.  Br. Albert looks around and sees only brown eyes, so fervently hopes his own eyes are brown.  We, as outside observers (or as brown-eyed monks in First Abbey), know that Br. Albert is hoping for a false thing--that his own eyes are really red--but Br. Albert doesn't know that, and so he goes on fervently hoping despite our superior knowledge. Smiley
 
Next, consider Second Abbey, with two REMs, Brs. Albert and Bennie.  Br. Bennie looks around and sees all brown eyes except for Br. Albert, so fervently hopes his own eyes are brown and that Br. Albert is the only REM.  But in hoping his own eyes are brown, Br. Bennie hopes that he is an observer in First Abbey where Br. Albert is a lone REM; that is, Br. Bennie hopes Br. Albert hopes there are no REMs, and, being an expert logician, Br. Benie realizes he is hoping this about Br. Albert's hopes.  In so doing, Br. Bennie places himself in our position in First Abbey, reasoning about Br. Albert's hoped-for hopes even though he knows that reality is other than Br. Albert hopes for.
 
Now, let's move on to Third Abbey where there are three REMs, Brs. Albert, Bennie and Charles.  Br. Charles looks around and sees that Br. Bennie's and Br. Albert's eyes are red, so fervently hopes his own eyes are brown and Brs. Albert and Bennie are the only REMs.  But in so hoping, Br. Charles hopes Br. Bennie considers himself to be living in Second Abbey.  And just as in Second Abbey, the Br. Bennie of Br. Charles' hopes himself hopes Br. Albert considers himself to be living in First Abbey where there might not be any REMs at all.  That is, Br. Charles hopes Br. Bennie hopes Br. Albert hopes there are no REMs, and, being an expert logician, Br. Charles puts himself in our place in Second Abbey and realises he is hoping all this.
 
Likewise, in Fourth Abbey, Br. David is hoping Fourth Abbey's Br. Charles considers himself to be living in Third Abbey, etc., and, of course, in Fifth Abbey Br. Ernie is fervently hoping that Br. David is hoping that Br. Charlie is hoping that Br. Bennie is hoping that Br. Albert is hoping that there aren't any REMs.
 
It should be clear that the names don't matter; that every REM in Fifth Abbey puts himself in the place of Br. Ernie and hopes things about the hopes of the hopes of the hopes of the hopes of all his Brothers, and, being himself an expert logician, realizes all the things he is hoping.
 
In general, every REM hopes that his own eyes are brown, and in so hoping, implicitly hopes things about his red-eyed Brother's hopes--specifically that each of them is hoping exactly what they would be hoping if they each saw one fewer REM than he himself does--and, being an expert logician, is able to  reason out all the myriad things he is hoping, not only about his Brothers and about their immediate hopes, but about the hopes of the hopes of the hopes of ... of all his Brothers, right down to that completely hypothetical, many-times-removed lone monk who everyone knows doesn't exist in reality but everyone is hoping that someone is hoping that someone else is hoping that ... he hopes there are no REMs.
 
 
 
Now along comes the unsuspecting tourist and his unfortunate remark.  On the surface, it appears nothing changes--everyone already knew that there was at least one REM, of course--but let's look at what it does to Br. Ernie's hopes.
 
Still hoping he himself has brown eyes, but now realizing that no monk, however hypothetical could reasonably believe that there are no REMs, Br. Ernie has begun to hope that Br. David hopes that Br. Charles hopes that Br. Bennie hopes that Br. Albert now realises he has red eyes.
 
That is, in First Abbey, Br. Albert would now have had his hopes dashed.  He could no longer hope that there were no REMs, and, seeing that all his Brothers had brown eyes, would know that he must be the lone REM.
 
In Second Abbey, Br. Bennie could no longer hope that Br. Albert hoped there were no REMs; he (Br. Bennie) could only hope that Br. Albert now knows himself to have red eyes.  We the observers know that this still isn't true; Br. Albert is, in reality, hoping the same things about Br. Bennie, but Br. Bennie doesn't know that, so he goes on hoping.
 
Likewise in Third Abbey Br. Charles puts himself in the place of an observer at Second Abbey and hopes Br. Bennie hopes Br. Albert now knows he has red eyes; and so on up to Br. Ernie's hopes in Fifth Abbey.
 
But after the first night passes, Br. Enrie's hopes are changed again.  Now, in a hypothetical First Abbey, Br. Albert would have committed suicide.  In Second Abbey, Br. Bennie, noticing that Br. Albert didn't actually commit suicide, can no longer hope that he is an observer in First Abbey; he must accept the reality that he's been living in Second Abbey all along and that his own eyes are red.  In Third Abbey, Br. Charles, still hoping he is an observer in Second Abbey, now hopes that Brs. Bennie and Ablert both now realize they have red eyes.  And so on, right up to Br. Ernie in Fifth Abbey who hopes he is an observer in Fourth Abbey where Br. David hopes he is an observer in Third Abbey where Br. Charles hopes he is an observer in Second abbey where Brs. Bernie and Albert now realize they must have red eyes.
 
Likewise, after the second night Br. Ernies hopes must change yet again, now hoping that Br. David hopes that Brs. Charles, Bennie, and Albert now realize they are all REMs.
 
And after the third night the best Br. Ernie can hope is that Brs. David, Charles, Bennie and Ablert all now realize they must have red eyes.  (And, of course, in reality, Brs. David, Charles, Bennie and Albert are all hoping exactly the same thing about all the monks except themselves.)
 
Finally, after the fourth night, when still no one in Fifth Abbey commits suicide, Br. Ernie is forced to abnadon his last remaining hope and accept that he must have red eyes himself.
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Re: BROWN EYES AND RED EYES  
« Reply #140 on: Mar 22nd, 2006, 8:32am »
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In summary then, to begin with, every red-eyed monk in Fifth Abbey hopes he is living as a brown-eyed monk in Fourth Abbey where each of four red-eyed monks hopes he is living as a brown-eyed monk in Third Abbey where each of three red-eyed monks hopes he is living as a brown-eyed monk in Second Abbey where each of two red-eyed monks hopes he is living as a brown-eyed monk in First Abbey where a lone red-eyed monk hopes his own eyes are brown.
 
On the day the tourist makes his unfortunate remark this delicate skien of hopes begins to unravel.  Now every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where every REM hopes to be a BEM in Third Abbey where every REM hopes to be a BEM in Second Abbey where every REM hopes to be a BEM in First Abbey where a lone REM just found out his own eye color.
 
On the day after the remark, every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where every REM hopes to be a BEM in Third Abbey where every REM hopes to be a BEM in Second Abbey where two REMs just found out their own eye color.
 
On the third day, every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where every REM hopes to be a BEM in Third Abbey where three REMs just found out their own eye color.
 
And the fourth day, every REM in Fifth Abbey hopes to be a BEM in Fourth Abbey where four REMs just founf out their own eye color.
 
And finally, the fifth day, every REM in Fifth Abbey finds out his own eye color.
 
 
 
Whew! Smiley
 
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Re: BROWN EYES AND RED EYES  
« Reply #141 on: Mar 22nd, 2006, 8:22pm »
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It was a bit of a trick question..  The tendancy is that once you believe someone doesn't agree with you, you tend to try to refute everything that person says, closing off your mind to the fact that that person could be right about something.
 
The trick is, that if the two statements made above are true the process starts wether you use top down or bottom up logic.  So the trick is if you try to refute either of the above statements then you end up trying to say the riddle breaks down.  
 
Work with me on this I am being devils advocate now...  The riddle states that the monks have a vow of silence.  The riddle states the tourist makes the statement, "I see people with Red Eyes".  
 
Situation 1: The riddle does not say that all of the monks heard the comment.  And we know they have a vow of silence so they cannot tell the others what they heard.  Do you agree this situation could break the problem?
 
Situation 2:All of the monks are eating breakfast when the tourist makes the statement, and the statement is made in the same room as all of the monks.  Do all of the monks know for certain that every other monk heard the statement?  Does the process start?
 
Situation 3: What if one of the REMs is deaf?  What if one of the BEM is deaf?
 
Situation 4: What if the tourist tells each monk individually that he sees monks with red eyes?
 
Situation 5: What if the tourist is at the monestary for two days.  The tourist tells every monk he has told every other monk that he sees REMs.  What happens?
 
A few other misc comments.  RMSGrey's statement #1.  My issue with this is, I would imagine we are all pretty smart people. Smarter than your average joe certainly.  Anytime you put people in a situation you cannot count on anything.  but if this is a 'tenant' of logic problems, then forgive me, as it has been 13 years since i was in my logic class in college. Wink
« Last Edit: Mar 22nd, 2006, 10:41pm by Thallion » IP Logged
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Re: BROWN EYES AND RED EYES  
« Reply #142 on: Mar 22nd, 2006, 10:43pm »
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I have another one.
 
Situation 6:  What if when the tourist makes the fateful comment, all of the REM's hear the comment, but not all of the BEM hear it?
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Re: BROWN EYES AND RED EYES  
« Reply #143 on: Mar 23rd, 2006, 1:10am »
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Or what if a REM when he find out decides not to kill himself.
Or what if BEM decide to kill every REM they see?
 
You can almost always sidestep the intended puzzle and make something else of it. That's usually covered in the first pages of a thread Wink
 
 
S1/2/6- As long as all REM hear the statement at the same time and know they all hear it, the process will start. The BEM don't matter in this case.  
If not all REM hear it, or think they don't all hear it the process doesn't start. If some monks think all hear it, but in fact some don't, awfull things may start to happen.
 
S3- If X of the REMs is deaf, then the others if they know this will kill themselves on the N-Xth night, or if they don't know this will kill them on the Nth night. The deaf ones survive (not having a clue that anything went on) Of course after that mass uicide, they may in the latter case deduce there are X deaf REMs and seeing only X-1 conclude they are the last one. But that's assuming monks don't commit mass suicide for other reasons (otherwise they may not be sure that the red eyes thing was the cause).
 
S4- If each monk is told individually, then there is no common knowledge and the process doesn't start, because now they really don't learn anything new from his statement.
 
S5- second level knowledge is not enough to start the process. The tourist must also tell every REM that he told every REM that he told every REM... etc up to at least the level of the number of REM.
If there is just one REM, telling him there is at least one is enough to bring him to kill himself. If there are two, he must also tell both that he told the other one, so that they know the other one would kill himself if he were the only one. Etc.
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Re: BROWN EYES AND RED EYES  
« Reply #144 on: Mar 24th, 2006, 3:31am »
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On the second day after the tourist departed, Br. Zoe leaves his room and wonders:
Where is everybody?  It is breakfast time but nobody is there.  This place has become crazy!  The brothers have acted strangely lately.  In fact, since that guy came to visit.  He said something about red-eyed monks.  I don't understand what it was about.  There are no red-eyed monks on this island.
Hello?  Anybody there?
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Re: BROWN EYES AND RED EYES  
« Reply #145 on: Mar 24th, 2006, 8:16pm »
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Ok, I agree with your assessment of all of the cases except 5.
 
in 5, I have broken the agreement on night one conditional.  Since the tourist was at the monestary two days, monks could have found out on day 1 or day two, that uncertainty means the monks can't pick a night 1, thus they can't start eliminating numbers.  
 
Also I disagree that he would have to tell, that he told, that he told.  I believe if the tourist told each monk exactly this it would be sufficient, "I have told every monk this same bit of information, I see red eyed monks"  Thus every monk would have the knowledge that every other monk also knew it.  
 
Here is an interesting side effect of this.  If the tourist tells the monks the above statement (all in one day), but he doesnt tell them all, he got confused on who he told, they all look the same in those robes anyway.  Well..  all of the REMs who he has told will kill themselves on nigth N regardless of how many REM he doesn't tell.  all the REMs he hasn't told will live on.
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Re: BROWN EYES AND RED EYES  
« Reply #146 on: Mar 25th, 2006, 4:52am »
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on Mar 24th, 2006, 8:16pm, Thallion wrote:
Ok, I agree with your assessment of all of the cases except 5.
 
in 5, I have broken the agreement on night one conditional.  Since the tourist was at the monestary two days, monks could have found out on day 1 or day two, that uncertainty means the monks can't pick a night 1, thus they can't start eliminating numbers.  
Ah I missed that one.  
And it seems once again I interpreted somethign differently from you.
 
The statement
"I have told every monk this same bit of information, I see red eyed monks"
is different from
"I see red eyes, and I told every monk I see red eyes" Which is how I read  "The tourist tells every monk he has told every other monk that he sees REMs"
 
So in that case you have insufficient depth of knowledge. Interpreted your (intended) way, the statement seems to be recursive and has infinite depth.  
 
It's a big problem in such word problem to be precise and unambiguous. Which I suppose is why logicians invented formal logic notations.
 
Quote:
Here is an interesting side effect of this.  If the tourist tells the monks the above statement (all in one day), but he doesnt tell them all, he got confused on who he told, they all look the same in those robes anyway.  Well..  all of the REMs who he has told will kill themselves on nigth N regardless of how many REM he doesn't tell.  all the REMs he hasn't told will live on.
Yep, that seems right. Same as if those other REMs were deaf.
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Re: BROWN EYES AND RED EYES  
« Reply #147 on: Mar 25th, 2006, 5:20am »
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on Mar 22nd, 2006, 8:22pm, Thallion wrote:
RMSGrey's statement #1.  My issue with this is, I would imagine we are all pretty smart people. Smarter than your average joe certainly.  Anytime you put people in a situation you cannot count on anything.  but if this is a 'tenant' of logic problems, then forgive me, as it has been 13 years since i was in my logic class in college. Wink

The problem with throwing "real people" into a problem is that everyone has a different idea of how "real people" behave...
 
It's a sensible default assumption that people in logic problems, having grown up in a world built around logic, are going to be competent logicians...
 
 
@Towr:
"I have told every monk this same bit of information, I see red eyed monks" is less likely to be believed than "I am telling every monk this and that I have seen red eyes here"
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Re: BROWN EYES AND RED EYES  
« Reply #148 on: May 21st, 2006, 9:23pm »
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"There is at least one Colorblind REM."
 
Cronos (I think) was correct to say that this information would not lead to suicides.    
 
A CB monk knows that all the other monks might be CBREM.  Normal monks know that all the REMs they see might be CBREM.  So, no monk will act, as uncertainty remains.
 
This makes me wonder.  Assume that the tourist simply said to the monks all at once:
 
"At least one of you is a REM."
 
Then, the tourist realizes at some day M<N the suicide rule.  As Jeremy asked, what might be a very best solution to prevent suicides?
 
What if the tourist says:
 
"You all know there is at least one REM."
 
or
 
"All of you have for some time known that there is at least one REM."
 
Once we have opened up the possibility that the monks are not perfectly perceptive in sight, those variations may have meaning.
 
As another question, consider what would happen if the tourist says to all the monks at once:
 
"You are each aware of the physical state of every other monk."
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Re: BROWN EYES AND RED EYES  
« Reply #149 on: May 22nd, 2006, 9:31am »
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on May 21st, 2006, 9:23pm, fireashwinter wrote:
"There is at least one Colorblind REM."
 
Cronos (I think) was correct to say that this information would not lead to suicides.    
 
A CB monk knows that all the other monks might be CBREM.  Normal monks know that all the REMs they see might be CBREM.  So, no monk will act, as uncertainty remains.

Sounds good.  
Quote:
This makes me wonder.  Assume that the tourist simply said to the monks all at once:
 
"At least one of you is a REM."
 
Then, the tourist realizes at some day M<N the suicide rule.  As Jeremy asked, what might be a very best solution to prevent suicides?

I don't think there's any way to prevent at least some suicides from happening as long as the monks are aware of whether other monks have suicided (unless some later information comes along to change things)
Quote:
What if the tourist says:
 
"You all know there is at least one REM."
 
or
 
"All of you have for some time known that there is at least one REM."

Certainly the second is equivalent to saying "there are at least 2 REMs and no-one is colour-blind" - the first might count as "there is at least one"
Quote:
Once we have opened up the possibility that the monks are not perfectly perceptive in sight, those variations may have meaning.
 
As another question, consider what would happen if the tourist says to all the monks at once:
 
"You are each aware of the physical state of every other monk."

The statement will do nothing on its own - it's one of the assumptions of the original...
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