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Topic: BROWN EYES AND RED EYES (Read 80050 times) |
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PSesulka
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Re: BROWN EYES AND RED EYES
« Reply #100 on: Dec 8th, 2002, 7:49pm » |
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I get it now
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datarocks
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Re: BROWN EYES AND RED EYES
« Reply #101 on: Dec 9th, 2002, 8:39pm » |
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guys, help me here...
there was a very similar riddle to this one involving three men at a table each wearing one of two colored hats....
but i forget how it goes... coudl someone please post it if they know it?
thanks
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towr
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Re: BROWN EYES AND RED EYES
« Reply #102 on: Dec 10th, 2002, 1:33am » |
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But how can you know they use days to pace their counting? Obviously they should use something, because if they don't and there are for instance two REM, then one may think faster than the other that if the toher was the only one he would have killed himself, so I must be one, so I must kill myself and then does so, before the other has thought that much and consequently considers, "well some REM killed himself, and that was the only one I saw, so they're all dead now."
But they can't really discus how to pace their counting, yet, if they're smart, they know there has to be some pacing. So it should be forced on them from the outside by someone repeating 'there is (still) at least one REM', counting at each utterance.
So unless there is any pacing most monks that think too fast, and don't consider the pacing-paradox, will kill themselves. And any others would keep on living..
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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #103 on: Dec 10th, 2002, 5:50am » |
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The riddle says that red-eyed monks must kill themselves at midnight. So, that condition defines the discrete "rounds" in which this process takes place.
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S. Owen
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Re: BROWN EYES AND RED EYES
« Reply #104 on: Dec 10th, 2002, 6:04am » |
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on Dec 9th, 2002, 8:39pm, datarocks wrote:guys, help me here...
there was a very similar riddle to this one involving three men at a table each wearing one of two colored hats....
but i forget how it goes... coudl someone please post it if they know it? |
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There is a riddle under "Easy" called "Three Hats" that might be what you're looking for.
I remember that some guy actually played out a riddle like what you're talking about on the Johnny Carson show a long time ago.. it involved white and black dots on each of three persons' foreheads, and guessing one's own color dot. Heh, maybe that will lead you to some info on the internet about it.
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towr
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Re: BROWN EYES AND RED EYES
« Reply #105 on: Dec 10th, 2002, 8:37am » |
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on Dec 10th, 2002, 5:50am, S. Owen wrote:| The riddle says that red-eyed monks must kill themselves at midnight. So, that condition defines the discrete "rounds" in which this process takes place. |
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I guess I missed that..
[mental_note]Must learn to read the problems better..[/mental_note]
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UofLis#1
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Re: BROWN EYES AND RED EYES
« Reply #106 on: Mar 10th, 2003, 8:59pm » |
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 Wow icant believe i finally have the answer. A friend told me this riddle once and it took me forever i just couldnt figure it out. With the hints i did but it was because of this forum i have finally the answer and i know what. i didnt get the answer from him because he wouldnt tell me it was because of the hints on this forum that i got it.
Thanks all!!!!!!!!!!!!!!!
GO CARDS!!!!!!!!!!!
-Ford
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Lusus
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Re: BROWN EYES AND RED EYES
« Reply #107 on: Apr 26th, 2003, 1:13am » |
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Keep in mind, the curse and the suicide are two different things. After all, if Red Eyed monks existed on the Island for years, then every monk would've realized that suicide is not a part of the curse of the Red Eyes.
If there was one Red Eyed Monk, he may kill himself that night to avoid the curse. But if there were two REM's then as one of them who wakes up the next day and sees that the other hasn't killed himself yet I have two things to consider:
1) either there is another REM that he sees (logic leads to this other person being me), or...
2) This person bravely chose to endure the curse. Even before the tourist arrived, I don't know that he doesn't already know he's red eyed. He may have chosen to endure the curse long ago. It has already been seen a monk can live for years with the curse, why not live a few more?
This alternative view remains in situations where there are more than two REM's.
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Cage Hiu
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Re: BROWN EYES AND RED EYES
« Reply #108 on: May 14th, 2003, 9:55pm » |
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Alternate sloution
Lets say you are a monk (doesn't matter what eye color) , you can draw attention to yourself by stamping your feet or what not.... As the other monks are watching you, you remove one of your EYES.
Therefore, the other monks will understand the logic and follow suit... you see your eye color with your other eye. So therefore the next day, you are left with a temple full of one brown-eyed MONKS. Also all monks uphold their vow of slience
Yes I am well aware the sloution is not optimal to the Nth death on the Nth day logic but the premises are also fullfilled by THIS means... it also aviods the chance of MASS SUCIDE if the monks are tricked by the tourist.
-C.H
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Mike V
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Re: BROWN EYES AND RED EYES
« Reply #109 on: Aug 5th, 2003, 8:08pm » |
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I know it's been a long time, but the concept of colorblind monks (CBM) intrigues me.
A CBREM (colorblind redeyed monk) could indeed commit suicide. Although he never can see the other colors, if all monks assume all other monks can see color. If there is only one CBREM he will commit suicide on the n+1th day, unless n = 1, in which case he kills himself on the 3rd day (after everyone else kills themselves on the 2nd day).
The reason is that since no one knows about his colorblindness, except him, they'll follow the normal logic. If he doesn't kill himself on day one, they all assume they must be red eyed and kill themselves on day two. He then realizes what had happened and that he has red eyes. If there are more redeyes than just him, they follow normal logic and kill themselves on day n. Seeing that n-1 monks killed themselves on day n, our CBREM realizes that he must have red eyes and kills himself on day n+1 (leaving the other monks confused, but alive).
This gets more complicated with more CBMs (red or non), but follows similar logic.
A) One interesting case is what happens when all monks are colorblind (but each assumes he is the only one).
B) Or, as mentioned before, all but one are colorblind (and each still assumes others see just fine).
Finally, when you take away the assumption about the others seeing ok (perhaps by having the tourist say, 'also, at least one of you is colorblind') I find that there generally is not enough information for all the red eyes to kill themselves (by generally I mean I've tried several different amounts of info they get, such as an exact number of CBMs or an at least, and similarly for REM).
Ok, sorry, but I just found this site and I love this riddle, finally finally, I want to thank Jonathan_the_Red if he happens to still be around. While I understood how it worked without them, he gave the best explanations (3 unique ones at that!) I've ever heard for this phenomenon. I've been inspired to take some logic classes.
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S. Owen
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Monks and the stock market
« Reply #110 on: Aug 5th, 2003, 9:41pm » |
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Also on the subject of color-blind monks -- I read an interesting article about investor behavior that bears some relation to this aspect of the riddle, so I thought I'd share:
I think one of the original points regarding color-blind monks was that there is a big assumption in the canonical solution to the riddle -- that all monks are very smart and arrive at this canonical solution. There is a certain circularity to it. Some monk could be color-blind, or reason differently -- things other monks cannot know -- and then the canonical solution is no longer valid.
The statement of the riddle kind of waves this all aside, but it is interesting to think about -- what assumptions really underpin the canonical solution?
But, that's not quite the point of this post.
Consider a game played by a group of very smart people, in which each person chooses a number between 0 and 100 independently and in secret. After this, all numbers are revealed and averaged. The person whose choice is closest to 80% of that average wins. And everybody wants to win of course.
What happens? Well, a normal person might consider that the average will probably be about 50, and guess 40. But of course everyone is thinking that. So, 32? Same thing. Following the reasoning in this riddle (sort of), the solution is that all participants guess 0. I take it this is a classic problem.
The article I read presented this as a simplified version of the reasoning that underpins the stock market -- the goal is not really to guess which companies will do best, but rather to guess which companies most other investors think will do best.
While the solution to what happens in the game above is 0, of cousre, we all know that if we took 20 of our friends and tried this, that nobody would guess 0. It's not a matter of intelligence -- even though you know the solution, you would not be wise to guess 0!
Your optimal guess depends on your assessment of other participants' guesses, which in turn depends on their assessments and you don't know those. This, the article says, is what keeps the stock market moving.
So the color-blind monk question is, I think, an important one. Is the canonical solution really a solution at all? I haven't worked out whether the "very smart monks" assumption is sufficent to resolve this, and would like to hear from people on this question. In any event, in the stock market, it is no solution at all.
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BaH
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Re: BROWN EYES AND RED EYES
« Reply #111 on: Dec 28th, 2005, 1:45pm » |
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lolol, there is one rem, and he tries to commit suicide on day one and fails, on day 2 he is sad and grieving because of his failed suicide attempt, and the other monks think he is just grieving becasue he has realized he is the 'other' rem. night 2 everyone dies except the rem, who is incarcerated for mass murder.
day 3, everyone who has waded through this 5 pages (so far) of posts join him in the sanitarium.
seriously tho, there comes a point where the chain logically gets broken, a rem will be colorblind, or too daft to follow, or wimpy to carry through. however, if you are walking around the compound and see people walking around pointing at the rems, and apparently counting, and they point at you in the process 
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Thallion
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Re: BROWN EYES AND RED EYES
« Reply #112 on: Mar 18th, 2006, 11:39pm » |
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Hi, hope I am not dragging up an old sore with this one.
This logic problem bugs the tar out of me. It's my belief that this problem breaks down at larger numbers. I do agree the dieing on n days is the correct formula, but I donot believe it would be triggered by the statement made to a large population of REMs. Note I am talking the original problem here not the additional what if's (colorblind etc..)
The number of BEM is inconsequential in the solving, so I will only address the number of REMs.
We will say there are 10 REM's at the island.
Each BEM thinks the following:
There are Either 10 REM (and I am BEM) or there are 11 REM (and I am one of them).
Each REM thinks the following:
There are Either 9 REM (I am BEM) or there are 10 (and I am REM)
The REM's want to be BEM, so they are each HOPING there are only 9 REM's so they want the Other REM's to be considering the possiblity of either 9 or 8 REM's. (this is the start of the induction and sequencing giving as Nth night solution) However the problem is None of the other REMs are thinking these possabilities.
Making the jump to continue the cycle downward after noone is considering the possible outcomes in my opinion makes the statement of "I see red eyed monks" mean nothing to the monks. (they knew this already) There is nothing that says REMs should start dying after the statement.
The fact giving of confirming 0 REM's (which breaks the bottom tier If statement and theoretically starting the cycle) was already known by every single Monk.
We have proven beyond a doubt for small numbers the formula is obvious. But what I am failing to see is why everyone is granting that the cycle would start with the tourists statement.
However if the problem had been along the lines of:
"Monks who live with vows of silence, and no mirrors, have either Brown or Red eyes, none of them know their own eye color. One day their god decrees that the red eye monks are cursed and if you are a red eye monk you should kill yourelf at midnight." I would agree the cycle would be started.
The difference is not that they learn there are REMs, (they already knew that, though they donot know for sure if they are one or not) the difference is they learn they should kill themselves. Now they are watching the other monks to see if they kill themselves or not. The change of action starts the cycle not "new" knowledge. (they all already knew there were REMs.
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rmsgrey
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Re: BROWN EYES AND RED EYES
« Reply #113 on: Mar 19th, 2006, 8:19am » |
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OK, let's try this variation. There are 6 monasteries, numbered 0,1,2,3,4,5, each with that many REMs. The monks all know this, but don't know which monastery is which (and aren't allowed to visit to find out either)
When the inconsiderate tourist visits all 6 monasteries in one day, at 0 he says nothing, but at each other, he starts the process going.
At 1, the lone REM realises he's not at 0 after all, and duly suicides at midnight
At 2, each REM is hoping they're living at 1, so hopes to find the other dead in the morning. When they meet, they realise the awful truth and suicide the second night.
At 3, each REM hopes they're at 2, realises on the second morning that they're not, and suicides on the third night
At 4, the REMs hope they're at 3 (where there will be bodies on the third morning) and duly suicide on the fourth night.
Or taking things in the other direction:
At 5, all 5 REMs (call them A,B,C,D,E) are hoping they're living at 4. In particular, E hopes that A,B,C and D are the only REMs. So E hopes that D is hoping he lives at 3, in which case D would hope that C hopes he's living at 2, in which case C would be hoping that B hopes he's at 1 where A has just given up hope that he's living at 0.
So, before the tourist visits, E hopes that they're at 4 and that D hopes that they're at 3 and that C hopes that they're at 2 and that B hopes that they're at 1 and that A hopes that he's living at 0.
When the tourist makes his announcement, it's clear that no-one could possibly expect A to believe he could be at 0, so now E hopes that they're at 4 and that D hopes that they're at 3 and that C hopes that they're at 2 and that B hopes that they're at 1 and that A knows it.
The next day, it's clear that no-one could possibly believe that A's the only REM, so E hopes that they're at 4 and that D hopes that they're at 3 and that C hopes that thy're at 2 and that B and A know it.
The following day, E hopes that they're at 4 and that D hopes that they're at 3 and that C, B and A know it.
Next day E hopes that they're at 4 and that D, C, B and A all know it.
On the final day, E, D, C, B and A all realise they're at 5, much to F's relief the following morning.
It's not important that everyone knows there's at least one REM, or even that they know everyone knows everyone knows everyone knows there's at least one REM. The point is that the 5 REMs don't know that everyone knows everyone knows everyone knows everyone knows there's at least one REM until the tourist speaks...
The argument is much simpler to follow if you start at the bottom and work up - if you accept that the REMs at monastery k will suicide on the k'th night, then the REMs at monastery k+1, who are smart enough to also have realised this, will realise they can't possibly be at k when they wake up after the k'th night and there hasn't been a massacre. The only thing keeping REMs alive before the tourist visits is the fact that none of the monks at any monastery could know for sure which monastery they're at, so no monastery would do anything different from the others to enable it to be distinguished. The tourist's comment distinguishes 0 from 1. The events of the first night distinguish 1 from 2. The events of the second night distinguish 2 from 3. The events of the third night distinguish 3 from 4, and so on.
Any monks unsure whether they're at k or k+1 will find out when they discover what happens on the k'th night. If not, then what's the minimum number of REMs required for no-one to suicide?
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Thallion
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Re: BROWN EYES AND RED EYES
« Reply #114 on: Mar 19th, 2006, 8:25pm » |
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Ok let me state it simplier..
I agree if the process starts that for any N number of REMs they will kill themselves on the Nth Night. (and I said this above as well)
I donot agree that telling the monks there are REM's when they already know this would start the process given N is a large number of REMs. There is absolutely no change in the monks perceptions or knowledge level or proper etiquette. (It apprears to me the critical number is around 5 from below)
My postulation was that for larger numbers the chain would not start, with this information. For small numbers it obviously does.
Using a single monestary with 5 REMs (A,B,C,D,E)
A is hoping there are 4 REMs, thus He Hopes B is seeing only 3. Further hoping B hopes C sees only 2.
However A knows C cannot be seeing just 2 Because you know B is a REM. C can only hope there are just 3 REMs, because, A and C both see BDE, so A knows C cannot think there are only 2 REMs.
If everyone knows that no-one can hope there are just 2 REMs then In My Opinion the chain would not start when given the information as presented.
The monks would not jump to the Theoretical case of only 2 monks as they know NONE of the other monks can even HOPE that is a possability.
Thats where the chain breaks in a single monestary problem.
Having multiple monestaries in your case would occur as forecast by N Rems killing themselves on the Nth night at each monestary. (only of course if they communicated that monks had died at other monestaries on each night)
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towr
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Re: BROWN EYES AND RED EYES
« Reply #115 on: Mar 20th, 2006, 2:04am » |
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on Mar 19th, 2006, 8:25pm, Thallion wrote:| I donot agree that telling the monks there are REM's when they already know this would start the process given N is a large number of REMs. |
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But it isn't that piece of knowledge that starts the cycle.
The tourist's statement delivers more. He says that there's at least one, but to the monks it means that now everyone knows that everyone knows that everyone knows ..<arbitrarily many 'that everyone knows'>.. that there is at least one REM.
It's now common knowledge, whereas before it was not. Sure everyone knew there were a certain minimum, but it wasn't common knowledge.
It's like the puzzle of the three armies, one large one in the middle, and two smaller allies on opposite sides. The two allies have to agree when to attack, because they can't win on their own. But any messenger they send might be caught.
So you send one messenger over, but you don't know if he arrived untill they send one back. But they don't know if their messenger arrived untill you send another one back. So you can never be sure the last messenger arrived and will always be slightly unsure if the plan goes through or not.
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| « Last Edit: Mar 20th, 2006, 2:07am by towr » |
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Thallion
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Re: BROWN EYES AND RED EYES
« Reply #116 on: Mar 20th, 2006, 5:00am » |
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Ah, but that is not true. With 5 REM's all of them know that all of the others know there is more than 1 REM.
A believes he is not a REM. He knows the other 4 are REM's so he knows all of the other monks know that there are at least 3 REMs. This is fact. Regardless of his own Eye color. He Knows that all of the other monks see at least 3 REMs, as he sees the same monks.
Everyone already knew there is at least 1 REM. It was common knowledge that every monk saw more than 1 REM. Wether or not the fact is stated outloud is irregardless of the monks perception of the number of REMs (for a large enough N)
This is why I think 5 is the critical Number(though 4 is on the edge). If none of the monks can consider the 2 monk case, then I do not think the chain would start. And with 5 REMs every monk knows the other monks see at least 3 REMs.
Hence my current position that given a large number of REMs (5 seems enough to me) the given information will not start the chain reaction.
To put it another way:
Given N number of REMs.
Everyone knows that everyone else knows that there are AT LEAST N-2 REMs.
Derived from assuming they are not a REM, they know the other REMS see at least the same monks that they see in common. Namely not themselves and not the other viewer. Hence N-2.
If N-2 > 2 Then the cycle cannot start by the information given. As they all knew 1 was not a possability, hence nothing changes by seeing no-one commit suicide, as that is the same event that has happened every night for the last years and years.
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| « Last Edit: Mar 20th, 2006, 5:09am by Thallion » |
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towr
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Re: BROWN EYES AND RED EYES
« Reply #117 on: Mar 20th, 2006, 5:28am » |
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on Mar 20th, 2006, 5:00am, Thallion wrote:| Ah, but that is not true. With 5 REM's all of them know that all of the others know there is more than 1 REM. |
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Yes, but as I said, they don't know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they know that they etc..
Suppose we have a total of 3 REMs. Each of them obviously knows that every other monks must see at least one, regardless of wether he is one himself.
However, A doesn't know whether B knows if C can see a monk with red eyes, even though he himself does know that C sees one.
Because if A doesn't have red eyes (as he hopes), and A thinks that B thinks he doesn't have red eyes (as B would hope) than C wouldn't see any red eyes. And thus wouldn't kill himself unless someone came buy to tell them that at least one of them does in fact have red eyes.
And that works for all N > 0 You need all levels of knowledge, so the monks can reason about eachother's reasoning.
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Icarus
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Re: BROWN EYES AND RED EYES
« Reply #118 on: Mar 20th, 2006, 3:25pm » |
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The change in information that the tourist provides is in the statement "If there was only one REM, he would know he was an REM." Before the tourist's unfortunate remark, everyone knows this statement to be false. After the remark, everyone knows that it is true.
It is from the truth of this remark that the monks can derive the statement "For all K, if there were exactly K REMs, they will commit suicide on the Kth night". This derivation has nothing to do with how many REMs really exist. It is strictly hypothetical.
Only after this statement is derived does the real situation comes into play. If there are actually N REMs, then the REMs know that there are either N-1 or N REMS, while the BEMs know that there are either N or N+1 REMs. So they all wait for N-1 nights (the BEMs are actually waiting for night N). When nothing happens on night N-1, the REMs all realize there is one more REM than they can see.
It does not matter that everyone knows there are at least N-1 REMs. The information change and deduction are not based on what the actual monks know. It is based on what hypothetical numbers of REMs would know if the hypothetical situation were real.
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rmsgrey
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Re: BROWN EYES AND RED EYES
« Reply #119 on: Mar 20th, 2006, 6:34pm » |
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Quote: Using a single monestary with 5 REMs (A,B,C,D,E)
A is hoping there are 4 REMs, thus He Hopes B is seeing only 3. Further hoping B hopes C sees only 2.
However A knows C cannot be seeing just 2 Because you know B is a REM. C can only hope there are just 3 REMs, because, A and C both see BDE, so A knows C cannot think there are only 2 REMs. |
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A knows that C sees at least 3 REMs (BDE), and B knows that C sees at least 3 REMs (ADE), but A, who's hoping the only REMs are the ones he knows about (BCDE) hopes that B only knows about C seeing at least 2 REMs (DE). If A's hope were true, then B (only seeing CDE) would hope that C only saw DE.
A knows C cannot think that there are only 2 REMs, but doesn't know that B also knows this, so hopes that B hopes that C sees only 2 (DE).
Similarly, D sees 4 REMs (ABCE), and C knows D sees at least 3 REMs (ABE) but B only knows that C knows that D sees at least 2 REMs (AE) and A only knows about B knowing about C knowing about D knowing about the 1 REM (E) because A doesn't know there's anything to know about himself being one.
Finally, while B knows that C knows that D knows that E knows about 1 REM (A), A doesn't know that, so hopes B hopes that C hopes that D hopes that E doesn't know there are any REMs.
Once the tourist speaks, A knows that B knows that C knows that D knows that E knows that at least one of ABCDE is an REM.
Once the first night passes, A knows that B knows that C knows that D knows that E isn't the only REM, so A knows that B knows that C knows that D knows that at least one of ABCD is an REM.
After the second night, A knows that B knows that C knows that D saw a second REM as well as E so A knows that B knows that C knows that at least one of ABC is an REM
After the third night, A knows that B knows that C saw a third REM as well as DE, so A knows that B knows that at least one of AB is an REM
After the fourth night, A knows that B saw a fourth REM as well as CDE, so A knows that A is an REM.
On the fifth night, A kills himself (so do BCDE having followed the same logic with the letters assigned differently)
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Thallion
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Re: BROWN EYES AND RED EYES
« Reply #120 on: Mar 20th, 2006, 9:37pm » |
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on Mar 20th, 2006, 5:28am, towr wrote:Yes, but as I said, they don't know that they know that they know that they know that they know ...
Suppose we have a total of 3 REMs. Each of them obviously knows that every other monks must see at least one, regardless of wether he is one himself. |
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In a case of N-2 > 2 they DO know that they know that they know. 3 is not large enough for n-2 >2 to be true, so in your example case I agree the chain starts.
on Mar 20th, 2006, 6:34pm, rmsgrey wrote:
A knows that C sees at least 3 REMs (BDE), and B knows that C sees at least 3 REMs (ADE), but A, who's hoping the only REMs are the ones he knows about (BCDE) hopes that B only knows about C seeing at least 2 REMs (DE). If A's hope were true, then B (only seeing CDE) would hope that C only saw DE.
A knows C cannot think that there are only 2 REMs, but doesn't know that B also knows this, so hopes that B hopes that C sees only 2 (DE). |
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Agree - A knows BCDE, A Wants B to only know CDE(even though wrong), A wants B to want C to only see DE. A knows that is not possible.
Thus A knows (as BCDE all have the same knowledge) That Each other person already knows there are at least 3 REMs. The best case hope is that each of the others are hoping that the others are only seeing 2. But A knows the others know that everyone sees at least 2.
Hence N-2 > 2 became my theory of when the process breaks.
on Mar 20th, 2006, 3:25pm, Icarus wrote:The change in information that the tourist provides is in the statement "If there was only one REM, he would know he was an REM." Before the tourist's unfortunate remark, everyone knows this statement to be false. After the remark, everyone knows that it is true.
... (cut for length not due to content) |
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This is a very good explaination, One of these best I have seen. But this is not what the tourist says. Though I see where you can make that conclusion of his statement.
I can't really contradict that the process would start if that had been the statement made by the tourist. If though is the key word. Which is also the key to this whole problem.
IF brings in the theoretical cases, and indeed if you worked the problem UP and you said "I see Red Eyes, if there were only one person he would know he had them." Indeed the process would start. However It's my ascertation that people would not jump to the hypothetical if the statement was only "i see red eyes" as it is in the problem.
When the Tourist says, "I see Red Eyes" Each monk (given N of REMs with N-2 > 2 is true) would simply say to themselves, "DUH so do I"
Recursive thought is not a natural tendancy, and this problem is recursive logic when done from the monks perspective. Thats why given a large enough N the knowledge that there are REMs wouldn't change anything.
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| « Last Edit: Mar 20th, 2006, 10:21pm by Thallion » |
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Thallion
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Re: BROWN EYES AND RED EYES
« Reply #121 on: Mar 20th, 2006, 11:32pm » |
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Most people when I propose this solution get hung up on the chain of events, so let me be clear. This is a chain reaction problem. Given that the chain starts the chain Will follow the pattern N REMs will die on the Nth Night. No arguements from me.
The question becomes "what is truely required to start the chain?"
Chains start because of a state change. Adding more vinegar to a beaker wont do anything it's only when you add in baking soda does anything start to happen.
If everyone knows that everyone else knows there is at least 1 REM, then the statement there is at least 1 REM, does not constitute a change in state.
Hence my postulation of if N is sufficiently large the information given is not enough. In this case I use N-2>2 though you could simply say n>4. But I like the n-2 >2 Because it indicates the level of knowledge that each person would have, as well as the cut off point, where that knowledge is important. The first side is each person knows that every other person knows there are at least n-2 REMs as both the viewer and every other person will see that same n-2 set of people. The second side of the equation is the number that indicates when the information there is at least 1 REM becomes valueable. At only 2 REMs they hope the other doesnt see any and hence would provide that other person with the knowledge that they are the only one.
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| « Last Edit: Mar 20th, 2006, 11:44pm by Thallion » |
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towr
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Re: BROWN EYES AND RED EYES
« Reply #122 on: Mar 21st, 2006, 2:21am » |
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on Mar 20th, 2006, 9:37pm, Thallion wrote:| In a case of N-2 > 2 they DO know that they know that they know. 3 is not large enough for n-2 >2 to be true, so in your example case I agree the chain starts. |
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It doesn't matter wether it's 3 or 4 or 5 million.
if there are N REMs they consider that the other N-1 consider there may be only N-2 that consider there are N-2 ... etc untill you come to one REM that considers there is hopefull no REM.
And every REM has such a chain leading to zero. You can model any N REM as an N dimensional hypercube where each monk considers all worlds along one direction equally possible (the ones where he is and the ones where he isn't REM himself) starting at any of those world (where a subset of N has red eyes) you can move along the edges to a world where none of the monks has red eyes.
As long as any monk can hope that some monk hopes that ... etc that a monk hopes there is no REM, then they all live on happily doing what they do. The tourist destroys this hope by make a public announcement that there is at least one. So the no-REM world is no longer possible, and no path can lead there. And so the hypercube unravels step by step untill monks start killing themselves. (Each Kth steps all world with K-1 REMs are eliminated)
Quote:| This is a very good explaination, One of these best I have seen. But this is not what the tourist says. Though I see where you can make that conclusion of his statement. |
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And the monks being infinitely smart do exactly that. What the monks know is everything that follows logically from what is said and what they know and see.
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| « Last Edit: Mar 21st, 2006, 2:42am by towr » |
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Re: BROWN EYES AND RED EYES
« Reply #123 on: Mar 21st, 2006, 2:38am » |
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on Mar 20th, 2006, 11:32pm, Thallion wrote:| If everyone knows that everyone else knows there is at least 1 REM, then the statement there is at least 1 REM, does not constitute a change in state. |
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It does if it's a public statement, because it changes the beliefs about beliefs (etc)
Let's look at the case of 4 monks A,B,C,D
using some mathematical abbreviation (and for any permutation of A,B,C,D)
we have at the start:
KA(>=3) {A knows that there are >= 3 REM)
KAKB(>=2) {A knows that B knows ...)
KAKC(>=2)
KAKD(>=2)
KAKBKC(>=1)
KAKBKD(>=1)
KAKCKB(>=1)
KAKCKD(>=1)
KAKDKB(>=1)
KAKDKC(>=1)
KAKBKCKD(>=0)
KAKBKDKC(>=0)
KAKCKBKD(>=0)
KAKCKDKB(>=0)
KAKDKBKC(>=0)
KAKDKCKB(>=0)
Now the publis statement by the tourist adds the relevant knowledge update
K*(>=1)
K*K*(>=1)
K*K*K*(>=1)
K*K*K*K*(>=1)
where * is a wildcard for any A,B,C,D (so all 4,16,64,256 combinations respectively).
So the relvant change in knowledge is that the last 6 lines become
KAKBKCKD(>=1)
KAKBKDKC(>=1)
KAKCKBKD(>=1)
KAKCKDKB(>=1)
KAKDKBKC(>=1)
KAKDKCKB(>=1)
Because of this change people start dying. And it works for any N, it'd just be more work to write out. (not to mention I've already left out a lot)
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Re: BROWN EYES AND RED EYES
« Reply #124 on: Mar 21st, 2006, 4:52am » |
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To illustrate the hypercube model, here's a picture (for 4 REMs):
magenta connects worlds indistinguishable to the first REM
green connects worlds indistinguishable to the second REM
blue connects worlds indistinguishable to the third REM
red connects worlds indistinguishable to the fourth REM
(All worlds are also reflexive for all monks, of course, but it's cluttered enough as it is)
For any actual world you start in, you can get to 0000 in 4 steps or less along different colored edges. Meaning that some monk believes it's possible that some other monk believes it's possible that a third monks believes it's possible that the fourth believes there are possibly no REMs and won't kill himself because of that..
[edit]oops, there's a typo in the picture, there's two worlds named 1101, the left one should be 1100[/edit]
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| « Last Edit: Oct 24th, 2008, 2:18pm by towr » |
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