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Blue Eyes logic puzzle
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Ayvielle
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PostPosted: Thu Mar 05, 2009 3:26 am    Post subject: Blue Eyes logic puzzle Reply with quote

Lifted from the xkcd blog:

Quote:

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?


There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."


I know the solution and I think I understand it pretty well. If anyone wants to ask clarifying questions I'll do my best to answer them. Anyone care to have a go at it?
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TFBW
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PostPosted: Thu Mar 05, 2009 1:17 pm    Post subject: Reply with quote

What a truly bizarre thing for the guru to say. If everyone can see everyone else at all times, then everybody can see someone with blue eyes at all times, and knows that everyone else can also see someone with blue eyes. Or, to put it another way, the people with blue eyes know that there are 99 or 100 blue-eyed people, and the people without blue eyes know that there are 100 or 101 blue-eyed people. The guru's statement, "I can see someone with blue eyes," should provoke a reaction of, "tell us something we don't know, you useless guru."

I get the feeling that this particular proof may be controversial, like the prisoner paradox.
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PostPosted: Thu Mar 05, 2009 2:58 pm    Post subject: Reply with quote

TFBW wrote:
What a truly bizarre thing for the guru to say. If everyone can see everyone else at all times, then everybody can see someone with blue eyes at all times, and knows that everyone else can also see someone with blue eyes. Or, to put it another way, the people with blue eyes know that there are 99 or 100 blue-eyed people, and the people without blue eyes know that there are 100 or 101 blue-eyed people. The guru's statement, "I can see someone with blue eyes," should provoke a reaction of, "tell us something we don't know, you useless guru."

I get the feeling that this particular proof may be controversial, like the prisoner paradox.


I will say that despite what it looks like, the Guru is introducing new information for the others.
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Ayvielle
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PostPosted: Thu Mar 05, 2009 2:59 pm    Post subject: Reply with quote

Anonymous wrote:
TFBW wrote:
What a truly bizarre thing for the guru to say. If everyone can see everyone else at all times, then everybody can see someone with blue eyes at all times, and knows that everyone else can also see someone with blue eyes. Or, to put it another way, the people with blue eyes know that there are 99 or 100 blue-eyed people, and the people without blue eyes know that there are 100 or 101 blue-eyed people. The guru's statement, "I can see someone with blue eyes," should provoke a reaction of, "tell us something we don't know, you useless guru."

I get the feeling that this particular proof may be controversial, like the prisoner paradox.


I will say that despite what it looks like, the Guru is introducing new information for the others.


^ sigh, forgot to log in
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TFBW
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PostPosted: Thu Mar 05, 2009 3:42 pm    Post subject: Reply with quote

In that case I can say with confidence that I don't understand the problem. It's clear to me from the problem description that everybody can see several people with blue eyes, and everybody knows that everybody can see several people with blue eyes. The Guru's statement is thus clearly (to me) a sub-statement of already known facts. More precisely, each blue-eyed person knows (before the guru speaks) that the Guru can see 99 blue-eyed people, 100 brown-eyed people, and one person of unknown eye colour. The brown-eyed people have the same knowledge but with the 99 and 100 swapped. "I can see someone who has blue eyes" comes as no surprise to anyone: everyone already knows she can see a lot of people with blue eyes.

If the guru's statement does introduce new information, then I must not understand the environment correctly, so there's no point racking my brains over it. I'll wait until you post the solution and then see if I can figure out what vital fact I'm overlooking in the description.
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Ayvielle
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PostPosted: Thu Mar 05, 2009 11:06 pm    Post subject: Reply with quote

TFBW wrote:
In that case I can say with confidence that I don't understand the problem. It's clear to me from the problem description that everybody can see several people with blue eyes, and everybody knows that everybody can see several people with blue eyes. The Guru's statement is thus clearly (to me) a sub-statement of already known facts. More precisely, each blue-eyed person knows (before the guru speaks) that the Guru can see 99 blue-eyed people, 100 brown-eyed people, and one person of unknown eye colour. The brown-eyed people have the same knowledge but with the 99 and 100 swapped. "I can see someone who has blue eyes" comes as no surprise to anyone: everyone already knows she can see a lot of people with blue eyes.

If the guru's statement does introduce new information, then I must not understand the environment correctly, so there's no point racking my brains over it. I'll wait until you post the solution and then see if I can figure out what vital fact I'm overlooking in the description.


Hint: Say there are 5 people on the island - 2 with blue eyes, 2 with brown eyes, and the Guru with green eyes. It's still true that everyone can see someone with blue eyes, and everyone knows the Guru can see someone with blue eyes. But does everybody know that everybody knows the Guru can see someone with blue eyes? How does this generalize?
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Bezman
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PostPosted: Fri Mar 06, 2009 11:32 am    Post subject: Reply with quote

OK, I used Google and cheated. I agree to the solution for 1 and 2 and even 3 people, but start at FIVE people (+100 other-coloured +1 guru) - I need to think through that scenario to verify that it holds up.
Or can you do it for me in a PM and save me the effort?
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Ayvielle
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PostPosted: Sat Mar 07, 2009 5:37 pm    Post subject: Reply with quote

Bezman wrote:
OK, I used Google and cheated. I agree to the solution for 1 and 2 and even 3 people, but start at FIVE people (+100 other-coloured +1 guru) - I need to think through that scenario to verify that it holds up.
Or can you do it for me in a PM and save me the effort?


I sent you a PM, let me know if you're still confused.
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TFBW
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PostPosted: Sun Mar 08, 2009 8:12 am    Post subject: Reply with quote

I haven't looked at the official solution yet (via the Bezman method), or any discussion surrounding it (if there is any), but I'm leaning towards the view that the correct answer is "no one leaves". I think I can guess how a different conclusion might be reached thanks to a subtle error in logic.

In the near future I plan to post my thoughts on the matter, unless someone requests more time to think about it without external influence. My guess at the "proof" will be just that: a guess, as opposed to a critique of the actual proof.
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Bezman
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PostPosted: Sun Mar 08, 2009 12:51 pm    Post subject: Reply with quote

Ayvielle wrote:
Bezman wrote:
OK, I used Google and cheated. I agree to the solution for 1 and 2 and even 3 people, but start at FIVE people (+100 other-coloured +1 guru) - I need to think through that scenario to verify that it holds up.
Or can you do it for me in a PM and save me the effort?


I sent you a PM, let me know if you're still confused.

No, but I still have my doubts as to whether it holds up. Whether I can prove that it doesn't - that's another story... Smile
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Ayvielle
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PostPosted: Sun Mar 08, 2009 6:25 pm    Post subject: Reply with quote

TFBW wrote:
I haven't looked at the official solution yet (via the Bezman method), or any discussion surrounding it (if there is any), but I'm leaning towards the view that the correct answer is "no one leaves". I think I can guess how a different conclusion might be reached thanks to a subtle error in logic.

In the near future I plan to post my thoughts on the matter, unless someone requests more time to think about it without external influence. My guess at the "proof" will be just that: a guess, as opposed to a critique of the actual proof.


Could you PM me what you think the "different conclusion" and "subtle error in logic" are?
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TFBW
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PostPosted: Mon Mar 09, 2009 7:11 am    Post subject: Reply with quote

This is indeed a very tricky logic problem, and it perhaps reveals more about one's logical school of thought than anything else. Logic isn't just a single set of rules over which there is no disagreement, and peculiar problems like this tend to highlight the differences between different sets of rules. The major failing of the question is thus the assertion that the participants are "perfect logicians" without reference to the school of logic in which they are perfect.

My guess is that the person who phrased the question was very much inclined to inductive reasoning, where one establishes a base case and shows that every other case follows in a uniform manner. I base this guess on the phrasing of the guru's remark: the assertion that there is a blue-eyed person. This assertion is absolutely key to the inductive base case, where there is only one blue-eyed person. Under those circumstances, the blue-eyed person can deduce that he is the blue eyed person because he can observe that everyone else is not blue eyed. That's the easy part. The inductive part is much harder: it involves people considering the question "how will other people behave if I have blue eyes or not" and also knowing how other people will act when they ask themselves the same question.

The problem that I see with this approach is that it assumes people will be asking themselves whether they have blue eyes or not, rather than whether they have brown eyes or not, or green eyes or not. Each person is testing for one of two possible outcomes: each computes what will happen if his eyes are not blue, and concludes that his eyes are blue if this fails to eventuate. But this process requires that everyone be aware of what other people are testing: if some are testing for blueness, while others are testing for brownness, then the whole thing falls apart.

The question seems to assume that the guru's mention of the colour "blue" will be sufficient to make everyone ask themselves the "blue or not" question. I think that this is an invalid assumption, and it renders the proof unsound. Here's the reasoning behind that assertion.

As I said earlier, the guru's remark adds no new information: as a matter of logical deduction, everyone in the crowd already knows that the guru can see someone with blue eyes, and everyone knows that this knowledge is universal. (This condition can be proved when there are three or more blue-eyed people in the crowd.) In the same way, everyone in the crowd knows that the guru can see someone with brown eyes (there are more than three brown-eyed participants), and thus could also have said, "I see someone with brown eyes." People thus have two obvious and equally reasonable hypotheses to test: "I have blue eyes", and "I have brown eyes", both of which could be false. Nobody can be sure which of these hypotheses another is testing, and thus can not gain useful knowledge from anyone else's actions.

Under this line of reasoning, everyone stays on the island forever.

Even if we tidy up the problem by changing the rule such that the guru declares "those of you with blue eyes may leave; the rest must stay" (thus ensuring that everyone tests the "I have blue eyes" hypothesis), pure inductive reasoning still isn't entirely "logical". It works correctly, but relies on people behaving in ways that they know to be contrary to fact: i.e. "irrationally". Inductive reasoning works from the base case through to the general case, such that "if there were only one person with blue eyes, he would deduce this and leave immediately." The problem is that in cases with three or more blue-eyed people present, there isn't only one person with blue eyes, and everyone knows it! How can it be rational to extrapolate from the counterfactual like this? When there are four or more blue-eyed people, everyone knows that nobody will leave immediately, so the fact that nobody leaves on the first day reveals nothing! This is just like the guru's statement which revealed nothing. If everyone can deduce that nobody will leave on the first day, shouldn't they all just automatically skip the first day and proceed as though it were the Nth day, where N is the minimum number of days that everyone knows everyone else will have to wait otherwise? Or is it not possible to compute such a common minimum, given the unknowns?

Figure that one out for yourselves. My brain hurts enough from this nonsense already! I'll say one thing, though: in some ways this is not so much a "logic" problem as a problem in optimisation. The question then becomes, "what common set of rules can the participants follow such that those who can leave do so in the minimum possible time?" Logic still enters into it, of course, but I doubt that a logician would call it a logic problem when phrased this way: it's too mathematical.
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Bezman
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PostPosted: Mon Mar 09, 2009 10:54 am    Post subject: Reply with quote

TFBW wrote:
...The problem is that in cases with three or more blue-eyed people present, there isn't only one person with blue eyes, and everyone knows it! How can it be rational to extrapolate from the counterfactual like this? When there are four or more blue-eyed people, everyone knows that nobody will leave immediately, so the fact that nobody leaves on the first day reveals nothing! ...

We have come to the same conclusion, TFBW and me.
I also agree with his continued reasoning (inclusive of the "hurting brain" part! Laughing) - why they would not just skip forward if they knew.

Now that Brett has mentioned this much, I may add my 2 bits as well: if you start with the case of 1 person and induct to person 4, you have come to a case where case 1 is no longer valid.
I'm testing for this right now...poor brain.
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PostPosted: Mon Mar 09, 2009 5:25 pm    Post subject: Reply with quote

TFBW wrote:
The major failing of the question is thus the assertion that the participants are "perfect logicians" without reference to the school of logic in which they are perfect.


They are perfect logicians in that if there is a way to find out what their own eye colour is, they will find it. That's why they ask themselves whether or not their eyes are blue. Because that's what leads them to find out that they are. Based on the Guru's statement there is no way that anyone without blue eyes will ever be able to leave the island, and the islanders know this. As a result, "not blue" and "blue" are the only two relevant categories.

[quote="TFBW]As I said earlier, the guru's remark adds no new information: as a matter of logical deduction, everyone in the crowd already knows that the guru can see someone with blue eyes, and everyone knows that this knowledge is universal.[/quote]

In the case of 3 people with blue eyes, everyone knows that everyone knows that there is someone with blue eyes. But they do not know that everyone knows that everyone knows that everyone knows that there is someone with blue eyes. Here is why: In the case of 2 blue-eyed people, you can say "everyone knows there is someone with blue eyes" but you can't say "everyone knows that everyone knows there is someone with blue eyes" because each of the blue-eyed people only see 1 other blue-eyed person, and if their own eyes are not blue, that other person will not know there is a blue-eyed person. In the case with 3 people with blue eyes, each blue-eyed person can only see 2 blue-eyed people, and they have no way of knowing they have blue eyes - so they don't know that everyone knows that everyone knows that everyone knows that there is someone with blue eyes, until the Guru says something.

TFBW wrote:
Even if we tidy up the problem by changing the rule such that the guru declares "those of you with blue eyes may leave; the rest must stay"


This is not tidying up the problem at all - in fact, it is changing it fundamentally in that now no one can ever know their own eye colour.

[quote=TFBW[/quote]The problem is that in cases with three or more blue-eyed people present, there isn't only one person with blue eyes, and everyone knows it! How can it be rational to extrapolate from the counterfactual like this?[/quote]

It's not irrational to consider what would happen if something were the case, even if it's not true that it is the case. From someone else's perspective, it might be the case, so when things don't happen as predicted in that case, that person will realize it is not the case. It does give information in this case, because as I described above, each blue-eyed person can only see 2 other blue-eyed people, and if it were the case that there actually were only 2 other blue-eyed people, certain behaviour would occur. (i.e. they leave on the second night). It's because the perspective of each blue-eyed person lends us the n-1 case that knowing what would happen in the n-1 case is extremely important. Because when they don't actually behave along those lines, each blue-eyed person is able to conclude that there are not exactly 2 blue-eyed people, there is one more - namely, themselves. Similarly the 2 blue-eyed people case reduces to the 1 blue-eyed person case from the perspective of each blue-eyed person.

TFBW wrote:
When there are four or more blue-eyed people, everyone knows that nobody will leave immediately, so the fact that nobody leaves on the first day reveals nothing! This is just like the guru's statement which revealed nothing. If everyone can deduce that nobody will leave on the first day, shouldn't they all just automatically skip the first day and proceed as though it were the Nth day, where N is the minimum number of days that everyone knows everyone else will have to wait otherwise? Or is it not possible to compute such a common minimum, given the unknowns?


Except it is not true that everyone knows that everyone knows that everyone knows that everyone knows that no one will leave immediately. It is just like the guru's statement, but you are wrong in that it gives no information. Blue-eyed person #1 needs to know that blue-eyed person #2 knows that blue-eyed person #3 knows that blue-eyed person #4 knows that no one is leaving on the first night.

TFBW wrote:
Logic still enters into it, of course, but I doubt that a logician would call it a logic problem when phrased this way: it's too mathematical.


Haha, logic is a field within mathematics (also with philosophy, but both fields of study are very very similar).
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Ayvielle
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PostPosted: Mon Mar 09, 2009 5:26 pm    Post subject: Reply with quote

^ forgot to log in, again! argh! Razz
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Ayvielle
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PostPosted: Mon Mar 09, 2009 5:30 pm    Post subject: Reply with quote

Just to clarify on something just in case it's not clear:

I know TFBW's beef is that the islanders wouldn't consider a case that isn't true, but Bezman's beef is something else. Bezman, I know I've said it before in PMs, and I'm just not sure how else to explain it. The fact that "A is true" isn't important, the only thing that we need is "IF A THEN B". The latter statement is true whether or not the former statement is true, so long as whenever A is true, B is also true. There is nowhere in any of the proofs I've given you where I have to rely on A being true when it's not.
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PostPosted: Wed Mar 11, 2009 5:47 am    Post subject: Reply with quote

Okay, I've heard a variant on this problem where there are only three people and two colors. Let me see if I can reproduce it...

There are three people sitting so each can see both of the others, and someone puts a hat on each of them. So they can see the hats on the others, but not on themselves. They know the hats can come in red or blue. All three hats are red. After the hats are placed, they are told to raise one hand if they can see at least one red hat. Everyone can see two red hats, so they all raise a hand. Then they sit there for a long time, trying to figure it out. Finally the smartest one says he knows his hat is red.

His reasoning is thus: Label the people A, B, and C, where the smart guy is person A. Suppose A has a blue hat. Now consider the problem from the viewpoint of person B. B sees a red hat and a blue hat, and he sees that C has his hand raised. So C can see a red hat, but A's hat is blue, therefore the red hat that C can see belongs to B. So B would quickly be able to conclude that his hat was red. Similarly, C would be able to make the same conclusion. Since neither of them knows they have a red hat, they must be looking at the same thing A is looking at: Two red hats. Therefore A has a red hat, the same as the others.

Now, this is not purely deductive reasoning, because it involves the assumption of a reasonable level of competence on the part of B and C, but that doesn't seem like an unreasonable assumption to make under the circumstances. It's not a proof by the standards of pure mathematics, but I think it would hold up in court.

Okay, so does this extend? Let's see, with four people, all red hats, person A could speculate on his hat being blue, which wouldn't change anything, except... If A had a blue hat, then B could speculate about having a blue hat, and if A and B had blue hats, C and D could immediately deduce that their hats were blue. Therefore when they didn't do so, B could conclude that he had a red hat, and since he doesn't, A must have one. Here the logic gets shakier. A has to assume that B is a pretty sharp cookie, but not quite sharp enough to make the same deduction he himself is making.

For N hats, you've got A speculating about one hat, which would let B speculate about two hats, which would let C speculate about three hats, and so on.

Trying to work out how the "perfect logician" condition would affect things. Hmm, if they all make potential deductions instantly, then the fact that no one instantly knew they had a red hat would cause everyone to know that no one could make the deduction...and therefore they could conclude that there was no deduction to be made.

I'm not quite sure if the eye color problem is logically equivalent, and something still seems fishy. I'll have to think it through some more.
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TFBW
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PostPosted: Wed Mar 11, 2009 5:47 am    Post subject: Reply with quote

I was about to concede that I'd made an error, and that the puzzle was sound after all, but I've finally been able to put my finger on the error. (I have been spending WAY WAY too much time on this and I MUST STOP NOW.)

Here's my charitable explanation of the puzzle, showing why it works.

1. Everyone can reasonably suspect that their own eyes are not blue.
2. Everyone knows that someone is definitely wrong about this. (The guru establishes this fact.)

Each person can build a reasonable hypothesis, starting with "my eyes are not blue, and that blue-eyed guy hypothesises that his eyes are not blue and that that the next blue-eyed guy hypothesises that his eyes are not blue..." and so on. The key is that each person must be able to deduce what other people are hypothesising. Each person thinks, "my hypothesis is that I am not blue eyed; the blue eyed people are testing a hypothesis which will show whether my hypothesis is correct, and I just have to wait and watch for their result."

The practical outworking of this chain of reasoning is as follows. If there's only one blue-eyed person, he immediately deduces that his eyes are blue and leaves. Everyone else is watching him to see if he leaves, knowing that if he does not, then he's not the only blue-eyed person. If he's not the only blue-eyed person, they rightly assume, then the other blue-eyed person is ME! This scales up: a person who can see two other blue eyed people thinks "each of those guys is waiting for the other to leave: if they haven't left on the second night, then they're waiting for ME!" And so on for three, four, five, and the rest.

So, for the puzzle as given, one hundred blue eyed people leave on the hundredth night.

Here's my spanner in the works, showing why the puzzle is flawed.

1. Everyone can reasonably suspect that their own eyes are not BROWN.
2. Everyone knows that someone is definitely wrong about this. (Everyone knows that everyone else can see at least one brown-eyed person.)

Each person could reasonably hypothesise "my eyes are not brown" and formulate a test for it, just like the blue-eyed hypotheses above. Of course, this hypothesis can only be tested if the brown-eyed people are also testing the same hypothesis.

The trouble is that there are two obvious possible hypotheses to test -- "my eyes are not blue", or "my eyes are not brown" -- and each hypothesis is only testable if other people are testing the same hypothesis. Neither of these hypotheses is particularly better than the other: one results in blue-eyed people leaving, and the other results in brown-eyed people leaving. Unfortunately, nobody has any grounds for choosing one hypothesis over the other, or for thinking that anyone else will choose one hypothesis over the other, so nobody can construct a testable hypothesis.

Consequently, NO ONE LEAVES! The one hundred blue eyed people who left in the "working" example reached a true knowledge of their own eye colour through a fortunate systematic error: they failed to notice that there was a viable alternative hypothesis.

In light of this, review my "fix" where the guru says, "those of you with blue eyes may leave; the rest must stay."
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Ayvielle
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PostPosted: Wed Mar 11, 2009 6:30 am    Post subject: Reply with quote

No one can ever figure out that they have brown eyes. The Guru only mentions blue eyes, there is definitely no additional information about people with brown eyes and therefore they can never know their own eye colour. That's why testing a brown-eye hypothesis is useless and the logicians will recognize this. The only useful hypothesis to test is a blue-eye hypothesis because it's the only eye colour anyone will ever be able to know they have, if they have it.

The Guru saying "Those of you with blue eyes may leave; the rest must stay" is not a fix, it completely changes the problem and definitely doesn't give any information. The Guru does give information when she says "I can see someone with blue eyes" - the universal knowledge that everyone knows that everyone knows blah blah blah that everyone knows that there is someone with blue eyes.

I'm not sure I understand what you mean by "charitable explanation" but I'm glad you understand the solution now (?).
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TFBW
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PostPosted: Wed Mar 11, 2009 7:03 am    Post subject: Reply with quote

A "charitable explanation" is one that deliberately overlooks flaws -- assumes the best.

Clearly we'll need an independent judge to decide which of us is right. I don't intend to try to prove myself any more than I have already.
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PostPosted: Sat May 30, 2009 7:43 pm    Post subject: Reply with quote

All right, I'm going to start from base case and see if I can figure out where this falls apart. Keep in mind that the islanders don't know that there are an even number of eye colors, or even that blue and brown are the only 2 choices, aside from the guru's. In any case, there must always be at least one blue-eyed person.

In each case, let B represent blue eyes and X represent non-blue eyes.

Case 1: B
So, the "base case" is one person besides the guru. This person obviously has blue eyes. The guru says so, and the person realizes this and leaves.

2: B X
Next, two people. Assume only one has blue eyes. He can see the other does not, and leaves that night.

3: B B
Two people with blue eyes. Each can see the other has blue eyes. If they themselves have non-blue eyes, the other person would leave that day. Because they do not, the next day they both leave, because they both know they have blue eyes, as otherwise, one of them would not be there.

4: B X X
The first person leaves that night, because he sees no one else has blue eyes. In short, case 2 occurs.

5: B B X
The first and second people both expect the other to leave that night if they do not have blue eyes. Because they do not, and they know the third person does not have blue eyes, they know that they must. They both leave the second night. Each B expects the other B to act as if case 2 is true.

6: B B B
Much like the previous case, except that each expects the other two to leave on the second night. Because they do not, they all leave on the third night. Each person would expects to observe case 3 if they do not have blue eyes.

I'm still waiting for this to fall apart somewhere.

7: B X X X
This is again the base case, and B leaves that night.

8: B B X X
This is a essentially a repeat of cases 3 and 5.

9: B B B X
Repeat of 6.

10: B B B B
Each person, if they did not have blue eyes, would expect a repeat of cases 6 and 9. Because they do not, and no one leaves on the third night, they all must leave on the fourth night.

It is at this point that, although I lack a rigorous proof, I understand where the problem's "solution" lies.

TFBW basically got it, except for his spanner. The trouble with his complication is that there is no test for brownness, there is merely a test for "not-blueness". Consider case 8. If of the two X's, one has green eyes, and the other has red eyes, then the color brown is irrelevant. This is why only blue eyed people may leave; the puzzle, as stated, does not rule out the possibility that any individual may have a unique eye color. The guru may have picked any available color in the crowd, including her own, as everyone knows she does not know the color of her own eyes and the base case eliminates her as a factor. Whatever color she picks is the color they end up working with.

The "new information" the guru gives is not the new information she actually says. The guru says that there is at least one blue-eyed person. While everyone already knows this (assuming at least two blue-eyed people in the crowd), the guru's statement provides two things. First, which color to focus on, because as described above, you have to pick a color to establish the base case. Second, it provides a day zero for the crowd to begin testing their own blueness/not blueness.

The puzzle works because the guru provides the focus. Specifically, the people are testing for "not-blueness", and when their test fails, they know they have blue eyes. Because everyone is testing simultaneously, everyone figures it out at the same time.

It's also worth mentioning (and is obvious at this point) that although the puzzle says the people can't speak to each other or communicate, they do end up learning new information with the passage of each day. They essentially are communicating. Each person with blue eyes says, "Looking at all of you, it's not safe for me to leave yet." Additionally, when they do leave, they let everyone else know that their eyes are some color that is not blue. Of course, those people remain stuck on the island, because they are not aware of what other color choices exist.

The boat leaving once per day at midnight is also vital to the puzzle. An empty boat is the signal to examine a new case, because your previous case failed.


Furthermore, this problem can be generalized further. You can have an arbitrary number of colors and an arbitrary number of people with each color. The answer remains that for whichever color the guru chooses, everyone with that color leaves on day X (or X-1, depending on how you count), where X is the number of people who have it.


Finally, I'll point out that a group of "perfect logicians" would know not to test for something such as brownness, because they would know that the guru established a base case of blue. Very Happy
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TFBW
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PostPosted: Mon Jun 01, 2009 2:12 am    Post subject: Reply with quote

tuxedobob wrote:
Finally, I'll point out that a group of "perfect logicians" would know not to test for something such as brownness, because they would know that the guru established a base case of blue.

I quite agree that if they all simultaneously choose to use "blueness" as the test, and they all know that they have all made this tacit decision, then the problem can be solved as suggested by the official solution. I just have difficulty accepting the guru's utterance of the word "blue" as a purely logical basis for this decision. In fact, I confidently assert that it does not constitute a sound logical basis: it is a psychological basis.

As perfect logicians, they know that they must all share a common test criterion in order for their observations of each other to have a known meaning. As perfect logicians, they know that there are two distinct eye colours which are guaranteed to be common knowledge between all parties: blue and brown. As perfect logicians, they know that either of these colours is equally adequate as a test case. As perfect logicians, they know that the guru's statement of the bleeding obvious, "I see someone with blue eyes", does not add any new information, and thus does not act as a tie-breaker between these two perfectly equal candidates.

Psychologically speaking, on the other hand, everyone knows that a tie-breaker is required, so they might all infer from the Guru's statement that "blueness" is the test they should use. This would have worked equally well if the guru had said nothing other than the word "blue", or "blueness", but the so-called "logic puzzle" would still rely on shared psychology to guarantee that they would all get the hint.
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Caldazar
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PostPosted: Thu Jun 04, 2009 11:13 pm    Post subject: Reply with quote

yah I agree with TFBW on this one that the guru's statement doesn't add any new information from a scientific or logical basis because what she says everyone is already aware of. and saying that everyone who has blue eyes may leave would actually be a fix because then everyone would want to test for that as opposed to testing for brown. And you are also incorrect when you say that no one will ever be able to figure out if they have brown eyes because they would assuming that everyone decided to try for brown as opposed to blue.
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tuxedobob
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PostPosted: Thu Jun 25, 2009 4:44 pm    Post subject: Reply with quote

Well, TFBW is saying that if everyone tests for blueness, there is a solution to the problem. If not everyone tests for blueness, there is no solution.

Logically, wouldn't a group of perfect logicians choose the only option that results in an answer? The only base case that the puzzle establishes is that a person has blue eyes.
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Caldazar
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PostPosted: Thu Jun 25, 2009 11:34 pm    Post subject: Reply with quote

yes but once again everyone already knows this...and they could solve the issue yust as well by all trying for brown eyes...the statement in itself does not bring any new information to the case nor any true indication that they should all be testing for blue eyes instead of brown eyes.
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