## My write-up of the "Blue Eyes" solution (SPOILER A

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Very nice work bernerbrau!

bernerbrau wrote:
3. Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

3. On the first 98 of those nights, they *are* verifying something they already know (that there are 99 people with blue eyes), but they are also verifying something they don't know - whether the other 99 *also* know what they know.

I might phrase it with a bit more detail, that on each of the first 98 nights, say night N, they are verifying that the other 99 people also know what they know about the innermost N layers of the nested hypotheticals.
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### Re:

xkcd wrote:It's not just hand-waving. The inductive portion of the proof is as follows:

Given: the K case works. That is, if there are K blue-eyed people, they will all leave on the Kth night.

Suppose there are K+1 blue-eyed people.

Each blue-eyed person can tell by observation that there are either K or K+1 blue-eyed people, depending which they themselves are.

Since they know that K blue-eyed people will leave the Kth night, if they wait through the Kth night and no one leaves, they know there are not K blue-eyed people. The only other possibilitiy is that there are K+1, so they know it includes them. Therefore, they leave the next ((K+1)th) night.

Therefore: If there are K+1 blue-eyed people, they all leave on the (K+1)th night.

So, we've shown that if it's known that K blue-eyed people will leave by the Kth night, then K+1 blue-eyed people will leave by the (K+1)th night. Now start with your base case K=1 or whatnot and go along with the induction.

This is what I was thinking about, but I don't completely agree. I agree that on the (K+1)th night, the K+!th person knows he is blue eyed and leaves the island. However, the situation after this is not exactly the inductive hypothesis. The inductive hypothesis states that if the island has K people initially, they all leave on the Kth night. The situation now is that it had K+1 people until the K+1th night, and now K are left. This doesn't directly imply that the other K people leave, since the inductive hypothesis assumes they had K days with K people to figure their color out, but in this case they have had K+1 days with K+1 people. So saying that it all works out at this point without further justification would be assuming the inductive hypothesis for K+1 while trying to prove it for K+1.

Anyone know how to handle this?
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Brilliant, bernerbrau. However, I'm not quite satisfied with this:
bernerbrau wrote:
1. What is the quantified piece of information that the Guru provides that each person did not already have?

1. The piece of information is "if there were one person on the island with blue eyes, he/she would leave tonight."

I think you've just restated the same thing. In the case where there really is only one blue-eyed person on the island, the piece of information you've suggested doesn't allow them to leave tonight as they don't actually know whether there's any blue-eyed people on the island. I agree that, in the case of two or more people, the fact as you've stated will achieve the same job, but I think it's a piece of information that can be deduced from the Guru's statement, whereas the quantified piece of information is usually given to be...
Spoiler:
That the existence of blue eyes is common knowledge.

But really, this might not be as much of a distinction as I first thought it was.
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### Re: Re:

Ankit1010 wrote:I agree that on the (K+1)th night, the K+1th person knows he is blue eyed and leaves the island. [...] The situation now is that it had K+1 people until the K+1th night, and now K are left.

No, read it again: each one of the K+1 blue-eyed people look at the other K blue-eyed people, see them not leave on night K, and thus determine their blue eyes. So all of the K+1 blue-eyed people determine their eye colour on day K+1 and leave night K+1.

There is no "K+1th person"... the people aren't numbered. Rather, each blue-eyed person individually sees themselves as the K+1th blue-eyed person, and they all leave together on night K+1.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Goldstein wrote:Brilliant, bernerbrau. However, I'm not quite satisfied with this:
bernerbrau wrote:
1. What is the quantified piece of information that the Guru provides that each person did not already have?

1. The piece of information is "if there were one person on the island with blue eyes, he/she would leave tonight."

I think you've just restated the same thing. In the case where there really is only one blue-eyed person on the island, the piece of information you've suggested doesn't allow them to leave tonight as they don't actually know whether there's any blue-eyed people on the island. I agree that, in the case of two or more people, the fact as you've stated will achieve the same job, but I think it's a piece of information that can be deduced from the Guru's statement, whereas the quantified piece of information is usually given to be...
Spoiler:
That the existence of blue eyes is common knowledge.

But really, this might not be as much of a distinction as I first thought it was.

contrary they all know that there are at least 2 different colours on the island. blue and brown NOT including the guru. the information the guru gives them sets off the chain reaction that gives them a counting start. they ALL know without doubt that by day 102, any amount between 99 and 101 people will have left the island. they just wait. day 100 rolls around noone has left the previous night meaning that there are 100 blue eyes they can leave THAT night. when they do and on day 101 there are 100 less people the brown eyes all concluce their eyes arent blue.

make it easier imagine there was one red eye person on the island takin the place of a blue eye. but he treats himself like a blue eye because he has no knowldge otherwise due to the fact the universal knowledge is now a unified objective : everyone knows what everyones target is and what everyone is working towards and therefore knows everyones thinking process being perfect logicians. said red eye would wait and by the time day 100 rolls around 99 people have left the night before, meaning he cant conclude he was a red eye just that he wasnt blue. those are the two outcomes either he figures he is blue eyed or figures he isnt.

the entire puzzle can be remade with Just 100 people on the island who only have blue eyes. again how can one man be sure he isnt the odd man out? doing this process means EVERYONE comes to the same conclusion simultaneously because the guru saying out loud "i see blue eyes" means that EVERYONE can then assume that everyone else is going to be thinking identically : start counting the days. if you are the only one counting the days and day 100 rolls by and noone has left doesnt mean you are blue eyed. just means noone else was doing what you were doing. ie counting
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Goldstein wrote:Brilliant, bernerbrau. However, I'm not quite satisfied with this:
bernerbrau wrote:
1. What is the quantified piece of information that the Guru provides that each person did not already have?

1. The piece of information is "if there were one person on the island with blue eyes, he/she would leave tonight."

I think you've just restated the same thing. In the case where there really is only one blue-eyed person on the island, the piece of information you've suggested doesn't allow them to leave tonight as they don't actually know whether there's any blue-eyed people on the island. I agree that, in the case of two or more people, the fact as you've stated will achieve the same job, but I think it's a piece of information that can be deduced from the Guru's statement, whereas the quantified piece of information is usually given to be...
Spoiler:
That the existence of blue eyes is common knowledge.

But really, this might not be as much of a distinction as I first thought it was.

Hmm, I'd amend your piece of information a bit.

In the case of 1 person, it goes from "there are blue eyes on the island" to "there are blue eyes on the island, and everybody here knows it."

In the case of 2, it goes from "there are blue eyes on the island, and everybody here knows it." to "(there are blue eyes on the island, and everybody here knows it), and everybody here knows it."

In the case of 3, it goes from "(there are blue eyes on the island, and everybody here knows it), and everybody here knows it." to "((there are blue eyes on the island, and everybody here knows it), and everybody here knows it), and everybody here knows it."

Basically, the piece of information is that "everyone's depth of knowledge about everyone else's depth of knowledge about the blue eyes has just increased by one level."

I believe I have stated this equivalently and a little more simply in my answer I gave. I don't think I have restated what's already known. If there were one person on the island with blue eyes before the Guru spoke, he would not be able to leave because he would have no information about blue eyes other than that he might or might not have them. After she speaks, he knows that he is the only one with blue eyes, and he knows that he can leave tonight. He now knows that "if there were one person with blue eyes on the island, he would leave tonight," because he knows he has blue eyes and is leaving tonight. The others see the blue-eyed person and, by simple virtue of the fact that he is the only known blue-eyed person here, that "if there were one person with blue eyes on the island, he would leave tonight," because they see one person and know that he will leave tonight if they themselves do not have blue eyes. Thus they are able, with that information, to watch and see if he leaves tonight.

If you like, it can also be phrased, "if there is exactly one person here with blue eyes, he now knows his eye color." Taken as a logical implication, P -> Q, P represents "there is exactly one person here with blue eyes", and Q represents "one person here now knows his eye color". Both P and Q are false, but this is not sufficient to rule out the implication itself. It is this implication, not previously available, that allows 1 person to look at 2 other people and know that, if his own eyes are not blue, each of the others will wait for the other to leave tonight, and when they don't, leave tomorrow.

Note that the Guru does not actually say, "if there is exactly one person here with blue eyes, he now knows his eye color." Rather, that is the not-immediately-obvious piece of information that can be inferred from the statement that the Guru does make, because she can see every one of the 200 logicians.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

bernerbrau wrote:
Goldstein wrote:Brilliant, bernerbrau. However, I'm not quite satisfied with this:
bernerbrau wrote:
1. What is the quantified piece of information that the Guru provides that each person did not already have?

1. The piece of information is "if there were one person on the island with blue eyes, he/she would leave tonight."

I think you've just restated the same thing. In the case where there really is only one blue-eyed person on the island, the piece of information you've suggested doesn't allow them to leave tonight as they don't actually know whether there's any blue-eyed people on the island. I agree that, in the case of two or more people, the fact as you've stated will achieve the same job, but I think it's a piece of information that can be deduced from the Guru's statement, whereas the quantified piece of information is usually given to be...
Spoiler:
That the existence of blue eyes is common knowledge.

But really, this might not be as much of a distinction as I first thought it was.

Hmm, I'd amend your piece of information a bit.

In the case of 1 person, it goes from "there are blue eyes on the island" to "there are blue eyes on the island, and everybody here knows it."

In the case of 2, it goes from "there are blue eyes on the island, and everybody here knows it." to "(there are blue eyes on the island, and everybody here knows it), and everybody here knows it."

In the case of 3, it goes from "(there are blue eyes on the island, and everybody here knows it), and everybody here knows it." to "((there are blue eyes on the island, and everybody here knows it), and everybody here knows it), and everybody here knows it."

Basically, the piece of information is that "everyone's depth of knowledge about everyone else's depth of knowledge about the blue eyes has just increased by one level."

Do you know what common knowledge means in logic?
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Are you familiar of the tale 'The Emperor's New Clothes'?

There's a point in the story that's particularly relevant to this discussion: When the boy yells that the emperor is naked and the crowd starts laughing.

Before that point, everyone knew the emperor was naked, but the members of the crowd didn't possess common knowledge -- in other words, they didn't know that everyone else knew the emperor was naked. They could only hesitantly "ooh" and "ahh" at the fancy garb, or risk being labeled incompetent.

When the boy yelled out, the type of knowledge changed. In an instant, everyone had common knowledge.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Qaanol wrote:Do you know what common knowledge means in logic?

I did not. I do now. That makes Goldstein's statement correct then. It also makes talking about the problem easier.

Unfortunately I never really took a formal logic course. Discrete mathematics is as close as I got.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sheesh, don't tell people you took that course.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Yakk wrote:Sheesh, don't tell people you took that course.

No, you're thinking of discreet mathematics, which is quite a fun course. Or so I heard.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

phlip wrote:
Yakk wrote:Sheesh, don't tell people you took that course.

No, you're thinking of discreet mathematics, which is quite a fun course. Or so I heard.

Ohhhh I get it now Haha :p
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

The original statement of the puzzle says that there are 100 blue-eyed and 100 brown-eyed logicians on an island and a guru shows up and speaks ". . . on one day in all their endless years on the island." So even before the guru speaks every blue-eye knows that there are either 99 blue-eyes and 101 brown-eyes, or there are 100 blue-eyes and 100 brown-eyes. Likewise, the numbers for the brown-eyes are either 99 blue-eyes and 101 brown-eyes, or 100 of each. Furthermore this is common knowledge; all 200 logicians know these things (and nothing else of consequence to the puzzle), and they all know that everyone else knows the same things, and they all know that everyone else knows that everyone else knows the same things, etc, etc, etc.

Let's say they put all of this into a computer database: everything that anybody (and thus everybody) knows is captured in a particular bit pattern. Then the guru speaks. If this adds to the total information then the bit pattern must be changed accordingly.

But of course nothing changes; the guru told them nothing they hadn't known for all those endless years on the island. The fact that it was a guru that made the statement is irrelevant (unless some kind of Arabian Nights magic that we don't know about is involved), and it makes no difference whether a guru says it or a logician just happens to say it to himself. It seems reasonable to me to suppose that in all those endless years many people had noticed that there was at least one blue-eyed person on the island. But such a thought would not have started some kind of magic count down clock that would allow all the blue-eyes to escape 99 days later. Such a thought would not trigger an updating of the information database, and neither would the statement of the guru.

If it we allow some islanders to have information not available to others then other outcomes are possible:
1. A single blue-eye and any number of brown-eyes are on a deserted beach when a guru appears and says "I see a blue-eye". That night the lucky one departs the island.
2. There are two blue-eyes and some brown-eyes on the beach when the guru appears. Two night later the two of them depart.
3. There are three blue-eyes on the beach when the guru speaks. "Well DUH, guru", respond all the logicians in unison.

The bottom line is that the guru provided the islanders no information and after she spoke they accumulated new information at the rate of zero per day. At that rate they can accumulate it for 99 days or 99 years and it still adds up to zero.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sequim82 wrote:the guru told them nothing they hadn't known for all those endless years on the island. The fact that it was a guru that made the statement is irrelevant (unless some kind of Arabian Nights magic that we don't know about is involved), and it makes no difference whether a guru says it or a logician just happens to say it to himself. It seems reasonable to me to suppose that in all those endless years many people had noticed that there was at least one blue-eyed person on the island. But such a thought would not have started some kind of magic count down clock that would allow all the blue-eyes to escape 99 days later. Such a thought would not trigger an updating of the information database, and neither would the statement of the guru.

This is where what you're saying becomes definitely wrong, though your first paragraph is a little hazy about what the line is between common knowlege and statements that everyone on the island would agree with, and really, that is where you need to start trying to sort this out. Think through your "etc, etc, etc" with clarity and you'll find the piece of information that the guru adds.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Likewise, a good idea would be to read some posts in this thread. Many others have had your problem with common knowledge, and as such it has been explained many times.

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sequim82 wrote:The bottom line is that the guru provided the islanders no information and after she spoke they accumulated new information at the rate of zero per day. At that rate they can accumulate it for 99 days or 99 years and it still adds up to zero.
The guru gave the islanders a date to start counting from. He might as well have said, "Ok, everyone with blue eyes, off the island!" and the effect would be the same.

Then, once all the blue eyes were off, the brown eyes would realize that everyone realizes that there's more brown eyes than any other color, so that they should start counting next, starting from the date the blue eyes left.

Then 100 days later the guru would be left all alone, because he wouldn't know what colors his own eyes were. He must really hate all the other people on the island to be so keen on getting rid of them all.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Randomizer wrote:
Sequim82 wrote:The bottom line is that the guru provided the islanders no information and after she spoke they accumulated new information at the rate of zero per day. At that rate they can accumulate it for 99 days or 99 years and it still adds up to zero.
The guru gave the islanders a date to start counting from. He might as well have said, "Ok, everyone with blue eyes, off the island!" and the effect would be the same.

This is not sufficient for anyone to leave. Consider the case with only one person of each eye color. Your Guru might well have said “Ok, everyone with purple eyes, off the island!”

Randomizer wrote:Then, once all the blue eyes were off, the brown eyes would realize that everyone realizes that there's more brown eyes than any other color, so that they should start counting next, starting from the date the blue eyes left.

This is not common knowledge. Even if each person individually thinks it would make sense, they cannot know for sure that the others are thinking the same. On the other hand it would work if the Guru said, “There is at least one person with blue eyes on the island. Henceforth, the day after any group of people with a single eye color leaves the island, if there is one eye color that is the most common among those remaining, and that knowledge is common to a depth necessary to identify it uniquely as the most common color, then you may consider me to say there is at least one person with that color eyes on the island on that day.”
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

This is not sufficient for anyone to leave. Consider the case with only one person of each eye color. Your Guru might well have said “Ok, everyone with purple eyes, off the island!”
I wasn't considering that case. I was only considering the puzzle as is - ie there are 100 blue eyed, 100 brown eyed, and one green eyed person on the island.

Although, "Everyone with purple eyes, get off!" works if there are at least six people with purple eyes. (Maybe fewer, but I only worked it out for six):

There's 7 guys on the island, 6 are purple. The Green eyed guy thinks, "I see people 6 with purple eyes". He knows that there are either 6 or 7 purple eyed people on the island.
Therefore he knows that everyone else sees 5, 6, or 7, and they all know that there are 5 or 6, or 6 or 7 purple-eyed people on the island.

These purple eyed people all see 5 and know that there are 5 or 6 purple-eyed people on the island.
Therefore they know that everyone else sees 4, 5, or 6, and know that there are 4 or 5, 5 or 6, or 6 or 7 purple eyed people on the island. They also know that no one on the island sees fewer than 3 purple-eyed people, and that therefore everyone knows that there are at least 3 purplies on the island, not counting themselves.

Since everyone knows that no one sees fewer than 3 purple-eyed people on the island, they also know that everyone knows there are at least 3 pruple-eyed people on the island, and that everyone knows this. They also know that there is at least one person on the island with purple eyes, so when the guru gives the "go" signal "everyone with purple eyes off!", they start counting.

Day one - no one leaves
Day two - no one leaves
Day three - no one leaves - no one who saw three pruples would leave that day. Everyone knew that everyone else saw at least three purples
Day four - No one leaves - Anyone who only sees only three purplies knows there are only four on the island, and so leave. But there are more than four on the island. So everyone sees at least 4 and no one leaves.
Day five - no one leaves - purplies know there's either 5 or 6 since they see five. Anyone who saw four would leave that day. No one leaves
Day six - all the purplies leave. It's obvious to them that they are the sixth purple. The green-eyed guy is stuck on the island alone.

Even if each person individually thinks it would make sense, they cannot know for sure that the others are thinking the same.
How do they even know that people who know their own eye color would get off the island? It's assumed. This puzzle only works if the people are allowed to know certain things about other people's thought processes. The puzzle states: "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." (emphasis mine)

Otherwise the blue eyed people wouldn't have known they could get off the island. But since they are all perfect logisticians, and everyone knows this, if one of them comes up with a means of getting off the island, all of them are aware of it. Once the blue eyes are gone, no one has to say, "Brown eyes, your turn", they all know it's the brown eyes' turn.
Last edited by Randomizer on Mon Mar 28, 2011 12:30 am UTC, edited 1 time in total.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Randomizer wrote:Day four - No one leaves - Anyone who only sees only three purplies knows there are only four on the island, and so leave. But there are more than four on the island.
Day five - no one leaves - purplies know there's either 5 or 6 since they see five. Anyone who saw four would leave that day. No one leaves

But... you just said further up your post that they would only leave if there are 6 or more (possibly fewer, but you had not "proved" it). And yet, your proof relies on people leaving if there are only 4 or 5! If no-one would ever leave in the 5-person case, as you admit is possible, then in the 6-person case they can't deduce that there are 6 people merely from the fact that no-one ever leaves, as that cannot rule out the 5-person case as an option for those people.

And the same is true wherever you draw the line. If it works for n but doesn't work for n-1, but the islanders' only evidence that lets them solve the n-person case is that no-one leaves, then they can't leave as they haven't truly ruled out the n-1-person case.

Randomizer wrote:Otherwise the blue eyed people wouldn't have known they could get off the island. But since they are all perfect logisticians, and everyone knows this, if one of them comes up with a means of getting off the island, all of them are aware of it. Once the blue eyes are gone, no one has to say, "Brown eyes, your turn", they all know it's the brown eyes' turn.

"Everyone is a perfect logician, and this is common knowledge" doesn't imply superrationality. To be more clear: if I can prove statement X, using only information that person A knows, then I know that A knows that statement X is true. However, if I decide on thing Y by other means (eg logical deduction based on knowledge that I don't know if A has, or know A doesn't have; random or arbitrary selection; actions towards some "goal" like "I want to leave the island" that may not be shared by A; etc) then I cannot be sure that A has decided on thing Y as well. We're not strictly symmetrical, I just know that if they can logically deduce something, they will do so.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

This is addressed particularly to jestingrabbit and timm01994. I admit to being a little careless when I enumerated what each group could see (I forgot about the possibility of one red-eye) but that's not important because I made no argument relative to those details. What I am curious about is your concern about my use of the term "common knowledge". I was speaking only about the blue-eyes could see and what the brown-eyes could see and what each group could deduce about the size of the groups.

Let's say the three of us are standing together on the beach. I can see three of us, and I can see that you can see three of us, and I can see that he can three of us, and I can see that you can see that I can see three of us, and he can see that I can see that you can see that I can see three of us, etc, etc,etc. How many 'etc' are there before we come to the bifurcation between the knowledge about how many of us are standing on the beach that is common to the three of us, and statements about how many of us are standing on the beach that we all agree with? Just curious.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sequim82 wrote:How many 'etc' are there before we come to the bifurcation between the knowledge about how many of us are standing on the beach that is common to the three of us, and statements about how many of us are standing on the beach that we all agree with? Just curious.

To satisfy the formal definition of "common knowledge" in formal logic, you need an infinite number of 'etc'.

For the specific situation of the Blue-eyes problem, the inhabitants of the island have a very large number of 'etc' but they are precisely one short of what they actually need to solve the problem. The guru's announcement combined with how it is given extend the 'etc' to infinite.

Seriously, with a thread this long please do not presume that your argument has never been brought up before. Your argument or something like it has been brought up over and over again and again and again many many times. Pick a page of this thread at random and you will probably find an argument similar to yours somewhere on that page, quickly followed by a variety of explanations for why it is wrong and the original solution is correct. Please read some of these explanations before you continue trying to refute the established solution.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

douglasm wrote:
Sequim82 wrote:How many 'etc' are there before we come to the bifurcation between the knowledge about how many of us are standing on the beach that is common to the three of us, and statements about how many of us are standing on the beach that we all agree with? Just curious.

To satisfy the formal definition of "common knowledge" in formal logic, you need an infinite number of 'etc'.

For the specific situation of the Blue-eyes problem, the inhabitants of the island have a very large number of 'etc' but they are precisely one short of what they actually need to solve the problem. The guru's announcement combined with how it is given extend the 'etc' to infinite.

Seriously, with a thread this long please do not presume that your argument has never been brought up before. Your argument or something like it has been brought up over and over again and again and again many many times. Pick a page of this thread at random and you will probably find an argument similar to yours somewhere on that page, quickly followed by a variety of explanations for why it is wrong and the original solution is correct. Please read some of these explanations before you continue trying to refute the established solution.

Don't presume that I am not familiar with the arguments that have been brought up before and don't presume that I am impressed with your little lecture about "common knowledge" and don't presume that I am intimidated by your assertion of an "established solution" and don't presume to tell me under what conditions I can post on this or any other thread.
Sequim82

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

I presumed none of that. I requested that you do something sensible that you had presented substantial evidence of having not done, and I attempted to provide an explanation you had asked for. That's all.

Arguments nearly identical to yours in principle have been brought up and refuted dozens of times in this thread. If you really were familiar with these arguments and their refutations and understood them, you would not have posted what you did. If you were familiar with them but did not understand or disagreed with them, picking apart the refutations would be a much better approach than effectively restating what they refuted without really adding anything.

You assert that the Guru's announcement provides no new information. Why and how this is wrong has been explained dozens of times in this thread. Where is the error in each of these dozens of explanations?
douglasm

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sequim82 wrote:But of course nothing changes; the guru told them nothing they hadn't known for all those endless years on the island. The fact that it was a guru that made the statement is irrelevant (unless some kind of Arabian Nights magic that we don't know about is involved), and it makes no difference whether a guru says it or a logician just happens to say it to himself.

The italicized part is not quite right. The manner in which the statement is made is extremely important here.

A good way to illustrate this can be found just a few posts up from your post:

redrogue wrote:Are you familiar of the tale 'The Emperor's New Clothes'?

There's a point in the story that's particularly relevant to this discussion: When the boy yells that the emperor is naked and the crowd starts laughing.

Before that point, everyone knew the emperor was naked, but the members of the crowd didn't possess common knowledge -- in other words, they didn't know that everyone else knew the emperor was naked. They could only hesitantly "ooh" and "ahh" at the fancy garb, or risk being labeled incompetent.

When the boy yelled out, the type of knowledge changed. In an instant, everyone had common knowledge.

In "The Emperor's New Clothes", the boy's public announcement is key - once it happens, everyone relaxes because they now know aren't the only one who sees the emperor as naked, and then everyone can see everyone else relax, and therefore everyone knows everyone else knows that the emperor is naked, and then they all start laughing. If any member of the crowd had just happened to say it to himself (but so that nobody else could hear), it wouldn't have had the same effect. The important part was not the literal fact of the boy's statement, because everyone could already see that for themselves. It was the fact that the statement was public, and therefore allowed everyone to gain knowledge about other people's knowledge, namely that everyone else saw the emperor as naked too.

A similar thing is happening in the islanders puzzle. It's that the guru's statement is made in public view of everyone that's important, not the direct fact of of what he says, and it allows everyone update their knowledge of everyone else's knowledge. Only, it's not knowledge-of-knowledge-of-real-fact in this case, which is only 2 levels deep, but rather it is knowledge-of-knowledge-of-knowledge-of.... going 100 levels deep.

Most people's intuition cuts off at around 3-4 iterations of "A knows that "B knows that "C knows that "D knows that..."""", which is why most people think the induction breaks down at step 3-4.
Last edited by lightvector on Mon Mar 28, 2011 7:40 am UTC, edited 1 time in total.
lightvector

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sequim82 wrote:Don't presume that I am not familiar with the arguments that have been brought up before and don't presume that I am impressed with your little lecture about "common knowledge" and don't presume that I am intimidated by your assertion of an "established solution" and don't presume to tell me under what conditions I can post on this or any other thread.

Chuff wrote:I write most of my letters from the bottom

Goldstein

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Sequim82 wrote:Let's say the three of us are standing together on the beach. I can see three of us, and I can see that you can see three of us, and I can see that he can three of us, and I can see that you can see that I can see three of us, and he can see that I can see that you can see that I can see three of us, etc, etc,etc. How many 'etc' are there before we come to the bifurcation between the knowledge about how many of us are standing on the beach that is common to the three of us, and statements about how many of us are standing on the beach that we all agree with? Just curious.

The difference between this and the blue eyes problem is that when the three of you can see eachother, it becomes common knowledge that there are three of you. In the blue eyes problem, this is like asking if it is common knowledge that there are 200 people on the island.

Here's where the common knowledge comes in the problem, and where the information that the guru says is clear.

Person1 is blue eyed, but doesn't know it. So, before the guru speaks, he sees that there are 99 blue eyed people and 100 brown eyed people. He also knows that every blue eyed person sees at least 100 brown eyed people and at least 98 blue eyed people. Now, there are two possible cases. Either Person1 has blue eyes, or he doesn't. If he does, then there are 100 brown eyes and 100 blue eyes on the island. If he does not, then what would the other blue eyed people be thinking right now?

Well, Person2, who is also blue eyed and doesn't know it, would see 98 blue eyed people, 100 brown eyed people, and Person1. This is what Person1 knows that Person2 knows. But since Person2 doesn't know his own eye color, then Person1 knows he has blue eyes, but does not know that Person2 knows this. That's easy to understand, right?

So the knowledge of Person2 from Person1's perspective is that there are 98 blue, 100 brown, and Person1. This is different from the actual knowledge of Person2 (which is identical to Person1's knowledge) because Person1 doesn't know everything that Person2 knows, only the information that he knows both absolutely must know.

Now, from Person1's perspective, Person2 could be using the same logic on Person3. "Well, I know Person3 sees 97 blues, 100 browns, myself and Person1." This is not what Person2 knows about Person3, it's what Person1 knows Person2 knows about Person3. Person2 actually knows that Person3 sees 98 blues, and Person1 knows that Person3 sees 98 blues, but Person1 only knows that Person2 knows that Person3 sees 97 blues.

Person4? Well, Person1 knows that Person4 sees 98 blues. And Person2 knows that Person4 sees 98 blues. And Person3 knows that Person4 sees 98 blues. But Person2 doesn't know that Person3 knows that; he only knows that Person3 knows that Person4 sees 97 blues. And Person1 doesn't know that; he only knows that Person2 knows that Person3 knows that Person4 sees 96 blues.

And so on and so forth, until you get to the bottom. Person1 knows that Person100 sees 98 blues, but Person1 only knows that Person99 knows that Person100 sees 97 blues. And Person1 only knows that Person2 knows that Person3 knows that... Person100 knows that 100 brown eyes and nothing about blue eyes at all.

So what happens when the guru speaks? Well, that last case changes. Now Person1 still only knows that Person100 sees 98 blue eyes, but now he knows that Person2 knows that Person3... knows that Person99 knows that Person100 knows that there are 100 browns and at least 1 blue.

Does that all make sense? That's why before the guru speaks, there is no induction. The sequence of knowledge of knowledge of knowledge gets cut off at the 100th person. Once the guru speaks, it's absurd even to imagine a 100th layer where someone doesn't know there is at least 1 blue eyed person, because that person must have been there when the guru spoke, and therefore knows there is at least 1.

This is why this statement is wrong:
3. There are three blue-eyes on the beach when the guru speaks. "Well DUH, guru", respond all the logicians in unison.

In the 3 person case, it's even easier to grasp why it's not common knowledge until the guru speaks. Say it's just John, Paul, and George on the island, all with blue eyes, following the same rules as the original problem. John sees that Paul and George have blue eyes, but doesn't know his own eye color. So, to John, Paul may see either 2 blues or 1 blue. If Paul only sees 1 blue eyed person (George), then he would know there was at least 1 blue eyed person, but what if Paul was not blue eyed? Then George would know nothing of blue eyes. So John knows that Paul knows there is at least 1 blue, and Paul knows that George knows there is at least 1 blue, but John does not know that Paul knows that George knows there is at least 1 blue. So when Ringo (with green eyes) shows up and says "There is one blue eyed person here." in front of all of them, then now John knows that Paul knows that George knows there is at least one blue eyed person. That's what changes.

If you still disagree with the 3 person case, consider this: What's the difference between 3 blue eyed people, and 2 blue eyed people with 1 brown eyed person? Would they still get no knowledge from the guru? If the 2 person case gets information but the 3 person case doesn't, then what goes on in the brown-eyed person's head in the 2/1 case? Does he still think "Well DUH, guru"?
Goals:
1. Disprove something widely accepted to be true (In progress... I'm getting there.)
2. Have an action figure made of myself (With karate chop action!)
3. Win a dancing contest (COMPLETE)
Xias

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Nice, Xias. I realise similar explanations have probably been given before, but that's the clearest one I've ever read myself. It actually seems to make intuitive sense of something I previously considered an abstract logical construct.

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Xias wrote:
Sequim82 wrote:Let's say the three of us are standing together on the beach. I can see three of us, and I can see that you can see three of us, and I can see that he can three of us, and I can see that you can see that I can see three of us, and he can see that I can see that you can see that I can see three of us, etc, etc,etc. How many 'etc' are there before we come to the bifurcation between the knowledge about how many of us are standing on the beach that is common to the three of us, and statements about how many of us are standing on the beach that we all agree with? Just curious.

The difference between this and the blue eyes problem is that when the three of you can see eachother, it becomes common knowledge that there are three of you. In the blue eyes problem, this is like asking if it is common knowledge that there are 200 people on the island.

You could turn it into something relevant by introducing the possibility that you are a ghost, and that you don't know whether you are or not but anyone who sees you will know. If there are ghosts on the beach, then the non ghosts should run screaming in the middle of the night, but only if they know that no ghosts will not follow them, also screaming from the fear of ghosts. A kid walks by and says "I see dead people". That is a proper blue eyes analog.

Edit: fixed the typo.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jestingrabbit

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

jestingrabbit wrote:You could turn it into something relevant by introducing the possibility that you are a ghost, and that you don't know whether you are or not but anyone who sees you will know. If there are ghosts on the beach, then the non ghosts should run screaming in the middle of the night, but only if they know that non ghosts will not follow them, also screaming from the fear of ghosts. A kid walks by and says "I see dead people". That is a proper blue eyes analog.

There's a typo in there, I think it's what I've underlined. The living flee from the ghosts, but only if the ghosts will not follow them (thinking they, too, are alive)? That seems like the sensible construction, but it's not quite the same: Suppose there's only one ghost: then the ghost realizes that fact immediately, and therefore stays. The living people all say "well, I see one ghost, so the number of ghosts is either 1 or 2. I'd better run, in case the number is 1." And off they go on night 1, leaving the ghost behind. If there are 2, then the ghosts (only) flee on night 1, because they're the only 2 people who think there might be just one ghost; the living people know there are at least 2 ghosts, and they might be running off, so running away on night 1 isn't logically any safer than staying. And if there are n ghosts, then they all run screaming on night n-1, leaving the living people on the beach.

At least, that's the way I deduce it would work (which makes it an interesting variation, actually); tell me if there's a problem with my logic...
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EricH

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Do they have reason to be certain about the Guru's statement?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Yakk
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Technically no, it's come up before that the question really should have "The Guru is reliable" in that first paragraph of common-knowledge givens. I mean, it's given in the "it's not a 169" disclaimer that the solution doesn't depend on anyone lying, but "the Guru isn't lying" != "'the Guru isn't lying' is common knowledge".

Most people have been generally working under a version of the puzzle where that is given.
While no one overhear you quickly tell me not cow cow.

phlip
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Ok, I'm new to this topic, but I have perused a bunch of the previous discussion. Forgive me if my line of reasoning has been discussed already.

I understand the basic solution but I'm still trying to wrap my head around the follow up questions.

Let's assume the guru speaks on day 0, which we'll call Black Friday. The day prior is day -1 and so on. On Black Friday, Tomorrow is Saturday
And Sunday comes afterwards (days 1 and 2, respectively) (I apologize for that last sentence to those who know why I'm apologizing.)

What isn't stated is that on some day prior to Black Friday (day # <=0, the islanders established the rule that those who know their eye color must leave. There are several possibilities:

1) Every islander (or at least all the blue eyes) was informed of this 'rule' on the same day and it was 'common knowlege.' It would seem in this case, this is equivalent to the day the guru speaks, since everyone already know that at least one islander has blue eyes.

2) Every islander (or at least all the blue eyes) was informed of this 'rule' on the same day but they did not know when the others had been informed or were to be informed of the rule.

3) The island and its state of affairs have existed 'forever', infinitely into the past -- hence there was no 'starting point' prior to the guru speaking.

John.

I'm assuming here there are no births or deaths (or old age senility) involved at all...

It seems like once you know the island leaving rule, and you also know that everyone else knows the island leaving rule, and that they know all know that everyone else knows the island leaving rule, then the countdown clock can start for you, and you know the countdown clock will have already started the same day (or earlier) for everyone else.)
jvriezen

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Let's reduce the numbers so we've got just one blue-eyed person and one brown-eyed person, plus the Guru, before she makes her announcement. You can start a 'countdown clock' any day you like, but what good does it do you? What good does it do when there are 2 blue-eyed people? Or 3? Or 100?
Chuff wrote:I write most of my letters from the bottom

Goldstein

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

I hesitated before adding to this thread. On the one hand, it's 23 pages long and the answer is pretty well established, but on the other hand, the very fact that it's 23 pages long and people continue to ask questions is evidence that more input could be useful -- sometimes paraphrasing things in a different way can help make the solution more "intuitive".

In my opinion, the puzzle hinges on the following, which is difficult to intuitively wrap one's head around.

It's meaningful (though confusing) to make statements like the following:

"Alice knows such-and-such"
"Alice knows that Bob knows such-and-such"
"Alice knows that Bob knows that Carol knows such-and-such"
"Alice knows that Bob knows that Carol knows that Doug knows such-and-such"
and so on.

In everyday life, we rarely go beyond two levels or thereabouts. But the above statements are all meaningful, and they are all different. Certainly, confusion starts to set in somewhere around the third level, but that doesn't change the fact that each of these statements has a precise meaning.

(spoilerized because of length)

Spoiler:
Let's say Alice, Bob, Carol, and Doug all have blue eyes. Consider the following claims (made before the guru makes her announcement.)

"There are at least 100 blue-eyed people" -- true (we know it from our vantage point as outsiders)
"Alice knows there are at least 100 blue-eyed people" -- false, she can only see 99
"Alice knows there are at least 99 blue-eyed people" -- true
"Bob knows there are at least 99 blue-eyed people" -- true

"Alice knows that Bob knows there are at least 99 blue-eyed people" -- this is false. Here, we're still within the realm of being fairly intuitive. In fact, Bob does know there are at least 99 blue-eyed people, but Alice doesn't know that Bob knows that. All Alice knows for sure is that she can see blue-eyed Bob and 98 other blue-eyed people.

"Alice knows that Bob knows there are at least 98 blue-eyed people" -- true
"Bob knows that Carol knows there are at least 98 blue-eyed people" -- true

Things become more confusing at the next step, but it's a totally legitimate step and it's "just" logic. What does Alice know about what Bob knows about what Carol knows? It's a type of question we rarely ask in everyday life, but we could. Alice wonders, "What is Bob thinking?" She wonders more specifically, "What is Bob thinking about Carol?" or "What is Bob thinking about what Carol thinks?" or "What does Bob know about what Carol knows?"

The next paragraph I typed, I deleted because I started to confuse myself. It just goes to show how unintuitive and alien this type of reasoning is. But I maintain the following:

"Alice knows that Bob knows that Carol knows there are at least 98 blue-eyed people" -- false
"Alice knows that Bob knows that Carol knows there are at least 97 blue-eyed people" -- true

Roughly speaking, every time you "add a layer", the number decreases by 1. I admit that it's still not totally intuitive, but hopefully people are willing to at least agree that when you "add a layer", you get a different sentence with a different meaning.

Putting it another way:

There's the actual state of affairs on the island (we outsiders happen to know there are 100 blue-eyed people)
Then there's, say, Doug's mental model of the island, which is missing some facts about the island.
Then there's Carol's mental model of Doug's mental model of the island, which is missing some facts about Doug's mental model of the island.
Then there's Bob's mental model of Carol's mental model of Doug's mental model of the island, which is missing some facts about Carol's mental model of Doug's mental model of the island.

Confusing? Absolutely, yes. But so are constructions like "the day after the day after the day after the day after the day after tomorrow". Yes, it's confusing, and you start to lose track, but each time you add the words "the day after", it means something.
skullturf

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

skullturf wrote:
Spoiler:
Let's say Alice, Bob, Carol, and Doug all have blue eyes. Consider the following claims (made before the guru makes her announcement.)

"There are at least 100 blue-eyed people" -- true (we know it from our vantage point as outsiders)
"Alice knows there are at least 100 blue-eyed people" -- false, she can only see 99
"Alice knows there are at least 99 blue-eyed people" -- true
"Bob knows there are at least 99 blue-eyed people" -- true

"Alice knows that Bob knows there are at least 99 blue-eyed people" -- this is false. Here, we're still within the realm of being fairly intuitive. In fact, Bob does know there are at least 99 blue-eyed people, but Alice doesn't know that Bob knows that. All Alice knows for sure is that she can see blue-eyed Bob and 98 other blue-eyed people.

"Alice knows that Bob knows there are at least 98 blue-eyed people" -- true
"Bob knows that Carol knows there are at least 98 blue-eyed people" -- true

Things become more confusing at the next step, but it's a totally legitimate step and it's "just" logic. What does Alice know about what Bob knows about what Carol knows? It's a type of question we rarely ask in everyday life, but we could. Alice wonders, "What is Bob thinking?" She wonders more specifically, "What is Bob thinking about Carol?" or "What is Bob thinking about what Carol thinks?" or "What does Bob know about what Carol knows?"

The next paragraph I typed, I deleted because I started to confuse myself. It just goes to show how unintuitive and alien this type of reasoning is. But I maintain the following:

"Alice knows that Bob knows that Carol knows there are at least 98 blue-eyed people" -- false
"Alice knows that Bob knows that Carol knows there are at least 97 blue-eyed people" -- true

I disagree with your line: "Alice knows that Bob knows that Carol knows there are at least 98 blue-eyed people".

I agree the first level of thinking is correct, each person can see 99 blues and knows each of the blues can see at least 98 blues. From this, Alice knows that Bob and Carol can see at least 98 blue-eyed people. But at the same time, Bob knows that Alice and Carol can see at least 98 blue-eyed people. Because it's common knowledge that every individual knows that every other individual can see at least 98 blue-eyed people, Alice automatically knows that Bob knows that Carol knows there are at least 98 blue-eyed people.

Hence, my disagreement. I'm probably wrong though :S

I'm still siding with the argument that the guru does not give any new information.

Thus, Alice
Abhinav

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Damn, I figured it out after I posted my reply. I agree with you guys now, the solution makes sense.
Abhinav

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Again, say Alice, Bob, Carol, and Doug are all islanders with blue eyes.

The reality of the island is that there are 100 blue-eyed people and 100 brown-eyed people.

In Alice's mental model of the island, there are 100 brown-eyed people, 99 blue-eyed people, and one person of unknown eye color (Alice herself).

In Bob's mental model of Alice's mental model of the island,
there are 100 brown-eyed, 98 blue-eyed, and 2 unknown (Alice and Bob).

In Carol's mental model of Bob's mental model of Alice's mental model of the island,
there are 100 brown-eyed, 97 blue-eyed, and 3 unknown (Alice, Bob, and Carol).

Of course, Carol knows Bob's and Alice's actual eye colors. But she also knows that Bob and Alice don't know their own eye colors. AND, Carol is aware that in Bob's model of Alice's thought processes, both Bob's and Alice's eye colors are unknown. AND Carol doesn't know what color her own eyes are in Bob's model of Alice's thought processes.
skullturf

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

The one blue eyed person theorem works, but two people+ does not.

One person: If the guru says that there is one person with blue eyes, that person that sees none with blue will know it is him/herself. Everyone else sees one person with blue eyes and it ends there.

With two people that have blue eyes, this happens:
Blue eye #1 & #2:
They each see one person with blue eyes. (thats all the information they have from the guru, one with blue eyes)
They cannot tell the other that they have blue eyes.
The 2nd person with blue eyes can assume that since the first person with blue eyes isn't leaving, there must be another with blue eyes or he'd do what I wrote in "One person."
He can only think its himself, not know it.
Remember that each of the brown eyed people think exactly the same way as the blue eyed people but they just see one more blue eyed person. They would also *think* that the blue eyed people aren't leaving because they see others with blue eyes (potentially themselves).
No one would leave from this hypothesis. Your solution forgets that the brown eyed people are thinking the same thing as the blue eyed people.
KingMob

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

KingMob wrote:With two people that have blue eyes, this happens:
Blue eye #1 & #2:
They each see one person with blue eyes. (thats all the information they have from the guru, one with blue eyes)
They cannot tell the other that they have blue eyes.
The 2nd person with blue eyes can assume that since the first person with blue eyes isn't leaving, there must be another with blue eyes or he'd do what I wrote in "One person."
He can only think its himself, not know it.

No. He can know it after the first oportunity to leave the island passes. He knows that if there there was only one blue eyed islander, they would have left. He knows that he can see only one blue eyed islander. Therefore there are two blue eyed islanders and the only islander that could be the other blue eyed islander, the one whose eyes he cannot see, and therefore his own eyes must be blue.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jestingrabbit

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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

KingMob wrote:...

You're missing a bit of logic. Let Blue eyed person 1 = B1 and blue eyed person 2 = B2. Now what you don't realize is that both B1 and B2 only see one other person with blue eyes. The people with brown eyes see two people with blue eyes. So after the first night B1 thinks to themself "Why didn't B2 leave? It must be because there is another person with blue eyes - but I don't see another person with blue eyes" which leads to the conclusion that they have blue eyes.

The people with brown eyes can't make that conclusion because they see two people with blue eyes.

Edit: Ninja'd
double epsilon = -.0000001;

Dason

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