## 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

This point has been brought up and refuted in great detail many many many many many MANY times in this thread. Randomly select any page in this thread and scroll to a random point on it, and the chance that you just landed in the middle of a discussion of this exact point is very high. Please do us all the courtesy of reading some of these discussions and the explanations offered before asking us to explain it all over again just for you.
douglasm

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

ok, here's a point of confusion. The puzzle I have been considering has the blue eyed people leaving. The last comment from phlip induced me to reread this puzzler carefully and I realized that it is those people who know their own eye color (regardless of that color) who leave. This must be the source of the confusion, but I haven't had a chance to think it through yet.

I am still convinced that my understanding is correct if it is the blue eyed people who leave.

I'll ponder all this and post something tomorrow.

Thanks again!

Maybe we'll have two good puzzles where we thought we only had one!
infti

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

I think your puzzle that was told to you is slightly wrong. Mainly because the telling of it, and the solution is a bit awkward. The story, I believe, should read something like:

'There are 100 blue-eyed and 100 brown eyed people on an island. They are unable to communicate to one another, but are able observe everyone's appearance. One day, a sign appears next to a dock, stating - "every night starting tonight, a ferry will leave for the mainland. There are only blue and brown eyed people on the island. Anyone who knows their own eye color may get on the ferry". Everyone considers perfect logic and therefore will know how to determine their own eye color.'

The question: 'Who leaves the island and when?'

Answer - Everyone leaves on the 100th day.

The reasoning:

You are a particular member of the island. You can count 100 of one eye color, and 99 of the other. So since you can be either blue or brown eyed, consider each in turn. If you were brown eyed, consider the blue eyed observers. They will each count 101 brown eye and 98 blue eyes. Taking observer after observer out of the picture in the same way, each will realise that if there were only two, on the first night if the only other blue eyed person did not leave, they would both leave the following night. All of them can make the same assumptions, so going back to the arbitrary choice of observer in the beginning, they will know that if they were brown eyed and counted 99 blue eyes, all the blue eyed people would leave on the 99th day. Since there are equal numbers, they all know to leave on the 100th day.

In fact, regardless of the population numbers, everyone leaves on the day of the number of people of the minority group!
andrewmmc

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

andrewmmc wrote:There are only blue and brown eyed people on the island.

This part makes things very different to the original puzzle... in the original puzzle, noone on the island knows this. Which doesn't make either version wrong, just different.

Also, this is slightly more ambiguous, as it's not clear if this implies that there is at least one of each... strictly logically, that statement would still be true if everyone on the island had brown eyes, for instance... and then the induction would break down. It would be clearer if it said "There are both blue and brown eyed people on the island, and no others" or something like that.

andrewmmc wrote:Anyone who knows their own eye color may get on the ferry.

This part is problematic too... for the solution to work, it has to be "anyone who can figure out their own eye color must get on the ferry." Otherwise, we don't get any information from people failing to leave... if we can see 5 blue-eyed people and they don't leave on day 5, we can't deduce that our eyes must be blue, because it's also an option that they know their eyes are blue, but decided not to leave.

andrewmmc wrote:In fact, regardless of the population numbers, everyone leaves on the day of the number of people of the minority group!

I think that if the numbers are uneven, say, 5 blue-eyed people and 195 brown-eyed people, then the blue-eyed people would leave on day 5, and the brown-eyed people on day 6... How would you have the brown-eyed people leaving on day 5 too?
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

phlip wrote:Also, this is slightly more ambiguous, as it's not clear if this implies that there is at least one of each... strictly logically, that statement would still be true if everyone on the island had brown eyes, for instance... and then the induction would break down. It would be clearer if it said "There are both blue and brown eyed people on the island, and no others" or something like that.

Sure, if we don't know anything about the population - but we're talking about a specific example where all of the observers can see both colors. Saying only blue or brown ensures every observer knows their own eye color must be one of the two.

phlip wrote:I think that if the numbers are uneven, say, 5 blue-eyed people and 195 brown-eyed people, then the blue-eyed people would leave on day 5, and the brown-eyed people on day 6... How would you have the brown-eyed people leaving on day 5 too?

Well, the deduction is made from the inactions of a particular group the previous day, not an action. So if the blue eyes start moving to the ferry on day 5, all the brown eyes know they must be brown eyed on that day. I suppose it is a question of whether they have time to get on the ferry too!
andrewmmc

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

andrewmmc wrote:Sure, if we don't know anything about the population - but we're talking about a specific example where all of the observers can see both colors. Saying only blue or brown ensures every observer knows their own eye color must be one of the two.

Perhaps the most important lesson to learn from blue eyes is that its not enough that everyone knows something. Everyone has to know that everyone knows it, and everyone has to know everyone knows that everyone knows etc before they can start to draw the sort of inference that you are implying they can draw here.

andrewmmc wrote:Well, the deduction is made from the inactions of a particular group the previous day, not an action. So if the blue eyes start moving to the ferry on day 5, all the brown eyes know they must be brown eyed on that day. I suppose it is a question of whether they have time to get on the ferry too!

and its these sort of inexact, maybe they have time, maybe not, that lead me to believe that Randal's statement is as near optimal as I care to concern myself with.
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:Perhaps the most important lesson to learn from blue eyes is that its not enough that everyone knows something. Everyone has to know that everyone knows it, and everyone has to know everyone knows that everyone knows etc before they can start to draw the sort of inference that you are implying they can draw here.

We know, from the rules, that everyone on the island can make the instant logical deduction that they are either blue eyed or brown eyed.

jestingrabbit wrote:and its these sort of inexact, maybe they have time, maybe not, that lead me to believe that Randal's statement is as near optimal as I care to concern myself with.

-- the original story doesn't make it clear either... my intention was to have the same rules but adjust the story a little - a simple change that makes both the problem and solution a little more elegant. On first reading, I was surprised that the brown eyes were not taken into account in the solution - it seems to me that most good logic conundrums are neat and complete in the way they tie up the loose ends.

So since the author states the story was told to him by "some dude in the street name Joel," I wondered if it has changed form as it has been passed down. The point of a problem like this is to illustrate a point, in this case recursive logic, so its merits are based on how clearly it illustrates the point.
andrewmmc

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

andrewmmc wrote:The point of a problem like this is to illustrate a point, in this case recursive logic, so its merits are based on how clearly it illustrates the point.

But in particular, the version posted at the start of this thread has the added beauty of making the Guru's statement appear trivial and useless at first glance. While I don't want to get into a discussion of the mechanics of your proposed version, I do feel it's less elegant in that the sign by the dock much more obviously provides tangible information.
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Goldstein

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

andrewmmc wrote:
-- the original story doesn't make it clear either... my intention was to have the same rules but adjust the story a little - a simple change that makes both the problem and solution a little more elegant. On first reading, I was surprised that the brown eyes were not taken into account in the solution - it seems to me that most good logic conundrums are neat and complete in the way they tie up the loose ends.

In the original, there is a public announcement that a blue eyed person exists.

In your version there are 3 public announcements:

1. Everyone has either blue or brown eyes.
2. A blue eyed person exists.
3. A brown eyed person exists.

Two of your announcements are not necessary.
Ermes Marana

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### Some questions concerning the Blue Eyes problem

Ok, the problem's here for anyone who doesn't know, and the logical solution is here.

There is already one or two threads concerning this, but as they're already (insert hyperbole) long, I figure I have a better chance of expanding my knowledge here.

Note that I'm a first-term Yr 11 student - I'm not stupid, but I don't have a tertiary education and some of you guys are pretty freaking amazing. So if you intend to answer, dumb it down a bit, 'k?

Now, it doesn't say it in the question, but in the answer i'm pretty sure it assumes that everyone on the island knows that everyone else is a perfect logician. Is this true, or is there some crazy step that I don't know?

Here's my real questions. Firstly: are they simply assuming that the guru's telling the truth or is there something I've missed? I'm fairly sure it will be the latter, but if it's not I'm frankly dissappointed.

Next - as they can already see 99 blue-eyed people, why don't they skip to Theorem 98? Here's my reasoning - every blue-eye knows that every other blue-eyed person can at the least see 98 other blue-eyed people. As such, can't they instantly deduce that no-one will leave on the first 97 days? Following this, shouldn't they all know that everyone else will skip to step 98?

Finally, can anyone explain to me the answer to the follow-up question 1? I hypothesize it has to do with the answer to my first question, but either way I'm all ears.

Vesuvius

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### Re: Some questions concerning the Blue Eyes problem

Vesuvius wrote:they're already (insert hyperbole) long

In short:
* Yes, everyone knows that everyone else is a perfect logician. This is made clear in the problem statement - that everyone is a perfect logician is in the paragraph that says "Everyone on the island knows all the rules in this paragraph".
* It's not explicitly stated in the question that it's common knowledge that the guru speaks the truth. To be completely unambiguous, that the guru speaks the truth really should be in that "everyone knows this" paragraph. But then, it does say that "the answer ... doesn't depend on ... anyone lying or guessing", so it's reasonable to take it as read that the guru's statement is understood to be true. If the guru's statement was untrustworthy, after all, nothing would change, and noone would leave.
* Because each person doesn't know exactly how many blue eyed people the others see, so they don't know how many days to skip. Like, if there are 100 blue-eyed people, then you say they'll skip to day 98, but if there's 99 blue-eyed people they'll skip to day 97... but say I'm on the island, and I see 99 blue eyed people not including myself, and I don't know whether my eyes are blue or not (so I don't know if the total count is 99 or 100)... should I skip to day 97 or 98? Any other strategy of this kind falls into the same trap - it only works if everyone follows it, and you can't be sure everyone will follow it.

Seriously, just read the real blue-eyes thread, all those answers are in there. Especially that last point, which is in there about once or twice per page. Don't think that you're the first person to mistakenly believe the guru's announcement gives no information.
While no one overhear you quickly tell me not cow cow.

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

I think we need to be clear here on three things: Facts everyone knows, Facts everyone knows everyone else knows, and Hypotheticals.

Facts everyone knows:
There are at least 99 blue eyed people on the island

Facts everyone knows that EVERYONE knows:
There are at least 98 blue eyed people on the island

Corollary to Facts everyone knows that EVERYONE knows:
No one on the island could be under the impression that someone else on the island believes there is less than 97 people on the island.

The nested hypothetical theory breaks down once you get below:
Person A sees 99 blue eyed people.
Person A assumes hypothetical blue eyed person sees only 98 blue eyed people.
Person A assumes person B assumes person C sees only 97 blue eyed people.
Person A assumes person B assumes person C assumes person D sees only 96 blue eyed people.

Now please explain to me, keeping in mind the "Corollary to Facts everyone knows that EVERYONE knows", how someone who exercises perfect logic would continue down this line of logic?

There is a clear and obvious contradiction here between the two bold statements here.
For this solution to be correct every one of these perfect logicians are thinking:
No one on the island could be under the impression that someone else on the island believes there is less than 97 people on the island
AND
Person A assumes person B assumes person C assumes person D sees only 96 blue eyed people.

No one who exercises perfect logic immediately and all the time would ever act on a theory driven by hypothetical situations which are PROVABLY FALSE.
phenom

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

You seem to be under the impression that facts that everyone knows that everyone knows are the same as facts that everyone one knows that everyone knows that everyone knows that everyone knows that everyone knows etc. This is not the case.

What precisely are you proposing would happen? Who would leave when?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jestingrabbit

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

You seem to be under the impression that facts that everyone knows that everyone knows are the same as facts that everyone one knows that everyone knows that everyone knows that everyone knows that everyone knows etc. This is not the case.

What precisely are you proposing would happen? Who would leave when?

With 100 UPS and 100 bloo, noone would ever leave.

If every blue eyed person sees 99 other blue eyed people.
EVERY blue eyed person KNOWS that every other blue eyed person sees at least 98 other blue eyed people.
If you accept the two premises above, you MUST accept that:
EVERY blue eyed person KNOWS that every other blue eyed person knows that EVERY other blue eyed person sees at least 97 other blue eyed people.

Am I correct that the bold statement above is the part of my original post you're taking issue with?
Please explain to me how that statement is incorrect?
phenom

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

phenom wrote:
You seem to be under the impression that facts that everyone knows that everyone knows are the same as facts that everyone one knows that everyone knows that everyone knows that everyone knows that everyone knows etc. This is not the case.

What precisely are you proposing would happen? Who would leave when?

With 100 UPS and 100 bloo, noone would ever leave.

If every bloo eyed person sees 99 other bloo eyed people.
EVERY bloo eyed person KNOWS that every other bloo eyed person sees at least 98 other bloo eyed people.
If you accept the two premises above, you MUST accept that:
EVERY bloo eyed person KNOWS that every other bloo eyed person knows that EVERY other bloo eyed person sees at least 97 other bloo eyed people.

Am I correct that the bold statement above is the part of my original post you're taking issue with?
Please explain to me how that statement is incorrect?

The bold statement is true by my reckoning. But I don't see how it invalidates the nested hypotheticals.

What is the least number of blue people who would never leave the island? ie if there were k blue and 200-k brown, what is the least value of k that ends with noone leaving?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

jestingrabbit

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

The bold statement is true by my reckoning. But I don't see how it invalidates the nested hypotheticals.

What is the least number of bloo people who would never leave the island? ie if there were k bloo and 200-k UPS, what is the least value of k that ends with noone leaving?

To quote an earlier post:
So, A might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, B can see only 98 bloo eyed people.
So, B might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, C can see only 97 bloo eyed people.
So, C might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, D can see only 96 bloo eyed people.
So, D might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, E can see only 95 bloo eyed people.
So, E might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, F can see only 94 eyed people.

See this is not the way it works. It's skipping several steps in the logical process by attributing each of these steps to separate individuals, when really it is more like:
Step 1: So, A might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, B can see only 98 bloo eyed people.
Step 2: So A thinks that B might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, C can see only 97 bloo eyed people.
Step 3: So A thinks that B thinks that C might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, D can see only 96 bloo eyed people.
Step 4: So A thinks that B thinks that C thinks that D might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, E can see only 95 bloo eyed people.
Step 5: So A thinks that B thinks that C thinks that D thinks that E might say to itself "Hypothetically,
I might not have bloo eyes, and in that case, F can see only 94 eyed people.

Now the problem here is that since A knows that EVERYONE knows that there are at least 98 blue eyed people, and A knows that NOONE thinks that someone may think there might be less than 97 blue eyed people...

Person A (which, since every person will be following perfect logic here, applies to everyone) will NEVER reach step 3, because person A will NEVER think that someone may think that someone may think there are only 96 blue eyed people.

That is where the train of logic would fail in the 100/100/1 case.

And this translates to any number of blue eyed people above 4.

Case, 4 blue eyed people:
Each blue eyed person sees 3 other blue eyed people
Each blue eyed person assumes that the blue eyed people see 2 other blue eyed people.
Each blue eyed person knows that none of the other blue eyed people believe there is less than 1 other blue eyed people.

Since it's possible for someone to believe someone may believe that someone may see only 1 person, the recursive logic will never cause any contradiction.

With 5 blue eyed people, everyone will know that everyone knows that everyone sees at least 2 blue eyed people, in which case, noone will ever leave.
phenom

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

phenom wrote:Now the problem here is that since A knows that EVERYONE knows that there are at least 98 bloo eyed people, and A knows that NOONE thinks that someone may think there might be less than 97 bloo eyed people...

This is the problem. Person A sees 99 blue-eyed people and from that can deduce that blue-eyed person B can see at least 98 blue-eyed people. However, person A doesn't know the colour of their eyes. All that person A can say for certain is that person B can see at least 98 blue-eyed people. Person A can't make any assumptions about how person B sees them, so person A has to consider a situation in which there are 99 blue-eyed people and person B is one of them. This shouldn't be hard; just as we, looking in from the outside, can imagine that person A thinks person B might see no more than 98 blue-eyed people, we can imagine person A looking in from the outside (as person A doesn't know their eye colour and has to consider that it might not be blue), and imagining that person B thinks person C might see no more than 97 blue-eyed people.

This is the part that people have difficulty grasping, and why this thread is approaching 800 posts, an increasing fraction of which boil down to "Please read the rest of this thread". It's counter-intuitive, and that's why the Blue Eyes puzzle is a great puzzle, but it's not wrong.
Chuff wrote:I write most of my letters from the bottom

Goldstein

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

This is the problem. Person A sees 99 bloo-eyed people and from that can deduce that bloo-eyed person B can see at least 98 bloo-eyed people. However, person A doesn't know the colour of their eyes. All that person A can say for certain is that person B can see at least 98 bloo-eyed people. Person A can't make any assumptions about how person B sees them, so person A has to consider a situation in which there are 99 bloo-eyed people and person B is one of them. This shouldn't be hard; just as we, looking in from the outside, can imagine that person A thinks person B might see no more than 98 bloo-eyed people, we can imagine person A looking in from the outside (as person A doesn't know their eye colour and has to consider that it might not be bloo), and imagining that person B thinks person C might see no more than 97 bloo-eyed people.

This is the part that people have difficulty grasping, and why this thread is approaching 800 posts, an increasing fraction of which boil down to "Please read the rest of this thread". It's counter-intuitive, and that's why the bloo Eyes puzzle is a great puzzle, but it's not wrong.

Right, exactly.

Person A sees 99 blue eyed people.
Person A knows that he (person A) has either blue or brown eyes.
So person A knows that person B (any of the 99 blue eyed people) sees AT LEAST 98 blue eyed people.
And as a result, if person A assumes he (person A) has brown eyes, then person B will know that he (person B) he may have brown eyes and in that case the other 98 blue eyed people may only see 97 other blue eyed people.

This will go no further, because since person A sees 99 blue people, he knows that EVERYONE else sees AT LEAST 98 other people.
He knows that there is NO PERSON on the island who will see less than 98 people. In which case, there is NO PERSON on the island who may believe that someone else thinks there is less than 97 people.
phenom

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

He knows that there is NO PERSON on the island who will see less than 98 people.

that's true

In which case, there is NO PERSON on the island who may believe that someone else thinks there is less than 97 people.

that's not true

ED1T: I think I can give a quick explanation...although they know they don't SEE other people, SEEING is not the only way to come to conclusions
Lem0n

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

He knows that there is NO PERSON on the island who will see less than 98 people.

In which case, there is NO PERSON on the island who may believe that someone else thinks there is less than 97 people.

Edit because I was being redundant.

We can all agree that in the 100 blue eyed case, everyone sees at least 99 blue eyed people.
And we can all agree that each person KNOWS FOR A FACT, that everyone else sees at least 98 blue eyed people.

Those are irrefutable truths given this specific problem.

Based on these truths, NO ONE on the island would ever possibly think that anyone else on the island believes there are, for hyperbole's sake, only 50 blue eyed people on the island.
Do you see this?

This is true, becase EVERYONE knows that EVERYONE sees at least 98 blue eyed people.

When you're saying:
Person A sees 99 people and assumes that person B sees 98 people and assumes person C sees 97 people and assumes person D sees 96 people... It is a logical fallacy because person A knows that person B knows that person C knows that person D sees at least 97 people.
Last edited by phenom on Sun Mar 28, 2010 1:14 am UTC, edited 1 time in total.
phenom

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

they can take conclusions on what the other people are SEEING. I agree with you on that.

But they're perfect logicians who are allowed to use logic (TOGETHER with what they can SEN5E - seeing for instance) to take conclusions... and they know that the others are doing that too

ED1T: If you don't get that, try to imagine an island with 4 people. Would your theory still be valid?
Lem0n

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

ED1T: If you don't get that, try to imagine an island with 4 people. Would your theory still be valid?

If you see a couple posts up, i argue 4 is the highest number of blue eyed people on the island that the solution still holds for, and it is still in line with my theory.
phenom

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

How about this, you post the logical thought process that Person A goes through to come to the conclusion that person X may believe there are only 95 blue eyed people.

And at each step in the logical process, keep in mind that
A) Everyone sees at least 99 blue eyed people.
B) Everyone knows that EVERYONE else sees at least 98 blue eyed people
C) Since everyone is a perfect logician, everyone knows that everyone knows that EVERYONE else sees at least 98 blue eyed people.
phenom

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

do you understand that, while C is true, the fact that
Since everyone is a perfect logician, everyone knows that everyone knows that EVERYONE else sees at least 98 bloo eyed people.

does not imply what they know that they know? it just imply what they know that they SEE
and they don't need to use only vision to come to conclusions

I won't try to make the step-by-step to 95 people because it's a complicated problem and I'll need to be very careful to not commit any mistake
but if you want to start with 5 people and show me until when you agree, I can try to point why you can "keep going"
Lem0n

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

The top-down “A thinks B thinks C thinks…” approach is the way a perfect logician may actually reason, but it is quite difficult for we imperfect humans to grasp. We have trouble thinking more than a few layers deep.

But we can think from the bottom up really well.

If the puzzle were very similar but had only 1 blue-eyed person, clearly he or she would leave on the first night. This person would see no other blue-eyed people, but would also know there is someone with blue eyes, and would conclude his or her own eyes must be blue.

Spoiler:
If the puzzle were very similar but had only 2 blue-eyed people, we can easily understand what would happen. One of the blue-eyed people, call him or her A, would see exactly 1 other person with blue eyes, call that person B. Person A would think, “If my eyes are not blue then by the previous paragraph, person B will leave tonight.” Person B thinks the same thing. After neither one leaves on the first night they both think, “My eyes must be blue, because if they were not the blue-eyed person I can see would have left last night.” So they both leave on the second night.

Spoiler:
If the puzzle were very similar but had only 3 blue-eyed people, we can understand that too. Person A would see 2 blue-eyed people, B and C. Person A would think, “If my eyes are not blue, then B and C will leave on the second night as in the previous paragraph.” But B has the same thought about A and C, while C has the same thought about A and B. So they are all waiting to see what the other two do on the second night. When nobody leaves on the second night, they each conclude “My eyes must be blue” and they all leave on the third night.

Spoiler:
If the puzzle were very similar but had only 4 blue-eyed people, we can understand that too. Person A would see 3 blue-eyed people, and would think, “If my eyes are not blue, then the 3 blue-eyed people I see will leave on the third night as in the previous paragraph.” But B, C, and D have the same type of thoughts, so they are all waiting to see what the others will do on the third night. When none of them leave on the third night, each of them concludes, “My eyes must be blue,” and they all leave on the fourth night.

If the puzzle were very similar but had only 5 blue-eyed people, person A would see 4 blue-eyed people and think, “If my eyes are not blue, then the 4 blue-eyed people I see will leave on the fourth night as in the previous paragraph.” So each blue-eyed person is waiting to see what the others do on the fourth night. When none of them leaves then, they all conclude, “My eyes must be blue,” and they all leave on the fifth night.

If the puzzle were very similar but had only 6 blue-eyed people, person A would see 5 blue-eyed people and think, “If my eyes are not blue, then the 5 blue-eyed people I see will leave on the fifth night as in the previous paragraph.” So each blue-eyed person is waiting to see what the others do on the fifth night. When none of them leaves then, they all conclude, “My eyes must be blue,” and they all leave on the sixth night.

Spoiler:
We could keep going, but it is tiring to type out a whole paragraph for each integer. Let’s see if we can make any general statement. Suppose there is some integer N for which the similar puzzle with N blue-eyed people will have them all leave on the Nth night. So far we have seen this is the case for 1≤N≤6.

If the puzzle were very similar but had N+1 blue-eyed people, person A would see N blue-eyed people and think, “If my eyes are not blue, then the N blue-eyed people I see will leave on the Nth night because that is known to be what happens for N blue-eyed people.” So each blue-eyed person is waiting to see what the others do on the Nth night. When none of them leaves then, they all conclude, “My eyes must be blue,” and they all leave on the (N+1)th night.

This shows that, if the puzzle with N blue-eyed people results in them all leaving on the Nth night, then the puzzle with (N+1) blue-eyed people results in them all leaving on the (N+1)th night. That’s a pretty big ‘IF’ though. So let’s look at it closer.

Spoiler:
We know it is true for N=1, so the IF applies and it is true for N=2. The puzzle with 2 people has them both leave on the 2nd night.
We know it is true for N=2, so the IF applies and it is true for N=3.
We know it is true for N=3, so the IF applies and it is true for N=4.
We know it is true for N=4, so the IF applies and it is true for N=5.
We know it is true for N=5, so the IF applies and it is true for N=6. We showed this explicitly above.
We know it is true for N=6, so the IF applies and it is true for N=7. Now we’re breaking new ground: the puzzle with 7 people has them all leave on the 7th night.
We now know it is true for N=7, so the IF applies and it is true for N=8.
We now know it is true for N=8, so the IF applies and it is true for N=9.
We now know it is true for N=9, so the IF applies and it is true for N=10. The puzzle with 10 people has them all leave on the 10th night.
We now know it is true for N=10, so the IF applies and it is true for N=11.
We now know it is true for N=11, so the IF applies and it is true for N=12.
We now know it is true for N=12, so the IF applies and it is true for N=13.
We now know it is true for N=13, so the IF applies and it is true for N=14.
We now know it is true for N=14, so the IF applies and it is true for N=15.
We now know it is true for N=15, so the IF applies and it is true for N=16.
We now know it is true for N=16, so the IF applies and it is true for N=17.
We now know it is true for N=17, so the IF applies and it is true for N=18.
We now know it is true for N=18, so the IF applies and it is true for N=19.
We now know it is true for N=19, so the IF applies and it is true for N=20.
We now know it is true for N=20, so the IF applies and it is true for N=21. The puzzle with 21 people has them all leave on the 21st night.

Spoiler:
By the principle of mathematical induction, a set containing 1 and such that if N is in the set then N+1 is in the set, contains every positive integer. The set we are looking at is, “Positive integers N such that the similar puzzle with N blue-eyed people has them all leave on the Nth night.” We have shown explicitly that it includes 1, 2, 3, 4, 5, and 6. We established that having it true for N implies it is true for N+1. So it is true for every positive integer, including the case N=100 described in the original puzzle.

Therefore all 100 blue-eyed people will leave on the 100th night.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Lem0n wrote:do you understand that, while C is true, the fact that
Since everyone is a perfect logician, everyone knows that everyone knows that EVERYONE else sees at least 98 bloo eyed people.

does not imply what they know that they know? it just imply what they know that they SEE
and they don't need to use only vision to come to conclusions

I won't try to make the step-by-step to 95 people because it's a complicated problem and I'll need to be very careful to not commit any mistake
but if you want to start with 5 people and show me until when you agree, I can try to point why you can "keep going"

The can only use vision and their perception of what other people may be thinking

I am a person on the island
I see 99 blue eyed people
I assume I am brown eyed
I know that if I am brown eyed, the 99 blue eyed people each see 98 blue eyed people
I know that if the 99 blue eyed people assume they are brown eyed, they will know that the 98 blue eyed people they see, will all see 97 blue eyed people.

This is all fairly straight forward, correct?

Now the next step is a little more complicated.
I know that the 99 blue eyed people know that the the 98 people they see, all see at least 97 people.
Correct? If I am brown eyed, there are 99 blue eyed people that all see 98 blue eyed people, and the 99 blue eyed people assume that that 98 blue eyed people see 97 blue eyed people.

So if I know that all the (99) blue eyed people know that there at least 98 blue eyed people,
Then I know that all the (99) blue eyed people know that all the (98) blue eyed people they see, see at least 97 blue eyed people.

If I know this, then everyone else knows this.
So everyone on the island knows that everyone sees at least 97 blue eyed people, and everyone on the island knows that everyone knows this.
phenom

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

ok

# Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?
# 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?
Lem0n

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

phenom wrote:So everyone on the island knows that everyone sees at least 97 blue eyed people, and everyone on the island knows that everyone knows this.

Yes, but not everyone knows that everyone knows that everyone knows that there are 97 blue eyed people. Only the 100 brown eyed people know that.

But let me argue the conclusion in a different way. You claim that if there are 5 blue eyed people then they won't leave, but if there are 4 blue eyed people they will. Lets say the four blue eyed people would leave on night k4, and lets further assume that there are 5 blue eyed people. Now, they are each unsure of whether there are 5 or 4 blue eyed people but they know only those two states of affairs are possible. Now, night k4 will come and go, and no one will leave. A blue eyed person then knows that there are necessarily not 4 blue eyed people, else they would have left. They may therefore conclude that there are five blue eyed people, and that they are one of them. They know their own eye colour is blue, they leave that night, night k4+1.

By induction, I can show that 100 blue eyed people will leave on the 100th opportunity they are given. The nested hypotheticals are a valid way to get to the same conclusion, and they help answer the question "what information does the guru impart?" and others, but you can get away without them. I think the problem that you have is that you think that if everyone knows that everyone knows something, then this implies everyone knows that everyone knows that everyone knows it. It doesn't, but a perverse, sadistic part of me wants to see you try to prove that it does.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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

But let me argue the conclusion in a different way. You claim that if there are 5 blue eyed people then they won't leave, but if there are 4 blue eyed people they will. Lets say the four blue eyed people would leave on night k4, and lets further assume that there are 5 blue eyed people. Now, they are each unsure of whether there are 5 or 4 blue eyed people but they know only those two states of affairs are possible. Now, night k4 will come and go, and no one will leave. A blue eyed person then knows that there are necessarily not 4 blue eyed people, else they would have left. They may therefore conclude that there are five blue eyed people, and that they are one of them. They know their own eye colour is blue, they leave that night, night k4+1.

By induction, I can show that 100 blue eyed people will leave on the 100th opportunity they are given. The nested hypotheticals are a valid way to get to the same conclusion, and they help answer the question "what information does the guru impart?" and others, but you can get away without them. I think the problem that you have is that you think that if everyone knows that everyone knows something, then this implies everyone knows that everyone knows that everyone knows it. It doesn't, but a perverse, sadistic part of me wants to see you try to prove that it does.

Given that on night n+1, anyone who counts n blue eyed people will leave the island, yes you can prove that if n, n+1. On and on until 100 and prove that the solution is valid.
This is reliant on a 'hat problem' type solution where everyone on the island agrees to make their actions based on a pre-determined methodology.

In reality, with 4 or fewer blue eyed people, they will use that methodology based on logical reason, whereas with 5 or more blue eyed people, that methodology CANNOT be logically concluded.

Furthermore, it CAN be logically disproved. Apparently I've been unsuccessful at doing so.

If someone can disprove that:
A) Each blue eyed person sees 99 blue eyed people
B) Each blue eyed person logically concludes that each blue eyed person they see can see at least 98 blue eyed people
C) Each person knows that each blue eyed person (A) they see, sees at least 98 blue eyed people, and that the 99 people (A) know that those 98 blue eyed people each see at least 97 blue eyed people

Then I will accept that perhaps I'm wrong.

But we have to accept that everyone on the island sees at least 99 blue eyed people.
And that everyone on the island knows that everyone on the island sees at least 98 blue eyed people.

Accepting the nested hypotheticals upon which the solution is based, means accepting that everyone on the island (who are all supposed to be perfect logicians) believe both that everyone on the island knows that everyone on the island sees at least 98 blue eyed people ANNDDDDD that everyone on the island believes that someone on the island sees only 10 (completely arbitrary) blue eyed people.

This is VERY LOGICALLY UNSOUND
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

phenom wrote:A) Each blue eyed person sees 99 blue eyed people
B) Each blue eyed person logically concludes that each blue eyed person they see can see at least 98 blue eyed people
C) Each person knows that each blue eyed person (A) they see, sees at least 98 blue eyed people, and that the 99 people (A) know that those 98 blue eyed people each see at least 97 blue eyed people

D) Each person knows that each blue eyed person (A) they see, sees at least 98 blue eyed people, and that the 99 people (A) know that those 98 blue eyed people each see at least 97 blue eyed people, and that the 99 people also know that the 98 people each know that each of those 97 blue eyed people see at least 96 blue eyed people.

Put another way, suppose a person has blue eyes. He knows that if he does not have blue eyes, there are 99 blue-eyed people on the island, which would mean A, B, and C are all false. Thus, that person cannot know that A, B, or C is true unless he knows he has blue eyes.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

phenom wrote:In reality, with 4 or fewer blue eyed people, they will use that methodology based on logical reason, whereas with 5 or more blue eyed people, that methodology CANNOT be logically concluded.

So you say that if there are 4 blue-eyed people, they'll leave on day 4? But if there are 5 they'll never leave?

So, if I'm on the island, and I can see 4 blue-eyed people (other than myself), and day 4 passes, and noone leaves... what's stopping me from deducing my eyes must be blue?

Forget the nested hypotheticals, and answer that one simple question - what's stopping me?
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

phenom wrote:Given that on night n+1, anyone who counts n blue eyed people will leave the island, yes you can prove that if n, n+1. On and on until 100 and prove that the solution is valid.
This is reliant on a 'hat problem' type solution where everyone on the island agrees to make their actions based on a pre-determined methodology.

No, its not reliant on an agreed methodology. Its reliant on the fact that the islanders are perfect logicians, that is, that if it can be inferred by them it will be inferred by them. The blue eyed people see that no one leaves on night k4. The blue eyed islanders know that

((there are five blue eyed people) AND (I am one of them)) OR ((there are four blue eyed islanders) AND (the blue eyed islanders leave on night k4)).

When night k4 passes, they then know

NOT (the blue eyed islanders leave on night k4).

The only way to make the first statement true is then for the first clause in the OR to be true, because the second is false. From that they can conclude that they have blue eyes. They don't need to agree anything, they just have to be able to determine what would occur if there are only 4 blue eyed islanders. And they can and will, because they're perfect logicians.

Again, there's nothing wrong with your A, B and C, I just don't see how that invalidates the hypotheticals. You say

ANNDDDDD that everyone on the island believes that someone on the island sees only 10 (completely arbitrary) blue eyed people.

But that isn't what's happening. Someone is hypothesizing that someone is hypothesizing that someone is hypothesizing that ... someone is hypothesizing that there might only be 10 blue eyed people. Everyone knows that everyone can see at least 98 blue eyed people (and the brown eyed know that everyone can see at least 99) but, as you correctly identify, that doesn't mean that everyone knows that everyone knows that everyone can see 98 blue eyed people. Likewise, not everyone knows that everyone knows that everyone knows that everyone knows there are 97 blue eyed people on the island.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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

jestingrabbit wrote:
ANNDDDDD that everyone on the island believes that someone on the island sees only 10 (completely arbitrary) blue eyed people.

But that isn't what's happening. Someone is hypothesizing that someone is hypothesizing that someone is hypothesizing that ... someone is hypothesizing that there might only be 10 blue eyed people. Everyone knows that everyone can see at least 98 blue eyed people (and the brown eyed know that everyone can see at least 99) but, as you correctly identify, that doesn't mean that everyone knows that everyone knows that everyone can see 98 blue eyed people. Likewise, not everyone knows that everyone knows that everyone knows that everyone knows there are 97 blue eyed people on the island.

Right.

Say you're on an island, and you can see 97 blue-eyed people. You know that there's either 97 or 98 blue-eyed people on the island, total. You hypothesise that your eyes might not be blue, in which case there would be a person on the island who can only see 96 blue-eyed people. Such a person may or may not exist (if your eyes are blue), but you can't discount it.

Now, say I'm on the island (a new island, but the same rules), and can see 98 blue-eyed people, of which you are one. I know that there's either 98 or 99 blue-eyed people on the island, total. Now, I hypothesise that my eyes might not be blue, in which case you would see 97 blue-eyed people, and you would thus reason as above. So I'm hypothesising that you're hypothesising that there might be a person on the island who can only see 96 blue-eyed people. Now, I know for sure that such a person doesn't exist, and it's possible that you know that too (if my eyes are blue), but I can't discount the possibility that you can't discount it.

Now, say jestingrabbit on the island (again, a new island, but the same rules), and can see 99 blue-eyed people, of which you and I are two. He knows that there's 99 or 100 blue-eyed people on the island, total. Now, he hypothesises that his eyes might not be blue, in which case I can see 98 blue-eyed people, and would thus reason as above. So he's hypothesising that I'm hypothesising that you're hypothesising that there might be a person on the island who sees only 96 blue-eyed people. He knows that no such person exists, and he knows that I know that no such person exists, and he knows that you know that no such person exists... but he can't discount the possibility that I can't discount the possibility that you can't discount the possibility that such a person exists. Because while he knows that you know that no such person exists, he doesn't know if I know that fact (that you know that no such person exists)... and, indeed, as far as he knows, it's possible that I don't.

You see the pattern? This can be continued as far as we like. The language gets increasingly more painful with each layer, but it's still logically valid.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

Say you're on an island, and you can see 97 blue-eyed people. You know that there's either 97 or 98 blue-eyed people on the island, total. You hypothesise that your eyes might not be blue, in which case there would be a person on the island who can only see 96 blue-eyed people. Such a person may or may not exist (if your eyes are blue), but you can't discount it.

Now, say I'm on the island (a new island, but the same rules), and can see 98 blue-eyed people, of which you are one. I know that there's either 98 or 99 blue-eyed people on the island, total. Now, I hypothesise that my eyes might not be blue, in which case you would see 97 blue-eyed people, and you would thus reason as above. So I'm hypothesising that you're hypothesising that there might be a person on the island who can only see 96 blue-eyed people. Now, I know for sure that such a person doesn't exist, and it's possible that you know that too (if my eyes are blue), but I can't discount the possibility that you can't discount it.

Now, say jestingrabbit on the island (again, a new island, but the same rules), and can see 99 blue-eyed people, of which you and I are two. He knows that there's 99 or 100 blue-eyed people on the island, total. Now, he hypothesises that his eyes might not be blue, in which case I can see 98 blue-eyed people, and would thus reason as above. So he's hypothesising that I'm hypothesising that you're hypothesising that there might be a person on the island who sees only 96 blue-eyed people. He knows that no such person exists, and he knows that I know that no such person exists, and he knows that you know that no such person exists... but he can't discount the possibility that I can't discount the possibility that you can't discount the possibility that such a person exists. Because while he knows that you know that no such person exists, he doesn't know if I know that fact (that you know that no such person exists)... and, indeed, as far as he knows, it's possible that I don't.

This is a very good representation of the way the nested hypotheticals are being misused.

You are attributing the thought process of each step of the hypothetical to that individual themselves, when in reality, it's being thought through by the top person in the nest of hypotheticals and as such, has all the information that the top person in the nest of hypotheticals has.

The most important flaw in your logic is that in each step of your hypothetical, you are putting people into a new island in which there ARE only x-y blue eyed people on the island. This is where everyone seems to be misunderstanding the way the hypotheticals would logically be conducted by each of the people on the island.

1) You're on an island, you see 99 blue eyed people.
2) You hypothesize that you are brown eyed in which case the 99 blue eyed people all see 98 blue eyed people.
3) You hypothesize that the 99 blue eyed people will hypothesize that they too are brown eyed, in which case the 98 blue eyed people will all see 97 blue eyed people.
We all agree on these three statements it seems. At their fundamental base meaning, we all agree on the above.

Now do we agree that for the solution as presented to work, we have to prove that someone may hypothesize that someone may hypothesize .... .... that someone may hypothesize that that the blue eyed people may only see 95 blue eyed people?

I see 99 blue eyed people.

I hypothesize that I am brown eyed and that the 99 blue eyed people will all see 98 blue eyed people.

I hypothesize that the 99 blue eyed people will hypothesize that they are brown eyed people and as a result, the 99 blue eyed people will hypothesize that the 98 blue eyed people they see will hypothesize that they too are brown eyed and the 98 blue eyed people will all see 97 blue eyed people.

I hypothesize that the 99 blue eyed people will hypothesize that that 98 blue eyed people will hypothesize that the 97 blue eyed people will all see 96 blue eyed people.

Given that I know everyone sees at least 98 blue eyed people, and that I know everyone knows everyone sees at least 97 blue eyed people, I know no one will ever think someone sees 96 blue eyed people.
And given that I know no one will ever think someone sees only 96 blue eyed people, I know that no one will ever think someone thinks someone sees 95 blue eyed people.
And finally, given that I know that no one will ever think someone thinks someone sees 95 blue eyed people, I know that no one will ever think someone thinks someone thinks someone may only see 94 blue eyed people.

NOTE: I do realize that prior to this post I was accepting one too few iterations as legitimate. With 100 and 100, each blue eyed person can logically recurse himself to the possibility of only 96 blue eyed people.
In line with this realization, 5 blue eyed people will still hold. 6 is the lowest number of blue eyed people with which this solution will fail.
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

No, its not reliant on an agreed methodology. Its reliant on the fact that the islanders are perfect logicians, that is, that if it can be inferred by them it will be inferred by them. The blue eyed people see that no one leaves on night k4. The blue eyed islanders know that

((there are six blue eyed people) AND (I am one of them)) OR ((there are five blue eyed islanders) AND (the blue eyed islanders leave on night k5)).

When night k5 passes, they then know

NOT (the blue eyed islanders leave on night k5).

The only way to make the first statement true is then for the first clause in the OR to be true, because the second is false. From that they can conclude that they have blue eyes. They don't need to agree anything, they just have to be able to determine what would occur if there are only 5 blue eyed islanders. And they can and will, because they're perfect logicians.

If you see my last post, I've corrected myself. I misspoke when I said 5 wouldn't work, it will still logically work, 6 is the lowest number that it will fail. Ive edited your quoted post to represent this.

The part of your post that I disagree with is:
The blue eyed islanders know that

((there are six blue eyed people) AND (I am one of them)) OR ((there are five blue eyed islanders) AND (the blue eyed islanders leave on night k5)).

Now explain to me how the blue eyed islanders would know that, if there were five blue eyed islanders, they would all leave on night k5? and don't say 'because the 5 blue eyed islanders saw that no one left on night 4'
With 5 people it works as follows:
I see 4 (W) blue eyed people.
I know that the W see at least 3 (X) blue eyed people.
I know that the W know that the X see at least 2 (Y) blue eyed people.
I know that the W know that the X may believe the Y only see 1 (Z) blue eyed person.
In which case after night one, I have gained new information. More specifically, I know that someone gained more information:
I know that the W know that the X NOW KNOW that the Y know there are at least 2 blue eyed people
I know that the W know that the X NOW KNOW that the Y know there are at least 3 blue eyed people.
I know that the W know that the X NOW KNOW that the Y know there are at least 4 blue eyed people.
I know that the W know that the X NOW KNOW that the Y know there are at least 5 blue eyed people.
At which point, the blue eyed people all know there are 5 blue eyed people and that they are one of them, and they leave on the fifth night.

With 6 people it (doesn't) work as follows:
I see 5 blue eyed people.
I know that the 5 blue eyed people see at least 4 blue eyed people.
I know that the 5 blue eyed people know that the 4 blue eyed people see at least 3 blue eyed people.
I know that the 5 blue eyed people know that the 4 blue eyed people may believe the 3 blue eyed people only see 2 blue eyed people.
In which case, no one expected anyone to leave on night 1, everyone knew that NO ONE would leave on night 1. Furthermore, no one expected anyone to expect anyone to expect someone to leave.
Since no one expected anyone to leave on night 1, no one gained any new information on day 2 morning, and so no one will leave day 2 night.
Since no one expected anyone to leave day 2 night, no one gains any new information day 3 morning.
And so on and so on.
phenom

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

phenom wrote:You are attributing the thought process of each step of the hypothetical to that individual themselves, when in reality, it's being thought through by the top person in the nest of hypotheticals and as such, has all the information that the top person in the nest of hypotheticals has.

There's the information the top person has, and there is the information that he imagines that the nth person down has, which are not the same. Yes, it's the top person doing the reasoning, but when he reasons about what the nth person down will know and will be able to deduce, he is only determining what that person will be able to figure out from the information he is imagined to possess.

Imagine you were a cave man living in 40,000 BCE, and you saw lightning for the first time. What would you think was going on? Now you're actually not a cave man, and you presumably know what lightning is, but when you imagine yourself as a caveman, you deny yourself access to that information, and simulate reasoning without it. That's exactly what's going on here, and why the information the top person has is irrelevant when determining how the hypothetical person will reason. All that's relevant is the information the hypothetical person has.

NOTE: I do realize that prior to this post I was accepting one too few iterations as legitimate. With 100 and 100, each blue eyed person can logically recurse himself to the possibility of only 96 blue eyed people.
In line with this realization, 5 blue eyed people will still hold. 6 is the lowest number of blue eyed people with which this solution will fail.

Now what happens when I am one of the six blue-eyed people on the island, and nobody left on night 5? What stops me from deducing that since nobody left on night 5, my eyes must be blue?
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### Re: My write-up of the "Blue Eyes" solution (SPOILER A

phenom wrote:With 6 people it (doesn't) work as follows:
I see 5 blue eyed people.
I know that the 5 blue eyed people see at least 4 blue eyed people.
I know that the 5 blue eyed people know that the 4 blue eyed people see at least 3 blue eyed people.
I know that the 5 blue eyed people know that the 4 blue eyed people may believe the 3 blue eyed people only see 2 blue eyed people.
In which case, no one expected anyone to leave on night 1, everyone knew that NO ONE would leave on night 1. Furthermore, no one expected anyone to expect anyone to expect someone to leave.
Since no one expected anyone to leave on night 1, no one gained any new information on day 2 morning, and so no one will leave day 2 night.
Since no one expected anyone to leave day 2 night, no one gains any new information day 3 morning.
And so on and so on.

You admit it works with 5: if there are 5 blue-eyed people on the island, all will leave on night 5. The reason is not important. But you say it doesn't work with 6. If I'm one of 6 blue eyed people, I know that either my eyes are blue or they aren't. If they aren't, the blue-eyed people will leave on night 5. That you accept (again, the reasoning is not important). And, being a perfect logician, I know this. So if nobody leaves on night 5, I know my eyes are blue.

With this in mind, how is your 5 works but 6 doesn't position at all consistent?

The nested hypotheticals are not important for this reasoning in any way. The only real benefit to the explanation involving nested hypotheticals is that it answers the question of what new information the guru's statement conveyed, and what new information is being learned each night nothing happens.

Double-posting rather than editing since I'm guessing phenom already read my previous post, and may not notice an edit.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

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

I guess the confusion here is that the fact that "I know that A, B and C knows X and that they know that everyone else knows X" doesn't mean that "I know that A knows that B knows that C knows X"

do you disagree with that statement?
Lem0n

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

phenom wrote:You are attributing the thought process of each step of the hypothetical to that individual themselves, when in reality, it's being thought through by the top person in the nest of hypotheticals and as such, has all the information that the top person in the nest of hypotheticals has.

Well, sort of. What I'm saying is "in hypothetical situation X, person A would do thing 1" and then "in hypothetical situation Y, person B would see person A as potentially being in that situation X, and thus doing thing 1... so, in response to that possibility, person B will do thing 2" and then "in hypothetical situation Z, person C would see person B as potentially being in that situation Y, and thus doing thing 2... so, in response to that possibility, person C will do thing 3".

Essentially, building the nested hypotheticals up from the inside out, instead of vice-versa.

phenom wrote:The most important flaw in your logic is that in each step of your hypothetical, you are putting people into a new island in which there ARE only x-y blue eyed people on the island. This is where everyone seems to be misunderstanding the way the hypotheticals would logically be conducted by each of the people on the island.

I was very clear to say that, each time, I was talking about a brand new situation, completely divorced from the previous situation:
phlip wrote:(a new island, but the same rules)
That is, I'm not "putting new people on the island". All I'm saying is that "if the situation was X, people would act like Y... I can't prove that the situation isn't X, so people may be acting like Y"... where for each one, the previous paragraph is simply justification for the fact that, if the situation was X, people would act like Y.

But still: you haven't taken any shots at the inductive proof. To quote myself:
phlip wrote:So you say that if there are 5 blue-eyed people, they'll leave on day 5 (not just "can" leave, or "might" leave, but will leave)? But if there are 6 they'll never leave?

So, if I'm on the island, and I can see 5 blue-eyed people (other than myself), and day 5 passes, and noone leaves... what's stopping me from deducing my eyes must be blue?

Forget the nested hypotheticals, and answer that one simple question - what's stopping me?
(numbers changed from 4 and 5 to 5 and 6 to match your change of position... and extra clarifying parenthesis added.)

Remember that if there are two proofs of thing A, and you want to claim thing A isn't true, you need to find flaws in both proofs. Similarly, if there's a proof of A (the inductive proof) and a proof of not-A (your mistaken belief that the guru gives no new information) then either one of the proofs is flawed, or logic is inconsistent. And I think we can discount the "logic is inconsistent" option for now.

So if you want your position to hold water, you're going to need to say what's wrong with the quoted.

Oh, wait, I missed it, you did take a shot at it:
phenom wrote:Now explain to me how the blue eyed islanders would know that, if there were five blue eyed islanders, they would all leave on night k5?

Because you said:
phenom wrote:I misspoke when I said 5 wouldn't work, it will still logically work, 6 is the lowest number that it will fail.

That is, you specifically said that if there are 5 blue-eyed people, it'll work, and they'll leave. This is a fact that is independent of any facts about the actual situation on the island, and all the islanders are perfect logicians, so they all know that (there are five blue-eyed islanders --> the blue eyed islanders leave on night 5). And its contrapositive (the blue-eyed islanders do not leave on night 5 --> there are not five blue-eyed islanders).

Your post seems to suggest that if I'm a blue-eyed person who can see 4 blue-eyed people, we'll all leave... but if I'm a brown-eyed person who can see 5 blue-eyed people, then they won't leave... which is clearly ridiculous, as these are the same situation, just with a different choice of who "I" am (a choice which should not and does not affect the actual actions of anyone, and is merely a device for talking about the problem in a natural language).
While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

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

prediction of the next message: "oops, I misspoke when I said 6 wouldn't work, it will still logically work, 7 is the lowest number that it will fail."
Lem0n

Posts: 87
Joined: Tue Dec 15, 2009 8:15 pm UTC

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