Blue Eyes

i don't follow

it didn't say that, so the guru would be left behind, since they can not communicate to her.

she doesn't know what color her eyes are. even if she knows that there are only three colors.

The guru sees the ferryman's eyes that ere blue

Yet when the 100 brown eyed leave the blue eyed has enough information to go through the same cycle and will all leave after 100 days more

i don't know what you're trying to say, but:

the only way they can gain meaningful information from the guru is if there's only one blue-eyed person in the crowd. being perfect logicians, they would realize this and have a different crowd show up each day at noon or have a different set of people in the crowd close their eyes at noon.

then, she waits for the day that all but one blue eyed person has their eyes closed or aren't present. at this point, everyone would look around. the blue eyed person would know their eyes are blue since everyone else has brown eyes, the brown eyed people would know their eyes are brown since they can only see one blue person, and the rest of the blue eyed people (either within earshot but not present, or just closing their eyes) would then realize they are blue eyed by the fact that she made the announcement on the day that they had their eyes closed /didn't show up.

so everyone gets to leave that day.

correction:
>the brown eyed people would know their eyes are brown since they can only see one blue person
they would know because she made the announcement on that day and they can see a blue eyed person in the crowd.

Did you read my initial post? Imagine the scenario where there are only 3 people of each color and you'll see what I mean (unless I'm spectacularly wrong of course)

So, even though it was really fucking annoying to wrap my head around it, a blue-eyed person sees 99 blue-eyed people. Everyone with blue eyes should be thinking "the guru was talking about somebody else". So, when nobody leaves at the 99th night, each blue-eyed person will know that he also has blue eyes, right?

Beyond that, brown-eyed people can't ever leave the island because no one can be certain that he doesn't have a different eye color, as stated in the second paragraph, nowhere is it mentioned that they know that there are only three eye-colors.

i don't understand your post. can you explain it another way or more explicitly?

Imagine there's only 2 people with blue eyes (labeled A and B) and a guru. Now on day 1 A thinks "Option 1: I have brown or green eyes and then B will not see any blue eyed people thus A will leave. Option 2: I have blue eyes, thus B will not be able to deduce that B is alone, thus B will not leave tomorrow"

They both have the same logic, so next day they see each other and instantly realize they both have blue eyes.

Make sense?

except she didn't say how many people with blue eyes she could see, just that there is at least one person with blue eyes, which tells you nothing you didn't already know

not really.

>Now on day 1 A thinks "Option 1: I have brown or green eyes
okay
>and then B will not see any blue eyed people thus A will leave.

but B can not communicate this information back to A. how does this follow?

>The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island.
This means the Guru speaks only once, ever.

It's implicitly conveyed by B not leaving

I do think I'm missing something though because since there were 100 blue eyed people the guru didn't add any info

exactly. i don't think people are getting this.

nobody can be sure what color their eyes are unless only one blue eyed person is showing their eyes at the time she makes her announcement.

I'm not sure I agree that that is what she said. She said "someone." Singular.

So everybody, as they can all see more than one blue-eyed person, knows she is lying and presumably they kill her and toss her carcass onto the boat.

Also, on night 100 all the blue eyed people leave.

only the blue eyed people leave on the hundredth day

the red eye thing caught me up for a second, but it really only fucks with the brown eye people after thinking about it

can someone tell me what i'm missing, because none of the other replies make any sense whatsoever.

if all you ever saw was brown or blue eyes after the oracle said there is at least one blue eyed, you'd leave immediately since you know that the blue eyed person was you, if nobody leaves on the first day that means everyone sees at least two blue eyed people, on the second day if you only see one other blue eyed person that means that you have blue eyes and can leave, else you just repeat the process as many times as there are blue eyed people before you can leave

On the 100th day, everyone with blue eyes leaves. Then on the 200th day, all the brown eyed people leave.

shit, meant just brown eyes at the beginning

and i guess they only see at least one blue eyes meaning there is at least two blue eyes

I think what he meant to type is
>A thinks "I have brown or green eyes, B will not see any blue eyed people, B will leave because he deduced he is the lone blue eyed person"

They can't know what color eyes they have, the guru adds nothing they don't already know. Every single islander can see a person with blue eyes but can never know what color their own eyes are given the parameters listed in the OP. Should the parameters change to exact numbers of only 2 eye colors then yes, they could work with that. As it stands, they cannot.

okay, tell me how the first person leaves. the only way i can see people getting off the island is in my previous post:

>tell me how the first person leaves
all the blue eyed people leave together, since no one leaves on the first day, they know there can't be just one, since no one leaves on the second day, they know there can't be just two etc.

I'm not sure whether to be sad or amused that there are people on this board who are incapable of googling clear explanations of the answer to this riddle (and why you're wrong).
This thread is as bad as those Monty Hall threads.

Sorry m8, given the parameters there is no solution to this puzzle, there are several solvable variations, this isn't one of them. If you got an answer you made a mistake and are too stupid to know it, or are just trolling, either way fuk off m8.

The only sources of new information are:

1. Obviously the guru
2. The remaining people the next day

The guru's information is the base/initial case. The remaining people the next day act as the inductive step.

Sadly, only the blue eyed people escape on the 100th day. I cheated, but the puzzle is still fun.

The guru could have saved everyone but offered a riddle instead... I think the guru was a mathematician.

There are clear explanations online. You're just too stupid to understand them. Good luck with those ad hominem arguments up against inductive proofs, though.

my solution works, at least.
it does not require communication, and only assumes that they're all trying to get off the island and willing to work cooperatively.

i don't get most of the other replies, and suspect the op modified the question from its usual form, whatever that may be. at least monty hall is somewhat confusing.

actually, the only arguable flaw in my solution is that they might not know ahead of time whether she will say anything or what she will say. so it does also assume they know that she can/will confirm that a blue-eyed member of the crowd is present when appropriate, or at least that she can confirm/deny the presence of a certain eye color. in that case, the "logical" choice is blue, since that's the first color by alphabetical order.

even then though, she could just wait for the correct members to be in her presence just by accident, which will likely happen sooner or later, and then make the announcement.

Islanders form 1 long row. Taking turns looking face to face at the guru
>guru sees first blue eyed person around noun
Everyone knows eye colour now

they all leave the island, except the guru

OP here, there is no modification

so did i get it right?

i need to escape the island that is Veeky Forums

Is it some modal logic shit like, "you now know that the guru knows that there are some people with blue eyes, and the guru knows that you know..."?

This is a great explanation.

Kind of a dick move on the gurus part. I guess the guru didn't like that the brown eyed people were full of shit.

You're wrong because you have no reading comprehension.

The guru does not speak every day at noon. The guru speaks once on one single day and says what she says. She never speaks before or after that.

That is not what modal logic is.

To understand this problem, you need to use nested imagination (also known as "inception"), and to pay attention to the base case. The only ways to extract information are reality pigeonhole principle (the case where there is only one blue eyed person on the island), and the simulation reaction property (where people simulate reality, and predict people leaving or staying based on those simulations). In every simulation where there are more than one blue eyed person, it is guaranteed that the simulated people will not deduce anything about their own identity on the first day, so you need to nest enough simulations inside of simulations until you can extract informaiton on the first day. Obviously, simulations that expand the number of blue eyed people will never be helpful, as having multiple blue eyed people in a simulation nets no information on day one, so you need to focus your simulations on the only case that leads to information on day one: the single blue eyed person case. Let each of the blue eyed people explore the possibility of having brown eyes. If they do this, then they are imagining an island with 99 blue eyed people on it. Now, because this island is a simulation of reality, each of the 100 simulations can explore the possibility of the 99 blue eyed people assuming they each have brown eyes. This produces 99 new simulations, each with 98 blue eyes, etc, etc... Now, the inception simulations are considering one blue eyed person hypothetical, so, after day one, those simulations are resolved, leaving the two blue eyed person simulations with critical information. Waiting another day, the two blue eyed person simulations resolve, sending critical information to the three blue eyed person simulations, etc, etc. So, after 100 days, the blue eyed people will gain all the information they need to know they are blue eyed, and the brown eyed people leave on day 101 (if they had blue eyes, everyone would have waited longer).

>So, even though it was really fucking annoying to wrap my head around it, a blue-eyed person sees 99 blue-eyed people. Everyone with blue eyes should be thinking "the guru was talking about somebody else". So, when nobody leaves at the 99th night, each blue-eyed person will know that he also has blue eyes, right?

Actually no. If there was someone with blue eyes and only one other person in the crowd with blue eyes and both didn't leave, one can conclude that. Otherwise, if there are more than two people with blue eyes in the whole crowd, they can assert nothing.

Ugh. Nevermind, I am retarded and can't read.

Actually it pretty much is what the solution is.

It's a generalization of the muddy children problem for K = 100. And they do indeed need to have "common knowledge" to solve the puzzle.

I don't however think this is solvable in the same way since there isn't a query on the population as the problem is stated. So there is never any new information gained except common knowledge. I don't think anyone leaves the island ever.

In this problem the "common knowledge" is deduced by each individual person each day. Even though they cannot directly communicate to each other, they can still observe each others presence on the island. By people still being on the island, the common knowledge is updated on a daily basis.

Ah, that makes sense. You've convinced me.

Blue eyed people are superior so they'd end up killing everyone else off and have more resources to live off of and wouldn't want to leave

everyone with blue eyes leaves on the 100th (or Nth) day

if we imagined a new problem where there are only 2 blue eyed people, though the guru doesn't specify who she's talking about, both blue eyed people can see only one other person with blue eyes. They know that if their own eyes are not blue, the person with blue eyes will realize that they are the one with blue eyes and leave immediately, when neither of them leave on the first day they will know that their own eyes are also blue and leave on the 2nd day.
if we add a 3rd person with blue eyes, they will recognize the logical equation of the first paragraph among the two blue eyed people they can see, and when those two people do not leave on the 2nd day they will know that they also have blue eyes and leave on the 3rd day.
4 is when it starts getting tricky, now we have to think from the perspective of one person through the eyes of another; each blue eyed person sees 3 people with blue eyes, and they know that if their own eyes are not blue then each of those three blue eyed people sees only two people with blue eyes and will leave on the 3rd day, when they do not leave on the 3rd day they know that their own eyes are blue as well.

In theory this logic continues to work for any number of blue eyed people, so that if you observe N blue eyed people and they do not leave after N days then you also have blue eyes. The rest can never leave because all the blue eyed people leave after the same number of days as the number of blue eyed people they see.

ftp://ling.upenn.edu/facpapers/robin_clark/courses/lx106/slides/intro99.pdf

reminder to those who can't understand the question:

the guru speaks one time, not every day at noon. one single time

The guru only speaks one time, ever? Or is it once per day?

Night one: "I can see someone who has blue eyes".
Everyone guesses that they have blue eyes, meaning that everone with blue eyes leave the island. If you do not leave the island, you don't have blue eyes.

Night two: "I can see someone who has brown eyes".
Everyone guesses that they have brown eyes, meaning that everyone who has bbrown eyes leave the island.

Assuming that the guru tells the eyecolour of a person that is currently on the island and that is not the guru, it will only take 2 nights for everyone to leave the island.

Fuck me if I'm wrong, but shouldn't this be the solution?

>The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island.
>once
>on one day
>in all their endless years on the island
One time ever

EDIT: -2 nights for everyone but the guru to leave the island.

Oh, I guess I misread. I thought that it was once a day.

Fuck me if I'm wrong v2:

Night one: "I can see someone who has blue eyes".
Everyone guesses that they have blue eyes, meaning that everyone with blue eyes leave the island.

Night two:
Everyone guesses that they have the eyecolour of the guru, meaning that no one leaves.

Night three:
Everyone guesses that they have the eyecolour of an arbitrary non-guru islander, meaning that the only eyecolour that can be chosen is brown and that everyone on the island except the guru leaves.

Should the guru leave the island? Do they deserve it?

Wait shit fuck me I am too illiterate for this shit.
I have been a stupid user. I realise my mistakes.
Delete me.

...

If you're the only blue-eyed person on the island
>Oh shit, no one else has blue eyes so it must be me. Time to leave

If there is one other blue-eyed person
>If this guy is the only one with blue eyes, he should realize it and leave on the first day.
>Day two: He hasn't left yet, so he must see someone else with blue eyes. Since I can't see them, it must be me.

If there are two other blue-eyed people.
>If these are the only blue-eyed people on the island, they will both leave on the second day.
>Day three: They haven't left, that means there's one more blue-eyed person and it must be me. Time to leave.

If there are three other blue-eyed people.
>If these are the only blue-eyed people on the island, they will both leave on the third day.
>Day four: They haven't left, that means there's one more blue-eyed person and it must be me.

etc.

Alright, this is pretty simple.

In order to find the solution, let's imagine a group of 4 citizen on this island plus the Guru. We know, for fact, that 50% of the citizen are blue-eyed (BL) and 50% are brown-eyed (BR), however no citizen knows their own color. Nevertheless, each citizen knows the colors of all the other citizen. With this in mind, in our 4 citizen scenario, 2 would have blue eyes and 2 would have brown eyes. As long as they don't get any additional information, it is impossible to leave the island (other than guessing, which we'll exclude as a solution). For the sake of order, let's take a look at what each citizen knows so far:

>BL citizen:
2 are BR
1 is BL

>BR Citizen
2 are BL
1 is BR

Now comes the interesting part: after X number of days, at noon, the Guru has this vision and declares: "One person other than myself on this island has blue eyes."

This information is crucial for our purely logical residents. With this in mind, they ALL know for fact that at least one of them has blue eyes. The BR citizen can't really deduce anything right off the bat as they already know that there are at the very least 2 BL among them. However, the BL citizen so far only know that there is 1 BL on this island, meaning that, in case said citizen didn't see any other blue eyes, he would be able to leave the island since he would be able to conclude "Nobody has blue eyes, but I know there is at least one blue eyed person, thus I must be the person with blue eyes."; nevertheless, after midnight, on the next day, everybody notices that no citizen has left, meaning that now both BL citizen know that there must be at least 2 BL citizen. They both conclude that they are those two and leave with the following ferry (2nd). Since both left, each BR citizen knows for fact that there can't be any other BL citizen. Unfortunately, that is not enough for them to conclude anything about their own eyes. The BR residents will stay forever.

Part 1/2

>>If these are the only blue-eyed people on the island, they will both leave on the second day.
that is not a logical conclusion that can be drawn. they would not know they are the two blue eyed people, especially if you have blue eyes

>that is not a logical conclusion that can be drawn.
Yes it is, see >If there is one other blue-eyed person
>>If this guy is the only one with blue eyes, he should realize it and leave on the first day.
>>Day two: He hasn't left yet, so he must see someone else with blue eyes. Since I can't see them, it must be me.

> they would not know they are the two blue eyed people, especially if you have blue eyes
but if you have blue eyes then they aren't the only blue-eyed people. What are you trying to say?

Part 2/2

If we expand this scenario to have 6 citizen instead of 4, with groups of 3 for each color instead, we'll get the following initial situation:

>BL citizen
3 are BR
2 are BL

>BR citizen
3 are BL
2 are BR

Again, as before, after the Guru's prophecy, the BR citizen are unable to deduce anything and thus completely rely on the reasoning of the BL citizen. Now, the BL citizen can see that there must be are at least 2 BL citizen. If there were only 2 BL citizen in total, the same logic from before (4 citizen total) would apply because the BL citizen would only see 1 other BL member, which means that they would both leave with the second ferry. However, since nobody leaves with the second ferry, they all three conclude that there must be at the very least 3 BL citizen, thus they all three leave with the 3rd ferry (or on the 4th day). Once again, the BR citizen can still not conclude anything on their own eyes apart from knowing that it's certainly not blue.

In a nutshell, this concept can be continued indefinitely. So, to conclude, the solution to the initial question is:

All blue-eyed citizen leave on the 100th night after the Guru's declaration. None of the brown-eyed citizen leave, nor does the Guru.

...

Someone looking at your eyes means you have blue eyes

Easiest way to see it is as follows:

1. Since there are two people with blue eyes then each person with blue eyes sees one other person with blue eyes.

2. On the first night, no one leaves the island. This means that neither person with blue eyes knows that he has blue eyes.

3. This means there is another person with blue eyes. Since you can see every person on the island and they all have brown eyes except the one, it follows that you must have blue eyes.

4. The other agent reasons similarly and they both leave on day 2.

No this stops working when there are at least 3 blue eyed people (your third paragraph).

Why is that the case? On the first day, all three only see 2 eyes. On the second day, all three see the same. Now if there was only 2 blue eyes people, they both would have realized and left by the second day (as you already agree). So when that 3rd day comes, and they still see 2 people with blue eyes, they know there must be more than 2 blue eyed people (otherwise they would have left). Since there's more than 2 but they only see 2, they conclude they are the 3rd. This works for any amount of people, it will just take more and more days as you add more and more people.

Knowing that I am superior to these worms that share the island with me, I logically conclude that I must be part of the blue eyed Aryan master race.

Before green eyes speaks the magic words, I honor the wizard by cannibalizing the green eye's flesh, starting with the upper right earlobe. Naturally, I gain the wisdom of a 1000 men. I take a big nap afterwards because boy am I stuffed!

The ship gives me a lift in the evening and I ditch the subhumans on the island (dirty brown eyes) to join the many Gods on Mt Olympus and eternally party with my boy Dionysus and my slutty boo Aphrodite.

Just step into the shoes of one of the blue-eyed people in that situation and try to imagine it, it really doesn't work. Your reasoning is just "If there were 2 blue-eyed people..." which just doesn't scale in this case. I wish I could explain this better...

What a shitty guru

But how is what the guru says in any way new information? Before he ever gets to say a single word, everybody already sees at the very least 99 blue eyed people and 99 brown eyed people, and everyone knows that everybody else sees at least 98 of each color. So "1+ people have blue eyes" isn't new information for anyone. Anyone could have made that statement, and knew that the guru could make it.

It's okay m8, there's nothing wrong with having an IQ of 100.

This exactly. That's why trying to prove this riddle with only two blue-eyed people does not scale.

It does scale though. The guru saying "I see 1+ blue eyed people" gives everyone on the island a chance to synchronize their clocks. You need to remember that they can't talk to each other, so even something as simple as a starting point makes it possible.

If you understand why this works for 2 people, you can extrapolate. But I'm going to explain both cases.
>2 blue eyed people
on day 1, each blue eyed person only sees 1 other blue eyed person.
on day 2, each blue eyed person sees that the one blue eyed person hasn't left.
therefore there must be another blue eyed person, but since they can't see one they deduce it is them.
they both leave the island
>3 blue eyed people
on day 1, each blue eyed person sees 2 other blue eyed people.
They will assume the above scenario will occur.
by the end of day 3 the above scenario has not occurred, so each person knows there must be more than 2 blue eyed people (otherwise the scenario would have occurred)
If each person only sees 2 other blue eyed people but they know there is more than 2, they deduce that the last blue eyed person is himself.
You can continue this extrapolation for arbitrarily large numbers. As long as the previous scenario doesn't occur, that's enough information to deduce for the next scenario.

Think of it this way: everyone follows the rule "If I see n blue-eyed people, I will leave on night n+1, unless they've already left."

Every brown-eyed person will see all the blue-eyes and want to leave the night after every blue-eyed person wants to leave.

>They will assume the above scenario will occur.
This is the flaw in the reasoning. The "above scenario" cannot happen precisely because there are three blue-eyed people. When there are at least 3 blue-eyed people, there is no situation where any of them can say "They should have already left if I didn't have blue eyes".

This isn't really an explanation.

>This isn't really an explanation.
It's a strategy.

The exact same strategy, in fact, just written differently.

And if it's not obvious that it works, kill yourself.

Wait. Since apparently every blue eyed person can see 99 other blue eyed people they for a fact know that there are 99 or more blue eyed people.

Why does it need to be stated by the Guru that there is at least one blue eyed person among them?

>everyone knows that everybody else sees at least 98 of each color
But not everyone knows that everybody else knows that everybody else sees at least 98 of each color (and so on).
Blue-eyed person 1 thinks that if he is not blue-eyed then blue-eyed person 2 could think that she is not blue-eyed and that she also can't assume that every other blue-eyed person sees more than 97 blue-eyed persons.
If you continue that reasoning then the person 1 thinks that the person 2 thinks that ... the person 99 thinks that the person 100 can't see anyone with blue-eyes. After noone leaves on the first day, imaginary person 99 realizes that he must also have blue eyes because otherwise the person 100 would've left.

>But not everyone knows that everybody else knows that everybody else sees at least 98 of each color (and so on).
They do:
>... Everyone can see everyone else at all times....
>... Everyone on this island knows all the rules in this paragraph,

It is a fact that every blue-eyed person sees 99 other blue eyed people, and that every non-blue-eyed person sees 100 blue-eyed people. Discount their own eye color still leaves at least 98 blue-eyed people visible for everyone, and this is common knowledge.

There is no scenario in which everybody can't immediately deduce "at least 1 person has blue eyes, and everybody else knows this as well, and knows that everybody else knows".

>And if it's not obvious that it works, kill yourself.
It doesn't work. You should kill yourself for being satisfied with that answer, idiot.

Explain why not.

>But not everyone knows that everybody else knows that everybody else sees at least 98 of each color (and so on).
>They do:
(disregard everyone without blue-eyes)
Person A does not know that person B knows that person C sees at least 98 people with blue eyes because if person A does not have blue eyes then person B sees only 98 pairs of blue eyes.
The fact that everyone sees at least 98 blue-eyed people (A) is common knowledge, but the fact that everyone knows that everybody else knows A is not common knowledge.

>This is the flaw in the reasoning. The "above scenario" cannot happen precisely because there are three blue-eyed people
Which is precisely why each person is able to deduce that they are the third blue eyed person. That scenario only happens when there are 2 blue eyed people. So when it doesn't happen, they know there must be more than 2.

Imagine 2 different cases
>2 blue eyed people and 1 brown eyed person (plus the guru)
>3 blued eyed people and 0 brown eyed people (plus the guru)
The first case corresponds with the first scenario, the second case corresponds with the second scenario.

>The fact that everyone sees at least 98 blue-eyed people (A) is common knowledge, but the fact that everyone knows that everybody else knows A is not common knowledge.
Yes, it is. Because no matter who you are, you tell what everyone else can see for certain except for your own eye color.

"I don't know my own color. So if I were this, or any other, blue-eyed person over there, I'd see 98 blue eye pairs for certain, and maybe 99 if I'm blue. And if I were this, or any other, brown-eyed person over there, I'd see 99 blue eye pairs, and maybe 100 if I'm blue. 98 and 99 are both at least 1. The entire content of this paragraph is valid and available for everyone."

>"1+ people have blue eyes" isn't new information for anyone

Partial credit.

As you astutely pointed out, knowing that there is a blue eyed individual on the island is not new information for any of them.

Yet, by making the statement, the guru does provide new information that is required for the inductive step to be usable.

You tell me, what new information did the islanders gain from the guru's declaration?