I think there is a fallacy here actually. Some of the comments come close to touching it.
If there are 3 islanders with blue eyes, each can see 2 blue-eyed people. The information that there is at least 1 blue is therefore not new information, and can't change the outcome.
The outsider's comment is superfluous given that N > 3.
The important point is that the outsider's comment "synchronizes" the islander's information.
On Day 2, say you are a blue eyed person. You can see two other blue eyed people. Since these two people don't commit suicide during that day, you know there must be three blue eyed people with one of them being you.
Why? The visitor didn't synchronize anything. All of the villagers already knew there was at least 1 blue eyed person. Per the story, 99 or 100 depending on your perspective.
Everyone commits suicide at noon in view of everyone else. So the lowest possible number of blue eyed people go up one per day.
Here's a longer explanation:
Since people who know their eye color suicides at noon in plain sight of everyone else, we can set up induction.
If there's only one blue eyed person, they would suicide the first day at noon. Since everyone sees > 1 blue eyed people, they know this wouldn't happen. Since this doesn't happen, it's common knowledge that there is more than 1 blue eyed person. So if you only see one other blue eyed person, you must have blue eyes (but you don't know this yet because you see two blue eyed people).
On the second day, if there are only two blue eyed people, they would both suicide at noon. You see two blue eyed people. No one commits suicide because they either see two blue eyed people (if they have blue eyes) or three blue eyed people (if they don't). Now, everyone knows that there are more than two blue eyed people.
Since no one committed suicide the day before, you reason that there must be three blue eyed people, with the 2 people you see plus yourself. Thus, you know your eye color.
the argument i hear about induction is that it's only valid IF n=1 is a true situation, n=2 is a true situation, etc, up to n=k (where k is 100 here). But we are only given n=100, and we cannot say that n=1,2,3... is true. The best counter example is the surprise execution paradox: suppose you are a prisoner, about to be executed next week. However, a strange rule applies - if you are not surprised by the date of your execution, you get to go free. You reason that you cannot be executed on sunday, the last day, since by friday, you'd know, and thus not be surprised. by induction, friday is also not a possible day, since you'd know by thurs, etc. You conclude that you cannot be surprised, and this will not get executed. But come wed, you get executed, and is completely surprised!
This problem is different from the surprise paradox, which relies on probability changing each day, and a possibility that either (a) the executioner will be proven a liar by randomly guessing Sunday, or (b) you confidently guessing Wednesday and being right.
> Now, everyone knows that there are more than two blue eyed people.
That's incorrect, in a subtle way. The villagers (every single one) knew that before. It isn't discovered knowledge. Everyone (including the visitor) knows there are 99 or 100 because they can count them already, prior to the visitor.
So nobody is suiciding for 99 days. As an individual who sees 99 other blue eyes, why would you think or not think that you have blue eyes on the next day?
I meant strictly more, as in at least three.
If you know that there are more than two blue eyed people and you can only see two, you know that you have blue eyes.
If you can see 99 blue eyes, you know that all the other people see either 98 or 99 (if there's a total of 99 blue eyes) or 99 or 100 (if there's 100 blue eyes). However, if some people see 98 and know that there's at least 99 people with blue eyes, then they would all suicide. Since they don't, you know everyone sees either 99 or 100 blue eyes. Since you see 99, you must have blue eyes.
As a villager, you already know everyone sees either 99 or 100 blue eyes. You see 99, but you don't know if everyone else sees 99 or 100 (if they have blue it's 99 if they dont it's 100). You knew it before the visitor, and after his statement.
This does not help in deducing if you have blue eyes.
> then they would all suicide
No they wouldn't. Nobody should suicide at all. The visitor saying someone has blue eyes, was already known by every member of the village and every member of the village saw more than 1 person with blue eyes prior. Nothing is different because he said something.
> Since you see 99, you must have blue eyes
I don't know there's 100, so I don't know I have blue eyes.
This is a solved problem. When a proof is written, you should challenge the proof, not just reassert your intuition.
You are ignoring islander A's view of B's view of C's view of ... Z's view of situation.
Think about 3 islander's, where A thinks "if I have green eyes, then B and C are in a situation equivalent to me not being here, and B sees exactly 1 blue eyed person C, and B knows that 'C sees 0 or 1 blue eyed people, but then C learned that 0 is impossible'"
Is the outsider not needed for this line of reasoning to work assuming that the total number of people doesn't change? All the 100 blue eyed would commit suicide after a hundred days after the creation of the island since they work out they have blue eyes. I think the main challenge in this puzzle is the lack of assumption on what kind of reasoning the islanders are capable of doing and what the initial conditions are.
Consider the case of 2 blue eyed people, Ben and Betty. Everybody knows there is at least one blue eyed person. But does everybody know that? That is, does everybody know that (everybody knows there is one blue eyed person)?
From Ben's perspective, Betty might be the only blue-eyed person, so Ben does not know if Betty knows there is at least one blue eyed person. This is important because Ben needs to correlate Betty's actions with her knowledge; he can't do that if he doesn't know what her knowledge is.
After the traveler's announcement, Ben knows that Betty knows there is at least one blue eyed person. This is the new information.
And with three people, we add another layer: Ben does not know if Betty knows that Bobby has blue eyes. And so on.
The visitor's announcement is special because it is infinitely layered. Everyone knows one person has blue eyes, and everyone knows that everyone knows, and so on, ad infinitum. This is called common knowledge: https://en.wikipedia.org/wiki/Common_knowledge_(logic)).
Though to be fair, this isn't explicitly stated: the traveler "addresses the entire tribe" but it's critical that everyone in the tribe knows this is common knowledge. Merely addressing the entire tribe is not enough: if the traveler's announcement was in the form of a BCC mass mailing, there would be no new information and so no suicides.
"synchronize" is a bit vague. It can be made more precise. Simply put, it sets the initial condition in the recursion. Before someone says "there is a blue eyed person here", the initial condition in the recursion does not hold. The islanders are not allowed to say it so the visitor was the first one to make the statement. If indeed one of the islanders said it (whether blue or brown eyed) the condition would hold as well and the recursion would follow too. However none of them can talk about it.
The problem is the initial leap of logic - why should any tribe member think that the foreigner is talking about them at all? The foreigner says 'there are other blue-eyed people' and everyone can see that there are indeed other blue-eyed people. There's no reason for anyone to think that they might be in the blue group.
If there was some reason to start a count, sure, but there isn't. The foreigner just states that there are other blue-eyed people, which the tribe already knows.
Reduce the problem to 2 blue-eyed people and it should be more clear. If Alice and Bob are the only people with blue eyes, then yes, Alice might think that the speaker was only talking about Bob. But when Bob fails to sacrifice himself the next day, that must be because he also sees a person with blue eyes, and that can only be Alice.
It's different with only two people than it is with multiple. When there's only one other person and they don't act the next day, then the logic works. But when there are numerous people and there is no reason to start counting, then it doesn't extrapolate out.
Put more clearly: Foreigner says "blue eyes!". n is 2. If you have blue eyes and you don't see the other blue-eyed person top themselves, you conclude that you also have blue eyes as does the other person, and it's belly-opening time.
But when n is 100, that initiation is missing. No-one tops themselves the next day? Hardly suprising - there are a hundred of the buggers. There is no reason to start counting, no "hang on, why didn't ol' blue eyes over there kick the bucket?".
And, given that these are humans and not cronjobs, no-one is going to be patiently waiting day in, day out for 100 days to see when they'll all kill themselves. Drama can't even fester, since the tribesfolk are forbidden from talking about the issue. Life goes on, and over three months later, the foreigner's statement wouldn't cause a bloodbath :)
TL;DR: if you want to make a fancy logic puzzle, don't use 'humans with bizarre rules' as your subjects. :)
You aren't applying all possible logical reasoning about the situation (such as the characters recursive mental models of each other's mental models), unlike the hypothetical completely rational characters.
I think you've missed my point. At low n, yes, that recursion works. But once you get to the point where you can see that each [other] blue-eye can see multiple blue-eyes, there's no real reason to start counting. There's nothing to logically deduce if the people aren't mentally counting, as keeping that mental count is a fundamental part of the algorithm. If n is 7, for example, then a blue-eyed tribesperson would see that the other blue-eyes can see [at least] the remaining 5. And as such, there is no reason to start counting. There's no "hey, that guy didn't do anything!" to start the process going, as there would be with low n. The algorithm is separated from it's starting conditions.
Also, the characters are definitively not 'completely rational', as they're also described as being 'devout' :)
In the original, you know the other islanders can see a minimum of 98 BP (blue eyed people). And you know that they know that you know that they can see at least 97 BP.
Given that, the statement that there is at least 1 blue eyed person doesn't provide new info. The argument that this "synchronizes" them is a little strange. For all we knew, they were in the middle of counting to 100 days prior to this statement due to the aforementioned common knowledge that everyone already knew there were blue eyed people.
They wouldn't be counting to 100 days, because there's no point where everyone would start counting together (no one is allowed to reveal anything about eye color).
So you know there's no reason for all the blue eyed people to commit suicide together, so you can't reason anything out about the total number of blue eyed people.
As per a comment in the original article, they could choose to "sync" when one of their children is first told the rules about not knowing your own eye color. This changes the way that person would behave, and is a completely reasonable time for each islander to reevaluate the logical cascade.
Close, but not quite. As others have pointed out, that's not new information either. The information given is the knowledge that everyone knows that everyone knows...that at least one person has blue eyes, to infinite depth. The key is that this knowledge was one level short of allowing the chain reaction to start prior to the announcement.
This comment is going to be long, so buckle in. A few different ways of thinking about this issue:
1. I note that you say that when N = 3, there's no new information given. But when N = 2, all islanders are still aware that there is at least one person with blue eyes.
I'm guessing that despite that fact, you see how in the N = 2 case, the two people with blue eyes can infer they have blue eyes. So the leap of intuition you should've had there is that somehow, new information is being conveyed, even though every islander already knows that there is at least one person with blue eyes.
2. But let's just tackle this with straightforward inductive logic. For now, we aren't going to worry about what new information is given, we're just going to do induction.
If I see that N people have blue eyes, then I know that either there are N people with blue eyes, or there are N+1 people with blue eyes, and I have blue eyes. If after N days, the people I see with blue eyes haven't killed themselves, then I know that there must be N+1 people with blue eyes, and I have blue eyes, ergo I must kill myself.
This holds with N=0 and N=1 and N=2, so it on some level doesn't matter what the logic is, induction works.
3. But okay, let's figure out what information is actually given. Let's say that on a 100 person island, I have blue eyes and Alice and Bob also have blue eyes. So from my perspective, I know that there are two possible scenarios:
a. Alice, Bob, and I have blue eyes.
b. Only Alice and Bob have blue eyes.
Let's consider situation b. In situation b, what does Alice think? Since we're stipulating that I have brown eyes, Alice sees that Bob has blue eyes and that's it. Therefor, Alice thinks that there are two subscenarios of situation b:
b1: Alice and Bob have blue eyes.
b2: Only Bob has blue eyes.
Hypothetical Alice in hypothetical situation b considers situation b2. She thinks about what Bob would know in situation b2. He would see that nobody besides potentially him has blue eyes. So he would kill himself when the visitor explained that at least one person had blue eyes.
If Bob does not kill himself, then hypothetical Alice in situation b knows that she is not in situation b2. There is only one other subscenario of situation b, and that is b1, in which both Alice and Bob have blue eyes. So Alice would kill herself on day 2 (as would Bob).
If Alice does not do that, then Alice clearly did not consider herself to be in situation b. The only alternative to situation b is situation a, in which Alice, Bob, and I have blue eyes. So we all kill ourselves on day 3.
And obviously since all of this is symmetrical -- neither Alice, Bob, or I have any privileged information, that means that if I can see three people with blue eyes, and they don't kill themselves on day 3, then the only other valid scenario is that I also have blue eyes, so on day 4, I kill myself.
The information added is information about what we know that other islanders know about people's eyes, not what we know about people's eyes.
> Now suppose inductively that n is larger than 1. Each blue-eyed person will reason as follows: “If I am not blue-eyed, then there will only be n-1 blue-eyed people on this island, and so they will all commit suicide n-1 days after the traveler’s address”
But no one knows n!
So I am not following the logic in the slightest.
Since no one knows n the entire logic sequence never gets started.
Specifically:
The only information the traveler gave them is that n >= 1. If n = 1 then the blue eyed person gained new information, which is how the inductive reasoning started, and I follow the logic that far.
But if n > 1 then there is no new information and the inductive process never starts because no one knows n, no one. There is not a single person who knows n.
New information is gained every day there isn't a ritual self-sacrifice. Start with the simplest scenario that his words imply and work up every day from there.
One blue eyed person could look around, see only brown (or other color) eyes, and figure out they must have been the one talked about, right? Boom, gone.
Two blue eyed people would look around and see the other blue eyed person. However when the conditions for "one blue eyed person" aren't met, they would see the next day that there are 998 non-blue-eyed people and reason that the other blue eyed person must have seen a blue eyed person and it would have to be them. Boom. Gone.
Three blue eyed people would look around and see two other blue eyed people. After the first two days they'd realize that since the conditions of the 2 blue eyed person scenario weren't met, there are 3 blue eyed people, and since they can see 997 other-eyed people, they must be the third in that scenario. Boom. Gone. And so on and so on.
The key of the problem includes the statement that all the islanders have seen all the other islanders and know their eye colors.
From there we have the simple case of n = 1: suppose Bob is the only blue-eyed islander. He's seen everyone else's eyes, knows nobody else has blue eyes, and so can conclude from n >= 1 that in fact n = 1 and he is that one.
For the case of n = 2: suppose Alice and Bob are the two blue-eyed islanders. Alice knows Bob is blue-eyed but doesn't know she is. Bob knows Alice is blue-eyed but doesn't know he is. There are two possibilities: n = 1 or n = 2, since they have seen every other person's eyes and know there are no other blue-eyed people. Each can watch the other's behavior on the first day. But since neither one knows their own eye color yet, neither one commits suicide that day. Then each one realizes that for the other not to commit suicide, it must be the case that n > 1. Since they know every other person's eye color, the only possible explanation is "n = 2 and I'm the other one". They commit suicide on day 2.
From there we can just apply the same reasoning every time. If n = 3, Alice, Bob and Carol each look for the other two to commit suicide on the second day (from the reasoning for n = 2). When they don't, the only possible conclusion is "n = 3 and I'm the third one" and they all commit suicide on day 3.
That line of reasoning then works with any successive value for n: if there are n blue-eyed islanders, each looks for the other n - 1 islanders to commit suicide on day n - 1. When that doesn't occur, the only possible conclusion those people can reach is "there's one more blue-eyed islander, and it's me". Then all n of them commit suicide on day n.
But the flaw is assuming that they will start counting at one.
The blue eyed person number 3 can see that there are 2 others at least, so they know for sure that n != 1, since n != 1 they have no reason to expect anyone to commit suicide.
The sequence gets started by the initial statement, and although in the case of n > 1 it seems as if it doesn't impart any new information, it does.
The distinction that matters to get the chain started is the difference between "I personally know there is at least one blue-eyed islander" and "everybody knows there is at least one blue-eyed islander and I know everybody knows it and they know I know it and I know they know I know it... (so on)".
Prior to that you can't know that other islanders could deduce all the things you can deduce from available information, and they can't know that you could deduce all the things they could. Afterward, you know exactly what they can deduce from the available information, they know what you can deduce, and you know that it produces a guaranteed conclusion, by day n, that n is the number of blue-eyed islanders.
Given the propensity of island based logic puzzles to define certain classes of people as always/sometimes lying or telling the truth, it's always bothered me that this puzzle never states that the foreigner never lies or is always believed.
> One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.
Why would it matter if he's a liar/truth teller in the classic island logic puzzle sense? It seems irrelevant to me because the islanders believe him fully and the puzzle is about figuring out the effects his statement has.
You are correct and thanks for the correction, wow do I feel dumb for missing that. Maybe I can save a small scrap of face in that at least occasionally this problem contains no such clause, e.g. the xkcd version although I should read that one again before submitting this reply.
Wouldn't it be better to use the XKCD version [1] of the problem? That version was published 2 years earlier, doesn't make light of suicide and there's a ton of useful discussions [2] about the logic of it in the forums.
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[ 2.5 ms ] story [ 84.1 ms ] threadIf there are 3 islanders with blue eyes, each can see 2 blue-eyed people. The information that there is at least 1 blue is therefore not new information, and can't change the outcome.
The outsider's comment is superfluous given that N > 3.
So now you know your eye color.
Why? The visitor didn't synchronize anything. All of the villagers already knew there was at least 1 blue eyed person. Per the story, 99 or 100 depending on your perspective.
Here's a longer explanation: Since people who know their eye color suicides at noon in plain sight of everyone else, we can set up induction.
If there's only one blue eyed person, they would suicide the first day at noon. Since everyone sees > 1 blue eyed people, they know this wouldn't happen. Since this doesn't happen, it's common knowledge that there is more than 1 blue eyed person. So if you only see one other blue eyed person, you must have blue eyes (but you don't know this yet because you see two blue eyed people).
On the second day, if there are only two blue eyed people, they would both suicide at noon. You see two blue eyed people. No one commits suicide because they either see two blue eyed people (if they have blue eyes) or three blue eyed people (if they don't). Now, everyone knows that there are more than two blue eyed people.
Since no one committed suicide the day before, you reason that there must be three blue eyed people, with the 2 people you see plus yourself. Thus, you know your eye color.
This is a bit long but I hope it's clear.
That's incorrect, in a subtle way. The villagers (every single one) knew that before. It isn't discovered knowledge. Everyone (including the visitor) knows there are 99 or 100 because they can count them already, prior to the visitor.
So nobody is suiciding for 99 days. As an individual who sees 99 other blue eyes, why would you think or not think that you have blue eyes on the next day?
If you can see 99 blue eyes, you know that all the other people see either 98 or 99 (if there's a total of 99 blue eyes) or 99 or 100 (if there's 100 blue eyes). However, if some people see 98 and know that there's at least 99 people with blue eyes, then they would all suicide. Since they don't, you know everyone sees either 99 or 100 blue eyes. Since you see 99, you must have blue eyes.
This does not help in deducing if you have blue eyes.
> then they would all suicide
No they wouldn't. Nobody should suicide at all. The visitor saying someone has blue eyes, was already known by every member of the village and every member of the village saw more than 1 person with blue eyes prior. Nothing is different because he said something.
> Since you see 99, you must have blue eyes
I don't know there's 100, so I don't know I have blue eyes.
You are ignoring islander A's view of B's view of C's view of ... Z's view of situation.
Think about 3 islander's, where A thinks "if I have green eyes, then B and C are in a situation equivalent to me not being here, and B sees exactly 1 blue eyed person C, and B knows that 'C sees 0 or 1 blue eyed people, but then C learned that 0 is impossible'"
From Ben's perspective, Betty might be the only blue-eyed person, so Ben does not know if Betty knows there is at least one blue eyed person. This is important because Ben needs to correlate Betty's actions with her knowledge; he can't do that if he doesn't know what her knowledge is.
After the traveler's announcement, Ben knows that Betty knows there is at least one blue eyed person. This is the new information.
And with three people, we add another layer: Ben does not know if Betty knows that Bobby has blue eyes. And so on.
The visitor's announcement is special because it is infinitely layered. Everyone knows one person has blue eyes, and everyone knows that everyone knows, and so on, ad infinitum. This is called common knowledge: https://en.wikipedia.org/wiki/Common_knowledge_(logic)).
Though to be fair, this isn't explicitly stated: the traveler "addresses the entire tribe" but it's critical that everyone in the tribe knows this is common knowledge. Merely addressing the entire tribe is not enough: if the traveler's announcement was in the form of a BCC mass mailing, there would be no new information and so no suicides.
If there was some reason to start a count, sure, but there isn't. The foreigner just states that there are other blue-eyed people, which the tribe already knows.
Put more clearly: Foreigner says "blue eyes!". n is 2. If you have blue eyes and you don't see the other blue-eyed person top themselves, you conclude that you also have blue eyes as does the other person, and it's belly-opening time.
But when n is 100, that initiation is missing. No-one tops themselves the next day? Hardly suprising - there are a hundred of the buggers. There is no reason to start counting, no "hang on, why didn't ol' blue eyes over there kick the bucket?".
And, given that these are humans and not cronjobs, no-one is going to be patiently waiting day in, day out for 100 days to see when they'll all kill themselves. Drama can't even fester, since the tribesfolk are forbidden from talking about the issue. Life goes on, and over three months later, the foreigner's statement wouldn't cause a bloodbath :)
TL;DR: if you want to make a fancy logic puzzle, don't use 'humans with bizarre rules' as your subjects. :)
Also, the characters are definitively not 'completely rational', as they're also described as being 'devout' :)
Given that, the statement that there is at least 1 blue eyed person doesn't provide new info. The argument that this "synchronizes" them is a little strange. For all we knew, they were in the middle of counting to 100 days prior to this statement due to the aforementioned common knowledge that everyone already knew there were blue eyed people.
So you know there's no reason for all the blue eyed people to commit suicide together, so you can't reason anything out about the total number of blue eyed people.
They can't discuss it. That would be a difficult social experiment, that would risk your life.
1. I note that you say that when N = 3, there's no new information given. But when N = 2, all islanders are still aware that there is at least one person with blue eyes.
I'm guessing that despite that fact, you see how in the N = 2 case, the two people with blue eyes can infer they have blue eyes. So the leap of intuition you should've had there is that somehow, new information is being conveyed, even though every islander already knows that there is at least one person with blue eyes.
2. But let's just tackle this with straightforward inductive logic. For now, we aren't going to worry about what new information is given, we're just going to do induction.
If I see that N people have blue eyes, then I know that either there are N people with blue eyes, or there are N+1 people with blue eyes, and I have blue eyes. If after N days, the people I see with blue eyes haven't killed themselves, then I know that there must be N+1 people with blue eyes, and I have blue eyes, ergo I must kill myself.
This holds with N=0 and N=1 and N=2, so it on some level doesn't matter what the logic is, induction works.
3. But okay, let's figure out what information is actually given. Let's say that on a 100 person island, I have blue eyes and Alice and Bob also have blue eyes. So from my perspective, I know that there are two possible scenarios:
a. Alice, Bob, and I have blue eyes. b. Only Alice and Bob have blue eyes.
Let's consider situation b. In situation b, what does Alice think? Since we're stipulating that I have brown eyes, Alice sees that Bob has blue eyes and that's it. Therefor, Alice thinks that there are two subscenarios of situation b:
b1: Alice and Bob have blue eyes. b2: Only Bob has blue eyes.
Hypothetical Alice in hypothetical situation b considers situation b2. She thinks about what Bob would know in situation b2. He would see that nobody besides potentially him has blue eyes. So he would kill himself when the visitor explained that at least one person had blue eyes.
If Bob does not kill himself, then hypothetical Alice in situation b knows that she is not in situation b2. There is only one other subscenario of situation b, and that is b1, in which both Alice and Bob have blue eyes. So Alice would kill herself on day 2 (as would Bob).
If Alice does not do that, then Alice clearly did not consider herself to be in situation b. The only alternative to situation b is situation a, in which Alice, Bob, and I have blue eyes. So we all kill ourselves on day 3.
And obviously since all of this is symmetrical -- neither Alice, Bob, or I have any privileged information, that means that if I can see three people with blue eyes, and they don't kill themselves on day 3, then the only other valid scenario is that I also have blue eyes, so on day 4, I kill myself.
The information added is information about what we know that other islanders know about people's eyes, not what we know about people's eyes.
But no one knows n!
So I am not following the logic in the slightest.
Since no one knows n the entire logic sequence never gets started.
Specifically:
The only information the traveler gave them is that n >= 1. If n = 1 then the blue eyed person gained new information, which is how the inductive reasoning started, and I follow the logic that far.
But if n > 1 then there is no new information and the inductive process never starts because no one knows n, no one. There is not a single person who knows n.
One blue eyed person could look around, see only brown (or other color) eyes, and figure out they must have been the one talked about, right? Boom, gone.
Two blue eyed people would look around and see the other blue eyed person. However when the conditions for "one blue eyed person" aren't met, they would see the next day that there are 998 non-blue-eyed people and reason that the other blue eyed person must have seen a blue eyed person and it would have to be them. Boom. Gone.
Three blue eyed people would look around and see two other blue eyed people. After the first two days they'd realize that since the conditions of the 2 blue eyed person scenario weren't met, there are 3 blue eyed people, and since they can see 997 other-eyed people, they must be the third in that scenario. Boom. Gone. And so on and so on.
From there we have the simple case of n = 1: suppose Bob is the only blue-eyed islander. He's seen everyone else's eyes, knows nobody else has blue eyes, and so can conclude from n >= 1 that in fact n = 1 and he is that one.
For the case of n = 2: suppose Alice and Bob are the two blue-eyed islanders. Alice knows Bob is blue-eyed but doesn't know she is. Bob knows Alice is blue-eyed but doesn't know he is. There are two possibilities: n = 1 or n = 2, since they have seen every other person's eyes and know there are no other blue-eyed people. Each can watch the other's behavior on the first day. But since neither one knows their own eye color yet, neither one commits suicide that day. Then each one realizes that for the other not to commit suicide, it must be the case that n > 1. Since they know every other person's eye color, the only possible explanation is "n = 2 and I'm the other one". They commit suicide on day 2.
From there we can just apply the same reasoning every time. If n = 3, Alice, Bob and Carol each look for the other two to commit suicide on the second day (from the reasoning for n = 2). When they don't, the only possible conclusion is "n = 3 and I'm the third one" and they all commit suicide on day 3.
That line of reasoning then works with any successive value for n: if there are n blue-eyed islanders, each looks for the other n - 1 islanders to commit suicide on day n - 1. When that doesn't occur, the only possible conclusion those people can reach is "there's one more blue-eyed islander, and it's me". Then all n of them commit suicide on day n.
But the flaw is assuming that they will start counting at one.
The blue eyed person number 3 can see that there are 2 others at least, so they know for sure that n != 1, since n != 1 they have no reason to expect anyone to commit suicide.
So the whole sequence never gets started.
The distinction that matters to get the chain started is the difference between "I personally know there is at least one blue-eyed islander" and "everybody knows there is at least one blue-eyed islander and I know everybody knows it and they know I know it and I know they know I know it... (so on)".
Prior to that you can't know that other islanders could deduce all the things you can deduce from available information, and they can't know that you could deduce all the things they could. Afterward, you know exactly what they can deduce from the available information, they know what you can deduce, and you know that it produces a guaranteed conclusion, by day n, that n is the number of blue-eyed islanders.
> One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.
Why would it matter if he's a liar/truth teller in the classic island logic puzzle sense? It seems irrelevant to me because the islanders believe him fully and the puzzle is about figuring out the effects his statement has.
[1] https://xkcd.com/blue_eyes.html
[2] http://forums.xkcd.com/viewtopic.php?f=3&t=3