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I read the Blue Eyes puzzle here, and the solution which I find quite interesting. My questions:

  1. What is the quantified piece of information that the Guru provides that each person did not already have?

  2. Each person knows, from the beginning, no fewer than 99 blue-eyed people to be on the island. Then how is considering the 1 and 2-person cases relevant, if each person can dismiss these 2 cases immediately as possibilities?

  3. Why must they wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

EDIT: Most answers seem to concentrate on question 1 which I understand partly: but I remain confused because of different answers. Can someone answer questions 2 and 3?

NNOX Apps
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    It's a good puzzle, but certainly does not qualify as the hardest logic puzzle in the world. – Ittay Weiss Sep 10 '13 at 10:03
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    A useful reference is http://en.wikipedia.org/wiki/Common_knowledge_(logic) – Andreas Caranti Sep 10 '13 at 10:47
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    I've read the verions with the monks and the gnomes before. – Raskolnikov Sep 10 '13 at 10:55
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    Previously: [Is there no solution to the “hardest” puzzle?](http://math.stackexchange.com/q/238288/856) –  Sep 10 '13 at 15:08
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    @IttayWeiss: it’s not hardest in the sense of lengthy computation, and it’s certainly not the hardest if you include puzzles requiring specialist knowledge; but it’s surely a contender for “the conceptually hardest puzzle accessible to non-mathematicians”. – Peter LeFanu Lumsdaine Sep 10 '13 at 16:35
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    Maybe after 99 answers, someone will post a correct and comprehensible solution. – nbubis Sep 10 '13 at 18:47
  • @IttayWeiss, what is an example of a harder logic problem? – Kenshin Sep 11 '13 at 01:11
  • @chris the main reason I consider this puzzle to be on the easy side is that once you reduce it to 2 and then 3 people, the analysis is immediate and gives the solution. Puzzles that don't allow for such reductions tend to be more difficult. For instance, the island with infinitely many logicians (who believe in the axiom of choice and possess amazing eye-sight and incredible memory, the witch who intends to place either a black or white hat on each logician's head, then ask them to at once state a color. If only finitely many of them get their own color wrong, the all live. Make them live. – Ittay Weiss Sep 11 '13 at 05:58
  • @Ittay: Make them all stand in line, with as rule that there neighbours must both have the same coloured hat. They will now ling up with black and white hats alternately. – Dorus Sep 11 '13 at 13:14
  • @dorus I don't follow you. – Ittay Weiss Sep 11 '13 at 13:48
  • I'm not sure if your rules would allow this. But let's say you let them all hold hands, and they can only hold the hands of 2 others with the same coloured hat. They can now form circles (or one VERY long line), where every logician know his hat has the same colour as his neighbour's neighbour. – Dorus Sep 11 '13 at 14:18
  • The point of Ittay's puzzle is that they have to agree on a strategy *in advance*, applicable to any assignment of hat colors and without further communication. – Ryan Reich Sep 11 '13 at 19:00
  • @aufkag: Your question 2 shows how hard it is to formulate such puzzles clearly. The description makes it clear to me that the islanders do *not* know in advance how many of them have blue eyes; but it doesn't to you, apparently. – reinierpost Sep 12 '13 at 09:19
  • @reinierpost It's not my question. My username is there because I slightly edited the question. That's all. – Řídící Sep 12 '13 at 09:25
  • @aufkag: ouch. sorry – reinierpost Sep 12 '13 at 22:23
  • I think I can answer question 2, but question 3 is really bugging me - why isn't some equivalent of Randal's Theorem 98 common knowledge? I know that everyone else can see the eye colours of everyone around them, and I know that there are 99 other people with blue eyes, each of whom sees either 98 or 99 blue-eyed people themselves. So why don't we accept that everyone has sufficient evidence to establish an effective Theorem 98 off the bat? – Paul Ross Sep 24 '13 at 22:18
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    Actually, I think I get it now - everybody knows that there are at least 98 people with blue eyes, but not everybody knows that everybody knows that, and it takes 98 days after the Guru's utterance to establish that. Everybody *does* know that everybody knows that there are at least *97* people with blue eyes, but it also takes 98 days to establish that everybody knows *that*. – Paul Ross Sep 26 '13 at 16:16
  • The whole puzzle is flawed nonsense. Suppose there were exactly two blue eyes: they each see one, everyone else sees two. On the first night the failure of either to leave lets them deduce they must also have blue eyes so they both leave the second night, and by induction N leave on the Nth night. No oracle is required. – TheMathemagician Oct 23 '14 at 14:21
  • @TheMathemagician those two with blue eyes would depart on the second day, since they only say one people with blue eyes. The others say two, so "would have departed" on day three, but they didn't since the other two already departed on day two. – o0'. May 03 '15 at 15:44
  • @TheMathemagician I don't see how anyone not leaving lets anyone deduce anything except that the people involved don't know the color of their own eyes, if they stayed. Just because someone stays doesn't tell them anything about their own eye color, given the knowledge there is at least one person with blue eyes - it only says that the person that had blue eyes couldn't figure out they had blue eyes and could leave. Given this example, a brown-eyed person could equally think they have blue eyes, when they learn there is at least one among them that has blue eyes. It doesn't make them right. – vapcguy Jun 02 '16 at 18:22
  • All the answers so far have ignored the implications of (1) the fact that any given person knows that there cannot be fewer blue-eyed people than the number that he can see, and (2) the common knowledge that everyone knows that there cannot be fewer than the number they can see less 1. Please see my answer to a similar question on Puzzling.SE here: http://puzzling.stackexchange.com/a/37673/20907 – Jed Schaaf Jul 14 '16 at 17:17
  • https://puzzling.stackexchange.com/q/236/9366 – NNOX Apps May 07 '17 at 03:40

24 Answers24

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Here's the story of one blue-eyed islander. The Guru said she saw someone with blue eyes. He looked around and thought "Hey, I don't see anyone with blue eyes. I guess she means me." And so he left right away.

Here's the story of two blue-eyed islanders. The Guru said she saw someone with blue eyes. They looked around and thought "OK, I see someone with blue eyes. I guess she means him," and they stayed. But the next day came, and they thought "Hey, that blue-eyed guy didn't figure it out. I guess he must have seen someone else with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." And so they left together on the second day.

Here's the story of three blue-eyed islanders. The Guru said she saw someone with blue eyes. They looked around and thought "OK, I see two people with blue eyes. I guess she means one of them," and they stayed. A day passed, and nobody left, and they thought to themselves "OK, this is the day those two guys figure it out." But another day passed, and nobody left. The blue-eyed people thought "Wait; those two guys didn't figure it out yet. I guess they must have seen another person with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." And so they left together on the third day.

Here's the story of four blue-eyed islanders. The Guru said she saw someone with blue eyes. They looked around and thought "OK, I see three people with blue eyes. I guess she means one of them," and they stayed. A day passed, and nobody left, but they were not worried; they knew it would take a couple of days. A second day passed, and nobody left, and they all thought to themselves "OK, this is the day those three guys figure it out." But another day passed, and nobody left. The blue-eyed people thought "Wait; those three guys didn't figure it out yet. I guess they all must have seen another person with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." And so they left together on the fourth day.

...and this is why they have to wait the full 99 days. It's not important that the Guru can see someone with blue eyes, unless there's only one islander. What's really important is that, given that the Guru can see someone with blue eyes, "those blue-eyed guys" should be able to figure it out among themselves, and that takes a specific amount of time for a given number of blue-eyed islanders. It's only when they can't do this for a number of islanders that doesn't include you that it becomes clear you must have blue eyes too.

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    But WHY doesn't brown eye leave after 100 days as well? Each islander sees at least 99 brown eyes, KNOWING that each of those knows that there are at least 98 brown eyes in the group? Every single brown eyed person knows that there are 99 or 100 brown eyes in the group, so that is common knowledge as well. – SinisterMJ Sep 10 '13 at 16:31
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    Because the brown-eyes see the blue-eyes leaving the island, and think "Hey, those blue-eyed guys figured it out. I guess that means there aren't any others, so my eyes aren't blue. But what color are they? There are lots of colors; just look at the Guru's green eyes. I still don't know." And so they stayed. – The Spooniest Sep 10 '13 at 16:45
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    This seems to me to be an explanation of the answer to the puzzle, but not an answer to the three enumerated questions the asker is asking about. – Kevin Sep 10 '13 at 17:39
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    I was attempting to illustrate the answers to the three questions. In the first question, the Guru doesn't give the islanders any new information: the new information comes from the fact that, given what the Guru said, they still can't figure it out. The answer to the second question is that the 1-person and 2-person cases establish how to determine if you have blue eyes (though you need a 3-person case to really get it going). These then go on to the third question: you need the full 99 days because that's how long it takes for people to figure it out if you don't have blue eyes. – The Spooniest Sep 10 '13 at 18:10
  • @Anton: In the original version of this riddle (as I heard it long ago), all the islanders are either brown- or blue-eyed. And they kill themselves when they learn their eye-color, rather than leave the island. In that version, all the islanders end up dead. – BlueRaja - Danny Pflughoeft Sep 10 '13 at 19:06
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    @TheSpooniest: Saying that the Guru did not give new information and that the new information came from the fact that, given what the Guru said, they still can't figure it out is self-contradictory. If there were no new information, then there would be nothing new to figure out, and the fact that indeed nothing was figured out would not be new information either. The speaking of the Guru **does** provide new information, but that information does not have to be contained in the message itself. – Marc van Leeuwen Sep 14 '13 at 15:33
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    @MarcvanLeeuwen I think that's what The Spooniest meant. The guru's *message* doesn't give any new information to the islanders, but the new information comes from the fact that they can't figure it out from the time the guru said it. – Joe Z. Oct 04 '13 at 05:12
  • This answer is incredibly flawed. Only when it is 1 person is it true, since the guru obviously only has 1 person they could be referring to. When it's 2 people, just because the real blue-eyed person doesn't leave does NOT mean the one that isn't can or will leave. By the rules, only a blue-eyed person can leave, and if they can see a blue-eyed person, they'll know the other person should leave, but know nothing of their own eyes or ability to leave. If the person you see has brown-eyes, yes, you can infer you have blue & should leave. The remaining guy/gal still won't know their own color. – vapcguy Jun 02 '16 at 18:36
  • With 3 people, this statement: "Wait; those 2 guys didn't [leave]. I guess they must have seen another person with blue eyes, but I don't see anyone else with blue eyes. I guess that means me." is incredibly ridiculous. If no one leaves because they don't know their own color, no one can just assume they didn't leave because they saw someone else w/blue eyes. A person also can't infer "guess it's me" and leave. No one knows their own eye color unless *everyone* they see has some other color than blue. If the other 2 have blue & don't leave, it doesn't mean you are blue & why they didn't leave. – vapcguy Jun 02 '16 at 18:48
  • @vapcguy Let us assume that there are *n* blue-eyed people on the island. If you don't have blue eyes, you will see *n* blue-eyed people, but if you do have blue eyes, you will see *n - 1*. That's how the puzzle works: blue-eyed people aren't waiting for enough people. And so when the time comes for *n - 1* people to figure it out, but "they" don't, the blue-eyed people realize there must be more blue-eyed people than they thought. For each of these people, there is only one person whose eyes they can't account for: their own. And that is how they figure it out they have blue eyes. – The Spooniest Jun 07 '16 at 18:03
  • @TheSpooniest Nice explanation, but they would have to know exactly how many have blue eyes to know they have the color, too (or that they don't have it). Sure, they can count. But if they count 99 blues and 100 browns, they'd have to know there were equal numbers of both from the start to know they are the 100th blue eyed person (and not the 101st brown-eyed) and could leave. If *n* is a given for this problem, I would not see why we'd need the guru's information - we could just count everyone and use your formula to deduce if we should leave or not. – vapcguy Jun 07 '16 at 18:22
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    @vapcguy: You don't have to know exactly how many there are, because there are only two possibilities: the same number that you see (if you do not have blue eyes), or one more than this number (if you have blue eyes). You know how long it would take "those blue-eyed guys" to figure it out if you do not have blue eyes, so when they fail to figure it out in that amount of time, only one possibility remains, and that possibility implies that you have blue eyes. – The Spooniest Jun 17 '16 at 20:05
  • @TheSpooniest Yeah, that's what I see being said. But I don't believe, assuming we had 11 people with ourselves as the 11th man, just because 10 other guys with blue-eyes don't leave in 10 days means "ohup, guess I must have blue eyes, I can leave!". Just means no one on the island has figured out that they have blue eyes. – vapcguy Jun 20 '16 at 13:07
  • @vapcguy But the rules of the riddle already state that everyone on the island is good at logic: if it is possible for them to figure out that they have blue eyes, they will do so. Because you can account for everyone's eyes *except* for your own, blue-eyed people see one fewer than the actual number. So if "they" can't figure it out on the day you think they "should", that means there must be at least blue-eyed person you haven't accounted for. But there's only one person you can't account for -yourself- so there can only be one "extra" blue-eyed person, and that person must be you. – The Spooniest Jun 20 '16 at 13:50
  • @TheSpooniest *So if "they" can't figure it out on the day you think they "should", that means there must be at least 1 blue-eyed person you haven't accounted for* -- that's the part that isn't logical to me, at all. Just because the blue-eyed people don't leave by however-many-days-there-are-blue-eyed people, or 1 gazillion days, it doesn't mean the individual in question can logically deduce they have blue eyes, just because others haven't left (after all, how could those others figure out *they* have blue eyes and leave?) That person-*me*-could still have brown eyes just the same as them. – vapcguy Jun 20 '16 at 16:38
  • I reiterate: it is explicitly stated in the rules of the riddle that as soon as it is possible for someone to figure out that he has blue eyes, he will do so. Therefore, it is safe to assume that if "those blue-eyed guys" have not yet figured it out, it is because they *cannot* have figured it out yet. Up until the day when the number of blue-eyed people you see "should" figure it out, that's to be expected. But if they *still* can't figure it out on that day, then you must not be seeing all of the blue-eyed people. And the only way that's possible is if you have blue eyes. – The Spooniest Jun 20 '16 at 19:05
  • @TheSpooniest So I guess we are making the assumption that everyone sees exactly only one person each day, no duplicating, and remembers each color exactly as he/she saw them? I would think that would have to be an explicit rule. And on that last day, they would have to know/learn/whatever that they have, indeed, seen/recorded everyone. And they would have to know that there are exactly half-and-half of each color -- otherwise it breaks down and no one can leave, if say, everyone had blue-eyes. – vapcguy Jun 21 '16 at 15:16
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    @vapcguy: Not at all. The rules explicitly state that everyone on the island can see everyone else (not themselves) at all times, and keeps a count of the colors they've seen. The reason I told the stories of 1, 2, 3, and 4 islanders is because the puzzle works by inductive reasoning. One blue-eyed person is a special case: *he doesn't know blue eyes exist* until the Guru mentions them, so she must mean him. For any number higher than 1, you know how long it "should" take for the number of people you see to figure it out. So when they can't, there must be someone else, and that has to be you. – The Spooniest Jun 21 '16 at 15:38
  • @vapcguy: This works even if everyone on the island has blue eyes. You see 199 people with blue eyes, and you are the 200th person. But when "those blue-eyed guys" don't figure it out on the 199th day, you know there must be more people with blue eyes. Since the only person you can't account for is yourself, you must be the 200th blue-eyed person. At the same time you are coming to this conclusion, so is everyone else, because, just like you, they saw 199 blue-eyed people who failed to figure it out at the right time. So everyone leaves the island together on the 200th day. – The Spooniest Jun 21 '16 at 15:43
  • @TheSpooniest Assuming that once they account for everyone, they are "greedy" (for lack of a better term) enough to aspire for more blue-eyed people, and when they don't see them, and the others haven't left yet, automatically come to the conclusion that they must be the final person because the others were waiting for "the others". I get the reasoning now, but it's not totally foolproof, to me. They'd have to know they found everyone and it'd have to assume that, when "the others" didn't leave, it was because they saw blue-eyes. To me, that could go on forever-there's no diff in day 1 or 200. – vapcguy Jun 21 '16 at 15:58
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    @vapcguy: The rules state that everyone can see everyone at all times. So they do know that they've seen everyone. When "the others" don't leave, it must be because they saw an additional set of blue eyes, because if your eyes were any other color, they'd have already figured out they had blue eyes and left. – The Spooniest Jun 21 '16 at 17:00
  • @TheSpooniest This is the first time anyone ever said something that made sense on this! They are just constantly looking for blue eyed people and thinking they haven't left because that person saw one (and that person saw one, and that person saw one... etc.)! Thanks for your patience with me! – vapcguy Jun 21 '16 at 17:58
  • In the the story of four blue-eyed islanders you say "A day passed, and nobody left, but they were not worried; they knew it would take a couple of days." but why do they "knew"? They just know that there is a logical path that allows to come to the conclusion in 3 days (for 3 islanders) but how do they know that there is no other possible line of thoughts that allows to come to the conclusion in less then 3 days? – Marco Disce Aug 18 '19 at 15:24
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I'll take up the challenge in nbubis's comment (even though there are not yet $99$ answers), and try to give a precise answer. And since this is a mathematics rather than a philosophy site, I'll try to use some formulas to describe what is going on.

As has been noted, the technical notion of common knowledge is important here. Clearly there is in this problem need to distinguish between (the truth of) a proposition and the fact that some person knows this proposition to hold. In fact there is no need distinguish individuals (and actually only the blue-eyed ones really matter), and it suffices to state when everybody (in the problem) knows a proposition. So if $P$ is any proposition, I will note $E(P)$ a new proposition which states that everybody knows $P$ to be true. Since everybody applies logic flawlessly $E(P)$ implies $P$, but $P$ does not imply $E(P)$. And $E(P)$ does not imply $E(E(P))$ either, which is yet a new proposition; it will be convenient to abbreviate it $E^2(P)$, and define $E^n(P)$ similarly for all $n\in\Bbb N$. Finally some things (like the general state of affairs on the island, including the logico-compulsive behaviour of its inhabitants) are common knowledge; I will write $C(P)$ for $\forall n\in\Bbb N:E^n(P)$.

Let $n$ denote the number of blue-eyed inhabitants. Now for instance according to the problem statement $n=100$ is true, but $E(n=100)$ is false; in fact none of the inhabitants know that $n=100$. But on the other hand $E(99\leq n\leq 101)$ is true (all inhabitants know that $99\leq n\leq 101$, even though the way they know this differs). I'll focus on lower bounds; while $E^2(n\geq99)$ is false (while everybody knows that $n\geq99$ holds, the blue-eyed inhabitants don't know that the other blue-eyed ones know this), $E^2(n\geq98)$ does hold. Similarly for all $i$ one has $E^i(n\geq100-i)$ but not $E^i(n>100-i)$. The new information provided by the public statement of the Guru is $C(n>0)$; this implies $E^i(n>0)$ for all$~i$, of which the instance relevant to the problem is $E^{100}(n>0)$, which was not previously true.

While this points to the key factor in the explanation of the riddle, it is somewhat more challenging to describe in detail what happens with the state of knowledge during the $100$ days before the blue-eyes finally leave the island. For that I will denote by $L(i)$ the statement "on night$~i$, some islanders leave". By the problem statement it is always common knowledge when this happens, but I will still write $C(L(i))$ or $C(\lnot L(i))$ to emphasize this common-knowledge status.

The problem statement gives us the following fact, which is in fact common knowledge:

For any $i\geq0$ and $k>0$, one has $n=k\land C(\lnot L(i))\land E(n\geq k)\to C(L(i+1))$.$\quad(*)$

In words, if $n$ is actually $k$, and on some day no islanders have left (yet), and everybody knows that $n\geq k$, then some islanders will leave the next night. This is because the $k>0$ blue-eyed islanders see $k-1$ others, and know that there must be at least $k$ of them. The following is true, and (therefore, as its proof is based on logic only) common knowledge:

Lemma. For all $l,k\in\Bbb N$ one has $E^{l+k}(n>0)\land C(\forall i\leq k:\lnot L(i))\to E^l(n>k)$.

This states informally that with sufficiently general knowledge (i.e., a sufficient power of $E$ applied to it) of the fact that $n>0$, it will after $k$ successive nights of nobody leaving be clear to all that $n>k$, but this new fact will have lost $k$ of its levels of $E(\cdot)$. One could simplify the lemma and its proof considerably by replacing the powers of $E$ by $C$, and given that the Guru indeed provides $C(n>0)$, this would suffice to explain what actually happens. However the refined statement is helpful in understanding for instance why $E^{99}(n>0)$, which is true without the Guru speaking, will not suffice to bring anybody into action. I admit that the lemma does not very well express the temporal element of the problem; it implicitly supposes that the information contained in its hypothesis was available before $(*)$ had the first occasion to be applied, i.e., before night$~1$ (but not before night$~0$, as $C(\lnot L(0))$ represents the given initial state).

Proof by induction on $k$, uniformly in $l$. For $k=0$ the conclusion is among the hypotheses; there is nothing to prove. Now assume the statement for $k$, and also the hypotheses $E^{l+k+1}(n>0)\land C(\forall i\leq k+1:\lnot L(i))$ of the statement for $k+1$ in place of $k$. The second part of the hyposthesis implies the weaker $C(\forall i\leq k:\lnot L(i))$, so we can apply the induction hypothesis with $l+1$ in place of $l$, and get its conclusion that $E^{l+1}(n>k)$. We instantiate $(*)$ with $(i,k):=(k,k+1)$, giving $$ n=k+1\land C(\lnot L(k))\land E(n\geq k+1)\to C(L(k+1)), $$ which implies (because $C(\lnot L(k+1))\implies \lnot L(k+1)\implies\lnot C(L(k+1))$) $$ C(\lnot L(k))\land E(n>k)\land C(\lnot L(k+1))\to n\neq k+1. $$ If $H$ is the hypothesis of this last statement, we actually know $E^l(H)$ (from our assumptions and the conclusion of applying our induction hypothesis). This allows us to conclude $E^l(n\neq k+1)$, which together with $E^l(n>k)$ gives $E^l(n>k+1)$, completing the proof.

Now to the detailed description of what happens; our $100$ blue-eyes wait until they know that $n\geq100$ before $(*)$ forces them to leave. The lemma for $l=1$ and $k=99$ says this will happen provided $E^{100}(n>0)$ holds and $\forall i\leq99:C(\lnot L(i))$. Our Guru provides $C(n>0)$ and hence $E^{100}(n>0)$, and $C(\lnot L(0))$ holds from the problem statement. One still needs to wait for the $99$ other instances of $C(\lnot L(i))$ to provide the prerequisite facts for action.

Summarising, one has the following answers to the questions.

$1$. What is the quantified piece of information that the Guru provides that each person did not already have?

This is $C(n>0)$, and it is its instance $E^{100}(n>0)$ that is really new information, and necessary for any action to take place (higher powers are also new information, but $E^{100}(n>0)$ alone gets things moving). Note that this requires the statement of the Guru be public (giving the information separately to individual inhabitants would have no effect; indeed it is not new information to them), and moreover the fact that it is public must be public (a television broadcast would not suffice if the inhabitants could have some doubt about whether everybody was watching), and this again must be known to everybody, and so forth $100$ levels deep. (One really needs a very strong problem statement to ensure this. If any inhabitant had a doubt whether another inhabitant might maybe have some doubt whether ... some inhabitant was really paying attention to the Guru, the logic would fail.)

So there is genuinely new information in the making of the statement by the Guru, but it is not contained in the message she brings itself, but in the fact that it causes that (everybody is aware that)$^{100}$ there are people with blue eyes.

$2$. 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?

While everyone knows that say $n>10$, wrapping it in a sufficient number of applications of $E(\cdot)$ makes it untrue. It is these wrapped-up statements that play a role in the reasoning.

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

Every night brings its new information, namely $C(\lnot L(i))$. While most of the times $\lnot L(i)$ itself was already known to everyone, the fact that it becomes common knowledge is genuine new information, and again this is essential for the problem.

Final remark. I note that I have used the rule that if $P\to Q$ holds, then $E^l(P)$ implies $E^l(Q)$. This might seem suspicious, as $E$ does not commute with all logical connectives, notably $E(P\lor Q)$ does not imply $E(P)\lor E(Q)$. Although I am not aware of all rules of the formalism, the one I applied is intuitively valid, by the "infallibly logical" nature of the inhabitants: if $P$ in fact implies $Q$, this will not escape their attention, and anyone who in addition knows $P$ to hold will therefore also know $Q$ to hold; in particular if everyone knows $P$ then they will also all know$~Q$.

Marc van Leeuwen
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    Imagine the Guru instead said: "Ask me any question you like". Someone asks "True or false, are their any blue-eyed people on this island?" Before the Guru answers, everybody knows what the answer will be (I think?). Does that mean that the useful information becomes available just by asking the right question!? If the Guru was struck down by lightning just before answering the question, would the blue-eyed people still leave after 100 nights? Like this: "The Guru was going to answer honestly. The answer would have been True, if she had survived. So we will all pretend she said "True" – Aaron McDaid Oct 26 '14 at 20:26
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    @AaronMcDaid: No. As I said in several places, it is not the information given itself, but the fact that everyone is aware that this becomes public knowledge that counts. By killing the Guru before she can speak, this extra information is not given, and nothing changes. It is similar to nobody leaving the first night: everyone knows that it will happen, but it still has to actually happen (in a public way) in order to provide new information. – Marc van Leeuwen Nov 29 '14 at 10:11
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    Would it be sufficient for all the islanders to gather around in a circle (where it is obvious that everyone can hear what is being said), and for one of the blue-eyed people to declare that $n > 0$? If not, why, and would it be any different if the declarer were brown-eyed? – Tony May 06 '15 at 01:02
  • @Tony: (1) that would be a different problem, but (2) that scenario would violate the explicitly stated and essential hypothesis that the islanders cannot communicate (and certainly not about eye colour) among each other. So it is rather pointless to even think about that problem. – Marc van Leeuwen May 07 '15 at 04:46
  • I would say that if some bit of information $x$ can be deduced on a purely logical basis then all the islanders can get that information, and also (since they know the others are all perfect logicians) all the islanders can deduce $E(x)$ by pure logic, and the same holds for $E^k(x)$ for any $k$. This should apply also to $x$="there is someone with blue eyes" or $x$="if asked the Guru would say there is someone with blue eyes for sure". – Marco Disce Feb 21 '16 at 13:28
  • @MarcoDisce: Neither of your statements follow from pure logic, i.e., without any privately known information based (for instance) on observation. On the other hand, I did indicate that certain things are purely logical and therefore automatically common knowledge. – Marc van Leeuwen Feb 21 '16 at 15:57
  • Ok but my argument coluld still hold if you replace "pure logic" with "pure logic plus observation". – Marco Disce Feb 21 '16 at 17:15
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    @MarcoDisce: No that is not true, and dealt with explicitly in my answer. By direct observation everybody can see that $n\geq99$, so $E(n\geq99)$ is true. However $E(E(n\geq99))$ is not true (initially). – Marc van Leeuwen Feb 21 '16 at 20:21
  • To build on @MarcvanLeeuwen's comment, ***E*** ( ***E*** ( _n_ ≥98)) is initially true (as is ***E*** ( ***E*** ( _n_ <101)) ), so ***E*** ( ***E*** ( _n_ ≥98)) can be assumed on the first day. ***E*** ( _n_ ≥99) will be demonstrated on the second day. This leads to ***E*** ( ***E*** ( _n_ ≥99)) becoming true on the third day, so on the fourth day, ***E*** ( _n_ =100) will be true. – Jed Schaaf Jul 15 '16 at 00:00
  • By simple observation $n\geq99$ is true, so $E(n\geq98)$ is true, thus $E^2(n\geq97)$ must also be true. I don't see how $E^3()$ adds anything to the proof. The guru's statement $C(n\geq0)$ is already contained within $E^2(n\geq97)$, which is already known to be true, so the only difference is that it synchronizes everyone's internal countdown. – Jed Schaaf Jul 15 '16 at 00:39
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    @JedSchaaf. You misunderstood the answer I gave. It is not the case that $E(n\geq99)$ is demonstrated on the second day; it is true from the start. But $E^2(n\geq99)$ does not become true on the third day (it is not implied by the initial facts, plus $C(n\geq0)$, $C(\lnot L(1))$ and $C(\lnot L(2))$), nor does it for a long time. As the lemma sates for $(l,k)=(2,98)$, the initial facts, plus $E^{100}(n\geq0)$, $C(\lnot L(1))$, $C(\lnot L(2))$, ..., $C(\lnot L(98))$ finally imply $E^2(n>98)$, which $E^2(n\geq99)$. So the statement $E^2(n\geq99)$ only becomes true after $98$ nights of no action. – Marc van Leeuwen Jul 16 '16 at 06:31
  • Your second comment seems to build on the confusion from the first. There is nothing special about the passage from $E^2$ to $E^3$, and $C(n>0)$ is not contained within (or implied by) $E^2(n\geq 97)$; since $C(n>0)$ involves _arbitrarily high_ powers $E^k(n>0)$, it can never be deduced from some finite power like $E^2(...)$. Common knowledge can only hold by virtue of a hypothesis, it cannot be deduced from explicit instances $E^k(...)$. – Marc van Leeuwen Jul 16 '16 at 06:45
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    Ah, ok. I found where my confusion lies. Another proposition is needed: "a blue-eyed person knows that", represented by $B(x)$. With this new proposition, and granted $n=100$, then $E(n\geq99)$, $E(B(n\geq98))$, and $E(B(B(n\geq97))$. Each successive day increases the minimum known possible value of $n$, so on the 4th day, if no blue-eyed people have left earlier, then all the blue-eyed people will realize they have blue eyes and will leave, because $E(n\leq101)$. – Jed Schaaf Jul 16 '16 at 18:47
  • And $E(B(n\leq100))$. – Jed Schaaf Jul 16 '16 at 21:42
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I think the answer to Question 1 is that after the Guru has spoken they all know that they all know that they all know that they all know that (repeat as many times as you like) someone has blue eyes. Previously they did not know that, and the statement is only true when it contains at most 99 "they all know that"s.

Derek Holt
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    That's wrong: All people always knew that there are at least 99 people with blue eyes and that everyone knows that. – Keinstein Sep 10 '13 at 11:56
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    That's true, but it doesn't contradict what I wrote! – Derek Holt Sep 10 '13 at 12:16
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    @Keinstein, It is true that the islanders know both of those things. But those two facts alone are not sufficient to qualify as common knowledge. They need to know: "1) there are blue eyed people on the island. 2) Everyone is aware of fact 1. 3) Everyone is aware of fact 2. 4) Everyone is aware of fact 3... X) everyone is aware of fact X-1." for an X of any size. Prior to the guru's statement, each islander only knows facts 1 through N, which falls infinitely short of qualifying as common knowledge. – Kevin Sep 10 '13 at 12:18
  • But that answers the original puzzle not the three questions given above. The guru didn't tell anything about $X$. So she didn't tell them how large $X$ ist. These three questions cannot be answered by recursion. – Keinstein Sep 10 '13 at 12:30
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    @Keinstein I disagree. I have answered Question 1, by specifying something that they all knew immediately after the Guru spoke that they did not know before he spoke. – Derek Holt Sep 10 '13 at 13:25
  • And I disagree with you that this is new information. I must confess that I was wrong when I thought that its common knowledge that 99 people have blue eyes; they know can be shure the others know about 98 people. I's the same problem as the Random832's answer. – Keinstein Sep 10 '13 at 13:37
  • @Keinstein: The guru didn't give them all the X's in Kevin's comment, he only gave them 1). But after the first night, they all know 2) as well, since no one has left. And since they all know 2), then after the second night they all know 3). And since they know 3).... etc. That is the whole point of this riddle. – BlueRaja - Danny Pflughoeft Sep 10 '13 at 15:09
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    @Keinstein: suppose there were 3 people, not 99. Then everybody already knew that “(1): at least one person has blue eyes”, and that “(2): everybody knows fact (1)”, but they *didn’t* already know that “(3): everybody knows fact (2)”. This statement (3) is the new knowledge the guru gives. Similarly, in the case with 99 people, the 99th fact in this series is the new information the guru gives. – Peter LeFanu Lumsdaine Sep 10 '13 at 16:33
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    I know that 98 people know fact (1). Then I know fact (2). Now the external information from the game rules come into play: “As I came to that conclusion everyone can come to this conclusion. And if he does he will.”¹ Thus I know fact (2), too. As I came to that conclusion everyone came to this conclusion. So, everyone knows fact (2) and so on… Note that without ¹ the puzzle didn't work at all. It's not the problem that I don't understand the recursion you use, I just don't see any error in my argumentation ;-). Most statements use depth first, while I use Breadth-first search. – Keinstein Sep 10 '13 at 17:05
  • "they all know that they all know that they all know that they all know that (repeat as many times as you like)". This is called [Common Knowledge](http://en.wikipedia.org/wiki/Common_knowledge_(logic)). It is stronger than [Mutual Knowledge](http://en.wikipedia.org/wiki/Mutual_knowledge_(logic)), which exists whenever multiple agents know the same thing. – Matthew Piziak Sep 10 '13 at 17:10
  • @DerekHolt a small question if the 100 blue eyes all leave on the 100th night does that mean that the 100 brown eyes all leave on the 101th night? – Songo Sep 10 '13 at 17:39
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    @Songo It isn't common knowledge that the remaining people all have brown eyes; I believe it's only common knowledge that they do not have blue eyes. Of course, if every islander knew they had either blue or brown eyes, then I think the answer is yes, but otherwise, no. – Andrew D Sep 10 '13 at 17:43
  • @AndrewD aah I see. Good point ;) – Songo Sep 10 '13 at 17:45
  • @Keinstein: The relevant meta-reasoning must be "If I came to this conclusion, everyone who *shares the relevant knowledge* has come to the same conclusion." If X sees $99$ blue-eyed people, then X concludes that (1) $b\ge 99$, (2) everybody knows that $b\ge 98$, and (3) everybody who knows that $b\ge 99$ (that's not everybody!) knows that everybody knows that $b \ge 98$. X only knows that everybody knows that everybody knows that $b\ge 97$. Each additional "everybody" reduces the minimum number of blue-eyed people by one... there's no common knowledge until the guru speaks. – mjqxxxx Sep 11 '13 at 15:33
  • X concludes $b≥3$ that means everyone knows $b≥2$. That means everyone knows that everyone knows that ö – Keinstein Sep 12 '13 at 05:51
  • X concludes $b≥3$ that means everyone knows $b≥2$. That means everyone knows that everyone knows that $b≥1$. That's the starting point. The conclusion “if everyone knows that everyone knows that $b≥1$” ⇒ “I know that” has been given in the description of the puzzle. $b≥1$ is a global observable while $b>99$ is only a local one. This if this conclusion doesn't work also the guru may not establish the same common knowldge as it needs the same conclusion. Every proof that there might not have been common knowledge about $b≥1$ needs communication (they are out of sync) which is impossible. – Keinstein Sep 12 '13 at 06:02
  • Here are two interesting philosophical problems involved: Strictly speaking complete induction is not a proof but an algorithm to construct a proof. It depends on a premise that could not be verified by the people. Another interesting point is that common knowledge about a false fact ($b=1$ contradicts $b≥97$) may be used to start this algorithm. In other words: A fact in a possible world establishes knowledge about the real world. BTW.: One chain is not enough to describe a decision tree. On the other hand common knowledge may use any logical and factual means available, not only chains. – Keinstein Sep 12 '13 at 06:40
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    This is a great concise answer, and I believe it's correct. The mind-bending thing about the problem, for me, is when I think "how can the fact that she sees a blue-eyed person not be common knowledge??" (common knowledge meaning everyone knows it and everyone knows that everyone knows it, etc.)... but the extremely counterintuitive fact is, it's *not* common knowledge. To try to correct my intuition, I think the best strategy is to meditate on the the 3-blue-eyed case, and if I ever get to the point where I really *get* it (hasn't happened yet), go on to the 4-blue-eyed case. – Don Hatch Aug 14 '14 at 19:22
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    Everyone is defining "common knowledge" mathematically. Common knowledge is gained through observation. If there are 1, 2, 3, 99, or n people around, I will know their eye color through seeing it - I will not need the guru to tell it to me. It will not be common knowledge to know what my own color is, however - he could see blue just like I can in the others' around me - and then he says he sees blue (confirming my observations of n people), but it's not NEW knowledge because I can see others' eyes just like him & it doesn't tell me what my color is, unless I do not see anyone with blue eyes. – vapcguy Jun 02 '16 at 19:02
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The guru starts the doomsday clock. Before the guru speaks, there is no "day 1". Without the common reference time, every blue eyed person (BEP) lives happily with the knowledge that there must be either 99 or 100 BEPS. But there is no way to decide which is true. The common reference time is the key to the apparent paradox. Without it, there is no expectation for the timely behavior of others.

The guru's statement essentially informs everyone on the island, "you better hope all of the X BEPs that you see leave X days from today otherwise it means you have blue eyes". To a BEP, X=99. Otherwise X=100.

Here is slightly different view of where the recursion comes from vs that from the wikipedia page.

Every BEP knows there are either 99 or 100 BEPs. And they all know that every BEP they see either sees 98 or 99.

Alice BEP (like all BEPs ) knows there are either 99 or 100 BEPs.

Bob BEP knows that Alice is considering either the hypotheses {98,99} or {99,100} (Bob himself knows the true number is not 98, he hopes Alice is not considering 100) -- range=[98:100]

Carol BEP has the same view of what Alice is thinking as Bob does. Carol hopes that Bob thinks Alice is thinking [97:99] but realizes that if Carol herself has blue eyes, then Bob's Alice range is [98:100]. range=[97:100]

Dave BEP hopes Carol's Bob's Alice range is [96:99].

and so on....

On the 99th day, every BEP realizes that lastmost person in the chain of hope has not left. So they all leave.

Mark Borgerding
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    I think this is the correct insight. The impetus the guru provides is synchronizing the system. This is not well covered in the usual formulation of the problem however. The common knowledge thread of this problem seems like a bit of a red herring: even before the guru it is reasonable to assume of I know that you know etc already is in place. – SEngstrom Oct 07 '14 at 02:20
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    @SEngstrom No it's not a reasonable assumption. You can assume that verybody knows that everybody knows that... repeated 98 times ... there is at least one blue eyed person. You can not assume that everybody knows that everybody knows that... repeated 99 times ... there is a blue eyed person. And that last step of the reasoning is needed. – Taemyr Oct 09 '14 at 11:52
  • @MarkBorgerding This is almost correct, but most of the chain can be granted on the first day, so every BEP's own range will be [99:100] and every BEP knows that any other BEP's range is either [98:99] or [99:100]. – Jed Schaaf Jul 15 '16 at 00:08
  • Grr. Can't edit after 5 minutes.... @MarkBorgerding This is almost correct, but most of the chain can be granted on the first day, so every BEP's own range will be [99:100] and every BEP knows that any other BEP's range is either [98:99] or [99:100]. 'C' has the same knowledge as 'A' and 'B', so the only possible ranges are [97:98], [98:99] or [99:100]. Each successive day eliminates 97, 98, and then 99 from the possibilities. All the BEPs leave the 4th day. – Jed Schaaf Jul 15 '16 at 00:15
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The question doesn't ask for the solution to the puzzle, which it already linked to.


The first paragraph of the linked puzzle ends with:

[...] Everyone on the island knows all the rules in this paragraph.

The whole paragraph is crucial, but two strongly interacting aspects may be overlooked. First, "[t]hey are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly." This means that they will update their knowledge logically and act accordingly. Second, "[e]veryone on the island knows all the rules in this paragraph" is also a rule in the paragraph. It also refers to itself. That implies that everyone knows that everyone knows that [infinite repetition] everyone knows that everyone on the island knows all the rules in this paragraph. This is called common knowledge (which is much stronger than, say, universal knowledge: everybody knows $P$). Combined with the first aspect, this is sometimes called common knowledge of rationality or CKR, which is often used in game theory (although its full power usually isn't needed, as in this case).

What is the quantified piece of information that the Guru provides that each person did not already have?

"I can see someone who has blue eyes[,]" in itself already was universal knowledge. Its public announcement makes it common knowledge. This, together with the repeated non-leaving of islanders, launches a cascading set of common knowledge that will eventually include that (and which) 100 islanders have blue eyes. (The public observation, i.e., all islanders observe all islanders observe all islanders observe all islanders ... not leaving the island, can, technically, be viewed also as a public announcement.)

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?

To get the blue-eyed islanders to realize that they are blue-eyed, they need to have i) common knowledge that at least 99 islanders have blue-eyes, and that, after that realization, ii) still nobody left. To get the common knowledge thing going it needs to pass through the 1 and 2-person cases.

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?

It is only after 99 nights, that they know that 99 nights wasn't sufficient for any blue-eyed islanders to figure out that they had blue-eyes themselves (notwithstanding CKR). After only 98 or less nights, this was still an uncertainty, not deducible and therefore not known. The islanders aren't "simply verifying something that they already know"; they are stepwise turning knowledge into common knowledge that is necessary for the last step.


NB: I believe the puzzle is more-or-less identical to (and therefore an adaptation of) "Muddy Children" (Fagin et al. 1995; Geanakoplos 1992), which is a textbook example in modal logic.

Keywords: epistemic modal logic, public announcement logic (PAL), dynamic logic of public observation, common knowledge

Řídící
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  • I do not think that the claim marked * is true. Consider the case where the guru announces "98 people who have blue eyes": everyone has known this before (universal knowledge), because everyone sees at least 99 blue-eyed people. Moreover, everyone knows that everyone knows: blue-eyed people see 99 other blue-eyed people, so these must each see at least 98 blue-eyed people. But there it stops. It is not true that everyone knows that everyone knows that everyone knows that the guru sees 98 blue-eyed people. This, however, is a necessary condition for "common knowledge". – Ansgar Esztermann Sep 12 '13 at 08:35
  • You say: He could also have been saying "I got some sand in my eyes, I don't see anything[,]" and still the solution would be the same. Accepting this, it follows by symmetry that the blue-eyeds and brown-eyeds will act the same, contradicting the original answer. A way out of this contradiction would be to say that the original answer is wrong, and all brown-eyeds and blue-eyeds leave on the 100th day. – Fillet Sep 12 '13 at 08:46
  • The symmetry is that there are "100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes)." The original guru statement could possibly break the symmetry, as it refers to blue eyes. If the guru statement is "I have sand in my eyes", then you have no reason to break the original symmetry. The world views are also symmetric, blue sees 99 blue and 100 brown, brown sees 99 brown and 100 blue. – Fillet Sep 12 '13 at 09:13
  • @AnsgarEsztermann I tossed it out. My bad. – Řídící Sep 12 '13 at 14:07
  • @Fillet I tossed it out. My bad. – Řídící Sep 12 '13 at 14:07
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    This is the best answer in my opinion, thanks! – Philip Feb 16 '14 at 03:56
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Just work out the case where there are 2 people, then 3 people, then 4 people. It's the same principle, just more mind-boggling, for higher $n$. When there are just 2 people the situation is pretty much clear. When there are 3 people, does each know that everybody knows that everybody knows that there are people with blue-eyes? (there was no typo in what I wrote). To make it clearer, give the people distinct names and ask yourself: if John has blue eyes, does he know that Jeff knows that Ted knows that there are people with blue eyes. Then answer the question: if John does not have blue eyes, does he know that Jeff knows that Ted knows that there are people with blue eyes. The answers are different. But, the answers become trivially 'yes' if it becomes common knowledge that there are blue-eyed people.

Ittay Weiss
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    It has always been common knowledge that there are blue-eyed people. So the last sentence is not correct. – Keinstein Sep 10 '13 at 11:29
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    @Keinstein: Though everybody always knew there were blue-eyed people, it was not in a technical sense [common knowledge](http://en.wikipedia.org/wiki/Common_knowledge_(logic)) until the Guru spoke. – Marc van Leeuwen Sep 10 '13 at 12:33
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    No. Everyone has seen that there are more than one people with blue eyes. Thus everyone knew that there are people with blue eyes. Everyone knows further that everyone has seen someone with blue eyes. Thus, everyone knows that everyone knows that there are people with blue eyes. As all know that all know this logic. The recursion goes ad infinitum. So it was common knowledge that there have been people with blue eyes. That has been emphasised in the linked version provided by A Googler. The only thing they didn't know was the exact number which became clear after 99 nights. – Keinstein Sep 10 '13 at 12:46
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    @Keinstein: You are wong. Suppose for simplicity there are just three pairs of blue eyes. Everybody knows there are blue eyes (they see two pairs). Everybody also knows that everybody knows this, because they know the two pairs they see can see each other. But nobody among the blue-eyed _knows_ that the previous sentence it true (i.e. that everybody knows that everybody knows), because if those two pairs of blue eyes were the only ones, each of them would see only one pair, and the argument used (I can see two pairs that can see each other) would not apply for them. – Marc van Leeuwen Sep 10 '13 at 14:04
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    This is true for 3 pairs. Ok. But it doesn't imply that it is true for more pairs especially for 100. Otherwise I'd like to see a proof that constructs a real contradiction for $n∈\mathbb N$. We know that there cannot be a circle of knowing the explicit color. But that is not a contradiction against a circle of knowing that someone knows that there is at least one person with blue eyes. So suppose that there has been an external source of information without an initial moment that implanted the common knowledge that there is at least one blue-eyed person. Why should everyone know the truth? – Keinstein Sep 10 '13 at 20:31
  • @MarcvanLeeuwen I think I have understood your algorithm. It involves communication between the people in the same way as it happens after the guru spoke: people leave the island. As this needs a notion of day zero this won't happen before the Guru spoke. Thus, even if there was common knowledge available beforehand it would not have been usable to find out the own eye-color. see also the last edit to my answer. – Keinstein Sep 11 '13 at 02:05
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With more than one blue-eyed islander, the guru's statement on its one is obvious to everyone, so in isolation it provides no information. As a result, no one heads for the ferry that night. However, without any more words being spoken, each passing day results in more information.

On day one, the guru's statement alone says "There is at least one blue-eyed person".

On day two, the guru's statement, plus the fact that the boat left empty, says "There are at least two blue-eyed people" (for if not, then someone would have left).

On day three, the guru's statement plus two observations of the boat, says "There are at least three blue-eyed people".

Now, the blue-eyed people can all see 99 others, so up to day 99 they still cannot deduce any more than they can see. And each is in the position that he knows there are at least 99 blueys, but he doesn't know if any of the blueys he can see knows that. On the 100th day, however, they have the one extra piece of information that is enough to complete the deduction.

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  • You agree, that people know in advance that nothing happens until day 99? – Keinstein Sep 10 '13 at 12:19
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    I think the real challenge is that there were also more than $99$ days _before_ the Guru spoke yet nothing happened. Therefore _something_ must have changed due to the Guru speaking; it _did_ provide new information. – Marc van Leeuwen Sep 10 '13 at 12:50
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    It didn't provide new information so much as a coordinating action to begin counting ferry departures. The people were sitting in a state of N=99 or N=100 blue eyed people. – Kyle Hale Sep 10 '13 at 16:04
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    @KyleHale it must have provided new information, since in the absence of new information, how can anything change? – Ittay Weiss Sep 10 '13 at 18:32
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    Suppose that a given person on the island can see k people with blue eyes. The information given by the guru is then, "if k blue-eyed people are gone after k nights, then my eyes are brown, otherwise my eyes are also blue." – Jaycob Coleman Sep 10 '13 at 20:03
  • Well, I will revise my answer slightly: the only information that is pertinent of all the words she used is "blue," which states which people should start counting off. (This also helps eliminate the problem's possibility of having red eyes.) But her chief value is doing the group communication the others are unable to do. What information does the starting pistol at a race provide? Whatever your answer is, that's what the Guru provided as well - synchronicity among common actors who are unable to synchronize on their own. – Kyle Hale Sep 10 '13 at 21:35
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    I will also add: the only way the islanders have to communicate with each other is to get on the ferry or not. The Guru's information is "Today is the day you would get on the ferry if you were the only person with blue eyes." All logic flows from there. Which you can see why it isn't really information, it's more of a ... decision? Again, it's firing the starting pistol. – Kyle Hale Sep 10 '13 at 21:55
  • @JaycobColeman your comment looks like a better answer than any of the others at this point. How about posting it as an answer? – Don Hatch Aug 13 '14 at 18:46
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    @KyleHale The precise piece of information that the Guru adds is "From this point on it is common knowledge that there is at least one blue eyed person". Both the added information and the firing gun is needed. If she had said "I see at least 99 blue eyed persons" the information added would be "From this point on it's common knowledge that there are at least 99 blue eyed persons", and the blue eyes would leave on day 2. – Taemyr Oct 09 '14 at 12:01
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Sorry this answer became so long. If you want the two-minute answer, just read the Terminology then skip down to the Answers to the Questions. You can then fill in the details as desired.

Terminology

  • Let $A, B, C, A_i$ denote the blue-eyed islanders.
  • Let $A_i^*$ denote the proposition that $A_i$ has blue eyes (which does not imply that $A_i$ knows this).
  • We'll use the standard symbols and rules of propositional logic.
  • $A\leadsto P$ means islander $A$ knows that $P$ is the case, where $P$ is a proposition. The operator "$\leadsto$" is right-associative and has higher precedence than the logical implication operator "$\Rightarrow$".
  • $\cal O$ is the proposition that there is at least one blue-eyed islander. This is what the Guru announces on day 1.

1 Blue-eyed Islander

To introduce use of this notation, let's briefly go over what happens when there is 1 blue-eyed islander $A$.

Day 1

After the Guru makes her announcement, it is the case that $$A\leadsto{\cal O}.\tag{1.1.1}$$ Since $A$ sees no islanders with blue eyes, she concludes it's her: $$A\leadsto A^*.\tag{1.1.2}$$ This fact entitles $A$ to leave the island on day 1, so we are done with the case of the 1 Blue-eyed Islander.

2 Blue-eyed Islanders

Before the Guru makes her announcement, the following statements can be said about blue-eyed islanders $A$ and $B$: $$A\leadsto{\cal O}\\ B\leadsto{\cal O}.\tag{2.0.1}$$ This follows simply from the fact that they can each see one other blue-eyed person on the island.

Day 1

After the Guru announces $\cal O$, it is the case that $$A\leadsto B\leadsto{\cal O}\tag{2.1.1}$$ as well as the one other permutation of this: $B\leadsto A\leadsto\cal O$. The above reads "$A$ knows that $B$ knows that $\cal O$." It is important to grasp at this point that this fact wasn't the case prior to the Guru's announcement. Even though both $A$ and $B$ knew there was at least one blue-eyed person on the island, $A$ didn't know that $B$ knew this, because for all $A$ knows, she may have brown eyes.

(2.1.1) can be written as $$A\leadsto B\leadsto(A^*\vee B^*).\tag{2.1.2}$$ The substitution ${\cal O}\mapsto A^*\vee B^*$ is valid in this particular context, because everyone else on the island other than $A$ and $B$ do not have blue eyes, which is known to both $A$ and $B$, and $A$ knows that $B$ knows this.

(2.1.2) can be written as $$A\leadsto(\neg A^*\Rightarrow B\leadsto B^*)\tag{2.1.3}$$ which will be useful for Day 2. To prove this, the following axiom is needed:

Knowledge Conjunction Axiom $$((A_i\leadsto P)\;\wedge\;(A_i\leadsto Q))\;\;\Leftrightarrow\;\; (A_i\leadsto(P\wedge Q))$$ $A_i$ knows $P$ and $A_i$ knows $Q$ if and only if $A_i$ knows $P$ and $Q$.

The proof for (2.1.3) is of primary importance, as it readily generalizes to any number of blue-eyed islanders and days, so a detailed proof is given here for the interested reader.

  1. $A\leadsto(\neg A^*\Rightarrow(B\leadsto\neg A^*))$
  2. $A\leadsto((B\leadsto\neg A^*)\Rightarrow(B\leadsto\neg A^*))$
  3. $A\leadsto(((B\leadsto\neg A^*)\Rightarrow(B\leadsto\neg A^*))\wedge (B\leadsto(A^*\vee B^*)))$
  4. $A\leadsto((B\leadsto\neg A^*)\Rightarrow((B\leadsto\neg A^*)\wedge (B\leadsto(A^*\vee B^*))))$
  5. $A\leadsto((B\leadsto\neg A^*)\Rightarrow(B\leadsto( \neg A^*\wedge(A^*\vee B^*))))$
  6. $A\leadsto((B\leadsto\neg A^*)\Rightarrow(B\leadsto B^*))$
  7. $A\leadsto(\neg A^*\Rightarrow B\leadsto B^*)$

Step-by-step justifications:

  1. $A$ knows that if she doesn't have blue eyes, then $B$ will know this.
  2. $P\Rightarrow P$ tautology.
  3. Knowledge Conjunction Axiom of step 2 with (2.1.2).
  4. $((P\Rightarrow P)\wedge Q)\;\Rightarrow\;(P\Rightarrow(P\wedge Q))$ tautology applied to step 3.
  5. Knowledge Conjunction Axiom applied to step 4.
  6. Disjunctive syllogism applied to step 5.
  7. Knowledge Conjunction Axiom applied to steps 1 and 6, and transitivity of $\Rightarrow$.

This just delineates in detail what many people can reason without the symbolic logic, which is the fact that $A$ knows that if she doesn't have blue eyes, then $B$ will know he does. The value of this formalism is that it extends readily into more complicated scenarios where our intuition may have trouble keeping up.

Day 2

No one left the island on Day 1, so no one knew they had blue eyes. In particular, $$\neg(B\leadsto B^*)\tag{2.2.0}$$ otherwise $B$ would have left. This fact is publicly known, so in particular $A$ knows it: $$A\leadsto\neg(B\leadsto B^*)\tag{2.2.1}.$$ Combining this with (2.1.3) via the Knowledge Conjunction Axiom gives $A\leadsto((\neg A^*\Rightarrow B\leadsto B^*)\wedge\neg(B\leadsto B^*)).$ Modus tollens yields $$A\leadsto A^*.\tag{2.2.2}$$ This is $A$'s ticket off the island, so she leaves today. These arguments for both days are symmetric in $A$ and $B$, so apply to $B$ as well. Both blue-eyed islanders leave the island on Day 2.

3 Blue-eyed Islanders

Before the Guru makes her announcement, $C\leadsto\cal O$, $B\leadsto\cal O$, and $A\leadsto\cal O$. In addition, $$A\leadsto B\leadsto{\cal O}\\ B\leadsto C\leadsto{\cal O}\\ C\leadsto A\leadsto{\cal O}.\tag{3.0.1}$$ For example, $A$ knows $B$ knows $\cal O$, because $A$ knows $B$ knows $C^*$.

Day 1

After the Guru announces $\cal O$, it is the case that $$A\leadsto B\leadsto C\leadsto{\cal O}\tag{3.1.1}$$ as well as the $3!-1=5$ other permutations of this in $(A,B,C)$. It is important to pause at this point and understand that this was not true prior to the Guru's announcement. Especially if one wishes to understand what quantified information the Guru is actually providing that each person didn't already have, this is it. Even though everyone knew that everyone else knew $\cal O$, that's only 2 levels deep. It required the Guru's public announcement to get to the 3rd level. $A$ did not know that $B\leadsto C\leadsto\cal O$ prior to her announcement.

Though the symbols provide the formalism, one informal but ituitive notion to consider is that of each person's "world"—the information available to a person. The world as seen through the eyes of $A$ is one in which there are 2 other blue-eyed people $B$ and $C$. Now consider the world of $B$ as considered by $A$. In this world, there is only 1 person whom with certainty has blue eyes: $C$. Moreover $C$ does not know if anyone else has blue eyes; $C$ does not know $\cal O$ until the Guru speaks, whose announcement penetrates through these worlds so that even in this doubly layered consideration, $C\leadsto\cal O$. In other words, $A\leadsto B\leadsto C\leadsto{\cal O}$. (Why am I reminded of the movie Inception?)

Moving forward as before, from (3.1.1), $$A\leadsto B\leadsto C\leadsto(A^*\vee B^*\vee C^*)\tag{3.1.2}$$ using the substitution ${\cal O}\mapsto A^*\vee B^*\vee C^*$. This is just representing the fact that any of $A$, $B$, or $C$ might be the one the Guru was talking about, which everyone knows that everyone knows etc. to arbitrary depth.

It follows from (3.1.2) that $$A\leadsto B\leadsto(\neg(A^*\wedge B^*)\Rightarrow C\leadsto C^*). \tag{3.1.3}$$ The proof for this takes on an analogous structure as the proof for (2.1.3) above.

Day 2

As before, no one left on Day 1, so it is concluded that $\neg(C\leadsto C^*)$ (as well as for $A$ and $B$). Since this is just as public as the Guru's announcement, everyone knows everyone knows etc. this to arbitrary depth. In particular: $$A\leadsto B\leadsto\neg(C\leadsto C^*)\tag{3.2.1}.$$ This fact, and its 5 other permutations, were not true until it was publicly observed that no one left on the ferry the prior midnight. Combining this with (3.1.3) via the Knowledge Conjunction Axiom, and applying modus tollens as before, yields: $$A\leadsto B\leadsto(A^*\vee B^*).\tag{3.2.2}$$ Now we are beginning to see a pattern here. This can be written as $$A\leadsto(\neg A^*\Rightarrow B\leadsto B^*)\tag{3.2.3}$$ which is identical to (2.1.3).

One might wonder, since $A$ knows that $\neg(B\leadsto B)$, because $B$ did not leave last night, can it be deduced from (3.2.3) that $A\leadsto A$?

The answer is no, but to see this it must be noted when certain knowledge was obtained. It would be more precise to write (3.2.3) as $$A\leadsto_2(\neg A^*\Rightarrow B\leadsto_2 B^*)\tag{3.2.4}$$ where $\leadsto_k$ denotes knowledge on Day $k$. The knowledge described in (3.2.3) was not known until Day 2, after observing that $C$ did not board the ferry, which means $\neg(C\leadsto_1 C^*)$, or that $C$ did not know she had blue eyes on Day 1. Similarly, $$A\leadsto_2\neg(B\leadsto_1 B^*).\tag{3.2.5}$$ Thus modus tollens cannot be applied to (3.2.4) and (3.2.5) because $B\leadsto_1 B$ and $B\leadsto_2 B$ are two different propositions.

Day 3

No one left again the previous night, so $$A\leadsto\neg(B\leadsto B^*).\tag{3.3.1}$$ Combined with (3.2.3), $A\leadsto A^*$ so $A$ can now leave the island. Same reasoning applies to $B$ and $C$, so all 3 islanders leave on Day 3.

$n$ Blue-eyed Islanders

Assume $2<n$. Prior to the Guru's announcement, it is a fact that $$A_1\leadsto A_2\leadsto\cdots \leadsto A_{n-2}\leadsto A_{n-1}\leadsto{\cal O}\tag{4.0.1}$$ including all other combinations and permutations of this chain of equal or lesser length, out of the the $n$ blue-eyed islanders $A_i$ for $i\in\{1,2,...,n\}$. Note that (4.0.1) includes only $n-1$ islanders, not $n$. This is because $A_1$ can imagine the world through $A_2$'s eyes, who looks through $A_3$'s eyes, ..., who looks through $A_{n-2}$'s eyes, who looks through $A_{n-1}$'s eyes, who gazes upon $A_n$ but in this world no information is available to guarantee any other blue-eyed person is on the island. So it cannot be concluded in this $(n-1)$-nested world that $A_n\leadsto\cal O$. Without the Guru's announcement, the longest chain of distinct blue-eyed islanders that can be stated is one which includes no more than $n-1$ islanders, such as (4.0.1).

Day 1

Once the Guru announces $\cal O$, the chain can now include all $n$ blue-eyed islanders. The guru's statement is equivalent to $n!$ statements, which are all the permutations of $A_i$ in $$A_1\leadsto A_2\leadsto\cdots \leadsto A_{n-1}\leadsto A_n\leadsto{\cal O}.\tag{4.1.1}$$ In anyone's world, no matter how deep the levels, the knowledge of $\cal O$ is always available.

Substituting as before for $\cal O$, this becomes $$A_1\leadsto A_2\leadsto\cdots \leadsto A_{n-1}\leadsto A_n\leadsto\bigvee_{i=1}^n A_i^*.\tag{4.1.2}$$ Using analogous steps to prove (2.1.3) from (2.1.2), it follows from (4.1.2) that $$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-2}\leadsto A_{n-1}\leadsto \left(\neg\bigvee_{i=1}^{n-1}A_i^*\Rightarrow A_n\leadsto A_n^*\right) .\tag{4.1.3}$$

Day 2

As in previous scenarios, since $A_n$ in particular didn't leave, it is publicly known that $\neg(A_n\leadsto A_n^*)$. Everyone already knew this so what new information is there? The new information may be expressed as another set of chain statements, of all $n!$ permutations in $A_i$ of $$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-2}\leadsto A_{n-1}\leadsto\neg(A_n\leadsto A_n^*).\tag{4.2.1}$$ This wasn't the case until the previous ferry left with no passengers.

Combining (4.1.3) and (4.2.1) together, using the Knowledge Conjunction Axiom and modus tollens yields $$A_1\leadsto A_2\leadsto\cdots \leadsto A_{n-2}\leadsto A_{n-1}\leadsto \bigvee_{i=1}^{n-1} A_i^*.\tag{4.2.2}$$ Following the same pattern as before, it can be deduced that $$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-3}\leadsto A_{n-2}\leadsto \left(\neg\bigvee_{i=1}^{n-2}A_i^*\Rightarrow A_{n-1}\leadsto A_{n-1}^*\right) .\tag{4.2.3}$$

Day $k$

Assume $1<k\le n$. On the previous night $A_{n-k+2}$ didn't leave, so it is publicly known that $\neg(A_{n-k+2}\leadsto A_{n-k+2}^*)$. The new information that wasn't previously available allows for all permutations and combinations in $A_i$ of identical length of the following statement to be made: $$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-k}\leadsto A_{n-k+1}\leadsto\neg(A_{n-k+2}\leadsto A_{n-k+2}^*).\tag{4.3.1}$$ Combined with the conclusions from the previous day, it is the case that $$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-k}\leadsto A_{n-k+1}\leadsto\bigvee_{i=1}^{n-k+1} A_i^*.\tag{4.3.2}$$ This can be proven from induction on $k$, which is omitted for brevity but is of the same form as the proof for (2.1.3). Also if one follows how (4.2.3) was derived from (4.2.2), (4.2.1) and (4.1.3) then this will also outline how one can prove this via induction.

Day $n$

(4.3.2) shrinks by one islander on each passing day, until finally when $k=n$ we are left with $$A_1\leadsto A_1^*.\tag{4.4.0}$$ Since these arguments have been symmetrical in all the $A_i$, $$\forall i\in\{1,2,...,n\}:A_i\leadsto A_i^*.\tag{4.4.1}$$ On Day $n$, all blue-eyed islanders leave the island.

Answers to the Questions

1) What is the quantified piece of information that the Guru provides that each person did not already have?

All $100!\approx 9.3\times 10^{157}$ permutations in $A_i$ of the statement $$A_1\leadsto A_2\leadsto\cdots \leadsto A_{99}\leadsto A_{100}\leadsto{\cal O}.\tag{5.1.1}$$ Everyone already knows ${\cal O}=\bigvee_{i=1}^{100}A_i^*$. That is not the value of the Guru's announcement. It is that everyone knows that everyone know that everyone knows ... that $\cal O$ is the case, that is the new information provided by the Guru's announcement that wasn't previously known. In contrast, if the Guru were to tell all the islanders in private the same fact $\cal O$, no islanders would be able to leave the island. So it is not simply the information content of her words that we must look at; there is additional information in knowing that everyone else heard her too. Correspondingly, each day that passes provides a new piece of information that is comparably subtle. When an islander doesn't leave, it is like another public announcement, which includes more information than just the fact that $A_i$ didn't leave last night. It is the knowledge that everyone else knows too. This knowledge is quantified in the answer to question 3 below. Eventually, after 100 days, these additional pieces of information will shrink (5.1.1) down to a fact that the islander can act upon. Namely, $A_i\leadsto A_i^*$ for all $i\in\{1,2,...,100\}$.

2) 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?

Yes, everyone knows that there are no less than 99 blue-eyed people on the island. The key concept to this problem is the recursive nature in which islanders deduce what they can, knowing only the information that they bring with them as they consider the world through one anothers' eyes. As $A_1$ does this from the perspective of $A_2$ who sees through the eyes of $A_3$, ..., who sees through the eyes of $A_{98}$, $A_{98}$ is left only to gaze upon and consider what $A_{99}$ and $A_{100}$ can possibly know within a world of such limited information. In this nested world 98 levels deep, we cannot take for granted that islanders $A_1$ through $A_{98}$ have blue eyes, just as we cannot take for granted that $A_1$ has blue eyes when we consider only her point of view on all the rest of the 99 blue-eyed islanders. Therefore considering the logic of a 2 blue-eyed-inhabited island is a worthwhile consideration. When 2 days go by in which $A_{99}$ and $A_{100}$ don't leave the island, then $A_1$ knows $A_2$ knows ... knows $A_{97}$ knows $A_{98}$ knows that someone else other than $A_{99}$ and $A_{100}$ have blue eyes. Now there's only 98 days to go.

If this escapes the intuition, then consider the case of 1, 2, and 3 blue-eyed islanders, and allow the logical formalism as deliniated above to provide the scaffolding that extends the intuition.

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

Because they're not. Every day that passes, new information is provided that wasn't previously known. It's not as simple as a statement that islander $A_k$ didn't leave, because yes that was already known and anticipated. There is additional information in the knowledge that everyone else knows that everyone else knows etc. that $A_i$ did not leave the island. To be precise, on day $k$, for $k>1$, the new facts that weren't previously the case are all $100!/(k-2)!$ permutations and combinations in $A_i$ of $$A_1\leadsto A_2\leadsto\cdots\leadsto A_{100-k}\leadsto A_{101-k}\leadsto\neg(A_{102-k}\leadsto A_{102-k}^*).\tag{5.1.2}$$ With each passing day $k$, it is these facts that whittle away at the chain of knowledge (5.1.1) setup by the Guru for each islander on Day 1.

Matt
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  • Sorry to deprive you of the honour of having the fourth longest post on the site with a score of $0$, but I had to upvote ;-) – John Gowers Oct 09 '13 at 08:35
  • Thanks! I came to the party after it ended, but still wearing my party hat. – Matt Oct 09 '13 at 16:40
  • Great attempt at an explanation, but I don't agree at all with your Day 2 assessment for 2 islanders. There is nothing mathematical that says just because no one left on Day 1, that *B* can leave just because *A* did not, just because "someone" has blue eyes and everyone knows someone has blue eyes. *B* will not *know* they have blue eyes anymore than *A* would -- especially if both had blue eyes - neither would leave, because they'd be thinking the other should. – vapcguy Jun 20 '16 at 12:58
  • @vapcguy If B had brown eyes, then A would have left on day 1 because A would see that no one else had blue eyes, and therefore it must be her. Since A didn't leave on day 1, B realizes that he himself must not have brown eyes, and therefore has blue eyes, so he leaves on day 2. Same goes for A. – Matt Jun 20 '16 at 21:17
  • @Matt Yes, if A looks around, sees no blue-eyed people, she can figure out she can leave because she must be the blue-eyed person. Conversely, if A doesn't leave because she sees a blue-eyed person, it gives B the right to go because it must be B. Perfect explanation in a 2-party system-I don't think you explained it this well above-thanks! But, I do take issue that B can't figure out he had blue eyes just because of A's failure to figure out she has blue eyes (if she did). A has to look at B, see B has brown eyes, and then leave. If A does not, but has blue eyes, it doesn't mean B has blue. – vapcguy Jun 21 '16 at 14:52
  • Seems like people are saying B thinks: "Oh, A didn't leave, so guess I have blue-eyes [leaves]". It's not really that simple if A and B can *both* have blue-eyes. B would see A's blue eyes and not leave, A would see B's blue-eyes and not leave. They wouldn't then just both leave on Day 2. – vapcguy Jun 21 '16 at 15:07
  • And as you get more complex/greater numbers, there would be more interactions/people/varying combinations. They could see a duplicate person they saw before on each day of this path to the total. And how could they remember who they've seen? Do we assume they are marking them off? And if everyone only leaves at the end, how would they make their eliminations so as to *know* they *must* have blue-eyes and leave (or *must* have brown-eyes, and therefore have to stay)? – vapcguy Jun 21 '16 at 15:07
  • Excellent answer that provides an in-depth, (fairly) easy-to-read proof and clearly answers all the questions. I do think that the islanders only need consider three "levels" deep, because they can each observe the actual number of blue-eyed people or the number minus 1, and can infer the minimum possible number that a blue-eyed person can see. Thus, the counting can start at that point, and all the blue-eyed people will leave on the 4th day, unless there are only 1, 2, or 3 blue-eyed people. (See my answer to a similar question on Puzzling.SE: http://puzzling.stackexchange.com/q/236/20907) – Jed Schaaf Jul 15 '16 at 05:18
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The passage of time is important input because an event happens every night, and that event provides information to every islander what the others know or do not know. Whether or not anyone leaves on a given night, the information content changes. By not leaving, everyone has communicated clearly, "I do not know my eye color".

When the guru speaks, he polarizes the group into two, let's call them groups S and groups T. Group S is blue-eyed, and so they see one fewer blue-eyed people than group T.

Everyone's question is then, do I belong to group S (snappy), or group T (tardy)?

And executes this algorithm: "I will leave the island on night X, where X is the number of blue people I see".

So for instance if there are 50 blue-eyed people, then group S is planning to leave on night 49, and group T on night 50.

Nobody knows whether he or she is part of group S or group T, but this comes to light on night 49 when group S leaves, leaving group T.

Of course, group T's travel plan is thereby wrecked! On night 49, everyone in group S knows they are in group S, and thereby know that their eyes are blue, and those in group T also know that they are group T. But all they know is that their eyes are not blue, which does not amount to knowing their own eye color, and so they must stay on the island forever.

So, why do the islanders go through this charade of waiting out all these days? Well, the algorithm requires it. They cannot simply trim 49 days out of the wait because that would require a collective decision. To trim 49 days of waiting you have to know that the two days in question are 49 and 50, of which 49 is the minimum. Everyone who plans to leave on night X knows that people with a different plan are either targetting X-1 or X+1, but does not know which! So there is no way to avoid having to count the days: nobody knows what "bias offset" value to subtract from the number of nights to shorten the waiting game.

If you look at this another way, counting the nights and leaving is a way for the islanders to communicate a message to all the other islanders. By waiting until night X and bailing from the island, each islander expresses the message "I saw X blue-eyed people on the day the guru spoke". The value of X is significant and so the days must be counted out earnestly; no shortcuts. It is this key piece of communication which triggers the exodus of blue-eyed people. If on day zero everyone were allowed to speak to say how many blue-eyed people he or she sees, then all the blue-eyed people could leave that same night, because only two numbers would be spoken (e.g. 49 and 50). And those uttering 49 would all know that they are part of group S.

Kaz
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1.) The quantified piece of new information that the Guru provides is not 'at least one person has blue eyes' (except in the $n=1$ case), since everyone knew that already. In fact, this quantified piece of information is rather complicated.

If there is one islander, then the new information is exactly 'there is at least one person on the island with blue eyes'. Then the one islander knows that that one person must be them.

If there are two people on the island, then it's not news to the first islander that there's someone on the island with blue eyes: they can see that their friend has blue eyes. However, after a day has passed and their friend hasn't left, they know the following piece of information that they didn't know before:

The fact that there's at least one person on this island with blue eyes was not news to the other person on the island.

If there are three people on the island, then the new piece of information, with brackets added to show structure, is

The fact that (the fact that there's at least one person on this island with blue eyes was not news to the other islanders) was not news to the other islanders.

And so on.

In fact, it occurs to me that you can state the new piece of information for $n$ islanders rather simply if you don't mind hiding the detail of the islander's logical deductions. It is:

The information that the Guru can see at least one islander with blue eyes was not enough to convince the other islanders that they had blue eyes in $n-1$ days.

Or even:

I have blue eyes.

But I suppose that those don't really count as 'quantified'.

I think that it's because the new piece of information is so complicated (and only becomes more complicated as we increase the number of islanders) that this puzzle seems so counter-intuitive.

2.) This time, we'll start with $n=100$, and work back down. If I am on the island, and I look round, I see $99$ people with blue eyes. If you're on the island with me then, as far as I know, you might be seeing $98$ people with blue eyes and one person (me) with brown eyes. This is because I don't know my own eye colour. So the $n=98$ case becomes relevant.

Let's suppose that I am absolutely convinced that I have brown eyes, and that everyone else on the island is convinced that they have brown eyes, at least until they're proved wrong. I look at you, looking at $98$ people with blue eyes, and think, 'Haha! Each of those $98$ people is looking around and seeing $98$ people with blue eyes, but you probably think that you have brown eyes, so you think that those people can only see $97$ people with blue eyes.'

In other words, in my imagination, there are $99$ people on the island with blue eyes, and in my imagination of your imagination, there are only $98$. Then in my imagination of your imagination of someone else's imagination, there are only $97$. Eventually, we get to some sub-sub-...-sub-imagination where there are only one or two people with blue eyes even though all of the islanders in fact know that there are at least $99$.

Don't worry if you have trouble getting your head round that one. The human mind isn't designed to handle so many conceptual layers - that's why tools such as induction are so useful.

John Gowers
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The reason it's not "common knowledge" beforehand that there is someone with blue eyes is:

Let's simplify the case to four people other than the guru: Two (Alice and Bob) with blue eyes, and two (Carol and Dave). It is true that everyone can, in fact, see at least one person with blue eyes. Alice can see Bob, Bob can see Alice, and Carol and Dave can see both of them. However, Alice does not know that Bob can see anyone with blue eyes, or vice versa. Alice and Bob also do not know that Carol and Dave can see two people with blue eyes.

Now extend it to six people. Erin has blue eyes, Frank has brown eyes. Erin can see that Alice and Bob both have blue eyes, but as far as she knows, either one of them can only see one person [i.e. the other one respectively], so she does not know that Bob knows that Alice knows that there is someone with blue eyes.

Now extend it to eight people. Blue4 can see Alice, Bob, and Erin all have blue eyes, but likewise does not know that Erin knows that Bob knows that Alice knows there is someone with blue eyes. Now extend it to ten people: Blue5 does not know that Blue4 knows that Erin knows that Bob knows that Alice knows. Now twelve, fourteen, etc.

This same logic, by the way, extends to why you can't shortcut and say that "people know in advance that nothing happens until day 99", because the Guru has not said there are two people with blue eyes, and [etc] does not know that Blue4 knows that Erin knows that Bob knows that there are two people with blue eyes, until the first day has passed. (since it is common knowledge that if there had been only one, they would have left immediately)

Random832
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  • The induction starts at 3 blue eyed people not with 2. – Keinstein Sep 10 '13 at 13:29
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    @Keinstein how does that mean it's not useful, informally, to describe the case with two? – Random832 Sep 10 '13 at 13:31
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    Lets take A, B, C, D, E, F where A, B, C are blue and D, E, F not. Then A nows that B sees C and that C sees B, B knows that C sees A and A sees C, C knows that A sees B and B sees C. The other three know that. Thus everyone knows that everyone sees someone blue. – Keinstein Sep 10 '13 at 13:43
  • Doesn't matter, since my point is about what people know about what other people know: i.e. that C doesn't know that A knows that B sees C. Describing the A/B case is useful for explaining it this way, since the facts of A/B are equivalent to what C knows in A/B/C. – Random832 Sep 10 '13 at 13:54
  • That was a little bit short, A knows that B sees C, B knows that, A nows that and so on. Thus, A and B have common knowledge about C. – Keinstein Sep 10 '13 at 14:12
  • Adding a 4th blue one, then A and B share common knowledge about C and G. And they also share common knowledge together with C about the existence of one person as well as with G (about the existence of one blue person). In short: any two people out of A, B, C, D share common knowledge about the existence of at least one person with blue eyes. They know that either of them shares a similar knowledge about (possibly) another person. – Keinstein Sep 10 '13 at 14:26
  • Anyway: there is a much better proof: Add the statement: “It is common knowledge that there exists at least one person with blue eyes.“ To the original puzzle. The solution would be the same. So that cannot be the anwer. On the other hand I'm still convinced that there is a minimal number ≤10 such that it gets common knowledge that there exists at least one person with blue eyes. – Keinstein Sep 10 '13 at 14:37
  • No, because then D doesn't know that (for example, you can rearrange these in any order) C knows that B knows that A knows about D. What D can know in A/B/C/D is equivalent to the facts of A/B/C, which include that C only knows the facts of A/B (and that A only knows the facts of B/C, and B only knows the facts of A/C). This can be extended indefinitely, though it becomes factorially more difficult to fully describe. – Random832 Sep 10 '13 at 15:19
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    I'd love to meet these Alice and Bob, they're all over the mathematics world, even in modern physics. – MyUserIsThis Sep 10 '13 at 16:14
  • About A means: “that A knows that there is at least one person with blue eyes”. Suppose you have blue eyes and don't know it. As soon you know that everyone of the other blue-eyed people knows about everyone that he knows that there is at least one person with blue eyes, everyone can conclude that the set of people {that don't know that everyone knows that there is least one person with blue eyes} is empty. As everyone knows that this set is empty, everyone knows that everyone knows that everyone knows that there is at least one person with blue eyes. This recursion works for its subsets, too. – Keinstein Sep 10 '13 at 16:16
  • You're forgetting to add an extra level of meta-knowledge per person. Yes, "everyone can conclude that the set of people {that don't know that everyone knows that there is least one person with blue eyes} is empty.", but that doesn't imply that everyone can conclude that _the set of people who cannot conclude __that___ is empty, and so on. – Random832 Sep 10 '13 at 17:23
  • @Random832 After some additional thoughts I came to the conclusion: It's useless to argue about whether it is necessary that it had to be common knowledge beforehand. It is sufficient to know that it may have been common knowledge beforehand. In that case we can assume that it was. And then there must be another reason. If there were only 3 people and A knows that B knows that C knows that A knows that there is at least one person with blue eyes. Then that is not a contradiction. How they know that is not defined at least in Wikipedia. It tells them nothing about the color of special person. – Keinstein Sep 10 '13 at 20:17
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The answer to question 1, if we assume no one ever knew their eye color since the beginning of time on that island and that no one ever left the island, is that the Guru told every blue-eyed islander that their eyes were blue. Because only after she says that can the thought process below be followed by blue-eyed islanders. What she said has no effect on the islanders with other eye colors. After 100 days, no one else can leave until the Guru mentions another color.

Regarding questions 2 and 3, to illustrate why we need to wait 100 days, we need to consider what each blue-eyed islander thinks every other blue-eyed islander perceives. When we do that, we can clearly see why we need to go down to the case of 1 and 2 blue-eyed islanders.

Each blue-eyed islander thinks:

  • I can see 99 blue-eyed islanders. If I am not blue-eyed, then:
    • Each of the 99 blue-eyed islanders I can see will see 98 blue-eyed islanders. If each of the 99 consider the possibility that they are not blue-eyed, then they will each think that:
      • Each of the 98 blue-eyed islanders they see will see 97 blue-eyed islanders. If each of the 98 consider the possibility that they are not blue-eyed, then they will each think that:
        • Each of the 97 blue-eyed islanders they see will see 96 blue-eyed islanders. If each of the 97 consider the possibility that they are not blue-eyed, they will think that:
          • ...
            • Each of the 2 blue-eyed islanders they see will see 1 blue-eyed islander. If each of the 2 consider the possibility that they are not blue-eyed, they will think that:
              • The one blue-eyed islander they see will leave today.
            • If no one leaves today, then the 2 they see will leave on the 2nd day.
          • ...
        • If no one leaves on the 96th day, then the 97 they see will leave on the 97th day.
      • If no one leaves on the 97th day, then the 98 they see will leave on the 98th day.
    • If no one leaves on the 98th day, then the 99 I can see will leave on the 99th day.
  • If no one leaves on the 99th day, then I am blue-eyed and I shall leave with the 99 I see on the 100th day.
andy
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What is the quantified piece of information that the Guru provides that each person did not already have?

The Guru provides the common knowledge that there is somebody with blue eyes. Basically, every person knows that $\exists$ somebody with blue eyes, and every person knows that every person knows that, and so on ad infinitum.

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?

Because for this person (person A), there are two possibilities:

  • He is blue eyed (100:100)
  • He is brown eyed (99:101)

So A can't leave immediately.

For the second one, let's pick a person B in person A's visualization. B sees only 98. In B's mind (remember, B is a figment of A's imagination while visualizing option 2), there are two possibilities:

  • He is blue eyed (99:101)
  • He is brown eyed (98:102)

So, A knows that B can't leave immediately.

Now, in B's mind in A's mind, there are two possibilities for a person C to leave:

  • He is blue eyed (98:102)
  • He is brown eyed (97:103)

Again, two possibilities, A knows that B knows that C can't leave.

This keeps going, till we reach person Y in an Inception-esque layered imagination of the rest, who has two options:

  • He is blue eyed (2:198)
  • He is brown eyed (1:199)

If it is the latter, he can see a person Z and imagine his options:

  • He is blue eyed (1:199)
  • He is brown eyed (0:200) This is not possible, as there has to be at least one blue eyed person.

Z has only one option, then. But we can't just pick this because this option only comes into play if Y knew that he was brown eyed, which only comes into play if X knew he was brown eyed, and so on all the way back up to B.

So, on the first day, because Z doesn't leave, the idea of there only being one blue eyed person is ruled out.

Wasn't it ruled out from the beginning? Yes, however it could not be ruled out from the minds of the others while visualizing.

Remember, A knows that if he had brown eyes (9:101), then B would be considering the option that he has brown eyes too (98:101), even if that doesn't add up from A's perspective. We have to take into account what options are being considered.

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?

They don't know it. Each blue person knows that it's either 99:100 or 100:100, but can't pick. They need to wait to ensure that it works.


A simpler problem

Let me try to work out the second part again for 6 islanders, where the actual distribution is (3:3). The bullets are chronologically ordered, the nesting is Inception-esque imagination.

  • Blue eyed person A knows that there are at least 2 blue eyed people. He knows that if he is blue eyed, the other blue eyes are thinking the same as him. But this isn't the only possible case. The lack of any definite knowledge means that they don't leave (yet). If he considers the case that he has brown eyes, he visualizes that:
    • Blue eyed person B knows that there is at least 1 blue eyed person. If there were 2, then both would be thinking the same thing, and again, due to lack of definite knowledge, they can't leave. It there is one, he visualizes that:
      • Blue eyed person C sees 0 blue eyed people. Concludes that he must be The One. Leaves.
  • On the first day, no one leaves. On the second day,
    • Blue eyed person A knows that there are at least 2 blue eyed people. He knows that if he is blue eyed, the other blue eyes are thinking the same as him. The lack of any definite knowledge means that they don't leave (yet). However, if he considers that he has brown eyes, he visualizes that:
      • Blue eyed person B knows that there is at least 1 blue eyed person. If there were 2, then both would be thinking the same thing. If there is one (C), then B knows that C would have left the day earlier. That didn't happen. So.. there must be two. As both are thinking the same thing, both leave.
  • On the second day, no one leaves. On the third day,
    • Blue eyed person A knows that there are at least 2 blue eyed people. He knows that if he is blue eyed, the other blue eyes are thinking the same as him. If there are only two blue eyed people, then they would have left the day earlier. Which didn't happen. So he must have blue eyes, and as all three are thinking the same thing, they all have blue eyes. The three leave.
  • The brown eyes leave the next day.

Note that while initially it is clear that the number of blue eyes $>=2$, the number of blue eyes can be less from the point of view of someone when another is considering the option that he is brown eyed. The lower numbers only come in due to this "but what is he thinking" thought that gets nested.

Manishearth
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    If the guru add no information, then nothing can change. Imagine that it is a written law that all blue-eyes people must leave. A law known to anybody. Without the guru's words nothing will happen and no-one will leave. So the Guru must add some information (look-up complete knowledge). – Ittay Weiss Sep 10 '13 at 18:36
  • @IttayWeiss The new knowledge is that "nobody else left". – Manishearth Sep 10 '13 at 18:41
  • that isn't it either, since the guru did not say that. The question is: at the instance the guru speaks, what new information does at least one of the people have that (s)he did not have before. The answers can't be 'nothing' since then behavior can't change (and if the guru said nothing, then nobody ever leaves). So the answer must be something, and it must be some piece of information not knows before guru speaks, yet known immediately afterwards. – Ittay Weiss Sep 10 '13 at 18:49
  • @IttayWeiss The new information is "nobody left". See the last section, it becomes clearer as the thought process is outlined carefully. Information need not be externally injected. – Manishearth Sep 10 '13 at 18:56
  • "nobody left" is not new information. Suppose the guru had said "If I told you right now that blue eyed people exist, would you expect anybody to leave by tomorrow?". The answer (with 100 blued eyed people) is 'no, we all know that that will not supply anybody with enough information to leave'. So, "nobody left" is not new information either. Note that there must be some new information conveyed by the *words* of the guru (not by time elapsing or anything like that), since if there is no change in the information state, then there can be no change in behavior. – Ittay Weiss Sep 10 '13 at 19:16
  • @IttayWeiss Again, _see the last section of my answer_. The information isn't directly useful at that stage, however when each mind is considering the possibilities (and is considering other minds considering the possibilities), it does come into play. No, there need not be any new info conveyed by the words. – Manishearth Sep 10 '13 at 19:21
  • no info implies no change in behavior. If you claim otherwise, please explain how can two identical states of knowledge lead to different behavior. The guru's words certainly do convey new information to each an every person who hears them. To see that, think what would have happened if the guru instead privately delivered the message that blue eyed people exist to each of the people. In that case, nobody would ever leave and the guru's words then do not convey new information. It is the fact that the guru spoke openly that conveys new info. – Ittay Weiss Sep 10 '13 at 19:25
  • @IttayWeiss I don't claim otherwise. I refute the claim that what the guru says is the only info brought. The time axis matters. – Manishearth Sep 10 '13 at 19:29
  • ok, then your answer 'nothing' to the question "what is the quantified piece of information that the guru conveyed" and your apparent insistence in the comments that the words did not convey new information, are confusing (me, at least) as to what it is you claim. Of course, time passing plays a role, but the question was what do the guru's words convey (what new information). To make it clear, assume it is common-knowledge written law that all blue eyed people must leave. Without the guru's words nobody will leave. With the guru's words behavior changes. So, these words must conveye info. – Ittay Weiss Sep 10 '13 at 19:34
  • @IttayWeiss Again, I refute the claim that the information must be provided by the guru. The information provided by the guru only really comes into play when there are only brow eyed people left. The information that takes 100 days to build up is provided by time. – Manishearth Sep 10 '13 at 19:46
  • you'll have to explain then how no new information leads to different behavior. I'm going to stop repeating myself on this matter though. – Ittay Weiss Sep 10 '13 at 20:02
  • @IttayWeiss I repeat: I say that the guru is not providing _new_ info; this _does not mean_ that there is no new info coming to the minds of the others. The guru can be replaced by a placard that lists the rules and says "Initially, y'all had atleast one blue eyed member". I assert that the knowledge that nobody reacted the day before is the new knowledge here. Again: I say that the guru is not providing info, I do not say that no new info is available. – Manishearth Sep 10 '13 at 20:06
  • then consider the situation where the rules are common knowledge and the only thing the guru says is "blue eyed people exist". Just that utterance either conveys new information or it does not. If it does not, then the behavior will be the same whether the utterance is made or not. If the behavior changes (which is the case!), then what new information did that utterance convey. That is the question (and the answer is: "complete knowledge" as answered already by several others posters so this discussion is really getting to be useless). – Ittay Weiss Sep 10 '13 at 20:17
  • It conveys information initially, yes. Afterwards it does nothing. I agree that the guru provides information on day 1. But it conveys nothing on the next few days. Again, you are assuming that any new information _must_ be conveyed by the guru. I am not making that assumption. I ... don't think Complete Knowledge is relevant here. – Manishearth Sep 10 '13 at 20:31
  • complete knowledge is not just relevant here, it is precisely what's going on here. Of course the guru's words don't provide any new information after day 1. They only provide new information the instance they are spoken. New information is *also* created with every day that passes and the actions (or lack thereof) taken by the people. Again, the question was "what is the new information conveyed by the guru's words". You first claimed 'nothing' now you say the do convey information (and, correctly yet irrelevantly to the discussion, that new information is also generated by other means). – Ittay Weiss Sep 10 '13 at 20:38
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    @IttayWeiss Ah, so we do agree :p I thought we were talking about incremental knowledge on each day; that's where the confusion seemed to be coming from. I'll correct that. – Manishearth Sep 10 '13 at 20:40
  • @zyx Ah, the version of the puzzle I've known is different and the guru speaks every day. I edited the answer and added a bit about common knowledge. – Manishearth Sep 13 '13 at 19:54
  • You might be thinking of the Muddy Children puzzle, which is equivalent but has the children's father, in the role of 'guru', speaking every round. – zyx Sep 13 '13 at 20:14
  • @zyx Nah, same puzzle, I just had heard of a version where the guru speaks every day (and I had not verified this with the xkcd one), with no other differences. That doesn't change anything, but I thought that this was the point the OP was confused about "every new day, does the guru provide any info?". Anyway, that got cleared up. – Manishearth Sep 13 '13 at 20:17
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Without outside information, every islander can prove, for every $k$, that

"Island logic" proves $P(k) \implies P(k+1)$

where $P(k)$ is the statement

(it is true and is common knowledge that) if nothing happens after $k$ nights then every person can see at least $k$ blue-eyed persons

The guru provides common knowledge [arbitrary finite chains of "X knows that Y knows that Z knows ..."] of $P(1)$, which implies all higher $P(k)$ and therefore a relationship between number of nights and number of blue eyes. After $k$ nights it forces common knowledge of the fact that everyone can see at least $k$ blue-eyed persons. At $k=99$, this reveals the exact eye-color distribution to all islanders.

Proof of $P(k) \to P(k+1)$: After the $k$th night, every person is known to see at least $k$ blue-eye persons. Anybody who sees only $k$ blue persons knows that those blue-ers see $k-1$ blue on each other and $1$ other (himself), and leaves the island the next night. Thus, $k+1$ nights of inactivity implies that everyone can see at least $k+1$ blue-eye persons. Everyone knows that everyone else can perform this deduction, so that $P(k)$ true and commonly known implies the same for $P(k+1)$.

Now, to the numbered questions.

  1. The islanders do not have unlimited-depth common knowledge of the fact that there is at least $1$ blue-eyed person, nor of $P(1)$, before the guru speaks.

  2. The inductive reasoning that proves $P(k)$, if unwound, contemplates a hypothetical island where an islander (seeing only $k-1$ blue-eyes) has non-blue eyes and thinks about another islander (seeing only $k-2$ blues) contemplating an island in which he has non-blue eyes, etc, arriving eventually at the $1$-person situation which finally has to confront the guru's statement deciding that case.

  3. The extent of what is common knowledge is increasing every night, until critical mass is reached.

zyx
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Here is my take on question 2. Let's take $n=4$, the first non-obvious case, and look at the perspective of islander 1. He thinks, "I am not sure what my eye color is, so assume it isn't blue. I see three blue-eyed people, but they all also think they are not blue-eyed. Among them is my good friend islander 2." He begins to imagine what islander 2 thinks, creating a fantasy where he (islander 1) is not blue-eyed and putting himself in the head of islander 2.

His fantasy self thinks, "Now, I see islander 1, but he is not blue-eyed, so I only see two blue-eyed people. Among them is my friend islander 3." The fantasy islander 1 taking the role of islander 2 now begins to think what islander 3 is thinking, creating a fantasy where neither islanders 1 nor 2 has blue eyes and taking the role of islander 3.

His fantasy self's fantasy self thinks, "Now, I see both islanders 1 and 2, neither of which is blue-eyed, so I see only one blue-eyed person, the hated islander 4. I hope that guy picks up on the fact that he is the only blue-eyed person here! "

But the real islander 4 is two levels deep in his own fantasy and is waiting for one is the others to "realize" they are the only one with blue eyes. Since everyone is waiting, nothing happens that day.

So after one day, when islander 4 does not realize he is the only one with blue eyes, the fantasy of the fantasy of islander 1 realizes he is wrong and thus had blue eyes. Thus, the fantasy of islander 1 (playing the role of islander 2) reconsiders and on the second day thinks that today, both islanders 3 and 4 will realize they are the only ones with blue eyes.

However, the real islanders 3 and 4 are one level into their respective fantasies that two of the others will figure it out. Everyone is waiting, so no one does anything.

So after two days, the fantasy of islander 1 (as islander 2) realizes that he is wrong and actually has blue eyes. So islander 1 himself reconsiders and expects that today is the day the others all realize they are the only the with blue eyes.

But all of the others came out of their own fantasies as well, and are expecting that the other three "figure out" they are the only blue-eyed islanders today. Everyone is waiting, so no one does anything.

Three days have passed, all the fantasies have been dissolved, and yet no one has left. The only remaining assumption that islander 1 entertained was that he himself is not blue-eyed. He realizes that this must be false, and leaves the island. Everyone else has the exact same thought process, so they leave too.

So that's why the base cases are important even though none of them are realistic: until the last day, every islander is in some fantasy within a fantasy within...and in those fantasies, the island really does have fewer blue-eyed people. They don't communicate directly about their eye color, but by their actions and the common knowledge that they must be thinking identically, they do tell the others what they have deduced. Eventually they collectively deduce that everyone expects all the others to have blue eyes, so since they are perfect logicians, they realize they have blue eyes.

The fantasies are necessary because from the start, they all are in denial and all expect the others to be in denial as well, and it takes a few rounds of observation for the fact that it is denial to penetrate all levels of their consciousness.

As for question 1, the guru only plays the role of a reference point to synchronize their fantasies. Otherwise no one would know where in their deductions the others were. They are always trying to do deductions, but not knowing what the others' actions mean, they don't get anywhere until someone says "hey, think about it together."

As for question 3, I think this is already explained, since on the first 99 days, they are not verifying what they knew; they are actually escaping from some kind of Inception delusion.

Ryan Reich
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  • This is dual to the other one: "The inductive reasoning that proves $P(k)$ [the statement 'it is true and common knowledge that if $k$ nights have passed uneventfully, then everyone can see at least $k$ blue-eyes'], if unwound, contemplates a hypothetical island where an islander (seeing only $k−1$ blue-eyes) has non-blue eyes and thinks about another islander (seeing only $k−2$ blues) contemplating an island in which he has non-blue eyes, etc, arriving eventually at the 1-person situation which finally has to confront the guru's statement deciding that case.". – zyx Sep 13 '13 at 18:38
  • Looks like we had the same idea at the same time. – Ryan Reich Sep 14 '13 at 05:11
  • *Dual* was intended almost literally; your solution is the primal one, that describes the underlying coordinates (inner thoughts of the islanders) whereas the other works with observables. That was my motivation for writing out that answer; to avoid reference to islanders' private thought processes or knowledge states. Nice coincidence that the solutions were posted at the same time, but I didn't notice that and wasn't referring to the relative timing of the posts, only the logical relationship between the two. Actually, I think that your answer is undervoted. – zyx Mar 09 '16 at 17:12
  • @xyz That's a fascinating perspective. Also, thank you. Perhaps people give up on it because it's long. – Ryan Reich Mar 10 '16 at 07:12
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Let $n$ be a natural number and let us think of a statement: $P(n)$: "On the day $n$ all $n$ blue-eyed people leave." Now the piece of information the Guru provides is very important, as it makes $P(1)$ true. Indeed, if there was only one blue-eyed person and the Guru said nothing, that person could not deduce he/she is the only one with blue eyes, like described in the solution (I am assuming, however, that there is no other way to deduce this).

So with the information Guru provided, we know, that $P(1)$ holds, so now we can use mathematical induction to prove, that $P(n)$ holds for every natural number $n$. When proving this, we can use the similar steps and thoughts that are used in the solution for the case of two blue-eyed people.

Jiri Sedlacek
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  • This doesn't answer question 1.: What information has been provided by the Guru? $P$ is common knowledge as all statements are common knowledge. $1$ is not the information that the Guru provided. This number is a conclusion that arrived one day after her statement. To which knowledge refers this conclusion and which information has been added? – Keinstein Sep 10 '13 at 11:44
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    The statement $P(n)$ is wrong. It is in fact a non-statement for all $n\neq100$ as it makes a false side-remark that there are $n$ blue-eyed people. It is a side-remark because it is not part of what $P(n)$ states, or else $P(n)$ would just state "there are $n$ blue-eyed people and they all leave on day $n$", something that cannot be proved by induction simply because it is false for $n\neq100$. You probably meant $P(n)$ to be "if there are $n$ blue-eyed people then they will all leave on day $n$", but this doesn't work either with induction (the induction hypothesis gives a vacuous truth). – Marc van Leeuwen Sep 10 '13 at 12:43
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  1. I think the information that the Guru provides, is that all blue eyed people have to leave. This starts the whole process of inductive reasoning.

  2. The 1, 2 cases are relevant, because if I had blue eyes and I didn't know, I'd think others would leave at 99th day since each of them would reason other 98 would leave on 98th day, and so on. The number goes down with each blue guy's hypothetical inductive reasoning, where he thinks there could also be one less blue eyed people since he is not sure about himself.

  3. On the first $n-1$ nights, they are not verifying. They are waiting, for the remaining $n-2$ blue guys to reason out their stuff. They know for sure nothing will happen on first $n-1$ days, because if there are really $n-1$ blue eyed people, they would each know nothing will happen on the first $n-2$ days and so on. This wait is unavoidable. Only interesting night would be the $n-1$th night, when everyone will leave if others saw only $n-2$ blue eyed, but won't if they also saw $n-1$ blue eyed (i.e. if the person thinking is himself blue eyed).

KalEl
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I believe the key question for a tribesman to ask after the Guru's statement is:

If I am not a blue, when would the blues I see be forced to leave?

They would then reason that the blues they see would ask the same question with one less blue:

If I am not a blue, when would the blues I see be forced to leave?

This continues recursively without a clear answer until the 2 blue's question is reached, in which case the 1 blue would be forced to leave the first night. Before the guru had spoken, that sole blue would NOT have left the island because of the knowledge of the color of their eyes. This is the key to why the guru's statement starts the countdown!!!

Update:

Here are the answers to your questions:

  1. That signifies the day that 1 blue would have left.
  2. We don't consider the 1 and 2-person cases directly. We consider the 99 case, which necessitates considering the 98 case, which necessitates considering the 97 case, .. which necessitates considering the 2 and 1-blue cases.
  3. Blues know that either the other blues will leave on the 99th night (only 99 blues) or they are blue (and will leave on the 100th). If there were only 99, those blues would know that either the other blues will leave on the 98th night or they are blue. Etc..
Briguy37
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Ok may be we can see it as recursion with base case 2 blue eyed people. So lets talk about the case of three blue eyed persons. As everyone is highly logical persons they can easily think about the 2 eyed base case. So lets take one of the three blue eyed person. He can see other 2 blue eyed persons, So as he can easily deduce the 2 eyed situation so he knows that the second day the other two must leave. But as the case they don't leave. So he can easily say he also have blue eyes. So they all leave on 3rd day. Similarly. in case of 4 blue eyed persons. Lets take one person and he can easily deduce the situation of three. So if there are M blue eyed persons then at Nth day all will leave.

0

I think the rationale here is correct, but I do think the solution is wrong. I think our island critters can deduce that they should leave after 2 days.

All blue-eyed people see 99 people with blue eyes, and all brown eyed people see 100 people with blue eyes. They will reason as in the solution, based on common knowledge. However, they don't have to wait 98 days, because everybody knows that nothing will happen the first 98 days. This is based on common knowledge. So logically, they can all be skipped. So on day 1, nobody leaves, on day 2, all people with blue eyes leave.

Maarten
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  • This answer is incorrect. Your conclusion that "on day 2, all people with blue eyes leave" contradicts your earlier statement that "everybody knows that nothing will happen the first 98 days." While this latter statement is true, it is not common knowledge. Blue-eyed people do not know that blue-eyed people know that nothing will happen the first 98 days. This is because (immediately after the Guru's statement) "There are 99 blue-eyed people" is consistent with blue-eyed people's knowledge, (con't) – Eric M. Schmidt Jul 05 '18 at 02:48
  • (continued) so the statement "It is consistent with blue-eyed people's knowledge that there are 98 blue-eyed people" is also consistent with blue-eyed people's knowledge, and hence finally the statement "It is consistent with blue-eyed people's knowledge that all blue-eyed people leave the 98th night." is consistent with blue-eyed people's knowledge. – Eric M. Schmidt Jul 05 '18 at 02:50
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I'll add one answer to the 99 answers to appear here :)

I'll simulate a situation with $3$ islanders, named X, Y and Z. By symmetry, every predicate of the form $P(X, Y, Z)$ that holds will also hold under any permutation of $X$, $Y$ and $Z$, so I will not present the permuted statements.

Let's start after the Guru speaks:

$X$ thinks that

  1. If I don't have blue eyes, then $Y$ and $Z$ would be the only people with blue eyes. $Y$ would think that
    1. If I don't have blue eyes, then $Z$ would see no one with blue eyes, and leave the next day.
    2. So, if $Z$ does not leave the next day, I must have blue eyes, and we both leave on the second day.
  2. So, if $Y$ and $Z$ do not leave in two days, I must have blue eyes, and we all leave on the third day.

Now go back to before the Guru speaks. Why would this line of reasoning not work?

The obstruction is the innermost reasoning, where only one person has blue eyes. The sequence of reasoning starts by assuming "If I don't have blue eyes", and that propagates to the next person until there is only one person left. Note that this is all in one person's head. In this particular instance, where we have $3$ people, $X$'s reasoning in 1.2 above relies on the fact that

$X$ knows that $Y$ knows that $Z$ knows $B$

where $B$ is the statement "there is at least one person with blue eyes".

Note that this statement is not true before the Guru speaks. $X$ only knows that $Z$ knows $B$. $X$ does not know that $Y$ knows that $Z$ knows $B$, because $X$ doesn't know $X$'s eye color, and $X$ cannot assume that $Y$ knows $Y$'s eye color either.

So, the Guru does add the information: $X$ knows that $Y$ knows that $Z$ knows $B$.

Some may argue that 3 people is such a special case. Let's consider when we add another person. Let's say $W$ is added to the system. $W$ also has blue eyes.

How can we reason that

$X$ knows that $Y$ knows that $Z$ knows that $W$ knows $B$

is not true before the Guru speaks?

Again, $X$ doesn't know $X$'s eye color. When $X$ thinks for $Y$, $X$ knows that $Y$ knows that $Z$ knows $B$, because $X$ knows that $Y$ can see that $Z$ sees $W$ with blue eyes. Note that we need $W$'s blue eyes for $X$'s reasoning for $Y$. The fact that $Z$ also sees $X$ with blue eyes is known to $Y$, but is not known to $Y$ in $X$'s mind.

Permuting people, we know that

$Y$ knows that $Z$ knows that $W$ knows $B$

because $Y$ sees that $Z$ sees $W$ and $X$ with blue eyes. But $Y$ in $X$'s mind does not know this, because it relies on the fact unknown to $X$: that $X$ has blue eyes.

I believe this should be enough to convince most people that this situation generalizes to an arbitrary number of people. This is an answer to question 1: the consequence of Guru's speaking is that everyone knows that everyone knows that ... (repeat as many times as desired) that $B$ holds.

As some people have suggested, one variant of the problem is when the Guru tells everyone personally about $B$. This would yield no new information, and nothing will ever change.

Another interesting variant is if the Guru is specific and says that "$X$ has blue eyes" instead of just "someone has blue eyes". The consequence is obvious: $X$ will leave the next day. Everyone else will retain their behavior. It is somewhat counterintuitive because the sentence "$X$ has blue eyes" seems to give more information than "someone has blue eyes". One explanation of this situation that I can think of is that the statement "$X$ has blue eyes" actually gives no more information except to $X$, and the rules of the game force $X$ to leave the next day, so $X$ cannot give back information to others after $X$ leaves. In a sense, more information on the first day could lead to less information on all the following days because of the island rules.

Questions 2 and 3 are kind of answered, but I can try to elaborate more.

The reason smaller cases are relevant is because the strategy that each person derives from following "If I don't have blue eyes, then the next would think..." repeatedly until the one person case is reached. In a way, one can say that the small cases help everyone form a strategy. This can sound ignorant or deep, depending on what you think my understanding of the problem is. To me, it is quite an appropriate answer to the question.

To answer question 3, suppose everyone had a chance to discuss the strategy before the problem starts (without seeing each other's eyes). They could try to agree on cutting down the number of days if they can agree what the number is. Unfortunately, this is impossible. The problem is this "common number" has to be built from another relevant number, and each person only has as a source the number of blue-eyed people he/she sees. Since not everyone sees the same number of blue-eyed people, not everyone deduces the same number of days to cut down.

I will demonstrate why my strategy of cutting down the number of days will not work. (I do think my strategy is quite general though.) Suppose $f(x)$ is the number of days a person who sees $x$ blue-eyed people will cut down. $f$ should be a non-decreasing function, and $f(x) \le x - 1$ for every positive integer $x$. If there are $n$ blue-eyed people, these blue-eyed people will cut down $f(n - 1)$ days, while those without blue eyes will cut down $f(n) \ge f(n - 1)$ days. If $f(n) = f(n - 1)$, then it's a jackpot, and everyone cuts down the same number of days. Otherwise, those without blue eyes will cut down more days than those with blue eyes, and end up leaving on the same day as (or even before) those with blue eyes. So, to make it safe for all $x$, we must have $f(x) = f(x - 1)$ for all $x$. Since $f(1) = 0$, we end up with $f(x) = 0$ for all $x$.

Tunococ
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Let's skip over the original solution, I'm going to assume you read that. I'm going to answer the questions stated:

Imagine we have 2 people on this island of 201. Bob, who has blue eyes, and jane, who has brown eye's.

1) What is the quantified piece of information that the Guru provides that each person did not already have?

The Guru gave information about the colour of bob's and Jane's eye's. Namely, Bob and Jane do not know if they have blue, brown, red or green eye's. They still don't know anything about there own eye colour on day 1, but following the original solution, they can deduce it on day 100.

2) 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?

Imagine that Bob tries to be smart. He decide that no-one will leave the first 99 days, since there are 99 blue-eyed people excluding himself. Jane also tries to be smart, she decide that no-one will leave the first 100 days, since there are 100 blue-eyed people excluding herself. But now what? They can't skip 99 or 100 days, because then the difference between what a blue-eyed and a brown-eyed person would do it no more. Hence they can't deduce anything from the other not leaving on day 1.

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

Because they might be verifying something they already know, but they also communicate something they can not communicate otherwise. Bob know he can leave on day 100. Jane know she can leave on day 101. There is no way for them to communicate this information any other way.

Ps. How bummed out must Jane be to see 100 blue-eyed people leave on day 100. Perhaps the Guru can be nice and say "I can see someone who has brown eyes.", so that history can repeat itself and 100 brown-eyed people can leave another 100 day's later.

Dorus
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  1. What is the quantified piece of information that the Guru provides that each person did not already have?

    • Certainty and possibility opening. Everyone knew each other's eyes color, up until the Guru stated someone had blue eyes. That someone could be you, and that is reciprocal to all players. He gave "a hint" that your eye color was blue, a hint you could eventually check.
  2. 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?

    • Although they aren't possibilities, they are logical building blocks. The 1-person case is singular, but the 2-person case allows for the 3, 4, 5, n people cases to be inferred from it.
  3. Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

    • Rephrasing: Why do a n group of people have to wait n-1 nights if on the first n-2 nights they are simply verifying something they know? The fact is, as stated on 1., they don't know their eye's colors, they have a hint that their eye's color is blue. As stated on 2., the process is based in inference, so, they need to check that no one figured their eye color during the n-1 nights to finally conclude that their eye color is, by the Guru's hint, blue. Since this process is common to all players, after n-1 nights everyone figures, by deduction, that each one's eye color is blue and as such, exit the island.
Doktoro Reichard
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  1. The quantifiable piece of information provided is that at least one person on the island has to have blue eyes. Since they are perfect, logical machines, they will count how many individuals have blue eyes, and come to the conclusion that, if no other person has blue eyes, then they must, and they must leave the island.

  2. If, however, they see at least one person with blue eyes, they have only the evidence that one person has blue eyes, and thus, shall remain. If, however, they see TWO blue-eyed people, they must assume that these two blue-eyed people are assuming that the other is the only one remaining, and thus shall wait for the two to leave. If, however, they see THREE blue eye people, they must assuem that each blue-eyed person is assuming there are two blue-eyed people, and that each of those two blue-eyed people will leave. Because they are perfectly logical beings, they can make no presumption of their own eye color, but cannot be aware of it either. But they are aware that other individuals can observe eye colors, and that they will base their own presumptions on those eye colors that they see, so they will presume that their own eye color is excluded, and that each blue-eyed individual also excludes their own eye color.

  3. Is actually a short-form explanation of 2. In much, much shorter form, it is because they must exclude themselves from the count, even if they have blue eyes. Ironically, not a single person on the island will know their own eye color until the last day becsause of this exlusion requirement (They MUST stay if they do not have blue eyes), because they assume every blue-eyed person is exluding themselves as well.

Zibbobz
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    the question is what new piece of information do the guru's words provide. Since they each know blue-eyed people exist, your answer does not address that. – Ittay Weiss Sep 10 '13 at 18:39
  • They all already know blue-eyed people exist, IF there is more than one blue-eyed person. If there are none that they can see, they must be the one. It's an absolute quantity. At LEAST one, but possibly more. They haven't been told how many, just that at least one must exist, because presumably the oracle cannot lie. A more interesting question might ask what would happen if the oracle CAN lie. I suppose a better way to phrase the first conclusion is that at least one blue-eyed person can be observed by the oracle. – Zibbobz Sep 10 '13 at 18:47
  • but each of them already knows that there is at least one person with blue eyes since there *are* lots and lots of blue-eyed people. – Ittay Weiss Sep 10 '13 at 19:13
  • And they can come to the conlusion that if they are not blue-eyed, then that many blue eyed people must exist, but if they ARE, then there must be that many +1, all information that they could already conclude. It is not the fact that blue eyed people exist that they were informed of, nor the number that exist, but that there is a quantity that can be observed and includes as many people as either they can see, or they can see plus one. Without the knowledge, they could only conclude that a certain number exists, and they cannot conclude what anyone else knows either. – Zibbobz Sep 10 '13 at 19:27
  • I don't understand your distinction between they can conclude that "there is a quantity that can be observed and included as many people as either they can see, or they can see plus 1" and that without the knowledge "they could only conclude that a certain number exists, and they cannot conclude what anyone else knows either". Of course with and without the information they can conclude quite a lot about their own knowledge and the knowledge of others. For instance, "everybody knows blued eyed people exist" is a statement known by everybody. So, I think the distinction you aim at is unclear. – Ittay Weiss Sep 10 '13 at 19:39
  • @IttayWeiss It's an incredibly subtle distinction. Before, they were only aware of an exact quantity of blue eyes that THEY can see, and while there's a possibility that their eyes might also be blue, they have no confirmation of that without the Oracle's statement. It confirms to them that such a theory is possible, and conclusively true if they wait long enough to find out - without this information, they have no logical reason to believe it is true other than probability - which on an island of pure logic is not enough. – Zibbobz Dec 28 '14 at 19:30