[This is a (lightly edited) repost of an old blog post of mine, which had attracted over 400 comments, and as such was becoming difficult to load; I request that people wishing to comment on that puzzle use this fresh post instead. -T]
This is one of my favorite logic puzzles, because of the presence of two highly plausible, but contradictory, solutions to the puzzle. Resolving this apparent contradiction requires very clear thinking about the nature of knowledge; but I won’t spoil the resolution here, and will simply describe the logic puzzle and its two putative solutions. (Readers, though, are welcome to discuss solutions in the comments.)
— The logic puzzle —
There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).
Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).
One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.
One evening, he addresses the entire tribe to thank them for their hospitality.
However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.
What effect, if anything, does this faux pas have on the tribe?
Note 1: For the purposes of this logic puzzle, “highly logical” means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.
Note 2: Bear in mind that this is a logic puzzle, rather than a description of a real-world scenario. The puzzle is not to determine whether the scenario is plausible (indeed, it is extremely implausible) or whether one can find a legalistic loophole in the wording of the scenario that allows for some sort of degenerate solution; instead, the puzzle is to determine (holding to the spirit of the puzzle, and not just to the letter) which of the solutions given below (if any) are correct, and if one solution is valid, to correctly explain why the other solution is invalid. (One could also resolve the logic puzzle by showing that the assumptions of the puzzle are logically inconsistent or not well-defined. However, merely demonstrating that the assumptions of the puzzle are highly unlikely, as opposed to logically impossible to satisfy, is not sufficient to resolve the puzzle.)
Note 3: An essentially equivalent version of the logic puzzle is also given at the xkcd web site. Many other versions of this puzzle can be found in many places; I myself heard of the puzzle as a child, though I don’t recall the precise source.
Below the fold are the two putative solutions to the logic puzzle. If you have not seen the puzzle before, I recommend you try to solve it first before reading either solution.
— Solution 1 —
The foreigner has no effect, because his comments do not tell the tribe anything that they do not already know (everyone in the tribe can already see that there are several blue-eyed people in their tribe).
— Solution 2 —
100 days after the address, all the blue eyed people commit suicide. This is proven as a special case of
Proposition. Suppose that the tribe had n blue-eyed people for some positive integer n. Then n days after the traveller’s address, all n blue-eyed people commit suicide.
Proof: We induct on n. When n=1, the single blue-eyed person realizes that the traveler is referring to him or her, and thus commits suicide on the next day. Now suppose inductively that n is larger than 1. Each blue-eyed person will reason as follows: “If I am not blue-eyed, then there will only be n-1 blue-eyed people on this island, and so they will all commit suicide n-1 days after the traveler’s address”. But when n-1 days pass, none of the blue-eyed people do so (because at that stage they have no evidence that they themselves are blue-eyed). After nobody commits suicide on the 
As the above two solutions give contradictory conclusions, at most one of them is correct. Which one (if any) is the correct solution, and what is the precise reason that the other solution is invalid?
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8 August, 2017 at 1:51 pm
nmks
So, this is a continuation from my post above. it is LONG. Sorry about that.
Suppose there are 14 people. All of whom are super intelligent and logical, as in all the other versions of this puzzle.
When they were put on the island (all at the same time) they were told the rules (again, all at the same time), namely that they can leave only if and when they discover their own eye-color, and that a boat would come there every morning to give them the opportunity to leave. (Plus all the usual stipulations about not discussing eye color, etc, as in all the other versions of this puzzle.)
Suppose 6 of them are blue-eyed: A, B, C, D, E, F, and 8 are brown-eyed: G, H, I, J, K, L, M, N.
(And below I will use “green” as the color each may imagine him/her self to be)
Blue A sees 5 blues (B, C, D, E, F) and 8 browns.
Blue A thinks:
“If I am green, then B sees 1 green (A) and 4 blues (C, D, E, F ). (And 8 browns)”
Blue A also thinks:
“If I am green, and if B also thinks s/he (i.e., B) is green, then B thinks that C sees 2 greens (A, B) and 3 blues (D, E, F ). (And 8 browns)”
Blue A also thinks:
“If I am green, and if B also thinks s/he (i.e., B) is green, and if B also thinks that C thinks s/he (i.e., C) is green, then B thinks that C thinks that D sees 3 greens (A, B, C), and 2 blues (E, F). (And 8 browns)”
Blue A also thinks:
“If I am green, and if B also thinks s/he (i.e., B) is green, and if B also thinks that C thinks s/he (i.e., C) is green, and if B also thinks that C thinks that D thinks s/he (i.e., D) is green, then B thinks that C thinks that D thinks that E sees 4 greens (A, B, C, D), and 1 blue (F). (And 8 browns)”
Blue A also thinks:
“If I am green, and if B also thinks s/he (i.e., B) is green, and if B also thinks that C thinks s/he (i.e., C) is green, and if B also thinks that C thinks that D thinks s/he (i.e., D) is green, and if B also thinks that C thinks that D thinks that E thinks that s/he (i.e., E) is green, then B also thinks that C thinks that D thinks that E thinks that F sees 5 greens (A, B, C, D, E), and no blues. (And 8 browns)”
So, yes, I agree that this kind of induction results in a hypothetical situation where someone sees NO blue eyes at all.
What happens now?
Scenario 1:
Although they all know that there are blue-eyed people on the island, and although they also know that everyone else on the island knows that every (real, non-hypothetical) person on the island knows there are blue-eyed people on the island, they nevertheless do not leave because no one has announced that there are blue-eyed people among them.
Thus every morning when the boat arrives they all look longingly at the boat operator and think “I wish that boat operator would say out loud that there is at least one blue-eyed person here. Then we could start the induction at n=1 and on day 6 we could all leave the island.”
But the boat operator sails away without saying anything. So they all sigh and sit and wait for the next day’s boat visit. And again the boat operator says nothing. And silently they all lament the fact that the boat operator does not announce what they already know. And so they all sit on the island and grow old and die there.
Scenario 2:
When they were first dropped off on the island and had the rules explained to them, they all (being super intelligent and logical) immediately imagined the possibility of Scenario 1 and thought: “If we are going to hope that the boat operator will one day make the statement that there are blue-eyed people here (something we all know anyway) we could end up on this island forever.”
So they reasoned as follows:
“Because we are all intelligent, and we know that everyone else is equally intelligent, we are just going to do without that statement.”
(***Readers, please don’t give up here, but please continue reading, since this is the crucial argument coming up***)
“In fact,” they continue to reason, “because we all know that saying the statement out loud does not tell any of us anything we don’t already know, we are going to act AS IF that statement had been made. So yes, we WILL start the induction at n=1, but we will ASSUME that the statement had already been spoken out loud.
And why, one may ask, can they confidently assume that? Here is their reasoning:
“After all, if saying that statement out loud does not tell us anything we don’t already know, how does it change the situation? Well, it allows us to consider the hypothetical situation where n=1. It is as if that hypothetical n=1 “hears” the statement and says “aha, that must be me” and leaves the island on day 1.
But we know that that is only an imaginary situation, that the hypothetical n=1 does not actually “hear” the statement, and so we can go ahead and consider that hypothetical situation without actually HEARING the words spoken out loud. We don’t need to hear the words, since we KNOW what they would be, and the rest takes place in our minds anyway.
So they start the induction at day 1, and at day 6 they all leave the island.
The remaining 8 brown-eyed people then use the same reasoning and also leave, 8 days after the blue-eyed ones left.
If anyone cares to comment I would be particularly grateful if you could address the issue of “hearing” the words spoken out loud, since what I am trying to decide is whether in most solutions of this puzzle one might be incorrectly conflating the “hearing” of words, with the “thinking” of them in a situation where they (the words) only benefit a hypothetical situation (i.e., a situation where they cannot be heard anyway).
And as I said earlier, I do not necessarily think I am right, I am working through this. I am also not a mathematician (but often wish I was when I see what fascinating things are discussed on a blog like this, for example!)
8 August, 2017 at 7:07 pm
David Ricketts
Ok, now all you need to do is go through the “Blue A thinks…” steps where the final step results in Blue A knowing for a fact he has blue eyes. (Why don’t you reduce the blue eyes to just 4 individuals…that would be easier. If it works for 4, we can assume it will work for 6.)
8 August, 2017 at 9:25 pm
David Ricketts
NMKS, your logic is flawed here. Look at the two premises of your first IF/AND step. Both premises must be true for your argument to have a chance of being valid. However if the first is true the second cannot be true. If the second is true then the first cannot be true.
Please do the “Blue A thinks” routine as you did earlier, and post the step by step results. Just use 3…A, B and C. If it works for 3 it will work for any number greater than 3.
23 August, 2022 at 6:21 am
drph1il
The statement gives information that they don’t already know. The statement gives the information: “everyone knows that everyone knows that everyone knows that … everyone knows that there are at least 1 blue eyed person”.
If they could just hallucinate any statement, then everyone would go to the boat on day 1 and say “I have [red|turquoise|…|golden|purple] eyes” dreaming of one of the statements “there’s at least one person with [red|turquoise|…|golden|purple] eyes”.
If they are instead given the statement: “one can only have blue or brown eyes” it doesn’t help. If they heard that statement and didn’t see anyone with blue eyes they would not go. Likewise, people would think “what if b thinks that c thinks that d … that f sees 0 blues” and realize that the person would not do any action if they saw 0 blues.
8 August, 2017 at 8:43 pm
nmks
Actually, I think it works for 3 blue-eyed people (but no fewer).
One blue-eyed person on the island. The blue-eyed person sees NO blue-eyed people. Nothing can happen without an announcement.
Two blue-eyed people on the island. Each of them sees 1 blue-eyed person, but does not know that that blue-eyed person can see a blue-eyed person. Nothing can happen without an announcement.
Three blue-eyed people on the island. Each of them sees 2 blue-eyed people, but the difference is that now each one also knows that each of those 2 can see at least 1 blue-eyed person. At this point (and with more blue-eyed people) it is no longer necessary to hear the announcement read out loud. Because in the real situation the people know all they need to know by looking around, while the more deeply nested levels are hypothetical, and only take place in each person’s mind. And it is not necessary to hear something said out loud in order for things to take place in the mind, as long as the real, non-hypothetical, levels are satisfied by the information taken in by looking.
So they can now go ahead and use induction, arguing as follows:
If there was just one blue, AND that blue knew about the existence of blues, s/he would leave on day 1.
If there were 2 blues, AND they knew about the existence of blues, they would leave on day 2.
If there were 3 blues, AND they knew about the existence of blues, which they do, they would leave on day 3.
And indeed, because no one left on day 2, each one realizes s/he must also be a blue, and all 3 leave on day 3.
8 August, 2017 at 9:29 pm
David Ricketts
NMKS, your logic is flawed here. Look at the two premises of your first step. Each premise must be true for your argument to be valid. However if the first is true the second cannot be true. If the second is true then the first cannot be true.
Please do the “Blue A thinks” routine as you did earlier, and post the step by step results. Just use 3…A, B and C. If it works for 3 it will work for any number greater than 3.
8 August, 2017 at 9:54 pm
nmks
Thanks. I will do that tomorrow. Meanwhile it’s not quite clear to me what you mean by “Look at the two premises of your first IF/AND step. Both premises must be true for your argument to have a chance of being valid. However if the first is true the second cannot be true. If the second is true then the first cannot be true.”
Are you referring to “If there was just one blue, AND that blue knew about the existence of blues, s/he would leave on day 1.”
If so, I probably did not express it well, but it is just another way of saying what all the versions of this puzzle say: that if there was just one blue, and that blue did NOT know about the existence of blues (e.g. because no outsider had made the statement) then that one blue could NOT leave. I’m just removing both negatives: if that one blue KNEW about the existence of blues (e.g. because an outsider HAD made the statement) then he/she COULD leave.
9 August, 2017 at 3:45 am
David Ricketts
You are saying that for your scenario to work, a single blue eyed person must also know that a blue eyed person existed. However, by doing so you have introduced an impossibility into the beginning of your chain of events. For a single to know that he would have to be told.
Another aspect of this is that in your scenario you accept that three can leave only if two do not leave…and you accept that two can only leave if one does not leave. However, one could never leave because he could never know a blue eyed even existed. In your scenario the one blue eyed knows that a blue eyed exists only because you say he knows…not because the conditions of your scenario allow him to know.
9 August, 2017 at 9:45 am
nmks
Thank you. You replied: “you have introduced an impossibility into the beginning of your chain of events. For a single to know that he would have to be told.”
My reply to that: The single blue-eyed scenario is hypothetical. The single blue-eyed person is hypothetical. How can a hypothetical person be “told” anything? How can a hypothetical person hear?
The nested process takes place in the minds of the islanders, and they know without actually having to HEAR the announcement, and can apply their knowledge to the hypothetical situations.
9 August, 2017 at 12:51 pm
David Ricketts
NMKS: I think you will determine that the nesting process you refer to will not give you a solution to the puzzle you have created. I think you will arrive at that determination when you attempt to go through the “Blue A thinks” process for the 3 blue eyed islanders….A, B and C.
Just as an aside regarding your two questions… “How can a hypothetical person be told anything? How can a hypothetical person hear?” The answer is.. “The exact same way you have your hypothetical islanders give knowledge to the hypothetical people in the hypothetical situations you say the islanders create to solve the puzzle.”
9 August, 2017 at 9:55 am
leopold
nmks: What you say is not quite right I’m afraid. The case of three blues is no different in principle from two blues (or even one blue).
In the case where there are 3 blues, (a) everyone knows that there are at least 2 blues (since everyone can see at least that many), and (b) everyone knows that everyone knows that there is at least one blue (since even a blue can see 2 blues, so knows that everyone knows there is at least one blue (since s/he knows that everyone can see at least that many)).
But it is NOT true that (c) everyone knows that everyone knows that everyone knows that there are some blues. For each blue (call him ‘A’) knows only that everyone knows that there is at least one blue: A knows this because he can see that everyone – even one of the blues he can see – can see at least one blue. Call the blues A can see ‘B’ and ‘C’: then A can see that B can see C and also that C can see B, so A knows that everyone knows there is at least one blue. But A doesn’t know that everyone knows that everyone knows that there is a blue – even though this is actually true – since he (A) has no reason to suppose for example that B knows that everyone knows there is a blue (for as far as A knows, he (A) might be a brown, in which case B, who doesn’t know that _he_ (B) is blue, also could not know C could know there was a blue).
(c) becomes true only when there is a public announcement _witnessed_ _by_ _all_ that there are some blues, and (c) becomes true then not because of the content of the announcement (which everyone already knows), but because of the fact of the announcement being witnessed by everyone, and by the fact that the witnessing is itself simultaneously witnessed by everyone present, and so on indefinitely (this is the crucial point).
And that is why the announcement is necessary for the required chain of inferences to be made (eventually) by each blue that s/he _is_ in fact a blue. It cannot get off the ground without it.
9 August, 2017 at 10:12 am
nmks
Thanks, leopold. Our latest posts happened almost at the same time, so you may not have seen my reply to David Ricketts immediately before your post. My reply there contains a question which I would love to see answered.
9 August, 2017 at 10:55 am
leopold
nmks: Ii you mean the question about how a hypothetical person can be told anything and so on, I would say that the islanders are all reasoning about one another’s reasoning, asking themselves what the others could infer from what has happened or not happened. That is really all that is going on. I wonder whether you aren’t taking the idea of a hypothetical person too literally. Anyway no such thing figures in the original problem so I am not sure where it comes from (maybe someone has used the term to explain something, but I am not going to go back over the whole thread to find out where this might be!)
10 August, 2017 at 5:08 am
David Ricketts
NMKS: On Tuesday you said you would go through the “Blue A thinks” exercise on Wednesday. Did you decide not to do it?
10 August, 2017 at 6:50 am
nmks
Sorry, David; I found I didn’t have time to sit down and do the systematic calculation, so I just posted a few quick, additional thoughts. Will do the calculation this afternoon.
Meanwhile, another quick thought. A reply to your reply to my question about hypothetical:
Your reply:
Just as an aside regarding your two questions… “How can a hypothetical person be told anything? How can a hypothetical person hear?” The answer is.. “The exact same way you have your hypothetical islanders give knowledge to the hypothetical people in the hypothetical situations you say the islanders create to solve the puzzle.”
My reply to that:
Yes, the whole thing is hypothetical, but, within that, I am distinguishing between the “real” situation, which is an island with 3 blue-eyed people, and the “hypothetical” situation where those 3 blue-eyed people start imagining scenarios where there are just 1 blue-eyed person, or just 2 blue-eyed people.
Okay. back later today.
10 August, 2017 at 10:38 am
nmks
Okay. Back again. I found that I had started writing it up for 4 blue-eyed people, not 3, so I will continue that rather than make changes. As you said, if it works for 3 it works for 4, and therefore vice versa.
Without the “announcement” that there is at least one blue-eyed person on the island:
If A thinks A is green, then A thinks that B sees 1 green (A) and 2 blues (C, D).
If A thinks A is green, and if A also thinks that B thinks B is green, then A thinks that B thinks that C sees 2 greens (A, B) and 1 blue (D).
If A thinks A is green, and if A also thinks that B thinks B is green, and if A also thinks that B thinks that C thinks C is green, then A thinks that B thinks that C thinks that D sees 3 greens (A, B, C) and zero blues.
Conclusion: without an “announcement,” it is possible for every blue to imagine a situation where another blue may think that someone does not see any blues. Hence the induction (starting with “1 blue-eyed person”) cannot begin and no one can leave the island.
HOWEVER, I put the word “announcement” in quotation marks because what I’m trying to get at, is that information can be taken in in different ways. In this case, the fact that everyone has seen, and everyone knows that everyone else has seen, the same information, is no different from everyone having heard it. So the announcement is visual instead of aural. But it is still an “announcement.”
(The reason the same argument will not work for fewer than 3 blue-eyed people is because in that scenario a “visual announcement” would not provide sufficient information. Everyone has to see not only 1 blue-eyed person, but everyone has to see that everyone else can see one 1-blue-eyed person. From there, they can proceed to apply that information to their mental induction process, that will get them off the island.)
10 August, 2017 at 12:50 pm
David Ricketts
NMKS: You just say they can figure it out. You do not actually demonstrate it. You say: “From there, they can proceed to apply that information to their mental induction process, that will get them off the island.” However, they cannot use the same mental induction process that was used with the aural announcement because that process was built on a base step of a single blue eyed person knowing he was the only one to exist due to the aural announcement. I was not able to solve the puzzle using the “visual announcement” method you presented. Perhaps you can. Please use your “visual announcement” and go through the “Blue A thinks” mental induction routine. Describe each step of A’s reasoning with the final step causing him to determine (know) he definitely has blue eyes. It would be easier if you just use 3 blue eyed people.
10 August, 2017 at 1:50 pm
nmks
David Ricketts: Reformulating the “If A thinks A is green, and if A also thinks that B thinks B is green etc” series would be of no use, because I am not saying that that is how A figures out that he/she has blue eyes. How he/she figures it out is by considering the situations if there were only 1 blue-eyed person, then if there were only 2, then if there were only 3. And that part is the same as what hundreds of other writers have already explained many times. I am not arguing with that part of the argument at all.
The part I am arguing with is that the islanders have to “hear” the information before the induction can begin, and I am hoping to show that SEEING it, as long as everyone also sees that everyone else sees it, is the same as hearing it.
Because: the only people that “hear” it are the ones that are on the island in the REAL situation (by that I mean the situation with 4 blue-eyed people on the island (or whatever number we decided to work with for this puzzle). Each one then takes this information and applies it IN THEIR MIND to hypothetical situations in which there are fewer than 4 blue-eyed people on the island.
That is what I am am trying to get it: They apply it IN THEIR MINDS. The “people” in the hypothetical situations – that is, situations that are created in the minds of the people in the “real situation” – did not hear anything. Only the people in the “real” situation heard it.
So how does that differ from those same “real” people SEEING it (as long as they also see that everyone else sees it) and similarly applying it in their minds to the same hypothetical situations?
10 August, 2017 at 2:59 pm
David Ricketts
Ok, it appears to me that you recognize that “hearing” provides a solution and “seeing” does not provide a solution. Your question seems to seek the reason why that’s the case. Unfortunately, the answer goes into the world of mathematics.
This puzzle is not solved by “inductive reasoning”(IR). It is solved by “mathematical induction”(MI) reasoning, which is actually a form of “deductive reasoning”(DR). IR only suggests a proof. DR and MI actually provide a proof. MI requires that a statement can be proved correct for the first of any set of natural numbers and then using induction it can also be true for the following numbers of the set. In your case the blue eyes are a set of 4. MI requires a proof for the first number of your set, which is 1. “Hearing” allows a proof for 1 whereas “seeing” does not. By hearing it the base case of “1” is established so that the inductive portion of MI can then be applied for any number of islanders greater than 1 to determine exactly how many there are and who they are. When there are enough blue eyed persons (3) for everyone to be able to think..”ok, we all know that blue eyed people exist”, there is nothing about that knowledge which allows a proof for the first person in the set. Therefore, MI cannot be used with the “seeing announcement”.
If this did not help you understand why the “seeing announcement” will not work but the “hearing announcement” will work to solve the puzzle, I suggest you search “mathematical induction” in wikipedia. Perhaps that explanation would explain it better than I have done.
https://en.wikipedia.org/wiki/Mathematical_induction
10 August, 2017 at 7:14 pm
nmks
Thank you, David Ricketts, for your explanations and patience. I will read the link re Mathematical Induction, look up the other kinds of induction, and also look around at what other online discussions are saying about this puzzle. I will no doubt return to it some time in the near future.
I have been trying on and off for several months now to find an explanation of why the statement has to be “heard.” When I first found out about the puzzle I was quickly convinced by the standard reasoning where one needs someone to say it out loud, but later, as I tried to imagine it as as a real-life situation, I started to have doubts. I wonder if this has ever been tried in practice, as an experiment, e.g., where people can leave a room instead of an island, using colored stickers, or hats, instead of eye-color, and giving everyone a starting time as well as periodic opportunities to leave the room (say every minute instead of every day).
12 August, 2017 at 2:09 am
leopold
nmks: you asked “how does [HEARING] differ from those same “real” people SEEING it (as long as they also see that everyone else sees it) and similarly applying it in their minds to the same hypothetical situations?” The answer lies in your parenthesis “(as long as they also see that everyone else sees it)”. The first thing to note is that “see[ing] that everyone else sees it”, is not sufficient for all the required deductions to go through. They need to be able, not only to “see that everyone else sees it” but also to see that everyone else sees that everyone else sees it, and furthermore to see that everyone else sees that everyone else sees that everyone else sees it, and so on, all the way up to 100 (if that is the number of blue-eyed islanders) nested iterations of “everyone sees that”.
Now SEEING (say) 99 blue-eyed islanders does not allow this; but HEARING (or rather WITNESSING) a public announcement in which everyone is informed of something they already know: that there is at least one blue, does allow it. And that is because the announcement is public, which allows each blue-eyed islander to deduce not only that there is at least one blue (which he knew anyway), but also that everyone knows that there is at least one blue (which he knew too, before the announcement), but also that everyone knows that everyone knows that there is at least one blue (indeed, he knew that too, before the announcement), but also…., and so on, ….. right up to (finally) that everyone knows that everyone knows that every one knows that everyone knows that … [with 100 occurrences of “everyone knows that”, and THIS he did NOT know before the announcement.
I think that even in the relatively simple case of just TWO blue-eyed islanders, you can already see that HEARING a public announcement whose content is something you already knew by looking gives you more information than merely SEEING.
(If there is exactly one blue eyed islander, he cannot see any others. Hearing that there is (at least) one allows him to deduce that there is one, and it is he. So already in this case HEARING gives you more information than SEEING, but you might well be sceptical that this very trivial case has any relevance for cases where there are more blues)
If there are exactly two blues, each one can see one blue, so he already knows there is at least one. But he doesn’t know that this other one knows there is at least one (since he doesn’t know that this other one can see one). Hearing that there is (at least) one doesn’t on the face of it tell him anything he didn’t already know, but when he hears this information, he also (and this is the key observation) witnesses that everyone else hears it, and now knows it, and this means that now he does know that the other one knows that there is at least one. There is no way he would be able to deduce this in the absence of a public announcement.
It is an instructive exercise to do this for the case of 3 blues. I wouldn’t try to do it for more than that though: if the point hasn’t been grasped by then, it will only become more obscure as you add more blues into the mix.
Remember that all the islanders are supposed to be perfect reasoners. That is essential for the puzzle to work, so trying it out “in practice” as you suggest would probably not be very illuminating (well, it might illuminate human psychology of course, but not the logic of what is going on here…)
15 August, 2017 at 7:20 pm
nmks
Thanks, leopold. I am going to hibernate for a while because I have a few extremely busy few weeks coming up. David Ricketts had provided a link to an article re Mathematical Induction (as opposed to other inductions) that I will read and ponder when I get a bit more time. As of now I am still unconvinced that hearing the announcement results in a different outcome from seeing it (as long as everyone sees that everyone else sees the same). But maybe I’ll change my mind after reading the mathematical induction articles. May be it has something to do with real and unreal situations. (I know math has all kinds of mysterious (to an non-math person) concepts, such as imaginary numbers, which I’m sure have their uses, and this puzzle may fall into one of those categories. I may be thinking too much in terms of real-life situations.)
Meanwhile, if you wish, maybe you would like to consider the following two (real-life) scenarios, and say which one is more likely to happen, given that all the people are highly intelligent, logical, and all want to leave the island. And knowing that when they were all put on the island. they had the rules explained to them, in a public announcement, (all at the same time), namely that they can only leave when they know their eye color..
Scenario 1:
Every morning when the boat arrives they all look longingly at the boat operator and think “I wish that boat operator would just say out loud that there is at least one blue-eyed person here. Then we could start the induction by considering n=1, n=2, etc. and eventually get off this island.”
But the boat operator sails away without saying anything. Same thing every day. And so they all sit on the island and grow old and die there.
Scenario 2:
As soon as they were put on the island they all looked at everyone else’s eyes, summed up the situation, and thought: if we wait around for someone to tell us what we already know we could end up sitting here forever. Since the only thing missing is a public statement that there are blue-eyed people here, we will proceed as if that statement had been spoken. So we can start the induction by considering n=1, n=2, etc., starting the next morning (which will be the first time the boat returns) and eventually we will get off this island.”
16 August, 2017 at 7:08 am
David Ricketts
NMKS: Your comment to Leopold got me thinking some more about the difference between hearing and seeing under your scenario.
Upon hearing the rules explained but not being told that a blue eyed person exists, they all know it is logical for every one of them to think that he himself might not have blue eyes. Once they are told by the outsider that a blue eyed person exists it is no longer logical for everyone to think that they “ALL” might think they don’t have blue eyes.
The difference I describe above is critical for the Mathematical Induction process to be used to allow them to leave the island.
21 August, 2021 at 8:19 pm
Darby L
The second scenario makes the most sense, not just to me, but presumably the perfect logicians on the island. If an announcement were made, it would be during the countdown period that had already begun, and it would be ignored. After that, the outsider would find no one to speak to. The only way the announcement would be helpful is if the islanders had been there without the knowledge or desire to leave. It seems that they could even arrive at different times, and as soon as an individual learns of the rules, and counts the others (who are surely in on the rules, they can count to that number of days (if everyone is perfectly accurate). Even if they pad the count with 10% extra days, they can be fairly sure that everyone will leave when enough time has passed. (In My Estimation)
21 August, 2021 at 8:27 pm
Darby L
Sorry about mixing up the various versions of this, but hopefully the logic fits this “negative” version, even though I doubt they would be putting in the same effort for a payoff of death.
16 August, 2017 at 8:07 am
nmks
Only have time for a quick note. My whole argument hinges on one thing: that the islanders will ALL think: “Since the only thing missing is a public statement that there are blue-eyed people here, we will proceed as if that statement had been spoken.” Maybe that seems illogical and maybe in a purely theoretical system that would not happen (as I might find out after reading more) But (meanwhile) wouldn’t it? The premise in the original puzzle was that everyone is logical and knows that everyone else is logical and knows that everyone else will come up with the most logical solution.Would not this be the most logical solution in this case?
This is why I prefer to deal with the version where the people want to leave the island, rather than the one where they commit ritual suicide. Because although the reasoning SHOULD be the same, I think it may somehow be different for the problem-solver if they think about how to find a way out of a bad situation (how not to get stuck on the island), rather than if they think about how a bad situation may arise (ritual suicide). Purely psychological; not affecting the logic within the problem, but MAYBE affecting the way a problem-solver might approach the problem, or at least be willing to consider other possibilities. Don’t know…
(Note that I am NOT introducing the kind of situations that one finds some people proposing whenever someone posts a logical problem, the kind that all amount to ignoring, or changing the original parameters. (I mean things like “but what if someone talks in their sleep about eye color” or such…) No. I think I’ve learned enough about logic problems to know that I must stick to the original parameters.)
16 August, 2017 at 9:42 am
David Ricketts
NMKS: You asked if it would be logical for them to think.. “Since the only thing missing is a public statement that there are blue-eyed people here, we will proceed as if that statement had been spoken.” Yes, I think they would think that. Next, it would be logical, I think, for them to all think…”OK…now how do we all proceed to a solution.” I believe you would say..”They all start counting days.” However, why are they counting days? The answer would be they are waiting to see if other blue eyed people they can see “do something” or “don’t do something”. Seeing the “something that is done” or seeing the “something that is not done” on a certain day during the counting process will tell each of the blue eyed people in an instant that they do have blue eyes. Unfortunately, since they were not told that a blue eyed person existed it was NOT EVEN POSSIBLE for the end of day 1 to have any meaning…therefore, day 2 could never have any meaning because day 1 could not have any meaning…nor day 3…nor day 4…etc. They would of course figure this out at the beginning and therefore not even begin the counting process because they would all know that their counting of days could never have any meaning to them.
16 August, 2017 at 11:31 am
Anonymous
David Ricketts: you seem to agree now that the people would think: “Since the only thing missing is a public statement that there are blue-eyed people here, we will proceed as if that statement had been spoken.” But that, I think, was the only hurdle, because “proceeding as if the statement had been spoken” means accepting that in the hypothetical situation where n=1, that solitary “1” would know about the existence of blue-eyed people (just as if the statement had been spoken.)
(Another thing I find interesting, though unrelated to the paragraph above, is that we are dealing with a third situation, not proposed by the original poster. Prof Tao had originally mentioned 2 possible outcomes, one where all commit suicide, the other where nothing happens. But I am suggesting, and we are discussing, a situation where something DOES happen, but without the announcement.)
16 August, 2017 at 1:47 pm
David Ricketts
Yes, I do agree it would be logical to think that because as you set the scenario… (1.) they all want to leave the island and (2.) an offer had been made to all of them at the same time by the boat operator. However, that’s where it ends because they would all quickly conclude they could not figure out an answer based on their common knowledge and they would then not even start the counting process.
You say that was the only hurdle…that is not correct…the other hurdle is to actually figure out who does and who does not have blue eyes. This is something you have never explained which is why I asked you long ago to go through the “Blue A thinks” process to show the “seeing” solution in detail…but you never did it. You say you believe they can figure it out by “seeing” as well as by “hearing”. If they can, you should be able to show it.
I know I am being presumptuous but I believe you have tried to figure out a “seeing” solution and failed in your attempt. It defies reason to believe you have not tried. I suspect you have accepted that “seeing” does not work. I think your only difficulty at this time is that you have not been able to figure out why it does not work…because it seems to you that it should work.
16 August, 2017 at 2:25 pm
nmks
(Don’t know why my last comment showed up as anonymous) Anyway, To proceed “as if the statement had been spoken” means to accept that in the hypothetical situation where there is only 1 blue-eyed person, that solitary blue-eyed person would know about the existence of blue-eyed people (just as he/she would if the statement had been spoken). I think that part should be clear.
In which case, if there are 4 blue-eyed people:
Every blue-eyed person will reason that if there was only 1 blue-eyed person on the island, and that person knew of the existence of blue-eyed people on the island (because of the “as if” ) that one blue-eyed person would leave when the boat arrives on day 1. But no one leaves on Day 1. Well, no-one expected anyone to leave on day 1 anyway, since they all know that n=1 is only a hypothetical situation.
Similarly with the next day: Everyone knows that if there were only 2 blue-eyed people, those 2 would leave on Day 2. (But again, no-one expects anyone to leave on day 2, since they all know that n=2 is also only a hypothetical situation.)
Third day. This is different because all the blue-eyed people can see 3 blue-eyed people. They know that if those 3 are the only blue-eyed people, they (those 3) will leave on Day 3. If they don’t leave on day 3, every blue-eyed person will think: they did not leave means they must be able to see another blue-eyed person. But who? I can only see 3 blues and lots of browns. Aha, I must be the other blue-eyed person. And so all 4 leave on Day 4.
16 August, 2017 at 3:33 pm
David Ricketts
NMKS: Yes, each of them can see the other 3 blue eyed people(BEP)…however, none of them knows that the others can see 3 BEP. Take “A” for example, he knows he can see 3 BEP but he knows he might not have blue eyes. If that’s the case then he knows the other 3 BEP’s can only see 2 BEP.
This is where the flaw in your reasoning shows up. You stated…”no-one expects anyone to leave on day 2, since they all know that n=2 is also only a hypothetical situation”. n=2 is hypothetical to them “individually” but n=2 is not hypothetical “collectively”. Blue A knows n=2 is a hypothetical for him but he also knows he might not have blue eyes so Blue A knows that for Blue B n=2 is a real possibility, not a hypothetical possibility. In A’s mind n=2 is a real possibility for C and D as well. For each one of them n=2 is a real possibility for the other 3. So, collectively n=2 is not hypothetical.
Again, I recommend you go through the “Blue A thinks” routine from beginning to end with the end being Blue A knowing for a fact that he has blue eyes. Solve it for Blue A and you solve it for all of them since they will all be thinking the same thing from their own perspective.
16 August, 2017 at 3:54 pm
nmks
Thanks. I really have to leave this for a few weeks as I don’t have time at the moment to work through that again.
I also first need to read up about Mathematical Induction, because there is something that I am missing. See, you wrote: “but I believe you have tried to figure out a “seeing” solution and failed in your attempt…I suspect you have accepted that “seeing” does not work. I think your only difficulty at this time is that you have not been able to figure out why it does not work…because it seems to you that it should work.”
That’s not quite the way it is because at this point I think it might work.
What I don’t understand is how you imagine the information getting transferred to the hypothetical situations. If the hypothetical situations (where n is less than 4) take place in the people’s minds (which obviously they do) then it should not make a difference whether the people doing those mental calculations came upon the information by seeing it or by hearing it (as long as they saw that everyone else saw). They apply the information that they know (and know that everyone knows) to the hypothetical situations (those nested situations where everyone knows that everyone knows that everyone knows that everyone knows….)
But I’ll stop here. I must read those articles you mentioned and catch up on other work before returning to this blog.
2 February, 2018 at 6:43 am
Kevin Buzzard
Terry — I asked this question to my 1st year undergraduate students in their “introduction to proof” course at Imperial College. As an optional part of that course the students can choose to answer the questions in Lean, some formal proof verification software. My student Chris Hughes formalised the question and then proved the formalised version in Lean. His proof is here: https://github.com/kbuzzard/xena/blob/21122ecb4ec9bb2a9d80a3a986b7c3d0fc3f089c/M1F/2017-18/Example_Sheet_05/blue_eyed_islander#L2
Kevin
2 February, 2018 at 10:46 am
nmks
Kevin, thank you so much for re-opening this discussion. I posted my thoughts on the blue-eyed puzzle here a couple of times and have received some replies and had nice discussions, but this is an opportunity to post again and see if anyone else will reply. I am sorry, I cannot understand the mathematical symbols that many people use, including those is your post, so mine is in plain English. And the premise of the puzzle is slightly different, though should not alter the reasoning. Okay, here goes:
An island (which we’ll call Island A) has 100 blue-eyed people and 1000 brown-eyed people. That means some of them see 99 blues and 1000 browns, and some see 100 blues and 999 browns. They are forbidden to discuss eye-color.
They all want to get off the island. They all know that everyone else also wants to get off the island. But they have no means of doing so.
One day a boatman arrives and says they can get off the island if and when they correctly name their own eye color. He says the boat will come back at the same time every morning starting tomorrow (which will count as Day One, i.e., the first day when someone could leave). Each time the boat arrives they will have a chance to say their eye color and leave the island if they are correct. (If they get it wrong they stay on the island forever).
However, he does NOT say anything else about eye color, and, in fact, never does, on any of his visits.
Now, many people in this discussion group (and other groups) have argued that an announcement such as “I see someone with blue eyes” is necessary before anyone can guess their eye color. Which means that, because the boatman did not make such a statement (and in fact never does), the islanders will remain on the island forever. Every time the boat arrives they think “Oh I wish that boatman would just say something like “I see someone with blue eyes,” which we already know anyway, so that we can start leaving.” But, because he never does, they never leave.
Now, suppose the boatman HAD said that he sees someone with blue eyes. What would happen? The same people, in this discussion group, argue that the whole process, (as discussed by MANY people already, so I don’t have to repeat it here) will begin, and on day 100 all 100 blue-eyed people will leave the island. (I am NOT going into the fate of the brown-eyed people in this discussion. One thing at a time. The fact is that those who think an announcement is necessary, argue that if such an announcement is made, all 100 blues would leave on day 100)
Are we agreed this far? If yes, I can proceed.
Imagine an identical situation on a different island, which we’ll call Island B. The islanders here are also all perfectly logical and know that everyone else is too, and also know that everyone wants to get off the island. Meaning, they all know that if the boatman had said “I see someone with blue eyes,” the blues would all leave on day 100
So each islander on Island B thinks (and knows that everyone else also thinks this because they are all logical and all want to get off the island) “If all is needed for us to start leaving the island is for that boatman to announce that someone has blue eyes, which we all already know, then we will simply act AS IF he had said it.” And on day 100 all the blue-eyed islanders leave.
Why am I not bothering with the usual argument that goes something like “but until the announcement is made, they don’t know that everyone else knows that everyone else knows that everyone else knows, etc, etc, etc, right down to one solitary islander, i.e., n=1”? Because those are hypothetical situations, and hypothetical islanders will not hear the statement anyway, because hypothetical people cannot hear. The only ones that hear the boatman’s statement are the real islanders, and they already know the content of the statement without it being spoken.
Now, MAYBE such hypothetical scenarios are necessary in the realm of computer programming or whatever, I don’t know because I am not a computer specialist, or a mathematician, or a scientist. And maybe that’s the kind of situation that is implied by the statement “all the islanders are logical.” In which case I take everything back. But, looking at the problem as presented, and taking “logical” in its everyday use, I still say the boatman’s announcement is not needed for the islanders to get off the island.
(Note that although I say I take the term “logical” in its everyday sense, I am still adhering to all the parameters of the original puzzle, i.e., I am not introducing any “derailers” that one sees now and then, such as “but couldn’t they build their own boat,” or anything of the sort.)
Comments would be much appreciated. But please not the kind that start “imagine if n=1” etc, because I have taken that into account. And I know that in situation n =1, or n =2, or maybe even n =3, this does not work, but I am discussing the original given situation where there are 100 blues and 1000 browns. All I am saying that if the process works when the boatman speaks, it also works when the boatman does not speak, because the only people that hear the boatman’s words are the real people, not the hypothetical people, and they already know they information that he conveys. Thank you.
2 February, 2018 at 11:32 am
Pace Nielsen
@nmks,
(1) There are ways to formalize “perfectly logical” so that the people on Island B do not think it is logical to act as if, contrary to reality, the boatman had made the announcement of blue eyes.
(2) There are ways to formalize “perfectly logical” so that they do pretend they all heard a public announcement.
(3) There are ways to formalize “perfectly logical” so that no (consistent) islander could have this property while simultaneously knowing they have this property.
I’m personally in the camp of (3).
If we allow that the islanders are only “perfectly logical” in a very limited sphere of knowledge and logic, I’m in camp (1). Apparently, this is the type of situation that Kevin’s student Chris was able to formalize in Lean.
I find it hard to picture a world where it is *logical*, rather than merely *practical*, to pretend something happened in order to achieve an end. But that’s the thing about these puzzles; they allow room for interpretation! If you interpret that sort of behavior as logical, then indeed the islanders might leave after 100 days with or without the announcement. However, they might also just leave immediately; depending on your conception of “perfect logic”.
2 February, 2018 at 12:22 pm
nmks
Pace Nieslen, thank you! I think this is the explanation I have been waiting for. It seems that it’s not my reasoning that’s at fault, but the fact that we are using “logical” in different ways. See, to me, as a lay person, i.e., not a mathematician, the meaning would be closer to the adjective “practical” that you mentioned above.
That is also why I changed the premise from the islanders having to commit suicide if they know their eye color, to them wanting to leave the island but only being able to to do so if they know their eye color. I realize that this is a tricky distinction because, of course, if a sequence of events if inevitable, the islanders would find it, regardless of their fate. I changed it because I thought it might prompt some people on this group to see it from my perspective. But I began to suspect that the reason for all the disagreements was in my understanding of the term “logical” because surely so many mathematicians would know what they are talking about!!! (as I said, maybe in computer programming or whatever, such hypothetical situations are acceptable and necessary)
I hope a few others might chime on on this. I would be interested to read some more comments. But, as for myself, I feel I can now put this problem to rest. My reasoning was fine within my use of the term “logical” which is not the same use that mathematicians/logicians use (no puns intended). Thank you for pointing this out so clearly. (This problem had been in my head for ages – I had even envisaged testing the hypothesis by having people at a dinner table with colored dots on their forehead and not being able to start eating until they guess their color.
I may later on ponder the last part of your post – namely would they all leave at once, or wait 100 days. I can see there’s something there that requires thinking through, but I will do that later. For now, I’m perfectly at ease. What a great feeling! Thank you again!!!
2 February, 2018 at 12:01 pm
David Ricketts
Nmks: They are forbidden to ever say anything about eye color, therefore they would never be able to tell the boatman their eye color. Let’s assume you modify your scenario to eliminate that problem. They all immediately know they are lacking sufficient information to ever determine their own eye color. However, they all know that over 90% of the islanders have brown eyes. Therefore, they each immediately tell the boatman that they have brown eyes. 1000 brown eyed people leave the island and 100 blue eyed people stay forever.
2 February, 2018 at 12:41 pm
nmks
David Ricketts, Yes, they could do that. But it would not help the blue ones and they also badly want to leave the island. I just replied to another poster that I am now very happy because I understand where the problem was. I mean, if everyone on this discussion group was on an island and wanted to leave, I’m sure we would all use my reasoning. That’s “logical” in the sense of “practical.” But apparently it’s not the way the term is used by many specialists, Your previous replies (a few months earlier) had got me thinking that the problem (i.e. problem of why we couldn’t agree) must lie somewhere in realm of what the term “logical” means to different people. So thank you, too, and everyone else who had replied to me in the past. It was a combination of the information gleaned from everyone’s posts that has resolved this problem for me. Best wishes to all! Math is great!!! (spoken by a non-mathematician!)
2 February, 2018 at 1:28 pm
nmks
David Ricketts, Yes, they could do that. But it would not help the blue ones and they also badly want to leave the island. I just replied to another poster that I am now very happy because I understand where the problem was. I mean, if everyone on this discussion group was on an island and wanted to leave, I’m sure we would all use my reasoning. That’s “logical” in the sense of “practical.” But apparently it’s not the way the term is used by many specialists, Your previous replies (a few months earlier) had got me thinking that the problem (i.e. problem of why we couldn’t agree) must lie somewhere in realm of what the term “logical” means to different people. So thank you, too, and everyone else who had replied to me in the past. It was a combination of the information gleaned from everyone’s posts that has resolved this problem for me. Best wishes to all! Math is great!!! (spoken by a non-mathematician!)
2 February, 2018 at 2:34 pm
David Ricketts
Nmks: From my own “practical” point of view I think what I said would actually happen. Practically, each individual knows he only has three options that make any sense at all since under your scenario there is no way to determine their own eye color using any kind of mathematical “logic”. 1. Do nothing in which case he does not leave but still has a chance to leave if a future opportunity ever arises. 2. Tell the boatman he has blue eyes and hopes he is correct. 3. Tell the boatman he has brown eyes and hopes he is correct. I think they would all pick number 3. Since they don’t know another opportunity will ever come but everyone does know there is a very high probability they have brown eyes. (P.S.-I’m glad you got an answer that helped you.)
2 February, 2018 at 7:08 am
Recent student successes | Xena
[…] problem! The question is here (M1F sheet 5) (question 6) (and also on Terry Tao’s blog here), and Chris’ solution is […]
5 February, 2018 at 4:34 am
xenaproject
Hi. Let me just make one comment about this discussion. When solving a mathematics problem using formal proof verification software, there is a very clear division of the task into two sub-tasks. The first sub-task is to change the mathematics problem into a computer programming problem written in the language of the software. And then the second sub-task is to write more code in the language of the software, which solves the computer programming problem.
For many maths problems of the kind which we give to undergraduates, doing the first sub-task is typically very easy. The computer program already has stuff like “the complex numbers”, and so if the original problem is, for example, to prove that the complex conjugate of a+b equals the sum of the complex conjugate of a and the complex conjugate of b, there is a completely unambiguous translation of this into the computer programming language, and the nice thing about modern proof verification software is that the translation of the problem *actually looks like mathematics still* — it literally becomes the statement “forall a b : complex_numbers, complex_conjugate(a+b)=complex_conjugate(a)+complex_conjugate(b)”. Actually it looks even better than that nowadays, because the software I’m using allows unicode, so e.g. instead of complex_numbers we really write a blackboard bold C.
However the islander question is more logically complicated, and so one has to really consider whether the translation into the computer programming problem actually correctly conveys the ideas which are either explicit or perhaps even implicit in the original LaTeX human-readable problem.
I believe that Terry’s comments on the problem in his post do an extremely good job of clarifying any ambiguities in this particular problem. However I am prepared to accept that in this particular instance some people may still believe that there is still a certain amount of subjectivity involved.
As for the computer code however, it runs, so I believe it.
Kevin
9 March, 2018 at 5:54 am
Ade...
From my point of view, concerning the blue-island puzzle, I think nobody will commit suicide. They cannot because the foreigner was the only one that discussed eye color and mentioned his one eye color. He did not mention anybody in particular who had the same eye color as his. After the foreigner’s comment they can only have the idea that some people have blue eyes–even though they already know that–and cannot discuss about it neither can they know theirs. If anybody thus commits suicide, it means somebody, one way, has broken the religious law because I will ask the person,” how did you know your eye color?”
9 March, 2018 at 2:17 pm
David Ricketts
Ade… All the blue eyed islanders can, and will, determine that they have blue eyes. They will all determine it at the same moment without any word being spoken by any person. Therefore, they will all commit suicide at the same time because their religion requires it.
9 March, 2018 at 7:48 pm
Anonymous
I quote you, “They will all determine it at the same moment without any word being spoken by any person. ” How? When? The only way they can know is if they run some numbers with the data they have–the 100 blue and 900 brown which I think they could have done before the visitor announced the presence of blue-eyed people. Hence, it is their choice to run the numbers on the data. The puzzle doesn’t state that they are curious to know their
Eye color. Hence, we can’t say whether they will do the calculations or not. Besides if they all start to commit suicide after the foreigner’s address, and they are curious to know their eye colors , they might not be as devout as we think because they would have known their eye colors before the visitor came.
9 March, 2018 at 7:50 pm
Ade...
The reply above was written by Ade…
9 March, 2018 at 10:53 pm
David Ricketts
The rules of the game says they are all devout and perfectly logical. The moment the visitor says he sees a blue eyed islander, the clock starts for all the islanders. Before the visitor spoke they never had a moment in time where the clock would start for all of them. To figure out who has blue eyes the common start date of the clock is all they needed. Being devout, the blue eyed islanders all commit suicide as required by their religion on the day they all figure out their eye color. The rest of the islanders live happily ever after…after the appropriate time of grief, of course.
9 March, 2018 at 11:59 pm
manuelshanghai
Solution 1 is correct. This is because a tribesperson is assumed to have no way of discovering his or her own eyecolor. As a consequence, it can never come to such a ritual suicide caused by eye color discovery.
I remind that, among others, the following assumptions are made in the puzzle:
a) “[…] each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces).”
b) “If a tribesperson does discover his or her own eye color, then[…]”
Assumption b) does not contradict to a) but, a) makes b) irrelevant. Solution 2 is wrong because it contratics to a) if the mentioned suicide in solution 2 is caused by those island people discovering their eyecolor(If they commit suicide for other reasons, then maybe solution 1 and 2 are not necessarily disjoint -but this kind of argument is maybe already of legalistic type and seems not to be in the interest of the puzzle presenter).
——
People might answer that the bracket “(there are no reflective surfaces)” means that a) only prohibits discovering by checking in a mirror-like surface. However, I see such discussions as to be not in the spirit of the puzzle because then a lot of opinion based interpretations or “legal” arguments arise. For example one could then reply “if somebody tells you such a fact, you still need to see it in order to know that it is really true and the guy didn’t lie to you or has a color weakness” etc. etc.
10 March, 2018 at 5:34 am
David Ricketts
a) was true until the visitor interjected his comments. His comments made a) no longer true because he started a clock ticking….and since all the islanders are perfectly logical, through the use of mathematical induction, the knowledge of their eye color was forced upon all the blue eyed islanders…and they all died.
10 March, 2018 at 7:46 am
nmks
David Ricketts is absolutely correct. And for newcomers to this problem – if you scroll up, there are numerous discussions, many of which describe the induction process in minute detail. Very interesting reading, with arguments both pro and con.
But, for another take on this problem, with a slightly altered scenario (namely that instead of ritual suicide, the islanders get a chance to leave the island (which they all want to do) if/when they discover their eye color) see my posts (e.g. by nmks on 2 February, 2018 at 10:46 am. In that one I do not explain the induction process, since that has been done so many times, but I go on from there with my other arguments). Part of my reason for changing it from ritual suicide to getting off the island is that the latter scenario gives the islanders an incentive for calculating their eye-color and also has well-defined sign posts (time posts?) each day during which they could leave (the boatman arriving). Sometimes this helps to visualize the whole thing.
My argument was that they would leave the island without the eye-color announcement as long as they had another announcement to start the clock ticking (again, this version gives the problem such a concrete sign post, whereas simply “living on the island” does not).
However, as David Rickets and Pace Nielsen and others have explained to me, it won’t work in the world of mathematics, – the flaw in my argument being my interpretation of the term Logical, which they use in a mathematical sense whereas I used it more in the sense of intelligent and practical. (And fair enough, I’m not a scientist, but I can see that when using math to design a machine, or computer app, etc, one needs that kind of logic that extends even into hypothetical situations.)
In my island solution, I just skipped those hypothetical situations. Acting AS IF the boatman HAD mentioned eye color, is not as crazy as it sounds. It is just fast forwarding over the first few days (like in chess where there are standard openings, people might decide to skip the first few moves of a standard opening and begin at, say, move 6 or whatever)
So glad this problem has been reopened…
24 April, 2018 at 6:54 pm
Hongbo Li
If I just change the situation a bit by adding one more assumption: people will do anything to survive only with the exception for being compelled to commit suicide by the religion. This is not a terrible assumption and it will completely alter the outcome: exactly one blue eyed tribe person will die (by suicide or homicide).
Reason:
If there are only one blue eyed person, it is automatically true because he will always die one way or the other.
If there are two blue eyed people, people will kill one of them (the other one will be able to live because it suffices to killing only one blue eyed. Killing more than one is unnecessary). The surviving blue eyed person will never know his own eye color. So is the same for the other brown eyed people.
If there are three blued eye people, which is a more interesting case. People still only need to kill one of them. The two surviving blue eyed people (let’s say, Alice and Bob) will be waiting for each other to commit suicide, but none of them will commit suicide. By symmetry, if they should commit suicide, they should do it at the same time. But they can’t do it at the same time because they have to wait for each other’s action to have enough information to know their own eye color.
This pattern should be able to turn into a prove with some minor effort.
18 June, 2018 at 11:00 pm
geepee
Both the answers are correct, and they are correct contingent on a single fact- the discovery of ‘n’! Till the time n is unknown, solution 1 holds and the moment n gets ‘discovered’, solution 2 comes in action. Talk about ignorance being bliss!!
Various shades of solutions are possible on the basis ‘legalistic’ arguments like eye color is spectrum or boolean and I actually could see most of them getting disproved if one were to focus on one trait of the individuals and the population as a whole- individuals are highly logical and the population is existing. This sets pretty rigourous boundaries for the solution.
What I love is that this serves almost like a mythological/ religious allegory for the pitfalls of discrimination.
17 November, 2018 at 2:57 am
Blue-eyed islanders (guest post) | Xena
[…] thought it was cute, but then forgot about it. Richard Thomas then ran into it in an old post on Terry Tao’s blog and suggested I put it onto an M1F example sheet. I did this, and knocked up a solution. Rohan […]
26 September, 2020 at 5:58 am
LANG JIANG
The proof has mistake of conceptual shift, it’s impossible to prove the N+1 day conclusion by recursion. Because if N>2 is True, then N=1 will NEVER be true, so the basis of recursion (N=1) is NOT true.
24 March, 2021 at 2:55 am
Christian
There is a concrete difference in what the people from the island know after the information.
Person 1 knows, that person 2 knows,… that person 99 knows, that person 100 knows that there is at least one person 100 knows that there is at least one person with blue eyes. Without the information from the foreigner, person 1 knows, that person 2 knows, that… person 99 knows that there is at least one person with blue eyes but nothing more.
27 March, 2021 at 2:56 am
Tommaso Monni
I think Solution 1 is incorrect. Indeed, it tells that the foreigner give an information the islanders already know: false as they do not know that there are only two colors until the foreigner speaks (to tell them there is a blue color). In other words, the foreigner information has an effect because it tells there is a color while before saying so the islanders could think (each of them alone could think) they have a different color with respect to what they see, a third color which is just the not-blue, not-brown color which should be thought as different for each of them.
I would add, sharing a thought that probably goes beyond the spirit of the puzzle that each of these third colors cannot even be compared with those seen.
It cannot be compared with those they see, starting an exclusion process, because this would mean again that they have information on the full (!) possible eye colors and this in turns would call for many other physical assumptions and probabilistic considerations, supporting Islander reasoning such as “I choose at random on a virtual uncountable set of colors, the set which is the bounded interval of the physical light frequency”; even though they would be allowed to do so, then a reasoning like ” I choose uniformly at random” will lead to zero chance of choosing a color that may relfects the fact that, in fact, one cannot even name (pronounce) that color (I could not until I for example pronounce an infinitely long word on a finite alphabet).
6 April, 2021 at 4:12 am
Leonidas
This crazy religious devotion makes the tribe’s fate purely unstable since they can only stay as long as nothing is mentioned publicly about the eye colors. It’s true that only an evident statement is made with no real information but this is still something said and this is still enough to ruin the house of cards. The triggering effect is not linked to the information transmitted (since there is no real information contents in this speech) it is linked to the condition that the whole tribe is there and they all hear this innocent sentence the same way and at the same time. (If only one were missing the effect would not come.) It is very easy to mislead the inattantive reader that nothing has happenned saying that the information is not new to anyone. Yes the informsation itself is not new but no one ever before declared this not at all new information in front of the whole community. This is the newness in the story – this is very new and this misfortunate condition is actually paving the direct road to the self extermination of these stupid believers. (A joke is coming to my mind. Uncle Joe leaves the factory – where he works – with a wheel barrow which is covered by a canvas sheet every day. Looking under the canvas the guard at the gate checks the wheel barrow all the time but it is always empty. He doesn’t understand what’s happening and one day finally he decides to talk about his confusion to Joe in a private manner. Joe i have the feeling that you are cheating us. I will not report it to anybody but tell me at last what you are stealing from this place day by day! And Joe replies: wheel barrow – my sweety – wheel barrow. I write this joke since something simillar happens in it as in the puzzle. The sentence said itself is empty but the fact that it’s rolling out causes a damage still. Proporly speaking quite a big damage in case of our tribe – more than the wheel barrows mean to the company.)
6 April, 2021 at 4:28 am
David B Ricketts
Correct…For the first time in the tribe’s history every member of the tribe has a starting time to begin counting days.
31 October, 2021 at 4:02 pm
Anonymous
Only solution 1 is plausible, given that the islanders are truly perfect logicians.
They would realize that no new information was given either about eye color or about knowledge of any persons on the island, and that everyone else also realizes this. Therefore there can be no possible change to the equilibrium state.
Any counterarguments to this simple line of reasoning?
31 October, 2021 at 5:19 pm
David Ricketts
The visitor gave them all a time to start counting days. They never had that until then. For them, that was “new information”. That was all they needed to figure out who had blue eyes.
31 October, 2021 at 9:24 pm
Matavovszky György
Absolutely right!
24 January, 2023 at 11:24 pm
Why Study Logic? | Gödel's Lost Letter and P=NP
[…] in material here) thought we had written about this puzzle on this blog, but it turns out to be a memory from Terry Tao’s […]