A note on today’s comic: Judging from my inbox, a lot of people have a different perspective on the GRY puzzle than me. First of all, I believe it’s a fairly old riddle. But more importantly, I think that originally the “English Language” bit wasn’t even a part of this riddle. It was just a head-breaker because there was no answer. But then someone who thought they were oh-so-clever came up with the idea that somewhere in the puzzle it could be read as asking what the third word in the actual phrase “the English Language” was. That’s a pretty lame interpretation. There have been lots of rewritings of the puzzle to try to make it work a little better, but they’ve never removed the lameness. Some of them more friendly to that stupid interpretation than others. I’d been hearing the angry/hungry puzzle for years (and talked about the interesting fact that there was no answer) before I ever heard someone try the “language” thing. Snopes has a good article on this puzzle.

I always hated those ‘lateral thinking’ puzzles — the ones where they said “oh HO, I never told you that the doctor’s mother was a midget and this was all happening a space station! Don’t you feel dumb now?‘ Because once I figured out that language was really, really flexible and imprecise, it seemed that the key to communication was just figuring out what they probably meant. And figuring out what they could possibly mean if you use all the wrong definitions and stuff is interesting, but I don’t think it teaches all that much more than bad communication. These should be kept very separate from actual logic puzzles, which are really neat. There’s a thriving section for them on the forums, and a lot of people found my site looking for the answer to the Blue Eyes puzzle (which is great because it doesn’t use trick language anywhere — indeed, I’ve rewritten it over and over to make it as unambiguous and clear as I can — and yet the answer is incredibly elusive).

0 thoughts on “Sideblaggin'

  1. I’m not convinced that the Guru provides anyone with any new information.

    Here’s a proof that everyone already knew the Guru’s information:

    From a blue-eyed person’s perspective:
    1. I can see 99 blue-eyed people.
    2. Select two of them, call them A and B.
    3. Everyone except for A can see that A has blue eyes.
    4. A can see that B has blue eyes.
    5. By 3 and 4 I know that everyone can see someone with blue eyes.

    The non-blue-eyed person’s argument is the same, except he can see 100 blue-eyed people instead of just 99. Since the argument holds for all people in the group, we know that everyone has come to the conclusion that everyone knows there is at least one blue-eyed person. Since this is the same information that the Guru’s statement reveals, there’s no new information.


  2. Here’s an alternative explanation, in narrative form:

    X walked along the shore of the beach. He’d been there for quite some time, and frankly was getting a little sick of it. Ever since he signed up for that stupid National Logic Conference he’d been stuck on this island with a bunch of lame ass-logicians. He couldn’t talk to them; no, that was against the rules. But even if he could, he wouldn’t want to. Logicians are pretty boring.

    The question he asked himself daily was this: why don’t I know the color of my eyes? Every morning I used to stand in front of the bathroom mirror and brush my teeth. Is it really possible that I’d never once glanced at my own reflection?

    But his days of leisurely teeth-brushing were long gone. For 14 years they’d been stuck here on this island, tasked with ascertaining their own eye color. No one else could remember their eye color either, so perhaps this inexplicable gap in his memory was not so uncommon among logicians. His wife used to joke that he would forget his own hat color if he couldn’t take it off and look at it. If only she were here now.

    Lunchtime. The island, devoid of civilization save for the 201 conference attendees, had plenty of natural vegetation; a cornucopia of tropical fruits and coconuts that had sustained them for so many years. They had a routine for lunch. Everyone gathered their own foodstuffs (collaboration requires communication, and is thus prohibited) and came to a cliff on the southeastern side of the island — it was where the helicopters had left them when they first arrived. It was also where the ferry would come to pick them up, once their task was complete. It was a link to the world they had once lived in. To their homes.

    “Ahem,” coughed one of the logicians. What the hell was this fool thinking? He knows there’s no talking allowed! Why open his mouth now? On closer inspection, X realized: this is the green-eyed guy. He wondered to himself if Greenie had finally realized his own eye color. But no, as a logician he knew it was impossible.

    The green-eyed man continued, “I — I’d like to say something. I see someone….”

    The crowd, still in shock, eagerly awaited the premise that might buy them a ticket back home.

    “I see someone… with blue eyes.”

    X reeled in disgust. Had he really waited so long, just to say that? Of COURSE someone has blue eyes, everyone can see that! Then a spark of logic revealed itself to him. If this was all the help he was going to get, he needed to take matters into his own hands….

    He made a snap decision to band together with the blue-eyes. Sure, there were more brown-eyes — 100 compared to the meager 99 of the blues — but it would be easier to move in a smaller group. Discreetly, he moved through the crowd, tapping each blue-eyed shoulder and whispering “meet me at north beach at sundown.” A deliberate rules violation, yes, but if they were going to get off this island they’d have to organize. Trick the ferry operators into letting them on the boat, then once on-board use whatever means available to incapacitate the crew and get to safety. It had to work the first time, there were no second chances.

    He shared his plan with the blues. Although reluctant to speak, they indicated through subtle body language and facial expressions that it was a good plan. Best of all, it was guaranteed to work. Even if the browns suspected something, there was no way they could communicate it to each other, let alone the ferry operators. There was just one problem…. Greenie.

    But X had a plan for that, too. Green had already demonstrated that he was too unpredictable to ignore. His random outbursts could easily compromise the entire operation. But they couldn’t just tie him up somewhere; the ferry crew took a tally every night and would be sure to notice if someone was missing. There was only one choice: they had to convince him he had blue eyes.

    After dinner, X pulled Green aside and led him to a secluded area of the woods. “Listen, I’m going to get off this island and I need your help. I’ll tell you the color of your eyes if you tell me mine.”

    Green hesitated for a moment, then spoke: “Blue”

    With a sigh of relief, X replied, “Oh thank God. By the way, you’re Blue too.”

    Months passed. There were many preparations to be made, and progress was slow due to the need for absolute secrecy. After 100 days, they were ready to make their move.

    As the ferry appeared on the horizon, X signaled his team. “It’s on.”

    They entered the ferry one by one, each saying their eye color: “Blue.” It was crucial that Greenie was last in line. In the resulting confusion when he discovers he’s *not* green-eyed, they would jump the guards and hijack the ferry. If all went well, it would be over in seconds.

    Green reached the front of the line. He hesitated… perhaps he could sense that something was wrong. Then, instead of naming his color, he whipped around and dashed into the woods.

    “NOW!” screamed X. He turned to his second-in-command, a blue-eye he affectionately called Y. “Don’t leave without me. I’ll make sure that green bastard doesn’t get away.”

    X ran after Green into the trees. He could barely make out his tattered logician garb through the brush. After a few minutes of close pursuit, Green suddenly stopped and turned around.

    Green sneered, “Did you really think it would be this easy?” X heard leaves rustling behind him. He was surrounded by brown-eyes carrying makeshift spears; there was no way out.

    “Let me introduce myself. I am El Guru. I have lived on this island for decades; I was here long before your little logic party arrived. In my solitude I’ve discovered true enlightenment, and I plan to share it with all of you. However, I cannot do that if you leave the island. Do you understand?”

    X understood, but it took a jab from a spear to make him admit it out loud. “Yes, El Guru. I underst-”

    *Whoosh!* X ducked out of the way of the incoming spear. He’d recognize that spear sound anywhere, that was Y’s spear, and he’d come to rescue him! The spear embedded itself in El Guru’s chest, letting out a spurt of blood and a loud “NOOOOOOOOOOOOooooooooooooo!”

    X turned to Y. “Thanks, I would have been dead without you!”

    “No problem. Let’s get out of here.”

    Dodging brown-eyed rocks and spears, they made their way back to the ferry. It was still there! The bodies of the ferry crew were floating face-down in the surf. X and Y leapt on-board and the blues pushed off into the deeper ocean.

    “X… there’s something I need to tell you,” said Y. In a flash, he’d pulled out a stone dagger and shoved it into X’s abdomen. “I… wear blue contacts. My eyes are actually brown. I’m sorry it had to end this way.”

    Before anyone knew what had happened, Y pulled out a coconut grenade and blew up the ferry, killing everyone on the boat.

    The End.

    Thus, the answer is: No one leaves the island.


  3. Well, I’m usually pretty bad at this, but there kind of seems to be a quick-and-easy answer, at least in my mind. See, in my head, if I counted 100 brown eyed people, and 99 blue eyed people, I’d think there’s a pretty good chance I was blue eyed. That’s not saying there wouldn’t be a chance there’s 101 brown eyed people and 99 blue eyed.

    Feel free to tear it apart.


  4. Hmmm…. The problem with the blues eyes riddle (not saying there actually is one, more likely a problem with me) is that everyone already knows that there are more than three blue eyes. Everyone keeps talking about “if there were one/two/three” but there simply aren’t. They don’t need to wait a day to see if there’s more than one, they don’t need to wait another day to know there’s more than two; they already know this, and they already know everyone else knows this.

    “However, they did NOT know that everyone else knew this.”

    Yes they did, because knew the rules of the game. Everyone knows evryone else is keeping track of eye color, and there are enough people that everyone knows there’s someone with blues eyes.

    Also, there a problem with the idea of perfect logicians, because they’d all be caught up in the whole “how do I know anything exists” dilemma. There is no logical proof of anything, even, as far as I can tell, the self. “I think therefore I am” is flawed. What if the truly correct and logical statement was “I think therefore I am not” and we were simply not capable of comprehending this even though it was true? So since no one can figure out if anything exists, no one can figure out their own eye color.

    “The third word is ‘gryphon’. The mistake was adding the ‘ends in’ part.”

    Actually, I think that the most common spelling is griffin. I only know of it being spelled “gryphon” in the Warcraft videogames.


  5. First, props to Matt for his theory. I thought it was great. The puzzle definitely didn’t mention that the people didn’t leave the island dead…

    I have to comment on the Blue Eyes puzzle, because it consumed several hours of my life. But I finally understood the answer. Let me give the explanation a shot, because none of the ones I read made much sense.


    Ok. So, as a blue-eyed person (though you don’t know it yet), you see 99 other blue-eyed people on the island (100 – your own = 99). However, as a perfect logician, you don’t know if EVERYONE knows this, and can’t yet be sure of the number. Once the Guru has confirmed aloud that she also sees blue-eyed people, you now know theat EVERYONE knows this. Thus the logic begins. Now, you know that the brown-eyed people are irrelevant, because the guru did not confirm to the group that there were brown-eyed people (if she had, no one would have left, because there are equal numbers of blue and brown eyes). You are only concerned with whether or not you are blue-eyed. So, if you are blue-eyed, all of the other blue-eyed people will also see 99 pairs of blue eyes. If you are *not* blue-eyed they will see 98 pairs (99 that you see -1). Now, if we reduce this to one blue-eyed person, they will look around and see no other blue eyed people. They will then leave the next night, knowing they have blue eyes. If there are two blue-eyed people, each will see one blue-eyed person, and assume that if they do not have blue eyes, the other person will know this and leave the island. If after one day, the other person has not left the island, they must assume that they also have blue eyes and will leave the next night. On it goes unil the 100th person, who waits 99 days (the number of blue eyes they see), to see if everyone has left. If, on day 99, everyone is still there, everyone else must also see 99 pairs of blue eyes, which means you also have blue eyes. Therefore, on day 100, all of the blue-eyed people leave the island because they figured out that they, too have blue eyes. On day 101, all of the brown-eyed people threaten the Guru with bodily harm to confirm that she also sees brown-eyed people. . . .


  6. Sorry, I just went back and looked at what I just wrote, and realized that if the Guru had said that she saw only blue and brown-eyed people, it would still have worked. Assuming you are a blue-eyed person, you see 99 blue and 100 brown eyes. If the blue-eyed people leave on day 99, you know you have brown eyes. If they are still there on day 99, you know you have blue eyes, and everyone leaves on day 100. Poor brown-eyed people. . . .


  7. I think the blue-eye puzzle cannot work this way.

    Step one: 1 am blue-eyed, and i can see 99 blue eyed persons
    Step two: i assume i am brown-eyed, that means, the 99 persons can see 98 blue-eyed persons.
    Step three: everone of them must assume, they are brown-eyed, and the other ones see 97 blue eyed persons.

    Because i am blue-eyed, they see 99 blue-eyed persons, so they have the same conclusion as i had in Step two.

    Or how does it work? It seems there can not be inductively added information, because the people can see each other everytime.


  8. I agree with allo.
    On (brief) consideration, these kinds of iterative explanations scaled up from the idea: “if I was the only blue eyed person…” etc etc etc

    I don’t believe that this would work as it clearly states that:”Everyone can see everyone else at all times ”

    Therefore there can be no watching to see if 99 people don’t do anything as you don’t know whether or not they are not moving because they see that you (and however many other people) have blue eyes or if they are not moving because they can see all the other blue eyed people (not you) the reaction (or non-reaction) of any blue eyed people that you can see offers no information as to the colour of your own eyes.

    I you were to have all of the islanders file past you one at a time and examine the reactions of each one as they saw your eyes (given one or two assumptions about the other peoples knowledge of everybody elses eyes) then it might work. But not under the conditions stipulated on this site.

    Does this make sense? Does anyone agree?


  9. Furthermore, I put forward the idea that as the only unique (as in eye colour) person on the island, and as I can’t see any reason why any of the brown or blue eyed people would figure out their own eye colour, then it is the guru who leaves on the next midnight.

    I can feel an explanation somewhere inside me, but I can’t find it…


  10. Further-furthermore, I apologise if I have simply re-stated something already torn apart in fierce argument on this very blog as I humbly admit to not reading every post here…


  11. Call me Mr Dumb-Ass but surely the answer is that every-one stands in a straight line in front of the guru? Therefore if the person at the front now knows their own eye colour!

    If this answer has already been posted, I apologise, but all the cod-logic in the varios posts was killing me!



  12. D0SBoots, I like your explanation a lot.

    The riddle itself is just genius. I’ve never found an answer to a riddle so satisfying (I couldn’t figure it out and peeked here, but through luck, I did guess the actual answer correctly :P)

    Here’s another explanation:

    1. You are on the island. You see one person with blue eyes, and some other people with brown eyes.

    The guru announces there’s someone with blue eyes. You think: “well, since everyone else has brown eyes, blue-eye is going to know it’s him”.

    When he doesn’t leave that night, you realize “he must’ve seen someone else with blue eyes… must be me!” and you leave together the next night.

    2. The ferry takes you to another island. This one has two blonds and some brunettes.

    The guru announces someone is blond.

    At first you think “duh, everyone including the blonds can see a blond”, but then you realize, “they’re in the same situation as me and blue-eye. They each think the guru is talking about the other one”. Sure enough, no one leaves that night.

    The next day, you think “Now each of them will know that the other saw a blonde, and leave together”. But no one leaves that night either!

    “Hmmm, they should have left like me and blue-eye. Unless… they saw another blond and thought the same thing as I did (that the two blonds would leave together on night 2). Wait, that means I must be blond too!” and the three of you leave that night.

    3. Much to your frustration, the ferry takes you to another island and gives everyone a hat (but you can’t see your own). There’s three guys with red hats, X, Y, and Z.

    The guru says “there is at least one person with a red hat”.

    “Oh, I’ve seen this before. X sees 2 red hats [note: your guy isn’t too bright at sensing the pattern here and figures he’s wearing a brown hat]. So X figures that Y and Z will leave on night 2 like me and blue-eye. (similarly, Y feels that X and Z will leave, and Z feels that X and Y will). But after no one leaves on night 2, he’ll realize he must also have a red hat, and leave like me and the blonds.”

    But of course after 3 nights, they haven’t left. “Oh, I see, I must be wearing a red hat too. So X actually saw 3 red hats and thought that Y, Z, and I would leave on night 3. Y thought that X, Z and I were going to leave, and Z thought that X, Y, and I were going to leave. Well, now that we know there’s four of us, we’ll leave tonight.”

    4. The ferry takes you to yet another island with 201 people blah blah blah


  13. Oh, and as for a lateral answer:

    One person left on the night of the guru’s announcement; the only blue-eyed person who didn’t wear sunglasses during the announcement.


  14. Rodd Heaton’s explanation, above, works pretty well. It could do with a little more explanation, however.

    If there is one person with blue eyes, and it’s you:
    1. The Guru said she saw someone with blue eyes. Since no one else has blue eyes, you know this must be you. You leave on the first night.

    If there are two people with blue eyes, and one of them is you:
    1. You can see one person with blue eyes. As per paragraph I, above, if this were the only person, you would expect that person to leave on the first night.
    2. That person doesn’t leave, and you, therefore, conclude that (since that person is also a perfect logician and hasn’t made a mistake) there are other people with blue eyes.
    3. You look around. Apart from that person, you can see no one else with blue eyes, so you conclude that you must be the other person with blue eyes.
    4. The other person would also have followed this thought process. You both, therefore, realize that you have blue eyes and leave together on the second night.

    If there are three people with blue eyes, and you are one of them:
    1. You see two people with blue eyes.
    2. If they are the only two, you do not expect them to leave on the first night but on the second (as per paragraph II, above).
    3. They do not leave on the second night, however (in fact, each of them expects you and the other blue-eyed individual to leave together on the second night if you are the only two with blue eyes). You must therefore conclude that there are other people, in addition to the two you see, with blue eyes.
    4. You look around. Apart from those two people, you can see no one else with blue eyes, and (since, according to point 3, there must be at least one other person with blue eyes) you conclude that you must be that single individual.
    5. The other two people with blue eyes would also have followed this thought process. All three of you, therefore, realize that you have blue eyes and leave together on the third night night.

    There are four people with blue eyes, and you are among them:
    1. This is, obviously, an extention of the logic presented above. If there were only three people with blue eyes, they would all have left on the third night, as per paragraph III.
    2. They don’t leave on the third night. But you look around and see no one else with blue eyes and, therefore, realize that you must be the final blue-eyed individual.
    3. You all leave together on the fourth night.

    You can see 99 people with blue eyes:
    1. Everyone realizes that this is going to take a while, and [insert witty comment].

    Ok, so, there are 100 blue-eyed people, so they all leave together on the 100th night.

    Rod Heaton is also correct that, at this point, the remaining people might threaten the Guru with bodily harm (in a non-verbal, physically-threatening way). However, Rod, these people are unable to communicate with each other, so instead of trying to force the Guru to confirm something, I think they would just be pissed that the Guru hadn’t initially said she saw brown eyes (or whatever color their own eyes are). But, hey, maybe the island was a beautiful tropical retreat that the remaining people didn’t really want to leave. Shit, they’ve survived “all their endless years on the island” so far; maybe it’s not that bad (except that there’s no more hot blondes with blue eyes).


  15. there is only one person with 2 blue eyes. if 99 blues have 1 eye and 1 has 2 eyes they will have to leave


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  17. Here’s how I see it, and I think this may clarify a few things who think the inductive argument doesn’t work (it isn’t really an inductive argument since there is no inductive hypothesis used, but the basic premise is still there):
    Let’s assume that I am a person on the island and let’s say I happen to have blue eyes. One of the following must be true for each individual on the island, “I have blue eyes,” or “I don’t have blue eyes”. The entire solution to this problem relies (in my opinion) on considering these two cases. Consider the case that there are two people (ignoring the Guru, as I will for each of the subsequent cases), one blue eyed the other brown. It is clear that the person with blue eyes can figure out they have blue eyes. Let’s now consider the case of four people, 2 blue eyed and 2 brown. As a blue eyed person, I see one person with blue eyes. Consider this: if I assume that I don’t have blue eyes, then we are down to the previous case where there was only one person with blue eyes, and they’d leave the island. They don’t leave the island, therefore my assumption must have been wrong and therefore I have blue eyes. Now bring this up to the case of 6 people, 3 with each colour of eyes. I assume I don’t have blue eyes. This brings us back to the previous case of only 2 people with blue eyes, who would both leave on the second night if they were the only two. If they don’t leave the island, then I too must have blue eyes and the three of us leave on night 3 (the others will follow exactly the same logic and therefore no one knows until we all know). This argument carries over inductively a little more elegantly (again, in my opinion) since for the case of n people with blue eyes (me being one of them), I assume I don’t have blue eyes, and it falls to the case of n-1 people with blue eyes. If they don’t leave on the n-1th night, then I must have blue eyes. Guru’s statement is also necessary since although it may be an obvious statement, it provides the base case that ultimately the process described above breaks down to.
    This argument makes a little more sense to me, because if we look at each individual on the island, they cannot know what all the others know, and so the individual can only use what they know and then consider the cases about what eye colour they have.


  18. In addition,
    I do have one problem with this riddle. I would say that this solution (which is essentially the same one as everyone else has, except of course those of you who have been witty enough to come up with humourous and possibly correct solutions) works assuming that everyone on the island is following this particular train of thought. Often times, different logic can be used to arrive at the same conclusion, although not always as cleanly or quickly. This doesn’t mean that one of these arguments is ‘more logical’ than the other, so perfect logicians might use different arguments. Let’s say (hypothetically) there is another way to determine your eye colour, but instead of taking you n nights to figure it out, it takes n+1 nights (or any number of nights other than n). Then those people on the island who are using the logic in my previous post get all screwed up because one blue eyed person takes n+1 nights instead. I’m not exactly sure how exactly this would affect the end result, but I’m pretty sure somehow it would. Unless of course someone can prove that this is the only solution (or, of course, that all solutions would take exactly n nights).


  19. The logic of the answer a lot of you is flawed because you’re forgetting an important element: everyone on the island can see everyone else on the island at all times. Therefore, this “if there’s 1 blue-eyed person, then X. If there’s 2 blue-eyed people, then Y” arguments don’t fit. Everyone knows that there are at least 99 blue-eyed people from the start. The Guru’s statement, in fact, does not add any new information, because, as per the rules, everyone else ALSO sees people with blue eyes, and has since the beginning.

    For the logical loop itself, also note that since they’re all perfect logicians, they would also consider this argument and, since each person doesn’t know their own eye color and only know that they see 99 or 100 people with blue-eyes (depending on their own color), then it’s equally possible that all the people with blue eyes but one would leave on the 99th day, the 100th not knowing they have blue eyes.

    FINALLY, it says that they’ve been on this island for “endless years”. If perfect logic could have given them an answer, those who could figure out their eye color with the information given would have already left the island.

    Sorry for any flaws in my own logic, I’m at work and have been working for about 12 hours. I’m tired. 😛


  20. Okay, I’ve had some time to sleep and re-evaluate my last post, and I have reached the conclusion that my final point is, in fact, the correct answer.

    No one leaves the island, ever.

    Here’s why:

    First, I’ll remind everyone of a couple points in the problem itself.

    1) Everyone on the island is able to see everyone else on the island at all times, and is actively noting everyone’s eye color.
    2) Everyone on the island is a perfect logician, able to reach the conclusion of a logical problem instantly. Everyone is also aware that everyone else on the island is a perfect logician.
    3) As stated in the problem, the inhabitants of the island have been there “endless years”. Considering the author says he wrote and re-wore the problem many times, meticulously evaluating the wordage used, this statement is not only important, it’s crucial in evaluating the correct answer.

    First, considering point 1, we can logically conclude that the Guru’s statement does not add any new information to what is already known by the island’s inhabitants (we’ll call this point 4), as everyone on the island is also able to see people with blue-eyes, regardless of their own eye color. Now, because of point 4 and, as stated, the fact that the inhabitants are unable to communicate in any way (and excluding variables that are not permitted by the parameters of the problem, such as reflective surfaces, etc.), it is now logical to conclude that the information available to the inhabitants now is the same information that was available to them when they first arrived on the island (point 5). Therefore, combing points 2 and 5, logic dictates that if there were a way to logically figure out what ones own eye color is with the available information, they would have figured it out instantly upon arrival on the island and left that very same night. Therefore, that in consideration with point 3, the only logical conclusion is that there is no way to figure out one’s own eye color with the given information.

    So, I’ll restate, the only possible answer is that no one leaves the island, ever. Remember that this is a hypothetical problem, and by logically solving it using ONLY the parameters set forth in the problem, the problem CLEARLY states that there is no other conclusion.


  21. One more thing: I’m aware the problem says the answer is not “no one leaves”, but there’s one thing that must be clarified.

    I googled the logic problem and found that it originally (apparently) used a variable to represent the number of blue-eyed people. In that instance, then yes, the circular n+1 theory does work. However, in the way that the problem is worded on this site, it does not. As stated here, it is not possible for anyone to leave, as per my post above.


  22. Well, you are wrong. 😉
    To be more specific: your points 4 and 5 are.
    We can not logically conclude that the Guru’s statement does not add any new information to what is already known.

    Because it does.

    Upon arrival the inhabitants did not have a rule to solve their problem.
    Without communicating they had not been able to settle on an eyecolor to “use” with their logic.
    With the Guru talking about blue eyes, the information given was not that there are blue-eyed persons on the island, but that everyone will begin to identify the blue-eyed ones using the mentioned logic (night after night waiting if all blue-eyed persons one can see are leaving).

    And as you mentionend in your second post: if the logic does work for n blue-eyed people it does work for 100 blue-eyed people.

    Summarized: Upon arrival they could not figure it out instantly, because they could not agree on an eyecolor without communication. Only now, with the Guru mentioning blue and starting the logic proces indirectly all needed information is given to the inhabitants.

    p.s. Sorry if the reasoning is a bit long winded. I am not a native speaker.

    Kind regards


  23. Btw.: “on one day in all their endless years on the island” could be the first day upon arrival. It has not to be AFTER endless years.
    And: “It doesn’t depend on tricky wording”


  24. To WraithSama:
    If you think of the solution as being inductive, the Guru’s statement provides the base case. Without the Guru, no base case, no induction, no solution and no one leaves. With the statements, the inhabitants can follow the logic of induction and the blue eyes get to leave.


  25. DanielS, that’s a genius observation! I was struggling with that part of the puzzle – the fact that the guru’s information is apparently something they all knew anyway (for a number of blue-eyed people greater than three). Now I understand that the guru’s statement is not so much a piece of information about their situation, as a piece of advice about how to go about solving the problem. It’s a way of agreeing on a standard way of solving the problem without communication between the [non-guru] islanders.


  26. Sorry, that should have been “they all knew that everyone knew anyway (for a number of blue-eyed people greater than two”. D’oh.

    Anyway, I have achieved blue-eyed puzzle enlightenment.


  27. Aaargh, this is my third post, and for that, I’m sorry. However, I just had a further thought: everyone on the island is a perfect logician, and if something can be deduced they will do it instantly. Therefore, they would know that this method of inductive reasoning is possible, and that all that is lacking is a common consensus on which eye-colour to use and a date to start the counting from – a ‘trigger point’ – i.e. the date of the event where the guru specifies an eye-colour. Now, they know full-well that everyone else on the island has perfect logic, thus they know that everyone else realises this too. They would also realise that the actual eye-colour specified is irrelevant, and in fact the reasoning could work independently for each different group of eye-colours on the island. One person could keep track of the number of days each different group had been on the island, and find out their own eye-colour by noticing when a group of a particular colour failed to leave on their ‘alloted’ night. Thus, they don’t really need a guru to tell them which eye-colour to keep track of. They would all realise this, because they are perfect logicians, and they would all realise that everyone had realised this, so it would be safe to assume that everyone started counting the days for every eye-colour the moment the guru spoke, regardless of what the guru said. The guru’s speech is simply a time-zero, nothing more.

    Thus, for the situation specified (100 blue and 100 brown) everyone leaves the island on the last (100th) night (except the guru, who will never learn their own eye colour in this particular situation).

    Is my reasoning correct?

    Even if there are only two groups, with equal numbers of members, this should still work, as a member of blue (say) will see 100 brown-eyed people, and will know that the brown-eyed group will certainly leave no earlier than night 100. He sees only 99 blue-eyed people, however, and knows that if they don’t leave on night 99 then he must have blue eyes. If there is anyone whose eye-colour is unique, then they will never be able to leave unless the guru specifically mentions their own particular eye colour, although they will eventually realise (when everyone else leaves) that they are unique.


  28. Yeah, it seems to me that you are right.
    Everyone except the guru should be able to leave on the 100th day.


  29. I had to lose a night of sleep before I convinced myself that the Guru information, which I’ll call P: “There is at least one blue-eyed person on the island”, actually adds new information to the system. I thought I was going to introduce something new to the thread, but I see now it’s already been described by D0SBoots and others. Anyway, here’s my explanation, maybe it helps someone.

    Let’s take a simpler situation where there are five blue-eyed people (I’ll call them “blues”) and a number of “non-blues”. We’ll call the five blues B1, B2, B3, B4 and B5.

    Before the Guru does his announcement, B1 is contemplating the number of blues on the island. Here’s what he is thinking:

    B1: Ok, so I see four blues [B2, B3, B4, B5]. Either I’m blue, in which case there are five blues, or I’m not, in which case there are four. I wonder what B2 is thinking. If I’m blue, he’ll be seeing four blues as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing three blues, and might be thinking something like this:

    “B2: Ok, so I see three blues [B3, B4, B5]. Either I’m blue, in which case there are four blues, or I’m not, in which case there are three. I wonder what B3 is thinking. If I’m blue, he’ll be seeing three blues as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing two blues, and might be thinking something like this:

    “B3: Ok, so I see two blues [B4, B5]. Either I’m blue, in which case there are three blues, or I’m not, in which case there are two. I wonder what B4 is thinking. If I’m blue, he’ll be seeing two blues as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing one blue, and might be thinking something like this:

    “B4: Ok, so I see one blue [B5]. Either I’m blue, in which case there are two blues, or I’m not, in which case there is one. I wonder what B5 is thinking. If I’m blue, he’ll be seeing one blue as well, and he’ll be thinking the exact same thing as I am. If I’m not blue, he’ll be seeing no blues, and might be thinking something like this:

    “B5: Ok, so I see no blues. There are either zero ore one blue on the island. I don’t know.”

    So, before the Guru announcement, B1 knows P; B1 knows that B2, B3, B4 and B5 know P; he knows that B2 knows that B3, B4, and B5 know that P, etc. recursively, but he can NOT say that he knows that B2 knows that B3 knows that B4 knows that B5 knows P!!


  30. this reminds me of an icelandic riddle. that is, with the part about language being the last word in “the english language”.
    the first part is a sentence with a lot of r’s, and then the last part is: how many r’s are in that? and the answer is none, becaus ethere’s only T, H and A in “that”…

    rómverskur riddari réðist inn í rómarborg(a roman knight invaded rome)
    rændi þar og ruplaði radísum og rófum(stole and robbed radishes and onions)
    hvað eru mörg err í því?(how many r’s are in that?)


  31. Simon: I think you have given the best explanation of this. YES, there is new information, and it’s NOT that there’s somebody with blue eyes (everybody, of course, knew that) and it’s NOT just a time-zero – well, it kind of is, but there’s more, though it’s very hard to grasp. When I trying to figure this out, I kept getting stuck at the 5-person case. I figured, if it works for 5, it works always. But I can see how four people can think that if there are three, each of those would think about if there were two and then leave… but after that, I get lost in each person thinking what each other person might be thinking. It does work, though. And the new information that the Guru gives doesn’t actually come about until 99 nights have passed and nobody leaves. NOW all of a sudden every blue-eyed person knows that there has to be more than the 99 blue-eyed people they can see.

    Matt: So awesome.

    Richard: No one else has given your prisoner question a shot, so here’s my idea. There’s good logic in the assertion that midnight on Jan 30th, he can’t be killed on Jan 31st, because there’s no days left. But he can’t extend that to midnight on Jan 29th, because there are two days left. Either day is fair game; he only can’t be killed on the 31st if the 30th has already passed! Until the 30th has gone by and he hasn’t been killed, the 31st isn’t safe, and either is the 30th. All he really knows is that he isn’t dead yet.

    Although even as I say that I’m getting sucked into wondering why he couldn’t be certain on the 30th that he was getting killed that day. They CAN’T leave it one more day. And by extension they wouldn’t set it for the 30th, because he would know on the 30th! So if he survives until the 29th, he would know it’s the 29th…

    Perhaps it’s that each day, he knows it has to be the day he’s going to die, because it can’t be any of the days afterwards. They screwed up by saying that, because every day he knows, and every day he’s wrong, and they simply lied to him. 😛

    @XKCD: I totally agree that you should clarify that nobody is colour-blind. My first thought was that everyone would assume if there was more than one blue-eyed person, then the Guru would have said so, so therefore there really is only one, and *everyone I’m looking at must actually have different coloured eyes – I must be colour blind – I must have blue eyes!!!* Then all the blue-eyed people would leave that night.

    By the same token, I kept getting caught up in motivations. Why on earth didn’t the Guru just say “I see an equal number of blue-eyed and brown-eyed people”??? Then everybody could go home that night. Obviously she must have some special motivation for saying what she did – she wants to go home herself, so somehow what she said will result in her leaving eventually. (My original hypothesis, before I figured it out, was the Guru would leave on day 100).

    There’s another version of this that says the not-talking-about-eyes-thing is a religious thing, which really makes a lot of sense, and that the “Guru” is really a tourist who slips up. (“Strange seeing people with my eye colour in this part of the world!”) Personally, I think this version helps a lot because it removes all question of motivation. (But I really do appreciate how much you’ve already removed ambiguity from the problem – it was very refreshing to have a problem that I KNEW didn’t have some stupid answer. Even though I kept going back to the colour-blind thing, I knew in my heart of hearts that this wasn’t the answer, so kept going.)

    Also, thanks so much for continuing to provide us all with the best comic ever. 🙂


  32. The answer: Any villagers who figure out their eye colour will leave the island, at midnight.

    With at least 200 people on the island, the fact that there exists a person with ‘blue’ eye colour is simply noise. If everyone is a perfect logician, then they all understand that everyone can see at least 99 people with blue eyes and at least 99 people with brown eyes. This isn’t new information. With the noise of a sample size of 200, there’s no reasonable logic to suggest you could figure out whether you’re blue eyed or brown eyed.

    The only way to solve the problem is to introduce more information. Since everyone is on an island, the best thing to do is look at your reflection in the water. Oh yeah! My eyes are yellow!


  33. I was playing around with the audio feature.

    And instead of saying “I am” it says “I A-M”.

    It made me giggle.


  34. It shouldnt be that hard.. after all, all they had to do was guess.. of course, being masters of logic, luck is one of the most illogical things to depend on.

    For example, a brown eyed man, seeing there are 99 brown eyed and 100 blue eyed people, would guess that he is brown eyed. If he got it right, he leaves that night. If not, assuming a person may try once a night, he guesses blue the next night.

    Since they dont die if they guess wrong, why not go ahead?

    The guru is for show, telling people what they can already see for themselves. Both brown eyed men and blue eyed men can see there are blue eyed people among them. all of them except for the guru may leave by the second night if they all tried.

    the guru looks pretty content to be staying on the island seems like it..


  35. Call me Mr Dumb-Ass but surely the answer is that every-one stands in a straight line in front of the guru? Therefore if the person at the front now knows their own eye colour!

    If this answer has already been posted, I apologise, but all the cod-logic in the varios posts was killing me!


  36. Oh, and as for a lateral answer:

    One person left on the night of the guru’s announcement; the only blue-eyed person who didn’t wear sunglasses during the announcement.


  37. the brown eyed people would all leave the day after the blue eyed people left.


  38. @Richard Brown – I tend to agree that this is an abuse of inductive reasoning that ends in a contradiction. Just like the prisoner who reasons that he can’t be executed because he won’t be surprised based on inductive logic. But he is executed and it does come as a surprise.

    @Simon – It’s unreasonable to suspect that each successive person could imagine that the next one could be thinking that there is one fewer blue-eyed person than the person “above” him in the chain imagines at minimum. In the extreme example of 100 blue-eyed islander, a blue-eyed seeing 99 other blue-eyed eventually has to think that another islander thinks that another islander could think, etc., that there might just be one other person with blue eyes, in light of the fact that there’s demonstrably many other islanders that person will see with blue eyes, and the person “above” him will see, and the person “above” him will see.

    @all who believe the blue-eyed islanders will eventually figure it out – please answer me two questions: (a) why does the Guru have to announce that at least one has blue eyes before they start counting off nights? (b) why can’t the brown-eyed islanders use the same inductive logic? After all, they all know that everyone knows that there is at least one brown-eyed prisoner, right?

    CAPTCHA: metamphetamines jubilant (teehee)


  39. Just my bit of reasoning: What keeps the people from simply telling the ferryman that they have blue eyes? If they get to ride the boat, they know they were right, and if they’re turned back, there’s always next night.


  40. This is fascinating, because I can see exactly how the chain of logic goes from the case of one blue-eyed person, to the case with a hundred, and yet it seems like it should break down somewhere (but where, exactly?). If the Guru were instead publicly asked (maybe there’s a sign with the question written on it, or a second Guru) whether she saw someone with blue eyes, every person would know that her answer must be yes, and every person would also know that everyone would know that… looking at the 99 or 100 blue-eyed people, nobody could possibly think she’d say no. So, her actual answer doesn’t seem to be required.

    What if she lied and said, “I see no blue-eyed people”? I’m almost sure nothing would change, since everyone would know it was a lie, and–since everyone could tell that everyone else had either the same or very similar information as they did (a blue-eyed person would know that the other blue-eyed people could see either 98 or 99 blue-eyed people, and a brown-eyed person would know that all the blue-eyed people could see either 99 or 100 other blue-eye people, so everyone knows that everyone else can see SOME blue-eyed people, no matter their own eye color)–they would also know that everyone else knew it was a lie.

    I think DanielS must be right… the Guru’s function is to break the communication barrier, if only very slightly, in order to provide a starting point.

    It reminds me of another puzzle I saw… I think it went like this:

    Alice and Bob are prisoners and logicians (and each knows the other is as good a logician as they are). The jailer puts a number on each of their heads (so that Alice can see Bob’s number and vice versa, but neither can see their own) and tells them that their numbers are positive integers, and that they’re also exactly 1 apart. The jailer sits them down in electric chairs and tells them that if they can guess their own number, they’ll both be set free, but if either says anything other than their own number–or try to communicate with each other–they’ll both be executed. If they say nothing, they’re just stuck there. The jailer asks them if either cares to guess what their number is, and if neither replies, asks again after a minute, repeating until someone says something. How do they get free?

    And the solution was fairly similar to the solution to this puzzle, and it goes:

    If Alice sees a 1 on Bob’s head, she immediately knows that her own number is 2, and say it the first time. If she sees a 2, she knows that her own number must be either 1 or 3, so she stays silent the first time. She knows that if her number is 1, Bob will know that his number is 2 and say so immediately, but if it’s 3, he won’t be sure whether his own is 2 or 4 and will be silent. So if she sees a 2 and Bob is silent, then when the jailer asks for the second time, she’ll say that her number is 3. And so on… if you number the opportunities to guess, then the strategy of each one is to wait until the opportunity with the same number as what they see, and then guess that number plus one. The jailer asking the question the first time maps to the Guru saying that she sees someone with blue eyes, sort of, and every question after that maps to waiting a day.


  41. i think that when the guru says “i see someone with blue eyes” everyone would automatically lock eyes on a blue eye, because so do they. therefore if people are staring you that morning, you must be blue eyed.


  42. I’m not 100% sure but I’m pretty sure its the Guru who leaves the island because it specifically says that no one may talk except for the Guru and the Guru is the only person who has green eyes. Therefore, when she talks, she realizes she is the Guru and has green eyes. And so she leaves the island!


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  45. I wonder whether it works like this:

    If they are allowed to look each other into the eyes
    (if that’s not considered communication),
    then, based on the information given, they can look each other in the eyes to tell what colour they have.

    First the brown eyed (or also the green eyed) look into the eyes of the blue eyed
    –> first night blue eyed leave;
    then, applying the same logic, the brown eyed can identify each other: they will all look at each other (50 identical pairs)
    –>second night brown eyed leave;
    and the green eyed who is left can guess, knowing that it is not blue or brown; the chances are depending on how many colours there are then.