tag:blogger.com,1999:blog-2744072865491516720.post537957412787411152..comments2023-05-03T06:35:33.259-04:00Comments on Higher Logics: Blue-eyed Islander Puzzle - an analysisSandro Magihttp://www.blogger.com/profile/05446177882449578817noreply@blogger.comBlogger40125tag:blogger.com,1999:blog-2744072865491516720.post-71143509739353646482017-10-26T11:28:03.596-04:002017-10-26T11:28:03.596-04:00The unexpected hanging paradox is not simpler, tha...The unexpected hanging paradox is not simpler, that's the point. It requires a formalization of natural language and its interpretation under temporal modalities. It's absurdly complex by comparison.<br /><br />Like I said, the blue eyed islanders solution is well accepted by the mathematical community as an instance of the common knowledge. I don't see any benefit in belaboring this point more than I already have.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-84028889236772934882017-10-26T11:22:16.034-04:002017-10-26T11:22:16.034-04:00I used the term "false induction" becaus...I used the term "false induction" because that example (the unexpected exam or hanging paradox) leads to a false result--a contradiction. Although the "blue eyes" induction may not be structurally identical to the hanging paradox, can we rule out some other problem where the seeming soundness of that induction creates an absurd result? It is a much more complex case than the unexpected hanging case. Complexity usually gives a flaw more hiding places. As you noted yourself, the hanging paradox has not yet been resolved (despite being much simpler.)CitizenJhttps://www.blogger.com/profile/16612825227302177808noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-28702456088973911912017-10-25T23:17:28.131-04:002017-10-25T23:17:28.131-04:00Does the blue eye puzzle entail a form of false in...<i>Does the blue eye puzzle entail a form of false induction?</i><br /><br />The scenario you describe is called the <a href="https://en.wikipedia.org/wiki/Unexpected_hanging_paradox" rel="nofollow">unexpected hanging paradox</a>. It's not a case of "false induction", whatever that may be. The problem with that paradox is in faithfully formalizing the judge's informal statements. No one yet knows how to do this, and so the apparent paradox is not yet resolved.<br /><br />This is not the case with the Blue Eyed Islanders puzzle. As I've detailed thoroughly in the comments here, the induction is simple and sound, and founded on the well-accepted principle of <a href="https://en.wikipedia.org/wiki/Common_knowledge_(logic)" rel="nofollow">"common knowledge"</a>.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-47565444296148891302017-10-25T13:45:26.615-04:002017-10-25T13:45:26.615-04:00In the (exam induction) case I gave earlier, an ab...In the (exam induction) case I gave earlier, an absurd conclusion results. This is at least a warning that induction processes must be looked at critically. In the case of the blue eyes induction, a seemingly absurd result is also reached, I.e., that 100 blue eyed people can deduce their own eye color from the guru's statement that she merely sees blue eyes. The problem implies that the 100 arrived on the island all at once. Otherwise, a population of 100 could not have accumulated, since a lesser number would already begun leaving or left (according to the arguments given by the solution). The guru only speaks once, not during an accumulation process. The induction process itself is a counter factual-something that never happened. There was no actual n=1 case, or n=2 case or any case except n=100. It is questionable to say that the guru actually communicates new information when 100 blue eyed are plainly visible, and based on an imaginary inductive process.<br /><br /> All this being said, I don't have strong convictions about my objections. But I think the failure of induction to give a reasonable result in one case should give pause about it in other cases, especially when an absurd seeming result occurs.CitizenJhttps://www.blogger.com/profile/16612825227302177808noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-48974436120199054212017-10-23T20:29:20.656-04:002017-10-23T20:29:20.656-04:00There are cases of "false induction". He...There are cases of "false induction". Here is one puzzle with false induction: A professor tells his class on a Friday that there will be an exam next week, but before the following weekend, and also that the exam will occur in the morning. He further adds that they won't know the day of the exam until the day it occurs. From this one student concludes, by the rules just given, that the exam cannot be given at all !<br /><br />His reasoning is that the exam cannot occur on Friday, since by Thursday afternoon the students will know the exam must be on Friday, contradicting the rule that they will not know until the exam day. Since Friday Was ruled out, Thursday becomes the last day the exam can be given. But the same reasoning now applies to Thursday, and it is eliminated as an exam day also. Continuing this way each day is eliminated. Yet obviously the students could be taken by surprise when the exam is suddenly given Tuesday morning.<br /><br />Does the blue eye puzzle entail a form of false induction?CitizenJhttps://www.blogger.com/profile/16612825227302177808noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-72955685154048872092017-08-18T08:03:33.167-04:002017-08-18T08:03:33.167-04:00Assuming you use the variant with only two eye col...Assuming you use the variant with only two eye colours, the islander's don't actually know that the island features only two eye colours, so your scenario can't happen.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-19337741491261306712017-08-06T04:14:57.650-04:002017-08-06T04:14:57.650-04:00And what if all 100 of them kill themselves, but t...And what if all 100 of them kill themselves, but then since the rest of the villagers know all the blue eyed villagers are gone, the rest of them must have brown eyes, then all of them will commit suicide the next day?Anonymoushttps://www.blogger.com/profile/13925044245309555441noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-80297711730220183672017-01-24T09:54:52.788-05:002017-01-24T09:54:52.788-05:00Schroedinger, you are incorrect as has been elabor...Schroedinger, you are incorrect as has been elaborated thoroughly in this thread. This mistake is pretty common, but the blue eyed islander is a very thoroughly analyzed mathematical problem and the posted solution is correct.<br /><br />On day 3, each blue eyed islander realizes that the other two blue eyed islanders themselves see 2 blue eyes, and thus their own eyes must also be blue. Every brown-eyed islander sees 3 blue eyes, so they remain.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-60223948400148995492017-01-24T01:40:09.197-05:002017-01-24T01:40:09.197-05:00On An island with 1 blue eyed person, given the sc...On An island with 1 blue eyed person, given the scenario, the person leaves/dies/walks into a volcano. Immediately. <br /><br />N=2:<br /><br />Day 1: <br /><br />Nobody leaves. They are expecting the other person to leave. <br /><br />Day 2:<br /><br />Since he other didn't leave, they can each surmise that they each also have blue eyes. From each viewpoint. (I think this is the critical flaw in the puzzle as posed. Not the inductive argument, nor the underlying logic. Just the puzzle. As posed.)<br /><br />N=3:<br /><br />DAY 1:<br /><br />A observes B&C; B observes A&C; C observed A&B. none can be certain of their own eye color. <br /><br />Day 2:<br /><br />Since none were certain on day 1; none have left. They continue to observe the other islanders, but can gain no further certainty. <br /><br />Day 3:<br /><br />Repeat day 2. <br /><br />the solution isn't valid for the specific situation described in the puzzle. I accept that it's meant to teach logic. Your inductive argument doesn't codify each islander's viewpoint. No islander can gain any more certainty from successive days because there is no additional certainty from waiting unless the islanders can communicate. Since they are forbidden to do so, their lack of exit is due to a lack of certainty from their viewpoint which is never resolved. <br /><br />I believe this is a failure of the puzzle to reflect the intended logical problem, or a failure of the logic to properly codify the problem as stated. <br /><br />Logic is useless in this context if it can't reflect the English expression of the problem. <br /><br />In my view. <br /><br /><br /><br />Schroedinger's Bullhttps://www.blogger.com/profile/08497731317409235498noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-26281713987818419472015-04-22T11:48:46.624-04:002015-04-22T11:48:46.624-04:00You wondered why "is it not optimal to assume...You wondered why "is it not optimal to assume you have blue eyes up unless everyone leaves the day before you would if it were true". In fact, that is entirely correct.<br /><br /><b>If</b> it is the case that, were you not blue yourself, all the blues you see would leave on a given day, <b>then</b> when you don't see them leave on that day, you can correctly conclude that you are a blue.<br /><br />The trick is establishing that <b>if</b>. I originally had a whole spiel that basically repeated stuff I said before about working from the top down, but it's easier to get this if we build up from the bottom up.<br /><br />Consider the case of 4 blues total. We agree that, absent the words of a foreigner, they won't ever leave, right? It's a stable situation. No one will leave an island of 4 blues "because it's Day 4", or for any other day afterward.<br /><br />Now consider the case of 5 total. Each of them knows there are either 4 or 5 blues. In the case of 4, then no one will leave on Day 4 (established in the previous paragraph). So observing other islanders <b>not</b> leaving provides zero information. Therefore, the 5 blues each have no reason to leave. Hence, 5 is just as stable as 4.<br /><br />Now consider the case of 6 blues. Each of them sees 5, and knows that 5 is stable. Thus, no information is gained by seeing nobody leave on Day 5. Because of this, the 6 blues will have no reason to deduce themselves to be blue, and they will not leave on Day 6. Hence, 6 is also stable.<br /><br />And so on for 7, 8, 9, 10…<br /><br />===========================<br /><br />Let's say someone (I'll call him Alex) disagrees with some step of this. For example, Alex agrees that 5 is stable, but that 6 blues would deduce their eye color and leave on Day 6.<br /><br />There's a very simple way to test this assertion. We give Alex a new randomly-chosen eye color and put him on the island. He sees 5 blues. On Day 5, none of those blues leaves. Should Alex leave on Day 6?<br /><br />Remember, this situation is logically consistent with there being 6 blues (including Alex). So his answer should be "Yes, having seen no one leave yesterday, I will leave today." But the situation is <i>also</i> consistent with there only being 5 blues, because 5 is stable. So if he left, he could very well be making a mistake.<br /><br />This basic argument can be re-applied by increasing the numbers by one. We've just established that 6 is stable, and can therefore repeat the argument to show that 7 is stable, and 8, and 9, and so on.Lenoxushttps://www.blogger.com/profile/10809085020841868387noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-13806204070651244652015-04-22T08:46:25.852-04:002015-04-22T08:46:25.852-04:00Ok, I'm beginning to see the problem, especial...Ok, I'm beginning to see the problem, especially since common knowledge has such a well defined meaning.<br /><br />The hump I can't get over is, why, when I know that everyone definitely sees a blue eyed person, and everyone will definitely know that everyone sees a blue eyed person, or more importantly, multiple blue eyed people, is it not optimal to assume you have blue eyes up unless everyone leaves the day before you would if it were true. If they don't leave, can't you be sure you do have blue eyes?Anonymoushttps://www.blogger.com/profile/07066633832251118226noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-30546474565755043222015-04-21T22:15:54.111-04:002015-04-21T22:15:54.111-04:00Craig: One problem is that you are using the term ...Craig: One problem is that you are using the term "common knowledge" to mean "everyone know that everyone knows". However, it doesn't; it actually means "everyone knows that everyone knows that everyone knows that everyone knows..." <i>ad infinitum</i>. And that makes a major difference, just like there's a big difference between just one and two layers ("everyone knows" vs "everyone knows that everyone knows").<br /><br />There is no reason to treat two layers as a logical "stopping point". Instead, when we add another layer in our consideration, we have to subtract one from the "commonly" known number. For example, say there are six blue-eyed people. In that case:<br /><br />Everyone knows that there are at least five blues.<br /><br />[Everyone knows] x 2 that there are at least four blues.<br /><br />[Everyone knows] x 3 that there are at least three blues.<br /><br />[Everyone knows] x 4 that there are at least two blues.<br /><br />[Everyone knows] x 5 that there is at least one blue.<br /><br />[Everyone knows] x 6 that there are at least zero blues. (A logically necessary statement under all circumstances anyway; you can't have a negative number of blue-eyed people).<br /><br />The human mind has a really hard time working past two layers here, which is why this puzzle is so famous.Lenoxushttps://www.blogger.com/profile/10809085020841868387noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-5806902961064018982015-04-21T21:10:56.424-04:002015-04-21T21:10:56.424-04:00It is pretty difficult to prove the lowest possibl...It is pretty difficult to prove the lowest possible case (for me at least, I have no formal training). That being said, I'm almost certain my argument requires n>5, but I don't think that will convince you. :)<br /><br />Say we go to n=10. the number of other islanders and their eye colors are irrelevant.<br /><br />I see 9 blue eyed people. <br /><br />Because I know that no one else can see less than 9, this must be common knowledge. There is no scenario where anyone can see few enough blue eyed people to reasonably assume it cannot be known by everyone else that there are blue eyed people.<br /><br />Even if I don't have blue eyes, I know that anyone who does will be able to see 8 blue eyed people. They can make the same inference, because no one can see less than 7 (same logic) and therefore everyone must know that everyone else knows.<br /><br />But I don't have to go any further, as a 9 seer (one who sees 9 blue eyed people,just to be clear), because I know no other scenario exists.<br /><br />The 8 seers, if any exist, won't have to go further than 6 seers, and 7 seers cannot exist, because I see 9, so no one, in reality, will ever use 7 as a starting point.<br /><br />Maybe I should add that the proof for why they leave is still the same, I'm just saying it's impossible to make"blue eyed people exist" common knowledge if it must necessarily be so with sufficient blue eyed people.<br /><br />The reason I think n > 5 is the lower bound is basically this: I see 5. I know they see ate least 4, so I know it's commons knowledge, they can't not see a blue eyed person.<br /><br />Their n becomes 5 essentially, and they can say, at worst, I know no one sees less than 3. Therefore no one can think there are less than 3. With three everyone sees blue eyed people, and everyone can see that everyone can see blue eyed people. <br /><br />But I am still reasoning as an n=6, so I know no one will ever see less than 4. I can rule out all scenarios after that as impossibilities, so while the 1 blue eyed person scenario is important to find the ideal day to leave, no one could possibly think the real life scenario is less than 3 blue eyed people, which would have blue eyed people as common knowledge.<br /><br />I'm going to stop myself here, as if I'm not explaining clearly enough yet why I think the lower bound is 6 I don't think I'm going to make it any better right now, and I'm also starting to convince myself it might be 7 :/<br /><br />It's definitely not higher than ten. Getting the ideal answer might be a bit out of my league though. I think it rests on whether 3 or 4 is necessary as a floor for common knowledge if no one can think there are less.Anonymoushttps://www.blogger.com/profile/07066633832251118226noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-81262069873936939352015-04-21T19:47:52.507-04:002015-04-21T19:47:52.507-04:00Craig, we're agreed that when there are only 1...Craig, we're agreed that when there are only 1, 2 or 3 blue eyes on the island, the foreigner adds important new information, correct?<br /><br />For n=1 blue eye, the foreigner states the existence of blue eyes, which the only blue eyed person then infers must be his own eye colour.<br /><br />For n=2, anybody seeing 1 blue eye provisionally assumes they're in n=1 until that's disproven on day 1, in which case they know they're in n=2.<br /><br />For n=3, anybody seeing 2 blue eyes provisionally assumes they're in n=2 until that's disproven on day 2, in which case they know they're in n=3.<br /><br />Now your conjecture is that when there are n>=4, the existence of blue eyes is already common knowledge. And yet, the template for induction is already established by this point, and you provide no reason to invalidate it. In order for your counterargument to work, this inductive argument must somehow be flawed, so you'll have to point out that flaw to convince me.<br /><br />Note that the amount of knowledge being added by the foreigner's statement becomes progressively more convoluted the more blue eyes there are, so while it might seem "obvious" to you that the knowledge is already present, you haven't proven it. Given how subtle the information is, I don't accept your informal argument at face value.<br /><br />Finally, others have posted <a href="https://terrytao.wordpress.com/2011/04/07/the-blue-eyed-islanders-puzzle-repost/#comment-51436" rel="nofollow">precisely your counterargument</a>, but they also describe the additional information which causes ultimately your argument to fail.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-69728211281920078362015-04-21T14:34:36.078-04:002015-04-21T14:34:36.078-04:00Sorry to dredge up this old post, but I stumbled u...Sorry to dredge up this old post, but I stumbled upon it, and I think I have something to add.<br /><br />Jim was onto something, but he failed to follow it through. It's not that there is no solution, but that, depending on the conditions, it either doesn't require an explorer, or only requires him to set a "day 0."<br /><br />With sufficient blue eyed people, there will be a "floor" for the real minimum of people anyone can see with blue eyes.<br /><br />Let's say there are 100 blue 100 brown. I have blue eyes. That means I see 99 blue eyed people.<br /><br />This tells me a few important things. I know there are blue eyed people, I know that everyone must know there are blue eyed people, and I know the minimum number of blue eyed people anyone can see is 98.<br /><br />So let's say I assume I don't have blue eyes. That means anyone with blue eyes will see 98 blue eyed people, and know that there are blue eyed people, that everyone else knows that there are blue eyed people, and that the minimum number of blue eyed people anyone can THINK THERE ARE is 97. <br /><br />This is the key: I, a blue eyed person, am now reasoning as another blue eyed person, and finding the floor because of it. The floor is 97, because they could see one less than me, if I'm not blue eyed, and could assume they don't have blue eyes. But no one will ever see less than 98 people with blue eyes, because I see 99, and they can only exclude themselves.<br /><br />That means the minimum number of people there can be with blue eyes is n-1, and the minimum anyone could think there were is n-2, where n=the number of blue eyed people the reasoner is seeing. As long as n-2 > 3, you can know that blue eyed people are common knowledge, and that everyone else can too.<br /><br />Then, you just leave on day n+1, provided all the blue eyed people haven't already left. They may leave one day before, at which point you'd know your eyes aren't blue.<br /><br />Then you can do the same for every other color (you'd start with the lowest number you can see) and as long as everyone else doesn't leave before you, you can figure out your eye color.Anonymoushttps://www.blogger.com/profile/07066633832251118226noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-42464310117434081672012-02-05T16:22:32.906-05:002012-02-05T16:22:32.906-05:00Sand Magi – I totally agree. I was being more face...Sand Magi – I totally agree. I was being more facetious/cynical with my last comment. In fact, this is probably my favorite puzzle! I even wrote a long piece dealing with common objections to the solution, here:<br /><br />http://forums.xkcd.com/viewtopic.php?t=80149&p=2874552<br /><br />It's just interesting how it doesn't translate into real human behavior due to some stumbling blocks about meta-reasoning we call seem to share.Lenoxushttps://www.blogger.com/profile/10809085020841868387noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-1791421759166676872012-02-05T13:49:59.272-05:002012-02-05T13:49:59.272-05:00@Lenoxus, I don't see how people's ability...@Lenoxus, I don't see how people's ability to reason is relevant to the logic puzzle. If we replaced people with robots in this puzzle, the same solution applies.<br /><br />The point of logic puzzles is to create a scenario and determine precisely what logic tells us should happen.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-12550765797596266562012-02-05T13:30:32.432-05:002012-02-05T13:30:32.432-05:00In a sense, I'm inclined to agree with what Ji...In a sense, I'm inclined to agree with what Jim had been saying earlier about humans vs computers. The many, many discussions I've read on this suggest that humans cannot intuitively think on more than 2 layers of recursive knowledge, ergo, no set of islanders could be "perfect logicians" like that in the first place. Most people think that<br /><br />"Everyone knows that everyone knows X"<br /><br />Is tha same hing as:<br /><br />"Everyone knows that (x infinity) everyone knows X."<br /><br />If our brains worked differently, we would have the same trouble with puzzles that required counting. "But when the library recieved another book, the total number of books can't possibly change – there were already more than 2, and more than 1 more than 2, etc! Effectively, numbers greater than 2 are equal."Lenoxushttps://www.blogger.com/profile/10809085020841868387noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-25735776310622568562011-09-17T16:44:46.197-04:002011-09-17T16:44:46.197-04:00This comment has been removed by the author.Brianhttps://www.blogger.com/profile/09270685900730961853noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-16224453533920002812011-09-17T16:38:07.807-04:002011-09-17T16:38:07.807-04:00Thanks for the reply.
I am open to the possibilit...Thanks for the reply.<br /><br />I am open to the possibility that I need to think about it more, but having spent so much time thinking about it already, I wonder whether I'll get a solid return on my time investment. Maybe I'll just shrug my shoulders and be okay with the fact that I may disagree with the majority on a certain logical puzzle.Brianhttps://www.blogger.com/profile/09270685900730961853noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-56311236930362397822011-09-17T12:24:29.126-04:002011-09-17T12:24:29.126-04:00Brian, the point is that a person does not know th...Brian, the point is that a person does not know their *own* hat colour. The argument hinges on this fact. His hat could be brown or blue, and the rule that they must kill themselves if they figure out their own hat colour is what establishes the induction.<br /><br />If there are only two people, one blue hat and one brown hat, someone declaring they see a blue hat means that person dies the next day. If there are 2 blue hats, they each assume that the other will kill themselves the following day. When the other person doesn't do so, they infer that they too must have a blue hat, and they both die the following day. And 3 people die on day 3, and 4 on day 4, and so on.<br /><br />The argument is subtle, but very simple. Even Jim and Rusty in the end acknowledged that this solution is correct. You need to think about this some more.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-2050421574355055262011-09-17T12:18:10.641-04:002011-09-17T12:18:10.641-04:00It's nice to know I'm not taking crazy pil...It's nice to know I'm not taking crazy pills.<br /><br />I stumbled across this puzzle and I also see precisely the error being identified by Jim and Rusty. Rusty's way of explaining the error is the best articulation of it, in my opinion: the logical formulation assumes that people are being added in a sequence, when in fact no such thing is occurring, so no new information arises.<br /><br />Further to the "common knowledge" argument, that also makes no sense. Five people with brown hats, five people with blue hats. Everyone knows there is at least one brown and at least one blue hat. If they are perfectly logical, they <i>also know</i> based on the number of people, that everyone else knows this too.<br /><br />It blows my mind how widely accepted this puzzle and solution is alleged to be. Seems to me like a case study in groupthink. I understand the solution, and people can only attack my objection by trying to suggest that I do not actually understand the solution. <br /><br />I guess when people relax their skepticism too much, and put too much faith in authority, anything is possible.Brianhttps://www.blogger.com/profile/09270685900730961853noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-58779056421110660522010-12-07T17:54:19.185-05:002010-12-07T17:54:19.185-05:00@jim, and @rusty, I just want to put in a word tha...@jim, and @rusty, I just want to put in a word that I was in precisely your shoes a few hours ago... I was sceptical of the "solution", thinking that when the number of blue-eyed people is more than 2, no new information is being added when the outsider announces "I see blue eyes." <br /><br />But Sandro is right... in fact new information *is* being added: after the announcement, everybody knows that everybody knows that (repeat n times) there is at least one pair of blue eyes. This is the technical meaning of "common knowledge". Although everyone already knew that at least one person had blue eyes, it was not <b>common knowledge</b> in this sense.<br /><br />It was the wikipedia article Sandro linked to that helped me understand this, along with the <a href="http://xkcd.com/solution.html" rel="nofollow">Randall Munroe's explanation</a>.Larshttps://www.blogger.com/profile/03240201453848389345noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-28155542534117341732010-09-13T23:09:22.804-04:002010-09-13T23:09:22.804-04:00Just noticed I provided a broken link. This is the...Just noticed I provided a broken link. <a href="http://higherlogics.blogspot.com/2008/02/blue-eyed-islander-puzzle-analysis.html?showComment=1274848552794#c3312848408206946794" rel="nofollow">This is the correct link</a> to the most detailed explanation.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.comtag:blogger.com,1999:blog-2744072865491516720.post-34140197421979642072010-08-24T23:18:44.623-04:002010-08-24T23:18:44.623-04:00But the islanders don't start out with one blu...<i>But the islanders don't start out with one blue-eyed person then add one more at a time. They start out with several. The "x leave/die on day x" reasoning, per se, is logical, but the adoption of it by three or more blue-eyed islanders is not.</i><br /><br />In fact it is perfectly logical, as explained many times in this thread. I have explained in step by step the induction proof that demonstrates why this is so. <a href="http://www.blogger.com/profile/05446177882449578817" rel="nofollow">This is perhaps the most detailed explanation</a>.Sandro Magihttps://www.blogger.com/profile/05446177882449578817noreply@blogger.com