And, this could easily happen with a recessive trait with close kin inbreeding. Maybe incest isn’t taboo on this island.
http://www1.cs.columbia.edu/~evs/intro/Oracle.html
and here’s another link to the problem. I think they’ll claim the answer to be on the nth day all of the islanders commit suicide. I think according to their logic however it would be on the (n +1 day)=mass suicide. I still think that if they all knew each other then they would all commit suicide the very first morning if there was only 1 person supposedly that had blue eyes. If there were more than one then nobody would die. Because they’d see that no one else had blue eyes and therefore the oracle/visitor had lied.
okay rfgdxm I’ll play nice. 
Whether they all kill themselves on the nth day or the (n+1)th day depends on whether you consider the day the oracle makes the announcement the first day or the day after the first day and on whether a person has to kill himself the day he finds out or the next day (for example, does he kill himself immediately, or does he wait till the next morning). However, these off-by-one considerations aren’t really at the heart of the issue. To be specific, in my original post, I specified that the person must kill himself on the next sunrise and that the nth day was the nth day after the oracle’s announcement.
If there is more than one blue-eyed person, each would see that blue eyes of the others. How would the oracle have lied?
Again you alter the hypothetical. In the hypotetical, the oracle is believed by all to be infallible. Perhaps the oracle first performs some miracles that were prophesized in their religion, and thus they all believe the oracle is a prophet of god.
But understand that when I was first presented this problem (some umpteen hundred years ago) the islanders all had blue eyes and the people with brown eyes were the ones who were supposed to die. I just figured political correctness and racial tensions changed the original set of variables.
Maybe I just had a smartass for a prof. He liked to throw curveballs at us ya see. Sorry 'bout that, carry on. 
This thread seems to have been burried for a while. I never really got a good answer to this question. Is it just that everyone thinks that the argument is correct, and there is no flaw?
Yes, I think the argument is generally sound. However, some of the assumptions made in the problem should be stated up front, otherwise it won’t work. I personally prefer the version with the monks and the ringing bells as posted by Sublight --it’s a much more clear and concise illustration of the concept.
…or you and bitwise make a pact and agree to play dumb. The next day, everybody on the island freaks out. They say the two of you alive and since they have brown eyes they all individually think that they are the only third person with blue eyes. 
Assuming they don’t run into each other, they all kill themselves. :smack:
I know, I know…it’s not part of the puzzle, but it helps understand why it works with N=2 without mass suicide. Ie, since N=2 means on the morning of day three the two people with blue eyes are dead. Everybody else lives. If N=3, then on the morning of day 4 all the people realize that the three with blue eyes have killed themselves.
Thus, the answer (I think as I didn’t read everybody else’s postings), is that it all hinges on WHEN people are to kill themselves. If they are to kill themselves at an appointed hour then I think what I wrote about is the solution. If they are just to kill themselves at sometime in the morning, I guess the solution would be that they do so on N day at the exact moment that “morning” ends. Thus, there is no mass suicide.
Make sense? Or am I just tired? :smack:
I don’t understand your logic. If one person can see six with blue eyes but doesn’t know if they are the 7th, that means that all others with brown eyes think the same way (assuming they are all equally perfect in logic, the only way the puzzle makes any sense). Everybody with blue eyes sees only 5 people with blue eyes and wonders if they are the 6th. Thus, assuming the oracle speaks on January 1 and sole blue eyed person would have to have died that night, by January 6, the guys with blue eyes all think, “Crap! I see five blues and the rest are brown. That means I have to die tonight with them.” The people with brown eyes all think on January 6, “Damn it. This is getting nerve racking. I see 6 people with blue eyes and it’s Day 6. If they are alive tomorrow morning then I’m the 7th person out of 7 with blue eyes on this stupid logic island with damn taboos and volcano rituals and oracles. I’m toast.” The next morning (January 7th) everybody with brown eyes wakes up and sees nobody with blue eyes. Yah! Party!
Right?
Here’s my thinking:
Jan. 1 → Oracle delivers the news. If N=0, everybody kills themself. If N=1, that poor slob kills himself as he doesn’t see any blue eyes and trusts the oracle.
Jan. N → If N /= 1 and N/= 0, people wake up and see (N minus 1) blue eyed guys walking around (if you have blue eyes). Before this day, you’ve only seen the date minus 1 blue eyed guys walking around. If you DON’T have blue eyes, you’re hoping this is the day that these guys kill themselves, otherwise you must have blue eyes, too. However, they save your from this error, by offing themselves that evening. Why? Again, if you have brown you figure you have one more day as you see N blue eyes. If you see N-1 blue eyes on day N, you know you’re the Nth one.
“Before this day, you’ve only seen the date minus 1 blue eyed guys walking around.”
That should read “If you have blue eyes, then before this day, you’ve only seen the date minus 1 blue eyed guys walking around. Thus, you know that (as do all blue eyed islanders) that on January N, the fact that N-1 blue eyeds are alive means you are one of them. You all die that night.”
Anybody? Am I wrong? I seem to be missing something obvious or do I have the solution?
But of course he does. Because on day 7 he knows that there are at least 7 blue eyed people (otherwise there would have been a bunch of suicides before that day) and he can clearly see that there are are no more 7 blue eyed people. Triskadecamus’ solution makes perfect sense to me.
Here is what I see as a potential flaw. To simplify things, let us only consider the case where there are two blue-eyed people.
The standard argument is as follows. On the first day, each blue-eyed person sees one other person with blue eyes, but doesn’t know that he himself has blue eyes. So they both wait till the next day. When they each see that the other has not committed suicide, they must conclude that they also have blue eyes. So they both killed themselves.
Implicit in this argument is the assumption that neither has enough information to conclude that he has blue eyes without waiting till the next day and observing that the other has not committed suicide. Why is that the case?
Whenever I ask this, people answer that they would need some other means of information like reflective surfaces or being able to talk about eye color, which are forbidden. Yes, communication and direct self-observation are forbidden in the problem statement to prevent easy ways of finding out one’s eye color. However, it is not clear to me that there is no way for the blue-eyed islanders to deduce that they have blue eyes, on the first day, using solely the information they do have, namely, the oracle’s announcement and their observations of the other blue-eyed islanders. You need to prove that no proof is possible for the islanders. How do you do that?
If there are only two blue eyed people and given the assumptions in this problem:
On day 1, if you’re one of the brown eyed people, at most you can conclude that there are two or three blue eyed people. If you’re one of the blue eyed people, at most you can conclude that there are one or two blue eyed people. So on day 1, no one can be sure if he has blue eyes on not. This reasoning is airtight as far as I can see, and there’s no need to find a proof to the contrary. If “x” has already been proven, why look for a proof of “not x”?
But you have not given a reason. You have simply stated that as fact. To me, it is intuitive that it should be true. However, it is not clear that it must be true.
OK, suppose that there are exactly two blue-eyed islanders, and I’m one of them. On the first day, the information I have is this:
I see one person with blue eyes.
It is public knowledge that at least one person on the island has blue eyes.
Now, if that other schlub were the only blue-eyed islander, the information I would have would be this:
I see one person with blue eyes.
It is public knowledge that at least one person on the island has blue eyes.
In other words, the information I have is consistent with a scenario wherein I don’t have blue eyes, so I cannot deduce, given that information, whether or not I have blue eyes. I can only deduce that if I have some other source of information about eye color, and the problem statement says that I do not have any other source of information about eye color.
It is not just intuitive; it is a fact in the sense that it is only logical conclusion given the problem’s assumptions. There is no reliance on intuition here. On day 1, if you’re one of the two blue eyed people, there is insufficient information to conclude your own eye color. Chronos’ explanation is a good one too.
OK, that makes sense; I think this can be generalized. It seems less clear in the induction step, but I think it can be done. I am trying to induce two statements simultaneously:
(1) On the nth day, if there are n blue-eyed, they will all commit suicide.
(2) On the nth day, if there are more than n blue-eyed people, they will all still be alive on the (n+1)th.
I think the problem is that (2) is assumed implicitly, and never made part of the induction argument.
Just saying “it is the only logical conclusion” is not an argument.
However, I agree with you now that it must be true. I think this statement of Chronos’s is the key.
Consider two situations:
- There are two blue-eyed people A and B.
- There is one brown-eyed person C and one blue-eyed person D.
C’s view of his situation is the same as A’s view of his. So if A can deduce that he has blue eyes, then C must be able to make the same deduction, which is erroneous.