If two people had blue eyes, they would each be expecting the other to leave on the first day, and when that didn't happen, would be able to deduce that they must have blue eyes as well. Through induction, we get 50 people leaving on the 50th day, because they each know of 49 people with blue eyes, and yet no one left on the 49th day.
Dennis Deems wrote:I don't understand why it's the 50th day. What do days have anything to do with it?
If two people had blue eyes, they would each be expecting the other to leave on the first day, and when that didn't happen, would be able to deduce that they must have blue eyes as well. Through induction, we get 50 people leaving on the 50th day, because they each know of 49 people with blue eyes, and yet no one left on the 49th day.
I don't understand any of this. Again with the days. Huh?
Greg Charles wrote:@Paul, I'm not sure what you mean by the hypothesis that there is one blueeyed islander.
Greg Charles wrote:@Paul, it quantizes the decision process. Otherwise the induction doesn't work. In the specific example I gave, all islanders would either see 49 people with blue eyes, or 50. No one leaves on the 49th day, which would tell each of the blueeyed islanders that there are 50 people with blue eyes, and since I (one of the 50) can only see 49, that means I'm one of them. When they all leave on the 50th day, the browneyed islanders would realize that they don't have blue eyes, but still wouldn't know what color their eyes are, so they would be stuck.
Spot false dilemmas now, ask me how!
(If you're not on the edge, you're taking up too much room.)
Paul Clapham wrote:The whole thing doesn't quite make sense to me either. Each person with blue eyes can easily tell that there are either 49 or 50 blueeyed islanders, and they can easily tell that the other islanders can draw the same conclusion. Likewise each person with brown eyes can tell that there are either 50 or 51 blueeyed islanders. So the hypothesis "There is 1 blueeyed islander" can never be entertained by any islander at any time.
Jeff Verdegan wrote:Yeah, I'm stuck on that bit too. Why does the Guru's statement that "there is at least one blueeyed" person matter? Everybody on the island already knows this, and everybody on the island already knows that everybody knows this
There are only two hard things in computer science: cache invalidation, naming things, and offbyone errors
(1) If a person discovers that there is at least one person with eye colour X, but cannot see any people with eye colour X, then that person can conclude that their eye colour is X.
(2) If nobody can conclude that their eye colour is X after the first day, then there must be at least two people with eye colour X. (And so if a person only sees one such person, they can conclude their eye colour is X.)
fred rosenberger wrote:
Jeff Verdegan wrote:Yeah, I'm stuck on that bit too. Why does the Guru's statement that "there is at least one blueeyed" person matter? Everybody on the island already knows this, and everybody on the island already knows that everybody knows this
what if there was only one person on the island with blue eyes? How would he know? All he knows is that 99 people have brown eyes. If he isn't told "There is at least one person with blue eyes", how does he know if his eyes are blue or brown?"
Paul Clapham wrote:Well, it's like this. (I think.) You have to look at the problem as if there is an archipelago of islands which have various numbers of people with various eye colours, who sit around thinking logically while they aren't harvesting their coconuts. And these people can only act based on theorems which are true on all islands (what we called "analytic" in my longago predicate logic course). And one of the theorems applicable in this question is:
(1) If a person discovers that there is at least one person with eye colour X, but cannot see any people with eye colour X, then that person can conclude that their eye colour is X.
That's the basic theorem on which the tower of induction is built which causes the 50 blueeyed people to leave on day 50. The next one up in the tower is:
(2) If nobody can conclude that their eye colour is X after the first day, then there must be at least two people with eye colour X. (And so if a person only sees one such person, they can conclude their eye colour is X.)
And so on by induction until the 50th day.
Now yes, it's true that on this particular island everybody can discover for themselves that there is at least one person with blue eyes and at least one person with brown eyes. But does this count as "discovery" for the purpose of Theorem 1? You think that it does. But for the premises of Theorem 1 to hold, that person must have learned of the existence of the person with eye colour X from some outside source, rather than from personal observation. So that implies that it doesn't.
Jeff Verdegan wrote:
Now comes the part that makes me reach for the scotch..
Now yes, it's true that on this particular island everybody can discover for themselves that there is at least one person with blue eyes and at least one person with brown eyes. But does this count as "discovery" for the purpose of Theorem 1? You think that it does. But for the premises of Theorem 1 to hold, that person must have learned of the existence of the person with eye colour X from some outside source, rather than from personal observation. So that implies that it doesn't.
I'm gonna have to mull that over for a while.
Paul Clapham wrote:
Here's my point of view: this puzzle was obviously invented by somebody considerably smarter than me (it has Smullyan written all over it but I haven't searched out its provenance). And therefore they would surely not have overlooked the obvious things which people are saying in this thread. Therefore those obvious things are most likely wrong. And so I'm going to look for reasons why they are wrong. So far that's the best I can come up with.
Jeff Verdegan wrote:
Paul Clapham wrote:Well, it's like this. (I think.) You have to look at the problem as if there is an archipelago of islands which have various numbers of people with various eye colours, who sit around thinking logically while they aren't harvesting their coconuts. And these people can only act based on theorems which are true on all islands (what we called "analytic" in my longago predicate logic course). And one of the theorems applicable in this question is:
(1) If a person discovers that there is at least one person with eye colour X, but cannot see any people with eye colour X, then that person can conclude that their eye colour is X.
Right. And we already know that on this island there is nobody who "cannot see anybody with eye color X", and, furthermore, everybody knows this fact.
That's the basic theorem on which the tower of induction is built which causes the 50 blueeyed people to leave on day 50. The next one up in the tower is:
(2) If nobody can conclude that their eye colour is X after the first day, then there must be at least two people with eye colour X. (And so if a person only sees one such person, they can conclude their eye colour is X.)
And so on by induction until the 50th day.
Yup, got all that.
Dennis Deems wrote:
Well I sure don't. I can't see what is suddenly known on day 50 that wasn't known on day 1. And I can't see how the fact of no one boarding the boat on day two means anything different than the fact of no one boarding the boat on day one.
Jeff Verdegan wrote:
So D3 rolls around, and, since nobody left on D2, everybody knows that everybody else saw at least 2 BLs. So the 3 BLs that are there look around, and they each see only 2 BLs, so they each know that they must be the third, so all 3 BLs leave on D3.
And so on.
eileen keeney wrote:
if X=1, the person with blue eyes knows that X=1.
fred rosenberger wrote:
eileen keeney wrote:
if X=1, the person with blue eyes knows that X=1.
And this is ONLY true if the Guru says "at least one person on the island has blue eyes". If he doesn't say that, then the one person with blue eyes doesn't know if X=1 or 0. Therefore, he can't know if he should go to the boat on the 1st day or the X+Y day.
fred rosenberger wrote:
eileen keeney wrote:
if X=1, the person with blue eyes knows that X=1.
And this is ONLY true if the Guru says "at least one person on the island has blue eyes". If he doesn't say that, then the one person with blue eyes doesn't know if X=1 or 0. Therefore, he can't know if he should go to the boat on the 1st day or the X+Y day.
There are only two hard things in computer science: cache invalidation, naming things, and offbyone errors
Jayesh A Lalwani wrote:On the first day, if no one left, all the blue eyed people should have deduced that they themselves are blue eyed and left on the second day.
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