I unerstand that and in that scenario the other person (who is the only one with blue eyes) would be the only one to know his eye color and leave? leaving us 2 there because we haven’t figured it out yet? i think that is right?
But what if we can’t consider him the only blue eyed person because we’re already lookin at 3 in total? is it still simply resolvable? or could I use this answer to still figure it out?
I think these are key to coming to an intuitive understanding of why the logic works.
How does he know for sure to start off with that there are exactly 99 people with blue eyes? at the starting point I only knew there were at least one?
I’ve read every post in this thread and still don’t get it. I understand how it works if there are 1 or 2 blue-eyed people, but not if there are more than 2.
(Besides, if all of these people were so logical, they would deduce that the rules were ridiculous, and would ignore them, and would all go on the boat. Assuming that they want to leave the island, of course. They might be perfectly happy there.)
He doesn’t know for sure that there are exactly 99 blues; he knows for sure that there are at least 99 blues (and possibly 100, if he himself is one). And he knows this by looking around and counting the people he sees with blue eyes.
So if instead the Guru had said, “I see someone with blue eyes and I see someone with brown eyes.” Does everybody eventually get to leave then (except the Guru) or nobody?
Without the Guru’s statement you wouldn’t even be able to prove P(1) in the induction proof. What the Guru says is key to the proof. Note that the Guru could’ve said that there are at least K blue-eyes as long as K<=N, the actual number of blue-eyes, and the same logic works. The Guru sets the induction proof (and the chain of inferences) in motion.
No, you haven’t missed the point. It’s supposed to be obvious. Despite everyone on the entire team knowing 19 out of 20 answers, and everyone knowing that everyone knows 19 answers, and everyone sharing 18 questions in common with everyone else on the team, it’s possible that the entire team will get everything wrong.
Why? Because their ignorance stacks up. They can’t change the ignorance of their predecessors, despite knowing the error of their answers. It’s the same as the islander puzzle.
The reason you can’t go through more than two people is because it’s simply “hard to think about”. The logic doesn’t change, it’s just that that’s usually the limit of our ability to think as others think. “I know” is easily interpreted. “I know you know” is similarly simple. “I know that you know that he knows” is a little tougher, but not much. But once you get to “I know that you know that he knows that she knows,” it starts to get complicated.
If you’re on the island and you see two blue-eyed people and they don’t leave on the second night, what does that mean? Like you, they both understand how it works for one or two people. But they haven’t left because they each see two. Meaning there must be a third. You.
Then, if you see three blue-eyed people and they don’t leave on the third, what does that mean?
Yes, but you’re describing the “how”, not the “why”. I am not getting, from your post, what explicit information the Guru is providing.
The non-inductive explanations, it seems to me, do answer that question.
Allow me to expand on my earlier pattern explanation. Imagine four islanders. They’re asked to write down what the person ahead of them is considering possible for the island’s eye pattern. They have to write all the possible patterns that the guy ahead of them wrote. They’re all blue, btw. D starts:
D: Either [1111] or [1110].
C: Either [1111, 1110] or [1101, 1100]
B: Either [1111, 1110, 1101, 1100] or [1011, 1010, 1001, 1000]
A: Either [1111, 1110, 1101, 1100, 1011, 11010, 1001, 1000] or [0111, 0110, 0101, 0100, 0011, 0010, 0001, 0000]
Now here’s the expansion. A reasons that B may have written [0111, 0110, 0101, 0100][0011, 0010, 0001, 0000], which is the case that A’s eyes are brown. B hasn’t written that, but A doesn’t know that. Of those eight, four have the erroneous attribute that B is a brown-eye. He wrote that because he’s ignorant of his own eye color. That set is:
[0011, 0010][0001, 0000]
B had to write that, A reasons, because he thought C might have written it. C didn’t write that, of course, but A doesn’t know the first slot’s wrong and B doesn’t know the second is wrong. B didn’t ever consider this a possibility either. It’s just that A doesn’t know that.
So to recap, A thinks it possible that B thought there’s a chance that C wrote [0011, 0010][0001, 0000]. What does that mean, again? That C thinks D could’ve written 0001 and 0000.
Now here’s the important part. The guru’s announcement says outright that “0000 is not a correct eye pattern”. Once the announcement has been made, A writes that B thinks that C wrote that D thinks either [0011,0010] or [0001, [del]0000[/del]].
This is significant because it narrows a set down to only one possibility. In that case, D would know his eye color because he’s deduced the pattern. When they wake up the next morning and D is still there, C (the one in A’s imagination ,not the real one) can no longer claim to believe D may have ever written [0001, 0000]
Now read what’s remaining in C’s assessment of D’s guess - [0011,0010]. Note that C’s eye color is the same in all patterns. That means C (remember, this is all A’s thoughts, not the thoughts of any actual ‘C’ person) should be able to deduce his eye color on the morning of the first day.
And yet, one the morning of the second day, C is still there. Wait, so that means C never actually assessed [0011, 0010][0001, 0000] at all. A now thinks that B will (possibly) reason this and realize that what C actually wrote could’ve been [0111, 0110][0101, 0100] in the first place. In all four cases, B’s eyes are blue and he can then leave the island on the next night.
But on the morning of the third day, B is still around. It’s then that A realizes that [0111, 0110][0101, 0100] was never written. What was really written way back in the beginning on B’s paper was [1111, 1110, 1101, 1100, 1011, 11010, 1001, 1000], not [0111, 0110, 0101, 0100, 0011, 0010, 0001, 0000]
Of course, everyone thinks in the first person, so they’re all reasoning from the point of view of A. So they’re effectively all character A simultaneously. And they’ve all just learned their eye color. So it’s time to pack up on the fourth night.
Huh? It is exactly what’s stated in the puzzle. The information is that there is at least one blue-eye on the island. If the Guru does not provide that information how else will all the islanders know it under all possible scenarios? They’re not allowed to communicate with each other in any way at all. Without the Guru’s announcement, if there’s only one blue-eye on the island he can’t conclude that he has blue eyes on day 1. If there are two blue-eyes neither can conclude that they’re the only blue-eyes on day 2. And so forth. Even without using induction the solution is intuitively straight-forward. Induction just formalizes the solution to work for an arbitrary number of blue-eyes.
The biggest flaw I can see with this riddle, as originally worded, is that there is no mention of whether or not the Guru is telling the truth, and furthermore, no rule about what the other islander’s know about the Guru’s truthfulness.
They are perfect logicians and therefore must understand the possibility that the guru’s statement is either true or false.
So, am I missing something here or was it just not worded well?
As far as I can tell, no one gets to leave the island because no one knows whether the Guru is telling the truth. It does say at the end that the answer does not rely on anyone lying, but it doesn’t say that the solution relies on the Guru telling the truth either.
I guess I am just nitpicking. Neat riddle though and the explanation and understanding of the solution is so much fun!
For N>2, the guru isn’t providing the information that there is at least one blue-eye on the island, because all of the islanders already have that information.
As far as I can tell the Guru is simply saying “okay people with blue eyes: start counting the nights as from tonight” otherwise they wouldn’t know when to start counting from. Is that right? Otherwise I totally understand the solution.
No. That has nothing to do with it. The point is that “0000…000” is no longer an option for the highest-level thought chain. Before the announcement, it’s present as an option, as I detailed a few posts ago.
True. The actual information he’s providing (even with N<=2) is that “Now everyone knows that everyone knows that…everyone knows there is a blue islander.” That’s the true catalyst.
Let me try saying in somewhat different words what others (particularly Chessic Sense) have already said: Before the Guru speaks, each islander believes that it is possible that another islander believes it is possible that another islander believes … [iterated an appropriate number of times] that nobody on the island has blue eyes. As long as this belief is out there, the induction can’t start. After the Guru speaks this belief is no longer possible because now everyone knows that everyone knows [to any level you want] that at least one islander has blue eyes. This is what starts the clock on the induction.
For N > 1, even.
You are correct; as worded, the problem implicitly assumes that it is common knowledge that the Guru speaks only truth.
On the one hand, I would say you are nitpicking here. On the other hand, there are other, quite similar problems, where I would say this exact kind of nitpickery is terribly important. So maybe I shouldn’t call it nitpickery at all.