I don’t see why you think everyone will die. No brown eyed islander can can prove that the statement “I have green eyes” is false.
Suppose there are 4 people, all with blue eyes. For simplicity of language, let’s say they all assume they have brown eyes until proven otherwise.
A thinks B sees 2 blues (namely, C and D).
A thinks B thinks C sees 1 blue (namely, D).
A thinks B thinks C thinks D sees no blues.
“Sees 2 blues (namely, C and D)” is surely equivalent to “thinks C sees 1 blue (namely, D)” and “sees 1 blue (namely, D)” is exactly the same as “thinks D sees no blues.”
If you disagree, exactly whose blue eyes do you think A thinks B thinks C thinks D sees?
Yeah, no kidding. But that’s only because you took it through 2 people- you asked what A thinks about C’s thoughts. No one thinks anyone sees a pattern with 2b/3r either. But A can’t rule it out if you ask him what he thinks of B’s thoughts about C.
I’m surprised that on page 5, you’re still not getting this. YES, B does think C could have written down brrrr, because E doesn’t know slot 5, D doesn’t know slot 4, C doesn’t know slot 3, and B doesn’t know slot 2. So by the time B.C.D.E rolls around, the only certain slot left is A.
All you’ve demonstrated is that C did not, in fact, write that down. We already know that.
It’s like you have B writing down “brrrr” and scratching it off because he thinks “well, no one could be thinking that”. But that’s because your imaginary B is answering a different question that he was asked. He’s trying to figure out what could be on C’s paper, not what C actually thinks or what D actually thinks or what E actually thinks. And there’s no way he could possibly rule out C thinking that D wrote down “brrrr”- the first two being definite answers, but unknown to B, the middle one being unknown to C, and the last two being due to the ignorance of the other two parties.
What B should do in your scenario is not scratch out “brrrr” but instead say “I know for a fact that this is wrong, and so do D and E, but C doesn’t know that they know. So I have to respect the fact that C may have made this error.” and he should write it down as a possible C answer.
Or, alternatively phrased, B should go “I know D didn’t actually write brrrr down, but that’s my knowledge because I know C’s eye color, just like D does. But there’s no way C could know that. C either wrote down bbrrr or brrrr, and I can’t tell you which one it is without knowing my own eye color. So it’s possible that she wrote brrrr.”
Consider the case that B actually is brown. Why would his logic differ? He doesn’t know the difference. So why would his paper look any different? By your reasoning, he would know the difference and brrrr is possible again. That’s granting B knowledge that he doesn’t actually have.
Imagine two islands, one that’s bbbbb and brbbb. They look the same to both Bs. To those two Bs, the world is identical. So how come one rules out brrrr and the other one doesn’t? Or if they both rule it out, then isn’t the brown-eyed B wrong? Because C actually does think that brrrr is possible?
Is this a joke? In case you’ve forgotten, the islanders can only have two eye colors. So yes, they can rule out green eyes because that would be impossible.
It takes no logic to deduce. It was stated as given information in the original puzzle.
The original puzzle does state that all the islanders have either blue or brown eyes, but it does not state that the islanders know that all the inhabitants (including themselves) have either blue or brown eyes (at least as Dr. Love posted it). How could they, given the taboo about discussing it?
It’s funny that, in a sense, what begbert2 is doing is failing a test for theory of mind, like one of those puppet show interviews psychologists give to little kids to see if they can recognize that other people might not have the same beliefs as them. (Granted, this is a much trickier test than your typical Sally-Anne skit, but the analogy still amuses me.)
No, because I can’t wrap my head around what the heck you’re talking about. It probably doesn’t help that I don’t care much.
What we have here is two ropes tied to the one rock - a ten foot one and a two-hundred foot one. We hold both ropes and drop the rock. You’re pointing at the long rope and saying that based on it the rock will reach a hundred feet or more. I know better because there’s a short rope, so it doesn’t matter what esoteric approach you use to show that the long rope is infinitely long.
Also, let me ask again what the killing process actually is? I seem to vaguely recall some esoteric explanation about how on each day the lack of a death causes all the villagers to learn something, thus incrementally whittling down the possibility tree one little step at a time.
Of course, for this explanation or any like it to be true, that requires the villagers to think that anybody ever believed that somebody was going to die that day in the first place. If everyone knew that it was impossible for anyone to die that day, then no additional information can possibly have been imparted by the lack of deaths.
Okay, so we’ve established that each villager may use the induction to deduce that the day person dies is the same as the number of blues they see, plus 1. This the inductively provable end result and is true regardless of what color the villager-in-question’s eye color is, and regardless of what esoteric thought processes they might be using to arrive at their conclusions.
So. A villager looks around and sees 4 blues, which tells him that everyone in the tribe sees, at a minimum, 3 blues. (The full range of possibilities includes 3, 4, or 5 blue eyes seen, dependent on the number of blues and the eye color of the viewer.)
Having established that everyone in the village sees 3 or more blues, our villager can be certain that each person has performed the same calculation regarding the people they see, and so everyone in the tribe knows that nobody in the tribe sees less than two blues. (The full range of possibilities has expanded to 2, 3, 4, or 5, dependent on the same factors.)
So. Having established that nobody on the tribe thinks that anybody in the tribe sees less than 2 blues, he knows that nobody in the tribe thinks anybody will die on days 1 or 2, because for that to happen somebody would have to actually be seeing 0 or 1 blues. Not thinking that people think that people think they see 0 or 1 blues; somebody would have to actually be seeing 0 or 1 blues. Which our villager just proved nobody does.
Which means that everybody knows for an absolute fact nobody will be dying on days 1 or 2. Which means that nobody will employing any esoteric thought processes that rely on learning of the lack of deaths on those days to whittle down possibilities.
That, or you’re failing a test of “being able to think laterally”. You’ve been taught one long-rope explanation for the villagers’ thought processes, and are unable to think outside of the box to accept proofs that the explanation you have been taught is incorrect, or proofs that the explanation you have been taught cannot possibly be correct (because nothing is learned by anyone on days 1 and 2).
Every post I make in this thread is contractually required to have at least one error. In this case it was doing this step and having the villager to bother to think about what anyone else might be thinking.
It’s a pretty dumb mistake too, since I immediately go on to firmly restate that the day of their deaths can be calculated from what they actually see, without regard to their thoughts at all.
So, a person who sees 4 blue eyes knows that nobody sees less than 3 blue eyes, and based on that alone knows that nobody will die on days 1, 2, or 3.
begbert2, do you agree with the following statements:
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The blue-eyed islanders all commit suicide on day 1 if and only if there is exactly 1 blue-eyed islander.
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If an islander sees N blue-eyed islanders, there are at most N+1 blue-eyed islanders.
As I see it, this is all you need to prove the induction. The fact that nobody learns anything on day 1, 2, or 658,284,184,184 (if it’s a big island) is irrelevant. The inductive statement then is
“The blue-eyed islanders all commit suicide on day N if and only if there are exactly N blue-eyed islanders.”
You’re late to the party; I bought the induction at post 199.
What I don’t buy is the favored incorrect (and extremely implausible) explanation for it. The islanders can do induction too after all.
To say the villagers “learn” something by observing that no one dies on Day 1 is, I would say, not necessarily the right way to put it; I agree that the islanders cannot “learn” anything prior to Day n, so long as there are at least n blue-eyed islanders. However, higher-order knowledge statements which weren’t true before become true afterwards.
The relevant phenomenon is like this:
Suppose it’s Tuesday and I happen to know that Sally has a crush on Tom, but I don’t know whether Tom knows yet. However, I do know Sally intends to tell Tom on Wednesday. Even if I lock myself into a cave shut off from the outside world on Tuesday, the following phenomenon happens, despite my being unable to “learn” anything while in the cave:
On Tuesday night, “I know that Tom knows that Sally has a crush on him” is false.
On Wednesday night, “I know that Tom knows that Sally has a crush on him” is true.
In this sense, then, higher-order statements about knowledge can switch from false to true without requiring one to be able to “learn” anything. This phenomenon is what occurs on Day 1, Day 2, etc., and which may have been glibly described as “learning” by others previously in this thread, despite the valid objection you raise. (However, on the penultimate day of my life, in the island problem, I do finally reach a point where, as I am uncertain whether I have brown or blue eyes, I become uncertain whether others will kill themselves off today or tomorrow, and thus upon learning that they do not kill themselves off today, learn that I have blue eyes and therefore will kill myself tomorrow. That this happens is clear from the fact that I eventually kill myself, as you accept, as I would only kill myself if I happened to learn my eye color (an instance of genuine learning))
all I have to say is that I’m glad other people were as confused (or more) as I was by the islanders riddle. I felt really dumb when I read it and didn’t get it.
As for puzzles, I always liked the ones like:
How many of each animal did Moses take on the Ark?
If a rooster lays an egg on a peaked roof, which way will the egg roll?
If a plane crashed on the border of Spain and France, where would the survivors be buried?
How many months have 28 days?
I never cared for the ones along the lines of your first three myself (there’s not much fun for me in mere “Ha, you didn’t realize that my question makes presuppositions which are highly nonstandard or just plain inaccurate and all I really wanted you to point out was this observation”), though of course others find them fun, and they certainly are far more in keeping with the idea of lateral thinking than most of the math problems in this thread. But I’m not sure what the expected answer for your number 4 is (infinitely many, one for each non-leap year? That seems to me mere gotcha-ism on the unfairly unspecified nature of what counts as distinct vs. equivalent months (i.e., is there a single month January, or a distinct January for each year?, and so on). But perhaps there is another answer you’re looking for).
But islander Sally doesn’t intend to tell islander Tom anything on Wednesday, or any other day, and they all know it. Nothing in the universe changes from day to day, until we get to the day before the villager knows they will have to kill themselves.
So you’ve just highlighted how their set of higher-order knowledge statements can’t change from day 1 to day 2. If the possibilities were going to start collapsing into one another they’d do so immediately, in the first nanosecond of day 0, as each islander logically anticipates their thoughts of future days.
Number 4 seems to be a semi-standard lateral thinking puzzle to me; you have to think outside the box to get the correct answer (which I presume is 12, or “all of them”), very comparably to the connect-nine-dots-with-four-straight-lines puzzle. The main difference is that in a fully-formed lateral thinking puzzle, the wrong answers are clearly wrong, so nobody will ever propose them and thus be slapped back with a ‘gotcha’.
ETA: on reflection, based on this critera the islander problem is a bad lateral thinking puzzle too, with its ‘gotchas’.
Actually, I want to modify my last post slightly: the switching of higher-order knowledge statements from false to true all happens with the missionary’s original statement “I see someone here has blue eyes”, from which no one perhaps can “learn” anything new, for some sense of “learn”, but which nonetheless has an effect, as described.
From then through just prior to the the penultimate noon, though, not even this kind of change happens. You are correct about that. However, the change in epistemological state I have previously been arguing for in this thread is indeed one which occurs; to wit, the one which occurs with the missionary’s statement kicking off Day 1 (which has been the main bone of contention). Note that in my previous reply to Dr. Love concerning it, I did take pains to point out why we should perhaps not consider this “learning”, though it is nonetheless a pertinent change in epistemological state. Apart from that, I have never made any claims about each new day bringing new knowledge, except for at the very end.
(I actually was writing this post before your last post, but it is in agreement with it)
So they postulate all these long chains of impossible possibilities (infinitely many of them, by the way, because they can wonder how many blues A thinks B thinks C thinks A thinks B thinks C thinks A thinks B thinks C sees), and then after engaging in these infinite thought processes they instantly collapse and discard them all? That seems like a roundabout way to get to Rome.
Though it does propose an alternate answer to the riddle:
“They all die on day 100.”
“No, they’re all too busy lost in endlessly recursive chains of thought and so never get around to it. Or rather, they wouldn’t have got around to it, but being all so lost in thought they died of thirst on day 4 instead.”
The key point to me is that the truth value of “X knows [whatever]” is necessarily independent from X’s eye color. If X is in a position to claim [whatever] with certainty in one setup, then he must remain in such a position even if we change his eye color and keep the rest of the setup the same.
And, accordingly, the truth value of “A knows that B knows that C knows that [whatever]” is similarly independent from A’s eye color, B’s eye color, and C’s eye color. If one setup is such that such a statement holds, then even after changing A’s eye color, B’s eye color, or C’s eye color, keeping the rest of the setup the same, such a statement must still hold. Do you accept this?
What I find most strange about people who don’t understand this puzzle, is that they almost always agree it works for 3 blue-eyed villagers.
That is, with 3 blue-eyed villagers A thinks B thinks C sees no blues (before the stranger’s announcement).
But the reasoning involved in the 3-person case is the same as for any higher number. If you aren’t going to understand the reasoning, you should get stuck at 2, not at 3.
With 4 people, A thinks B thinks C thinks D sees no blues (before the stranger’s announcement).
Whose blue eyes could A possibly think B thinks C thinks D sees? There aren’t blue eyes floating around on their own. If you don’t think it works, then just point out whose blue eyes they are.
A thinks B sees 2 blues (namely, C and D).
A thinks B thinks C sees 1 blue (namely, D).
A thinks B thinks C thinks D sees no blues.
I think this is a pretty simple way to put it. Each line follows easily from the previous one.
No, I retract that; I was momentarily confused.
There is a change of higher-order knowledge that occurs on Day 1, Day 2, and so on. To wit, prior to the Day 1 noon, “A knows that B knows that C knows that … Penultimate guy knows that there are at least two blues” is false, and after the Day 2 noon, it becomes true.
It’s just that higher-order knowledge about the particular statement “There is at least one blue” never changes after the missionary’s original announcement, because after the missionary’s announcement it has already become, in the technical sense, common knowledge. (Everyone knows that everyone knows that everyone knows that … it holds). The Day n killings each likewise make “There are/are not more than n blues” common knowledge, and in that sense effect similar changes in epistemological state.