I believe that 4 believes that 3 sees 2 and 1.But then we're ignoring the rest of what I see, as 5.
I believe that 4 believes that 3 believes that 2 sees 1.
I believe that 4 believes that 3 believes that 2 believes that 1 sees none.
The outsider's pronouncement doesn't add any information. Everyone already knew that there was at least one blue eyed person.Your problem is that you're assuming that these two statements are equivalent, when in fact they are not.
Does someone have a really simple, plain English solution?Forget about 100 people. Let's say it's you and Cindy on this island, and Cindy's got blue eyes.
That's true, but everyone already know that everyone knew there was at least one blue eyed person.That's not correct. It really has nothing to do with the issue.
What they needed was an agreed upon time to start the inductive process.
Of course they did.No, they didn't.
OK, if they only kill themselves the next day if they KNOW their own eye color, then no one dies.Well, no. You're right that the browns do not die, using the phrasing of the question in this post, since, as you point out, they still don't know if they've got brown eyes; all they know is they don't have blue eyes.
Any way, if you break it down into 4 people it seems easy to solve.No, sorry, you're still wrong. They're not "assuming", they're deducing.
4 villagers, green, blue, blue, brown.
day 1: they all notice that the other villager(s) with blue eyes don't kill themselves
day 2: nothing happens, since they still don't know what their own eye color is. they cant assume it's blue based on the comment. move along...
So, am I correct in thinking that every single brown-eyed person has plans to kill themselves on day 101 that they are able to call off because of the mass suicides on day 100?If they know that everybody either has brown eyes or blue eyes.
“All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).”That's an instruction to disregard the sort of mental gymnastics you keep bringing up. You're not meant to regard it as some sort of social thought-experiment, as Tao says in the web page it's expositing the Common Knowledge concept of logic and set theory.
“Deadlock can be avoided if certain information about processes is available in advance of resource allocation.”dgebellak : The part that confused me the most was how the visitor gave them any new information.
Day one rolls around and everyone on the island can see at least see at LEAST three blueys. Let's call them A, B, and C. Any given observer will reach the following conclusions:That holds true...until after 2 days when 2 of those 3 haven't done the deed.
A can see both B and C so he has no reason to suspect that he's a bluey
B can see both A and C so he has no reason to suspect that he's a bluey
C can see both A and B so he has no reason to suspect that he's a bluey
Because of this mutual deadlock nobody is expected to commit suicide and nobody does.
But in (iii) is where I think a mistake is made. If there are 3 or more BEP, such as in (iii), then each BEP knows something they do not in (i) or (ii), which is that every BEP (regardless of their own eye colour) knows that there is at least one other BEP.Yes, but they don't know that every BEP person knows that every BEP knows that there is at least one BEP.
The visitor didn't give them new information really, just an unspoken common point in time to pace everything off of.He definitely gave them new information.
Hypothesis: The m blue-eyed people I can see committed suicide m days after quonsar stirred shit up. This means that each one firmly concluded that there were m blue-eyed people on the island including himself. If I had blue eyes they would have concluded that there were m+1 blue-eyed people on the island the way I would if I had blue eyes. Therefore I must not have blue eyes.It's counter-intuitive like most induction proofs and it's only true in the contrived narrowly-defined parameters of the puzzle but it's genuine induction.
Basis Step, m=1: The 1 blue-eyed person I can see committed suicide 1 day after quonsar stirred shit up. This means that he firmly concluded he was the only blue-eyed person on the island. If I had blue eyes he would have concluded that there were 2 blue-eyed people on the island the way I would if I had blue eyes. Therefore I must not have blue eyes.
I, B, see three blue-eyes. [Recall, in A's thought experiment, A has supposed they themself are not blue-eyed. The hypothetical B will thus know that A's eyes are not blue.] I will perform a thought experiment. Suppose I am not blue-eyed. Then C will reason thus:I, A, have finished my thought experiment. If I have non-blue eyes, either C, D and E will kill themselves on day 3, or B, C, D and E will kill themselves on day 4. Since I have more information than B did in my thought experiment, I know that C, D and E will not kill themselves on day 3. So if I have non-blue eyes, B, C, D and E will kill themselves on day 4. If they do not do so, then I have blue eyes, and I will kill myself on day 5 (along with B, C, D and E).I, C, see two blue-eyes. [Recall, A is supposing they themself are not blue-eyed, and in A's thought experiment, B is supposing they themself are not blue-eyed.] I will perform a thought experiment. Suppose I am not blue-eyed. Then D will reason thus:I, B, have finished my thought experiment. If I have non-blue eyes, either D and E will kill themselves on day 2, or C, D and E will kill themselves on day 3. Since I have more information than C did in my thought experiment, I know that D and E will not kill themselves on day 2. So if I have non-blue eyes, C, D and E will kill themselves on day 3. If they do not do so, then I must have blue eyes, and I will kill myself on day 4 (along with C, D and E).I, D, see one blue-eye. [A has supposed they themself are not blue-eyed, in A's thought experiment B is supposing they are not blue-eyed, and in B's thought experiment, C is supposing they themself are not blue-eyed.] I will perform a thought experiment. Suppose I am not blue-eyed. Then E will reason thus:I, C, have finished my thought experiment. If I have non-blue eyes, E will kill themself on day 1, or D and E will kill themselves on day 2. Since I have more information than D did in my thought experiment, I know that E will not kill themself on day 1. So if I have non-blue eyes, D and E will kill themselves on day 2. If they do not do so, then I must have blue eyes, and I will kill myself on day 3 (along with D and E).I, E, see no blue-eyes. But I know there is one. I must be it. I will kill myself tomorrow.I, D, have finished my thought experiment. If I have non-blue eyes, E will kill themself on day 1. Thus, if E does not kill themself on day 1, I must have blue eyes, in which case I will kill myself on day 2 (along with E).
...It is Tao's proof which assigns the assumption to each BEI:
All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).
for the purposes of this logic puzzle, “highly logical” means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.
Each blue-eyed person will reason as follows: “If I am not blue-eyed, then..."Factoring in uncertainty, this is n = 3
1) In his thought experiment, B has proposed a C who can only see two blue-eyes.No, tkolar, he (being A) does not know that B knows that no one on the island can be seeing less than three blue-eyes.
2) In reality, he knows that B knows that no one on the island can be seeing less than three blue-eyes.
"I, C, see two blue-eyes."And now we have a problem. A [who is thinking all of this] knows for a fact that C can see at least three sets of blue eyes. A also knows that B -- having started with the exact same starting conditions -- knows the same thing.
2) In reality, he knows that B knows that no one on the island can be seeing less than three blue-eyes.Just in case it is not still clear why this is a false statement, I want you to think about it a little more:
with 99 BEIs, there's a whole lotta suppositions to assume, and whose implications to work out.First, so what? Second, induction is a standard method that allows you to work out many, many implications in one fell swoop, without needing to work them all out individually.
Is there a solution that doesn't rely on inducting from n=1 or 2 or 3 or assuming that the basic pattern remains true of large numbers?Yes. Manually work out all of the implications, each one individually, exactly as samw did.
2) In reality, [A] knows that B knows that no one on the island can be seeing less than three blue-eyes.No, tkolar, he (being A) does not know that B knows that no one on the island can be seeing less than three blue-eyes.
A has run into a situation where his hypothetical B is using the existence of a "only-sees-two-blueys" C in a thought experiment, when he knows that the real B would never accept the existence of an "only-sees-two-blueys" C.No. He does not know that.
If I am not blue-eyed, then there will only be n-1 blue-eyed people on this island, and so they will all commit suicide n-1 days after the traveler’s addressIs there a solution besides manual calculations that doesn't rely on that assumption? My essential question is given uncertainty and a large n, can it be shown that there is no stalemate where A cannot choose between two competing hypothesis?
Don't the first two blue-eyed citizens commit suicide on the second noon after the second one comes into consciousness?You mean if nobody makes an announcement like the stranger made? If so, then no.
Why is an outsider needed?Well, think about it from the point of view of one of them. Let's say the first.
So, the inductive proof assumes that the pattern holds without regard for the magnitude of nNo. No, it does not.
inductive proof isn't a mathematical proofNo offense, but try telling that to a mathematician. You should count yourself lucky if he only laughs behind your back.
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posted by trol at 7:55 AM on February 15, 2008