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There is a logic puzzle aiming on freeing same-color-eyed people from an island. The thing is that they must be certain of their own eye color so that they can leave. For that reason an external party must provide a [single] statement containing no piece of new information to the crowd.

This puzzle is described in the following link:

The answer provided is, basically (disregarding the eye color, which could be green or blue) : "At least one of you has blue eyes", or "I can see that at least one of you has blue eyes". It could even be "At least n-1 of you has blue eyes".;

However, when I introduced this puzzle to a few friends of mine, one of them mentioned that the information : "Your neighbor has blue eyes"

He argues this is no new information since it will require inferences. I argue that it means the following:

(1) For each one of you, there is a neighbor having blue eyes;

Which could mean the same thing as:

(2) To all of you, there is a neighbor having blue eyes;

And allows the following inference:

(3) For each one of you, there is someone with blue eyes;

I am not a big Math expert, so maybe I wouldn't be able to precisely represent that in symbols, but the discussion revolves on the following: if (1) has the same value as (2) & (3) and if 3) contains new information, then (1) & (2) also contain new information.

Since the information cannot be conveyed to a single person on the island, I am uncertain if that would be a valid solution regarding the novelty of information and regarding the apparent individual-based message.

What do you think:

1- Are the statements (1),(2) & (3) equivalent?

2- This I should maybe post on the philosophy forum: Is the statement (1) valid, since it seems to be directed to every single individual instead to the collection of individuals at once?

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  • $\begingroup$ Your statements for (1), (2) and (3) are rather imprecise and I'm not too sure what to make of them. What exactly do you mean by "For each one of you, there is a . . . . pair of blue eyes"? Do you mean to say that "the number of pairs of blue eyes" is equal to the number of people? The condition of "being neighbors with" is also a rather imprecise condition. What do you take to be "a neighbor"? $\endgroup$ – EuYu Mar 9 '16 at 17:11
  • $\begingroup$ Yes, I used "a pair of blue eyes" assuming every person has two eyes. The word "pair" just added extra difficulty. I might correct it. Sorry @EuYu. $\endgroup$ – ClayKaboom Mar 10 '16 at 15:49
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It depends on whether the people are sitting in a circle at the time of the announcement, which is the set-up for some forms of the puzzle, and on whether "neighbor" means an adjacent person in the seating arrangement or only "another resident of the island".

For $n$ people all with blue eyes, sitting in a circle, a public announcement that "each of you has at least one [adjacent] neighbor with blue eyes" implies common knowledge that at least $\lfloor \frac{n}{2} \rfloor$ of them have blue eyes. By the same argument as the original puzzle this causes a complete exodus from the island at day $\lceil \frac{n}{2} \rceil + 1$, but if there are subtle additional implications specific to this variant maybe it could happen on an earlier day. I think it does not.

If they are just people in no particular arrangement, and "neighbor" means "some other resident of the island", the announcement is the same as "there are at least $2$ blue eyed people on the island".

The answer provided is, basically (disregarding the eye color, which could be green or blue) : "At least one of you has blue eyes", or "I can see that at least one of you has blue eyes". It could even be "At least n-1 of you has blue eyes".

The answer is that the new information is "common knowledge that at least $k=1$ people have blue eyes". This starts the induction argument at $k=1$ that leads to mass exodus at step $n$. If the announcement names a larger number of blue eyes such as $k=2$ or $n-1$ then the induction begins at at that value and this accelerates the process.

Common knowledge means more than everyone knowing some Fact, it is that all finite chains of "A knows that B knows that C knows that ... knows Fact" are true. The assumption is that public announcement of a Fact makes it common knowledge to those who are present at the announcement.

1) For each one of you, there is a neighbor having a pair of blue eyes;

2) To all of you, there is a neighbor having a pair of blue eyes;

3) For each one of you, there is a . . . . pair of blue eyes;

Unless they sit in an arrangement where one person in the center is everyone's "neighbor", (2) can only mean the same as (1). Statement (3) means that for each person there exists a different person with blue eyes this is equivalent to the existence of at least $2$ blue eyed persons.

Is the statement 1) valid, since it seems to be directed to every single individual instead to the collection of individuals at once?

If the statement is made as a public announcement then the chain of common-knowledge inferences begins. If it is only made privately to each person then if $n>2$ there is no new information, and nothing happens as a result.

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