A Blue-Eye Paradox: Solution Not Unique

[Zafa Pi]

Another way to see the difficulty is to note that a logical language cannot in general make statements about whether a statement in that language is provable. That's because we get an infinite regress when we try to refer to wffs in the language. To define a wff we need to give a recursive definition that starts with a fixed finite or infinite set of symbols for constants, variables, functions and predicates. But then how do we refer to a wff in the language? We run into the 'naming vs using' problem. A wff like '1+1=2' is not a reference to the wff '1+1=2', just as I am not my name. So we need to generate some new constant symbols to refer to wffs using the original set of language symbols. But that then means that our original assumption that we were starting with the full set of language symbols was incorrect.

Trying to define a formal language that can refer to itself ends up being like trying to define a set of all sets, and we know what sort of a muddle that leads to.

We can refer to our language in ordinary speech in our language, because ordinary speech is informal. But if we are to talk about 'perfect logicians' we can only use formal languages to specify what their perfection means, and then we get stuck in an infinite regress.

Perfect logicians is a red herring. Healthy young PhDs in math will do. You understand the spirit of the problem so do something constructive and give it your best formulation.

 

[andrewkirk]

I imagine Hilbert said something similar to Godel.

 

[maline]

But 'perfect logician' is unclear so let us expand C. It says

'Each person will deduce whatever can be deduced from B+C in FOPL under L'

C does not refer to a particular set of axioms. It says that "each person, at each stage, has some set of axioms, and will make all possible deductions from those axioms".
For the riddle to work, we must additionally postulate C', namely : " C is in fact included in each person's axiom set". We must also add C'' and C''', defined recursively. But this recursion need not be infinite: it is sufficient that the number of such axioms (including C) be equal to the number of monks. If for instance there are two monks, it is okay if A does not know whether B knows that A is a perfect logician.

 

[andrewkirk]

Now I've gone and confused myself! I'm still fairly sure that the problem cannot be properly stated in formal terms, and yet I think I've figured out a solution. One of those has to be wrong. I'll present my suggested solution, then I'll go back and try to work out if I can see a way to formally state the problem.

When I say a solution, I mean an explanation of why they all have to wait until the 100th day, and what new information is provided at each stage. It turns out that new information becomes available every night when the ferry departs empty, and also on the Oracle's pronouncement on the first day (but not on subsequent days).

I shall use unary predicate C(r) to indicate u\geq r, where u is the number of blue-eyed people, and unary predicate K to indicate 'everybody knows that'. I will use indices to signify repeated applications of K. Thus K^3C(7) means

'Everybody knows that (Everybody knows that (Everybody knows that (u\geq 7)))'

Say there are n blue-eyed people (I'll call each such person a 'blue' from now on). It is not hard to show that we people with eyes of other colours do not affect the calculations, so we will omit them from consideration.

Then each person can see n-1 blues. So we have
$$KC(n-1)$$
A given person P entertains two possibilities, that u=n or u=n-1. In the second case, P will be non-blue, and what do they know about what others know? Each other person P' will be able to see (n-2) blues. On the other hand if u=n each other person P' will be able to see n-1 blues. So P knows that regardless of their own eye colour, each other person P' can see at least n-2 blues. That is, P knows that KC(n-2). Hence we have, since P is arbitrary:
$$K^2C(n-2)$$
Now think about what P knows about what another person P' knows. P knows it might be the case that P is non-blue and if so, for each other person P' it appears possible that P' is also non-blue. For P' considering that case they realize that each person P'' in the group excluding P' and P can see n-3 blues, so they know that u\geq n-3. For P' considering the alternative, where P' is blue, P'' can see n-2 blues. So either way P'' knows u\geq n-2. So P knows that P' knows that P'' knows that n\geq n-3. Whence we have:
$$K^3C(n-3)$$
We can continue this process, noting that at each step the exponent of K and the argument of C add to n.

So the initial state of knowledge of all players can be written as a vector of statements:
$$KC(n-1),\ K^2C(n-2),\ K^3C(n-3),\ ...\ ,\ K^{n-2}(2),\ K^{n-1}(1),\ K^n(0)$$
The ends are the easiest to interpret. The left says that everybody knows u\geq n-1 and the right says that everybody knows - ad infinitum - that there are at least zero blues.

Then the oracle speaks. Everybody hears her, and everybody knows that everybody else has heard her, so it is now the case that, for any natural number r, we have K^rC(1). This enables us to strengthen the last component of the knowledge vector by increasing the argument to C, but does not affect any of the other components. The new knowledge vector is:
$$KC(n-1),\ K^2C(n-2),\ K^3C(n-3),\ ...\ ,\ K^{n-2}(2),\ K^{n-1}(1),\ K^n(1)$$
So the oracle has communicated new information, not about what people know directly but about what they know others know others know others know...

Next, the ferry leaves and, if n>1 then nobody is on it. So now everybody knows that there are at least two blues, and they know that everybody else knows that, and they can prefix the K operator to that as many times as they like. The new knowledge vector is:
$$KC(n-1),\ K^2C(n-2),\ K^3C(n-3),\ ...\ ,\ K^{n-3}(3)\ ,\ K^{n-2}(2),\ K^{n-1}(2),\ K^n(2)$$
The next piece of new info is when the ferry leaves empty on day 2, provided n>2. Then everybody knows (to the umpteenth power) that u\geq 3. So the new knowledge vector is:
$$KC(n-1),\ K^2C(n-2),\ K^3C(n-3),\ ...\ ,\ K^{n-4}(4)\ ,\ K^{n-3}(3)\ ,\ K^{n-2}(3),\ K^{n-1}(3),\ K^n(3)$$
This continues, and each day the knowledge vector changes, by the minimum argument to all C predicates being increased by 1.
The general form of the knowledge vector after an empty ferry departure on day r is a vector of n components, of which the jth is:
$$K^{j}C(\max(n-j,r+1))$$
Each day, the knowledge of what others know others know increases, even though the direct knowledge of each person about the number of blues does not change.

When the ferry departs empty on day n-2 the knowledge vector has arguments with a minimum of n-1. Now, for the first time, everybody knows that everybody knows (etc, etc) that u\geq n-1. But they still don't know if their own eyes are blue. They only discover that when the ferry departs empty on day n-1 and the knowledge vector arguments attain a minimum of n. So, for all r we now have K^rC(n).

So they all leave on the nth day's ferry.

Every day there was new info from the ferry departure. And the first oracle pronouncement was necessary, although the subsequent ones were not.

 

[Dr.AbeNikIanEdL]

And the first oracle pronouncement was necessary, although the subsequent ones were not.

What subsequent pronouncements do you mean?

 

[andrewkirk]

What subsequent pronouncements do you mean?

My mistake. I misremembered the problem statement. I thought it said the oracle made an announcement every day. But now that I check back, I see that she only makes the announcement on one day. So the days would have to be numbered starting with day 1 being the day she made her pronouncement.

 

[Ken G]

But this recursion need not be infinite: it is sufficient that the number of such axioms (including C) be equal to the number of monks. If for instance there are two monks, it is okay if A does not know whether B knows that A is a perfect logician.

And indeed, it need not be the number of monks, only the number of blue-eyed monks!

 

[maline]

And indeed, it need not be the number of monks, only the number of blue-eyed monks

Ah, in the version i heard they were all blue- eyed.
By the way, even if we did require an infinite recursive set of axioms- the union of all finite statements of the form "Each person has in his axiom sat the fact that each person has in his axiom set... that each person, at each stage, will make all possible FOPL deductions from whatever his axiom set is"- I think that would be a legitimate FOPL axiom schema, because each such axiom includes no self- reference.

 

[Ken G]

Ah, in the version i heard they were all blue- eyed.

I see, a more general version has some N blue-eyed people, and then suicide is on day N, etc.

By the way, even if we did require an infinite recursive set of axioms- the union of all finite statements of the form "Each person has in his axiom sat the fact that each person has in his axiom set... that each person, at each stage, will make all possible FOPL deductions from whatever his axiom set is"- I think that would be a legitimate FOPL axiom schema, because each such axiom includes no self- reference.

Yes, I think the idea with axioms is you don't necessarily have to list them all, it is enough to provide an algorithm that churns them out, and then the algorithm can be used in proofs. A mathematician might be able to say this in a more formally correct way.

 

[.Scott]

To attack this, we need to know exactly what the monks know:
1) They know how many other monks have blues eyes.
2) They know that the monks will follow the rules.
3) Implicit from the way the problem is presented, the monks recognize the blue-eyed monks. In other words, they would know if a different set of blue-eyed monks were preset on one day from the previous day.

Also, as most on this thread have already noticed, we don't need to deal with brown-eyed monks. We just need to know that there may be some.

Demystifier suggested that history may be important, so I will present the logical reasoning so that there is a build up of history before the priest shows up and makes his declaration.

Day 1: 1 blue (named "A") is placed on the island. He has no way of determining his eye color. If the priest shows up and declares that he sees a blue, the blue will kill himslef at the next opportunity. Generally speaking, when there is 1 blue on the island, a declaration of blue will cause that blue to die at the next opportunity. All monks, being perfectly logical, and knowing that all other blues are perfectly logical, will rely on this rule.

Day 2: a second blue ("B") is placed on the island. This is where the logic becomes critical. Each monk knows that there are 1 or 2 blues. So they also know that if the priest showed up, he would say that there was a blue. What they don't know is how the blue (the other monk) will react. If A sees B kill himself, then A knows that he (A) is not blue. On the other hand, if B does not kill himself, then the number of blues cannot be 1. Both A and B would see this and kill themselves at the next opportunity.

But since they know what the priest would say, do they really need the priest? The answer is yes, because although each one knows that there are 1 or 2 blue, they don't know whether the other thinks there are (0 or 1) or (1 or 2). So what is important is not that priest said anything newsworthy, but that monks saw every blue monk be informed of this.

So this creates a second completely reliable rule: If a priest declares that there is a blue and there are two blue, both will kill themselves on the second opportunity.

Day 3: a third blue ("C") is placed onto the island. He immediately deduces that there are either 2 or 3 blues on the island and that the other blues know that there are either (1 or 2) or (2 or 3) blues on the island. At this point, he can deduce nothing else.

But if the priest makes a declaration, then he knows that on the second opportunity the "2" theory will be tested. And that if no one kills themselves, that the answer must be 3.

But I like the "history" idea, so I will make another post to this thread dealing with it more comprehensively - probably not today. But unless I run into a priest making eye declarations (or another monk beets me to it), it will be soon.

 

[andrewkirk]

For the riddle to work, we must additionally postulate C', namely : " C is in fact included in each person's axiom set". We must also add C'' and C''', defined recursively. But this recursion need not be infinite: it is sufficient that the number of such axioms (including C) be equal to the number of monks.

I think you may have something there. I need to think about it in relation to axiomatising the blue-eyed problem but, at the moment - as Winnie the Pooh said - 'I try to think, but nothing happens'. Maybe later in the day, after I go out and get some fresh air.

Yes, I think the idea with axioms is you don't necessarily have to list them all, it is enough to provide an algorithm that churns them out, and then the algorithm can be used in proofs.

That's valid. An infinite set of axioms that is specified by a higher level formal construction is called an axiom schema. An example is the axiom schema of substitution:
$$\forall y\ (\forall x\ \phi)\rightarrow \phi^y_x$$
where $\phi$ is any wff and $\phi^y_x$ is the wff obtained by replacing all free occurrences of $x$ by $y$. That is, if $\phi$ is true for all $x$ then it remains true for any consistent substitution of something else for $x$, whether that be a simple constant or a complicated term.

There is one axiom in this schema for each wff $\phi$ and since there is a countably infinite number of wffs, there is a countably infinite number of axioms in this schema.

 

[lukesfn]

Perhaps I am confused, stupid and ignorant but…

I feel like there is flaw in the solution to the problem as posed because the monks must determine what the guru’s behaviour would be in a hypothetical situation which is not observed. I It seems to a hidden assumption which I am confused about whether or not is justified.

Examining the case of 4 blue eyed monks, once the guru makes the announcement, they all know that there are either 4 or 3 blue eye monks, but they don’t know if the guru made the announcement to 3 blue eye monks or 4 blue eye monks, so they can’t determine that if there where 3 blue eye monks the the guru would have said anything, so when they consider the possibility that there are only 3 monks as a hypothetical, they have to also have to consider the hypothetical that if there are 3 monks, then than the guru would not have spoken.

Even though all 4 know they know that the if are only 3 blue eye monks, they know the guru has made the announcement to 3 blue eye monks, they don’t know there are 3 blue eye monks, so they don’t know what the guru’s behaviour actually might be in that case.

For the puzzle solution to work, the monks must determine that all other monks would determine though a chain of recursive hypothetical determinations that in the hypothetical situation of there being only one blue eyed monk that the guru would have made the same announcement, but by the above logic, I am not sure they can determine that.

I’m I making sense? I assume some people here are smart enough to see what point I am trying to get at and can tell if it is flawed or not.

 

[andrewkirk]

@lukesfn You are none of those things, otherwise you would not be interested in this problem at all, let alone able to ask meaningful questions about it as you have.

When the guru makes the announcement on an island with four blue monks, what Monk 1 now knows to be the case, which he did not know to be the case earlier, is that Monk 2 knows that Monk 3 knows that Monk 4 knows that there is at least one blue.

 

[lukesfn]

When the guru makes the announcement on an island with four blue monks, what Monk 1 now knows to be the case, which he did not know to be the case earlier, is that Monk 2 knows that Monk 3 knows that Monk 4 knows that there is at least one blue.

I find it very easy to get confused about this problem, however, my first instinct is that this statement is both incorrect and irrelevant. Why wouldn't Monk 1 already know that? And why would that knowledge make any difference?

edit: sorry, I apologise if my tone sounded to aggressive

 

[Heinera]

I find it very easy to get confused about this problem, however, my first instinct is that this statement is both incorrect and irrelevant. Why wouldn't Monk 1 already know that? And why would that knowledge make any difference?

edit: sorry, I apologise if my tone sounded to aggressive

Yes, it is a bit more subtle than that. Consider the case with only two monks. It is not enough that one of the monks exclaims that "at least one monk has blue eyes." The other monk would of course now know that he has blue eyes and commit suicide, but the first monk would still have no idea about the color of his own eyes. It is only when an outsider makes the statement that the induction process will work.

 

[Andreas C]

I see what you mean. Yes, one needs to run back to RBB, but only for reductio ad absurdum. To prove that something cannot be, one considers the possibility that it can be and derives a contradiction.

But I don't think that there is any disagreement between you and me on that. The only question is whether we really need a prophet at Day 0, for the case BBB. I say we don't.

I used to be confused in the same way that you are, then I thought about it, and realized that it was simple. You have to use 3 individuals and pretend you are them (so that it's easier to understand):

Person A:
I see 2 blue eyed people. Let's pretend I have brown eyes, and let's get in Person B's shoes:

Person B: I see a person with brown eyes and a person with blue eyes. Let's pretend I have brown eyes and get into Person C's shoes.

Person C: I see 2 brown eyers. I might have brown eyes, maybe nobody has blue eyes.

Person B: I KNOW Person C has blue eyes, but he doesn't. I can't be sure that Person C knows there is at least one person with blue eyes. I would only be sure if I was sure I had blue eyes, but I am not.

Person A: Hmm, it turns out that if I have brown eyes, I can't be sure that Person B knows that Person C knows there is at least one person with blue eyes. I know C knows that, but B doesn't. I know for sure that B has blue eyes, but B doesn't, just like I don't know for sure if I have blue or brown eyes.

 

[Laurie K]

Person A: I see 2 blue eyed people. Let's pretend I have brown eyes, and let's get in Person B's shoes:

Person B: I see a person with brown eyes and a person with blue eyes. Let's pretend I have brown eyes and get into Person C's shoes.

Person C: I see 2 brown eyers. I might have brown eyes, maybe nobody has blue eyes.

Wouldn't C see the same as B as A sees B and C as both being blue eyed?

 

[andrewkirk]

I find it very easy to get confused about this problem, however, my first instinct is that this statement is both incorrect and irrelevant. Why wouldn't Monk 1 already know that? And why would that knowledge make any difference?

Read through post 94, to see why it makes a difference. That covers the general case.

For the specific case of four monks, here's what is learned when the oracle speaks.

Monk 1 (M1) doesn't already know (that M2 knows that M3 knows that M4 knows that there is at least one blue), because M1 believes M1 could be non-blue, and if that were the case then*:
... - M2 sees only two blues. And M2 believes M2 could be non-blue, and if that were the case then:
... - M3 sees only one blue. And M3 believes M3 could be non-blue, and if that were the case then:
... - M4 sees no blues. And, provided the oracle has not spoken yet, M4 believes M4 could be non-blue (and if that were the case then there would be no blues at all).

When the oracle says there is at least one blue, the underlined scenario is ruled out, so now M1 knows that M2 knows that M3 knows that M4 knows that there is at least one blue. Put more succinctly, in the terminology of post 6, we then have $K^4C(1)$, whereas before we only had $K^3C(1)$ and $K^4C(0)$.

* note the nesting in the next bit, denoted by indentation. It signifies the number of levels of people thinking about what other people think, and is crucial, like in the movie Inception.

EDIT: I've just noticed Andreas C's post a couple up from here. I think he does a good job of explaining the same sort of concept as I'm describing here, from a different angle involving putting oneself in somebody else's shoes.

 

[D H]

The story tells that one day, which I call day-zero, the prophet said what he said, but the story does not tell what other events were happening before that.

To which version of the story are you referring? There are many, many versions, dating back to 1953 at a minimum. Some variations have previously been discussed right here on PhysicsForums. (Aside, NUTS! I'm still not quite used to this new setup, and as a result I just lost half an hour of research and writing as a result.)

Nitpicking can be fun, but it can also distract. In this case, nitpicking distracts from the very key distinction between mutual knowledge versus common knowledge. Mutual knowledge (for example, every dragon knows that some dragons have green eyes) is not the same as common knowledge (for example, every dragon knows that every dragon knows that some dragons have green eyes).

There certainly are issues with the concept of "common knowledge." All properly stated versions of the puzzle make the distribution between mutual knowledge and common knowledge quite clear. This distinction is key to understanding the puzzle. Nitpicking jut gets in the way. Whether common knowledge is attainable in the real world is a perpetual issue.

 

[.Scott]

@lukesfnWhen the guru makes the announcement on an island with four blue monks, what Monk 1 now knows to be the case, which he did not know to be the case earlier, is that Monk 2 knows that Monk 3 knows that Monk 4 knows that there is at least one blue.

That is similar to it, but it's not it. It's more complicated than that. Before the guru makes his announcement, all the monks realize that there is not enough information for anyone to deduce their eye color, and that is why years can go by without any monk committing suicide.

I've taken a few shots at trying to construct an English sentence describing what four blue-eyed monks know, but it's a very awkward sentence. For 3 blue-eyed monks ("blues") it is this: Every monk know that there are 2 or 3 blues and that every other blue knows that there are 1 to 3 blues. They also know something about what every other monk might know about what every other monk might know, but I won't try to describe that - and for 3 blues, it doesn't matter. What matters is that without the guru, there is nothing to contradict any of this progression.

Then the guru makes his announcement and everyone knows that at the next suicide opportunity, every monk will know that it is not possible for any monk to think that any monk thinks that 1 blue is possible. With that possibility eliminated, the next suicide opportunity will eliminate the 2 blue possibility.

But I think I described it more simply in post 100. Basically, once the guru makes his announcement, it triggers a series of "rules" that the monks (with perfect logic) have already deduced.

 

[.Scott]

Try this one:
Suicides opportunities are only at midnights.

Given that all monks:
1) Recognize all other monk on the island that they share with them.
2) Always exercise perfect logic.
3) Will commit suicide if and only if they know they are blue-eyed.
4) Always know when a suicide has taken place on their island.
5) Know that all other monks are like themselves, always logical, always follow the rules, always recognize all other monks, always know when a suicide has taken place on the island they are on.

You, a perfectly logical monk, arrive on the island on day 100 and see 12 blue-eyed monks (blues). On day 103, the guru makes his declaration. Then on day 104, six blues commit suicide.

If the guru makes no additional announcements, what if any suicide activity will occur?

 

[Andreas C]

Wouldn't C see the same as B as A sees B and C as both being blue eyed?

Yes and no. I think you misunderstood what I said. I am talking about what Person B would be thinking IF Person A had brown eyes (which, for all Person A knows, may or may not be true).

 

[TheQuietOne]

I think they would either kill the man or make him commit suicide because them not asking questions about their color the ones with blue eyes would not know ( neither would the tribesmen with brown eyes). But seeing that he knows his own eye color they would either force him to commit suicide or, if he refused (tried to escape) they would kill him. (unless a brown eyed guy looks at his blue eyed friend and says "goodbye").

 

[lukesfn]

Monk 1 (M1) doesn't already know (that M2 knows that M3 knows that M4 knows that there is at least one blue), because M1 believes M1 could be non-blue, and if that were the case then*:
... - M2 sees only two blues. And M2 believes M2 could be non-blue, and if that were the case then:

Ok, I was wondering if that was your intended meaning of M1, M2, but ironically, I couldn't be sure until I got this extra information. I already understand all this, and I think you might be missing the point I am trying to make.

... - M3 sees only one blue. And M3 believes M3 could be non-blue, and if that were the case then:
... - M4 sees no blues. And, provided the oracle has not spoken yet, M4 believes M4 could be non-blue (and if that were the case then there would be no blues at all).

Ok, this is where assumption comes in which I can't see the justification for.
M4 is a hypothetical case where there is only one blue eyed monk. I am not sure the monks can assume the hypothetical monk would have heard a hypothetical guru speak.

From the gurus speech, the 4 monks know that the guru told 3 or 4 blue eyed monks that at least one of them had blue eyes. Why does that make M1 consider that if there where 3 Monks, M2 would consider that if there where 2 monks, M3 would consider that there is a guru, or that the guru told the 2 monks that at least one of them has blue eyes?

But I think I described it more simply in post 100. Basically, once the guru makes his announcement, it triggers a series of "rules" that the monks (with perfect logic) have already deduced.

My point is that this is not correct. It appears to me that a logic short cut has been taken making an unjustified assumption, and feel like I have already pointed it out in this post and more so in post 102.

You have to consider that the rules of how the guru work in hyperthentical cases may be unknown. The monks only know that a guru spoke to either 3 or 4 blue eyed monks. How can they know that all other monks would come up with the same strategy based on counting from a time when they imagine a guru speaking to 1 monk?

 

[Astudious]

As I said in a previous post, no you don't need a prophet for that case, because it is an extreme... watch out; not needing him does not mean you get something wrong if you include him. We can just say that in that particular case his existence is unecessary...
But in all other cases his presence is needed to reach a reliable answer... (as I mentioned the RRB state, in which without the prophet everybody'd die)...I guess the prophet makes the problem solvable in all cases then, and so he's necessary for the solution, and for some special cases his information is just repeating.

But I believe that the bbb case where all 3 islanders are b allows an extension to a very large number of cases.

To be precise, as I see it, we only need 3 or more islanders with b for the prophet's pronouncement to be useless. In that case, there is no new relevant common knowledge and they would have committed suicide on a count starting from finding out about the suicide law.

*****

Why? Consider one of the other islanders, X, besides the 3 (say A, B, C) who must have blue eyes - these other islanders may have blue or brown (let's go with red, r) eyes. Of course all the islanders know [someone has blue eyes] already (X sees A, B, C with blue eyes; A sees B, C with blue eyes; B sees A, C with blue eyes; etc.). Do all the islanders know that {all the islanders know that [someone has blue eyes]}?

Well, X knows A can see B, C with blue eyes, and X knows B can see A, C with blue eyes, etc. so X knows {A knows [someone has blue eyes]}, X knows {B knows [someone has blue eyes]}, etc. X knows all the other islanders besides the 3 (A, B, C) will make the same inference about A, B and C, so X knows that {all the islanders know that [someone has blue eyes]}. (So all islanders besides the 3 (A, B, C) know that {all the islanders know that [someone has blue eyes]}.)

A knows B can see C with blue eyes, so A knows that {B knows that [someone has blue eyes]}. By symmetry A knows that {C knows that [someone has blue eyes]} (here the someone is B), and B knows that {A knows that [someone has blue eyes]}, B knows that {C knows that [someone has blue eyes]}, so forth. Also, A knows X can see B, C with blue eyes, so A knows that {X knows that [someone has blue eyes]}. By symmetry, B knows that {X knows that [someone has blue eyes]} (here someone is A, C), and C knows that {X knows that [someone has blue eyes]} (here someone is A, B). Since the same as applies to X also applies to all islanders besides these 3, thus A, B, and C know that {all the islanders know that [someone has blue eyes]}.

So all islanders know that {all the islanders know that [someone has blue eyes]}.

So if there are at least 3 islanders with blue eyes, then all islanders know that {all the islanders know that [someone has blue eyes]} - no need for a prophetic announcement.

****

It is true that the prophetic announcement cannot hurt, but the problem is that in many cases (those above) it does not suffice as a time-marker for starting the count. Rather, the suicide law itself (along with the villagers first seeing each other) would be the marker for the time to start in such cases.

 

[andrewkirk]

I am not sure the monks can assume the hypothetical monk would have heard a hypothetical guru speak.

Neither the monks nor the guru's statement are hypothetical. What is hypothetical is the state of knowledge of those monks, plus the colour of M1's eyes.

Bear in mind though that these recent posts are just loose attempts to render accessible the more formal argument in post 94. If you are concerned about validity of assumptions, that needs to be addressed in relation to post 94, as any apparent assumptions in the more recent posts may just be artefacts of the informal language they use.

Having said that, even 94 is not completely formal, as the problem of how to axiomatise the deductive system used by the monks remains unsolved. I think I may have a lead on that, based on Russell's solution to his own set theory paradox, and am investigating what I can do with it.

 

[.Scott]

I feel like there is flaw in the solution to the problem as posed because the monks must determine what the guru’s behaviour would be in a hypothetical situation which is not observed. It seems to a hidden assumption which I am confused about whether or not is justified.

No, nothing like that is happening. At the root of it, the monks know that if there were only one blue-eyed monk, that monks would respond to the guru's declaration by committing suicide at the next suicide time (midnight or whenever). So, once the guru makes the declaration and a suicide time passes with no suicide, every monk has a new way of determining that the number of monks among them with blue eyes is not 1. And they know all other blue-eyed monks know this. And the logic proceeds from there as described in post #100.

Examining the case of 4 blue eyed monks, once the guru makes the announcement, they all know that there are either 4 or 3 blue eye monks, but they don’t know if the guru made the announcement to 3 blue eye monks or 4 blue eye monks, so they can’t determine that if there where 3 blue eye monks the the guru would have said anything, so when they consider the possibility that there are only 3 monks as a hypothetical, they have to also have to consider the hypothetical that if there are 3 monks, then than the guru would not have spoken.

You're right in the sense that you have created a straw horse argument and then disproved it. But you are distracting yourself. What you are thinking is the proposed solution (in post 100) is not. All the blues know is that all blues, whether it be 3 or 4, will see a new demonstration that it is not 1 at the next suicide opportunity - and that they will see a consequential demonstration that it is not 2 on the second opportunity. None of them know what will happen on the third opportunity. This is all the result of logical deduction - not strategy.
When the guru makes his statement that he sees a blue-eyed monk, none of these perfectly logical monks is distracted by the fact that they already have certain evidence of this.[/quote]

Even though all 4 know they know that the if are only 3 blue eye monks, they know the guru has made the announcement to 3 blue eye monks, they don’t know there are 3 blue eye monks, so they don’t know what the guru’s behaviour actually might be in that case.

They know that the guru would only have made his statement that there is at least one blue-eyed monk, truthfully. That is sufficient for the post #100 logic to be used.

For the puzzle solution to work, the monks must determine that all other monks would determine though a chain of recursive hypothetical determinations that in the hypothetical situation of there being only one blue eyed monk that the guru would have made the same announcement, but by the above logic, I am not sure they can determine that.

All monks know that all other monks are perfect logicians. The rules are deduced as I stated them in post #100. Which specific rule have I not proven to your satisfaction?[/quote]

 

[.Scott]

My point is that this is not correct. It appears to me that a logic short cut has been taken making an unjustified assumption, and feel like I have already pointed it out in this post and more so in post 102.

You have to consider that the rules of how the guru work in [hypothetical] cases may be unknown. The monks only know that a guru spoke to either 3 or 4 blue eyed monks. How can they know that all other monks would come up with the same strategy based on counting from a time when they imagine a guru speaking to 1 monk?

Only blue-eyed monks matter. Monks of all other eye color are just furnishings.
It should be stated in the problem that when the guru speaks, all blue-eyed monks are listening, and that all blue-eyed monks know that all other blue-eyed monks are listening.
What I am describing isn't a "strategy" that a monk can come up with. It is of a set of logical rules that all monks will discover because they are "perfect logicians". And, all monks know that the other monks also a perfect logicians so they know that all of them will also discover these rules.

I took another look at your post #102 and will reply to it above.

 

[Heinera]

So if there are at least 3 islanders with blue eyes, then all islanders know that {all the islanders know that [someone has blue eyes]} - no need for a prophetic announcement.

But "all islanders know that {all the islanders know that [someone has blue eyes]}" is not enough. The proposition "all the islanders know that" must be nested the same number of levels as there are number of blue eyed islanders for the induction to work.

So if there are exactly 3 islanders with blue eyes, we need "all the islanders know that {all the islanders know that {all the islanders know that [someone has blue eyes]}}".

 

[.Scott]

It is true that the prophetic announcement cannot hurt, but the problem is that in many cases (those above) it does not suffice as a time-marker for starting the count. Rather, the suicide law itself (along with the villagers first seeing each other) would be the marker for the time to start in such cases.

The guru announcement will serve as a trigger to the countdown because it eliminates the possibility on only one blue-eyed monk on the next suicide opportunity. That is all that is critical. Any incident that creates that will trigger the count down. Simply showing up on the island together won't do this.

Many of the recursion statements in this thread about monks knowing about other monks are either wrong or irrelevant - and are effectively straw horses in the discussion. For example, this is an argument against a straw horse:

So all islanders know that {all the islanders know that [someone has blue eyes]}.

So if there are at least 3 islanders with blue eyes, then all islanders know that {all the islanders know that [someone has blue eyes]} - no need for a prophetic announcement.

The simplest way of stating what happens when the guru makes his declaration is that it sets up a public demonstration that the number of blue-eyed monks is not 1. The logic that follows is described in post 100.