Blue-Eye Paradox: Solution Not Unique

[Lord Crc]

From X's prespective however, he (X) is either Blue or Brown eyed...
What would happen in the case that you had 1 Blue and 9 Brown eyed people [without the input of the prophet].

That is an entirely different scenario from the one described in the stack exchange post and thus it's hardly surpring it has a different result, no?

 

[maline]

The only question is whether we really need a prophet at Day 0, for the case BBB. I say we don't.

Why not? Please explain in detail what chain of inferences could allow any of the monks to discover his own eye color, without the prophet's input. If you like you can define day zero as when the suicide law was instituted.

 

[Zafa Pi]

I agree that prophet's information is important in RRR, RRB, and RBB cases. However, I think that prophet's information is not important* in the BBB case.

The prophets is not important in the case RBB. Everybody sees a B so no information was imparted by the prophet. Now I know that you have some fancy pants logic that says otherwise, using ten dollar words like "inference", but my logic is a perfectly good solution. I see that you brought up B. Russel, well of course, he is depicted on your icon. Well I told my solution to my mom, and she's a wonderful person and pretty damn smart too, and she agrees with my solution. So in the RBB case after the prophet speaks nothing new happens. That's my solution and I'm sticking to it.

I hope the monitors allow sarcasm.

 

[ChrisVer]

That is an entirely different scenario from the one described in the stack exchange post and thus it's hardly surpring it has a different result, no?

It is the same problem where the numbers are changed a little (ratios of Blue/Brown)... the concepts are still the same, since the inhabitants never knew those ratios to start with... For example for an inhabitant with blue eyes, the numbers could as well be 99 Blues and 101 Browns (considering himself a brown eye) or 100 Blues and 100 Browns... similiarily for a brown-eyed person (after replacing Blue<->Brown).

Someone has to give some numbers in the start, in order to get an answer to the logical question (because the total number of days depend on the total number of blue-eyed). Why would that affect the total logic that leads to the solution? This logic utilizes only what 1 observes and takes as information and the fact that they are robot-like taking decisions.

 

[ChrisVer]

Why not? Please explain in detail what chain of inferences could allow any of the monks to discover his own eye color, without the prophet's input. If you like you can define day zero as when the suicide law was instituted.

I think I understood that objection.. let's call day 0 the day that the rule becomes known to the 3B inhabitants...
Day 0: each B sees BB... he can't say that he is a !B
Day 1: Noone dies. Each B sees BB again... he still can't say that he is a !B
Day 2: Noone dies. Each B sees BB again... however if that B was a !B, the other two Bs would have died by this day***. He becomes certain he is a B.
Day 3: B dies... all of B die...

The prophet's input here is unnecessary as I mentioned to 1 post, I suspect due to the complete symmetry of the setup... In fact, I agree with Demystifier here; that in this special case the outcome is determined just by the rule itself... However, I disagree that this is a complete answer since there are cases where it cannot be applied without inconsistency to the game rules (so it's preferable to add an extra unnecessary to some cases parameter which makes the solution always existent for all, even those that it is not supposed to affect)*** if the mentioned B was a !B, one of the other two Bs would have been seeing a B(!B) and would have died by Day2.

 

[Lord Crc]

It is the same problem where the numbers are changed a little (ratios of Blue/Brown)... the concepts are still the same, since the inhabitants never knew those ratios to start with...

The initial conditions give are that there are 100 blue-eyed and 100 brown-eyed, so we know nobody on the island is thinking maybe he's the only blue-eyed and the other 199 are all brown-eyed. Everyone on the island knows the distribution is either 99 blue-eyed and 101 brown-eyed, or vice versa.

In either of these two cases (as with your 5/5 example), an arbitrary person X can see at least two blue-eyed people, regardless of their own color. This means that an arbitrary person X knows that any other person Y can see at least one blue-eyed person, regardless of Y's color.

This is why I don't see how the Oracle's fact changes anything.

That you can get other solutions by changing the initial conditions is irrelevant, just as the other solutions to $\lim_{x \to \infty} a^x$ are irrelevant if you ask me to find the limit when $a \in (0, 1)$.

 

[ChrisVer]

The initial conditions give are that there are 100 blue-eyed and 100 brown-eyed, so we know nobody on the island is thinking maybe he's the only blue-eyed and the other 199 are all brown-eyed. Everyone on the island knows the distribution is either 99 blue-eyed and 101 brown-eyed, or vice versa.

I am sorry I don't understand your point here...
I believe you turn the "paradox" in something extremely specialized. Well you can find a solution to this, but it won't be the general solution.
In particular the solution: $y(x)=5 e^{x}$ is a solution to the differential equation $\frac{dy}{dx} = y$ but it's a special case of a complete family of solutions that are $y(x) = a e^{x}$.
Not only that (which has nothing wrong in it), but by just looking at the particular data will lead you to something worse/wrong. As I demonstrated with the alternative initial condition state, you will reach an unreasonable result which violates the game rules.
So the presence of the prophet is 100% necessary to make all the cases solvable and so he well exists in the initial setup. It changes nothing in some cases, but also makes others solvable as well...

In the diff eq example above, you can end up with $5e^x$ being the solution to $\frac{dy}{dx} = y ~~,~~y(0)=5$ (taking a particular initial condition), but it's not the solution to $\frac{dy}{dx} = y ~~,~~y(0)=1$ or $\frac{dy}{dx} = y ~~,~~y(0)=9$... The $y=a e^x$ is by setting $a$ appropriately.

That you can get other solutions by changing the initial conditions is irrelevant, just as the other solutions to limx→∞axlim_{x \to \infty} a^x are irrelevant if you ask me to find the limit when a∈(0,1)a \in (0, 1).

Well that's like a flaw... I am not going to think a lot about it at the moment, to see whether you would have to approach the limit differently depending on whether $a$ is in (0,1) or !(0,1)... but I could take your example one step further and ask you if you'd find the limit differently if $a=0.0001$ or $a=0.98$ (randomly taken numbers).

 

[Lord Crc]

I am sorry I don't understand your point here...

My point is that initial conditions in the original problem means that every person can see at least two other persons with blue eyes, and from that it follows (or at least, my brain tells me so) that every person already knows at least one person has blue eyes. The logic used to deduce this cannot work for all initial conditions, but does (at least, my brain tells me so) for the initial conditions given in the original problem.

Well that's like a flaw... I am not going to think a lot about it at the moment, to see whether you would have to approach the limit differently depending on whether $a$ is in (0,1) or !(0,1)...

Well for non-negative $a$ it has three solutions, $0$ for $a \in [0, 1)$, $1$ for $a = 1$ and $\infty$ for $a \in (1, \infty)$. But my point was that if you tell me $a \in [0, 1)$, then the two other cases are irrelevant.

edit: I meant $a \in [0, 1)$, getting a bit tired here.

 

[Dr.AbeNikIanEdL]

That cannot be from the point of view of one of known B's, because they know what is X. So it can only be from the point of view of X himself. Now I, as external observer, know that X=B, so I know that the actual situation is BBB. But of course, X does not know it yet, so this is how his evolution looks like:
Day 0: XBB. X knows that the two B's will not kill themselves at Day 1.
Day 1: XBB. The two B's do not kill themselves, just as X predicted. X hopes that X=R, in which case the two B's will kill themselves at Day 2.
Day 2: The two B's do not kill themselves.
Day 3: X observes that B's did not kill themselves. Therefore BBB. Therefore X has to kill himself at Day 3. All 3 of them kill themselves at Day 3.

Could you proof that in the RBB case it follows that the two B's will kill themselves at Day 2? You seemed to agree that the prophet is needed for that case!

 

[andrewkirk]

another solution (the standard one) is that they will all commit suicides after 100 days

How does suicide come into it? I read carefully over the first link in the OP and glanced over the other two and couldn't see any mention of suicide. The key action is leaving the island. Why the refs to suicide and how do they relate to the problem as stated?

 

[Dr.AbeNikIanEdL]

Apparently, in an alternative formulation the rule is that they have to kill themselves as soon as they are sure they have blue eyes. Of course for the logic problem it is not important what action it actually is as long as it is a unique action visible to the others.

 

[ChrisVer]

How does suicide come into it? I read carefully over the first link in the OP and glanced over the other two and couldn't see any mention of suicide. The key action is leaving the island. Why the refs to suicide and how do they relate to the problem as stated?

The setup was somehow changed from the departing inhabitants of an island, to monks who listen to a prophet and commit suicide.

 

[disregardthat]

The initial conditions give are that there are 100 blue-eyed and 100 brown-eyed, so we know nobody on the island is thinking maybe he's the only blue-eyed and the other 199 are all brown-eyed. Everyone on the island knows the distribution is either 99 blue-eyed and 101 brown-eyed, or vice versa.

Let's say there are 3 blue-eyed monks on the island, A,B and C, and that the prophet has not spoken yet. Then everyone knows that there are either 3 or 2 blue-eyed monks. In the perspective of A, he must consider two cases. Either he has blue eyes, or he has brown eyes. If A has blue eyes, then B and C see two monks with blue eyes. If A has brown eyes, then B and C see one monk with blue eyes. Due to the latter possibility, A concludes that B and C individually can not exclude the possibility that they in fact too have brown eyes.

If we further investigate this last case, then the last monk can not exclude the possibility that he has brown eyes as well. In other words, A further concludes that B cannot conclude that C can exclude the possibility of having brown eyes. Similarly, A concludes that C cannot conclude that B can exclude the possibility of having brown eyes. And of course, B and C will mirror these conclusions.

So everyone does not know that everyone knows there are 3 or 2 blue-eyed monks. The correct statement is that everyone knows that everyone knows that there are 3,2 or 1 blue-eyed monks. And by the argument above, the only correct statement in a further nesting would be that everyone knows that everyone knows that everyone knows that there are 3,2,1 or 0 blue-eyed monks. But once the prophet claims there are at least one blue-eyed monk, everyone strengthens this last statement to everyone knows that everyone knows that everyone knows that there are 3,2,1 blue-eyed monks. This is because everyone knows that everyone heard the prophets claim. And everyone knows that everyone knows that everyone heard the prophets claim. And so on...

This is the explanation of precicely how the prophet is actually giving additional information to everyone. This may be extended to the cases where there are more monks on the island. It is similar, but of course much longer for 100 monks.

 

[andrewkirk]

I think that the problem is not well-specified, and hence has no solution. The reason is the vagueness of the term 'perfect logician'.

I take it that 'perfect logician' is intended to mean something like 'somebody that will deduce from a given set of axioms anything that can be deduced from them using first-order predicate logic (FOPL) in language L' . I choose FOPL because it is the most flexible and expressive logic one can use without having to make a whole bunch of other definitions and constraints in order to avoid paradoxes. We could consider it for other logics such as one of the many varieties of modal logic, but for the sake of concreteness and simplicity, let's stay with FOPL for now. The language L doesn't need to be specified for the purposes of this problem but I think we should assume it is countable, as I think there may be problems with uncountable languages.

The axioms are set out in the first paragraph of the problem statement, and include statements like 'the oracle speaks once a day ...' and 'those who have deduced that they have blue eyes leave on the day they make the deduction'. It also includes the 'perfect logician' statement. Let's denote all the other axioms collectively by B and the 'perfect logician' axiom by C. Then the set of axioms is B+C (where I use '+' here to indicate union). 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'​

Oh dear. There's still a C in there, which we remember was unclear. Let's expand that. that gives us:

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

Bother! There's still a C in there. Let's get rid of it by expanding again:

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

... and so on, ad infinitum

The consequence of this infinite regress is that there is no finite statement of the axiom C. Since an axiom must be a well-formed formula (wff) and one of the requirements of being a wff is that it be composed of a finite number of symbols, we find that there is no possible valid axiom C.

So the problem is not well-defined, and hence has no solution.

It's conceivable there may be some way around this using a higher-order logic. But my hunch is that, whatever logic we use, we are going to run into a Godelian problem that one cannot deduce, from inside any sufficiently rich logical theory, which wffs are theorems of the theory.

 

[ChrisVer]

I don't get it, what is wrong in your axiom "C"?
Also I don't see how this goes to infinity since there is a fixed and finite number of possible "each person".

 

[andrewkirk]

I don't get it, what is wrong in your axiom "C"?

It's self-referential and generates an infinite regress in trying to understand what it is saying.

C says 'each person will deduce whatever can be deduced from this set of axioms'

But this set of axioms includes C. So to work out what a person can deduce, we first need to know what they can deduce.

 

[ChrisVer]

I see finiteness in the 3 people example, and I understand that in the BBB case, it's the perfect logician assuming that the others are perfect logicians that leads to the solution. Of course if you keep on trying to divide this up, you will end up interconnecting all people [since they all die on the same day] and start looping between their connections, "repeating the same axiom".
Is a circle something finite or infinite? Seeing it as a circle it's finite (parametrized in x<[0,2pi] )... seeing it as a closed curve it can be infinite (parametrized in x<[-infty, infty] ).

 

[Zafa Pi]

The prophets is not important in the case RBB. Everybody sees a B so no information was imparted by the prophet. Now I know that you have some fancy pants logic that says otherwise, using ten dollar words like "inference", but my logic is a perfectly good solution. I see that you brought up B. Russel, well of course, he is depicted on your icon. Well I told my solution to my mom, and she's a wonderful person and pretty damn smart too, and she agrees with my solution. So in the RBB case after the prophet speaks nothing new happens. That's my solution and I'm sticking to it.

I hope the monitors allow sarcasm.

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 apologize for being snarky in post #73, but in case RBB and BBB it can be claimed that the prophet creates no new info when she says there is at least one B. So in either case one could claim the solution is that nothing new will happen. But you see them as different. How so (in the fewest words you can)?

 

[andrewkirk]

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.

 

[Ken G]

The prophet said them something that they already knew, so there is no reason to change anything in their behavior.

Actually, this isn't correct. It is true that the prophet said something they all already knew, but this does not mean new information was not conveyed. The whole crux of the situation is that it is essential that everyone in the tribe witness the prophet speaking to everyone else, so the new information is that the people now know what the other people know. There is no reason to ever have any more than two blue-eyed people, that version of the puzzle makes all clear. If there are two blue-eyed people, the solution really is pretty much unique, but you do have to assume that people can tell when other people are listening, and that the other people will draw the correct conclusions. But that's not a stretch when there are two people with blue eyes.

 

[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 $j$th 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 $n$th 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.