Blue-Eye Paradox: Solution Not Unique

[Heinera]

n=3: In this case the first sentence is not new, similarly to the case n=2. In addition, for n=3, even the second sentence is not new. (For instance, the first blue-eyer knows two blue-eyers (the second and the third one), he also knows that the second blue-eyer knows at least one blue-eyer (the third one), and finally he knows that the third blue-eyer also knows at least one blue-eyer (the second one).) Therefore nothing in the prophet's message is new in the case n=3. In other words, assumption (3) is wrong for n=3. Q.E.D.

Before I can adress this, I need to know what you think will happen in the case n=3. If the monks are told about the suicide rule at t=0, will they commit suicide at t=3 even without any statement from the prophet?

 

[stevendaryl]

Next observe that the two-sentence message above is equivalent to:
"At least one of monks has blue eyes. Now all monks know that."

Not quite. The public announcement is equivalent to arbitrarily (as many levels as there are monks) nested statements:

1. At least one monk has blue eyes.
2. Now all monks know that.
3. Now all monks know that all monks know that.
4. Now all monks know that all monks know that all monks know that.
5. etc.

A public announcement accomplishes this, as does your self-referential private announcement ("At least one monk has blue eyes, and this message will be sent to each monk"), but your pair of sentences doesn't accomplish that.

Now the crucial question is this. Is the information provided by the prophet new to the monks? Let us study case by case.

n=1: For this single blue-eyer, the first sentence in the letter "At least one of monks has blue eyes" is new.

n=2: In this case, the blue-eyers already know that "At least one of monks has blue eyes". They know it without the prophet. It is not new. But the second sentence "Now all monks know that" is something new.

n=3: In this case the first sentence is not new, similarly to the case n=2. In addition, for n=3, even the second sentence is not new.

That's true, but your two sentences are not equivalent to the original announcement. Let's call our three islanders "Alice", "Bob", and "Carol". Assume that all three have blue eyes. Before any announcement, the situation is this:

1. Alice has blue eyes.
   
2. Alice knows that at least Bob and Carol have blue eyes.
3. Alice knows that Bob knows that at least Carol has blue eyes.
4. Similarly for other permutations of the names

What Alice doesn't know is whether Bob knows that Carol knows that there is at least one person with blue eyes.

The possible eye-situations before the announcement are described by the following "possible worlds":

1. World 1: Alice, Bob and Carol all have blue eyes.
2. World 2: Bob and Carol have blue eyes, but Alice doesn't.
3. World 3: Only Carol has blue eyes.
4. World 4: Nobody has blue eyes.

• In world 1, Alice believes that the actual world is either world 1 or world 2.
• In world 2, Bob believes that the actual world is either world 2 or world 3.
• In world 3, Carol believes that the actual world is either world 3 or world 4.

1. Alice believes it's possible that the actual world is world 2.
2. Alice believes it's possible that Bob thinks it's possible that the actual world is world 3.
3. Alice believes it's possible that Bob thinks it's possible that Carol thinks it's possible that the actual world is world 4.

After the public announcement that there is at least one blue-eyed individual, the first two sentences are unchanged, but the third sentence becomes:

Alice believes it's possible that Bob knows that Carol knows that the actual world is world 3. (Because world 4 has been ruled out by the announcement).

 

[.Scott]

That is not correct (the "No"). If the guru says that "everybody knows there is at least one blue-eyed monk", a group of two monks would immediately (and independently) deduce they both had blue eyes, and could thus skip a time step. (Monk A reasons that if he had brown eyes, monk B couldn't know there was at least one blue-eyed monk, and same for B vs. A). The induction that takes us to n > 2 goes as before.

Furthermore, if the guru says "everybody knows that everybody knows there is at least one blue-eyed monk" (to a group of size n >= 3) he would move the suicide moment forward by two time steps, etc. This clearly shows that the nested levels of knowledge are relevant to the problem.

You're right. Those statements are functionally equivalent to "there's at least 2 blue-eyed" and "there's at least 3 blue-eyed".

 

[stevendaryl]

Using graphs, the situation can be described as follows:

The actual world is W1, where all three have blue eyes. The arrows labeled "Alice" indicate which worlds Alice thinks are possible. She thinks W2 is possible. In world W2, only Carol and Bob have blue eyes. The arrows labeled "Bob" indicate which worlds Bob would think were possible, if the actual world were W2. So if the actual world were W2, then Bob would think that W3 is possible. In world W3, only Carol has blue eyes. If the actual world were W3, then Carol would think that W4 is possible. In world W4, nobody has blue eyes. So before the announcement, it's possible (according to Alice's information) that the actual world is W2, but Bob thinks that he's in W3, and Bob thinks that Carol thinks that she's in W4.

After the announcement, the situation is changed to the following:

After the announcement, Carol knows that the actual world is not W4. So Alice thinks it's possible that we're in world W2, but that Bob believes we're in world W3. In world W3, Carol knows that she has blue eyes (since she knows that she isn't in world W4). So in W3, Carol commits suicide. If Carol doesn't commit suicide, then Bob figures out that the actual world is not W3. So now the situation is:

Alice thinks it's possible that we're in world W2. But in world W2, Bob would know that he's in world W2 (since W3 has been eliminated as a possibility). So in W2, Bob would know that he's in world W2, and commit suicide.

If that doesn't happen, Alice knows that we're in world W1.

 

[Demystifier]

Before I can adress this, I need to know what you think will happen in the case n=3. If the monks are told about the suicide rule at t=0, will they commit suicide at t=3 even without any statement from the prophet?

Yes they will.

 

[stevendaryl]

Yes they will.

Once again, suppose the islanders are Alice, Bob and Carol. They all have blue eyes, but Alice erroneously believes that she has brown eyes. How would she ever come to know that she was mistaken?

 

[Demystifier]

@maline and @stevendaryl from your responses I can conclude that the main controversy is whether or not the message
"At least one of monks has blue eyes. This message is sent to all monks."
is equivalent to
"At least one of monks has blue eyes. Now all monks know that."

I say it's equivalent, and you say it's not. You say that the first message contains more relevant information than the second one. But intuitively, it doesn't make sense to me. So I need to think more about it.

 

[lukesfn]

With the above graph, have you thought about worlds where no announcements where made? That question always confuses me, sometimes I think they can be safely ignored, but sometimes I think that has been assumed without a fool proof justification.

I was also considering similar thoughts to what demystifier put quite clearly in post 139.

I've been reading numerous standard solutions, and they often describe them selves as an inductive solution, but I keep getting confused when I try to pin point the exact points where inductive reasoning is needed, rather then deductive.

I am wondering that if for that for certain numbers of blue eyed people, the inductive reasoning feels so strong that it is mistaken for deductive reasoning.

 

[Demystifier]

Once again, suppose the islanders are Alice, Bob and Carol. They all have blue eyes, but Alice erroneously believes that she has brown eyes. How would she ever come to know that she was mistaken?

But we are talking about logical people. There is no any logical reason for Alice to believe that she has brown eyes.

 

[Demystifier]

With the above graph, have you thought about worlds where no announcements where made?

Yes. My answer is that, for $n\geq 3$, they will all kill themselves after $n$ days, counting from the day at which they all together learned about the suicide law. This assumes that they did learn the suicide law together at the same day. If they learned it in another way, then the answer depends on how exactly did they learn it.

 

[stevendaryl]

@maline and @stevendaryl from your responses I can conclude that the main controversy is whether or not the message
"At least one of monks has blue eyes. This message is sent to all monks."
is equivalent to
"At least one of monks has blue eyes. Now all monks know that."

I say it's equivalent, and you say it's not. You say that the first message contains more relevant information than the second one. But intuitively, it doesn't make sense to me. So I need to think more about it.

As I said, there are (at least) three levels of statements about Alice, Bob and Carol (the three islanders):

1. Alice's eye color
   
2. What Alice believes about her eye color.
3. What Alice believes about what Bob believes about his own eye color.
4. What Alice believes about what Bob believes about what Carol believes about her own eye color.

When the monk announces to everyone "There is at least one blue-eyed person", that doesn't affect level 1, nor level 2, nor even level 3, but it does affect level 4.

 

[stevendaryl]

But we are talking about logical people. There is no any logical reason for Alice to believe that she has brown eyes.

But there is no logical reason for her to conclude that she has blue eyes. As far as she knows, she MIGHT have brown eyes.

Once again, here are 4 possible worlds:

1. W1: All three have blue eyes.
2. W2: Only Carol and Bob have blue eyes.
3. W3: Only Carol has blue eyes.
4. W4: Nobody has blue eyes.

In W1, Alice believes that W2 is possible.
In W2, Bob believes that W3 is possible.
In W3, Carol believes that W4 is possible.

So Alice believes that it is possible that (Bob believes that it is possible that (Carol believes it is possible that (nobody has blue eyes))).

After the announcement, Alice no longer believes this.

 

[stevendaryl]

With the above graph, have you thought about worlds where no announcements where made?

You can consider such worlds, if you like, but the point is that nobody in the actual world would think that they were in that world. So there is no chain of arrows from the actual world to that world. So that world would not come into play in the reasoning.

 

[lukesfn]

After the announcement, Alice no longer believes this.

The correct number of days after the announcement, Alice no longer believes W2 is possible, but she already new W3 and W4 where not possible.

When the monk announces to everyone "There is at least one blue-eyed person", that doesn't affect level 1, nor level 2, nor even level 3, but it does affect level 4

The announcement affects a world that Alice already didn't believe existed, which is odd.

 

[Demystifier]

@maline and @stevendaryl from your responses I can conclude that the main controversy is whether or not the message
"At least one of monks has blue eyes. This message is sent to all monks."
is equivalent to
"At least one of monks has blue eyes. Now all monks know that."

I say it's equivalent, and you say it's not. You say that the first message contains more relevant information than the second one. But intuitively, it doesn't make sense to me. So I need to think more about it.

Let me try to explain why I might be wrong.
Suppose that I receive the letter
"At least one of monks has blue eyes. Now all monks know that."
What can I conclude from it? I can interpret it in two interestingly different ways:
The first interpretation is that all the others have received the same message as I did, namely
"At least one of monks has blue eyes. Now all monks know that."
The second interpretation is that all the others received a shorter message than I did, namely
"At least one of monks has blue eyes."
In the first interpretation, "that" refers to both sentences, so it involves self-reference.
In the second interpretation, "that" refers only to the first sentence, so it doesn't involve self-reference.

The first interpretation is equivalent to "At least one of monks has blue eyes. This message is sent to all monks." (Do you agree?)
But the second interpretation is not equivalent to it. So I might have been wrong in first taking the first interpretation (to achieve equivalence), but later taking the second interpretation (to show that it does not carry so much new information). If that was my mistake (which seems to be the case), then I admit that I was wrong and now accept that the standard solution is correct. So now it seems I know where my mistake was, but let me not make the final conclusion before thinking about it once again.

 

[stevendaryl]

I am wondering that if for that for certain numbers of blue eyed people, the inductive reasoning feels so strong that it is mistaken for deductive reasoning.

I think that you might be confusing "inductive reasoning" with the mathematical technique of "mathematical induction", which is a form of DEDUCTIVE reasoning.

Inductive reasoning is a matter of generalizing from a bunch of examples. Every swan I've ever seen is white, so I conclude "All swans are white". That's not logically valid, because there might be a black swan that I've never seen.

Mathematical induction amount to proving a statement about all positive integers by showing that it's true for $n=0$ and showing that it is possible to reduce the case for one number to the case for a smaller number. Mathematical induction is logically valid.

 

[stevendaryl]

The correct number of days after the announcement, Alice no longer believes W2 is possible, but she already new W3 and W4 where not possible.

Yes, Alice knows that W3 and W4 are not possible, but as far as she knows, Bob believes that W3 is possible, and as far as she knows, Bob believes that Carol believes that W4 is possible.

 

[stevendaryl]

Let me try to explain why I might be wrong.
Suppose that I receive the letter
"At least one of monks has blue eyes. Now all monks know that."
What can I conclude from it? I can interpret it in two interestingly different ways:
The first interpretation is that all the others have received the same message as I did, namely
"At least one of monks has blue eyes. Now all monks know that."
The second interpretation is that all the others received a shorter message than I did, namely
"At least one of monks has blue eyes."
In the first interpretation, "that" refers to both sentences, so it involves self-reference.
In the second interpretation, "that" refers only to the first sentence, so it doesn't involve self-reference.

Exactly right. Saying "At least one monk has blue eyes, and all monks received this message" is equivalent to:

1. At least one monk has blue eyes.
2. All monks know that 1 is the case.
3. All monks know that 2 is the case.
4. All monks know that 3 is the case.
5. Etc.

Each additional statement gives more information (up until you run out of monks). If a monk only receives message 1, he might think that he's the only one who received the message. If he receives messages 1&2, he might think that the other monks only received message 1. If he receives messages 1, 2, and 3, he might think that the other monks only received messages 1&2. Etc.

 

[.Scott]

Yes. My answer is that, for $n\geq 3$, they will all kill themselves after $n$ days, counting from the day at which they all together learned about the suicide law. This assumes that they did learn the suicide law together at the same day. If they learned it in another way, then the answer depends on how exactly did they learn it.

Learning about the suicide rule is not equivalent to a new public statement that there is at least one blue-eyed monk. The countdown starts when the statement made by the guru is combined with the daily suicide opportunity to publicly demonstrate a minimum number of blue-eyed monks that is greater than 1.

A simple statement of the rules would never, under any circumstances, resolve the blue-eyed monk count to any monk. In contrast, a statement that there is at least one blue-eyed monk would resolve the blue-eyed monk count to a monk if he was the only blue-eyed monk ... and he would communicate this to all other monks on the next scheduled suicide opportunity.

 

[lukesfn]

Mathematical induction amount to proving a statement about all positive integers by showing that it's true for $n=0$ and showing that it is possible to reduce the case for one number to the case for a smaller number. Mathematical induction is logically valid.

Yes, thank you, you are probably correct, I what little formal knowledge I have in this area has been long forgotten.

Also, I don't think anything else I am saying leads anywhere particularly useful apart from being things my intuition doesn't like.

I feel like either the instant application of perfect logic being something quite un-natural, can cause very unintuitive results in some cases, or instant application of perfect logic as a concept has some flaw in scope of this puzzle.

 

[stevendaryl]

Yes, thank you, you are probably correct, I what little formal knowledge I have in this area has been long forgotten.

Also, I don't think anything else I am saying leads anywhere particularly useful apart from being things my intuition doesn't like.

I feel like either the instant application of perfect logic being something quite un-natural, can cause very unintuitive results in some cases, or instant application of perfect logic as a concept has some flaw in scope of this puzzle.

I think the reasoning is completely air-tight.

Suppose that only Carol has blue eyes, and she receives the announcement that there is at least one blue-eyed person. Then she would know it was her, and would kill herself. So we conclude:

• Fact 1: If Carol has blue eyes and nobody else has blue eyes, she kills herself in one day.

This is logically equivalent to:

• Fact 1': If Carol has blue eyes, and she doesn't kill herself after one day, then somebody else also has blue eyes.

Now, suppose that only Carol and Bob have blue eyes and they receive the announcement. Bob doesn't know whether he has blue eyes, or not. But he knows Fact 1'. So if Carol doesn't kill herself after one day, then there are at least two blue-eyed people. Looking around, Bob would know that the second person was him. So he would kill himself on the second day. So we conclude:

• Fact 2: If Carol and Bob have blue eyes and Alice doesn't have blue eyes, then Bob kills himself on the second day.

This is logically equivalent to:

• Fact 2': If Carol and Bob both have blue eyes, and Bob does not kill himself on the second day, then Alice has blue eyes.

Now, suppose that Alice, Bob and Carol all three have blue eyes. After two days, Bob doesn't kill himself. Then Alice uses Fact 2' to conclude that she herself has blue eyes. So she kills herself on the third day. So we conclude:

• Fact 3: If Alice, Bob and Carol all have blue eyes, then Alice kills herself on the third day.

So the mathematical induction here is just the chain from Fact 1 to Fact 2 to Fact 3.

 

[D H]

Okay, well I'm assuming that the population is fixed (other than deaths by suicide).

That is a good assumption. There's no reason to make the problem more complex than it already is.

Not quite. For the puzzle to work, a blue-eyed islander has to consider it possible that his eyes are non-blue.

That there have to exist non blue-eyed islanders (or whatever) does not necessarily follow. An islander merely has to consider it to be possible that their own eyes are not blue.

(3) No other source of information, except by the prophet himself, provides sufficient information for all blue-eyers to eventually realize that they are blue-eyers.

then one arrives at the final result

(4) If there are n blue-eyers, they will realize that they are blue-eyers after n days, starting from the first day at which the information was given by the prophet.

Note that (1) and (2) are not sufficient to get (4). To get (4), one also needs (3). In other words, (3) is tacitly assumed in the standard solution that leads to the final result (4).

Your assumption #3 is very explicitly stated in the well-formulated versions of the riddle. The story has to be carefully crafted so as to make your assumption #3 explicit. For example, the island has no reflective surfaces, and communicating eye color (whether by talking, writing, or even surreptitious looks) is forbidden.

This why I have complaining about the pervasive use of non-standard versions of the riddle in this thread. It's akin to people trying to understand the twin paradox in special relativity by creating their own non-standard quintuplet paradox that misses the point. This is why this thread has become so long.

The other thing that has to be made clear is that the new knowledge imparted by the guru has to be common knowledge. Everyone has to have heard, understood, and believed the guru's statement.

 

[lukesfn]

I think the reasoning is completely air-tight.

Well, I have thought it all through from many angles, and I can't see any cracks my self, but I often wast large amounts of time on impossible problems before I really understand exactly why they are impossible. The only way I can see to attack the problem is to try to think of a reason why a Alice wouldn't apply the induction method, but any such argument seems unconvincing.

 

[Heinera]

Yes they will.

Ok, so you claim three blue-eyed monks will comit suicide at t+3 without any announcement. Imagine three blue-eyed monks, each sitting in solitude. They all know about the suicide rule, but have no idea about their own eye colour, so they just sit there. One day (time = t) they are all placed in the same room. According to your argument, this action alone should somehow trigger a logical process so that they all know they have blue eyes at time t+3. How?

 

[.Scott]

it seems notoriously difficult to formulate mathematically.

I believe I have it:
$T_n$ represents the statement that there are $n$ blue-eyed monks on the island.
$M_m$ represents the Monk number $m$.
$M_b$ represents every other blue-eyed monk. More specifically, for the 1st person monk $M_m$ given in the context of the $M_b$ usage, it is the set of all blue-eyed monks except $M_m$.
$B_m$ represents the statement that $M_m$ is blue-eyed.
$B$ represents the statement "I am blue-eyed". More specifically, for each $m$ given in the context of where $B$ is used, $M_m$ knows $B_m$.

$Q_{0,0}$ represents what a blue-eyed monk knows who sees no blue-eyed monks and has no guru clue.
$Q_{0,0} = (B\Rightarrow T_1)\wedge(\overline{B}\Rightarrow T_0)$

$Q_{n,0}$ represents what a blue-eyed monk knows who sees $n$ blue-eyed monks where $n>0$ and has no guru clue.
$Q_{n,0} = (B\Rightarrow (T_{n+1}\wedge(M_b$ knows $Q_{n,0})))\wedge(\overline{B}\Rightarrow (T_n\wedge(M_b$ knows $Q_{n-1,0})))$

$Q_{0,0.5}$ represents what a blue-eyed monk knows who sees no blue-eyed monks and has received the $\overline{T_0}$ public announcement from the guru and has not had a suicide opportunity.
$Q_{0,0.5} = B \wedge T_1$

$Q_{n,0.5}$ represents what a blue-eyed monk knows who sees $n$ blue-eyed monks where $n>0$ and has received the $\overline{T_0}$ public announcement from the guru but there has not been a suicide opportunity since then.
$Q_{n,0.5} = (B\Rightarrow (T_{n+1}\wedge(M_b$ knows $Q_{n,0.5})))\wedge(\overline{B}\Rightarrow (T_n\wedge(M_b$ knows $Q_{n-1,0.5})))$

$Q_{1,1}$ represents what a blue-eyed monk knows who sees 1 blue-eyed monks and there has been 1 suicide opportunity since the $\overline{T_0}$ public announcement.
$Q_{1,1} = B\wedge T_2\wedge(M_b$ knows $Q_{1,1})$
The other term to that, $\overline{B}\Rightarrow(T_1\wedge(M_b$ knows $Q_{0,1}))$, is dropped because $(T_1\wedge(M_b$ knows $Q_{0,1}))$, translated roughly as "there is only one blue-eyed monk and he knows he is dead", is demonstrated as false.

$Q_{n,1}$ represents what a blue-eyed monk knows who sees $n$ blue-eyed monks where $n>1$ and has received the $\overline{T_0}$ public announcement from the guru and there has been 1 suicide opportunity since then.
$Q_{n,1} = (B\Rightarrow (T_{n+1}\wedge(M_b$ knows $Q_{n,1})))\wedge(\overline{B}\Rightarrow (T_n\wedge(M_b$ knows $Q_{n-1,1})))$

These can be used to demonstrate that you need the declaration and that you need as many suicide opportunities as blue=eyed monks.
But more importantly, it shows the correct recursive statements.

 

[Demystifier]

Ok, so you claim three blue-eyed monks will comit suicide at t+3 without any announcement. Imagine three blue-eyed monks, each sitting in solitude. They all know about the suicide rule, but have no idea about their own eye colour, so they just sit there. One day (time = t) they are all placed in the same room. According to your argument, this action alone should somehow trigger a logical process so that they all know they have blue eyes at time t+3. How?

In the meantime I have changed my mind in the post #155, which you liked.

 

[.Scott]

Yes. My answer is that, for $n\geq 3$, they will all kill themselves after $n$ days, counting from the day at which they all together learned about the suicide law. This assumes that they did learn the suicide law together at the same day. If they learned it in another way, then the answer depends on how exactly did they learn it.

I believe your logic here is that learning the rules together and seeing blue-eyed monks is a suitable substitute for a declaration from the guru that he sees a blue-eyed monk. But that is not the case. The difference is that monks observing other monks can never reveal to any monk anything about his own eye color. In contrast, there is a condition ($T_1$) when such a declaration will reveal the eye color to a monk.
That is the critical difference - and why simply learning the rule together on the same day will not trigger the same progression as the guru.

If that was not you logic, tell me what it is.

 

[.Scott]

regardng "Okay, well I'm assuming that the population is fixed (other than deaths by suicide)."

That is a good assumption. There's no reason to make the problem more complex than it already is.

Our perfectly logical monks will make no such assumption. They will need a reason to believe that the blue-eyed monk population isn't changing. So the problem description needs to include either a static-population rule or a stipulation that the monks can recognize each other.

I prefer the stipulation that they recognize each other because, from my point of view, analysis of the original game is complete and it's time to move on to scenarios where not all monks have the same information - such as post #111.

 

[Demystifier]

If that was not you logic, tell me what it is.

In the meanwhile, I have changed my logic. See post #155.

 

[stevendaryl]

In the meanwhile, I have changed my logic. See post #155.

Well, your post #155 admitted the possibility that your reasoning was mistaken, but didn't go so far as to confirm or deny the standard conclusion.

 

[D H]

I prefer the stipulation that they recognize each other because, from my point of view, analysis of the original game is complete and it's time to move on to scenarios where not all monks have the same information - such as post #111.

With regard to what happens when six monks on the first day after the guru makes here statement? It breaks the chain. They might have broken the rule against looking at mirrors, they might have broken the about discussing eye color, or they might have committed suicide for some non-related reasons. Whichever is the case, the chain is broken. Terence Tao, in a comment on his version of the problem stated

"With these sorts of logic puzzles, there is always the implied assumption of “idealised conditions” (no deafness, colour-blindness, or other physical disabilities, no pirates to come and randomly abduct half the population on Day 17, no unexpected breakdown of causality or other laws of physics, and so forth). It is a trivial matter to “break” the logic puzzle using non-idealised conditions, but this is not the most interesting aspect to the puzzle."​

To be brutally honest, these side discussions are akin to discussing the relativistic triplet paradox when the discussants don't understand the twin paradox. Properly constructed versions of the problem showcase the difference between mutual knowledge and common knowledge. Throughout this thread, there hasn't been much discussion on those key differences.

This thread started on a bad footing, and it shows in that the discussion has gone on for nine pages. (Long threads typically are not a sign of a good discussion.) One way it started off on a wrong footing was that it did not create a base of common knowledge for the discussion. Which version of the problem should be discussed? Monks committing suicide? That's a rather non-standard version of the problem.

This lack of a common basis for a discussion is a bit strange given that the problem is about the difference between mutual knowledge and common knowledge. The opening post did link to the wikipedia article on common knowledge, but in my opinion, that article (like many at wikipedia) isn't that good. The article on common knowledge at the Stanford Encyclopedia of Philosophy, http://plato.stanford.edu/entries/common-knowledge/, is much better and is much more in depth.

 

[stevendaryl]

This lack of a common basis for a discussion is a bit strange given that the problem is about the difference between mutual knowledge and common knowledge. The opening post did link to the wikipedia article on common knowledge, but in my opinion, that article (like many at wikipedia) isn't that good. The article on common knowledge at the Stanford Encyclopedia of Philosophy, http://plato.stanford.edu/entries/common-knowledge/, is much better and is much more in depth.

The difference between common knowledge and mutual knowledge can be formally captured using indexed "knowledge operators". Let $K_i P$ mean "Person #i knows $P$". Then to say that $P$ is mutual knowledge is to say that, for each individual $i$,

$K_i P$

To say that $P$ is common knowledge is to say

For all $i$: $K_i P$
For all $i$ and $j$: $K_i (K_j P)$
For all $i, j$ and $k$: $K_i (K_j (K_k P))$
etc.

 

[.Scott]

With regard to what happens when six monks on the first day after the guru makes here statement? It breaks the chain. They might have broken the rule against looking at mirrors, they might have broken the about discussing eye color, or they might have committed suicide for some non-related reasons. Whichever is the case, the chain is broken.

Bear in mind that this problem is intended to reflect "history". All monks are following the rule and no monk is committing suicide unless he know he is blue-eyed.

 

[.Scott]

In the meanwhile, I have changed my logic. See post #155.

Does this mean you do not believe that day 1 will become a trigger to the suicide countdown - even with no messages or statements from the guru?

 

[Buzz Bloom]

Hi @Demystifier:

I confess that I have not read through all 174 previous posts. However I did randomly scan through about 20-30 of them. None of the ones I scanned mentioned the following point.

Before the public announcement that at least one person had blue eyes, the following was not common knowledge:

Every person knows that: every person knows that: every person knows that: . . . etc. at least one person has blue eyes.​

If I understand the problem statement correctly, the above must be common knowledge that they all have at the same time in order for all the N people with blue eyes to deduce, after N-1 days, that they all have blue eyes.

CORRECTION
I have come to realize that the above description of the common knowledge gained from the public announcement is incomplete.

Every person knows that at least one person has blue eyes, AND
every person knows that every person knows that at least one person has blue eyes, AND
every person knows that every person knows that every person knows that at least one person has

blue eyes, AND​

. . . etc.​

Note that before the announcement:
if there is one person with blue eyes, it is not true that every person knows that at least one person has blue eyes;
if there are two persons with blue eyes, it is not true that every person knows that every person knows that at least one person has blue eyes;
if there are three persons with blue eyes, it is not true that every person knows that every person knows that every person knows that at least one person has blue eyes;
etc.

Regards,
Buzz