Monks. (1 Viewer)

[gr_111]

no :\ . How can the process not occur if they all know that there is at least one blue eyed monk (or at least one brown eyed monk), and they are all as intelligent as we are assuming?

Come on, seanieg, explain...!

 

[gr_111]

This doesn't work without being told that there is atleast 1 blue eyed monk, which can't be true since you stated there is none, so the case of having no blue eyed monks can't happen.

yeah, fine, but i was just trying to account for every different configuration.

 

[seanieg89]

So this is the answer I was looking for:

By telling the monks "at least one of you has blue eyes", the stranger is actually telling the monks a sequence of facts, at least one of which is new knowledge to them.

S1) At least one monk on the island has blue eyes.

S2) Every monk on the island knows S1)

S3) Every monk on the island knows S2)

Etc.

No matter how many blue eyed monks there were to begin with (excluding the trivial case of zero when the stranger is lying), at least one of these facts is new to the monks.

Eg if there were exactly two monks with blue eyes, then every monk on the island knows that there is a blue eyed monk on the island (Every monk knows S1). But the two blue eyed monks will NOT know S2 until the strangers words.

 

[SunnyScience]

Is it bad that even with both solutions now, I still don't understand it? lol.

 

[seanieg89]

Is it bad that even with both solutions now, I still don't understand it? lol.

Not really, these things are designed to be pretty counter-intuitive. Think about it a bit and perhaps it will make more sense later.

 

[barbernator]

So this is the answer I was looking for:

By telling the monks "at least one of you has blue eyes", the stranger is actually telling the monks a sequence of facts, at least one of which is new knowledge to them.

S1) At least one monk on the island has blue eyes.

S2) Every monk on the island knows S1)

S3) Every monk on the island knows S2)

Etc.

No matter how many blue eyed monks there were to begin with (excluding the trivial case of zero when the stranger is lying), at least one of these facts is new to the monks.

Eg if there were exactly two monks with blue eyes, then every monk on the island knows that there is a blue eyed monk on the island (Every monk knows S1). But the two blue eyed monks will NOT know S2 until the strangers words.

ahhh yes

 

[someth1ng]

I don't think this works if there is more than 1 monk with blue eyes?

If you had 1 monk with blue eyes, he'd know because there are no other blue monks (he must know).

If you had 2 monks with blue eyes, each of the blue monks will look at each other and won't be able to determine if they have blue eyes or not?

 

[seanieg89]

I don't think this works if there is more than 1 monk with blue eyes?

If you had 1 monk with blue eyes, he'd know because there are no other blue monks (he must know).

If you had 2 monks with blue eyes, each of the blue monks will look at each other and won't be able to determine if they have blue eyes or not?

They will see each other alive the next day and deduce that they themselves must have blue eyes. Read the earlier post on the solution in terms of induction.

 

[someth1ng]

They will see each other alive the next day and deduce that they themselves must have blue eyes. Read the earlier post on the solution in terms of induction.

Right, that just clicked...lol

What about 3 monks?

 

[seanieg89]

Each of the three will observe that the two blue eyed monks they can see do no commit suicide on the second night. Hence all three will deduce they are blue and commit suicide on the third night.

 

[someth1ng]

Each of the three will observe that the two blue eyed monks they can see do no commit suicide on the second night. Hence all three will deduce they are blue and commit suicide on the third night.

So if there's three blue monks, the first one will see two but if those two don't suicide on the second night, there must be three.
That same goes if there's four blue monks? If there's no suicide on the third night, there must be four inc himself?


hmmm...interesting

 

[seanieg89]

Yes.

 

[gr_111]

So this is the answer I was looking for:

By telling the monks "at least one of you has blue eyes", the stranger is actually telling the monks a sequence of facts, at least one of which is new knowledge to them.

S1) At least one monk on the island has blue eyes.

S2) Every monk on the island knows S1)

S3) Every monk on the island knows S2)

Etc.

No matter how many blue eyed monks there were to begin with (excluding the trivial case of zero when the stranger is lying), at least one of these facts is new to the monks.

Eg if there were exactly two monks with blue eyes, then every monk on the island knows that there is a blue eyed monk on the island (Every monk knows S1). But the two blue eyed monks will NOT know S2 until the strangers words.

Sorry if i'm being annoying, but I still can't see how they need to know S2 for the action to start?

 

[sleepybearx]

dis is lyk s0 confuzing. who thought of this, what have you been doing with your life!?