Re: Second attempt ...
 

So, I took a bit of liberty when I stated when they left. All the monks were there on breakfast day 3.

Very standard riddle, but I think it's a very good one.

Re: Another logic riddle
 

how can these solutions work if one does not know the exact number 'some' is? You are all assuming that these monks know exactly how many BEMs there are. what if some was 10 or 4?

Each monk would look around and still see other blue eyed monks. they wouldn't know if they were one as well, unless they actually knew exactly how many there were?

if they knew how many there were? then they would all be gone after the first meal. For example, if they knew that there were 5BEMs each BEM would see only 4 BEMs and would know to leave.

Melch

STRANGE MetaSolution
 

Solution and Meta-Solution
White
<font color="white">
Solution:
The obvious answer is 7.
At the 7th meal each of the 7 reasons as follows:
If there's only the six BEM that I see, at the 6th meal they would have reasoned as follows and not shown up today:
I see 5 BEM. If I'm not Blue Eyed they should not be here today having reasoned as follows at the 5th meal:
I see 4 BEM. If I'm not Blue Eyed they should not be here today having reasoned as follows at the 4th meal:
I see 3 Bem. If I'm not Blue Eyed they should not be here today having reasoned as follows at the 3rd meal:
I see 2 BEM. If I'm not Blued Eyed they should not be here today having reasoned as follows at the 2nd meal:
I see another BEM. If I'm not Blue Eyed he should not be here having concluded he had blue eyes at the first meal.

MetaSolution:
But if there are 7 BEM they already know there are some BEMs before the stranger comes and tells them so. The stranger has provided no additional information, so why didn't the Monks figure this out long ago?

Answer: Before the stranger makes the announcement the string of logic cannot make the last step. A single BEM cannot conclude he has Blue Eyes until the stranger comes and announces the fact.

This is STRANGE because with 7 actual BEM's it's clear to them all that noone could possibly think he's a lone BEM, yet for the string of logic to work the information available to the Monks must be such that IF there were a Lone BEM he could conclude he was blue eyed after one meal.
</font>
End White

I think I need to see more examples of such "Strangeness" before my mind quits rebelling against this notion.

PairTheBoard

Re: STRANGE MetaSolution
 

It's not that strange. I will clarify your solution in white:

<font color="white">It's really a proof my mathematical induction. To prove something by mathematical induction, you need a base case, and an inductive step. Let P(n) = If there are n blue eyed monks, they will be gone at the nth meal after the strangers statement. Then a proof by induction requires:

P(1) = If there is 1 blue eyed monk, he will be gone at the first meal after the stranger's statement.

Clearly this is true because that blue eyed monk sees no one. This is what the stranger's statement is required to prove.

P(n) -&gt; P(n+1) = Assume that If there are n blue eyed monks, and they would be gone at the nth mea. Then if there are n+1 blue eyed monks, they would be gone at the n+1st meal.

This is basically what you said in your original problem, only in the other direction (sort of). All you have to do is state that if there were n+1 blue eyed monks, each of those monks would expect the other n to leave at the nth meal if they themselves did not have blue eyes. When none of them left, they all leave the next meal knowing that they must have been blue eyed for the others to have stayed.</font>

Re: STRANGE MetaSolution
 

ok. so maybe "STRANGE Meta-Solution" was a bit of an overstatement.

PairTheBoard

Re: Another logic riddle
 

You mean to tell me that none of these guys is able to look in a doggone mirror and ascertain his own eye color?
I'm going to file suit against the OP to try and get the time back that I spent reading this thread. [img]/images/graemlins/grin.gif[/img]

Defending Strangeness
 

White
<font color="white">
Thinking about it some more, I've still got the sense of Strangeness here.

Whether you show the solution by induction or straight forward backward reasoning, the key to what New Information the Stranger has provided when he tells a room full of Monks who can all see at least 6 Blue Eyed Monks, that there are some Blue Eyed Monks there - what a suprise - is the following. The stranger's comment has now made the following statement True.

"If there were one BEM he would leave after hearing the stranger's comment".

This statement was not true before the stranger's comment but now is true. Even though there is no single Blue Eyed Monk and everybody knows it.

The Monks are thinking, There is no single Blue Eyed Monk here, but IF there were and he heard the stranger's statement, this is what he would do. Therefore ...

I think this is something that could be played around with. It seems to me there's a priciple here that could be used in creating other puzzles. It certainly has a strange flavor to my sense of taste.
</font>
End White

PairTheBoard

A monk\'s breakfast
 

It's safe to say you messed up the description of the riddle pretty well.

[img]/images/graemlins/cool.gif[/img]