Another logic riddle
 

So, there are 20 monks. They only see each other three times a day, during breakfast, lunch, and dinner, where they sit around in a big circle facing everyone else.

Otherwise, they're in meditation.

These monks believe that being blue-eyed is a sin, and any blue-eyed monk must leave. They are also very logical, and all think at a very high level.

At breakfast, the monks welcome a stranger. At the beginning of the meal, he says "some of you are blue-eyed."

Two days later, not a blue-eyed monk is to be found at lunch, while every other monk stays.

How many blue-eyed monks are there? And how do they know to leave?

Re: Another logic riddle
 

First, being blue-eyed is not a sin.

Second, answer: <font color="white">6</font>

Re: Another logic riddle
 

And we are to assume that they all have oaths of silence?

Re: Another logic riddle
 

I'm going to go with <font color="white">none. The guest lied. If a monk had blue eyes and believed that any blue eyed monks should leave.. he would already have left.</font>

Re: Another logic riddle
 

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I'm going to go with none. The guest lied. If a monk had blue eyes and believed that any blue eyed monks should leave.. he would already have left.

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[ QUOTE ]
Another logic riddle

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Re: Another logic riddle
 

[ QUOTE ]
Another logic riddle

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Wow, I didn't read that part. It must be <font color="white"> 6 because there are six meals and that is oh so very logical .</font>

Re: Another logic riddle
 

[ QUOTE ]
And we are to assume that they all have oaths of silence?

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All the monks took an oath not to talk about being blue-eyed or anything having to do with that. They only see each other at meal-time.

The guy was telling the truth. The monks know that.

And, why do the blue-eyed monks know that they are blue-eyed?

Re: Another logic riddle
 

read below but i dunno how to write in white so dont look, it may be wrong im not too good at these logic riddles

Re: Another logic riddle
 

there are 2, the 1st monk looks around and sees only 1 blue eyed monk and knows he must be the other one, the 2nd monk sees this guy leave and doesnt see another monk with blue eyes therefore he realizes he must have been the other. there cannot be less than 2 because the stranger said SOME of you are blue eyed. <font color="white"> </font> <font color="white"> </font> <font color="white"> </font>

Re: Another logic riddle
 

I have to assume the monks don't know their own eye color.
I can't assume the stranger was telling the truth.

If there was no blue-eyed-monk (BEM), every other monk, seeing no pair of blue eyes, would assume the stranger was lying.

If there was only 1 BEM, every other monk, seeing only 1 pair of blue eyes, would assume they were the other -- and at the next meal there would be only 1 BEM.

If there were 2 BEMs, each would only see 1 other -- and at the first meal (breakfast day 1) conclude they were the other, and both would leave. None at lunch.

If there were 3 BEMs, each would see 2 others at the next meal (lunch day 1), and conclude the others were seeing more than 1 (or they would have left). That must be them. All 3 leave. None at dinner.

If there were 4 BEMs, each would see 3 others (dinner day 1). So, why didn't all three leave? There's only one explanation. They all see more than 2. There must be 4. That must be them. All four leave. None at breakfast day 2.

If there were 5 BEMs, each would see four others (breakfast day 2). Why didn't they leave? They must each also see 5. That must be them. All 5 leave. None at lunch day 2.

Etc.

6 = Lunch day 2, none at dinner, day 2.
7 = Dinner day 2, none at breakfast day 3.
8 = Breakfast day 3, none at lunch.

So the question is, how many times must they meet before they come to this conclusion? There are six meals between the breakfast with the stranger and lunch two days later.

The answer is that there may have been 2, 3, 4, 5, 6, 7, or 8 BEMs.

Since the problem states -
[ QUOTE ]
Two days later, not a blue-eyed monk is to be found at lunch, while every other monk stays.

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It doesn't state that it is the first time this pattern occurred.

-SheetWise

Re: Another logic riddle
 

The answer is:

<font color="white"> 7. At the first breakfast, if there is only one BEM, then he will leave because he will look around and not see any BEMs and will know that he must be the one.

If there were two BEMs they will both leave at the first lunch. Each BEM will only see one BEM, but they will know that there must be two because if there were only one he would have left at breakfast.

And so on until the 7th meal when 7 BEMs leave </font>

Re: Another logic riddle
 

[ QUOTE ]
7. At the first breakfast, if there is only one BEM, then he will leave because he will look around and not see any BEMs and will know that he must be the one.

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Can't be. The stranger said "some..." - if there is only one BEM, every other monk will assume they are the other and leave. The one BEM will assume the stranger was lying when he said "some of you are blue eyed". The next meal will have only one person -- the BEM.

Re: Another logic riddle
 

This is a very standard kind of riddle, which is probably most of the reason that jason_t answered quickly and with no explanation.

Here's something you might enjoy thinking about: suppose in your community of monks there are at least two blue-eyed monks. What information is the stranger putting into the system, then? It's clear that all of the monks already know that there are some blue-eyed monks.

Re: Another logic riddle
 

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suppose in your community of monks there are at least two blue-eyed monks. What information is the stranger putting into the system, then? It's clear that all of the monks already know that there are some blue-eyed monks.

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I don't think it's clear that all of the monks already know that there are some blue-eyed monks. Two of them know that there is at least one, the rest know that there are at least two.

I had to look up "some". Could this be (one of) my fatal flaw(s)? -

[ QUOTE ]
adj 1: quantifier; used with either mass nouns or plural count nouns to indicate an unspecified number or quantity; "have some milk"; "some roses were still blooming"; "having some friends over"; "some apples"; "some paper"

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adj 1: quantifier; used with either mass nouns or plural count nouns to indicate an unspecified number or quantity; "have some milk"; "some roses were still blooming"; "having some friends over"; "some apples "; "some paper"

Re: Another logic riddle
 

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I don't think it's clear that all of the monks already know that there are some blue-eyed monks. Two of them know that there is at least one, the rest know that there are at least two.

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Nobody here means some in the context you're taking it; take some to mean at least one. Alternatively, assume you have 3+ monks with blue eyes and then even your usage is satisfied.

Re: Another logic riddle
 

Answer:
<font color="white"> 7. If there was only 1 he'd be gone at lunch, if 2 they'd be gone at dinner, etc. </font>

Re: Another logic riddle
 

The stranger is clearly lying, because the monks would already know if they had blue eyes by looking in a mirror, a window, or a pool of water. Also even if there were no reflective surfaces the monks surely would told the other monks that they have blue eyes unless there is a vow of silence, which you also did not specify. If neither of these things are true, then there are (answer in white) <font color="white"> 7 monks by the principle of mathematical induction. </font>

Second attempt ...
 

Ooops.

On the revised assumption that some can mean one. [img]/images/graemlins/wink.gif[/img]
I still can't assume the stranger was telling the truth.

If there was no blue-eyed-monk (BEM), every other monk, seeing no pair of blue eyes, would assume it was them and leave. Nobody at lunch.

If there was only 1 BEM, he would realize who he was at breakfast, and not show up at lunch. The others, realizing there were some, and some saw no others, stay. None at lunch day 1.

If there were 2 BEMs, each would only see 1 other at breakfast -- and wait for them to leave before lunch, At lunch day 1, seeing the other return, conclude there was another, it was them, and both would leave. None at dinner day 1.

If there were 3 BEMs, each would see 2 others at breakfast, and wait for them to show up to lunch day 1, see the other return, realize they were 1 of 2, and leave before dinner day 1. When they show up at dinner day 1, conclude the two were seeing more than 1 other, Conclude it was them, and all 3 leave. None at breakfast day 2.

Etc.

4 = None at lunch, day 2.
5 = None at dinner day 2.
6 = None at breakfast day 3.
7 = None at lunch day 3.

So there may have been 2, 3, 4, 5, 6, or 7 BEMs, again, since the problem doesn't state lunch day 3 is the first time the pattern occurred.

Where did I screw up this time? [img]/images/graemlins/wink.gif[/img]

-SheetWise

Re: Second attempt ...
 

Correction. And .... the revised option of 1.

Re: Another logic riddle
 

The only thing that make sense is: <font color="white"> </font> 18 of the monks are blind. Thats the only thing which fit with the statemaent: a stranger says some are BEMs. the starnger comes with new info, the alarm goes off and two monks that are not blind will each know that they both are blue-eyed because the stranger said "some"

Alternative: 19 are blind. last one is blue-eyed. The word "some" is not taken into consideration.

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]