Eye color suicide
Mike Franklin
eyes must kill themselves" puzzle. In my version there were 38 blue eyed
and 47 brown eyed people. I said that 38 days after a visitor says
"There's at least one person on this island with blue eyes" the blue eyed
people kill themselves, which I believe to be the correct answer. My
mother had this point: Since the visitor said something that everybody
already knew, why did this cause a problem. I was unable to answer this,
and I hoped members of this group could.
So, what do you experts think?
Mike
Peter Renzland
I suspect it has to do with prividing a virtual/actual starting point.
How about a precise problem statement?
There are two traps in the problem as I know it
1. The case of all one colour
2. The abbreviation/shortcut trap (why start count at 1 when we can start
at 37?)
There is also the matter of effective protocol - what exactly is the
enumeration device?
--
----------------- _@_ {)/' (}, @ `\@ {)/' peter@ o
Peter Renzland TORONTO /\ /\_._,(_/ ()_/7 /\_._,(_\ PLANIX \_/
Je danse, donc je suis. ' \ /_\ /_\ /) /\ /_\ .~~~~~ _|_
----------------- /) /( / )( \ ' ) ( ` com~~~ |
Aaron Bergman
<franklin-011...@franklin.chinalake.navy.mil>,
fran...@llab.chinalake.navy.mil (Mike Franklin) wrote:
> Over Thanksgiving, I told the familiar "Whoever learns the
> color of their eyes must kill themselves" puzzle. In my
> version there were 38 blue eyed and 47 brown eyed people. I
> said that 38 days after a visitor says "There's at least one
> person on this island with blue eyes" the blue eyed people
> kill themselves, which I believe to be the correct answer. My
> mother had this point: Since the visitor said something that
> everybody already knew, why did this cause a problem. I was
> unable to answer this, and I hoped members of this group
> could.
>
> So, what do you experts think?
To save a little time, it's easier to talk about the 2 blue eyed
case. In that case, it is true that the statement "there is
atleast one blue eyed person on this island" is something that
everyone knows. The new information is that now everyone knows
that everyone knows that there is atleast one blue eyed person on
the island. For higher numbers of people, repeat as needed.
Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
Mike Franklin
Bergman <aber...@princeton.edu> wrote:
> To save a little time, it's easier to talk about the 2 blue eyed
> case. In that case, it is true that the statement "there is
> atleast one blue eyed person on this island" is something that
> everyone knows. The new information is that now everyone knows
> that everyone knows that there is atleast one blue eyed person on
> the island. For higher numbers of people, repeat as needed.
>
> Aaron
If there are three blue eyed's, then everyone already knows that everyone
knows that there is at least one blue eyed. So where's the new info?
Mike
Martin Julian DeMello
It's a matter of mental models
If there are three people, each imagine the other two see *two* blue eyed
people.
So say A sees B and C have blue eyes. "Aha," he thinks, "everyone knows
everyone knows there is at least one blue eyed person". But then he thinks
about it some more, and realises that B sees only C with blue eyes. So he
thinks B thinks C thinks that *no one* has blue eyes. For who would C see
with blue eyes? Only A and B. But A doesn't know he has blue eyes, so he
thinks C sees only B. And then he reasons that B doesn't know *he* has blue
eyes, and so even though C can see B, B doesn't know that.
Or something :)
(In summary -
C sees A and B
A thinks C sees B
A thinks *B thinks* C sees no one
)
This generalises to N people, which is why you need an unambiguous external
source that says 'there is one blue eyed person' - that provides you with a
stable 'everyone knows everyone knows everyone knows...' through any number
of levels of iteration.
--
Martin DeMello
Monwhea Jeng
> In article <abergman-8CEAD2...@cnn.princeton.edu>, Aaron
> Bergman <aber...@princeton.edu> wrote:
>
> > To save a little time, it's easier to talk about the 2 blue eyed
> > case. In that case, it is true that the statement "there is
> > atleast one blue eyed person on this island" is something that
> > everyone knows. The new information is that now everyone knows
> > that everyone knows that there is atleast one blue eyed person on
> > the island. For higher numbers of people, repeat as needed.
> >
> > Aaron
>
> If there are three blue eyed's, then everyone already knows that everyone
> knows that there is at least one blue eyed. So where's the new info?
>
> Mike
Yes, but now the new info is that everyone knows that everyone knows that
everyone knows that there is at least one blue-eyed person. They didn't
know that before. The "repeat as needed" in the original instructions
meant to add a new layer of "everyone knows" for each new person.
Momo
Willem-Jan Monsuwe
)If there are three blue eyed's, then everyone already knows that everyone
)knows that there is at least one blue eyed. So where's the new info?
The new info is that everyone knows that everyone knows that everyone knows
there is atleast one blue eyed person on the island.
So, in the N-person case, the new info is that repN(everyone knows that)
there is at least one blue-eyed person on the island.
Suppose there are 10 blue-eyed persons, then they all see 9 blue-eyed
persons. Each will try to reason when those 9 will figure this out, given
that he himself does _not_ have blue eyes. So he will assume each of those
9 persons sees _8_ persons, and will reason accordingly. (They don't see 8
persons, but he _assumes_ that they do.)
So if someone says 'There is at least one blue-eyed person' the new
information (with 10 blue-eyed persons) is that everyone meta-knows that
there are blue-eyed persons.
SaSW,
--
Willem (at stack dot nl)
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
Ben John
> Suppose there are 10 blue-eyed persons, then they all see 9 blue-eyed
> persons. Each will try to reason when those 9 will figure this out, given
> that he himself does _not_ have blue eyes. So he will assume each of
those
> 9 persons sees _8_ persons, and will reason accordingly. (They don't see
8
> persons, but he _assumes_ that they do.)
> So if someone says 'There is at least one blue-eyed person' the new
> information (with 10 blue-eyed persons) is that everyone meta-knows that
> there are blue-eyed persons.
>
Has anyone actually ever tried this experiment in real life? Or been part of
such a group of people trying to figure out the colour of their eyes/spots
on heads/number of hats etc..
I know in the puzzles the people are always very inteligent philosophers, so
how would you reason in a room containing a mixture of different people?
Richard Heathfield
It's probably not worth trying this with young children...
"Daddy's hat is green."
"No, you're not supposed..."
"No it isn't! It's blue!"
"Shh! Try to work out..."
"I'm bored."
"Mummy, what colour is my hat?"
"It's red, dear."
"NO! She's supposed to try to w..."
"That's nice, Mummy, we've got the same colour!"
"Give me my hat back! Daddy, he pinched my hat!"
--
Richard Heathfield
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R Answers: http://users.powernet.co.uk/eton/kandr2/index.html
Mike Franklin
(Willem-Jan Monsuwe) wrote:
> So, in the N-person case, the new info is that repN(everyone knows that)
Thanks. That makes sense to me. Now if only I can convince my mom.
Mike
Joe Slater
>(Willem-Jan Monsuwe) wrote:
>> So, in the N-person case, the new info is that repN(everyone knows that)
>> there is at least one blue-eyed person on the island.
fran...@llab.chinalake.navy.mil (Mike Franklin) wrote:
>Thanks. That makes sense to me. Now if only I can convince my mom.
The puzzle does make sense to me, but suppose that there are three
blue-eyed people on the island. In that case everybody *already* knows
that there must be at least one blue-eyed person, and know that
everybody else knows this.
I now both understand and do not understand the puzzle. Please
collapse my understanding.
jds
--
And now kind friends, what I have wrote,
I hope you will pass o'er,
And not criticize, as some have done,
Hitherto herebefore. (Julia Moore, "The Author's Early Life")
Dennis Yelle
[...]
> Martin "don't make my brown eyes blue" DeMello
Or, "Donuts make my brown eyes blue."
Dennis Yelle
--
I am a computer programmer and I am looking for a job.
There is a link to my resume here:
http://table.jps.net/~vert/
Hugo van der Sanden
>
> >In article <slrn92n06e...@snail.stack.nl>, wil...@snail.stack.nl
> >(Willem-Jan Monsuwe) wrote:
> >> So, in the N-person case, the new info is that repN(everyone knows that)
> >> there is at least one blue-eyed person on the island.
>
> fran...@llab.chinalake.navy.mil (Mike Franklin) wrote:
> >Thanks. That makes sense to me. Now if only I can convince my mom.
>
> The puzzle does make sense to me, but suppose that there are three
> blue-eyed people on the island. In that case everybody *already* knows
> that there must be at least one blue-eyed person, and know that
> everybody else knows this.
>
> I now both understand and do not understand the puzzle. Please
> collapse my understanding.
Everyone knows that there are blue-eyed people on the island. The only
reason nobody dies is because it is a taboo subject - noone has ever
mentioned the fact, and everyone knows the blue-eyed people are alive
only because of that.
As soon as someone breaks that taboo, and mentions that there is a
blue-eyed person on the island, the clock is ticking. Anyone who has
seen seven pairs of blue eyes on neighbours expects that in seven days
those neighbours - who he assumes all know of only six blue-eyed
neighbours - will kill themselves. On the eighth day he realises the
awful truth.
Does that help?
Hugo
Martin Julian DeMello
> The puzzle does make sense to me, but suppose that there are three
> blue-eyed people on the island. In that case everybody *already* knows
> that there must be at least one blue-eyed person, and know that
> everybody else knows this.
> I now both understand and do not understand the puzzle. Please
> collapse my understanding.
I stepped through the three person case upthread - here's a rehashing of it:
1. C sees A and B
2. A thinks C sees B
3. A thinks *B thinks* C sees no one
1 - Given
2 - Because A doesn't know his own eye colour, so he defaults to brown
3 - Because A thinks *he* is brown, and even though he knows B is blue, he
also knows that B does not know that.
This generalises to N people, which is why you need an unambiguous external
source that says 'there is one blue eyed person' - that provides you with a
stable 'everyone knows everyone knows everyone knows...' through any number
of levels of iteration.
--
Matthew T. Russotto
Joe Slater <joeDEL...@yoyo.cc.monash.edu.au> wrote:
}>In article <slrn92n06e...@snail.stack.nl>, wil...@snail.stack.nl
}>(Willem-Jan Monsuwe) wrote:
}>> So, in the N-person case, the new info is that repN(everyone knows that)
}>> there is at least one blue-eyed person on the island.
}
}fran...@llab.chinalake.navy.mil (Mike Franklin) wrote:
}>Thanks. That makes sense to me. Now if only I can convince my mom.
}
}blue-eyed people on the island. In that case everybody *already* knows
}that there must be at least one blue-eyed person, and know that
}everybody else knows this.
But not everyone knows that everyone knows that there's at least two
blue-eyed people on the island. The blue-eyed people have to consider the
possibility that the other blue-eyed people think there's only one
blue-eyed person on the island. When the announcement is made, the
blue-eyed people) know nothing will happen the first day. But the
second day, they know that if they don't have blue eyes, all the other
blue-eyed people will kill themselves.
}I now both understand and do not understand the puzzle. Please
}collapse my understanding.
IMO, the announcement basically starts a timer which provides
information on day n-1.
--
Matthew T. Russotto russ...@pond.com
"Extremism in defense of liberty is no vice, and moderation in pursuit
of justice is no virtue."
Joe Slater
>> I now both understand and do not understand the puzzle. Please
>> collapse my understanding.
Hugo van der Sanden <h...@crypt0.demon.co.uk> wrote:
>Everyone knows that there are blue-eyed people on the island. The only
>reason nobody dies is because it is a taboo subject - noone has ever
>mentioned the fact, and everyone knows the blue-eyed people are alive
>only because of that.
>
>As soon as someone breaks that taboo, and mentions that there is a
>blue-eyed person on the island, the clock is ticking. Anyone who has
>seen seven pairs of blue eyes on neighbours expects that in seven days
>those neighbours - who he assumes all know of only six blue-eyed
>neighbours - will kill themselves. On the eighth day he realises the
>awful truth.
>
>Does that help?
Not really. Let's say that there are at least three blue-eyed people
on the island. Suppose an islander broke the taboo by announcing that
there is at least one blue-eyed person on the island. Has he said
anything which is not known (and *known* to be known) by everyone
else?
Aaron Bergman
Slater <joeDEL...@yoyo.cc.monash.edu.au> wrote:
> Not really. Let's say that there are at least three blue-eyed
> people on the island. Suppose an islander broke the taboo by
> announcing that there is at least one blue-eyed person on the
> island. Has he said anything which is not known (and *known* to be known) by everyone
> else?
Imagine that you have blue eyes. You see two other people with
blue eyes. Now you know that they each know that someone has blue
eyes on the island, but you're not sure that they know that
everyone knows this. Once the statement is made, you know that
they know that everyone knows that (whew) someone on the island
has blue eyes. Once you know this, you know that they will kill
themselves at the end of the second day if they're the only two
blue eyed people on the island. When they don't, well, oh shit.
For four people, if you're one of the blue eyed people, you now
know that they know that there are blue eyed people on the
island. You even know that they know that everyone knows that
there are blue eyed people on the island. Not only that, you also
know that they know that everyone knows that there are blue eyed
people on the island. But, do you know that they know that
everyone knows that everyone knows there are blue eyed people on
the island? There are three of them, so if each of them sees two
people with blue eyes, they know that each of them knows that
everyone knows that there are blue eyed people on the island. But
then, if they were the only three, they'd be looking at two
people each, so they're not sure that those two people know that
everyone knows that everyone knows there are blue eyed people on
the island. So, if you don't know your own eye color, you don't
know that they know that everyone knows that everyone knows that
there are blue eyed people on the island. But, once the oracle
speaks, it communicates to everyone that there is atleast one
blue eyed person on the island. But it also communicates that
everyone knows that. And it communicates that everyone knows that
everyone knows that. And it communicates, well, you get the idea.
The statement actually conveys this infinite chain of
information. And that's new.
I take no responsibility for getting the number of "everyone
knows" in the above correct.
Joe Slater
>Imagine that you have blue eyes. You see two other people with
>blue eyes. Now you know that they each know that someone has blue
>eyes on the island, but you're not sure that they know that
>everyone knows this.
I *am* sure. Everybody on the island knows everybody else. As long as
there are at least two blue-eyed people on the island, it is
necessarily true that everybody knows that there is at least one
blue-eyed person. The missionary cannot give them any more
information.
I understand how the chain of logic works. I just don't see that it
can work for n>=3. When I approach it from n=1, n=2, n=3 ... then it
makes sense. When I approach it from n=3, n=4 ... then it doesn't make
sense.
Courtenay Footman
>>In article <slrn92n06e...@snail.stack.nl>, wil...@snail.stack.nl
>>(Willem-Jan Monsuwe) wrote:
>>> So, in the N-person case, the new info is that repN(everyone knows that)
>>> there is at least one blue-eyed person on the island.
>
>fran...@llab.chinalake.navy.mil (Mike Franklin) wrote:
>>Thanks. That makes sense to me. Now if only I can convince my mom.
>
>The puzzle does make sense to me, but suppose that there are three
>blue-eyed people on the island. In that case everybody *already* knows
>that there must be at least one blue-eyed person, and know that
>everybody else knows this.
>
>collapse my understanding.
>
Here is my attempt.
Three people, W, X, Y. Two eye colors, one B, one unmentioned.
People will ritually kill themselves at noon on the first day they
know they have B eyes.
At the start, W sees two B eyes. He knows that X sees one B eye.
He does not know if X knows if Y sees any B eyes.
Now the information that there is one B eye is given.
Now W knows that if, after noon has passed, and no one has killed
themselves, then X knows that Y sees at least one B eye. This is the
new information that is lethal.
To demonstrate the inductive process, lets add a fourth person, Z:
At the start, W sees three B eyes. He knows that X sees two B eye.
He knows that X knows that Y sees one B eye. He does not know if
X knows if Y knows if Z sees any B eyes.
Again, the information is given. This is interesting, because now
not only does W know that no will die after the first killing period,
he knows that everyone else knows that fact, as well. Thus, he knows
that no one will be surprised by the fact that no will die after the
the first noon.
Now W knows that, after the first noon, X will know that Y knows Z sees
at least one B eye. This is enough to start the lethal chain going, since
on the morning of the second day W, X, and Y know that at least one of the
three of them has a B eye, and that they all know that fact, so the
previous case holds. Clearly, this can be extended to n people.
Monwhea Jeng
> Aaron Bergman <aber...@princeton.edu> wrote:
> >Imagine that you have blue eyes. You see two other people with
> >blue eyes. Now you know that they each know that someone has blue
> >eyes on the island, but you're not sure that they know that
> >everyone knows this.
>
> I *am* sure. Everybody on the island knows everybody else. As long as
> there are at least two blue-eyed people on the island, it is
> necessarily true that everybody knows that there is at least one
> blue-eyed person. The missionary cannot give them any more
> information.
You're misreading the responses to your question -- specifically the
referent of the final "this" in Aaron Bergman's answer. As you say, "it is
neccessarily true that everyone knows that there is at least one
blue-eyed person." No one here is disputing that. But it is not true
that "everyone knows that everyone knows that there is at least one
blue-eyed person." Look at the two statements
A. Everyone knows that there is a blue-eyed person on the island
B. Everyone knows that everyone knows that there is a blue-eyed person
on the island
Now imagine that you are on the island, and see your two fellow
islander, both with blue eyes -- call them Kim and Larry, and yourself
Joe. You know Kim and Larry have blue eyes, but you do not know if you
have blue eyes. You certainly know A is true. But how do you know that
B is true? You don't. For to know that B was true, you would have to
know that "Kim knows that everyone knows that there is a blue-eyed
person on the island." (or in other words, "Kim knows A.") Now certainly
I, as an observer off the island, knows that "Kim knows A." But you,
on the island, do not know that "Kim knows A." Because, as far as you
know, you might have brown eyes. Then Kim would look around, and see two
people, one (you, Joe) with brown eyes, and one (Larry) with blue eyes.
Then Kim would think "Maybe I (Kim) have brown eyes. Then Larry would
see both myself (Kim) and this other guy (Joe) both with brown eyes.
Then Larry would be the only blue-eyed person on the island, and thus
would not know that there was a blue-eyed person on the
island. Therefore I do not know A." So you might think that Kim does not
know A. Note, of course, that Kim really does know A, but as one of the
three men on the island, you do not know "Kim knows A." So to sum up,
the new information is that Joe knows that Kim knows that Larry knows
that there is a blue-eyed person on the island.
Confused? Probably. Read the paragraph above again, paying close
attention to which thoughts are attributed to which islanders. Trust me,
it makes sense.
Momo
Russ Perry Jr
logic, at least after an explanation like those going on here, now.
But I still don't think it would work... People will either not take
the time to do the reasoning requiring, will not be competent to do so,
or will live in denial either way. :-)
--
//*================================================================++
|| Russ Perry Jr 2175 S Tonne Dr #105 Arlington Hts IL 60005 ||
|| 847-952-9729 slap...@enteract.com VIDEOGAME COLLECTOR! ||
++================================================================*//
Matthew T. Russotto
}>> I now both understand and do not understand the puzzle. Please
}>> collapse my understanding.
}
}Hugo van der Sanden <h...@crypt0.demon.co.uk> wrote:
}>Everyone knows that there are blue-eyed people on the island. The only
}>reason nobody dies is because it is a taboo subject - noone has ever
}>mentioned the fact, and everyone knows the blue-eyed people are alive
}>only because of that.
}>
}>As soon as someone breaks that taboo, and mentions that there is a
}>blue-eyed person on the island, the clock is ticking. Anyone who has
}>seen seven pairs of blue eyes on neighbours expects that in seven days
}>those neighbours - who he assumes all know of only six blue-eyed
}>neighbours - will kill themselves. On the eighth day he realises the
}>awful truth.
}>
}>Does that help?
}
}Not really. Let's say that there are at least three blue-eyed people
}on the island. Suppose an islander broke the taboo by announcing that
}there is at least one blue-eyed person on the island. Has he said
}anything which is not known (and *known* to be known) by everyone
}else?
I think it matters if he's brown-eyed or blue-eyed. Assume that
islanders (even renegade taboo-breaking islanders) always tell the
truth about such things. Then if a brown-eyed islander announces it
in the one-blue-eyed case, the blue eyed person kills himself. By the
same induction as the original problem, the same result will happen --
on the nth day, the n blue-eyed people kill themselves. If a
blue-eyed islander announces it, different things happen. In the case
of two blue-eyed people, the other blue-eyed person kills himself on
the first day -- but the original blue-eyed person can't determine his
eye color from this, therefore he fails to kill himself. Induction in
this case leads to all blue-eyed people but the speaker killing
themselves on the n-1st day. So by speaking, he's managed to give
information to all the blue-eyed people except himself.
zach...@gmail.com
> In article <abergman-8CEAD2...@cnn.princeton.edu>, Aaron
> Bergman <aber...@princeton.edu> wrote:
>
> > To save a little time, it's easier to talk about the 2 blue eyed
> > that everyone knows that there is atleast one blue eyed person on
> >
> > Aaron
>
>
Yes, but it is not the case that everyone knows that everyone knows that everyone knows (3 now) that there is at least 1 blue eyed person.
Curlytop
space-time continuum:
>> If there are three blue eyed's, then everyone already knows that everyone
>> knows that there is at least one blue eyed. So where's the new info?
>>
>> Mike
>
> Yes, but it is not the case that everyone knows that everyone knows that
> everyone knows (3 now) that there is at least 1 blue eyed person.
who posts to this group.
Zlfrys.
--
ξ: ) Proud to be curly
Interchange the alphabetic letter groups to reply
Eric Sosman
> zach...@gmail.com set the following eddies spiralling through the
> space-time continuum:
temporal dimension already exceeds twelve years ...
--
Eric Sosman
eso...@comcast-dot-net.invalid