
The blue-eyed islanders puzzle  - dwwoelfel
http://terrytao.wordpress.com/2011/04/07/the-blue-eyed-islanders-puzzle-repost/
======
lkozma
Solution 2 is correct, Solution 1 is wrong because there is always transfer of
information in the stranger's speech.

Suppose there was only 1 blue-eyed person:

    
    
      [before]: Not everyone knows there exist blue eyed people 
      (the blue eyed person doesn't know it)
      [after]: Everyone knows there exist blue eyed people.
    

Suppose there are 2 blue-eyed persons:

    
    
      [before]: Everyone knows there are blue eyed persons.
      Not everyone knows that everyone knows that there are blue eyed persons 
      (the 2 blue eyed don't know). 
      If Alice and Bob are blue eyed, Alice knows Bob is blue eyed, 
      but thinks that Bob thinks nobody has blue eyes.
      [after]: Everyone knows and knows that everyone knows etc.
    

Suppose there are 3 blue-eyed persons:

    
    
      [before]: Everyone knows there are blue eyed people.
      Everyone knows everyone knows there are blue eyed people.
      Not everyone knows everyone knows everyone knows there are blue eyed people.
      This last statement changes after the speech.
    

Similarly for any n:

    
    
      [before]: (everyone knows that)*(n-1) there are blue eyed people.
      [after]: everyone knows everything.
    

EDIT: reworded for clarity.

~~~
po
I understand the logic behind the solution and it makes full sense to me. What
I still struggle with is: what specific piece of information is being passed
by the foreigner. I get that it is "everyone knows everything" but for n > 1
everyone already knew the specific fact that was being communicated. What is
special about the announcement that drives it? It seems like it is actually
"the timer starts now!"

~~~
lkozma
The specific information which was not known and is known after the
foreigner's speech is that "everyone knows that everyone knows that everyone
knows that (repeated n or more times) there are people with blue eyes." This
was known only up to n-1 repetitions. To infer this new statement, the
islanders have to be aware that the others are listening, they understand,
they know that everyone understands, etc. etc.

~~~
po
_To infer this new statement, the islanders have to be aware that the others
are listening, they understand, they know that everyone understands, etc.
etc._

That's deeply unsatisfying because we are told that everyone is capable of
making the observation that is communicated right from the beginning for n >
1\. What prevents the group from acting as soon as the situation starts but
before the foreigner speaks?

The message adds information to the system but it is not that there is one
person with blue eyes, i.e. what was explicitly stated. I think it adds
information about the time at which an action will or will not occur. The
information passed is "you can expect something to happen to the people with
blue eyes tomorrow."

------
StavrosK
Does anyone know if the solution is posted anywhere? It seems that the only
thing the stranger does is to give them a reference point to figure out how
many blue-eyed people there are, and not much else.

~~~
ErrantX
As a logic puzzle it is very tenuous. The proposed solution (which is lower
down the blog post) is that because any blue eyed Islander is able to count
the number of blue eyed people (99), if on day 99 those people do not kill
themselves it means that there is one further blue eyed Islander, and it is
them.

Brown eyed Islanders see 100 blue eyes, and so know that if mass suicide
occurs on day 100 then they are not blue eyed.

It's only a short reprieve though; because either the browns know that only
two colours exist (and so on day 100 immediately know they are brown) or they
face the same conundrum as blue, and commit suicide on day 900.

The _problem_ with the puzzle is not really the logical reasoning (which is
sound, as described in the puzzle). The problem is that the Islanders, who are
described as very logical, should _know_ this information already. That they
are still alive when the foreigner arrives is illogical.

EDIT: to downvoters; consider the problem & solution very very carefully (or
read my subsequent comment). Initially it seems a little illogical that the
browns would die too, but it is ultimately the same basic problem that the
blues face :)

~~~
andrew1
> It's only a short reprieve though; because either the browns know that only
> two colours exist (and so on day 100 immediately know they are brown) or
> they face the same conundrum as blue, and commit suicide on day 900.

In the case where they aren't told that only two colours exist, why would they
kill themselves on day 900? Suppose instead that one of the brown eyed people
had green eyes, I don't see how the brown eyed people or green eyed person
could deduce the colour of their own eyes.

~~~
ErrantX
Imagine the problem like this. The island is populated by just 100 Blue eyed
people.

An individual can therefore see 99 blue eyes. Their logical reasoning is that
if they are _not_ blue eyed then on day 99 all of the blue eyed people will
commit suicide (because they can see 98 blue and 1 of another colour). Because
the others don't commit suicide on day 99 the only logical inference is that
_everyone else_ can see 99 blue eyes too. And so they too must be blue.

The reason the solution is so complex to comprehend is because the logical
inference has little to do with what eye colours exist in the tribe. It is,
ultimately, simple mathematics. If a tribe has n blue eyes, y brown eyes and z
green eyes any of the individuals in the tribe will eventually be able to
logically infer which colour they have, regardless of having any starter
information.

Another way to think of it is this; they know that there are either n people
with X eyes. Or n+1. Where n is the number of people they can see with X eyes
and the unknown factor is their own eye colour. For everyone without X eyes n
is greater than those with X eyes.

~~~
andrew1
You originally said that the people with brown eyes would commit suicide after
900 days. Can you explain to me how any of them can know for sure that they
don't have green eyes?

~~~
ErrantX
I have done; I was using the "blue eyes on their own" as an example to
demonstrate it.

I'll try again (this can take a long while to get your head round :D).

Once the Blue eyed people are dead each islander now knows that there are
definitely at least 899 brown eyed people, with the unknown factor their own
eye colour.

Their hypothesis must be as follows (imagine this is the logical reasoning of
ONE islander laying out the scenario):

\- If I do not have Brown eyes that means each of the brown eyed people can
see 898 browns. If that is the case, then on day 899 all of those people will
kill themselves.

\- If I did have Brown eyes then each of the brown eyed people can _also_ see
899 brown eyes. If that is the case then no one will kill themselves on day
899, therefore as I can only see 899 browns, I must have brown

Here is where most people struggle to understand: The number of brown eyes
each individual can see is _fixed_ at 899. The _only_ unknown factor is their
own eye colour. The solution is that everyone knows each other unknown factor.
You know that either n or n+1 brown eyes exist (where n is the number of brown
eyes they can observe). And so does everyone else (and, crucially, you know
they know).

If everyone's n matches yours, then on day n+1 you can infer what the unknown
factor is.

Another way of thinking about it is this; an individual can logically reason
that only two types of people exist on the island. Those who see 899 browns
(him, at the least), those who see 898 browns (potentially everyone else; on
the hypothesis that his eye colour differs). If on day 899 no one commits
suicide then each individual knows that only the former exists. And therefore
they have brown eyes.

(this is horribly hard to put across :S which is why it is such a delicious
problem! The main thing to remember is that to the knowledge of one individual
there is a known number of brown eyes, _and_ a figure for the minimum and
maximum number of brown eyes any other individual can see. If the number of
days exceeds the number of browns you can see, you must be brown)

~~~
andrew1
Suppose instead of being 900 brown eyed people there are X brown eyed people.

Suppose X is one, then the brown eyed person appears to be stuck, all the blue
eyed people are dead so he knows he doesn't have blue eyes, but he can't
determine what colour his eyes are (are they brown?, are they green?).

If X is two then we're left with two people after all the blues are dead. Each
one can see a brown eyed person, but again, this doesn't help them determine
what colour their own eyes are. Again, why can they not be green?

The case for X>2 seems similar to the case for X=2, there's no magical point
at which they can suddenly all say 'I don't have green eyes, I must have brown
eyes'

With all respect, I think _you_ are underestimating the problem. :)

~~~
ErrantX
_Suppose X is one, then the brown eyed person appears to be stuck_

Yes, the problem specifically doesn't work if the number of people with an eye
colour is below a certain amount.

 _The case for X >2 seems similar to the case for X=2, there's no magical
point at which they can suddenly all say _

In a scenario where there is, say, just brown and green (regardless of whether
they know this or not), then X>3 is enough.

Where X=3 or less there is not enough information to go on (you successfully
point out the flaws) and they could conceivably be any other eye colour. But
above that you know it is impossible for there to be an individual who can see
only one or two browns, therefore it can be inferred properly.

One of the main problems comes from the explanation of the issue which
involves the suggestion that you recurse the problem back to X=1, this is
inaccurate because there is always a _known_ absolute minimum people that
could have brown eyes (n, the brown eyed people you can see) _and_ a known
minimum number that any one person could see (n - 1). So long as for a brown
eyed person (n - 1) > 2 the the inference is possible.

Remember; it does not matter what alternative eye colour they might be (and
whether they can see it or not). Only that they are either brown or not brown.
The logical inference is based on what the others must be seeing.

Try jotting it down and then reasoning it out as an individual in the tribe.
Once it clicks it will seem simple :)

~~~
andrew1
OK, so we agree that if there are X <= 3 people then they're stuck and cannot
kill themselves.

You claim that if X = 4 and everyone has brown eyes then on day 4 they will
all kill themselves. Suppose I am one of these islanders. I can see three
people with brown eyes. On days one, two and three no one kills themselves. On
day four I get up and kill myself because I know that I have brown eyes.

Fine, now let's consider the case where there are three people with brown
eyes, and one person with green eyes:

As I think we've agreed above, the three people with brown eyes cannot infer
that they have brown eyes (i.e. the X < 3 case). So they are not able to kill
themselves.

But consider the green eyed person: he can see three people with brown eyes
and no one kills themselves on days one, two or three.

He is in _exactly_ the same situation at this point as the brown eyed person
we considered in the four-brown-eyed-people case. So at this point, by your
logic, he must know that he has brown eyes. Which is a contradiction.

I think you are wrong that induction is not involved in establishing eye
colour. (unless you can convince me in the X = 4 case that is :) ).

This isn't a problem I'm unfamiliar with it, a colleague asked me it when I
was being interviewed for my current job, and it gets rediscussed
periodically. I really do think you're wrong I'm afraid.

~~~
ErrantX
Umm, we defined X as the number of people with brown eyes. So on an island
with 4 people, three of whom have brown eyes, then X=3 and, yep, there is a
problem.

Where X=4 (i.e. there _are_ 4 people with brown eyes) it works.

In your case, where the number of people is 4, but X=3, then the fourth
possibly incorrectly infers that he has brown eyes and so, on his own, kills
himself. On the other hand they are "highly logical" so I argue they would
realise that there were too few people to know.

~~~
andrew1
But that's precisely the point, if they're all "highly logical" then in the
island-wtih-3-brown-and-1-green case the person with green eyes can't
correctly know their eye colour so can't kill themself. But they are in
exactly the same situation as a person with brown eyes in the island-with-
four-brown-people case. So if the green eyed person in the first case can't
deduce their eye colour, then neither can the brown eyed person in the second
case.

If you don't agree, please explain to me what extra piece of information the
brown eyed person in the second case has which means he can kill himself on
day four. As far as I can see they are both in the situation that they can see
three people with brown eyes, and no one has yet killed themselves. _What is
the extra piece of information that allows the brown eyed person in the second
case to deduce that he has brown eyes?_

