
The blue-eyed islanders puzzle (2008) - networked
https://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/
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murbard2
A dramatization [http://slatestarcodex.com/2015/10/15/it-was-you-who-made-
my-...](http://slatestarcodex.com/2015/10/15/it-was-you-who-made-my-blue-eyes-
blue/)

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gcr
Came here to post this story! It's so good.

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awesomepantsm
I think there is a fallacy here actually. Some of the comments come close to
touching it.

If there are 3 islanders with blue eyes, each can see 2 blue-eyed people. The
information that there is at least 1 blue is therefore not new information,
and can't change the outcome.

The outsider's comment is superfluous given that N > 3.

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alwaysmetara
The important point is that the outsider's comment "synchronizes" the
islander's information. On Day 2, say you are a blue eyed person. You can see
two other blue eyed people. Since these two people don't commit suicide during
that day, you know there must be three blue eyed people with one of them being
you.

So now you know your eye color.

~~~
awesomepantsm
Why can't they synchronize at any other moment? They already knew N is either
99 or 100 even before the visitor came.

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millstone
Consider the case of 2 blue eyed people, Ben and Betty. Everybody knows there
is at least one blue eyed person. But does everybody know _that_? That is,
does everybody know that (everybody knows there is one blue eyed person)?

From Ben's perspective, Betty _might_ be the only blue-eyed person, so Ben
does not know if Betty knows there is at least one blue eyed person. This is
important because Ben needs to correlate Betty's actions with her knowledge;
he can't do that if he doesn't know what her knowledge is.

After the traveler's announcement, Ben knows that Betty knows there is at
least one blue eyed person. This is the new information.

And with three people, we add another layer: Ben does not know if Betty knows
that Bobby has blue eyes. And so on.

The visitor's announcement is special because it is infinitely layered.
Everyone knows one person has blue eyes, and everyone knows that everyone
knows, and so on, ad infinitum. This is called common knowledge:
[https://en.wikipedia.org/wiki/Common_knowledge_(logic)](https://en.wikipedia.org/wiki/Common_knowledge_\(logic\))).

Though to be fair, this isn't explicitly stated: the traveler "addresses the
entire tribe" but it's critical that everyone in the tribe knows this is
common knowledge. Merely addressing the entire tribe is not enough: if the
traveler's announcement was in the form of a BCC mass mailing, there would be
no new information and so no suicides.

~~~
knughit
It isn't infinitely layered! It is N-layered, which is critical.

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curun1r
Wouldn't it be better to use the XKCD version [1] of the problem? That version
was published 2 years earlier, doesn't make light of suicide and there's a ton
of useful discussions [2] about the logic of it in the forums.

[1] [https://xkcd.com/blue_eyes.html](https://xkcd.com/blue_eyes.html)

[2]
[http://forums.xkcd.com/viewtopic.php?f=3&t=3](http://forums.xkcd.com/viewtopic.php?f=3&t=3)

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ars
> Now suppose inductively that n is larger than 1. Each blue-eyed person will
> reason as follows: “If I am not blue-eyed, then there will only be n-1 blue-
> eyed people on this island, and so they will all commit suicide n-1 days
> after the traveler’s address”

But no one knows n!

So I am not following the logic in the slightest.

Since no one knows n the entire logic sequence never gets started.

Specifically:

The only information the traveler gave them is that n >= 1. If n = 1 then the
blue eyed person gained new information, which is how the inductive reasoning
started, and I follow the logic that far.

But if n > 1 then there is no new information and the inductive process never
starts because no one knows n, no one. There is not a single person who knows
n.

~~~
ubernostrum
The key of the problem includes the statement that all the islanders have seen
all the other islanders and know their eye colors.

From there we have the simple case of n = 1: suppose Bob is the only blue-eyed
islander. He's seen everyone else's eyes, knows nobody else has blue eyes, and
so can conclude from n >= 1 that in fact n = 1 and he is that one.

For the case of n = 2: suppose Alice and Bob are the two blue-eyed islanders.
Alice knows Bob is blue-eyed but doesn't know she is. Bob knows Alice is blue-
eyed but doesn't know he is. There are two possibilities: n = 1 or n = 2,
since they have seen every other person's eyes and know there are no other
blue-eyed people. Each can watch the other's behavior on the first day. But
since neither one knows their own eye color yet, neither one commits suicide
that day. Then each one realizes that for the other not to commit suicide, it
must be the case that n > 1\. Since they know every other person's eye color,
the only possible explanation is "n = 2 and I'm the other one". They commit
suicide on day 2.

From there we can just apply the same reasoning every time. If n = 3, Alice,
Bob and Carol each look for the other two to commit suicide on the second day
(from the reasoning for n = 2). When they don't, the only possible conclusion
is "n = 3 and I'm the third one" and they all commit suicide on day 3.

That line of reasoning then works with any successive value for n: if there
are n blue-eyed islanders, each looks for the other n - 1 islanders to commit
suicide on day n - 1. When that doesn't occur, the only possible conclusion
those people can reach is "there's one more blue-eyed islander, and it's me".
Then all n of them commit suicide on day n.

~~~
ars
Thanks, that helps a lot.

But the flaw is assuming that they will start counting at one.

The blue eyed person number 3 can see that there are 2 others at least, so
they know for sure that n != 1, since n != 1 they have no reason to expect
anyone to commit suicide.

So the whole sequence never gets started.

~~~
ubernostrum
The sequence gets started by the initial statement, and although in the case
of n > 1 it seems as if it doesn't impart any new information, it does.

The distinction that matters to get the chain started is the difference
between "I personally know there is at least one blue-eyed islander" and
"everybody knows there is at least one blue-eyed islander and I know everybody
knows it and they know I know it and I know they know I know it... (so on)".

Prior to that you can't _know_ that other islanders could deduce all the
things you can deduce from available information, and they can't know that you
could deduce all the things they could. Afterward, you know exactly what they
can deduce from the available information, they know what you can deduce, and
you know that it produces a guaranteed conclusion, by day _n_ , that _n_ is
the number of blue-eyed islanders.

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eternauta3k
How does it work out if you do it in continuous time?

~~~
pervycreeper
No differently.

~~~
sopooneo
Then when does the mass suicide occur? At time n*0 after the pronouncement by
the visitor?

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Steko
Given the propensity of island based logic puzzles to define certain classes
of people as always/sometimes lying or telling the truth, it's always bothered
me that this puzzle never states that the foreigner never lies or is always
believed.

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tbrake
The puzzle does state he is fully believed.

> One day, a blue-eyed foreigner visits to the island and wins the complete
> trust of the tribe.

Why would it matter if he's a liar/truth teller in the classic island logic
puzzle sense? It seems irrelevant to me because the islanders believe him
fully and the puzzle is about figuring out the effects his statement has.

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Steko
You are correct and thanks for the correction, wow do I feel dumb for missing
that. Maybe I can save a small scrap of face in that at least occasionally
this problem contains no such clause, e.g. the xkcd version although I should
read that one again before submitting this reply.

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pervycreeper
Thread is a bit long. Which of the comments is particularly insightful/ worth
reading?

~~~
matroosberg
Prof. Tao himself links to the wikipedia article
[https://en.wikipedia.org/wiki/Common_knowledge_%28logic%29](https://en.wikipedia.org/wiki/Common_knowledge_%28logic%29)
which uses this as an example for Common Logic.

