

Blue Eyes - A Logic Puzzle (spoiler in the comments) - palish
http://xkcd.com/blue_eyes.html

======
shawndrost
I wrote about what information the guru gives recently:

<http://news.ycombinator.com/item?id=56196>

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palish
Normally I don't cross-post, but this was too interesting. Solution is here:
<http://xkcd.com/solution.html>

~~~
euccastro
[Edit: this was originally garbled in rot13 with a warning of SPOILER ALERT. I
have decoded it back to readable text since the spoiler warning has been added
to the post title.]

From the solution page:

"1. What is the quantified piece of information that the Guru provides that
each person did not already have?"

Indeed! I can understand the solution, but I don't know how the word of the
Guru matters. Rather, under the given rules, everyone should leave the island
exactly on the 100th midnight after the 'game' starts. Say, after they enter
the island, or after the rules are established.

~~~
palish
While amusing, the point of posting this to News.YC is to openly discuss the
solution :P

Upmodded for geekiness anyway though.

From what I understand, it's the fact that everyone knows _that everyone else
knows_ that someone has blue eyes that matters. You have to look at it with an
extra layer of indirection.

~~~
euccastro
Yes, but they don't need the guru to tell them that, do they? Do you mean the
green eye guru thing is just red herring?

[Edit:] Besides, I claim the solution is wrong. _Everyone_ but the guru needs
to leave on the 100th midnight. Everyone but the guru sees one group of 99, so
everyone should know to wait 99 days, then conclude that if the other folks in
that group haven't left it's because they see a group of 99 too.

The guru doesn't leave because he sees two groups of 100, so he will wait 101
days. By that time, everyone has left, which means there were indeed two
groups of 100 and herself had a different color.

Note that everyone but the guru sees a group of 1 (the guru), but this does
not mean that everyone must leave at day two unless the guru has left the
first day. That's because the guru doesn't know there is anyone with green
eyes. If he knew, he'd trivially conclude it had to be him, so he had to leave
the first day. None of this affects the other 200 at all.

[Edit^2:] I don't _really_ believe I'm right; I just fail to see what I'm
missing. I realise that the guru is there to somehow kickstart the induction
providing the base case, and that the previous paragraph somehow has the key
to this. But I fail to grasp how that is really necessary.

[Edit^3:] OK, it goes like this. We have to prove the thesis that _if I see
any group of n people with the same eye color, I can NOT conclude that I
belong to that group, after any number of days, until someone publicly reveals
that there is at least one person in that group_.

The paragraph right before Edit^2 explains the base case of a group of one.
Now we have to establish the induction step: if the thesis is true for a group
of n, does it necessarily follow for n+1? Yes, because if I am _not_ in the
group, each person in the n+1 group only sees _n_ other people with the same
eye color. Which, by the induction hypothesis (i.e., that _the thesis is true
for a group of n_ ), doesn't let them conclude anything, after any number of
days. Therefore, I can't infer anything from their failure to leave the island
after any number of days.

Makes sense?

~~~
shawndrost
"Everyone but the guru sees one group of 99"

No, the guru and the people with brown eyes see 100 people with blue eyes.

Edit: I see now that you really meant "a group of 99", not "99 blue-eyed
people". My comment doesn't apply. FWIW, I don't think it matters how many
brown-eyed people there are.

~~~
euccastro
Yes, I meant "a group of 99 people with the same color"; sorry for not making
it clear. And I see now that brown-eyed people don't matter, but the reason
why they don't matter turned out more involved than I originally thought.

