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[Sowelu]

I like this thread fine, but your attitude towards this problem is a little grating. It's like when people learn about thermodynamics, think of some case where they don't understand why a perpetual motion machine of some description wouldn't work, and assume that it would instead of actually asking a physics instructor.

I'm pretty sure this logic problem is not "Oh, find a logic instructor and they will demonstrate why this is the only possible result".  I'm pretty sure it's "They will bury their head in their hands and say it's a stupid application that relies on spurious logic".  There are practical problems that this approach will solve, and this isn't one of them, at least not as stated.  As goddamn stupid as it sounds, yes, starting to count from 1 IS a non sequitor.  There's no reason.  Other problems in this class are interesting because something COULD happen on day 2, and it doesn't, which is important information.  In this problem, nothing can happen on day 2.

It feels like a cheap extension to a -good- problem, invented by someone who wasn't actually a logician themselves.

Now, if someone has a proof that you can't shorten the time--then that would be interesting.  If I attack the problem hard enough that an (informal at best) proof of that pops out, then I'll be happy.  But right now the solution is pretty damn lacking.  It's ONE way to the answer, it shows ONE way that people can leave, but it doesn't prove that it's the only one...and when you phrase the question to say that the people are perfect logicians, you have to be more exhaustive.

Besides, while I don't want to make this an ad hominem--Randall Munroe is essentially a layman.  It's not like he has a PhD in logic.  Yeah, fun hobby, we all like XKCD, but I don't see any reason to trust that he's provided the only solution.

[G-Flex]

Besides, while I don't want to make this an ad hominem--Randall Munroe is essentially a layman.  It's not like he has a PhD in logic.  Yeah, fun hobby, we all like XKCD, but I don't see any reason to trust that he's provided the only solution.

And you seem to think that people who are qualified won't be able to tell you either.

[Sowelu]

Besides, while I don't want to make this an ad hominem--Randall Munroe is essentially a layman.  It's not like he has a PhD in logic.  Yeah, fun hobby, we all like XKCD, but I don't see any reason to trust that he's provided the only solution.

And you seem to think that people who are qualified won't be able to tell you either.

I'm sorry, did I miss a link to a rigorous proof that the provided solution is the one and only correct solution, that there is no faster way for them to gather that information?  I admit I might have missed a link somewhere, but I didn't see one when I just looked.  Do we have a resident expert authority on these forums?

[G-Flex]

We don't, but you said it's been "bothering you for years". After that much time, I'd think you would put a bit more rigorous research into it.

[Sowelu]

We don't, but you said it's been "bothering you for years". After that much time, I'd think you would put a bit more rigorous research into it.

Good point, I guess.  I'll have to do that later tonight & I'll shut up in the meantime.

[Earthquake Damage]

Sowelu brings up one good point:  the timing.

Why do they leave on day 100?  There's no good reason for it.  I think they could all treat day 1 as though it were day 98, simply because they know that everyone must see at least 98 blue-eyed people.  The minimum number anyone sees is 99, and if they each assume they're not blue-eyed until proven otherwise, they will know that 99 blues each seeing 98 blues cannot leave before day 98.  So rather than twiddle their thumbs for 97 days, they can skip straight to day 98.

The problem doesn't seem to ask how we can maximize the number of people who leave.  It just asks who will leave and when.  The answer, as far as I can tell, is "98 to 100 blue-eyed people will leave anywhere from day 1 to day 100".

Am I wrong?  If so, how?

[Sowelu]

They can't skip straight to day 98.  Blue-eyed guy says "I see 99 people with blue eyes, assume I'm brown eyed.  There are 99 blue-eyed people.  BUT.  Each of those blue-eyed people will see 98 blue-eyed people, and they will do the same thing I did.  They will also get into the heads of those other people, and the number drops to 97, and so on".

If a better solution exists, it needs to not be recursive.

Also doing research is hard because after something goes up on XKCD, every coffee-shop-Macbook-and-beret intellectual puts it up on their own blog.  The internet is full of chaff now, especially this long since it went up.  Ever try searching the web for a new word or phrase that got introduced on XKCD even an hour after the strip went up?  Yeah, ten thousand hits or so.

[Earthquake Damage]

They can't skip straight to day 98.  Blue-eyed guy says "I see 99 people with blue eyes, assume I'm brown eyed.  There are 99 blue-eyed people.  BUT.  Each of those blue-eyed people will see 98 blue-eyed people, and they will do the same thing I did.  They will also get into the heads of those other people, and the number drops to 97, and so on".

Surely it won't recurse to 1, though, since if I see 99 everyone else must see at least 98.  There is no person who sees only 97.

[G-Flex]

Check out the example on the solution page about what happens if there are three blue-eyed people. That should explain it well enough.

[Earthquake Damage]

Check out the example on the solution page about what happens if there are three blue-eyed people. That should explain it well enough.

I don't see why 100 people must wait so long before anything interesting can begin to happen.

Suppose I'm one of the islanders.  I see 99 people with blue eyes.  If I have blue eyes, then they all see 99 and we'll leave on day 100, right?  If I don't, they'll all see 98 and leave on day 99, right?  That's the soonest anyone can possibly leave using that method.  So why not skip straight to day 98?  Given the actual data available to me in this particular case, there is no n=97 case.  It's fictional.  Why wait an additional 97 days?

FYI, he doesn't address this question.  He merely poses it:

Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

[G-Flex]

Suppose I'm one of the islanders.  I see 99 people with blue eyes.  If I have blue eyes, then they all see 99 and we'll leave on day 100, right?  If I don't, they'll all see 98 and leave on day 99, right?  That's the soonest anyone can possibly leave using that method.  So why not skip straight to day 98?

Because the other islanders can't consolidate their information that way.

[Earthquake Damage]

Because the other islanders can't consolidate their information that way.

So the problem would be some may skip days while others don't.  So long as everyone follows suit, either way should work, right?  There's no way then, sans communication, to ensure they do.  Presumably the correct answer is to wait the full number of days before acting, but the system breaks if anyone guesses otherwise (i.e. that everyone will skip to day 98).  It still looks like it boils down to a coordination problem, and I don't see any premises that affect the decision.

[Ampersand]

The problem you all seem to be forgetting is that the problem is no about how a collection of people behave, but how it is possible for a single individual to deduce his own eye color given these premises. Everyone has to start with the state, "I do not know my eye color, and no one else knows their eye color."

Suppose there were two people with blue eyes, and two people with brown eyes. If the guru is not present to say "Someone here has blue eyes," There is no other information available to them that permits them to logically deduce their own eye color. They can only sit, and throw wild guesses into the air.

So, we have the guru say, "Someone here has blue eyes."

"Oh", blue-1 thinks, "It must be him. He'll leave tonight." the next morning, when Blue-2 is still there, he thinks "This person with blue eyes is still here, which means that he could not logically deduce that he was the one with blue eyes, which means that there is someone else with blue eyes. It can't be either of those two, so it must be me."

There is no way for anyone to logically deduce their own eye color without the Guru's statement. It is an absolute necessity.

Further, the reason they wait 98 days to confirm what they already know, that there are 99 people with blue eyes, is because they still cannot logically deduce their own eye color until the last day passes. No one will 'skip' days because doing so does not permit one to arrive logically at your own eye color.

Suppose I have blue eyes. I can see 99 people with blue eyes, and I know for a fact that each of them can see 98 people with blue eyes. I MAY have blue eyes, therefore, everyone else MAY see 99 people with blue eyes, as I do. The only way for me to find out my eye color for a certainty is to wait those 99 days and see if everyone is there in the morning, because in this circumstance, the only way for anyone to know the color of their own eyes is to know what other people see. You cannot skip ahead, because doing so will not enable you to prove anything about yourself.

[Leafsnail]

How would you skip?  Noone has any way to know whether they're meant to go or not.  Heck, if I pose a "first person" version of the problem to you:

Exactly the same as the question, except you can see 93 people with blue eyes, 34 with brown eyes and the guru (who has green eyes).  What do you do?

[Sowelu]

Let me rephrase the question, and slightly change the answer, in a way that makes it make at least a *little* more sense.  Still not where I want to go with it, but still.

Instead of "I see someone with blue eyes", the guru says "I like pie", and everyone decides that since the guru said something, it's time to star thinking.  Or, better yet...

100 blue-eyed, 100 brown-eyed, no guru.  Everyone meets for the first time on day 1, so there IS new information on that day.  After a hundred days, everyone leaves at once.

**********

Okay.  Here's a puzzle.  It's stupid and simple but surprisingly it's (barely) related, plus it at least sounds kind of like Monty Hall's bastard child:

There's a huge pile of envelopes that contain various amounts of money, from one dollar up to 65 thousand (give or take).  Someone gives you two envelopes and tells you to pick one.  Before you open it, they tell you that the other one has either half as much or twice as much, and offers to let you change your selection.  What do you do?  Would your answer change if you could open the first envelope and look inside before you switched?