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Famously the hardest logic puzzle in the world: Blue Eyes
https://xkcd.com/blue_eyes.html
The answer
The answer is >!the 100 people with blue eyes leave on the 100th night.!< Double spoiler tagged so it hopefully works on every Lemmy client.
For the logic, imagine if it were actually just three people: the guru, one brown eyes, and one blue eyes. For the sake of clarity, I'll speak through the perspective of a blue eyes in these examples. Guru says "I see someone with blue eyes." Brown eyes also sees someone with blue eyes. But I see no one with blue eyes. I deduce that I must have blue eyes, and leave that same night.
Now >!imagine there are two blue eyes, two brown eyes, and the guru. Guru says "I see someone with blue eyes." Brown eyes both see two sets of blue eyes, but I only see one set of blue eyes. I figure if the other person sees no other blue eyes, they'll leave the first night. They don't, which can only mean that they also saw someone with blue eyes, which must be me. We both leave the second night.!<
You >!can expand this logic all the way out to 100, so on the 100th night, all the blue eyes leave.!<
I can't see how this expands from your last case of 2 blue eyes to any more blue eyes?
When there are two blue eyed people (and you can see one of them) then the guru saying they see a blue eyed person has value because you can wait and see what the only person with blue eyes does. If they do nothing it's because the gurus statement hasn't added anything to them (they already see someone with blue eyes). And this in turn tells you something about what they must see - namely that you have blue eyes.
But how does this work when there are 3 people with blue eyes?
There isn't anyone who might see no blue eyes. And you know this, because you see at least 2 sets of blue eyes. When no-one leaves on day 1 it's because they're still not sure, because there's no circumstance where the gurus statement helped anyone determine anything. So nothing happening on day 1 doesn't add any useful info to day 2.
So the gurus statement doesn't seem to set anything in motion?
Same way it expands to two: When there are three blue eyes, then each of them guesses they might have brown or something and there could be only two blue on the island, in which case as described those two would have left on the second night.
But they didn't... So there must be three total. Same with 4, the 3 you can see would have left on night 3 if each of them saw the other two not leave on night 2...
Leaving or not is the only communication, and what the guru really did was start a timer. It has to start at 1 even though everyone can see that there's more than one simply due to the constraints of the riddle - if the guru were allowed to say 'I see at least 50 blue eyed people' then it would start at 50 because there's no other fixed reference available. Everyone knows there's either 99 or 100, but they don't know which of those it is, so need a way to count to there. They also think everyone else can see anything from 98 to 101 depending, so it's not as straightforward as thinking the count could start at 99.
I don't think that's right.
Let's try it out:
Basic case: 1 brown, 1 blue. Day 1. Guru says I see someone with blue eyes, blue eye person immediately leaves. End
Next: 2 brown, 2 blue.
Day 1; Guru speaks. It doesn't help anyone immediately because everyone can see a blue eyed person, so no one leaves first night.
Day 2; The next night, everyone knows this, that everyone else can see a blue eyed person. Which tells the blue eyed people that their eyes are not brown. (They now know no-one is looking at all brown eyes). So the 2 blue eyed people who now realise their eyes aren't brown leave that night on day 2. The end
Next case: 3 brown, 3 blue (I'm arbitrarily making brown = blue, I don't think it actually matters).
Day 1, guru speaks, no-one leaves.
Day 2 everyone now knows no-one is looking at all brown. So if anyone could see only 1 other person with blue eyes at this point, they would conclude they themselves have blue. I suppose if you were one of the three blue eyed people you wouldn't know if the other blue eyed people were looking at 1 blue or more. No-one leaves that night.
Day 3 I suppose now everyone can conclude that no-one was looking at only 1 blue, everyone can see at least two blue. So if the other blue eyed people can see 2 blues that means you must have blue eyes. So all blue eyed people leave Day 3?
Hmm. Maybe I've talked my way round to it. Maybe this keeps going on, each day without departure eliminating anyone seeing that many blue eyes until you get to 100.
It just seems so utterly counterintuitive that everyone sits there for 99 nights unable to conclude anything?
Yeah. It does seem counterintuitive, but it's a result of the uncertainty that what they see is what others do. So they have to communicate a number, and the only way they can is leaving or not each night to count up to it.
I thought about it more and concluded that if the guru had said "I see only blue and brown eyed people" then everyone (but her) could leave the island using the same logic, regardless of how many of each color there was (greater than zero of course because otherwise she wouldn't see that color). Same for any number of colors too as long as she lists them all and makes it clear that's all of them and doesn't include herself.
spoiler
I think the problem is underspecified: it doesnt define what happens if they get it wrong. If they get it wrong are they barred from exiting? Or can they make a guess every night? Whats stopping them from all saying blue and the boat leaving with 100 blue eyed people on night 1? Or do they only get one guess?Also it would have saved more people if the guru had simply stated the counts they daw because everyone could count the other people and then see what color was missing (their own eye color) and they could have all left night 1.
The everybody is a perfect logician part means no one gets it wrong and no one make a guess.
They also don't say the color of their eyes when they leave. They can just use logic to figure it out.
There are some other problems with this versions wording though. Usually you need to specify that the speaker doesn't lie.