• 10 Posts
  • 826 Comments
Joined 2 years ago
Cake day: June 30th, 2023
  • FourPacketsOfPeanuts@lemmy.worldtoAsk Lemmy@lemmy.world•Give me some of your hardest riddles? (with solutions in spoilers)
    1·
    19 hours ago

    Thanks. Yes, I’m managing to absorb it now.

    Though the hardest part is getting an intuition about why the “algorithm 1” “algorithm 2” thinking happens at all when they’s a group of 100 people and everyone can see at least 99 blue eyed people. I get the induction, but why does anyone think ‘well algorithm 1 people would leave first night’ when there obviously can’t be anyone in this group. The only immediate question on everyone’s mind is “are there 99 blue eyed people (what I see) or 100 (me included)?”

  • FourPacketsOfPeanuts@lemmy.worldtoAsk Lemmy@lemmy.world•Give me some of your hardest riddles? (with solutions in spoilers)
    8·
    3 days ago

    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.

    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?

  • FourPacketsOfPeanuts@lemmy.worldtoAsk Lemmy@lemmy.world•Give me some of your hardest riddles? (with solutions in spoilers)
    11·
    3 days ago

    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?