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Riddle Me This

1567810

Posts

  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    Here's the answer:
    This only works if you know that every single person is doing the same thinking you are.

    The guru says that he sees someone with blue eyes. Ok, that means you either have Blue Eyes or you do not.

    If there were two people on the island, then the person with blue eyes would know by the end of the second day and would leave.

    If there were 10 people on the island with five blue eyes, it would take 5 days for everyone to leave.

    Basically, if you see nobody else with blue eyes, then you know you have blue eyes.

    If you see one person with blue eyes, and he doesn't disappear after the first day, then you know there are at least two people with blue eyes. If you see three people with blue eyes, and they do not disappear on the third day, then you know you must also have blue eyes.

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  • AMP'dAMP'd Registered User regular
    edited November 2010
    smart people love logic puzzles because they propose worlds in which everyone is equally genius-level

    AMP'd on
    [SIGPIC][/SIGPIC]
  • redheadredhead Registered User
    edited November 2010
    Here's the answer:
    This only works if you know that every single person is doing the same thinking you are.

    y-yes? this is true, which is why the puzzle specifies that everyone on the island is a perfect logician. i dunno, was that not said in the first version posted here? not sure what you're getting at

    redhead on
  • Captain KCaptain K Registered User regular
    edited November 2010
    I think there's just something in the rules of the process that I must be misunderstanding--when you know, without a doubt, what your own eye color is, you leave the island. Everyone on the island will instantly know something if it is logically possible to deduce it. Everyone on the island knows that there is someone on the island with blue eyes.

    that's it, right? I can't connect the dots at all, even if I play the "pretend there are only 1, 2, 3, 4, 5 people on the island and so on" game... it feels like there's no way to connect them at all.

    but I don't want to cheat because I know I'll feel like a baller if I figure it out

    Captain K on
  • redheadredhead Registered User
    edited November 2010
    AMP'd wrote: »
    if you get the solution the way it's meant to be gotten, the guru provides the information that makes your first assumption possible

    this was to me, right? could you explain in more detail? i must be dim because i still can't quite get how the guru's statement catalyzes the whole process.

    redhead on
  • Aroused BullAroused Bull Registered User
    edited November 2010
    Captain K wrote: »
    walrus and bull

    that doesn't explain it to me


    you've made some mental leap that's not clear in those words, even in the example with 2 and 2

    This is the answer:
    THE ANSWER
    Assume there's just one person with blue eyes (there's not, but it helps). Obviously that person leaves the first night, since they can see everyone else has brown eyes and deduce that theirs are blue.
    Say two people have blue eyes. Being perfectly logical, they each realise that if they don't have blue eyes, the other person is the only blue-eyed person on the island. So that person will leave the first night. If that person doesn't leave the first night, they know they have blue eyes too, so they both leave the second night.
    The sequence continues, until you get 100 people leaving on the 100th night.

    Aroused Bull on
  • redheadredhead Registered User
    edited November 2010
    there are a bunch of logic puzzles around this "level"--the pirate one, the hats one, and this one are probably the most common in my experience--and i'd say this is the hardest. so yeah, if you solve it with no help you definitely get to feel like a baller.

    redhead on
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    redhead wrote: »
    Here's the answer:
    This only works if you know that every single person is doing the same thinking you are.

    y-yes? this is true, which is why the puzzle specifies that everyone on the island is a perfect logician. i dunno, was that not said in the first version posted here? not sure what you're getting at

    The way the fact pattern is set up, it's a throw-away at the setup. It should be the focal point of the entire discussion.

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  • SLyMSLyM Registered User regular
    edited November 2010
    For people who know the answer:
    The oracle let's you make the assumption that if there were 2 they would leave on the second day. If it wasn't for him you couldn't figure that out.

    SLyM on
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  • SimBenSimBen Registered User regular
    edited November 2010
    Okay some friends just came over and I think we solved it.
    Okay so if you see nobody with blue eyes, then you know you have blue eyes and you can leave.

    If you see one person with blue eyes, and that person doesn't leave, you know that person sees someone else with blue eyes and it has to be you, so you both leave the next night.

    If you see two people with blue eyes, and neither of them leaves the first night, it's because they at least see each other. If neither of them leaves the SECOND night, it's because they've each seen two people having blue eyes, so the third person has to be you, and you leave on the third night.

    If there's a hundred people on the island with blue eyes, then all hundred of them leave after a hundred days.

    SimBen on
    sig.gif
  • SLyMSLyM Registered User regular
    edited November 2010
    SimBen wrote: »
    Okay some friends just came over and I think we solved it.

    Okay so if you see nobody with blue eyes, then you know you have blue eyes and you can leave.

    If you see one person with blue eyes, and that person doesn't leave, you know that person sees someone else with blue eyes and it has to be you, so you both leave the next night.

    If you see two people with blue eyes, and neither of them leaves the first night, it's because they at least see each other. If neither of them leaves the SECOND night, it's because they've each seen two people having blue eyes, so the third person has to be you, and you leave on the third night.

    If there's a hundred people on the island with blue eyes, then all hundred of them leave after a hundred days.

    WINNAR!

    SLyM on
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  • Captain KCaptain K Registered User regular
    edited November 2010
    well I just read the answer, I never would have figured that out, I definitely didn't understand the way it worked

    Captain K on
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    I edited my spoiler to be more informative.

    But the crux of the whole thing is that everyone else is thinking the same thing as you. It doesn't work if there is any question about that.

    Munkus Beaver on
    Steam name: munkus_beaver
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    Blizzard thing: munkus#1952
    Humor can be dissected, as a frog can, but it dies in the process.
    Twitter which gives health updates and the like: https://twitter.com/MunkusBeaver
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  • AMP'dAMP'd Registered User regular
    edited November 2010
    redhead wrote: »
    AMP'd wrote: »
    if you get the solution the way it's meant to be gotten, the guru provides the information that makes your first assumption possible

    this was to me, right? could you explain in more detail? i must be dim because i still can't quite get how the guru's statement catalyzes the whole process.

    you've read the xkcd solution, yeah?
    the first theorem, that if there is only one person on the island with blue eyes he will leave on the first night, says that the guy will look around, now that he knows someone has blue eyes, see no one with blue eyes, and then leave. but he only looks around because the guru said that someone has blue eyes

    the second theorem is only true because the first one is true, and so on, up to the 99th theorem. if these people don't all start looking around at the same time, there's no way for them to synchronize when they realize how many people have blue eyes and then leave.

    AMP'd on
    [SIGPIC][/SIGPIC]
  • redheadredhead Registered User
    edited November 2010
    AMP'd wrote: »
    redhead wrote: »
    AMP'd wrote: »
    if you get the solution the way it's meant to be gotten, the guru provides the information that makes your first assumption possible

    this was to me, right? could you explain in more detail? i must be dim because i still can't quite get how the guru's statement catalyzes the whole process.

    you've read the xkcd solution, yeah?
    the first theorem, that if there is only one person on the island with blue eyes he will leave on the first night, says that the guy will look around, now that he knows someone has blue eyes, see no one with blue eyes, and then leave. but he only looks around because the guru said that someone has blue eyes

    the second theorem is only true because the first one is true, and so on, up to the 99th theorem. if these people don't all start looking around at the same time, there's no way for them to synchronize when they realize how many people have blue eyes and then leave.

    augh, this is still fucking with my brain. i can't stop thinking "but they all already knew there was someone with blue eyes! they could just look around and see that!"

    give me a second here, i'm sure i can make this make sense to myself

    redhead on
  • redheadredhead Registered User
    edited November 2010
    OH ok got it

    edit: thanks!

    redhead on
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    Captain K wrote: »
    well I just read the answer, I never would have figured that out, I definitely didn't understand the way it worked

    2 people on the island, 1 person with blue eyes

    They see the other person does not have blue eyes. They leave that night.

    3 people on the island, 2 people with blue eyes. Person 1 and 2 have blue eyes, person 3 does not.

    Person 1 sees that person 2 has blue eyes. He knows that person 2 will leave after tonight if they are the only person with blue eyes. This is because they would see nobody else with blue eyes and therefore would be the only person with blue eyes and therefore would be able to leave. Person 2 does not leave tonight. Therefore the only answer is that Person 1 also has blue eyes. They leave the next night.

    If there were 100 people with blue eyes, and Person 1 was one of them, then after the 99th night he would know that he also had blue eyes. Otherwise, everyone else would have figured out on the 99th night that they were the only ones with blue eyes and they would have left then.


    The things that do not matter in this puzzle: how many people that are there who do not have blue eyes, the total number of people.

    Munkus Beaver on
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    Blizzard thing: munkus#1952
    Humor can be dissected, as a frog can, but it dies in the process.
    Twitter which gives health updates and the like: https://twitter.com/MunkusBeaver
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  • AMP'dAMP'd Registered User regular
    edited November 2010
    redhead wrote: »
    OH ok got it

    edit: thanks!
    the key is not just that they're all thinking the same thing, it's that they're all thinking the same thing at the same time

    AMP'd on
    [SIGPIC][/SIGPIC]
  • Captain KCaptain K Registered User regular
    edited November 2010
    yeah I mean I understood the process once the answer was posted but somehow it was not at all clear to me before

    Captain K on
  • redheadredhead Registered User
    edited November 2010
    AMP'd wrote: »
    redhead wrote: »
    OH ok got it

    edit: thanks!
    the key is not just that they're all thinking the same thing, it's that they're all thinking the same thing at the same time

    yeahhh

    i've solved this riddle before, forgotten it, been told it again and re-solved it, then told it to other people a bunch of times

    and i never quite understood that til now

    redhead on
  • FishmanFishman This'll be the year that we won't forget Registered User regular
    edited November 2010
    The other part of the eyes riddle is the night after all the Blue eyed people leave, all the people with Brown eyes leave; they know their only options are Blue or Brown and there's no green, pink purple or Chartreuse coloured eyes, and so can get on the ferry the next day, leaving the guru all alone, weeping with his sightless no-eye kicking at the sand in loneliness and wishing he'd just kept his damned mouth shut to begin with.

    Fishman on
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    That's unbelievably cool. Your new name is cool guy. Let's have sex.
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    Fishman wrote: »
    The other part of the eyes riddle is the night after all the Blue eyed people leave, all the people with Brown eyes leave; they know their only options are Blue or Brown and there's no green, pink purple or Chartreuse coloured eyes, and so can get on the ferry the next day, leaving the guru all alone, weeping with his sightless no-eye kicking at the sand in loneliness and wishing he'd just kept his damned mouth shut to begin with.

    No, the brown eyed people don't get to ever leave, because they don't know if they are brown or some other color.

    They rot in a brown eyed, muted world.

    Munkus Beaver on
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    Humor can be dissected, as a frog can, but it dies in the process.
    Twitter which gives health updates and the like: https://twitter.com/MunkusBeaver
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  • Lord DaveLord Dave Registered User regular
    edited November 2010
    > stab oracle

    Lord Dave on
    Denny's is for winners.
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  • BahamutZEROBahamutZERO ROLLING STAAAAAAAAAAAAART Registered User regular
    edited November 2010
    wait did the riddle specify that the oracle never states the same person's eye color more than once?

    BahamutZERO on
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    Nah, the oracle says that and then Lord Dave stabs the oracle and the oracle is dead.

    The oracle's only purpose is to start the series of logical events that every other person is going through.

    Munkus Beaver on
    Steam name: munkus_beaver
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    Blizzard thing: munkus#1952
    Humor can be dissected, as a frog can, but it dies in the process.
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  • BahamutZEROBahamutZERO ROLLING STAAAAAAAAAAAAART Registered User regular
    edited November 2010
    oh right the oracle only speaks once

    BahamutZERO on
  • MrMonroeMrMonroe Registered User regular
    edited November 2010
    I wouldn't call this the hardest logic problem in the world

    I would call this the hardest logic problem in the world:

    "Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which."

    MrMonroe on
  • FishmanFishman This'll be the year that we won't forget Registered User regular
    edited November 2010
    Fishman wrote: »
    The other part of the eyes riddle is the night after all the Blue eyed people leave, all the people with Brown eyes leave; they know their only options are Blue or Brown and there's no green, pink purple or Chartreuse coloured eyes, and so can get on the ferry the next day, leaving the guru all alone, weeping with his sightless no-eye kicking at the sand in loneliness and wishing he'd just kept his damned mouth shut to begin with.

    No, the brown eyed people don't get to ever leave, because they don't know if they are brown or some other color.

    They rot in a brown eyed, muted world.

    I guess it depends on the rules. They way I usually encounter it is:
    - People on the island have brown eyes or blue eyes
    - People on the island are perfectly logical and rational
    - People on the island know all of the above facts (including the fact that their eyes are either one or the other).

    But yeah, if I go around pointing that out, I don't get to make a sad guru joke... spoilsport.

    Fishman on
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  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    It's not the hardest logic problem in the world, not hardly.

    It's just a logic extrapolation.

    It's not exactly mensa-level difficult once you get the premises straight.

    Munkus Beaver on
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  • Captain KCaptain K Registered User regular
    edited November 2010
    SHIT mrmonroe

    Captain K on
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    Uh, ok. I know how you answer that when you have 2 gods. You ask them what the other one would say about themself. The one telling you the truth would tell you a lie, saying that they are the truthful one. The one telling you a lie would tell a lie about the truthful one, saying they are the lying one.

    I don't know how you do it with three and not knowing yes or no.

    Munkus Beaver on
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  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    Ok, let me think this out.

    You have god 1, 2, and 3.

    You ask god 1 whether god 3 would say god 3 is telling the truth.
    You ask god 2 whether god 3 would say god 3 is telling the truth.
    You ask god 3 whether god 1 is telling the truth.

    Munkus Beaver on
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    Blizzard thing: munkus#1952
    Humor can be dissected, as a frog can, but it dies in the process.
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  • MrMonroeMrMonroe Registered User regular
    edited November 2010
    I guess I'm supposed to put in the clarifications

    1) It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
    2) What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
    3) Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
    4) Random will answer 'da' or 'ja' when asked any yes-no question

    Pay attention to number 3.
    Random does not answer "randomly." It is random whether he is a truth teller or a liar at any given moment

    MrMonroe on
  • BedigunzBedigunz Registered User regular
    edited November 2010
    I hate this riddle

    Bedigunz on
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  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    My thinking so far:
    Ok, I would ask god 1 if he was the random god with the flip of the coin in his head showing heads. I would then ask him if he was the random god with the coin in his head showing tails.

    The answer to the first question would be Yes for L god and R god, no for T god. The answer for question 2 would be Yes for L god and No for T god and R god.

    So if the god changed his answer, then I would know he was the random god. If he did not change his answer, I would know he would either be the truthful god or the lying god.
    Am I on the right path here?

    Munkus Beaver on
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    Humor can be dissected, as a frog can, but it dies in the process.
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  • MrMonroeMrMonroe Registered User regular
    edited November 2010
    You are on the right path, but I think your first question should be modified a little.
    you're on the right track using a conditional question, but you need to be... less specific... when you're talking about the coin in his head.

    There's also multiple ways to answer this problem, including one that is more efficient than the others.

    MrMonroe on
  • Munkus BeaverMunkus Beaver Registered User, ClubPA regular
    edited November 2010
    I think I got it.
    Question 1: Are you the random god who tells lies? The answer to this question will be no for everyone but the lying god.
    Question 2: Are you the random god who tells the truth? This would be yes for the random god and yes for the lying god.

    Therefore: If the answer changed, then I know that the first response means no and the second response means yes. If he gives the same answer, then I know he is either the lying god or the truthful god

    If he changed his answer, then I would ask the 2nd god if the 1st god was the random god. I would know the truth about the first god and I would know what Yes and No were in their language, so it would reveal all of their identities.

    If he did not change his answer, I would ask the 2nd god if he was the random god who tells the truth. If his response matches the previous answe

    Munkus Beaver on
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    WiiU: munkusbeaver and Nintendo ID (3DS thinger): 0619-4510-9772
    Blizzard thing: munkus#1952
    Humor can be dissected, as a frog can, but it dies in the process.
    Twitter which gives health updates and the like: https://twitter.com/MunkusBeaver
    Please give to the Crohn's and Colitis Foundation of America: http://www.ccfa.org/
  • the wookthe wook Registered User regular
    edited November 2010
    It's not the hardest logic problem in the world, not hardly.

    It's just a logic extrapolation.

    It's not exactly mensa-level difficult once you get the premises straight.

    Mensa level difficult isn't that difficult

    the wook on
  • EdcrabEdcrab Registered User regular
    edited November 2010
    skettios wrote: »
    Edcrab wrote: »
    A fairly good mathematical one that I've found is either (a) solved instantly or (b) takes people a while to get:


    3 3 7 7

    With those four numbers, make 24. You must use each number only once

    You may only add, subtract, divide, multiply, and use brackets. You may not use powers or any tricksy shit like putting 3 and 7 together and saying it's now 37.
    3x7 = 21 + the other 3 = 24

    Close, but you have to use all the numbers! That's what makes it a puzzle rather than a sum

    Answer:
    (3/7+3)*7

    That is, divide three by seven, add three to the result, and then multiply that total by seven.

    Edcrab on
    cBY55.gifbmJsl.png
  • bsjezzbsjezz Registered User regular
    edited November 2010
    hey did anybody else read that good story i wrote

    i was happy with that

    bsjezz on
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