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Posers and Puzzles

Posers and Puzzles

  1. Standard member forkedknight
    Defend the Universe
    08 Sep '11 00:31 / 2 edits
    There exists an island on which all inhabitants of the island are perfection logicians. If a conclusion can be logically deduced, all will do it immediately.

    No-one can communicate in any way with any other inhabitant of the island (no talking, to sign language, etc.)

    No one on the island can see their own eyes or knows their own eye color. However, all inhabitants of the island can see and knows the eye color of all other inhabitants of the island.

    All inhabitants of the island have one of three eye colors:

    Some number (N) have green eyes
    Some number (M) have brown eyes
    One inhabitant, the Guru, has red eyes

    M and N are both >= 100

    Every night, just before midnight, a ferry comes to the island, and all island inhabitants that have figured out their own eye color board the ferry to leave the island.

    Up until today however, no one has left the island. Today (Day 0) at noon, the Guru stands up in front of all the inhabitants of the island and is allowed to speak one phrase:

    "I can see someone with green eyes"

    Who leaves the island, and on what day(s)?

    *edit* This puzzle is not my own. I will credit the source after the answer is revealed.
  2. Subscriber AThousandYoung
    It's about respect
    08 Sep '11 02:08
    Assuming the Guru knows someone has red eyes, he will be the one to leave, because he knows nobody else has red eyes.
  3. 08 Sep '11 04:56
    Originally posted by forkedknight
    There exists an island on which all inhabitants of the island are perfection logicians. If a conclusion can be logically deduced, all will do it immediately.

    No-one can communicate in any way with any other inhabitant of the island (no talking, to sign language, etc.)

    No one on the island can see their own eyes or knows their own eye color. Ho ...[text shortened]... ?

    *edit* This puzzle is not my own. I will credit the source after the answer is revealed.
    I know this one, and I think you need to specify that the guru says this on day 0 and every day following (and of course, as a guru, is truthful).
  4. Standard member forkedknight
    Defend the Universe
    08 Sep '11 05:42 / 5 edits
    Originally posted by AThousandYoung
    Assuming the Guru knows someone has red eyes, he will be the one to leave, because he knows nobody else has red eyes.
    The Guru doesn't know his own eye color, and no-one knows how many (total) of each color of eyes there are, nor that there are 3 (or only 3) colors of eyes.

    i.e., if I am Green-eyed inhabitant #14, I can see N-1 green-eyed people, M brown-eyed people, and 1 red-eyed person. I have no idea if my own eyes are green, brown, red, or possibly yellow.

    (This all applies before day 0.)
  5. Standard member forkedknight
    Defend the Universe
    08 Sep '11 05:43 / 1 edit
    Originally posted by JS357
    I know this one, and I think you need to specify that the guru says this on day 0 and every day following (and of course, as a guru, is truthful).
    The Guru says this once, and only once.
  6. 08 Sep '11 06:03
    Originally posted by forkedknight
    The Guru says this once, and only once.
    OK, I think you are right, after all, a good puzzle that way too. Sorry.
  7. Standard member wolfgang59
    Infidel
    10 Sep '11 00:41
    Originally posted by forkedknight
    There exists an island on which all inhabitants of the island are perfection logicians. If a conclusion can be logically deduced, all will do it immediately.

    No-one can communicate in any way with any other inhabitant of the island (no talking, to sign language, etc.)

    No one on the island can see their own eyes or knows their own eye color. Ho ...[text shortened]... ?

    *edit* This puzzle is not my own. I will credit the source after the answer is revealed.
    If there were only one green-eyed person then that person would not be able to see any green-eyed people and would therefore deduce his own eye colour and leave on day 1. (The first midnight ferry)

    if there were 2 green-eyed people then they could each see only one green-eyed person and when that person did not leave (as above logic) would deduce that they had green-eyes and both leave on day 2.

    Etcetera etcetera.

    Therefore ALL the green-eyed people leave on day N.

    The remaining people would know that they have either brown or red eyes but cannot leave without further information.
  8. Standard member forkedknight
    Defend the Universe
    28 Sep '11 03:39 / 2 edits
    That's correct, assuming the first ferry comes on day 1 (I referred to it as day 0)

    I borrowed this problem from XKCD, although I think he over-explains his description and makes it sound too complicated.
    http://www.xkcd.com/blue_eyes.html