2 thoughts on “The Hardest Logic Puzzle in the World [Allegedly]

  1. The solution can be seen with a reasonable argument.
    let k=no. of green eyed people
    If k = 1,
    the person will recognize that they alone have green eyes
    (by seeing only non-green eyes in the others) and leave at the first dawn.
    If k = 2, no one will leave at the first dawn. The two green-eyed people,
    seeing only one person with green eyes, and that no one left on the
    1st dawn (and thus that k > 1), will leave on the second dawn. Inductively,
    it can be reasoned that no one will leave at the first k-1 dawns if and only
    if there are at least k green-eyed people. Those with green eyes, seeing k-1
    green-eyed people among the others and knowing there must be at least k, will
    reason that they must have green eyes and leave.
    so, k=100, k-1=99, so they all will leave on 100th dawn.

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