teaster1.txt

not-inept (Jun, 2018)

I was watching brooklyn nine nine and figured I have a go at the brain teaser posed in s02 e19. I'm likely not the first or last to solve it, but it was fun :D

Question: Given 1 2 3 4 5 6 7 8 9 10 11 12, one of the 12 weighs more or less than the others with only 3 comparisons

A(1 2 3 4) vs B(5 6 7 8), with C(9 10 11 12) to the side
1a. Is balanced
    The difference is in (9 10 11 12)
    X(1 2) vs Y(11 12)
    2a. Is balanced
        We know X(1 2) is balanced, so the difference is in (9 10)
        XX(1) vs YY(9)
        3a. Is balanced
            10 is the guilty number
        3b. Is not balanced
            9 is the guilty number       
    2b. Is not balanced 
        We know X(1 2) is balanced, so the difference is in (11 12)
        XX(1) vs YY(11)
        3a. Is balanced
            12 is the guilty number
        3b. Is not balanced
            11 is the guilty number
1b. Is not balanced
    Let's assume A > B (if not, just switch the groups as they are arbitrary)
    This means either group A has a heavier element, or group B has a lighter element. A cannot contain a lighter element, and B cannot contain a heavier element. C is known to be neutral.
    The difference must be in A(1 2 3 4) or B(5 6 7 8)
    X(1 2 5) vs Y(3 4 6)
    2a. Is balanced
        The difference is in (7 8)
        XX(1) vs YY(7)
        3a. Is balanced
            8 is the guilty number
        3b. Is not balanced
            7 is the guilty number
    2b. Is not balanced
        Each side contains two known heavier group elements, and one known lighter group element.
        If either side is heavier, then it must be because either its heavier group elements contain a heavier element, or the other side's lighter group element is lighter.
        The difference is in (1 2 6) or (3 4 5). We will assume (1 2 6), but the procedure is the same for either side.
        XX(1 6) vs YY(11 12)
        3a. Is balanced
            2 is the guilty number
        3b. XX > YY
            We know that the number originally in set B (6) must be lighter if it is different. Therefore it must be the number in the check originating from set A.
            1 is the guilty number
        3c. XX < YY
            We know that the number originally in set B (6) must be lighter if it is different. Therefore it must be the number in the check originating from set B.
            6 is the guilty number