Newsgroups: bit.listserv.politics
Date:         Fri, 13 Nov 1992 12:07:55 CST
From:         Kip Shaw <KIS100F@SMSVMA.BITNET>
Subject:      Re: Stalinist Purge of the Bad Apple: the solution revealed

>>There are 12 identical looking apples.
>>11 of them weigh exactly the same.
>>The 12-th of them is defective (such as: liberal or conservative) in the
>>sense that it is heavier or lighter than the rest of the apples.
>>The apples are shuffled and get randomly arranged.
>>An ordinary 2-pan balance is provided.
>>Exactly 3 weighing chances are given.
>>Determine the defective apple AND also tell whether the defective apple
>>is heavier or lighter (ie liberal or conservative).
>>(over)
>
>Your Logic Czar feels that it is incumbent upon him to show how simple
>the problem really is. Can't have all of our other Czars and elected
>officials worrying their pretty heads about it.
>
>Choose 8 apples. Weigh four of them against the other four.
>
>    Case 1: They balance against each other. Then the defective
>            one is among the other four.
>
>         Then take three of those four, and balance them against
>         three of the apples that you know are good ones.
>
>              Case 1.1 They balance. Then the defective one is
>                        remaining. Check it against a good one
>                        to see whether it is heavy or light.
>
>              Case 1.2 The good three are heavier. Then we know
>                       the defective apple is light and is one of
>                       three. Balance one of those three against
>                       another.
>
>                 Case 1.2.1 They balance. Then the defective one
>                            is the third, and is light.
>
>                 Case 1.2.2 They don't balance. Then the lighter
>                            of the two is defective.
>
>              Case 1.3  The good three are lighter. Then proceed
>                        as in case 1.2, but the bad apple is heavier.
>
>       Case 2: They don't balance. Call the heavier four apples
>               "Batch A," the lighter four "Batch B," and the
>               remaining four "Batch C." Label the Batch A apples
>               "A1" "A2" A3" and "A4," and likewise for the B Batch.
>               (Don't bother labeling the C apples. We know they
>                are good apples.)
>
>         Now put A1, A2, B1, and B2 in one pan, and A3, B3, and two
>         of the C apples in the other.
>
>         Case 2.1 The side with the C apples is light. Then, either
>           A1 or A2 is a defective heavy apple, or B3 is a defective
>           light apple. So weigh A1 against A2.
>
>           Case 2.1.1 They balance. Then B3 is defective and light.
>
>           Case 2.1.2 A1 is heavier. Then it is defective and heavy.
>
>           Case 2.1.3 A2 is heavier. Then it is defective and heavy.
>
>        Case 2.2  The side with the C apples is heavy. Then, either
>           either B1 or B2 is defective and light, or A3 is defective
>           and heavy. So weigh B1 vs B2.
>
>           Case 2.2.1 They balance. Then A3 is heavy.
>
>           Case 2.2.2 B1 is lighter. Then it is defective and light.
>
>           Case 2.2.3 B2 is lighter. Then it is defective and light.
>
>   Case 3  A1, A2, B1, and B2 together BALANCE against A3, B3, and two C's.
>
>           Then, either A4 is heavy or B4 is light. So just weigh A4
>           against one of the C's. If it is heavy, it is the defective
>           apple. If it balances, then B4 is the defective (light) one.
>
>That was fun.
>But your Logic Czar has no delusions of genius, have no fears.
>
>Now back to our regularly scheduled arguments.
>
>Jamie

Hello Jamie,
The solution is wonderful and has a Perot-like ring of simplicity with it!
Kip Shaw.
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