Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC
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$ W* l1 q; X9 Z# \3 W5.39.217.76First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.
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3 ]* D- ^( Z1 d; M& M' q( T$ w公仔箱論壇From the first weight, let's say AAAA is heavier then BBBB, then either the problem ball is heavier among AAAA, or lighter among BBBB, but all C balls are normal.tvb now,tvbnow,bttvb( {6 B9 L& j0 u/ S$ g" ]+ G
tvb now,tvbnow,bttvb5 Y/ M5 p( v. A9 O. p; f# L2 A
2nd weight....AAAB ^ ACCC, there will be 3 scenarios:. [( n8 ~) a% a2 `1 M' f' ?
(1) if AAAB is heavier then ACCC, for sure the problem ball is a heavier ball and it's among AAA. Take 2 of these and weight against each other, the heavier ball is the problem, if balance then it's the remaining ball.TVBNOW 含有熱門話題,最新最快電視,軟體,遊戲,電影,動漫及日常生活及興趣交流等資訊。6 d3 {) Q/ f5 C4 H) \+ u
(2) if ACCC is heavier then AAAB then weight the A ball in ACCC against a normal C ball, if balance then the problem ball is the B ball in AAAB and it's a lighter ball. If heavier then this is the problem ball. (note it cannot be lighter)- I. U3 t" Z+ E* R1 l9 z& o
(3) if balance then the problem ball is one of the 3 B balls not touched in the 2nd weight and it's a lighter ball. Take 2 of these B balls and weight against each other. If balance then the other B ball is the problem. If not balance then the lighter ball is the problem. |