Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC5 r' G- D5 `# B: k" a% s) N6 h
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First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.tvb now,tvbnow,bttvb- L6 Q1 H% V# G6 p- }3 o" }
- R* @% O& U Z: T6 C6 [. C! L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.公仔箱論壇6 |$ ~$ N4 a' y7 J: u/ \5 C
tvb now,tvbnow,bttvb0 m" i& b- R* E: \9 G9 K4 R
2nd weight....AAAB ^ ACCC, there will be 3 scenarios:
& \) W$ c. a/ I& D. A5 W3 k# lTVBNOW 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。(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. B/ C. q3 v( E: T$ j
(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)
$ F- d& p+ H" F: }2 Y# K- Stvb now,tvbnow,bttvb(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. |