Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC: R4 J: R' y, s, \
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。) V9 A0 }' M$ e+ F, u
First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.
% E; x c! i' j' ?( K' F: _
& l( D; C' h( L, V" e6 pTVBNOW 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。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,bttvb1 m$ d+ |3 D$ v* G$ ^9 k3 y& v
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。 i2 q7 r: X: I/ @
2nd weight....AAAB ^ ACCC, there will be 3 scenarios:
! p g. o$ O6 A(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.公仔箱論壇' j5 @1 q( T$ s% g& b" N
(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)公仔箱論壇! K- z9 w* R s( z& P$ |
(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. |