Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC
$ @: `. a# B* b6 b3 @+ i. qTVBNOW 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。
+ A: I3 t7 W% v, [! W% T2 c5 S8 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。First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.
+ A8 X4 `9 U; u! l3 C
7 b: f- {9 o# t: n6 Q) ^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.* k9 ], h! g" A6 x. C6 w( C; E/ o
% S# E/ u. c; a8 R7 {3 v2nd weight....AAAB ^ ACCC, there will be 3 scenarios:
, h1 L( B0 b: H0 L0 }% {- q. mos.tvboxnow.com(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.
! ?5 L3 {" V3 h) ]tvb now,tvbnow,bttvb(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)
0 p8 T1 _; W5 b) ^公仔箱論壇(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. |