TxHockeyGuy
Contributor
bob1dp:Sorry about the strange formatting but the board doesnt do an outline very well. Man was this a tough one. He is my take on it.
Take three groups of four: A, B, and C. Weigh group A against group B
1. If group A = group B the counterfeit is in group C. Take 3xC and weigh it against 3xA. Set aside one C.
A. If group 3xA=group 3xC then the set aside C that wasn't weighed contains the counterfeit. Weigh one A against un-weighed set aside C to get weight of counterfeit C.
B. If group 3xA>group 3xC then one of 3xC is the counterfeit and light. Weigh C.1 against C.2 set aside C.3. If C.1 = C.2 then C.3 is the counterfeit and is light. If C.1 > C.2 then C.2 is the counterfeit and is light. If C.1 < C.2 then C.1 is the counterfeit and is light.
2. If group A > group B, then group C contains no counterfeit coins and that either group A has a heavy counterfeit or group B has a light counterfeit. Set aside 1xA and 2xB. Move 2xA to group B, Move 1xB to group A. Add 1xC to group A. You are now weighing 1A*+1B+1C against 2A+1B* and setting aside 1xA and 2xB.
A. If 1A*+1B+1C >2A+1B* which was the original uneven weighing. Then 1A* or 1B* are the counterfeits because they are the only element that didn't change. Weigh 1B* against 1C and if 1B*=1C then 1A* is the counterfeit and heavier. If 1B*<1C then it is B* and it is lighter
B. If 1A+1B+1C =2A+1B Then set aside 1xA or 2xB Are the counterfeit. Weigh 2xC against (1xA + 1xB) and set aside B. If (1xA + 1xB) weighs heavy 1xA is the counterfeit and is heavy, if it weighs equal the set aside B is the counterfeit and is light. If it weighs is light then the 1xB being weighed is the counterfeit and light.
C. If 1A+1B+1C <2A+1B which is the opposite of the original mismatch. The 1B in group A and 2A in group B are the counterfeits. Weigh 2xC against 1A + 1B and set aside A. If it weighs heavy A is the counterfeit and is heavy, if it weighs equal the set aside A is the counterfeit and is heavy. If it weighs light B is the counterfeit and light.
I had a little trouble following your answer but I believe it is correct. I was going to write out my solution on here but found a nice online example.
http://home.att.net/~numericana/answer/weighing.htm#weighing