SPOILER!!: three predefined weighings The question was: how to find the one odd-weighted coin out of 12 with three weighings of a balance scale. The location of the coins must be determined beforehand. -- SPOILER -- The three weighings are labeled N = 1, 2, 3. For each weighing: - if the left scale goes down then the value of that weighing is W(N)=(3^N), - if it balances then the value is W(N)=0, - otherwise the value is W(N)=-(3^N). Label the coins 1,2,3,..,12 and conduct the weighings as follows: weighing 1:   left: 1, 4, 5, 11   right: 2, 7, 8, 10 weighing 2:   left: 2, 3, 4, 7   right: 5, 6, 11, 12 weighing 3:   left: 5, 6, 8, 9   right: 7, 10, 11, 12 Add up the values from the three weighings: SUM=W(1) + W(2) + W(3). The number of the odd coin is: abs(SUM). The weight difference is given by: sign(SUM) (positive=>heavier) unless the coin is one of {7, 10, 11, 12} in which case it's indicated by sign(-SUM). Here's a related question: obviously there are many other permutations of positions that also solve this problem, can you find a formula that, for all possible permutations, links the heavier coins that will yield negative SUMs (obviously the answer is {7, 10, 11, 12} for the above solution)? Piers Haken Thursday, February 07, 2002 Unrelated question - as there are so few Piers Hakens in the world - mind if I ask if you are the Piers Haken who went to Dulwich College up to 1990? Matt Clark Thursday, February 07, 2002 Your solution did not work for me. When N goes from 1..3, the value of W(N) needs to be: W(N) = (L < R) ? 3^(N-1) : (L > R) ? -(3^(N-1)) : 0 Also, I was getting the same result for coin 5 and coin 7. The pairings I came up with are: weighing 1: left: 1, 4, 7, 11 right: 2, 5, 8, 10 weighing 2: left: 2, 3, 4, 6 right: 5, 7, 11, 12 weighing 3: left: 5, 7, 8, 9 right: 6, 10, 11, 12 which changes the "reversed set" to {6, 10, 11, 12} broken down it looks like: coin 1 = +1+0+0 coin 2 = -1+3+0 coin 3 = +0+3+0 coin 4 = +1+3+0 coin 5 = -1-3+9 coin 6 = +0+3-9 (== -6) coin 7 = +1-3+9 coin 8 = -1+0+9 coin 9 = +0+0+9 coin 10= -1+0-9 (== -10) coin 11= +1-3-9 (== -11) coin 12= +0-3-9 (== -12) David Woodruff Tuesday, February 19, 2002 A variant of this one was the first interview question I was ever asked. There is a much simpler solution. Divide the 12 coins into 3 groups of 4. Weigh two of the groups. It will be obvious what to do next. Steven T Abell Monday, March 18, 2002   Fog Creek Home