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Motovated Design & Analysis Newsletter - February 2010

December 2009 Puzzle Answer - the full explanation

Balance Scales If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So....split the coins into 3 groups of 4.

Weigh coins 1, 2, 3, 4 against coins 5, 6, 7, 8.

1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).

1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.

1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.

1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.

1.2. Now if (at first weighing) the side with coins 5, 6, 7, 8 are heavier than the side with coins 1, 2, 3, 4. This means that either 1, 2, 3, 4 is light or 5, 6, 7, 8 is heavy. Weigh 1, 2 and 5 against 3, 6 and 9.

1.2.1. If (when we weigh 1, 2 and 5 against 3, 6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.

1.2.2. If (when we weigh 1, 2 and 5 against 3, 6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.

1.2.3. If (when we weigh 1, 2 and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.

1.3 If (at the first weighing) coins 1, 2, 3, 4 are heavier than coins 5, 6, 7, 8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1, 2, 3, 4 with 5, 6, 7, 8.

So the minimum number of iterations (to always identify the fake coin) is 3