More counterfeits

Not 100% but I think I'm close

Three piles of 4, pick any two. compare. If equal, the counterfeit is in the third pile. Discard both piles and pick two coins from the third pile. If equal, the counterfeit is one of the other two. Remove one, and replace with one of the remaining two. If balanced, the one left is counterfeit. If unbalanced, the new coin is counterfeit.

If the original two piles do not balance, the counterfeit is on the scale, but you don't know which side (heavy or light). You could use the third pile to figure it out, but then you are at 4 weighings. You could split a pile to figure out if it contained the odd coin, but 4 weighings if it doesn't... sort of stuck here...

S

Starting right, with 4 v 4

And a hint

After the first weighing, you will have identified either four or eight genuine coins. The genuine coins can be useful in subsequent weighings.

Re: And a hint

The first part is not too difficult if you think about it, but the second part is quite a bit more involved. I didn't answer for two reasons:

1) I have seen this before and know the first one.

2) Haven't figured out the second part yet ;-)

Re: And a hint

"The first part is not too difficult if you think about it"

Speak for yourself! I found it quite difficult :-(

Steve: You investigation after the first weighing showed a balance successfully identifies the counterfeit coin, but doesn't tell whether it is light or heavy, which is one of the requirements.

Hints and solution

Solution in pieces, so if you are still working on it, read only as far as the hints you want.

For convenience, number the coins 1-12.

First of all, is it even possible. Well, one test is the number of possible results (12 possibilities for the counterfeit x 2 possibilities for the weight discrepency (light or heavy) = 24). Each weighing givew us three possibilities, so in three weighings, there are 3^3 = 27 possible results. I don't think this says it is possible, but at least there is potential.

Hint 1: Steve's start of weighing 1,2,3,4 v 5,6,7,8 is the correct start.

Hint 2: Using "Known good" coins in the balance can help.

Hint 3 (Partial solution, for case when first weighing is balanced): For weighing two, balance 9 and 10 versus 11 and 1. If they are balanced, we know that 12 is bad, and we balance it versus a known good coin to determine whether it is heavy or light. If the 9/10 side is heavy, we know that either 9 is heavy, 10 is heavy, or 11 is light. Third weighing is 9 v 10. If balanced, 11 is the counterfeit and is light. If unbalanced, the heavier one is the counterfeit. The obvious parallel situation occurs with the opposite result of the second weighing.

Hint 4 (Direction for rest of solution of first part): Suppose the left cup is lighter than the right cup. For the next step, we should divide the 8 potential answers into groups of 3, 3 and 2. For example, we want to have three answers in the case when the left cup is lighter or equal to the right cup, and 2 answers when the left cup is heavier than the right cup. That means that 3 out of the 8 coins should be left on the place where they started, three should be taken away, and 2 should change places. We may use coins 9, 10, 11, and 12 to supplement weighing, so we would have an equal number of coins in each cup. For example, we can have coins 1, 2, and 5 be left in their places, coins 3, 4 and 6 be removed from the balance, and coins 7 and 8 change cups. In this case, we need to add three good coins, so our second weighing is: 1, 2, 7, and 8 in the left cup; 5, 9, 10, 11 in the right cup. If the cups are equal, then the fake coin is among 3, 4 or 6. If the left cup is lighter, then the fake coin is among 1, 2, and 5, and if the left cup is heavier, then the fake coin is among 7 or 8, and for each number we know if it is heavier or lighter.

Part 2 solutions:

1,2,3,4 v 5,6,7,8

1,8,9,4 v 10,6,11,3

10,2,7,8 v 12,4,6,11

And someone with a creative flair came up with the solution of lettering the coins with the letters "FAKE MIND CLOT" , and use the weighings:

MA DO v LIKE

ME TO v FIND

FAKE v COIN

Disclaimer: Part 2 solutions were found in the course of looking for a clearer explanation of Part 1 solution.