Four coins, no reference, two weighings?

Question

You have four coins, exactly one of which is fake. The fake is either heavier or lighter than a genuine coin; you do not know which. There is no extra genuine coin. You have a two-pan balance. There are 4 × 2 = 8 possibilities, and two weighings have 9 outcomes, so counting does not forbid it. Can you always identify the fake coin and whether it is heavy or light in two weighings? Prove your answer.

Accepted Answer

No. Every first weighing leaves some branch with more than three possibilities, and one weighing cannot separate more than three. First weighing: one against one. Say A against B, with C and D aside. - If they balance, both A and B are genuine, and the fake is C or D, heavy or light: 4 possibilities. The last weighing has only 3 outcomes. Impossible to finish. First weighing: two against two. Say A and B against C and D. The scale cannot balance (the fake is on the scale), so it tilts. - If the left side is heavier, the fake is A heavy, B heavy, C light, or D light: 4 possibilities. Again one…

Four coins, no reference, two weighings?

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