In the famous
"The Odd Coin" problem you are given twelve coins, exactly one of which is lighter or heavier than the other coins. You are to determine the counterfeit coin, and whether it is lighter or heavier than the other coins, in just three weighings with a balance.
Can you solve this problem with the additional restriction that you must decide what coins go on each pan for all three weighings before any weighing takes place?
(In reply to
re(2): Another way to answer it by McWorter)
Well, neither does mine...you could mark each of the coins with what the results of the weighings. Assume the coin is heavy and mark what the weighing result would be on the heads side. Assume the coin is light and mark what the result would be on the tails side. For example, coin 5 would be marked RLL on the heads side, and LRR on the tails side. When you see the results of the weighings, just find the coin with the corresponding marking. The side you find the marking on (heads or tails) tells you if that coin is heavy or light.
Or...you could just take the results and look at the weighing plan--it wouldn't be hard to figure out which coin fits the results, and whether it's light or heavy, although it may take a few minutes. The table or the pre-marking are faster strategies.
All these solutions are essentially equivalent.
Have you got a completely different solution in mind, McWorter?
re(3): Another way to answer it
