A Random Scale Solution

I recently posted the following puzzle:

Puzzle. We have 32n identical-looking coins. One of the coins is fake and lighter than the other coins, which all are real. We also have three scales: two normal and one random. Find the fake coin in the smallest total number of weighings.

Here is my son Sergei’s solution. Divide the coins into nine groups of equal size and number the groups in ternary: 00, 01, 02, 10, 11, 12, 20, 21, and 22. On each scale we put three groups versus three groups. On the first scale we compare the three groups that start with 1 with the three groups that start with 2. For the second scale we do the same using the last digit instead of the first one, and for the third scale we use the sum of two digits modulo 3. Any pair of scales, if they are assumed to be normal, would point to one out of nine groups as the group containing the fake coin.

If all three pairs of scales agree on one group, then this is the group containing the fake coin. Thus in three weighings, we reduce the number of groups of coins by a factor of nine. If the pairs of scales do not agree, then the random scale produced a wrong weighing and thus can be found out. How do we do that? We have three out of nine groups of coins each of which might contain the fake coin. We compare two of the groups on all three scales. This way we know exactly which group contains the fake coin and, consequently, which scale generated a wrong weighing. If we know the random scale, we can speed up the rest of the process of finding the fake coin. Thus in the worst case we require 3n+3 weighings.

The big idea here is that as soon as the random scale shows a wrong weighing result it can be found out. So in the worst case, the random scale behaves as a normal scale and messes things up at the very end. Sergei’s solution can be improved to 3n+1 weighings. Can you do that?

The improved solution is written in a paper Взвешивания на «хитрых весах» (in Russian) by Konstantin Knop, that is published in Математика в школе 2009-2. The paper contains an even stronger solution that provides a better asymptotics.

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