All subsets requires 2^12 weighings. The beautiful thing about error correcting codes is finding a minimal set of spanning items. And in this case, binary will not suffice, since you can only distinguish 2^3 = 8 items. The third position, one most people don't think if, is the "off balance" set, which gives three choies.
Then 3 weighings, 3 choices per, gives 27 possible outcomes, which is enough to distinguish which of 12 items is off, as well as whether or not it's higher or lower in weight (another factor of 2), for 24 choices. So there is a little extra in the 3=27-24 information theoretical bounds.
Using this idea, you can generate all sorts of hard coin problems, vastly harder than this one :), by mining error correction tables like http://www.codetables.de/ and converting to word problems.
Then 3 weighings, 3 choices per, gives 27 possible outcomes, which is enough to distinguish which of 12 items is off, as well as whether or not it's higher or lower in weight (another factor of 2), for 24 choices. So there is a little extra in the 3=27-24 information theoretical bounds.
Using this idea, you can generate all sorts of hard coin problems, vastly harder than this one :), by mining error correction tables like http://www.codetables.de/ and converting to word problems.