Solution win if there's no strictly better solution, aka. no solution that take less or same amount of items in mapping table, less or same amount of instrucions, and at least one of amounts of items and instructions is less.
I understand this statement as
A solution is measured in two ways: the number of items in the mapping table, and the number of instructions. A solution wins if no other solution is strictly better; solution A is strictly better than solution B if A is strictly better in one measure and at least as good in the other, i.e.
(A<B && A<=B) || (A<=B && A<B)(assuming lower score is better in both measures).
I'm worried about this winning criterion because it permits many "winners" and we don't know which measure to primarily optimize for. The worst part is that, given an existing "winning" solution, one can tweak a little bit and post another "winning" solution, and then we have "another winner", not "the latter winning against the former".
On the other hand, l4m2 claimed that multiple winners is not a problem because we already have something like "shortest code in each language wins", and we can get a collection of, say, "an exact bound of table size for each instruction count".
Is the proposed winning criterion valid?