Non-Adaptive 12 Identical Balls
I just figured it out. It should be possible because each weighing has 3 outcomes so 3*3= 27 outcomes and there are only 24 different possibilities for the answer.
Let’s say you have balls ABCDEFGHIJKL
1st weighing ABCDE v FGHIJ
2nd weighing GCJE v AIDK
3rd weighing BEI v GHL
S= same, L= left side heavier, R= right side heavier for the outcomes AL = A is lighter and AH = A is heavier (for example HHH = EH)
F, K, and L are only weighed once, so if their side is heavier or lighter and the other two weighings are the same, then you know if they are heavier or lighter.
E and I and G are weighed 3 times, LLL= EH and RRR=EL and RRL= IH and LLR = IL and LRL = GH and RLR= GL
All other balls are weighed twice. If their side is heavier or lighter both times then that ball is heaver or lighter. The key to this is that none of the balls weighed twice are weighed with any of the other twice weighed balls.
All combinations except the impossible SSS, RLL, and LRR give unique answers which is also 27 possible outcomes- 3 impossible ones = 24 total outcomescontext