8 NonAdaptive 12 Identical Balls 
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NonAdaptive 12 Identical Balls
ok now in a nice structure:
now we only have to reduce the problem to this first:
case 1:
case 2 in the assumption 1234>5678 else switch the balls names:
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NonAdaptive 12 Identical Balls
I Have an answer: if there are 2 balls: 'a' and 'b' left which are not in balance then weigh 'a' vs a third ball 'c' and then there are 2 cases: 'a' vs 'c' is in balance then 'b' is odd else 'a' is odd. This will always be our last weighing of balls. So i will construct 2 weighings that will lead to this case. The first weigh is balls 1 to 4 vs balls 5  8 to split 2 cases: there is balance > problem in 9  12 > case 1 there is no balance > problem in 1  8 > case 2 the second weigh depends on the case: case 1: weigh 9  10 vs 1  2 > if in balance 2 balls left, if not in balance 2 balls left > OK case 2: weigh 1  2  7 vs 5  6  3 (and lets asume 1234 heavier then 5678 else rename balls in the following) this leads to 3 cases: balance > 2 balls left : 4 and 8 => ok
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NonAdaptive 12 Identical Balls
wait how does that latest part seem so nice? context 
NonAdaptive 12 Identical Balls
wait how does that latest part seem so nice? context 