8 NonAdaptive 12 Identical Balls 
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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 
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
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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