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### SMURF

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 Non-Adaptive 12 Identical Balls wait how does that latest part seem so nice? context Non-Adaptive 12 Identical Balls wait how does that latest part seem so nice? context Non-Adaptive 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 `````` left heavier -> 2 balls left : 1 and 2 => ok right heavier -> 1 ball left : 3 `````` context Non-Adaptive 12 Identical Balls ok now in a nice structure: `````` 2 balls, 'a' and 'b' left -> third ball 'c' -> weigh 'a' vs 'c' if 'a' > 'c' -> 'a' odd else 'b' odd `````` now we only have to reduce the problem to this first: `````` weigh 1-2-3-4 vs 5-6-7-8 if in balance -> case 1 else -> case 2 `````` case 1: `````` weigh 9 - 10 vs 1 - 2 if in balance -> 2 balls left 11 & 12 else 2 balls left : 9 & 10 `````` case 2 in the assumption 1234>5678 else switch the balls names: `````` weigh 1-2-7 vs 5-6-3 if in balance -> 2 ballse left : 4 & 8 else if left heavier -> 2 balls left : 1 & 2 else 1 ball left : 3 `````` context