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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
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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 
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Non-Adaptive 12 Identical Balls

wait how does that latest part seem so nice?

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Non-Adaptive 12 Identical Balls

wait how does that latest part seem so nice?

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