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Non-Adaptive 12 Identical Balls
I think I have come up to a solution. Please check and try to find failing case if any (though I am pretty sure there shouldn't be any.) ^{ spoiler}don't read this Divide the balls in 3 sets of 4 balls each: A, B, C *First Weighing: A1, A2, A3, A4 vs B1, B2, B3, B4 *Second Weighing: A1, A2, B1, B2 vs A3, C1, C2, C3 *Third Weighing: A1, B1, B3, C1 vs A2, B2, B4, C2 Explanation: We can consider case wise: *Case 1: all weighing are equal or First two weightings are equal: C4 as it was never taken for weightings, and all other are defect-less.
*Case2: First Weighting is equal; second is unequal:
Case 3: First weighting is unequal; Again without loss of generality assume that Left was heavier. One of balls As can be heavier or one of the balls in B can be lighter. Case 3.1 in second weighting left is Heavier. That can be only because of balls A1 or A2. So, one of them is odd. In the third weighting they are in opposite pans, so whichever pan is heavier that ball is odd and heavy. *Case 3.2: In second weighting right is heavier: That can either be because one of the balls B1 or B2 is ligher or ball A3 is heavier In the third weighting we dont have ball A3 and balls B1 and B2 are in opposite pans. So, whichever pan is ligher that ball is odd. If the pans are equal that means A3 is odd and heavier. *Case 3.3 In second weighting pans are equal. That means none of the A1, A2, A3 or B1, B2 are defective. In the third weighting we have B3 and B4 in opposite pans, whichever pan is lighter that ball is defective. If the pans are equal, then A4 is odd and heavier. This covers all the cases. Hope you enjoy the solution. :) context |