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Tricky Puzzle (12 Balls)
Asked in Google, Facebook, Amazon, Udaan, Nutanix, Netflix, American Express

# Problem Statment

You have 12 balls identical in every way except that one of them weighs slightly less or more. You have a balance scale. Find the Minimum number of comparisons to find the defected ball.

# Solution Approach

Let us number our balls from 1 to 12, so we have {1,2,3,4,5,6,7,8,9,10,11,12}. Divide these balls into three Groups: G1 = {1,2,3,4} G2 = {5,6,7,8} G3 = {9,10,11,12} Weigh G1 and G2, Now Three cases rise,
weight(G1) = weight(G2)
weight(G1) > weight(G2)
weight(G1) < weight(G2)
This means the defected ball is in G3
So, suspected balls are {9,10,11,12} and confirmed balls are {1,2,3,4,5,6,7,8}.
Now make two more groups named, G4 = {8,9} and G5 = {10,11} Note: Ball no. 8 is not defected one. Now weigh G4 and G5 and three cases rise,
a) weight(G4) = weight(G5) Our answer is ball 12. b) weight(G4) > weight(G5) Either {9} is heavy or {10,11} is light, so weigh {10} and {11} if both are equal then, ans={9}, else ball with less weight will be answer. c) weight(G4) < weight(G5) Either {9} is light or {10,11} is heavy, so weigh {10} and {11} if both are equal then, ans={9}, else ball with heavy weight will be answer. The Maximum comparison in each case for this scenario is 3.
This means either G1 is heavier or G2 is lighter. So, Suspected balls are {1,2,3,4} and {5,6,7,8} and confirmed balls are {9,10,11,12} Now make two groups, G8 = {1,2,5} and G9 = {3,6,9} Now weigh them and three cases arise : a) weight(G8) = weight(G9) So, now suspects balls are {4,7,8} and rest are confirmed balls. Since weight(G1)> weight(G2) so, either {4} is heavy or either of {7,8} is light. Weight ball {7} and ball {8} , if equal answer is ball {4} else answer is ball with lighter weight. b) weight(G8) > weight(G9) So either of {1,2,5} is heavy or {3,6} is lighter. Point to think here: Ball {5} and {3} are no more suspected because, if {5} was heavy then, weight(G2) would have been greater and similarly if {3} has been lighter then also weight(G2) would have been greater. So now we can say either of {1,2} is heavy or {3} is lighter. Weigh ball {1} and ball {2} , if equal answer is ball {3} else answer is ball with heavier weight. c) weight(G8) < weight(G9) So either of {1 , 2 , 5 } is light or {3 , 6 } is heavier. Point to think here: Ball {1,2 } can not be light since the weight of G1 is more and also ball {6} can not be heavy as the weight of G2 is less. So either {5} is lighter or {3} is heavier. Weigh ball {5} with ball {9} , if equal answer is ball {3} else answer is ball {5}. The Maximum comparison in each case for this scenario is 3.
This means either G1 is lighter or G2 is heavier. So, Suspected balls are {1,2,3,4} and {5,6,7,8} and confirmed balls are {9,10,11,12}
Now make two groups, G10 = {1, 5 , 6 } and G11 = {2 , 7 , 9} Now weigh them and three cases rise : a) weight(G10) = weight(G11) So, now suspects balls are {3 ,4,8} and rest are confirmed balls. Since weight(G1)<weight(G2) so, either {3 ,4} is lighter or either of {8} is heavy. Weigh ball {3} and ball {4} , if equal answer is ball {8} else answer is ball with lighter weight. b) weight(G10) > weight(G11) So either of {1 , 5 , 6 } is heavy or {2,7} is lighter. Point to think here: Ball {1} and {7} are no more suspected because, if {1} was heavy then, weight(G2) would have been lesser and similarly if {7} has been lighter then also weight(G2) would have been lesser. So now we can say either of {5,6} is heavy or {2} is lighter. Weigh ball {5} and ball {6}, if equal answer is ball {2} else answer is a ball with heavier weight. c ) weight(G10) < weight(G11) So either of {1 , 5 , 6 } is lighter or {2,7} is heavier. Point to think here: Ball {5} and {6} are no more suspected because, if they would have been lighter then weight(G2) would have been less. Similarly, ball {2} is not heavier, as weight(G2) is more. So now we can say either {1} is lighter or {7} is heavier. Weigh ball {1} and ball {9} if equal answer is ball {7} else answer is a ball with heavier weight.
The Maximum comparison in each case for this scenario is 3.