Problem 4
A man has 12 billiard balls and a pair of scales. All the balls look identical but one
has a different weight to the others. Find this ball with a minimum number of
weighings
Solution
The answer is it can always be done in 3 weighings.
Weigh the balls in group of four.
xxxx xxxx remaining xxxx
If groups balance then the false ball in in the third group
- 1. If two groups balance then false ball is in third group.
so we have nnnn nnnn ffff (f means possbly false)
Weigh three from false group against three normal
nnn fff.
- if they are equal then we know the remaining ball is false and the third weighing wil reveal whether it is heavier or lighter.
Take two of the three false balls and weight. If they are equal then third ball is false ball.
If not, then we know, it is heavyer or lighter ball.
- if they are not equal, then we will know from the scales whether the false ball is heavier or lighter.
Take two of the possibly false balls and weigh against each other. If they are equal then the third ball is the false ball.
If they are not equal then, as we know whether the false ball is heavier or lighter, we can find the false ball.
- 2. If the two groups do not balance then we have the situation
llll hhhh nnnn
where
l means that this ball is possibly false and light
h means that this ball is possibly false and heavy
n means that the ball is normal.
Now weigh
lllhh against lnnnn
- if equal then have only two balls left which are possibly heavy. Weigh again and select the heaviest.
- if not equal then there are two cases:
- in case a) the false ball must be one of the three possible light ones. Weigh any two and if equal select the third or take the lightest.
- in case b) the false ball must be one of the possible heavy ones. Weigh and take the heaviest.