Told you, I love sharing!
Sunday, 7 January 2018
Puzzles | Find the faulty ball from 10 balls in minimum steps
There are 10 balls and one ball is faulty among them, determine the faulty ball in minimum steps. The faulty ball may be lighter or heavier than other 9 balls.
Divide balls into 3 groups (A, B, C) and each group contain 3 balls each, and one extra ball (X).
Compare groups A and B, and then compare B and C
1. If A = B = C then X is the faulty ball.
2. If A != B and B != C
then the faulty ball is in B,
else if A != B
then the faulty ball is in A,
else if B != C
then the faulty ball is in C.
We can identify the faulty group and the faulty ball is heavier or lighter based on
if the faulty group is HEAVY/LIGHT compared to other groups.
3. After finding out the group which contains the faulty ball and the properties whether
the faulty ball is HEAVY or LIGHT.
Choose any two balls (b1, b2) from the faulty group (the last ball left is b3).
If b1 = b2
then the faulty ball is b3.
No weight comparison required
If faulty group is HEAVY
then pick the HEAVY ball is faulty
If faulty group is LIGHT
then pick the LIGHT ball is faulty.
Max comparisons required are 3.
Share to Twitter
Share to Facebook
Share to Pinterest
Post a Comment
Post Comments (Atom)