Finding the Odd Ball

1. Finding the Odd Ball Antonia Ferguson November 10, 2009

2. THE PROBLEM • There are 12 balls; 11 physically identical balls and 1 that is a different weight/density. You have a scale balance that can be used three times. Your goal is to identify the odd ball, and determine whether it is heavier or lighter than the other 11.

3. Solution • First, list all of the 3 digit combinations of 0,1 and 2 • 3 digits represent 3 uses of the balance. • 33 = 27 possible combinations where a combination = 3 digit numbers of {0,1,2}. • There are 9 combinations with a 0 in the first place, 9 with a 1 in the first, and 9 with a 2 in the first, and so forth for the next 2 places. • Thus we effectively are determining the grouping of balls for each weighing.

4. Combinations of 0,1 and 2

5. Reasoning • By grouping the balls with the 3 digit numbers, we ensure that the groups of balls for each weighing will be constructed in such a way as to exclude all balls but one by the end of the 3 balances. • This is difficult to see/understand until the end of the process, so we will return to this concept later in detail.

6. Narrow the Range to 12 • We have a list of 27 combinations giving 3 groups of 9 for each weighing – but we need 12 combinations that give 3 groups of 4 for each weighing, and we need to do this in a way that leaves ordering we can use to identify the odd ball. • From the list of 27 cross out 15 • 000, 111, and 222 (scale gives same reading regardless of other 3 balls in group). • Numbers whose first digit change is not 01, 12, or 20 (ensures 4 balls in all 3 groups for all 3 weighings) • Example of accepted: 011, 112, 200 • Example of rejected: 211, 021, and 101

7. 12 Remaining Combinations

8. 1st Weighing • Weigh all balls with a 0 in the first position against the balls with a 2 in the first position • ABCD vs. IJKL • If the 0 side is heavier write 0, if the 2 side is heavier write 2, and if the scale is balanced write 1 • We assume the odd ball is heavier, but a later step will correct for this if it is lighter.

9. 2nd Weighing • Weigh the balls with 0 in the second position against the balls with 2 in the second position. • AIJK vs. FGHL • Write 0 if the 0 side is heavier, 2 if the 2 side is heavier, or 1 if the scale is balanced.

10. 3rd Weighing • Weigh the balls with 0 in the third position against the balls with 2 in the third position. • BFIL vs. DEHK • Write 0 if the 0 side is heavier, 2 if the 2 side is heavier, or 1 if the scale is balanced.

11. What happens if a number is not on list? • If a number is not on the list of 12 change the 0’s to 2’s and 2’s to 0’s • This indicates the ball is lighter, so the numbers must be flipped • Originally getting 020 means that ball K is lighter (202) • This happens because throughout the problem you are assuming the odd ball is heavier

12. Algorithm • (3^N-3)/2; N= number of weighings • This equals the maximum number of balls that can be solved per N weighings • Subtract three because 3 impossible combinations • Divide by 2 because we cannot directly differentiate heavier and lighter on a balance. • Can solve • 3 balls in 2 weighings • 39 balls in 4 weighings • 120 balls in 5 weighings

13. Conclusion • We have solved the problem using a technique called enumeration-an ordered listing of the elements of the indexed set. • The restrictions and structure-criteria we place on the set and its elements ensure a ‘well-ordered set’, which in this case meant a set that grouped itself into 9 unique subsets. • ‘unique’ is used to mean exclusionary, such that the groups are ordered so that only one possible ball can be present in all 3 that weigh heavier. • Not only way to solve

14. QUESTIONS?!?!

15. Works Cited • http://mathforum.org/library/drmath/view/56766.html • http://www.mathsisfun.com/pool_balls_solution.html