Finding the Odd Ball

Antonia FergusonNovember 10, 2009

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.

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.

Combinations of 0,1 and 2

000 100 200001 101 201002 102 202010 110 210011 111 211012 112 212020 120 220021 121 221022 122 222

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.

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

12 Remaining Combinations

Ball ‘A’ 001

‘B’ 010

‘C’ 011

‘D’ 012

‘E’ 112

‘F’ 120

‘G’ 121

‘H’ 122

‘I’ 200

‘J’ 201

‘K’ 202

‘L’ 220

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.

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.

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.

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

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

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

QUESTIONS?!?!

Works Cited

http://mathforum.org/library/drmath/view/56766.html

http://www.mathsisfun.com/pool_balls_solution.html