| C | Positive pan | Negative pan |
| 1 | 4, 5, 10, 11 | 1, 2, 8, 13 |
| 3 | 2, 4, 5, 12 | 3, 6, 11, 13 |
| 9 | 6, 8, 10, 12 | 5, 9, 11, 13 |
For Greenblatt's expanded problem, label the coins zero through 13,
reserving zero for the known honest coin, and use this table:
| C | Positive pan | Negative pan |
| 1 | 0, 4, 5, 10, 11 | 1, 2, 7, 8, 13 |
| 3 | 2, 4, 5, 7, 12 | 0, 3, 6, 11, 13 |
| 9 | 0, 6, 8, 10, 12 | 5, 7, 9, 11, 13 |
Add the values of the individual comparisons. If the sum is odd, change its sign. All coins are genuine if the sum is zero. Otherwise, the magnitude of the sum is the label of the counterfeit, and its sign, adjusted as described, indicates heavy (+) or light (-).
Equal-arm balance problems are a natural fit to trinary numbers (base three). Here, I have used the digits -1, 0, and +1, but the place values are still 1, 3, and 9. For example, the number 13 is 1 + 3 + 9, so the coin 13 is always on the same pan. The number 11 is -1 + 3 + 9, so it is on the same pan as 13 for the 3 and 9 comparisons, but on the opposite pan for the 1 comparison. The number 12 is 0 + 3 + 9, so it doesn't appear in the 1 comparison at all.
The odd coins need to be switched in sign to produce as many plusses as minuses, but even then the scheme won't work if the number 7 is included. Having a known good coin to balance the difficult 7 (notice that 0 and seven always appear on opposite pans) shows why Greenblatt's expanded version can work. Note that three comparisons have 27 outcomes, since one comparison has three. With 13 coins, there are 27 possible states: 13 possibilities for heavy, as many more for light, and the 27th, that all are genuine. There can be no more information garnered from three comparisons.
A simpler equal-arm balance problem: given such a balance, and the need
to weigh out integer quantities 1, 2, 3, ... 40. What is the minimum number
of calibrated weights that make an adequate set?
Copyright © 2000 by Jerry Avins