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A number of logic puzzles exist that are based on the balancing of similar-looking items, often coins, to determine which one is of a different value within a limited number of uses of the balance scales. These differ from puzzles where items are assigned weights, in that only the relative mass of these items is relevant.
A well-known example has nine (or fewer) items, say coins (or balls), that are identical in weight save for one, which in this example we will say is lighter than the others—a counterfeit (an oddball). The difference is only perceptible by using a pair of scales balance, but only the coins themselves can be weighed.
Is it possible to isolate the counterfeit coin with only two weighings?
To find a solution to the problem we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one we can compare any two coins, leaving the third out. If the two coins tested weigh the same, then the lighter coin must be one of those not on the balance - otherwise it is the one indicated as lighter by the balance.
Now, assume we have three coins wrapped in a bigger coin-shaped box. In one move, we can find which of the three boxes is lighter (this box would contain the lighter coin) and, in the second weighing, as was shown above, we can find which of the three coins within the box is lighter. So in two weighings we can find a single light coin from a set of .
Note that we could reason along the same line, further, to see that in three weighings one can find the odd-lighter coin among 27 coins and in 4 weighings, from 81 coins.
The twelve-coin problem
A more complex version exists where there are twelve coins, at least eleven of which are identical. If there is one that is different, it is not known whether it is heavier or lighter than the others. This time the balance may be used three times to determine if there is a unique coin; and if there is, to isolate it and determine its weight relative to the others. (This puzzle and its solution first appeared in an article in 1945.) The problem has a simpler variant with 3 coins in 2 weighings, and a more complex variant with 39 coins in 4 weighings.
There is more than one solution to this problem. One is easily scalable to a higher number of coins by using base-three numbering: labeling each coin with a different number of three digits in base three, and positioning at the n-th weightings all the coins that are labeled with the n-th digit identical to the label of the plate. Other step-by-step procedures are similar to the following. It is less straightforward for this problem, and the second and third weightings depend on what has happened previously, although that need not be the case (see below).
- Four coins are put on each side. There are two possibilities:
- 1. One side is heavier than the other. If this is the case, remove three coins from the heavier side, move three coins from the lighter side to the heavier side, and place three coins that were not weighed the first time on the lighter side. (Remember which coins are which.) There are three possibilities:
- 1.a) The same side that was heavier the first time is still heavier. This means that either the coin that stayed there is heavier or that the coin that stayed on the lighter side is lighter. Balancing one of these against one of the other ten coins will reveal which of these is true, thus solving the puzzle.
- 1.b) The side that was heavier the first time is lighter the second time. This means that one of the three coins that went from the lighter side to the heavier side is the light coin. For the third attempt, weigh two of these coins against each other: if one is lighter, it is the unique coin; if they balance, the third coin is the light one.
- 1.c) Both sides are even. This means the one of the three coins that was removed from the heavier side is the heavy coin. For the third attempt, weigh two of these coins against each other: if one is heavier, it is the unique coin; if they balance, the third coin is the heavy one.
- 2. Both sides are even. If this is the case, all eight coins are identical and can be set aside. Take the four remaining coins and place three on one side of the balance. Place 3 of the 8 identical coins on the other side. There are three possibilities:
- 2.a) The three remaining coins are lighter. In this case you now know that one of those three coins is the odd one out and that it is lighter. Take two of those three coins and weigh them against each other. If the balance tips then the lighter coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is lighter.
- 2.b) The three remaining coins are heavier. In this case you now know that one of those three coins is the odd one out and that it is heavier. Take two of those three coins and weigh them against each other. If the balance tips then the heavier coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is heavier.
- 2.c) The three remaining coins balance. In this case you know that the unweighed coin is the odd one out. Weigh the remaining coin against one of the other 11 coins and this will tell you whether it is heavier or lighter.
With some outside the box thinking, such as assuming that there are authentic (genuine) coins at hand, a solution may be found quicker. In fact if there is one authentic coin for reference then the suspect coins can be thirteen. Number the coins from 1 to 13 and the authentic coin number 0 and perform these weighings in any order:
- 0, 1, 4, 5, 6 against 7, 10, 11, 12, 13
- 0, 2, 4, 10, 11 against 5, 8, 9, 12, 13
- 0, 3, 8, 10, 12 against 6, 7, 9, 11, 13
If the scales are only off balance once, then it must be one of the coins 1, 2, 3 which only appear in one weighing. If there is never balance then it must be one of the coins 10–13 that appear in all weighings. Picking out the one counterfeit coin corresponding to each of the 27 outcomes is always possible (13 coins one either too heavy or too light is 26 possibilities) except when all weighings are balanced, in which case there is no counterfeit coin (or its weight is correct). If coins 0 and 13 are deleted from these weighings they give one generic solution to the 12-coin problem.
If two coins are counterfeit this procedure will in general not pick out either of these but rather some authentic coin. For instance if both coins 1 and 2 are counterfeit then either coin 4 or 5 will be wrongly picked out.
In a relaxed variation of this puzzle, one only needs to find the counterfeit coin without necessarily being able to tell its weight relative to the others. In this case, clearly any solution that previously weighed every coin at some point can be adapted to handle one extra coin. This coin will never be put on the scales, but if all weighings are balanced it will be picked as the counterfeit coin. It is not possible to do any better, since any coin that is put on the scales at some point and picked as the counterfeit coin can then always be assigned weight relative to the others.
- Grossman, Howard (1945). Scripta Mathematica XI.
Niobe, the protagonist of Piers Anthony's novel With a Tangled Skein, must solve the twelve-coin variation of this puzzle to find her son in Hell: Satan has disguised the son to look identical to eleven other demons, and he is heavier or lighter depending on whether he is cursed to lie or able to speak truthfully. The solution in the book follows the given example 1.c.