In an old problem, you are given 12 coins that are identical except that one of them is slightly heavier or lighter than the other coins. You are also given a balance that can be used to compare the weights of two coins or two sets of coins. Using the balance just three times, can you determine which coin’s weight is different, and whether it is heavier or lighter than the other coins?

According to Wikipedia’s “Balance Puzzle” article, this puzzle has been around since 1945. There are different ways to solve it, but some of the methods are cumbersome to describe, involving flow charts and choosing weighings that depend on the results of previous weighings. When I first encountered this puzzle, I worked out a solution in which three pre-planned weighings would always yield the answer. I reasoned that there are 24 possible answers to the puzzle: 12 coins x 2 ways the odd coin may be different (i.e., heavier or lighter). One weighing can have three possible outcomes: the left side of the balance is heavier (call this result L), neither side is heavier (=), or the right side is heavier (R). A series of three weighings can therefore have 3x3x3 = 27 possible outcomes, such as LLR, L=L, RR=, etc. Since this is more than the number of possible answers, it seemed to me that I might be able to assign each of the 12 coins to a unique set of locations for the three weighings—locations being the left side of the balance, the right side of the balance, or off the balance.

(Others have come up with this solution as well. I don’t claim to have been the first, but I had not seen this solution when I first saw the puzzle.)

I made a list of all 27 outcomes and crossed off === (all three weighings balancing), since that might help identify the coin that is different but would not indicate whether the coin was heavier or lighter. My next insight was that the remaining 26 outcomes could be paired into 13 sets of opposite outcomes such as LLL and RRR, or RL= and LR=. Either one of a pair of opposite outcomes will identify the same coin as being different, but one will mean the coin is heavy and the other that the coin is light. For example, if we have one (and only one) coin that is on the left side of the balance in all three weighings, the result LLL means that it is the heavy coin while the result RRR means that it is the light coin.

Here is the full list of possible outcomes, paired up:

LLL RRR

LLR RRL

LRL RLR

LRR RLL

LL= RR=

LR= RL=

L=L R=R

L=R R=L

=LR =RL

=LL =RR

L== R==

=L= =R=

==L ==R

How many coins should take part in each weighing? If we draw two rectangles side by side to contain the coins that will be weighed in the first weighing, and two more rectangles under each of them to represent coins in the next two weighings, the answer soon becomes apparent. Number the coins 1 through 12. Choose a pair of opposite outcomes, such as LLR and RRL, for coin 1. For these outcomes to correspond to coin 1 being heavier or lighter, coin 1 must appear either (i) on the left balance in weighings one and two, and the right balance in weighing three, or (ii) on the right balance in weighings one and two, and the left balance in weighing three. Either way will work fine, and either way will mean writing the number 1 into three of the six rectangles. Similarly, if we assign the outcomes L== and R== to coin 10, we write 10 into the rectangle representing either the left side or the right side of the scale’s first weighing, and nowhere else. Looking at all 13 possible pairs of outcomes, and considering that we will only be using 12 of them, we can see that the either 24, 25, or 26 numbers will be entered into the rectangles, depending on which outcome pair we omit. Clearly 25 cannot be right, since at least one weighing would be between unequal numbers of coins, which would yield a worthless result. Could we have one weighing of five coins against five, and two other weighings of four against four, for a total of 26? No, because there are only nine coins that can have an outcome requiring them to be present in each particular weighing, and so it is not possible to have five coins on each side of the balance.

Therefore, we must fill each rectangle with four numbers to reach the total of 24, which means that the outcome pair we omit must be one of the first four (those with no equals signs among the outcomes). Suppose we eliminate the first outcome pair (LLL RRR) and assign the next 12 outcome pairs in order to coins 1 through 12.

Put coin 1 on the left, left, and right sides of the balance for the three weighings, in that order. (It could instead be put on the right, right, and left sides.) Put coin 2 on the left, right, and left sides. And so on. If one side of a balance is filled up, choose the member of the outcome pair that puts the coin on the other side of the balance.

Here’s is one way the coins might be assigned to the weighings to yield a unique answer no matter what the results of the three weighings:

1. LLR RRL

2: LRL RLR

3: LRR RLL

4: LL= RR=

5: LR= RL=

6: L=L R=R

7: L=R R=L

8: =LR =RL

9: =LL =RR

10: L== R==

11: =L= =R=

12: ==L ==R

First weighing

1,2,3,4 vs. 5,6,7,10

Second weighing

1,4,5,11 vs. 2,3,8,9

Third weighing

2,7,8,12 vs. 1,3,6,9

Each set of weighing results determines a unique answer. E.g., if the scale balances =LR (equal, left side heavier, right side heavier), in that order, then coin 8 is the odd one out, and is light.

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