Balance puzzle

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For the WPC puzzle with numbers, see Balance (puzzle).

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. These differ from puzzles where items are assigned weights, in that only the relative mass of these items is relevant.

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[edit] Premise

A well-known example has nine (or fewer) items, say coins, that are identical in weight save for one, which in this example we will say is lighter than the others—a counterfeit. The difference is only perceptible by using a balance, but only the coins themselves can be weighed, and it can only be used twice in total.

Is it possible to isolate the counterfeit coin with only two weighings?

[edit] Solution

The solution is based on dividing into groups of three. Two groups are balanced against each other; if one is lighter it contains the counterfeit coin; if they balance perfectly, all the coins weighed are authentic, and the counterfeit is in the third group.

Therefore, for the first weighing, three coins are put on each side of the balance, and the counterfeit coin is either one of the three on the lighter side, or one of the three that hasn't been weighed if the scale balances. Put one of these three coins on either side of the balance; if one is lighter it is the counterfeit, and if they are equal the counterfeit is the one not on the balance.

[edit] The Twelve-Coin Problem

A more complex version exists where there are twelve coins, eleven of which are identical and one of which is different, but it is not known whether it is heavier or lighter than the others. This time the balance may be used three times to isolate the unique coin and determine its weight relative to the others.

[edit] Solution

The procedure is less straightforward for this problem, and the second and third weighings depend on what has happened previously.

  • 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.

[edit] In literature

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.

[edit] External links