Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Fall 2004 MATH025.1001
Twelve coins and a balance
We are given 12 coins and a balance.
One of the coins is counterfeit; it has a weight different from the other
coins.
Find a way to use the balance three times to detect which coin is counterfeit,
as well as whether it is heavier or lighter than the valid coins.
We solve the problem by first solving some simpler problems.
- Suppose we have 1 coin instead of 12.
Then we know that this only coin must be the counterfeit one.
It is impossible to determine heavier or lighter, because we have no good coin
to compare with.
- Suppose we have 2 coins instead of 12.
In one weighing we know that one coin is heavier than the other.
For example, coin 2 is heavier than coin 1.
So IF coin 1 is counterfeit, then it must be lighter; and IF coin 2 is
counterfeit, then it must be heavier.
Otherwise we never know which one is counterfeit.
- Suppose we have 3 coins instead of 12.
We found that the following works with only two weighings.
At the first weighing, say W1, we put coin 1 on the left, say L, coin 2
on the right, say R, and we don't use coin 3, say n.
At the second weighing, say W2, we put coin 1 on the left, say L, coin 3
on the right, say R, and we don't use coin 2, say n.
We can put this in a table as follows:
- Suppose we have 4 coins instead of 12.
Then we could have shown that we can not solve the problem in two weighings.
So the previous case for 3 coins is somehow optimal when we permit only two
weighings.
- We construct the solution for 12 coins and three weighings, from the
case for 3 coins and two weighings as follows.
Note that all rows contain at least one L or R; no two rows are equal; and no
two rows are duals, that is, convert into one another when we write Rs for Ls
and Ls for Rs.
Also note that each column contains an equal number of Ls and Rs.
- Recall that we can not solve the 4 coin problem in two weighings.
But we can when we add a fifth `good' coin.
If we have 4 coins, one of which is a counterfeit with a different weight,
and we have a fifth coin, a valid coin say G, then we can find in two weighings
which of the coins in counterfeit, as well as whether it is lighter or heavier:
Last updated: September 2004
Comments & suggestions:
wimr@mscs.mu.edu