Marquette University

Suppose we have a balance, and a pile of identical looking coins. Unfortunately, one of the coins is fake. The fake coin is identical but for its weight.

• We are given 3 identical looking coins, and a balance. Exactly one coin is fake, and is lighter than the good coins.
  • Question: Use one weighing with the balance to find the lighter coin.
  • Answer: Number the coins by 1, 2, and 3. Put coin 1 on the left scale, and put coin 2 on the right scale. Engage the balance.
    • If the left scale goes up, then coin 1 is fake.
    • If the right scale goes up, then coin 2 is fake.
    • If the scales remain level, then coin 3 is fake.
  Note that a weighing gives at most 3 possible different answers. So 1 weighing can not always work to find a fake coin from among a pile of 4.
• We are given 9 identical looking coins, and a balance. Exactly one coin is fake, and is lighter than the good coins.
  • Question: Use two weighings with the balance to find the lighter coin.
  • Answer: Number the coins from 1 to 9. Put coins 1 through 3 on the left scale, and put coins 5 through 6 on the right scale. Engage the balance.
    • If the left scale goes up, then coin 1, 2, or 3 is fake. Apply one more weighing as above to find which of these three.
    • If the right scale goes up, then coin 4, 5, or 6 is fake. Apply one more weighing as above to find which of these three.
    • If the scales remain level, then coin 7, 8, or 9 is fake. Apply one more weighing as above to find which of these three.
    We can fix the two weighings in advance, as in the following table. So the solution below is better, because we decide on how to do the second weighing, before we know the outcome of the first. In the table below, the first weighing we call W1. The second weighing we call W2. We write L in a column to indicate that the corresponding coin is put on the left scale of the balance. We write R in a column to indicate that the corresponding coin is put on the right scale of the balance. We write n in a column to indicate that the corresponding coin is not used in the weighing.

    coin	W1 W2
    1 2 3 4 5 6 7 8 9	L L L R L n R L R R R n n L n R n n

    The 9 rows present all possible 9 ways to write down a pair of elements from L, R, and n. That is the maximum we can expect from two weighings. With 10 coins, two weighings can not guarantee that we find the fake, lighter, one.
• Suppose we have 27 identical looking coins, and a balance, such that exactly one coin is fake, and is lighter than the good coins. Can you see how to find the fake coin in 3 weighings? Just expand the table above.
• Mathematics. Yes. Suppose we have 3^n identical looking coins, and a balance, such that exactly one coin is fake, and is lighter than the good coins. Can you see how to find the fake coin in n weighings?

• Suppose we have 1 coin instead of 12, and as many weighings with the balance as you wish. Then we know that this only coin must be the counterfeit one, without doing any weighings at all. 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, and as many weighings with the balance as you wish. 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. But we don't know which of these two cases is the truth.
• Suppose we have 3 coins instead of 12, and only two weighings. Here is a solution. 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:

  coin	W1 W2
  1 2 3	L L R n n R

  In class we checked that this works. The table is such that (1) each column has an equal number of Ls and Rs; (2) no row consists of all n's; (3) no two rows are identical; and (4) no two rows are duals of one another, that is, they still differ when in one of them we replace all Ls by Rs, and all Rs by Ls.
  This puzzle lacks a certain perfection: (1) On one hand, the problem world has 6 possibilities, with one of 3 coins maybe being in 2 situations (lighter or heavier). (2) On the other, the two weighings could potentially distinguish 9 possibilities, by 3 times 3 possible outcomes of the two weighings.
• Let us try to make the puzzle above slightly less imperfect. Suppose we have 4 coins instead of 12. Now the problem world has 8 possibilities, with one of 4 coins maybe being in 2 situations (lighter or heavier). In class we showed 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.
• When we revise the puzzle with 4 coins a little, we are able to make it slightly less imprefect: Solution: Add a fifth `good' coin to the story of 4 coins above. So we have 4 coins, one of which is fake with a different weight, and we have a fifth coin, a good coin say G. So still 8 possible worlds and 9 possible distinctions. Now we can find in two weighings which of the coins in counterfeit, as well as whether it is lighter or heavier:

  coin	W1 W2
  1 2 3 4 G	L L R n n R L R R L

  So we do better, although there still is a gap of 1 between the 8 possible worlds of 4 coins, one of them lighter or heavier, versus the 9 potential distinctions that two weighings can make. What if we rephrase the problem by saying that there is at most one fake coin among the 4 coins? Hmmmmmm.
• We construct the solution for 12 coins and three weighings, from the case for 3 coins and two weighings as follows.

  coin	W1 W2	W3
  1 2 3	L L R n n R	L L L
  4 5 6	L L R n n R	R R R
  7 8 9	L L R n n R	n n n
  10 11 12	L R R L n n	n L R

  Note that this table satisfies the same conditions (1), (2), (3), and (4) as the previous, smaller, table with three coins.
• Recall that we can not solve the 4 coin problem in two weighings. We can not solve the problem with 13 coins and three weighings either.
  Just as with 4 coins plus a good coin G, there is a solution with 13 coins plus a good coin G. Find it.
  What if there are 13 coins of which at most one coin is fake? Hmmmmmm.