## Monday, 13 June 2011

### Mathematical poetry 2.

The following poem is about a certain Professor Felix Fiddlesticks:

F set the coins out in a row
And chalked on each a letter, so,
To form the words "F AM NOT LICKED"
(An idea in his brain had clicked).
And now his mother he'll enjoin:
MA DO LIKE
ME TO FIND
FAKE COIN
-- Cedric A.B. Smith

Now this may appear to not make much sense, but it is in fact a really clever solutions to the 12 coin problem:

Imagine you are given 12 coins, and a set of weighing scales. 11 of the coins have the same weight, but one has a different weight. To make matters worse, you do not know if it is heavier of lighter than the others. The problem is to find out which coin is different, and whether it is lighter or heavier, using at most three weighings on a pair of scales.

Before you see the answer and I explain the connection between the puzzle and the poem, have a go at solving the problem yourself. Here is a flash applet which allows you to play that game yourself:
________________________________________________________________
________________________________________

Did you solve it? Did the poem help? If you managed to do it, give yourself a pat on the back. Otherwise, let me explain. Firstly, do as the poem says; on the coins write one of the single letters F, A, M, N, O, T, L, I, C, K, E and D. Now, we weigh the coins
MADO against LIKE
METO against FIND
FAKE against COIN

in each of the weighings we have two possibilities. Either, the pans balance or they don't. If any of the weighings do balance, we immediately know that all 8 coins are genuine and the dodgy coin is in the four we haven't weighed. Conversely, if the pans do not balance then we know that the four coins not on the scales are genuine.

As a specific example, suppose that, in our weighings, the right pan is always lower. We can then logically deduce that the coin "I" is the dodgy one and that it is heavier, since coin "I" is the only coin that appears in the right balance in all three weighings. If you are interested in the entire solution to this problem look at the last post in this Dr. Math forum post.