Problem 06

Problem 06

If two classical dices are thrown, the sum has generating function

F(x) = x2 + 2x3 + 3x4 + 4x5 + 5x6 + 6x7 + 5x8 + 4x9 + 3x10 + 2x11 + x12 = x2 (1 + x)2 (1 + x + x2)2 (1 - x + x2)2.

One can write other positive integers on the 12 faces to obtain the same sum distribution. Find the (unique) other way.

Solution

We are looking for a1, ..., a6, b1, ..., b6, such that F(x) = (xa1 + xa2 + ... + xa6)(xb1 + xb2 + ... + xb6) = G(x)H(x).

Since we need positive integers, x must divide both factors. Since G(1) = H(1) = 6, the factors 1 + x and 1 + x + x2 must also appear exactly once in both G(x) and H(x). Hence, the only choice is whether to put the two factors 1 - x + x2 together or not.

The first option yields G(x) = H(x) = x (1 + x) (1 + x + x2) (1 - x + x2) = x1 + x2 + x3 + x4 + x5 + x6, which yields the classical dice. The second option yields the unique other solution; one dice with {1, 3, 4, 5, 6, 8} and one dice with {1, 2, 2, 3, 3, 4}:

G(x) = x (1 + x) (1 + x + x2) (1 - x + x2)2 = x1 + x3 + x4 + x5 + x6 + x8,
H(x) = x (1 + x) (1 + x + x2) = x1 + x2 + x2 + x3 + x3 + x4,

Remarks

Students often wonder why they should study generating functions. Artificial counting problems don't convince them. And I can't blame them. Alas, most routine exercises on this subject are just counting problems. This is a different kind of problem, with for most an unexpected result. I googled this and it turns out to be known as the Sicherman dice.