# Solution to puzzle 41: Crazy dice

We will use a generating function to represent each die.

Standard dice can be represented by the generating function f(x) = x1 + x2 + x3 + x4 + x5 + x6.
Here, the exponent represents the score on the die face; the coefficient is the number of ways each score can be obtained.

The sums that can be obtained by throwing two standard dice correspond to multiplying their generating functions:
(x1 + x2 + x3 + x4 + x5 + x6)2 = x2 + 2x3 + 3x4 + 4x5 + 5x6 + 6x7 + 5x8 + 4x9 + 3x10 + 2x11 + x12.
Here again, each coefficient represents the number of ways each score (exponent) may be obtained.

Any alternative dice (that have the same number of ways of obtaining each sum as the standard dice) must have generating functions whose product is the same as for the standard dice, above.  So the problem is reduced to finding an alternative factorization, into two factors, of (f(x))2; a more amenable task.

 f(x) = x(1 + x + x2 + x3 + x4 + x5) = x(x6 − 1)/(x − 1) = x(x3 + 1)(x3 − 1)/(x − 1) = x(x + 1)(x2 − x + 1)(x2 + x + 1)

Therefore (f(x))2 = x2(x + 1)2(x2 − x + 1)2(x2 + x + 1)2

An alternative factorization must redistribute these terms into two factors, g(x) and h(x), which will be the generating functions for the alternative dice.

We have two further constraints:

• Each face has at least one dot.  This corresponds to g(x) and h(x) each having a minimum exponent of 1.  Therefore g(x) and h(x) must each get one copy of the factor x.
• Each die has six faces.  This corresponds to a requirement that g(1) = h(1) = 6.  Looking at the factorization of f(x), (x + 1) yields 2, (x2 + x + 1) yields 3, while the remaining two factors yield 1.  Therefore g(x) and h(x) must each get one copy of (x + 1) and (x2 + x + 1).

This leaves only the two (x2 − x + 1) factors to distribute among g(x) and h(x).
Clearly, if we give one copy to each function, we get g(x) = h(x) = f(x), yielding the standard dice.
The only alternative is to give both factors to one of the functions:

 g(x) = x(x + 1)(x2 + x + 1) = x + 2x2 + 2x3 + x4 h(x) = x(x + 1)(x2 + x + 1)(x2 − x + 1)2 = x + x3 + x4 + x5 + x6 + x8

This yields unique alternative dice of {1,2,2,3,3,4} and {1,3,4,5,6,8}.