Dice Generating Function

Solution to Non-Standard Dice Problem Using Generating Functions 6

The generating function for a single die is a polynomial P( x) = ∑ an x n where n =1

n

the term an x indicates that the integer n occurs an times. For a standard die, 6

a1 = a2 = a3 = a4 = a5 = a6 = 1 so P( x) = ∑ x n . The generating function that n =1

expresses the total of two dice is the product of the generating functions for each die. For standard dice, that generating function is P( x)2 = x 2 + 2 x3 + 3x 4 + 4 x5 + 5 x6 + 6 x7 + 5 x8 + 4 x9 + 3x10 + 2 x11 + x12 To find non-standard dice that satisfy the problem we must find two generating functions Q(x) and R(x) different from P(x) that satisfy the conditions: 1. The coefficients of Q(x) and R(x) must be non-negative integers. 2. The sum of the coefficients of each polynomial must be 6. 3. Q( x) R( x) = P( x)2 To solve this problem, factor P( x) 2 into prime factors as follows: P( x) = x( x + 1)( x 2 + x + 1)( x 2 − x + 1) P( x)2 = x 2 ( x + 1)2 ( x 2 + x + 1)2 ( x 2 − x + 1)2 Since the sum of the coefficients of a polynomial is the value of the polynomial at 1, condition 2 is equivalent to Q(1) = 6 and R(1) = 6 . Thus each polynomial must contain a factor x + 1 and a factor x 2 + x + 1 . Also, each polynomial cannot have a constant term, so it must have a factor x. There are only two possibilities left, depending upon how the x 2 − x + 1 factors are distributed. One yields the standard dice, the other is Q( x) = x( x + 1)( x 2 + x + 1) = x + 2 x 2 + 2 x3 + x 4 R( x) = x( x + 1)( x 2 + x + 1)( x 2 − x + 1)2 = x + x3 + x 4 + x5 + x6 + x8 (or the reverse) and yields the only non-standard dice. The die corresponding to Q(x) has numbers 1, 2, 2, 3, 3, 4. The die corresponding to R(x) has numbers 1, 3, 4, 5, 6, 8.