Question

Why does evaluating polynomials with n points using the Fast Fourier Transform take O(n log n) time? I am specifically talking about implementing a divide and conquer algorithm that divides the polynomial A(x) into its even powers and odd powers and then uses recursion.

Answer 1

Let T(n) be time used by the FFT algorithm to evaluate a polynomial of degree n at n points.

The algorithm splits

A(x)=xB(x^2)+C(x^2),

i.e. into two polynomials: odd and even coefficients. For example: 3x^3 + 2x^2 + 9x + 7 is split into x(3x^2 + 9) + (2x^2 + 7).

Originally you wanted to compute 3x^3 + 2x^2 + 9x + 7 at points a,b,c,d.

Now you want to compute 3x+9 and 2x+7 at points a², b², c², d². Later you will combine that to get values of 3x^3 + 2x^2 + 2x + 7 at a,b,c,d.

The crucial idea: since you use roots of unity, half of the values in a², b², c², d² are the same. Suppose that a²=c² and b²=d².

So you need to compute 3x+2 and 2x+7 at points a², b².

This means you reduced an instance of size N into two instances of size N/2 and O(N) postprocessing.

FFT repeats this process recursively. This is the same recursion equation as for mergesort, which is O(N log N) complexity.