Given all roots, how do I find the coefficients of a polynomial in time faster than O(n^2)?

Question

Given all the roots of a polynomial, I have to figure out an algorithm that generates the coefficients faster than O(n^2). I\'m having trouble approaching this problem. I\'m

Answer 1

kraskevich basically nailed it. Some details are missing, and it would be too long to fit into the comment field. Here are the details.

Basically, you want to set up as a polynomial multiplication problem. Your input would be p1,...pN where pj(x) = (x-rj).

Here's pseudo-code:

function multiply2Poly(p1, p2)

   // here, use FFT, multiply and use IFFT back


function multiplyPoly(p[1],...p[N])
    if (N==1) return p[1]
    if (N==2) return multiply2Poly(p[1],p[2])
    else {
         return multiply2Poly(multiplyPoly(p[1],...p[N/2]),multiplyPoly(p[1+N/2],...,p[N])
    }


function getCoef(r[1],...r[N])

    return multiplyPoly((p[1]=x-r[1]),...(p[N]=x-r[N]));

And for the FFT part:

Observe that if two polynomials are:

p1 = a[0]+a[1] x + ...+a[n] x^n
p2 = b[0]+a[1] x + ...+a[n] x^n

Then p1 * p2 = c[0] + c[1] x + ...+c[n] x^n

where C = A [x] B where [x] = convolution. A = (a[0],...,a[n]), B = (b[0],...,b[n]) and C = (c[0],...,c[n]).

Then use FFT and the convolution theorem to speed this up.

C = A [x] B = IFFT{ FFT{A} * FFT{B} } where * here is just multiplication.

runtime of IFFT = runtime of FFT = O(n log n).
multiplication run time is n, so the total run time is O(n log n).

The total run time of multiplyPoly is then:

T(N) = R(N/2) + T(N/2)*2

And R(N/2) = O(n log n) is the run time of multiply2Poly as described above.

so

T(N) = T(N/2) * 2 + O(n log n)

and now the Master theorem gives the O(n (log n)^2)