1 / 9

Polynomials and Fast Fourier Transform Chapter 30, pp.823-848 new edition

Polynomials and Fast Fourier Transform Chapter 30, pp.823-848 new edition. Polynomials. Polynomial in coefficient representation A(x ) = a 0 + a 1 x +a 2 x 2 +…+ a n-2 x n-2 + a n-1 x n-1 Operations over polynomials: polynomial degree n = highest nonzero coeff addition = O(n)

beatrice-west
Download Presentation

Polynomials and Fast Fourier Transform Chapter 30, pp.823-848 new edition

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Polynomials and Fast Fourier TransformChapter 30, pp.823-848 new edition CS8550

  2. Polynomials • Polynomial in coefficient representation A(x) = a0 + a1x+a2 x2+…+ an-2 xn-2+ an-1 xn-1 • Operations over polynomials: • polynomial degree n = highest nonzero coeff • addition = O(n) • multiplication = O(n2)!!! Bad -- too slow! • evaluation (finding the value in a point) = O(n) !!! Good -- Horner’s rule = stack-based A(x0 ) = a0 + x0 (a1 + x0 (a2 + …+ x0 (an-2 + x0 (an-1))…) • Point-value representation • (x0 ,y0 ), (x1 ,y1 ), …, (xn-1 ,yn-1 ) - n point-values are sufficient CS8550

  3. Point-value representation • Interpolation = getting polynomial coefficients from point-value representation • Theorem (unique interpolation): for any set of distinct n point-value pairs, ! polynomial of degree less than n • Operations over polynomials: • addition = O(n) • multiplication = O(n)!!! Good -- • how? • evaluation (finding the value in a point) = O(n2)!!! Bad -- too slow! • IDEA: Use both representations!!! CS8550

  4. Simple Transformations • Coefficient => point-value (evaluation) • just O(n) per point • O(n2)!!! Bad -- too slow! • Point-value => coefficient (interpolation) • Lagrange’s formula n-1  yk  (x-xj) / (xk-xj) k=0 jk jk • O(n2)!!! Bad -- too slow! • GOAL • both transformations in O(n log n) CS8550

  5. O(n log n) Multiplication • Double-degree bound • 2n point-value pairs • O(n) • Evaluate • Compute point point-value representations using FFT • in (2n)-roots of unity • O(n log n) • Point-wise multiply • Multiply the values for each of 2n points • O(n) • Interpolate • Compute coefficient representation of product using FFT • O(n log n) CS8550

  6. Complex Roots of Unity • Point-value representation in complex roots of 1 = DFT = discrete Fourier Transform • Complex n-th root of 1: wn = 1 • Complex numbers: i =  -1 • The principal n-th root is wn1= e2 i /n = cos(2/n)+ i sin (2/n) • n roots of n-th power : wn0 = 1, wn1 = principal , wn2 = wnwn, …, wnn-1 • Properties: • Cancellation: wdndk= wnk • Halving: if n is even then the squares of nth roots are n/2-roots • Summation: n-1  (wnk ) j= 0 j=0 CS8550

  7. Fast Fourier Transform • Problem: • Given polynomial A(x) = a0 + a1x+a2 x2+…+ an-2 xn-2+ an-1 xn-1 • Find values in roots of unity • FFT = divide and conquer: A[0](x)= a0 + a1x+a2 x2+…+ an-2 xn/2-1 A[1](x) = a1 + a3x+a5 x2+…+ an-1 xn/2-1 • A(x) = A[0](x2)+ xA[1](x2) • Recursive procedure: • Evaluate A[0](x) and A[1](x) in points (wn0 )2 , (wn1 )2 ,…, (wnn-1 )2 • n/2 roots each • Combine the results • T(n) = 2T(n/2) +(n) = (n log n) (master theorem) CS8550

  8. Inverse FFT Interpolation y = (y0 y1 y2...yn-1), a = (a0 a1 a2...an-1), x = (x0 x1 x2…xn-1) Vn = Vn (x) Vandermonde matrix y = a Vn (x) Replace x = (x0 x1 x2…xn-1) => wn = (wn0w n1 w n2 …wnn-1) DFT = y = a Vn (wn) a = y Vn-1 (wn) CS8550

  9. Efficient FFT CS8550

More Related