10/16/01 - 10/18/01 CS8550

Polynomials and Fast Fourier Transform

Chapter 30, pp.823-848 new edition

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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

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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!!!

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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 jk jk– O(n2)!!! Bad -- too slow!

• GOAL – both transformations in O(n log n)

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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)

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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 : wn

0 = 1, wn1 = principal , wn

2 = wnwn, …, wnn-1

• Properties:– Cancellation: wdn

dk = wnk

– Halving: if n is even then the squares of nth roots are n/2-roots– Summation: n-1 (wn

k ) j = 0 j=0

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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 (wn

0 )2 , (wn1 )2 ,…, (wn

n-1 )2

• n/2 roots each

– Combine the results

• T(n) = 2T(n/2) +(n) = (n log n) (master theorem)

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Inverse FFT Interpolationyy = (y0 y1 y2...yn-1), aa = (a0 a1 a2...an-1), xx = (x0 x1 x2…xn-1)

VVn = VVn (xx) Vandermonde matrix

y y = aa V Vn (xx)

Replace xx = (x0 x1 x2…xn-1) => wwn = (wn0

w n1 w n

2 …wnn-1)

DFT = y = DFT = y = aa V Vn (wwn)

a = a = y Vy Vn-1 (wwn)

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Efficient FFT