FFT: Fast Polynomial Multiplications

Transcription

1 FFT: Fast Polynomial Multiplications Jie Wang University of Massachusetts Lowell Department of Computer Science J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 1 / 20

2 Overview So far we have learned five basic algorithm design techniques: (1) sorting and searching; (2) divide-and-conquer; (3) greedy selection; (4) dynamic programming, and (5) linear programming. Another common technique is to use math tricks (actually DP and LP are in this category). Fast Fourier transform for computing polynomial multiplications is a typical math trick. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 2 / 20

3 Polynomials General form: n 1 A(x) = a 0 x 0 + a 1 x + a n 1 x n 1 = a j x j. A(x) has degree k if its highest nonzero coefficient is a k, denoted by degree(a) = k. The degree of a polynomial of degree-bound n may be any integer between 0 and n 1. Unless otherwise stated, all polynomials we will consider have degree-bound n. j=0 J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 3 / 20

4 Evaluation and Summation Point evaluation (Horner s rule): A(x 0 ) = a 0 + x 0 (a 1 + x 0 (a x 0 (a n 2 + x 0 (a n 1 )) )). Runtime of point evaluation: Θ(n). Summation: n 1 A(x) + B(x) = (a j + b j )x j. Runtime of summation: Θ(n). j=0 J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 4 / 20

5 Multiplication and Convolution Multiplication: C(x) = A(x)B(x), where C(x) has degree-bound 2n 1, C(x) = c j = 2n 2 j=0 c j x j j a k b j k. k=0 c = (c 1, c 2,..., c 2n 2 ) is called the convolution of a = (a 0, a 1,..., a n 1 ) and b = (b 1, b 2,..., b n 1 ), denoted by c = a b. Runtime: Θ(n 2 ). Note: Since C(x) belongs to polynomials of degree-bound 2n, for computational convenience, we will treat C(x) as a polynomial of degree bound 2n (rather than 2n 1). J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 5 / 20

6 Representations Coefficient representation: A polynomial of degree-bound n can be represented by (a 0, a 1,..., a n 1 ). Point-value representation: {(x 0, y 0 ),..., (x n 1, y n 1 )}, where y i = A(x i ) and x i x j if i j. Given a polynomial, computing a point-value representation is straightforward and the runtime is Θ(n 2 ). J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 6 / 20

7 Interpolation Interpolation: The process from point-value representation to coefficient representation. Lagrange s formula: For any set {(x 0, y 0 ),..., (x n 1, y n 1 )} of n point-value pairs with pairwise different x k, there is a unique polynomial A(x) of degree-bound n such that y k = A(x k ) for k = 0,..., n 1, where n 1 j k A(x) = y (x x j) k j k (x k x j ). k=0 This indicates the uniqueness of interpolating polynomials. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 7 / 20

8 Point-value Representations and Operations Suppose Addition: A :{(x 0, y 0 ),..., (x n 1, y n 1 )} B :{(x 0, y 0),..., (x n 1, y n 1)}. C(x) = A(x) + B(x) : {(x 0, y 0 + y 0),..., (x n 1, y n 1 + y n 1)}. Multiplication needs extended point-value representations on 2n points: Then A :{(x 0, y 0 ),..., (x 2n 1, y 2n 1 )} B :{(x 0, y 0),..., (x 2n 1, y 2n 1)}. C(x) = A(x)B(x) : {(x 0, y 0 y 0),..., (x 2n 1, y 2n 1 y 2n 1)}. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 8 / 20

9 Multiplications J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 9 / 20

10 Discrete and Fast Fourier Transforms (DFT, FFT) Complex root of unity: Let ωn, 0 ωn, 1..., ωn n 1 ω n = 1, where be complex roots of Recall that e iu = cos(u) + i sin(u). ω k n = e 2πik/n, k = 0, 1,..., n 1. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 10 / 20

11 Lemmas Lemma 30.3 (Cancellation lemma) ω dk dn = ωk n for any integers n 0, k 0, and d > 0. Proof. ω dk dn = (e2πi/dn ) dk = (e 2πi/n ) k = ω k n. Corollary ω n/2 n = ω 2 = 1. Lemma 30.5 (Halving lemma) If n > 0 is even, then (ω k+n/2 n ) 2 = (ω k n) 2. Proof. By Corollary 30.4 that ωn n/2 = ωn, k and so (ω k+n/2 ω k+n/2 n = ω 2 = 1, we have n ) 2 = (ωn) k 2. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 11 / 20

12 Summation Lemma Summation lemma n 1 (ωn) k j = 0 j=0 for any integer n > 0 and any integer k 0 with k not divisible by n. Proof. n 1 (ωn) k j = (ωk n) n 1 ωn k 1 j=0 = (ωn n) k 1 ω k n 1 = (1)n 1 ω k n 1 = 0. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 12 / 20

13 The DFT Let n 1 A(x) = a j x j, j=0 n 1 y k = A(ωn) k = a j ωn kj. y = (y 0, y 1,..., y n 1 ) is the DFT of a = (a 0, a 1,..., a n 1 ), denoted by y = DFT n (a). j=0 J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 13 / 20

14 The FFT Assume that n is a power of 2. Divide-and conquer. Rewrite A(x) as A(x) = A [0] (x 2 ) + xa [1] (x 2 ), where A [0] (x) = a 0 + a 2 x + a 4 x a n 2 x n/2 1, A [1] (x) = a 1 + a 3 x + a 5 x a n 1 x n/2 1. Only need to evaluate two half-size polynomials A [0] (x) and A [1] (x) at (ω 0 n) 2, (ω 1 n) 2,..., (ω n 1 n ) 2, and combine the results. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 14 / 20

15 An Important Observation By the halving lemma, there are only n/2 distinct values in (ωn) 0 2, (ωn) 1 2,..., (ωn n 1 ) 2, where (ω 0 n) 2 = (ω n/2 n ) 2, (ω 1 n) 2 = (ω 1+n/2 n ) 2,..., (ωn n/2 1 ) 2 = (ωn n 1 ) 2. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 15 / 20

16 Recursive FFT Runtime: T (n) = 2T (n/2) + Θ(n) = Θ(n log n). J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 16 / 20

17 Line 11 For y 0, y 1,..., y n/2 1, line 11 yields y k = y [0] k + ω k n y [1] k = A [0] (ω k n/2 ) + ωk n A [1] (ω k n/2 ) = A [0] (ω 2k n ) + ω k n A [1] (ω 2k n ) = A(ω k n). by the cancellation lemma J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 17 / 20

18 Line 12 For y n/2, y n/2+1,..., y n 1, line 12 yields y k+n/2 = y [0] k ωn k y [1] k = y [0] k + ω k+n/2 n y [1] k = A [0] (ωn 2k ) + ωn k+n/2 A [1] (ωn 2k ) since ω k+n/2 n = ω k n by the cancellation lemma = A [0] (ωn 2k+n ) + ωn k+n/2 A [1] (ωn 2k+n ) since ωn 2k+n = ωn 2k = A(ωn k+n/2 ). J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 18 / 20

19 Interpolation Rewrite y = DFT n (a) as The (j, k) entry of the inverse of the Vandermonde matrix V n is ωn kj /n. Proof. n 1 [Vn 1 V n ] jj = a = DFT 1 n (y). k=0 (ω kj/n n n 1 (ωn kj ) = k=0 ω k(j j) n /n = { 1, if j = j 0, otherwise.. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 19 / 20

20 Coefficients For j = 0, 1,..., n 1, a j = 1 n n 1 y k ωn kj. k=0 Modify recursive FFT to replace a with y and ω n with ωn 1, we can computer DFT 1 n in Θ(n log n) time. This means that we can transform the coefficient representation to the point-value representation in O(n log n) time, and so is the other direction. Theorem 30.8 (Convolution theorem) For any two vectors a and b of length n, where n is a power of 2, a b = DFT 1 2n (DFT 2n(a) DFT 2n (b)), where a and b are padded with 0s to length 2n and denotes the componentwise product of two 2n-element vectors. J. Wang (UMass Lowell) FFT: Fast Polynomial Multiplications 20 / 20