The Fast Fourier Transform: A Brief Overview. with Applications. Petros Kondylis. Petros Kondylis. December 4, 2014

Transcription

1 December 4, 2014

2 Timeline Researcher Date Length of Sequence Application CF Gauss 1805 Any Composite Integer Interpolation of orbits of celestial bodies F Carlini Harmonic Analysis of Barometric Pressure A Smith , 8, 16, 32 Correcting deviations in compasses on ships J D Everett Modelling underground temperature variations C Runge n K Harmonic Analysis of Functions K Stumpff n K & 3 n Harmonic Analysis of Functions Danielson & Lanczos n X-ray diffraction in crystals LH Thomas 1948 Integer /w relatively prime factors Harmonic Analysis of Functions I J Good 1958 Integer /w relatively prime factors Harmonic Analysis of Functions Cooley & Tukey 1965 Composite Integer Harmonic Analysis of Functions S Winograd 1976 Integer /w relatively prime factors Complexity theory for Harmonic Analysis

3 Discrete Transform De nition (Discrete Transform) Matrix-vector-multiplication y = Mx y a y 1 1 ω n ωn 2 ω n 1 0 n y 2 = 1 ωn 2 ωn 4 ω 2(n 1) a 1 n a ωn kj 2 y n 1 1 ωn n 1 ωn 2(n 1) ω n (n 1)(n 1) a n 1 where x = (a 0,, a n 1 ), y = (y 0,, y n 1 ) C n, ω n = e 2πi/n (primitive nth root of unity) and The direct evaluation requires O(n 2 ) complex multiplications and additions

4 To Divide is to Conquer Map the original problem in such a way that the following inequality is satis ed: cost(subproblems) + cost(mapping) < cost(originalproblem) The division can be applied recursively to the subproblems leading to a reduction of the order of complexity Question: But How? Answer: Consider Subsets of the initial sequence Take the DFT of these subsequences Reconstruct the DFT of the Initial subsequence

5 To Divide The original problem: n 1 y k = a j (ωn) k j, k = 0,, n 1 j=0 Let P l, l = 0,, r 1 be the partition of {0, 1,, n 1} de ning r different subsets of the input sequence The partition gives: r 1 y k = a j (ωn) k j, k = 0,, n 1 j P j l=0 r 1 y k = ω j l n a j (ωn) k j j l j P j l=0 The chosen partition depends on the nature of the problem (input sequence)

6 To Divide Example: For input lengths equal to powers of 2 ie n = 2 s, s N we can have a partition P l, l = 0, 1 where P 0 = {0 : 2 : n 1}, P 1 = {1 : 2 : n 1} y k = y k = y k = n 1 j=0:2 n 1 j=0:2 n/2 1 j=0 a j (ω k n) j + ω k n n 1 j=1:2 a j (ωn 2k ) j/2 + ωn k a 2j (ω k n/2 )j ± ω k n k = 0,, n/2 1 a j (ω k n) j 1 n 1 j=0:2 n/2 1 j=0 a j+1 (ω 2k n ) j/2 a 2j+1 (ω k n/2 )j

7 To Divide (again)

8 To Conquer Algorithm 1 Recursive-FFT(a) 1: n = alenght 2: if n == 1 then 3: return a 4: ω n = e 2πi/n 5: ω = 1 6: a [0] = (a 0, a 2,, a n 2 ) 7: a [1] = (a 1, a 3,, a n 1 ) 8: y [0] = Recursive FFT(a [0] ) 9: y [1] = Recursive FFT(a [1] ) 10: for k = 0 to n/2 1 do 11: t = ω y [1] k 12: y k = y [0] k + t 13: y k+n/2 = y [0] k t 14: ω = ωω n 15: return y Complexity: T(n) = 2T(n/2) + Θ(n) = Θ(nlgn)

9 The Butter y Operation r k = y [0] k + ω k y [1] k r k+n/2 = y [0] k ω k y [1] k

10 The FFT circuit: Example The input is arranged in bit reversed order Unique path between each input a k and each output A(ω j ) On the path between a k and A(ω j ) the labels add up to jk mod 8

11 Multiplying Polynomial The product of two degree-d polynomials is a polynomial of degree 2d: A(x) = a 0 + a 1 x + + a d x d and B(x) = b 0 + b 1 x + + b d x d C(x) = A(x)B(x) = c 0 + c 1 x + + c 2d x 2d c k = a 0 b k + a 1 b k a k b 0 = Complexity: Θ(d 2 ) k a i b k 1 i=0

12 Representing Polynomials Two ways of representing polynomials Coefficient form a 0, a 1,, a d Values A(x 0 ), A(x 1 ), A(x d ) at d + 1 distinct points Multiplying polynomials in the value representation takes linear time C(x k ) = A(x k )B(x k ), k = 0,, 2d

13 Are we there yet? Evaluation takes Θ(n 2 ) time Multiplication takes Θ(n) time Interpolation takes Θ(n 2 ) time This joke isn't funny (or is it)!

14 Evaluation as in Linear Transformation as in Fast Transformation A(x 0 ) A(x 1 ) A(x 2 ) A(x n 1 ) 1 x 0 x 2 0 x n x 1 x 2 1 x n 1 1 = 1 x 2 x 2 2 x n x n 1 x 2 n 1 x n 1 n 1 a 0 a 1 a 2 a n 1 Is it OK if I x k = ω k n? By all means! A(ω 0 n) A(ω 1 n) A(ω 2 n) A(ω n 1 n ) ω n ωn 2 ωn n 1 = 1 ωn 2 ωn 4 ωn 2(n 1) 1 ωn n 1 ωn 2(n 1) ω n (n 1)(n 1) a 0 a 1 a 2 a n 1 But this I can do in Θ(nlgn) time (you don't say!)

15 Some kind of Magic (what???) Lemma The columns of matrix M are orthogonal to each other ω n ωn 2 ωn n 1 M = 1 ωn 2 ωn 4 ωn 2(n 1) 1 ωn n 1 ωn 2(n 1) ω n (n 1)(n 1) Proof Take the inner product of any columns j and k of matrix M 1 + ω k j + ω 2(k j)) + + ω (n 1)(k j) = (1 ω n(k j )/(1 ω k j ) For j k it evaluates to 0 For j = k it evaluates to n

16 Can we go back now please The orthogonality property can be summarized in the single equation MM = ni M = nm 1 M = M(ω 1 ) Inversion Formula: M(ω n ) 1 = 1 n M(ω 1 ) M is the conjugate transpose of M But ω 1 is also an nth root of unity and therefore multiplication by M(ω 1 ), ie interpolation, is itself just an FFT operation, but ω replaced by ω 1

17 Polynomial Multiplication in a Nutshell