fast fourier transform

Princeton University • COS 423 • Theory of Algorithms • Spring 2002 • Kevin Wayne

Fast Fourier Transform

Jean Baptiste Joseph Fourier (1768-1830)

These lecture slides are adapted from CLRS.

2

Fast Fourier Transform

Applications. Perhaps single algorithmic discovery that has had the greatest

practical impact in history. Optics, acoustics, quantum physics, telecommunications, systems

theory, signal processing, speech recognition, data compression. Progress in these areas limited by lack of fast algorithms.

History. Cooley-Tukey (1965) revolutionized all of these areas. Danielson-Lanczos (1942) efficient algorithm. Runge-König (1924) laid theoretical groundwork. Gauss (1805, 1866) describes similar algorithm. Importance not realized until advent of digital computers.

3

Polynomials: Coefficient Representation

Degree n polynomial.

Addition: O(n) ops.

Evaluation: O(n) using Horner's method.

Multiplication (convolution): O(n2).

11

2210)(

nn xaxaxaaxp

11

2210)(

nn xbxbxbbxq

1111100 )()()()()(

nnn xbaxbabaxqxp

))))(((()( 12210 nn axaxaxaxaxp

2211

0011000 )()()()()()(

nnn

j

k

jkjk xbaxbaxbababaxqxp

4

Polynomials: Point-Value Representation

Degree n polynomial. Uniquely specified by knowing p(x) at n different values of x.

)( where ,),(x,),,(x),,(x 11-n1100 kkn xpyyyy

x

y = p(x)

xj

yj

5

Polynomials: Point-Value Representation

Degree n polynomial.

Addition: O(n).

Multiplication: O(n), but need 2n points.

Evaluation: O(n2) using Lagrange's formula.

)( where ,),(x,),,(x),,(x 11-n1100 kkn xpyyyy

),(x,),,(x),,(x 111-n111000 nn zyzyzy

1

0 )(

)(

)(n

kkj

jk

kjj

k xx

xx

yxp

),(x,,),(x),,(x 12121-2n111000 nn zyzyzy

)(z where ,),(x,),,(x),,(x 11-n1100 kkn xqzzz

6

Best of Both Worlds

Can we get "fast" multiplication and evaluation?

Yes! Convert back and forth between two representations.

coefficient

Representation

O(n2)

Multiplication

O(n)

Evaluation

point-value O(n) O(n2)

FFT O(n log n) O(n log n)

1-n10

1-n10

,,b,b

a,,a,a

b

2-2n10 ,,,c cc

)(,),(),q(x

)(,),(),p(x

1-2n10

1-2n10

xqxq

xpxp

)(,),(),r(x 1-2n10 xrxr O(n)

point-value multiplication

O(n2)

coefficient multiplication

O(n log n)evaluationFFT

interpolationinverse FFT

O(n log n)

7

Converting Between Representations: Naïve Solution

Evaluation (coefficient to point-value). Given a polynomial p(x) =a0 + a2x

1 + . . . + an-1 xn-1, choose n distinct

points {x0, x1, . . . , xn-1 } and compute yk = p(xk), for each k using

Horner's method. O(n2).

Interpolation (point-value to coefficient). Given n distinct points {x0, x1, . . . , xn-1 } and yk = p(xk), compute the

coefficients {a0, a1, . . . , an-1 } by solving the following linear system

of equations. Note Vandermonde matrix is invertible iff xk are distinct.

O(n3).

1

2

1

0

1

2

1

0

11

211

12

222

11

211

10

200

1

1

1

1

nnnnnn

n

n

n

y

y

y

y

a

a

a

a

xxx

xxx

xxx

xxx

8

Fast Interpolation: Key Idea

Key idea: choose {x0, x1, . . . , xn-1 } to make computation easier!

Set xk = xj?

1

2

1

0

1

2

1

0

11

211

12

222

11

211

10

200

1

1

1

1

nnnnnn

n

n

n

y

y

y

y

a

a

a

a

xxx

xxx

xxx

xxx

9

1

2

1

0

1

2

1

0

132

132

132

132

)5()5()5(51

)17()17()17(171

)5()5()5(51

)17()17()17(171

nnn

n

n

n

y

y

y

y

a

a

a

a



Set xk = xj?

Use negative numbers: set xk = -xj so that (xk )2 = (xj )

2.

– set xk = -xn/2 + k

– E = peven(x2) = a0 + a2172 + a4174 + a6176 + . . . + an-217n-2

– O = x podd(x2) = a117 + a3173 + a5175 + a7177 + . . . + an-117n - 1

– y0 = E + O, yn/2 = E - O

10



Set xk = xj?

Use negative numbers: set xk = -xj so that (xk )2 = (xj )

2.

– set xk = -xn/2 + k

Use complex numbers: set x k = k where is nth root of unity.

– (xk )2 = (-xn/2 + k )

2

– (xk )4 = (-xn/4 + k )

4

– (xk )8 = (-xn/8 + k )

8

1

3

2

1

0

)1)(1()1(3)1(21

)1(3963

)1(2642

1321

1

3

2

1

0

1

1

1

1

11111

nnnnnn

n

n

n

n a

a

a

a

a

y

y

y

y

y

11

Roots of Unity

An nth root of unity is a complex number z such that zn = 1. = e 2 i / n = principal n th root of unity. e i t = cos t + i sin t. i2 = -1. There are exactly n roots of unity: k, k = 0, 1, . . . , n-1.

0 = 1

1

2 = i

3

4 = -1

5

6 = -i

7

12

Roots of Unity: Properties

L1: Let be the principal nth root of unity. If n > 0, then n/2 = -1. Proof: = e 2 i / n n/2 = e i = -1. (Euler's formula)

L2: Let n > 0 be even, and let and be the principal nth and (n/2)th roots of unity. Then (k ) 2 = k .

Proof: (k ) 2 = e (2k)2 i / n = e (k) 2 i / (n / 2) = k .

L3: Let n > 0 be even. Then, the squares of the n complex nth roots of unity are the n/2 complex (n/2) th roots of unity.

Proof: If we square all of the nth roots of unity, then each (n/2) th root is obtained exactly twice since:

– L1 k + n / 2 = - k

– thus, (k + n / 2) 2 = (k ) 2

– L2 both of these = k

k + n / 2 and k have the same square

13

Divide-and-Conquer

Given degree n polynomial p(x) = a0 + a1x1 + a2 x

2 + . . . + an-1 xn-1.

Assume n is a power of 2, and let be the principal n th root of unity. Define even and odd polynomials:

– peven(x) := a0 + a2x1 + a4x

2 + a6x3 + . . . + an-2 x

n/2 - 1

– podd(x) := a1 + a3x1 + a5x

2 + a7x3 + . . . + an-1 x

n/2 - 1

– p(x) = peven(x2) + x podd(x

2)

Reduces problem of– evaluating degree n polynomial p(x) at 0, 1, . . . , n-1

to– evaluating two degree n/2 polynomials at: (0)2, (1)2, . . . , (n-1)2 .

L3 peven(x) and podd(x) only evaluated at n/2 complex (n/2) th roots of

unity.

14

FFT Algorithm

if (n == 1) // n is a power of 2 return a0

e 2 i / n

(e0,e1,e2,...,en/2-1) FFT(n/2, a0,a2,a4,...,an-2)(d0,d1,d2,...,dn/2-1) FFT(n/2, a1,a3,a5,...,an-1)

for k = 0 to n/2 - 1 yk ek + k dk

yk+n/2 ek - k dk

return (y0,y1,y2,...,yn-1)

FFT (n, a0, a1, a2, . . . , an-1)

O(n) complex multiplies if we pre-compute k.

)log()()()2/(2)( nnOnTnOnTnT

15

Recursion Tree

a0, a1, a2, a3, a4, a5, a6, a7

a1, a3, a5, a7a0, a2, a4, a6

a3, a7a1, a5a0, a4 a2, a6

a0 a4 a2 a6 a1 a5 a3 a7

"bit-reversed" order

16

Proof of Correctness

Proof of correctness. Need to show yk = p(k ) for each k = 0, . . . , n-1, where is the principaln th root of unity.

Base case. n = 1 = 1. Algorithm returns y0 = a0 = a0 0 .

Induction step. Assume algorithm correct for n / 2.– let be the principal (n/2) th root of unity

– ek = peven(k ) = peven(2k ) by Lemma 2

– dk = podd(k ) = podd(2k ) by Lemma 2

– recall p(x) = peven(x2) + x podd(x

2)

1

0)(

n

j

jkj

k ap

)(

)(

)( 2odd

2even

k

k

kk

kkk

p

p

p

deyk

)(

)()(

)()(

)()(

2/

2odd

2/2even

2odd

2/2even

2odd

2even

2/

nk

nknknk

knkk

kknk

p

pp

pp

pp

deyk

kk

k

17

Best of Both Worlds



coefficient

Representation

O(n2)

Multiplication

O(n)

Evaluation



1-n10

1-n10

,,b,b

a,,a,a

b

2-2n10 ,,,c cc

)(,),(),q(x

)(,),(),p(x

1-2n10

1-2n10

xqxq

xpxp

)(,),(),r(x 1-2n10 xrxr O(n)


O(n2)




O(n log n)

18

Forward FFT: given {a0, a1, . . . , an-1 } , compute {y0, y1, . . . , yn-1 } .

Inverse FFT: given {y0, y1, . . . , yn-1 } compute {a0, a1, . . . , an-1 }.

1

3

2

1

0

)1)(1()1(3)1(21

)1(3963

)1(2642

1321

1

3

2

1

0

1

1

1

1

11111

nnnnnn

n

n

n

n a

a

a

a

a

y

y

y

y

y

1

3

2

1

0

1

)1)(1()1(3)1(21

)1(3963

)1(2642

1321

1

3

2

1

0

1

1

1

1

11111

nnnnnn

n

n

n

n y

y

y

y

y

a

a

a

a

a

Inverse FFT

19

Great news: same algorithm as FFT, except use -1 as "principal" nth root of unity (and divide by n).

)1)(1()1(3)1(2)1(

)1(3963

)1(2642

)1(321

1

1

1

1

11111

nnnnn

n

n

n

nF

)1)(1()1(3)1(2)1(

)1(3963

)1(2642

)1(321

1

1

1

1

1

11111

1

nnnnn

n

n

n

n nF

Inverse FFT

20

Inverse FFT: Proof of Correctness

Summation lemma. Let be a primitive n th root of unity. Then

If k is a multiple of n then k = 1. Each nth root of unity k is a root of

xn - 1 = (x - 1) (1 + x + x2 + . . . + xn-1),if k 1 we have: 1 + k + k(2) + . . . + k(n-1) = 0.

Claim: Fn and Fn-1 are inverses.

otherwise0

mod01

0

nknn

j

jk

otherwise0

if1

1 1

0

)(

1

0

1

ii

n

nFF

n

j

jii

n

j

ijij

iinn

21

Inverse FFT: Algorithm

if (n == 1) // n is a power of 2 return a0

e-2i/n

(e0,e1,e2,...,en/2-1) FFT(n/2, a0,a2,a4,...,an-2)(d0,d1,d2,...,dn/2-1) FFT(n/2, a1,a3,a5,...,an-1)

for k = 0 to n/2 - 1 yk (ek + k dk ) / n

yk+n/2 (ek - k dk ) / n

return (y0,y1,y2,...,yn-1)

IFFT (n, a0, a1, a2, . . . , an-1)

22

Best of Both Worlds



coefficient

Representation

O(n2)

Multiplication

O(n)

Evaluation



1-n10

1-n10

,,b,b

a,,a,a

b

2-2n10 ,,,c cc

)(,),(),q(x

)(,),(),p(x

1-2n10

1-2n10

xqxq

xpxp

)(,),(),r(x 1-2n10 xrxr O(n)


O(n2)




O(n log n)

23

Integer Arithmetic

Multiply two n-digit integers: a = an-1 . . . a1a0 and b = bn-1 . . . b1b0.

Form two degree n polynomials. Note: a = p(10), b = q(10).

Compute product using FFT in O(n log n) steps. Evaluate r(10) = a b. Problem: O(n log n) complex arithmetic steps.

Solution. Strassen (1968): carry out arithmetic to suitable precision.

– T(n) = O(n T(log n)) T(n) = O(n log n (log log n)1+ )

Schönhage-Strassen (1971): use modular arithmetic.– T(n) = O(n log n log log n)

1

0)(

n

j

jjxaxp

1

0)(

n

j

jjxbxq

)()()( xqxpxr

Princeton University • COS 423 • Theory of Algorithms • Spring 2002 • Kevin Wayne

Extra Slides

25

Polynomials: Representation and Addition

Degree n polynomial. Coefficient representation.

Point-value representation.

Addition. O(n) for either representation.

1

0)(

n

j

jjxaxp

)(that such

),(x,),,(x),,(x 11-n1100

kk

n

xpy

yyy

1

0)(

n

j

jjxbxq

)(zthat such

),(x,),,(x),,(x 11-n1100

kk

n

xq

zzz

1

0)()()()(

n

j

jjj xbaxqxpxr

),(x,),,(x),,(x 111-n111000 nn zyzyzy

26

Polynomials: Evaluation and Multiplication

Multiplication (convolution). Brute force.

O(n2).

Use 2n points.O(n).

Evaluation. Horner's method.

O(n).

Lagrange's formula.O(n2).

))))(((()( 12210 nn axaxaxaxaxp

1

0 )(

)(

)(n

kkj

jk

kjj

k xx

xx

yxp

),(x,),,(x),,(x 12121-n111000 nn zyzyzy

j

kkjkj

n

j

jj

bac

xcxqxpxr

0

22

0)()()(

fast fourier transform

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