FFT(Fast Fourier Transform)

p2.

FFT1

0 1 1

10 1 1

2 2 3 2 20 1 2 2 3 2 2

0

( ) .....

( ) .....

( ) ( ) ( )

= c .....

nn

nn

n nn n

j

j k j kk

A x a a x a x

B x b b x b x

C x A x B x

c x c x c x c x

c a b

•Coefficient representation: How to evaluate A(x0)?

p3.

Horner’s rule:

Point-value representation:

0 0 0 1 0 2 0 2 0 1( ) ( ( ..... ( )...))

(n)n nA x a x a x a x a x a

0 0

1 1 n-1 1 k

A point-value representation of a polynomial A(x) of

degree-bound n is a set of n point-value pairs {(x , ),

(x , ),.....(x , )}, where y ( )n k

y

y y A x

0 0 1 1 2n-1 2 1

' ' '0 0 1 1 2n-1 2 1

' ' '0 0 0 1 1 1 2n-1 2 1 2 1

:{(x , ), (x , ),.....(x , )}

:{(x , ), (x , ),.....(x , )}

:{(x , ), (x , ),.....(x , )}

n

n

n n

A y y y

B y y y

C y y y y y y

p4.

Thm1

0 0 1 1 n-1 1

k

For any set {(x , ), (x , ),.....(x , )} of n point-value

pairs, there is a unique poly A(x) of degree n-1, such that

y ( ) for k=0, 1,...., n-1

n

k

y y y

A x

Pf:2 1

0 0 0

2 11 1 1

2 11 1 1

0 0

1 1

1 1

1

1 if X = then

1

y

y

n

n

nn n n

n n

x x x

x x x

x x x

a

aX

a y

p5.

n-point interpolation:Lagrange’s formula:

0 0

1 11

1 1

det( ) ( ) 0 if 's are distinct

Thus, 's can be solved uniquely.

y

y

k jk

n

kj

j

n

a

aX

X x x x

a

a y

1

0

( )

( )( )

jnj k

kk k j

j k

x x

A x yx x

p6.

0 1 1

0 1 1

, ,....,

, ,....,n

n

a a a

b b b

0 02 2

0 02 2

2 1 2 12 2

( ), ( )

( ), ( )

,

( ), ( )

n n

n n

n nn n

A B

A B

A B

02

02

2 12

( )

( )

( )

n

n

nn

C

C

C

0 1 2 2, ,...., nc c c

Evalation

( log )n n

2

Ordinary Mult

( )n

Pointwise Mult

( )n

Interpolation

( log )n n

Point-valuerep.

Cof.rep.

p7.

Thm2The product of 2 polynomials of deg-bound n can

be computed in time (nlog n), with both the input

and output representation in coefficient form

0

( )mod

n 2 ik/n iu

2 i/n

0 1 2 1

Complex roots of unity:

1, e for k=0, 1, ..., n-1 e cos sin

e , the principal n-th root of unity

, , ,..

=?, =

.,

1

=

n

n

nn n

j k j k nn n

n

n

n n

n

n

u i u

1 1nn

p8.

Lemma 3 (Cancellation Lemma)n, k, d: non-negative integers,

Cor. 4 n: even positive integer

dk kdn n

Pf:

2 / 2 /( ) ( )dk i dn dk i n k kdn ne e

/ 22 1n

n

p9.

Lemma 5 (Halving lemma)n: even positive integer

The squares of the n complex n-th roots of unity are n/2

complex (n/2)th roots of unity.

2 2/ 2,

/ 2 2 2 2/ 2

/ 2

( ) where {0}

( )

and have the same squre

k k kn n n

k n k n k kn n n n

k k nn n

k Z

Pf:

Lemma 6 (Summation lemma)n-1

kn

j=0

, {0}, | , ( ) 0jn Z k Z n k Pf:

k nn-1k n nn k k

j=0 n n

( ) 1 ( ) 1( ) 0

1 1

n kj

p10.

DFT-1

0 1 -1

0

-1

0 1 -10

0 1 -1

0 1 n-1

Evaluate ( ) , ,....., ,

Assume n is a power of 2

Let , ,...., , and ( )

, ,...., is the DFT of the coefficient

vector a=<a ,a ,...., a

nj n

j n n nj

nk kj

n k n j nj

n

A x a x at

a a a a y A a

y y y y

,

DFT ( )ny a

p11.

Interpolation at the complex roots of unity:

02 1

1

2 4 2( 1)2

1 2( 1)1

1 1 1 1 y

1 y

y = 1

1

nn n n

nn n n

n nn n n ny

0

1

2

( 1)( 1)1

,

1

, ( )

n nn

kjn n k j n

n

a

a

a

a

y V a V

a V y

p12.

Thm 7 1

,, 0,1,....., 1, ( ) /kjn j k nj k n V n

11 1 1

, ' , , '0

1'

0

1( ' )

0

, ( ) ( ) ( )

1 =

1 = -------(*)

if j=j', then (*)=1, if j

n

n n n n n j j n j k n k jk

nkj kjn n

k

nk j jn

k

V V I V V V V

n

n

1 1

0 0

j', then by lemma 6, (*)=0

-(n-1) < j'-j <n-1, and n

1,

|

(j'-j

)n n

k j k jj k n k j n

k j

a y y an

Pf:

p13.

FFT[0] 2 / 2 1

0 2 4 2

[1] 2 / 2 11 3 5 1

[0] [12 2]

( ) .....

( ) .....

(*) ( ) ( ) ( )

nn

nn

A x a a x a x a x

A x a a x a x a x

A x x A xA x

[0] [1]

0 2

0 1

1 2 1 2

1

1. evaluating ( ) and ( ) at

Thus ev

( ) , ( ) ,....., (

aluating A(x) at , ,....., reduce

)

2. combining the results according to (*)

to

nn n n

nn n n

A x A x

p14.

[0] [0]/ 2

[1] [1]/ 2

[0] 2 [1] 2

[0] [1

/ 2 [0] 2 / 2 [1]/

]/ 2 / 2

[

2

[0] 1]

( )

( )

( ) ( ) ( )

= ( ) ( )

Le

( )

=

t

) (

(

kk n

kk n

k k k kk n n n n

k k

k n k n k nk

kn n

n n n

n

k

n

kk n

y A A

y A

y A

y A A A

A

A

A

y y

[

2

[0] / 2 [1]/ 2 / 2

[0] / 2 [ 0] [1] ]1

)

= ( ) ( )

=

k nn

k k n kn n n

k nk k

kk n kn y

A

y

A

y y

2 /n

[0]0 2 2

[1]1 3 1

[0] [0]

[1]

Recursive-FFT(a)

{ n=length[a]; /* n: power of 2 */

if n=1 the return a;

;

=1

a ( , ,....., );

a ( , ,....., );

y Recursive-FFT(a );

y Recurs

i n

n

n

e

a a a

a a a

[1]

[0] [1]k k k

[0] [1]k+n/2 k k

n

ive-FFT(a );

for k=0 to (n/2 - 1) do

{

y y y ;

y y y ;

= ; }

}

( ) 2 ( / 2) ( )

= (nlog n)

T n T n n

p16.

Thm 8 (Convolution thm)

-12 2 2

: power of 2

, : vectors of length

=DFT (DFT ( ) DFT ( ))n n n

n

a b n

a b a b

0 0 0 1 1 0 0 2 1 1 2 0

k=0

, , ,......

j-th elt: j

k j k

a b a b a b a b a b a b a b

a b

Componentwise product

Efficient FFT implement[0]ky

[1]ky

kn

[0] [1]kk n ky y

[0] [1]kk n ky y

0 1 2 3 4 5 6 7( , , , , , , , )a a a a a a a a

0 2 4 6( , , , )a a a a 1 3 5 7( , , , )a a a a

2 6( , )a a1 5( , )a a 3 7( , )a a0 4( , )a a

3

011

a 7

111

a5

101

a1

001

a6

110

a2

010

a4

100

a0

000

a

s

s-1

s-1 s-1 s-1

:

for s=1 to (lg n) do

for k=0 to n-1 by 2

combine the two 2 element DFT's in

A[k..k+2 1] and A[k+2 ..k+2 1]

i

Idea

s snto one 2 element DFT in A[k..k+2 1]

[0]y

[1]y

s

2 /m

FFT-Base(a)

{ n = length[a];

for s=1 to (lg n) do

{ m=2 ;

;

for k=0 to n-1 by do

{ =1;

for j=0 to (m/2 - 1) do

{ t= [ / 2];

m

i me

A k j m

u=A[k+j];

A[k+j]=u+t;

A[k+j+m/2]=u-t;

; }

} } }m

p19.

s 2 /

Iterative-FFT(a)

{ Bit-Reverse-Copy(a, A);

n=length[a];

for s=1 to (lg n) do

{ m=2 ; ; 1;

for j=0 to (m/2 -1) do

{ for k=j to n-1 by m do

i mm e

{ t = [ / 2];

u = A[k];

A[k]=u+t;

A[k+m/2]=u-t; }

}

}

}

m

A k m

k

Bit-Reverse-Copy(a,A)

{ n=length[a];

for k=0 to n-1 do

eg

rev(011

A[rev(k

)=110, rev(00

)] = a

}

1)=100

0a

1a

2a

4a

3a

5a

6a

7a

02

02

02

02

04

14

04

14

08

18

38

28

0y

1y

2y

4y

3y

5y

6y

7y

FFT circuit

8

In general, depth: (lg n)

size : (nlg n)

n

S=1 S=2 S=3