ln 3.2 polynomials and fft

8/2/2019 LN 3.2 Polynomials and FFT

1/22

FFT(Fast Fourier Transform)


2/22

p2.

Polynoms


3/22

p3.

Polynomial addit ion and mult iplicat ion1

0 1 1

1

0 1 1

2 2 3 2 2

0 1 2 2 3 2 2

0

( ) .....

( ) .....

( ) ( ) ( )

= c .....

n

n

n

n

n n

n n

j

j k j k

k

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 representat ion:

How to evaluat e A(x0)?


4/22

p4.

Different Representations

Coefficient representation & Horners rule:

Root & Scale representation:

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


5/22

p5.

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 1

1 1 1

2 1

1 1 1

0 0

1 1

1 1

1

1if X = then

1

yy

n

n

n

n n n

n n

x x x

x x x

x x x

aa

X

a y

Vandermonde Matrix


6/22

p6.

n-point interpolation:

Lagranges formula:

0 0

1 11

1 1

det( ) ( ) 0 if 's are distinct

Thus, 's can be solved uniquely.

y

y

k j

k

n

k

j

j

n

a

aX

X x x x

a

a y

1

0

( )( )

( )

jnj k

k

k k j

j k

x xA x y

x x


7/22

p7.

Complexity Analysis in 3 representation


8/22

p8.

0 1 1

0 1 1

, ,....,

, ,....,

n

n

a a a

b b b

0 0

2 2

0 0

2 2

2 1 2 1

2 2

( ), ( )

( ), ( )

,

( ), ( )

n n

n n

n n

n n

A B

A B

A B

0

2

0

2

2 1

2

( )

( )

( )

n

n

n

n

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.


9/22

p9.

Thm2

The 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

n

n n

j k j k n

n n

n

n

n n

n

n

u i u

1 1nn


10/22

p10.

Lemm a 3 (Cancellat ion Lemma)

n, k, d: non-negat ive int egers,

Cor. 4 n: even posit ive integer

dk k

dn n

Pf:

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

/ 2

2 1n

n


11/22

p11.

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 k

n n n

k n k n k k

n n n n

k k n

n n

k Z

Pf:

Lemma 6 (Summat ion lemma)n-1 k

n

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


12/22

p12.

DFT-1

0 1 -1

0

-1

0 1 -1

0

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=


13/22

p13.

Interpolation at the complex roots of unity:

02 1

1

2 4 2( 1)

2

1 2( 1)1

1 1 1 1y

1y

y = 1

1

n

n n n

n

n n n

n nn n n n

y

0

1

2

( 1)( 1)1

,

1

, ( )

n nn

kj

n n k j n

n

aa

a

a

y V a V

a V y


14/22

p14.

Thm 71

,, 0,1,....., 1, ( ) / kj

n 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 j

k

nkj kj

n n

k

nk j j

n

k

V V I V V V V

n

n

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

-(n-1) < j'-j


15/22

p15.

FFT[0] 2 / 2 1

0 2 4 2

[1] 2 / 2 1

1 3 5 1

[0] [12 2]

( ) .....

( ) .....

(*) ( ) ( ) ( )

n

n

n

n

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

n

n n n

n

n n n

A x A x


16/22

p16.


17/22

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


18/22

p18.


19/22

p19.


20/22

Eff icient FFT im plement[0]

ky

[1]

ky

k

n

[0] [1]k

k n k

y y

[0] [1]k

k n k

y 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 a 1 5( , )a a 3 7( , )a a0 4( , )a a

3

011

a7

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

do combine the two 2 element DFT's in

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

Idea

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

[0 ]y

[1]y

Butterfly operation


21/22

p21.

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 m

m 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


22/22

0a

1a

2a

4a

3a

5a

6a

7a

0

2

0

2

0

2

0

2

0

4

1

4

0

4

1

4

0

8

1

8

3

8

2

8

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

ln 3.2 polynomials and fft

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