Polynomials and FFT

Lecture 11

Prof. Dr. Aydın Öztürk

Polynomials

A polynomial in the variable x is defined as

where are the

coefficiens of the polynomial.

1

0)(

n

j

jj xaxA

110 ,, naaa

Polynomials(cont.)

A polynomial A(x) is said have degree k if its highest non-zero coefficient is ak.

Any integer strictly greather than the degree of a polynomial is a degree bound of that polynomial.

Therefore, the degree of a polynomial of degree bound n may be any integer between 0 and n-1.

Polynomial addition

If A(x) and B(x) are polynomials of degree bound n, their sum is a polynomial C(x) s.t.

C(x)= A(x)+ B(x)

For all x underlying their field F that is if

jjj

jn

jj

jn

jj

jn

jj

bac

xcxC

xbxBxaxA

where

)(

then

)( and )(

1

0

1

0

1

0

Polynomial addition(cont.)Example:

4674)(

then

542)(

91076)(

If

23

3

23

xxxxC

xxxB

xxxxA

Polynomial multiplication

If A(x) and B(x) are polynomials of degree bound n, their product C(x) is a ppolynomial of degree-bound 2n-1 s.t.

C(x)= A(x) B(x)

for all x underlying their field F.

j

kkjkj

jn

jj

jn

jj

jn

jj

bac

xcxC

xbxBxaxA

0

22

0

1

0

1

0

where

)(

then

)( and )(

Polynomial multiplication(cont.)Example:

45867520441412

)()()(

then

542)(

91076)(

If

23456

3

23

xxxxxx

xBxAxC

xxxB

xxxxA

Representation of polynomials

Coefficient representationA coefficient representation of a polynomial

of degree bound n is a vector of coefficients

)(1

0

jn

jj xaxA

),,( 110 naaa a

Coefficient representation(cont.)A coefficient representation is convenient for

certain operations on polynomials.

Example: Evaluating A(x) at x0. Evaluation takes time Θ(n) when Horner’s rule is used:

))))((()( 1020201000 nn axaxaxaxaxA

Point-value representation

A point-value representation of a polynomial A(x) of degree-bound n is a set of poin-value pairs.

s.t. all of the xk are distinct and

With Horners’ method, an n-point evaluation takes Θ(n2).

1,,1,0for )( nkxAy kk

)},(,),,(),,{( 111100 nn yxyxyx

Point-value representation(cont.)

Addition based on point-value representation:

If we have a point value representation for A(x)

and for B(x)

Then a point-value representation for C(x)= A(x)+ B(x) is

)},(,),,(),,{( 111100 nn yxyxyx

)},(,),,(),,{( '11

'11

'00 nn yxyxyx

)},(,),,(),,{( '111

'111

'000 nnn yyxyyxyyx

Point-value representation(cont.)

The time to add two polynomials of degree bound point-value representation is Θ(n).

Point-value representation(cont.)

Multiplication based on point-value representation:

If we have a point value representation for A(x)

and for B(x)

then a point-value representation for C(x)= A(x)B(x) is

)},(,),,(),,{( 12121100 nn yxyxyx

)},(,),,(),,{( '1212

'11

'00 nn yxyxyx

)},(,),,(),,{( '111

'111

'000 nnn yyxyyxyyx

Point-value representation(cont.)

The time to multiply two polynomials of degree bound point-value representation is Θ(n).

Fast multiplication of polinomials

Can we use the linear-time multiplication method for polynomials in point-value form to expedite polynomial multiplication in coefficient form?

The answer hinges on our ability to convert a polynomial quickly from coefficient form to point-value form and vice-versa.

Fast multiplication of polinomials

We can choose the evaluation points carefully, we can convert between representations in Θ(nlg n) time.

If we choose complex roots of unity as the evaluation points, we can produce a point-value representation by taking the Discrete Fourier Transform(DFT) of a coefficient vector

The inverse operation can be performed by taking the inverse DFT of point-value pairs in Θ(nlg n) time.

Fast multiplication of polinomials

110

110

,,

,,

n

n

bbb

aaa

)(),(

)(),(

)(),(

12

122

12

12

02

02

nn

nn

nn

nn

BA

BA

BA

)(

)(

)(

122

12

02

nn

n

n

C

C

C

1210 ,, nccc Ordinary multiplication

Θ(n2 )

Θ(nlg n)

Pointwise multiplication

Θ(n)

Interpolation time

Θ(nlg n)

Coefficient

representation

Point-value

representation

The DFT and FFT

In this section we define• Complex roots of unity,• Define the DFT • Show how the FFT computes the DFT

Complex roots of unityA complex nth root of unity is a complex number ω s.t.

The n roots are:

To interpret this formula we use the defination of exponential of complex number:

1n

1,...,1,0for ,/2 nke nik

)sin()cos( uiueiu

Complex roots of unity

The value is called the principal root of unity.

All of the other complex nth roots of unity are powers of .

The n complex nth roots of unity are:

nin e /2

,,,, 110 nnnn

n

Complex roots of unity

Lemma-1: For any integers n ≥ 0 and d > 0,

Proof:

kn

kni

dkdnidkdn

w

e

e

)(

)(/2

/2

kn

dkdn

Complex roots of unityCorollary: For any even integer n > 0,

Lemma-2: If n > 0 is even then the squares of the nth roots of unity are the (n/2)th roots of unity.

Proof: We have (by lemma-1)

122/ n

n

2

2

2

222/

)(

)(

kn

kn

nn

kn

nkn

nkn

kn

kn 2/

2)(

Complex roots of unity

Lemma-3: For any integer n ≥ 1 and nonnegative integer k not divisible by n,

Proof:

0)(1

0

n

j

jnk

0

1

1)1(

1

1)(

1

1)()(

1

0

kn

k

kn

knn

kn

nkn

n

j

jnk

The DFT

We wish to evaluate a polynomial

of degree bound n at

Without loss of generality, we assume that n is a power of 2.

(We canalways add new high order zero coefficient as necessary)

1

0)(

n

j

jj xaxA

,,,, 110 nnnn

The DFT

Assume that A(x) is given in coefficient form

For k=0,1, ..., n-1 we define

The vector is called the Discrete Fourier Transform(DFT) of the coefficient vector We also write y=DFTn(a).

1

0

)(n

j

kjnj

knk

a

Ay

),,( 110 naaa a

),,( 110 nyyy y),,( 110 naaa a

The FFT(cont.)Let

It follows that

12/1

2531

]1[

12/2

2420

]0[

)(

)(

n

n

nn

xaxaxaaxA

xaxaxaaxA

)()()( 2]1[2]0[ xxAxAxA

The FFT(cont.)The problem of evaluating A(x) at

reduces to

1. Evaluating the degree-bound n/2 polynomials

and at the points

as

2. Combining the results.

110 ,,, nnnn

)(]0[ xA

)(]1[ xA 212120 )(,,)(,)( nnnn

)( knk Ay

Recursive FFT

vector.a be toassumed is 14

13

12

11

12/010

)FFT(-RECURSIVE9

)FFT(-RECURSIVE8

),,,(7

),,,(6

15

4

3

12

][1

)FFT(-RECURSIVE

]1[]0[2/

]1[]0[

]1[]1[

]0[]0[

131]1[

220]0[

/2

yy

yyy

yyy

nk

ay

ay

aaaa

aaaa

e

a

n

alengthn

a

n

kknk

kkk

n

n

nin

return

do

tofor

returnthen

if

Running time of RECURSIVE-FFT

We note that exclusive of the recursive calls, each invocation takes time Θ(n).

The recurrence for the running time is therefore

)lg(

)()2/(2)(

nn

nnTnT

Interpolation

We can write the DFT as the matrix product y = Va that is

1

3

2

1

0

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

)1(3963

)1(2642

132

1

3

2

1

0

1

1

1

1

11111

nnn

nn

nn

nnn

nnnnn

nnnnn

nnnnn

n a

a

a

a

a

y

y

y

y

y

Interpolation(cont.)

Theorem: For j,k=0,1, ..., n-1, the ( j,k) entry of the inverse of matrix is

Given the inverse matrix V -1, we have that

is given by

nkjn

kj /),( )(1 yDFT

n

1

0

1 n

k

kjnkj y

na

Interpolation(cont.)

By using the FFT and the inverse FFT, we can transform a polynomial of degree-bound n back and forth between its coefficient representation and a point-value representation in time Θ(n lg n).

Interpolation(cont.)

Theorem: For any two vectors a and b of length n is a power of 2,

Where the vectors a and b are padded with zero’s to length 2n and

. denotes the componentwise product of two 2n element vectors.

))()(( 221

2 baba nnn DFTDFTDFT

Example

Multiply the following polynomials.

Run time:

432

2

2

67731

)()()(

321)(

21)(

xxxx

xBxAxQ

xxxB

xxxA

)( 2n

Example

Multiply the polynomials in

The Discrete Fourier Transform(DFT) of the coefficient vectors:

DFT(a)=

[4.000, (-0.309 - 2.126i), (0.809 + 1.314i), (0.809 - 1.314i), ( -0.309+ 2.126i)]

DFT(b)=

[6.000, (-0.809 - 3.665i), (0.309 + 1.677i), (0.309 -1.677i), (-0.809 + 3.665i)]

Run time:

))log( nn

)0,0,3,2,1(

)0,0,2,1,1(

b

a

Example

DFT(a)∙DFT(b)=

[24.00, (-7.545 + 2.853i), (-1.954 + 1.763i), ( -1.954 - 1.763i), (-7.545 - 2.853i)]

6) 7 7 3 1(

))()((1

baba DFTDFTDFT