1 fast multiplication 1.polynomial multiplication 2.toom–cook method 3. primitive roots of unity...

1

Fast Multiplication

1. Polynomial multiplication2. Toom–Cook method3. Primitive Roots of Unity 4. The Discrete Fourier Transform 5. The FFT Algorithm 6. Schönhage-Strassen’s method7. Conclusion

2

Polynomial 表示式 :225)( xxxp

{(-1,-4),(0,-5),(1,-2)}或者

52

5522

2

21

1

5

2

4

)01)(11(

)0)(1(2

)10)(10(

)1)(1(5

)11)(01(

)1)(0(4)(

2

222

222

xx

xxxxx

xxxxx

xxxxxxxp

by Lagrange Interpolating Polynomial

(p(-1), p(0), p(1))

(-5, 1, 2)evaluation interpolation

1. Polynomial multiplication

3

),(,),,(),,( 2211 mm yxyxyx

Lagrange Interpolating Polynomial

m

i

m

j ji

ji

mmmm

mm

m

m

m

m

ij

xx

xxyxp

xxxxxx

xxxxxxy

xxxxxx

xxxxxxy

xxxxxx

xxxxxxyxp

1 1

121

121

23212

312

13121

321

)(

)()(

)())((

)())((

)())((

)())((

)())((

)())(()(


4

Given coefficients (a0,a1,a2,…,an-1) and (b0,b1,b2,…,bn-1) defining two polynomials, p() and q(), and number x, compute p(x)q(x).

1

0

)()()(n

i

ii xcxqxpxC

i

jjiji bac

0

Polynomial Multiplication


5

(a0,a1,a2,…,an-1)(b0,b1,b2,…,bn-1)


選 2n+1 個點 (p(w0), p(w1),…, p(w2n) )(q(w0), q(w1),…, q(w2n) )evaluation

multiplication

(p(w0)*q(w0), p(w1)*q(w1),…, p(w2n)*q(w2n) )

interpolation

(c0,c1,c2,…,cn,…,c2n-1)

P(x)=a0+a1x+…+an-1xn-1

q(x)=b0+b1x+…+bn-1xn-1



6

(5,1)(1,2)


選 3 個點 (p(0)=5, p(1)=6, p(2)=7 )(q(0)=1, q(1)=3, q(2)=5 )

evaluation

multiplication

C(0)=5, C(1)=18, C(2)=3

interpolation

(5,11,2)

P(x)=5+xq(x)=1+2x


22

222

21155112

2

3521823

2

5

)12)(02(

)1)(0(35

)21)(01(

)2)(0(18

)20)(10(

)2)(1(5)(C

xxxx

xxxxxx

xxxxxxx


7


Evaluation O(n) Total O(n2)

慎選特殊的點可以降低運算

8

Polynomial Evaluation Horner’s Rule:

Given coefficients (a0,a1,a2,…,an-1), defining polynomial

Given x, we can evaluate p(x) in O(n) time using the equation

Eval(A,x): [Where A=(a0,a1,a2,…,an-1)] If n=1, then return a0

Else, Let A’=(a1,a2,…,an-1) [assume this can be done in constant time] return a0+x*Eval(A’,x)

1

0

)(n

i

ii xaxp

)))((()( 12210 nn xaaxaxaxaxp


9

2.Toom–Cook methodToom-k : a simplification of a description of Toom–Cook polynomial multiplication described by Marco Bodrato in 2007.

Marco Bodrato. Towards Optimal Toom–Cook Multiplication for Univariate and Multivariate Polynomials in Characteristic 2 and 0. In WAIFI'07 proceedings, volume 4547 of LNCS, pages 116-133. June 21-22, 2007. author website

Splitting ( 將整數轉成多項式 )Evaluation ( 找出多項式對應的點 )Pointwise multiplication ( 點値相乘 )Interpolation ( 由點找出多項式 )Recomposition ( 將多項式轉成整數 )

five main steps:

給定兩整數 M, N ，擬計算 M*N

10

2.Toom–Cook method:Toom-3Toom-3 : 例子說明 *

M = 12 3456 7890 1234 5678 9012 N= 9 8765 4321 9876 5432 1098

切成 3 等份，取 base B=108

m0 = 56789012 m1 = 78901234 m2 = 123456 n0 = 54321098 n1 = 43219876 n2 = 98765

p(x) = m0 + m1x + m2x2

q(x) = n0 + n1x + n2x2

(1) Splitting ( 將整數轉成多項式 )

11


(2) Evaluation ( 找出多項式對應的點 )

p(0) = m0 = 56789012 = 56789012 p(1) = m0 + m1 + m2 = 56789012 + 78901234 + 123456 = 135813702 p(−1) = m0 − m1 + m2 = 56789012 − 78901234 + 123456 = −21988766 p(−2) = m0 − 2m1 + 4m2 = 56789012 − 2×78901234 + 4×123456 = −100519632 p(∞) = m2 = 123456 = 123456 q(0) = n0 = 54321098 = 54321098 q(1) = n0 + n1 + n2 = 54321098 + 43219876 + 98765 = 97639739 q(−1) = n0 − n1 + n2 = 54321098 − 43219876 + 98765 = 11199987 q(−2) = n0 − 2n1 + 4n2 = 54321098 − 2×43219876 + 4×98765 = −31723594 q(∞) = n2 = 98765 = 98765

p(x) = m0 + m1x + m2x2

q(x) = n0 + n1x + n2x2

12


(3) Pointwise multiplication ( 點値相乘 )

r(0) = p(0)q(0) = 56789012 × 54321098 = 3084841486175176 r(1) = p(1)q(1) = 135813702 × 97639739 = 13260814415903778 r(−1) = p(−1)q(−1) = −21988766 × 11199987 = −246273893346042 r(−2) = p(−2)q(−2) = −100519632 × −31723594 = 3188843994597408 r(∞) = p(∞)q(∞) = 123456 × 98765 = 12193131840

r(x)=p(x)q(x)

13


(4) Interpolation ( 由點找出多項式 )

14



15


(4) Interpolation ( 由點找出多項式 )one sequence given by Bodrato

16



17


(5) Recomposition ( 將多項式轉成整數 )

18

2.Toom–Cook method:

)()3

(5)T( nOn

Tn )()3

(9)T( nOn

Tn

ComplexityToom-3 reduces 9 multiplications to 5, and runs in Θ(nlog(5)/log(3)), about Θ(n1.

465). In general, Toom-k runs in Θ(c(k)ne), where e = log(2k-1) / log(k), ne is the time spent on sub-multiplications, and c is the time spent on additions and multiplication by small constants

)()T( 5log3nOn )()T( )12(log kknOn )()T( 1 nOn

)log2()T( lg2 nnOn n (in Knuth’s book)

19

Polynomial multiplication

Evaluation O(n) Total O(n2)

慎選特殊的點可以降低運算 -Primitive Roots of Unity

執行 DFT 得到只需 O(nlogn)

,)A(,)A(,)A(,)1A( 32 ,,,,1 32

20

3. Primitive Roots of Unity A number is a primitive n-th root of unity, for n>1, if

n = 1 The numbers 1, , 2, …, n-1 are all distinct

Example 1: Z*

11:

22=4, 62=3, 72=5, 82=9 are 5-th roots of unity in Z*11

2-1=6, 3-1=4, 4-1=3, 5-1=9, 6-1=2, 7-1=8, 8-1=7, 9-1=5 Example 2: The complex number e2i/n is a primitive n-th root

of unity, where

x x^2 x^3 x^4 x^5 x^6 x^7 x^8 x^9 x^101 1 1 1 1 1 1 1 1 12 4 8 5 10 9 7 3 6 13 9 5 4 1 3 9 5 4 14 5 9 3 1 4 5 9 3 15 3 4 9 1 5 3 4 9 16 3 7 9 10 5 8 4 2 17 5 2 3 10 4 6 9 8 18 9 6 4 10 3 2 5 7 19 4 3 5 1 9 4 3 5 110 1 10 1 10 1 10 1 10 1

1i

21

Inverse Property: If is a primitive root of unity, then -1=n-1

Proof: n-1=n=1 Cancellation Property: For non-zero -n<k<n,

Proof:

Reduction Property: If w is a primitve (2n)-th root of unity, then 2 is a primitive n-th root of unity.

Proof: If 1,,2,…,2n-1 are all distinct, so are 1,2,(2)2,…,(2)n-1

Reflective Property: If n is even, then n/2 = -1. Proof: By the cancellation property, for k=n/2:

Corollary: k+n/2= -k.

01

0

n

j

kj

01

11

1

1)1(

1

1)(

1

1)(1

0

kk

k

k

kn

k

nkn

j

kj

)1)(2/(0 2/2/02/02/01

0

)2/( nnnnn

j

jn n

3. Primitive Roots of Unity

22

4.The Discrete Fourier Transform Given coefficients (a0,a1,a2,…,an-1) for an (n-1)-degree polynomi

al p(x)

The Discrete Fourier Transform is to evaluate p at the values 1,,2,…,n-1

We produce (y0,y1,y2,…,yn-1), where yj=p(j)

1

1

0

11

211

11

211

10

200

1

1

0

1

1

1

)(

)(

)(

nn

nnn

n

n

n a

a

a

xxx

xxx

xxx

xp

xp

xp

1

1

0

11211

12

11

1

0

1

1

1 1 1 1

)(

)(

)1(

nnnnn

n

nn a

a

a

py

py

py

23

Matrix form: y=Fa, where F[i,j]=ij.

The Inverse Discrete Fourier Transform recovers the coefficients of an (n-1)-degree polynomial given its values at 1,,2,…,n-1

Matrix form: a=F -1y, where F -1[i,j]=-ij/n.

1

0

n

i

ijij ay

Fa

1

1

0

11211

12

11

1

0

1

1

1 1 1 1

)(

)(

)1(

n

nnnn

n

nn a

a

a

py

py

py

註 : the matrix F 中 column 彼此互相 orthogonal 滿足 FF*=nI ，其中 F* 為 F 之 conjugate

4.The Discrete Fourier Transform

24

The DFT and inverse DFT really are inverse operations Proof: Let A=F -1F. We want to show that A=I, where

If i=j, then

If i and j are different, then

1

0

1],[

n

k

kjki

njiA

Property)onCancellati(by 01

],[1

0

)(

n

k

kij

njiA

1111

],[1

0

01

0

nnnn

iiAn

k

n

k

kiki


25

Convolution The DFT and the

inverse DFT can be used to multiply two polynomials

So we can get the coefficients of the product polynomial quickly if we can compute the DFT (and its inverse) quickly…

Pad with n 0's Pad with n 0's

[a0,a1,a2,...,an-1] [b0,b1,b2,...,bn-1]

DFT DFT

[a0,a1,a2,...,an-1,0,0,...,0] [b0,b1,b2,...,bn-1,0,0,...,0]

[y0,y1,y2,...,y2n-1] [z0,z1,z2,...,z2n-1]

ComponentMultiply

inverse DFT

[y0z0,y1z1,...,y2n-1z2n-1]

[c0,c1,c2,...,c2n-1]

(Convolution)


26

Convolutions127356

1 2 7

3 5 6×6 12 42

5 10 35

3 6 21

3 11 37 47 42

(3,11,37,47,42)Convolutions


27

Convolutions1273561 2 73 5 6×6 12 42

5 10 35

3 6 21

3 11 37 47 42

(37,50,53)acyclic convolutions

+ 3 11

37 50 53

1 2 73 5 6×6 12 42

5 10 35

3 6 21

3 11 37 47 42

(37,44,31)negacyclic convolutions

- 3 11

37 44 31


28

Convolution theorem

CyclicConvolution(X, Y) = IDFT(DFT(X) · DFT(Y))


29

5.The Fast Fourier Transform

The FFT is an efficient algorithm for computing the DFT The FFT is based on the divide-and-conquer paradigm:

If n is even, we can divide a polynomial

into two polynomials

and we can write

30

7

6

5

4

3

2

1

0

884942352821147

4236302418126

3530252015105

282420161284

21181512963

1412108642

765432

1

1

1

1

1

1

1

1 1 1 1 1 1 1 1

M

a

a

a

a

a

a

a

a

7

5

3

1

6

4

2

0

884935217422814

4230186362412

3525155302010

282012424168

21159318126

1410621284

753642

1

1

1

1

1

1

1

1 1 1 1 1 1 1 1

a

a

a

a

a

a

a

a

7

5

3

1

4418126

1284

642

3

2

6

4

2

0

4418126

1284

642

7

5

3

1

44211593

141062

753

6

4

2

0

4418126

1284

642

1

1

1

1 1 1 1

0 0 0

0 0 0

0 0 0

0 0 0 1

1

1

1

1 1 1 1

1 1 1 1

1

1

1

1 1 1 1

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a


Given coefficients (a0,a1,a2,…,a7) and w primitive 8-th root of unity

31

7

5

3

1

4418126

1284

642

3

2

6

4

2

0

4418126

1284

642

7

5

3

1

44211593

141062

753

6

4

2

0

4418126

1284

642

1

1

1

1 1 1 1

0 0 0

0 0 0

0 0 0

0 0 0 1

1

1

1

1 1 1 1

1 1 1 1

1

1

1

1 1 1 1

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a


32

7

5

3

1

6

4

2

0

884935217422814

4230186362412

3525155302010

282012424168

4

1

1

1

1

,1

a

a

a

a

a

a

a

a

7

5

3

1

44211593

141062

753

6

4

2

0

4418126

1284

642

1 1 1 1

1

1

1

1 1 1 1

a

a

a

a

a

a

a

a


33

1

1

1

1 1 1 1

M

18126

1284

642

n/2

令

7

5

3

1

2/

3

2

6

4

2

0

2/

7

5

3

1

2/

3

2

6

4

2

0

2/

M

0 0 0

0 0 0

0 0 0

0 0 0 1

M

M

0 0 0

0 0 0

0 0 0

0 0 0 1

M

M

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

nn

nn


34

5.The Fast Fourier Transform a0 a1 a2 a3 a4 a5 a6 a7

35The running time is O(n log n). [inverse FFT is similar]


T(n)=2T(n/2)+O(n)

36

(a0,a1,a2,…,an-1)(b0,b1,b2,…,bn-1)


選 w primitive m-th root of unity

(p(w0), p(w1),…, p(wm-1) )(q(w0), q(w1),…, q(wm-1) )Evaluation

FFTmultiplication

(p(w0)*q(w0), p(w1)*q(w1),…, p(wm-1)*q(wm-1) )

InterpolationIFFT

(c0,c1,c2,…,cn,…,cm-1)

p(x)=a0+a1x+…+an-1xn-1

q(x)=b0+b1x+…+bn-1xn-1


Polynomial multiplication

The running time is O(n log n).

(m2n)

37

6.Schönhage-Strassen’s method給定兩大數 M,N 。計算 M*N

(1) Splitting (將整數轉成多項式 )

令 B=2k, 則可產生 degree 最多 n-1 之兩個多項式

(2) Evaluation (找出多項式對應的點 )

p(x)=a0+a1B+…+an-1Bn-1

q(x)=b0+b1B+…+bn-1Bn-1

P=(a0,a1,a2,…,an-1)Q=(b0,b1,b2,…,bn-1)

))(,),(),1((),FFT(Q

))(,),(),1((),FFT(P1

1

m

m

qqq

ppp

選 w primitive m-th root of unity (m2n)

38

))()()(,),()()(),1()1()1(( 111 mmm qpCqpCqpC

(3) Pointwise multiplication (點値相乘 )

221210

111 ))),()()(,),()()(),1()1()1((IFFT(C(x)

n

n

mmn

xcxcc

qpCqpCqpC

(4) Interpolation (由點找出多項式 )

對上述點執行 the inverse Fourier transform

6. Schönhage-Strassen’s method

39

222210 BBN*M

nnccc

(5)Recomposition (將多項式轉成整數 )

令 x=B

222210C(x)

nn xcxcc


40


Complexity 分析

The running time is O(n log n) 單位 ?

The bound on the numerical errors α on the ci after the FFT process can be proved to be

where ε 1.e≅ -16

α ≤ 6n2B2log(n)ε

the number of digits of these basicnumbers should be of the order of log(B)+log(n)

41


,BB,B 121 N

)logloglog()( nnnONT

Complexity 分析

The running time is O(n log n(log(B) + log(n))2)

然而如果針對所有乘法都採用 recursive 觀念，即

得到 )log()()( nnONTNNT

42

7. Conclusion

Method* complexity integer

classical

Karatsuba 1962 10,000 位數 Toom–3 1963

Toom–Cook (knuth) 1966

40,000 位數

Schönhage-Strassen 1971

Fürer 2007

)O( 2n

)O( 46.1n

)O( 58.1n

)log2O( lg2 nn n

1522

1722

)logloglogO( nnn

)2logO( )(log* nOnn

1 fast multiplication 1.polynomial multiplication 2.toom–cook method 3. primitive roots of unity...

Documents