1

Karatsuba’s Algorithm for Integer Multiplication

Jeremy R. Johnson

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Introduction

• Objective: To derive a family of asymptotically fast integer multiplication algorithms using polynomial interpolation

– Karatsuba’s Algorithm– Polynomial algebra– Interpolation– Vandermonde Matrices– Toom-Cook algorithm– Polynomial multiplication using interpolation– Faster algorithms for integer multiplication

References: Lipson, Cormen et al.

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Karatsuba’s Algorithm

• Using the classical pen and paper algorithm two n digit integers can be multiplied in O(n2) operations. Karatsuba came up with a faster algorithm.

• Let A and B be two integers with– A = A110k + A0, A0 < 10k

– B = B110k + B0, B0 < 10k

– C = A*B = (A110k + A0)(B110k + B0)

= A1B1102k + (A1B0 + A0 B1)10k + A0B0

Instead this can be computed with 3 multiplications

• T0 = A0B0

• T1 = (A1 + A0)(B1 + B0)

• T2 = A1B1

• C = T2102k + (T1 - T0 - T2)10k + T0

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Complexity of Karatsuba’s Algorithm

• Let T(n) be the time to compute the product of two n-digit numbers using Karatsuba’s algorithm. Assume n = 2k. T(n) = (nlg(3)), lg(3) 1.58

• T(n) 3T(n/2) + cn 3(3T(n/4) + c(n/2)) + cn = 32T(n/22) + cn(3/2 + 1) 32(3T(n/23) + c(n/4)) + cn(3/2 + 1) = 33T(n/23) + cn(32/22 + 3/2 + 1) … 3iT(n/2i) + cn(3i-1/2i-1 + … + 3/2 + 1) ... c3k + cn[((3/2)k - 1)/(3/2 -1)] --- Assuming T(1) c c3k + 2c(3k - 2k) 3c3lg(n) = 3cnlg(3)

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Divide & Conquer Recurrence

Assume T(n) = aT(n/b) + (n)

• T(n) = (n) [a < b]

• T(n) = (nlog(n)) [a = b]

• T(n) = (nlogb

(a)) [a > b]

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Polynomial Algebra

• Let F[x] denote the set of polynomials in the variable x whose coefficients are in the field F.

• F[x] becomes an algebra where +, * are defined by polynomial addition and multiplication.

babac

xc

xbxa

ik

mk

nkiij

jikik

knm

kk

jn

jj

im

ii

xBxAxC

xBxA

),min(

),0max(

0

00

,)()()(

)(,)(

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Interpolation

• A polynomial of degree n is uniquely determined by its value at (n+1) distinct points.

Theorem: Let A(x) and B(x) be polynomials of degree m. If A(i) = B(i) for i = 0,…,m, then A(x) = B(x).

Proof.

Recall that a polynomial of degree m has m roots.

A(x) = Q(x)(x- ) + A(), if A() = 0, A(x) = Q(x)(x- ), and deg(Q) = m-1

Consider the polynomial C(x) = A(x) - B(x). Since C(i) = A(i) - B(i) = 0, for m+1 points, C(x) = 0, and A(x) must equal B(x).

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Lagrange Interpolation Formula

• Find a polynomial of degree m given its value at (m+1) distinct points. Assume A(i) = yi

• Observe that

yi

m

iij ji

jx

xA

0 )(

)()(

yyki

m

iij ji

jk

kA

0 )(

)()(

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Matrix Version of Polynomial Evaluation

• Let A(x) = a3x3 + a2x2 + a1x + a0

• Evaluation at the points , , , is obtained from the following matrix-vector product

aaaa

A

A

A

A

3

2

1

0

321

321

321

321

1

1

1

1

)(

)(

)(

)(

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Matrix Interpretation of Interpolation

• Let A(x) = anxn + … + a1x +a0 be a polynomial of degree n. The problem of determining the (n+1) coefficients an,…,a1,a0 from the (n+1) values A(0),…,A(n) is equivalent to solving the linear system

a

aa

n

nn

n

n

nA

A

A

3

1

0

1

1

1

1

0

1

0

1

0

...

...1

............

...1

...1

)(

...

)(

)(

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Vandermonde Matrix

n

nn

n

n

nV

...1

............

...1

...1

),...,(

1

1

1

1

0

1

0

0

)()det( iji

jV

V(0,…, n) is non-singular when 0,…, n are distinct.

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Polynomial Multiplication using Interpolation

• Compute C(x) = A(x)B(x), where degree(A(x)) = m, and degree(B(x)) = n. Degree(C(x)) = m+n, and C(x) is uniquely determined by its value at m+n+1 distinct points.

• [Evaluation] Compute A(i) and B(i) for distinct i, i=0,…,m+n.

• [Pointwise Product] Compute C(i) = A(i)*B(i) for i=0,…,m+n.

• [Interpolation] Compute the coefficients of C(x) = cnxm+n + … + c1x +c0 from the points C(i) = A(i)*B(i) for i=0,…,m+n.

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Interpolation and Karatsuba’s Algorithm

• Let A(x) = A1x + A0, B(x) = B1x + B0, C(x) = A(x)B(x) = C2x2 + C1x + C0

• Then A(10k) = A, B(10k) = B, and C = C(10k) = A(10k)B(10k) = AB

• Use interpolation based algorithm:

– Evaluate A(), A(), A() and B(), B(), B() for = 0, = 1, and =.– Compute C() = A()B(), C() = A() B(), C() = A()B()

– Interpolate the coefficients C2, C1, and C0

– Compute C = C2102k + C110k + C0

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Matrix Equation for Karatsuba’s Algorithm

• Modified Vandermonde Matrix

• Interpolation

ccc

C

C

C

2

1

0

100

111

001

)(

)1(

)0(

BABBAA

BA

ccc

11

1010

00

2

1

0

100

111

001

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Integer Multiplication Splitting the Inputs into 3 Parts

• Instead of breaking up the inputs into 2 equal parts as is done for Karatsuba’s algorithm, we can split the inputs into three equal parts.

• This algorithm is based on an interpolation based polynomial product of two quadratic polynomials.

• Let A(x) = A2x2 + A1x + A0, B(x) = B2x2 + B1x + B, C(x) = A(x)B(x) = C4x4 + C3x3 + C2x2 + C1x + C0

• Thus there are 5 products. The divide and conquer part still takes time = O(n). Therefore the total computing time T(n) = 5T(n/3) + O(n) = (nlog

3(5)), log3(5) 1.46

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Asymptotically Fast Integer Multiplication

• We can obtain a sequence of asymptotically faster multiplication algorithms by splitting the inputs into more and more pieces.

• If we split A and B into k equal parts, then the corresponding multiplication algorithm is obtained from an interpolation based polynomial multiplication algorithm of two degree (k-1) polynomials.

• Since the product polynomial is of degree 2(k-1), we need to evaluate at 2k-1 points. Thus there are (2k-1) products. The divide and conquer part still takes time = O(n). Therefore the total computing time T(n) = (2k-1)T(n/k) + O(n) = (nlog

k(2k-1)).

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Asymptotically Fast Integer Multiplication

• Using the previous construction we can find an algorithm to multiply two n digit integers in time (n1+ ) for any positive .

– logk(2k-1) = logk(k(2-1/k)) = 1 + logk(2-1/k)

– logk(2-1/k) logk(2) = ln(2)/ln(k) 0.

• Can we do better?

• The answer is yes. There is a faster algorithm, with computing time (nlog(n)loglog(n)), based on the fast Fourier transform (FFT). This algorithm is also based on interpolation and the polynomial version of the CRT.