Karatsuba’s Algorithm for Integer Multiplication September 4, 1997 Karatsuba’s Algorithm for Integer Multiplication Jeremy R. Johnson

September 4, 1997 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.

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

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)

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]

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.

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).

Lagrange Interpolation Formula Find a polynomial of degree m given its value at (m+1) distinct points. Assume A(i) = yi Observe that

Matrix Version of Polynomial Evaluation Let A(x) = a3x3 + a2x2 + a1x + a0 Evaluation at the points , , ,  is obtained from the following matrix-vector product

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

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

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.

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

Matrix Equation for Karatsuba’s Algorithm Modified Vandermonde Matrix Interpolation

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) = (nlog3(5)), log3(5)  1.46

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) = (nlogk(2k-1)).

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.