Karatsuba’s Algorithm for Integer Multiplication

1. Karatsuba’s Algorithm for Integer Multiplication Jeremy R. Johnson

2. 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.

3. 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

4. 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)

5. 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]

6. 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.

7. 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).

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

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

10. 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

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

12. 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.

13. 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

14. Matrix Equation for Karatsuba’s Algorithm • Modified Vandermonde Matrix • Interpolation

15. 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

16. 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)).

17. 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.