Applied Symbolic Computation September 4, 1997 Applied Symbolic Computation (CS 680/480) Lecture 6: Integer Multiplication, Interpolation, and the Chinese Remainder Theorem Jeremy R. Johnson May 9, 2001 May 9, 2001 Applied Symbolic Computation

Applied Symbolic Computation September 4, 1997 Introduction Objective: To show the relationship between the Chinese Remainder Theorem and Interpolation. To derive a series of fast integer multiplication algorithms using interpolation. Karatsuba’s Algorithm Polynomial algebra Polynomial version of the Chinese Remainder Theorem Interpolation Vandermonde Matrices Toom-Cook algorithm Polynomial multiplication using interpolation References: Lipson, Cormen et al. May 9, 2001 Applied Symbolic Computation

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 May 9, 2001 Applied Symbolic Computation

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) ...  cn[((3/2)k - 1)/(3/2 -1)] --- Assuming T(1)  c  2c(3k - 2k)  2c3lg(n) = 2cnlg(3) May 9, 2001 Applied Symbolic Computation

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] May 9, 2001 Applied Symbolic Computation

Applied Symbolic Computation 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. May 9, 2001 Applied Symbolic Computation

Polynomial Algebra mod a Polynomial A(x)  B(x) (mod f(x))  f(x)|(A(x) - B(x)) This equivalence relation partitions polynomials in F[x] into equivalence classes where the class [A(x)] consists of the set {A(x) + k(x)f(x), where k(x) and f(x) are in F[x]}. Choose a representative for [A(x)] with degree < deg(f(x)). Can choose rem(A(x),f(x)). Arithmetic [A(x)] + [B(x)] = [A(x) + B(x)] [A(x)] * [B(x)] = [A(x)B(x)] The set of equivalence classes with arithmetic defined like this is denoted by F[x]/(f(x)) May 9, 2001 Applied Symbolic Computation

Applied Symbolic Computation Modular Inverses Definition: B(x) is the inverse of A(x) mod f(x), if A(x)B(x)  1 (mod f(x)) The equation A(x)B(x)  1 (mod f(x)) has a solution iff gcd(A(x),f(x)) = 1. In particular, if f(x) is irreducible, then F[x]/(f(x)) is a field. By the Extended Euclidean Algorithm, there exist u(x) and v(x) such that A(x)u(x) + B(x)v(x) = gcd(A(x),f(x)). When gcd(A(x),f(x)) = 1, we get A(x)v(x) + f(x)v(x) = 1. Taking this equation mod f(x), we see that A(x)v(x)  1 (mod f(x)) May 9, 2001 Applied Symbolic Computation

Polynomial Version of the Chinese Remainder Theorem Theorem: Let f(x) and g(x) be polynomials in F[x] (coefficients in a field). Assume that gcd(f(x),g(x)) = 1. For any A1(x) and A2(x) there exist a polynomial A(x) with A(x)  A1(x) (mod f(x)) and A(x)  A2(x) (mod g(x)). Theorem: F[x]/(f(x)g(x))  F[x]/(f(x))  F[x]/(g(x)). I.E. There is a 1-1 mapping from F[x]/(f(x)g(x)) onto F[x]/(f(x))  F[x]/(g(x)) that preserves arithmetic. A(x)  (A(x) mod f(x), A(x) mod g(x)) May 9, 2001 Applied Symbolic Computation

Constructive Chinese Remainder Theorem Theorem: If gcd(f(x),g(x)) = 1, then there exist Ef(x) and Eg(x) (orthogonal idempotents) Ef(x)  1 (mod f(x)) Ef(x)  0 (mod g(x)) Eg(x)  0 (mod f(x)) Eg(x)  1 (mod g(x)) It follows that A1(x) Ef(x) + A2(x) Eg(x)  A1(x) (mod f(x)) and  A2(x) (mod g(x)). Proof. Since gcd(f(x),g(x)) = 1, by the Extended Euclidean Algorithm, there exist u(x) and v(x) with f(x)u(x) + g(x)v(x) = 1. Set Ef(x) = g(x)v(x) and Eg(x) = f(x)u(x) May 9, 2001 Applied Symbolic Computation

Applied Symbolic Computation 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). May 9, 2001 Applied Symbolic Computation

Lagrange Interpolation Formula Find a polynomial of degree m given its value at (m+1) distinct points. Assume A(i) = yi Observe that May 9, 2001 Applied Symbolic Computation

Polynomial Evaluation, Interpolation and the CRT Since A(x) = Q(x)(x- ) + A(), A(x)  A() (mod (x- )) If   , then gcd((x- ),(x- )) = 1. Therefore, we can apply the CRT to find a quadratic polynomial A(x) such that A() = y and A() = z. A(x) = A() E(x) + A() E(x), where E(x) = (x - )/( - ) E(x) = (x - )/( - ) Observe that E() = 1 and E() = 0, so that E(x)  1 (mod (x- )) and E(x)  0 (mod (x- )). The equivalent results hold for E(x) May 9, 2001 Applied Symbolic Computation

Applied Symbolic Computation Multifactor CRT The CRT can be generalized to the case when we have n pairwise relatively prime polynomials. If f1(x),…,fn(x) are pairwise relatively prime, i.e. gcd(fi(x),fj(x)) = 1 for i  j, then given A1(x),…,An(x) there exists a polynomial A(x) such that A  Ai(x) (mod fi(x)). Moreover, there exist a system of orthogonal idempotents: E1(x),…,En(x), such that Ei(x)  1 (mod fi(x)) and Ei(x)  0 (mod fj(x)) for i  j. A(x) = A1(x)E1(x) + … + An(x)En(x) May 9, 2001 Applied Symbolic Computation

Lagrange Interpolation and the CRT Assume that 0, 1,…, n are distinct and let fi(x) = (x- i). Then gcd(fi(x),fj(x)) = 1 for i  j. Let Then Ei(x)  1 (mod fi(x)) and Ei(x)  0 (mod fj(x)) for i  j, and Lagrange’s interpolation formula is A(x) = A (0)E0(x) + … + A(n )En(x) May 9, 2001 Applied Symbolic Computation

Matrix Version of Polynomial Evaluation Let A(x) = a3x3 + a2x2 + a1x + a0 Evaluation at the points , , ,  is obtained from the following matrix-vector product May 9, 2001 Applied Symbolic Computation

Applied Symbolic Computation Vandermonde Matrix V(0,…, n) is non-singular when 0,…, n are distinct. May 9, 2001 Applied Symbolic Computation

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 May 9, 2001 Applied Symbolic Computation

Matrix Interpretation of Interpolation The previous system has a solution when the 0,…, n are distinct. The solution can be obtained using Lagrange interpolation. In fact, the inverse of V(0,…, n) is obtained from the idempotents May 9, 2001 Applied Symbolic Computation

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. May 9, 2001 Applied Symbolic Computation

Interpolation and Karatsuba’s Algorithm Let A(x) = A1x + A0, B(x) = B1x + B, 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 May 9, 2001 Applied Symbolic Computation

Matrix Equation for Karatsuba’s Algorithm Modified Vandermonde Matrix Interpolation May 9, 2001 Applied Symbolic Computation

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 May 9, 2001 Applied Symbolic Computation

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)). May 9, 2001 Applied Symbolic Computation

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. May 9, 2001 Applied Symbolic Computation