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

2. 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.

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.