Integer Multiplication

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1 Integer Multiplication in almost linear time Martin Fürer CSE 588 Department of Computer Science and Engineering Pennsylvania State University 1/24/08 Karatsuba algebraic Split each of the two factors of length n into two pieces of length n/2 each. Interpret the pieces as coefficients of linear polynomials. Evaluate the polynomials at 3 points (because the product polynomial has 3 coefficients). Karatuba and Ofman use 0, 1 and (see next page). Multiply corresponding values (recursively). Interpolate, i.e., find the unique polynomial with these three values at the 3 points.

2 Evaluation at infinity A polynomial p(x) = d j=0 a jx j can be homogenized to q(x, y) = m 1 j=0 a jx j y d j. (Every monomial has degree d.) For all r R, evaluating p(x) at r means computing p(r) = q(r, 1). Evaluating p(x) at can be defined as computing q(1, 0). Another simple Karatsuba-Ofman type algorithm is obtained by evaluating at 1, 0 and 1. For both versions, the running time is O(n log 2 3 ) = O(n ) by the Master Theorem (if the method is applied recursively). The exponent of integer multiplication The algebraic interpretation (evaluation, multiplication of values, interpolation) of the algorithm of Karatsuba and Ofman easily leads to a faster algorithm based on higher degree polynomials. This generalization is due to Toom [1963]. We define the exponent of integer multiplication as the infimum of real numbers α, such that integer multiplication can be done in time O(n α ). Trivially, 1 α 2, but by Karatsuba and Ofman α log 2 3. We present a version of Toom s result just intended to show that α = 1.

3 Algorithm A k Input: factors a, b N of length n, i.e., 0 a, b < 10 n. If n < n 0, then compute the product with school multiplication and exit. Chop the two factors into k pieces of length n/k each. Assume k n, i.e., k divides n. The factor a defines a j by a = k 1 j=0 a j10 jn/k with 0 a j < 10 n/k. (Similarly for b.) Let m = 2k 1 and define a j = 0 for k j < m. The factors a and b are now represented by the polynomials p a(x) = k 1 j=0 a jx j and p b (x) = k 1 j=0 b jx j. Multiply the two polynomials p a(x) and p b (x) to obtain p(x) with the algorithm of the next page. Let p(x) = m 1 j=0 c jx j. Then c = ab = m 1 j=0 c j10 jn/k Multiplication of Polynomials The input consists of two polynomials p a(x) and p b (x) Evaluate both polynomials at 0, 1,..., m 1. Multiply corresponding values. Interpolate, i.e., find the unique polynomial p(x) of degree m 1 whose value at x j = j is p a(x j )p b (x j ) for j = 0, 1,..., m 1.

4 Evaluation is a linear function Interpolation is unique Evaluation of a polynomial p a(x) = m 1 j=0 a jx j (with input (a 0,..., a m 1 ) T ) at m distinct (constant) points is a regular linear function. p a(x 0 ) 1 x 0 x x m 1 0 p a(x 1 ) 1 x 1 x x m 1 1 p a(x 1 ) = 1 x 1 x x m p a(x m 1 ) 1 x m 1 xm x m 1 m 1 }{{} Vandermonde Matrix V The determinant of V is (x j x i ) 0. 0 i<j<m a 0 a 1 a 2. a m 1 Complexity of arithmetic operations Additions of numbers of length n can be done in linear time. Multiplications with a constant factor can be done in liner time. Divisions (with remainder) by a constant can be done in liner time. Result: Evaluation and Interpolation in Algorithm A k can be done in linear time.

5 Bounds on the Values Recall: m = 2k 1 is a constant. x i j = j i has length i log 10 j. Values have length less than m log 10 m + n/k < m log 10 m + 2n/m. Multiplication Time Let M(n) be the time spent by Algorithm A k to multiply two numbers of length n. We get the following recurrence. { constant if n < n 0 M(n) m M(m log 10 m + 2n/m) + O(n) otherwise Without the (small) term m log 10 m the recurrence had an immediate solution. We still obtain a good solution if m log 10 m ɛ 2n/m for some positive real constant ɛ. This inequality holds for n sufficiently large, i.e., for n 1 2ɛ m2 log 10 m.

6 Simplified Recurrence Equation The previous page shows that just choosing n 0 appropriately simplifies the recurrence equation. For n 0 = 1 2ɛ m2 log 10 m we have { constant if n < n 0 M(n) m M((1 + ɛ)2n/m) + O(n) otherwise Using the Master Theorem The Master Theorem provides the solution The exponent is M(n) = O(n log b m ) with base b = m 2(1+ɛ) log b m = ln n ln b = ln m ln m ln 2 ln(1 + ɛ) < ln m ln m ln 2 ɛ We want to show M(n) = O(n 1+δ ) for an arbitrary given real δ > 0. Thus we want to choose ɛ and m such that ln m ln m ln 2 ɛ 1 + δ

7 Choice of m and ɛ We want ln m ln m ln 2 ɛ 1 + δ This inequality holds if ln m > ln 2 + ɛ (which holds for m 3 and ɛ < ln 3 ln 2) and ln m (ln m ln 2)(1 + δ) ɛ(1 + δ) or ɛ δ ln m ln δ Remember, ɛ will just determine n 0, but ɛ has to be positive, i.e., we have to choose m such that δ ln m > (1 + δ) ln 2, i.e., m δ > 2 1+δ or m > 2 1 δ +1. Summary We want to show that the exponent of integer multiplication is 1. To do so, we have to provide an O(n 1+δ ) time algorithm for every real number δ > 0. We can choose m = 2 2 1/δ + 1, and run Algorithm A k for k = (m + 1)/2. A better choice of m is a much bigger value, allowing a much smaller value of ɛ, and the O(n 1+δ ) behavior kicking in for a smaller value of n 0. A better practical algorithm would select k (and thus m = 2k 1) as a function of n instead of a constant depending on δ, requiring a more complicated analysis.

Chapter 1 Divide and Conquer Algorithm Theory WS 2016/17 Fabian Kuhn

Chapter 1 Divide and Conquer Algorithm Theory WS 2016/17 Fabian Kuhn Formulation of the D&C principle Divide-and-conquer method for solving a problem instance of size n: 1. Divide n c: Solve the problem