Contents - Student subdomain for University of Bath

226 APPENDIX B. EXCURSUS
B.3.2
Karatsuba’s method and sparse polynomials
The use of the Karatsuba formula and its variants for sparse polynomials is
less obvious. One preliminary remark is in order: in the dense case we split
the multiplicands in equation (B.6) or its equivalent in two (the same point for
each), and we were multiplying components of half the size. This is no longer
the case for sparse polynomials, e.g. every splitting point of
(a 7 x 7 + a 6 x 6 + a 5 x 5 + a 0 )(b 7 + b 2 x 2 + b 1 x + b 0 ) (B.7)
gives a 3–1 split of one or the other: indeed possibly both, as when we use x 4
as the splitting point.
Worse, in equation (B.5), the component multiplications were on ‘smaller’
polynomials, whereas, measured in number of terms, this is no longer the case.
If we split equation (B.7) at x 4 , the sub-problem corresponding to (a+b)∗(c+d)
in equation (B.5) is
(a 7 x 3 + a 6 x 2 + a 5 x 1 + a 0 )(b 3 + b 2 x 2 + b 1 x + b 0 )
which has as many terms as the original (B.7). This difficulty led [Fat03] to
conclude that Karatsuba-based methods did not apply to sparse polynomials.
It is clear that they cannot always apply, since the product of a polynomial
with m terms by one with n terms may have mn terms, but it is probaby the
difficulty of deciding when they apply that has led system designers to shun
them.
B.3.3
Karatsuba’s method and multivariate polynomials
TO BE COMPLETED
B.3.4
Faster still
It is possible to do still better than 3 k = ( 2 k) log 2 3
for multiplying dense polynomials
f and g with n = 2 k coefficients. Let α 0 , . . . , α 2n−1 be distinct values.
If h = fg is the desired product, then h(α i ) = f(α i )g(α i ). Hence if we write
˜f for the vector f(α 0 ), f(α 1 ), . . . , f(α 2n−1 ), then ˜h = ˜f ∗ ˜g, where ∗ denoted
element-by-element multiplication of vectors, an O(n) operation on vectors of
length 2n. This gives rise to three questions.
1. What should the α i be?
2. How do we efficiently compute ˜f from f (and ˜g from g)?
3. How do we efficiently compute h from ˜h?
The answer to the first question is that the α j should be the 2n-th roots of unity
in order, i.e. over C we would have α i = e 2πij/2n . In practice, though, we work
over a finite field in which these roots of unity exist. If we do this, then the