Faster polynomial multiplication over finite fields using cyclotomic coefficient rings

Abstract

We present an algorithm that computes the product of two n-bit integers in O(n log n (4 √ 2) log * n ) bit operations. Previously, the best known bound was O(n log n 6 log * n ). We also prove that for a fixed prime p, polynomials in Fp[X] of degree n may be multiplied in O(n log n 4 log * n ) bit operations; the previous best bound was O(n log n 8 log * n ).

Faster integer multiplication using short lattice vectors

“…(vi) Reduce each product from (v) to a collection of forward and inverse DFTs of length S over Z/q ′ Z, and recurse. The structure of this algorithm is very similar to that of [11]. The main difference is that it is not necessary to explicitly split the coefficients into chunks in step (iv); this happens automatically as a consequence of storing the coefficients in θ-representation.…”

“…8] to the recurrence in Proposition 3.9. Alternatively, it follows by the same method used to deduce [11,Cor. 3] from [11,Prop.…”

“…Finally, Harvey and van der Hoeven [11] proposed using FFT primes, i.e., primes of the form q = a · 2 k + 1, where a is small. They reduce multiplication in Z/qZ to multiplication in Z[y]/(y m + a) via a straightforward splitting of the integers into m chunks, where m is a power of two.…”

“…Then in Sections 3 and 4 we prove Theorem 1.1, using an algorithm that is structurally very similar to [11]. We make no attempt to minimise the implied big-O constant in Theorem 1.1; our goal is to give the simplest possible proof of the asymptotic bound, without any regard for questions of practicality.…”

Fast Gröbner basis computation and polynomial reduction for generic bivariate ideals

“…During the next step of the loop, the term to be reduced is X 15 Y 5 marked by a diamond; this time it is reduced against G 8 which implies an update of Q 8 . For the second task, one has to rewrite the equation to maintain sufficient precision (loop in lines [11][12][13][14][15][16][17][18]; this is possible because a new quotient is now entirely known. In our example on Figure 4.b), we just completed the computation of Q 10 , so we find S 8 , S 9 such that Q 10 G 10 = S 8 G 8 + S 9 G 9 and we perform the replacement in equation (9); this is represented by the grey arrows in the picture.…”