Discrete weighted transforms and large-integer arithmetic

Abstract

It is well known that Discrete Fourier Transform (DFT) techniques may be used to multiply large integers. We introduce the concept of Discrete Weighted Transforms (DWTs) which, in certain situations, substantially improve the speed of multiplication by obviating costly zero-padding of digits. In particular, when arithmetic is to be performed modulo Fermât Numbers 22"1 + 1 , or Mersenne Numbers 29-1 , weighted transforms effectively reduce FFT run lengths. We indicate how these ideas can be applied to enhance known algorithms for general multiplication, division, and factorization oflarge integers.

Faster polynomial multiplication over finite fields using cyclotomic coefficient rings

“…By cutting the input integers into many small chunks, we convert to a multiplication in (Z/P Z)[X]/(X N −1) for a suitable N ≈ 2n/ lg P . One technical headache is that n is not necessarily divisible by N ; following [17], we deal with this by adapting an idea of Crandall and Fagin [9]. Next, by the Chinese remainder theorem, we reduce to multiplying in F p [X]/(X N − 1) for each p separately.…”

“…If N and P are positive integers such that N | n and lg P > 2n/N + lg N , then we may reduce the given problem to multiplication in (Z/P Z)[X]/(X N − 1), by cutting up the integers into chunks of n/N bits. In this section we briefly recall a variant [17, §9.2] of an algorithm due to Crandall and Fagin [9] that achieves the same reduction without the assumption that N | n. Assume that N n and lg P > 2⌈n/N ⌉ + lg N + 1, and that we have available some θ ∈ Z/P Z with θ N = 2. (This θ plays the same role as the real N -th root of 2 in the original Crandall-Fagin algorithm.)…”

A new truncated fourier transform algorithm

“…Given the images fi = f mod Φi, one can evaluate f at the roots of Φi by way of a Discrete Weighted Transform (DWT), which comprises an affine transformation followed by an FFT [3].…”

Faster integer multiplication using short lattice vectors

“…They convert multiplication in Z/qZ to multiplication in Z[y]/(y m −1), where m is a power of two. Because k is not divisible by m, the process of splitting an element of Z/qZ into m chunks is somewhat involved, and depends on a variant of the Crandall-Fagin algorithm [7].…”