Prime Numbers

504 Chapter 9 FAST ALGORITHMS FOR LARGE-INTEGER ARITHMETIC
This recursion formula for cyclic convolution can be proved via polynomial
algebra (see Exercise 9.42). The recursion relation together with some astute
algebraic observations led [Nussbaumer 1981] to an efficient convolution
scheme devoid of floating-point transforms. The algorithm is thus devoid of
rounding-error problems, and often, therefore, is the method of choice for
rigorous machine proofs involving large-integer arithmetic.
Looking longingly at the previous recursion, it is clear that if only we
had a fast negacyclic algorithm, then a cyclic convolution could be done
directly, much like that which an FFT performs via decimation of signal
lengths. To this end, let R denote a ring in which 2 is cancelable; i.e., x = y
whenever 2x =2y. (It is intriguing that this is all that is required to “ignite”
Nussbaumer convolution.) Assume a length D =2 k for negacyclic convolution,
and that D factors as D = mr, withm|r. Now, negacyclic convolution is
equivalent to polynomial multiplication (mod t D + 1) (see Exercises), and as
an operation can in a certain sense be “factored” as specified in the following:
Theorem 9.5.24 (Nussbaumer). Let D =2 k = mr, m|r. Then negacyclic
convolution of length-D signals whose elements belong to a ring R is
equivalent, in the sense that polynomial coefficients correspond to signal
elements, to multiplication in the polynomial ring
S = R[t]/ t D +1 .
Furthermore, this ring is isomorphic to
where T is the polynomial ring
T [t]/ (z − t m ) ,
T = R[z]/ (z r +1).
Finally, z r/m is an m-th root of −1 in T .
Nussbaumer’s idea is to use the root of −1 in a manner reminiscent of our
DWT, to perform a negacyclic convolution.
Let us exhibit explicit polynomial manipulations to clarify the import of
Theorem 9.5.24. Let
x(t) =x0 + x1t + ···+ xD−1t D−1 ,
and similarly for signal y, withthexj,yj in R. Note that x ×− y is equivalent
to multiplication x(t)y(t) in the ring S. Now decompose
x(t) =
m−1
j=0
Xj(t m )t j ,
and similarly for y(t), and interpret each of the polynomials Xj,Yj as an
element of ring T ;thus
Xj(z) =xj + xj+mz + ···+ x j+m(r−1)z r−1 ,