On Fri, 1 Oct 1999, Steinar H. Gunderson wrote:
> >In this way, two exponents could be tested using the same number of
> >FFT/IFFT operations as is currently being used to test 1 exponent.
> 
> Unless _I'm_ the one mixing up things here (very likely...), a possible
> error might be that you're mixing DFT and FFT? Aren't those two different
> algorithms?

The FFT (Fast Fourier Transform) algorithm is simply a fast way of
computing the DFT (Discrete Fourier Transform), so the two terms are
interchangeable as far as I know.  Sorry about the confusion.

Mike

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