Thankyou Steinar.

It ain't "rocket science" but very pretty none the less.
I'll check it out on 2^11-1 in Basic when I feel like some programming
coming on:)

BTW "i.e." <> "e.g"

David


----------------------------------------
> Date: Tue, 14 Nov 2006 12:04:06 +0100
> From: [EMAIL PROTECTED]
> To: [email protected]
> Subject: Re: [Prime] Shifting the residue
> 
> On Tue, Nov 14, 2006 at 04:37:02AM +0000, david eddy wrote:
> > I am wrestling to understand why this shifting works, assuming it doesn't 
> > require
> > understanding of the FFT. Can anyone help?
> 
> We are doing arithmetic modulo 2^p-1 (say p=7 for simplicity), which means
> our operations are inherently circular -- if we touch bits >= p, it's just
> the same as folding them down below (ie. if you add 256, which is bit number
> 8=p+1, that's the same as adding 2, which is bit number 1). Given that
> squaring by FFT give us this kind of circular operation by default (it's a
> property of the FFT), and our only other operation is the subtraction of 2
> (which is also circular, since it's also done mod 2^p-1), all our operations
> are indeed circular.
> 
> But then it doesn't really matter how we store the number in memory; storing
> a0 a1 a2 a3 (or whatever, depending on how many bits we need) is exactly the
> same as storing a1 a2 a3 a0, a2 a3 a0 a1, or a3 a0 a1 a2. The operations are
> exactly the same, except that we have to remember how much we've shifted by
> when subtracting 2 (since that has to go on the right bits; subtracting 2 is
> certainly not the same as subtracting 4, 8, or 16), and we keep this shift
> all the way down the test. (The end result we test for, zero, is the same no
> matter what our shift is, of course.)
> 
> /* Steinar */
> -- 
> Homepage: http://www.sesse.net/
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