Re: Montgomery Reduction for GF(2)[x] ?

From: Colin Andrew Percival (cperciva_at_sfu.ca)
Date: 05/31/03 

Date: 31 May 2003 01:51:40 GMT

Tom St Denis <tomstdenis@iahu.ca> wrote:
> 1. for t from 1 to k do
> 2. if the lsb of p(x) is one then
> 3. p(x) = p(x) + v(x)
> 4. p(x) = p(x) / x

  How is that better than:
1. for t from k to 1 do
2. if the msb of p(x) is one then
3. p(x) = p(x) + v(x) * x^(t-1)

Colin Percival

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