On Wed, 2022-02-23 at 09:53 +1100, David Harvey wrote:
> On Tue, 2022-02-22 at 23:23 +0100, Marco Bodrato wrote:
> > Ciao David,
> >
> > Il Mar, 22 Febbraio 2022 10:55 pm, David Harvey ha scritto:
> > > On Tue, 2022-02-22 at 22:39 +0100, Marco Bodrato wrote:
> > > > > E.g, in this case we could
On Tue, 2022-02-22 at 23:23 +0100, Marco Bodrato wrote:
> Ciao David,
>
> Il Mar, 22 Febbraio 2022 10:55 pm, David Harvey ha scritto:
> > On Tue, 2022-02-22 at 22:39 +0100, Marco Bodrato wrote:
> > > > E.g, in this case we could try a top-level B^66 - 1 product, split in
> > > > B^33+1 and
Ciao David,
Il Mar, 22 Febbraio 2022 10:55 pm, David Harvey ha scritto:
> On Tue, 2022-02-22 at 22:39 +0100, Marco Bodrato wrote:
>> > E.g, in this case we could try a top-level B^66 - 1 product, split in
>> > B^33+1 and B^33-1; then the former suits your new algorithm well, but
>> > the former
On Tue, 2022-02-22 at 22:39 +0100, Marco Bodrato wrote:
>
> > E.g, in this case we could try a top-level B^66 - 1 product, split in
> > B^33+1 and B^33-1; then the former suits your new algorithm well, but
> > the former can't be recursively split (at least not on a B boundary). If
>
> I fully
Ciao,
Il Mar, 22 Febbraio 2022 8:04 pm, Niels Möller ha scritto:
> Marco Bodrato writes:
>> Simply, if a multiplication mod B^{3n}+1 is needed, the code computes
>> - a product mod B^{n}+1
>> - a product mod B^{2n}-B^{n}+1
>> - with CRT, the desired result is obtained.
> Ok, and can the
Marco Bodrato writes:
> Nothing special...
> Simply, if a multiplication mod B^{3n}+1 is needed, the code computes
> - a product mod B^{n}+1
> - a product mod B^{2n}-B^{n}+1
> - with CRT, the desired result is obtained.
Ok, and can the second product be computed more efficiently than full
Ciao,
Il 2022-02-21 01:37 Torbjörn Granlund ha scritto:
I am too busy to examine the code to see what you've done. Perhaps you
could outline the algorithms here?
Nothing special...
Simply, if a multiplication mod B^{3n}+1 is needed, the code computes
- a product mod B^{n}+1
- a product mod
