Re: Reimplementing the cubic sieve faster

Laël Cellier <[email protected]>
Newsgroups gmane.comp.mathematics.pari.user
Message-ID <[email protected]>
And if there’s many more numbers ? Turns out then that solving on the 
first 160 bits in subexponential time is prefered…

Also what does it means to have a number below a factor base with finite 
fields having several elements ?

Le 10/06/2025 à 19:21, Watson Ladd a écrit :
> Cado-nfs should be eating the prime field for breakfast at that size
>
>
>
> On Tue, Jun 10, 2025, 9:47 AM Laël Cellier <[email protected]> 
> wrote:
>
>     The problem in my use case is the prime field is 500 bits large
>     while the suborder is around 160Bits large : so it’s a bit too
>     large for pollard rho and ʙʙɢꜱ
>
>     Le 10/06/2025 à 18:43, Watson Ladd a écrit :
>>
>>
>>     On Tue, Jun 10, 2025, 9:33 AM Laël Cellier
>>     <[email protected]> wrote:
>>
>>         As as supplemental question, is it possible to shrink the
>>         factor base if
>>         we know the discrete logarithm is below a specific bound ?
>>
>>
>>     No. But two grumpy giants and baby can be of use here if the
>>     bound is small.
>>
>>
>>         Or more generally, to speed up the algorithm beside in the
>>         end solving
>>         the linear system modulo ((P−1)÷suborder) ?
>>
>>         Le 03/06/2025 à 00:20, Bill Allombert a écrit :
>>         > On Mon, Jun 02, 2025 at 11:58:31PM +0200, Laël Cellier wrote:
>>         >> Problem, is in my case it doesnt integrate with Pari-ɢᴘ.
>>         > You can always use extern(), thats does not seem to be a
>>         practical problem.
>>         >
>>         >> How do I solve the linear system ?
>>         > This is the hard part.
>>         > cado-nfs sparse linear algebra is much faster than PARI.
>>         > In fact PARI quadratic sieve would probably be fast enough
>>         > if cado-nfs sparse linear algebra was used.
>>         >
>>         >> Stupid question, but when you write about picking a
>>         triplet such a+b+c=0, do
>>         >> you mean picking them mod p ?
>>         > No, a,b,c are much smaller than p, so |a+b+c| < p.
>>         >
>>         > Cheers,
>>         > Bill.
>>
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