James Wanless on Sat, 12 Mar 2011 17:07:28 +0100


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Re: Polrootsmod query


Hello Bill,
Thanks very much once again for your valued help. pls see below for response(s)...
J

On 12 Mar 2011, at 15:06, Bill Allombert wrote:

On Sat, Mar 12, 2011 at 02:03:40PM +0000, James Wanless wrote:
Just realized I (probably?) sent this to the wrong list - I'll try
and get it right in future!

Why is PARI's implementation of polrootsmod so fast??? :)))
I've implemented my own version using GMP and Algorithm 1.6.1 of
Cohen ACCANT, but PARI's seems order(s) of magnitude faster.
What algorithm does polrootsmod use, pls? Maybe Cantor-Zassenhaus
split etc., ie Cohen Algorithms 3.4.x, using matrices?? I've
looked at PARI's code and haven't (yet?) understood it, despite
noticing a few references to 'Berlekamp'...

Hello James,

THe relevant C function is FpX_roots_i.

I am not totally clear (?) - would it be feasible to work from 'factormod' or 'factorcantor' and easily generate the roots from the results these give? [math question as opposed to coding question]. If so, would this be of a similar (or even better?) efficiency algorithm- wise, even though going via this route, ie via _factorization_, rather than just root-finding directly, a more complex algorithm involving matrices is needed?

PARI use Cantor-Zassenhauss and follow essentially Algorithm 1.6.1.

Interesting... I wonder why your version is so much quicker than mine, then.

Most of the time is spend in the computation of (X mod (p,P))^p
(see remark (2) page 38). For that you need one of the algorithms
of section 1.2 (binary powering).

Yes I believe I am already doing this correctly (I _was_ aware of remark 2, and have carefully checked I am not falling foul of that)

You also need to implement fast polynomial arithmetic over Fp[X].

Maybe that's it... would you say, in general, that there's a lot of optimization going on in PARI that could be having a large positive effect in this respect?


Note that PARI 2.4.3 is faster than PARI 2.3.5.

Which might seem to suggest 'yes' to my question immediately above.


Cheers,
Bill.