David R. Kohel on Sat, 03 Oct 1998 12:54:31 +0800


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Re: Problem with ellap


This problem has been around for a long time, and I always lift to 
char 0 first.  In version 2.0.alpha that I have installed, it don't 
know how to get the correct answer without doing the lift:


                      GP/PARI CALCULATOR Version 2.0.alpha
                      (i586 running linux 32-bit version)
                  (readline enabled, extended help available)

                             Copyright 1989-1997 by
          C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.

Type ? for help.

   realprecision = 28 significant digits
   seriesprecision = 16 significant terms
   format = g0.28

parisize = 10000000, primelimit = 1000000, buffersize = 30000
(12:44) gp > e = ellinit([0, 2, 0, 1, 1] * Mod(1, 5));
time = 0 ms.
(12:44) gp > ellap(e, 5)
time = 0 ms.
%2 = 0
(12:44) gp > ellap(e, 5, 1)
time = 0 ms.
%3 = 0
(12:44) gp > ellap(lift(e),5)
time = 0 ms.
%4 = -4

--David

> The default algorithm used by ellap appears to produce incorrect results.
> (The curve below is not supersingular at 5, so we know right off the bat
> that 0 must be wrong).
> 
> Roland
> 
>                        GP/PARI CALCULATOR Version 2.0.11 (beta)
>                     i586 running linux (ix86 kernel) 32-bit version
>                       (readline enabled, extended help available)
> 
>                               Copyright (C) 1989-1998 by
>               C. Batut, K. Belabas, D. Bernardi, H. Cohen and M. Olivier.
> 
> Send bug reports, suggestions and patches to pari@math.u-bordeaux.fr
> Type ? for help.
> 
>    realprecision = 28 significant digits
>    seriesprecision = 16 significant terms
>    format = g0.28
> 
> parisize = 20000000, primelimit = 500000, buffersize = 30000
> ? e = ellinit([0, 2, 0, 1, 1] * Mod(1, 5))
> %1 = [Mod(0, 5), Mod(2, 5), Mod(0, 5), Mod(1, 5), Mod(1, 5), Mod(3, 5),
> Mod(2, 5), Mod(4, 5), Mod(2, 5), Mod(1, 5), Mod(0, 5), Mod(2, 5), Mod(3,
> 5), 0, 0, 0, 0, 0, 0]
> ? ellap(e, 5)
> %2 = 0
> ? ellap(e, 5, 1)
> %3 = -4
>