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Very long computation
https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1587		 Page 1 of 1
Author:  Sotto [ Tue Dec 12, 2006 6:13 pm ]
Post subject:  Very long computation
Hi,

I'm trying to make computations with Singular like the following one. They are very very long (after two days I'm still waiting...). Is there something that I can do to speed the computation?

Thanks, Nicola

**********************

ring r = ( 0, p1, p2, x, y ), (u1, u2, u3, u4 ), lp;


// For every generator n/d of your field you construct a polynomial
// n(u1,u2,u3,u4)-n(p1,p2,x,y)/d(p1,p2,x,y)*d(u1,u2,u3,u4). The u1,
// u2,u3,u4 are new variables whereas the p1,p2,x,y are parameters
// (meaning we calculate these polynomials over Q(p1,p2,x,y)[u1,u2,u3,u4].


poly F = (2*u1*u2+u4*u3^3+u4^3*u3)*(u4*u1+u2*u3)-(2*p1*p2+y*x^3+y^3*x)*(y*p1+p2*x) ;


poly G = (u3*u1+u4*u2)*(2*u1*u2+u4*u3^3+u4^3*u3)^2-(x*p1+y*p2)*(2*p1*p2+y*x^3+y^3*x)^2 ;


poly H1 = 1/2*u1^2+1/2*u2^2+1/8*u3^4+3/4*u3^2*u4^2+1/8*u4^4-1/2*p1^2-1/2*p2^2-1/8*x^4-3/4*x^2*y^2-1/8*y^4 ;


poly H2 = (u1*u2+1/2*u3*u4*(u3^2+u4^2))^2-(p1*p2+1/2*x*y*(x^2+y^2))^2 ;


poly FF = 2*(u4*u3^5+4*u4^3*u3^3+2*u1*u2*u3^2+u4^5*u3+2*u1*u2*u4^2)*(2*u1*u2+u4*u3^3+u4^3*u3)-2*(y*x^5+4*y^3*x^3+2*p1*p2*x^2+y^5*x+2*p1*p2*y^2)*(2*p1*p2+y*x^3+y^3*x) ;


ideal H = H1, H2, F, G ;

ideal J = groebner( H );

reduce( FF, J);
Author:  hannes [ Wed Dec 13, 2006 8:43 pm ]
Post subject:  Re: very long computation with parameters
It is often faster to avoid parameters:
instead of the ring r=(0,p1, p2, x, y ), (u1, u2, u3, u4 ), lp;
define ring R=0,(u1,u2,u3,u4, p1,p2,x,y),(lp(4),dp);
do your groebner calculation in that ring R and simplify
the result (i.e. remove all but one element with the same leading
monomial w.r.t. u1..u4.

Hans Schoenemann
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