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Singular versus Maple
https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1676		 Page 1 of 1
Author:  DZDjokovic [ Sat Jun 28, 2008 3:46 am ]
Post subject:  Singular versus Maple
Hello,
I have an ideal "jed" with 9 generators in the polynomial ring over
the rationals in 8 variables, with pure lexicographic ordering.
Maple 11 finds its groebner basis in less than a minute (really in
about 20 sec) while the Singular 3-0-4 command (with prot option on)
ideal gb=groebner(jed);
takes more than 2 hours. They both produce gb of size 8 and
I am convinced that at least gb[1] is the same.
The example has not be concocted for this purpose, it appeared
in my work. I expect Singular to be more efficient in this kind of
computation than Maple and so I am very much puzzled by this example.
If anybody is interested, here is the file that generates "jed".
Thanks,
DZ Djokovic

LIB "matrix.lib";
proc ReD(poly p) { return( subst(p,j,0) ); }
proc ImD(poly p) { return( ReD((-j)*p) ); }
proc KonjMat(matrix v) { return( subst(v,j,-j) ); }
proc HermTr(matrix v) { return( transpose(KonjMat(v)) ); }
proc UnInv(matrix X) { def Y=HermTr(X);
ideal i=trace(X*X)/2,trace(X*Y),trace(X*X*X)/3,trace(X*X*Y),
trace(X*X*Y*Y),trace(X*Y*X*X*Y*Y);
ideal L=ReD(i[1]),ImD(i[1]),i[2],ReD(i[3]),ImD(i[3]),
ReD(i[4]),ImD(i[4]),i[5],i[6]; return(L); }
ring R=(0,j),(x,r,y,z,u,p,v,q),lp;
minpoly=j2+1;
matrix A[3][3]=2+3j,6,0, 0,1-3j,1, 4,0,-3; pmat(A);
matrix B[3][3]=u+j*p,z,0, 0,v+j*q,x+j*r, y,0; B[3,3]=-trace(B); pmat(B);
ideal iA=UnInv(A); ideal iB=UnInv(B); int m=size(iA); m;
ideal jed=iB[1]-iA[1];
for(int i=2; i<=m ; i++ ) { jed=jed,iB[i]-iA[i]; }
def s="ideal jed="+string(jed);
size(s);
ring S=0,(x,r,y,z,u,p,v,q),lp;
execute(s);
size(jed);
vdim(jed);
option(prot);
ideal gb=groebner(jed);
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