From: ganter@milton.u.washington.edu (M. Ganter)
Newsgroups: sci.math.symbolic
Subject: Re: Grobner basis calculation in MapleV
Keywords: elimination theory
Date: 15 Apr 92 18:11:40 GMT
Organization: University of Washington

avr@cs.purdue.EDU (Andrew V. Royappa) writes:

>Hi, I have a question about the Gr"obner basis calculation in Maple.

>Now I tried this in maple for the surface F(x,y,z) = x^2+y^2-z^2-1 = 0,
>a hyperboloid of one sheet.  (So I know the implicit form in advance).
>The parametric equations I used are

>	x = (u^2-v^2+1) / (u^2+v^2-1)
>	y = 2*u*v       / (u^2+v^2-1)
>	z = 2*u		/ (u^2+v^2-1)

>I formed the 3 polynomials as above, and computed the Grobner basis,
>expecting it to contain the polynomial x^2+y^2-z^2-1 (or perhaps a
>constant multiple of it). This is what happens in Maple V:
>================================================================

>> FL := [(u^2+v^2-1)*x-(u^2-v^2+1),(u^2+v^2-1)*y-2*u*v,(u^2+v^2-1)*z-2*u];
>FL :=

>   2    2           2    2        2    2                   2    2
>[(u  + v  - 1) x - u  + v  - 1, (u  + v  - 1) y - 2 u v, (u  + v  - 1) z - 2 u]

>_____________________________________________________________________________
>/* This sets up the ordering u > v > x > y > z that we want */
>> IL := [u,v,x,y,z];
>                             IL := [u, v, x, y, z]

>_____________________________________________________________________________
>> factor(gbasis(FL,IL,plex));
>       2      2                 2      2
>   [y u  + y v  - y - 2 u v, z u  + z v  - z - 2 u,

>                  2        2                                2
>       u v y - x v  + x - v  + 1 - u z, - u (- z v + y), z v  - z - u + u x,

>          2      2              2          2
>       u y  - x v  z + z x - z v  + z - u z , - (v - 1) (v + 1) (- z v + y),

>                           2        2    2
>       (v - 1) (v + 1) (- z  - 1 + x  + y )]

>_____________________________________________________________________________
>> 
Andrew, First let me say I'm not an expert in Grobner Basis but I've
been using it heavily for the last six months.

The generation of a Groebner basis is not a unique process.  
I tryed the same problem and got the same results!
Then I switched the variable ordering to:
   [ v, u, x, y, z ]
After factoring the results of gbasis, the basis set contained three
polynomials ALL with leading factors of "u".  The last polynomial contained
the hyperboloid.

Next, I tried using resultants.  Let's call your equations (1),(2) and (3).
I took the resultant of (1) and (2) eliminating u .  Generating eq. (4).
Then I took the resultant of (1) and (3) eliminating u. Generating eq. (5).
Lastly, I took the resultant of (4) and (5) eliminating v.  The result was
0 (zero).  Meaning that eq. (4) and (5) had common factors/roots.  After 
factoring and removing common roots. I formed the resultant of (4~) and
(5~) eliminating v. Generating eq. (6).  After factoring (6), extraneous
factors of (x+1) and (x-1) were still present.  In a very gross sense,
resultants should generate similar results as gbasis for tasks involving
variable elimination.

I have this strange feeling that the implicitization of the parametric
surface is NOT quite as straight-forward as Chris's book seems to imply.
You might check out some of Sederberg's articles in IEEE CG&A on this
topic....Good Luck .....

Mark Ganter/Univ. of Washington/Seattle,WA 98195/<ganter@u.washington.edu>
Newsgroups: sci.math.symbolic