20110328, 21:20  #12 
Dec 2008
1501_{8} Posts 
A paper by GiustiHeintzHägeleMoraisPardoMontaña provides a rather effective algorithm for the solving of multivariate polynomial systems by way of Hilbert's Nullstellensantz:
http://arxiv.org/pdf/alggeom/9608010v1 However, I am not so sure an actual implementation of the algorithm (based on the NewtonHensel lifting method) would be easy to do. Last fiddled with by flouran on 20110328 at 21:40 
20110329, 00:09  #13  
Aug 2006
3×1,993 Posts 
Quote:


20110329, 00:12  #14  
Aug 2006
3·1,993 Posts 
Quote:


20110329, 01:48  #15  
"William"
May 2003
New Haven
2^{6}·37 Posts 
Quote:


20110329, 07:23  #16 
Mar 2011
2·3 Posts 
Thanks for all the ideas.
So far, I've tried using backtracking (with all the trimming I could find), but the results aren't satisfying at all (more than 24 hours and still no answer). There are exactly 15 variables in each equation, and the equations are symmetric, meaning each variable appears in exactly 7 equations. The variables are mixed in such way that there aren't any independent sets of variables and there are no repetitions of equations or dead variables. Does anyone know if the function bintprog (Matlab) should be able to assist in this case? Thanks again for your help 
20110329, 11:24  #17  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
couldn't you set two equations equal and eventually get an equation that replaces one variable with 2 then you can say variable=variable+variable or what ever the case may be. 

20110329, 12:52  #18 
Mar 2011
2×3 Posts 
I don't see how this is possible, since (and sorry if i didn't mention it) any 2 equations have at most 1 common variable

20110329, 13:18  #19  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
x+y+z = 4 a+x+b = 4 a+b=y+z also if x+y+z = 4 cx+d = 4 (c+d)(y+z) = 2*x so you can use the common variable somehow to set up more relations. 

20110329, 14:21  #20  
Jun 2003
The Texas Hill Country
10001000001_{2} Posts 
Quote:
I suspect that your variables and equations could be expressed in a matrix notation that would be much simpler. In addition, if the equations express only the structure of the constraints, it is highly likely that there are many symmetrical solutions. Many of the solution techniques will fail under these circumstances, but they can succeed if you eliminate the symmetrically equivalent solutions by adding a few more constraints. 

20110329, 14:38  #21 
Mar 2011
2·3 Posts 
Unfortunately I only have the equations...
Would it help if I uploaded them and/or their matrix representation? 
20110329, 16:10  #22 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

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