I have a polynomial system with $n+k$ unknowns ($n+k$ can be greater than 8), that is known to have a limited number of isolated solutions.

I want to solve this system numerically, but if I plug it in an algebraic computing system (Macaulay2 or Bertini) and try to solve it with homotopy continuation, the PC just remains blocked, and I can't find any solution.

I am trying to understand why this happens, with the following consideration: ALL the solutions are in some way equivalent (I mean that if $(x_1,..x_n)$, than the other solutions are just permutations of this array), so I am interested in computing just 1 solution, not all the possible solutions.

Here is my question

  1. Where is the bottleneck of the homotopy method, in the number of solutions or in the number of variables?
  2. The fact that I am interested only in one solution, may speed up the method just analyzing a single homotopy path?
  3. Is there a way in Macaulay2 to calculate only a single solution, and not all of them?

To give more details about the system, if I have $k+n$ unknowns ($k<n$), then I have $(n+1) + {n \choose 2} +{n \choose 3}$ equations and $k!$ equivalent solutions. The degree of the system is always 4, and I have one equation of degree 1, $n$ of degree 2, ${n \choose 2}$ of degree 3 and ${n \choose 3}$ of degree 4. The system can be solved easily with Groebner basis when $k=2$ and $k = 3$, but also computing Groebner basis for higher values of $k$ becomes very slow..


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    $\begingroup$ In case it helps people not native or not fluent enough in English (including myself): bottleneck = a ​problem that ​delays ​progress; from here: dictionary.cambridge.org/fr/dictionnaire/anglais/bottleneck $\endgroup$ – YCor Mar 22 '16 at 17:21
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    $\begingroup$ Perhaps it would be useful to provide more information on the system, e.g. "How many equations?", "What is the degree of the equations?" or even a link to an example system, which one could try to solve. $\endgroup$ – Moritz Firsching Mar 22 '16 at 19:31
  • $\begingroup$ for the benefit of a homotopy theorist, what is this "homotopy" method? Does it perchance involve integrating something on a sphere (or a simplex boundary)? $\endgroup$ – Jesse C. McKeown Mar 23 '16 at 2:41
  • $\begingroup$ @JesseC.McKeown It's a really neat method: try homotopy continuation method in your favorite search engine! $\endgroup$ – Moritz Firsching Mar 23 '16 at 7:50
  • $\begingroup$ OK, so the idea is you start with a function whose zeroes are easy to find, and perturb it in such a way that correcting the old zeroes is not-too-difficult... so, have you checked whether those assumptions are indeed correct? that it's possible to choose not-too-many perturbations, AND those perturbations have solutions near enough eachother AND the correct-the-solution step actually happens quickly? because it sounds like you've got a very complicated system of equations already, and even the First Step might be difficult. $\endgroup$ – Jesse C. McKeown Mar 23 '16 at 23:00

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