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Re: Homotopic algorithm to solve a system of equations

  • To: mathgroup at
  • Subject: [mg66875] Re: Homotopic algorithm to solve a system of equations
  • From: "aTn" <ayottes at>
  • Date: Fri, 2 Jun 2006 04:09:01 -0400 (EDT)
  • References: <e5mhml$kjl$>
  • Sender: owner-wri-mathgroup at

Ok, some of you wrote back to me wondering if such algorithms exist and
how they work. Here is a quick explanation.

The methods I'm referring to are know as  "probability one globally
convergent homotopy algorithms" (see L. Waston's great introductory
paper) that were developped to help solve a system of equations
(polynomial with real coefficients).

The basic result used in these algorithms is a parametrized version of
Sards theorem:

Let U be an open set in R^m and let p: U x R^n x [0,1[ ---> R^n be a
C^2 map. If p is transversal to zero (i.e. 0 is a regular value of p),
then the set of points 'v' in  the open set U such that p_v : R^n x
[0,1[ --> R^n : (x,t) ---> p(v,x,t) is transversal to 0, is of measure

The way to apply this result to equation solving is the following.
Consider a finite set of n polynomials with n indeterminates (and real
coefficients). To it corresponds a (smooth) function F:R^n---> R^n. The
idea is to find an open set U in R^m, a C^2 map p: U x R^n x [0,1[ -->
R^n which is transversal to zero, such that a solution of p_v(x,0) = 0
is known explicitly for all v in U, p_v(x,1)=  F(x) for all v in U and
all x in R^n, and finally for which p_v^(-1)(0) is bounded for all v in
U. Then (by the previous theorem) you can find a path in p_a^(-1)(0)
for some 'v' for which the differential of p_v is of full rank linking
a zero of p_v(. , 0) to a zero of p_v(. , 1) = F. We want the
differential to be of full rank so we can use algorithms such as the
Newton method to solve successive problems along the chosen path.

I'm looking for a package implementing known probability one globally
convergent homotopy algorithms. I hope this message makes my question


Etienne (aTn)

P.S.: I typed fast, so feel free to point out mystakes I might have
made :)

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