NLsolve.jl for different input and output dimensions?

Specific Domains Optimization (Mathematical)
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Hi all,

I am currently trying out NLsolve.jl to find roots of given functions, which works great at the moment, but I realized that I can’t seem to find a way to specify a dimension for the output different from the input… Is there a way to do so, or are there other packages that allow this?

Thanks!

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Hi, as far as I know, NLsolve.jl solves equations of the form

0=f_1(x_1, x_2, x_3, \ldots)\\ 0=f_2(x_1, x_2, x_3, \ldots)\\ 0=f_3(x_1, x_2, x_3, \ldots)\\ \ldots

If the number of equations does not match the number of variable x_i your system is either under- or overdetermined. Are you certain the problem you are looking at is actually a root-finding problem? (or perhaps an optimization?)

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Actually, I know a solution is not unique, but in a set of dimension at least k ; that is why I would like to solve n+k equations, to get the local unicity again. I don’t know if this is the more appropriate way though.

I tried making changes of coordinates to drop dimensions, but this didn’t work well in my particular case.

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NLsolve will only give you point solutions and therefore requires k=0.

I don’t know of a package which would give you the entire solutions set for k>0, however:

You could add some cost function to indicate which point solution (out of the complete solution set) you are interested in. Then solve it as an optimization problem with your n equations as side conditions.

EDIT: Or if you just want any solution you can just append k zeros to the output and NLsolve will hopefully converge to one solution out of your k-dimensional set.

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You might want to try this out: Interval constraint programming tutorial

What motivates your need? What problem do you want to solve?

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Characterising the entire solution set is a big part of the problem indeed… I was counting on NLsolve to find at least one root. I thought of the cost function, but doesn’t it require (if the equations f_j(x_i) = 0 are side conditions) the initial point to satisfy them already ?
As for the k zeros, my problem is overdetermined, so I should add k free variables to the input maybe. Didn’t try this one out, I will give it a shot.

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I will look at this, thanks !

I need to find the roots of a function in \mathbb{R}^{18} (I can put bounds, so that I have [-a; a]^{18} instead), but as I said, the roots are actually in continuous sets, and the function is costly to evaluate (about a second per call). I looked into NOMAD.jl, (blackbox optimization) but it didn’t succeed, so I am currently trying other methods.

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If the problem is overdetermined (but you are certain a solution exists) you could add some free variables to match the number of conditions?

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I looks like you could formulate this as a nonlinear least squares problem.

\min F(x), \mbox{ where } F(x) = \sum f_i^2(x)

and use an optimizer. If the input dimension is N and the output is M, the M > N is the overdetermined case and, while you may not have a solution, the optimizer will tell you something. In the underdetermined (M < N) case a least squares code would likely find one solution.

I am not clear on what kind of problem you have. I see that N = 18. Is M = 18-k?

I think you should try a nonlinear least squares code before trying interval methods or direct search methods like Nomad because gradient-based methods work better (most of the time) if your problem is smooth.

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I am in the overdetermined case (M = 18 + k > N), I add equations that a solution must satisfy. I will try a least squares method then !

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Just fyi, overdetermined means more equations than variables.

Underdetermined means more variables than equations.

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You should try to find a solver that does not add the artificial k equations, unless the added equations are not artificial and mean something for the problem. The risk in adding equations is that you can turn a problem with a zero residual into a problem without one.

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So I tried using this, but NLsolve doesn’t converge in this case for my problem. Currently trying it out on smaller test cases to see what’s going on.

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Because of the problem’s symmetries, I am sure that if a solution exists, it will belong to a continuous family of solutions. I tried changing variables to drop some and get the unicity this way, but it dramatically slows down the process - no clue why.

(I should maybe add that the function I am minimizing uses TaylorIntegration’s taylorinteg, and thus has no analytical form)