Learning an expression with four terms

[strifinopoulos]

I am trying to learn an equation of the form:

a*(x+y)+b*(x+y)^(2/3)-cx(x-1)/(x+y)^(1/3)-d*(y-x)^2/(x+y)

where a,b,c,d are constants

Things I have tried:
-Change variables to x, z=x+y, w=x-y and train on them
-define: ()^(2/3), ()^(1/3) and square as new unary operators (and put constraints on their complexities <3)
-increase d the maxdepth=maxsize=30
-increase the population_size
-increase the complexity of constants
-randomize the tree more frequently
-let the code run until loss<10

I am running out of ideas and the loss is still > 300 (the program halts since there is no improvement)
I would be grateful for any ideas.

[MilesCranmer]

What's the range of x and y?

Regardless, this looks like a pretty difficult expression to find. But it might be possible.

[strifinopoulos]

x and y are positive integer numbers. Is there any way to specify that there are four terms? that might be a good start.

[MilesCranmer]

One thing you could do is play around with the nested_constraints and constraints so that only + and - can be nested repeatedly. That will essentially force it to have a sum of simple terms.

[MilesCranmer]

The other option is to use this PR: MilesCranmer/SymbolicRegression.jl#143 which is not yet merged, and heed the comment here.

One easy way to get multiple expressions out is to enforce that the root node is a binary operator (or else, an infinite loss is returned), and is ignored. Then, tree.l is ϕ and tree.r is ψ, and you can do your evaluation on each, using separate calls to eval_tree_array. Sometimes it might delete the right child, but that's okay since those branches of the evolutionary tree would just die out, given their large loss.

You would basically have tree.l.l be term 1, tree.l.r be term 2, tree.r.l be term 3, and tree.r.r be term 4. You would to double-check that the tree is composed of two binary operators at the top levels though, before using tree.l.l.

However, you'll have to use the Julia backend to configure this, rather than PySR. Once I get around to merging it I will think about how to make it configurable in PySR though.

[MilesCranmer]

Also see https://astroautomata.com/PySR/backend/ for customizing the backend for use in PySR.