Consider function $f: R^n \rightarrow R$ - function of n variables. The function is a black box, but I have a priori knowledge of $k$ functions $g_i: R^n \times R \rightarrow R^n$ - set of function $g_i(x, \alpha)$ such that $\forall x, \alpha: f(x) = f(g_i(x, \alpha))$. I can consider this functions as $k$ parametrized equipotential curves (let's say, that $g_i(x, 0) = x$, so g_i(x) is curve, passing through x when alpha is zero and on which f(x) remains constant). My question is the following: is there any ways or approach to reduce degree of freedom of function f by k. Or, in other words, get "good" parametrization $x(y), y \in R^{n-k}$ ("good" in terms of some sense of beauty, I can not explain it mathimetically, because don't know what exactly I want).
It can be related to the first integrals of differential equation (as we can reduce the order of the system if we know some of its first integrals), but I don't know how to apply this approach in my case.
To clarify my question, let's consider an example:
Let's say, we have this (potentially unknown) function: $$f(x, y, z) = x^2 + y^2$$ But we know, that for any $(x, y, z)$ there exist equipotential surfaces, which are a combination of rotation in the plane XY and translation along the axis z. In terms of my question, there exist two functions: $$g_1(x, y, z, \phi) = (x * \cos(\phi) - y * \sin(\phi), x * \sin(\phi) + y * \cos(\phi), z) \\ g_2(x, y, z, \alpha) = (x, y, z + \alpha)$$
(I think, I chose pretty bad designations, but anyway). I do not understand, how by knowing only this two functions ($g_1$ and $g_2$) I can derive parametrization $$x(\alpha) = \alpha \\ y(\alpha) = z(\alpha) = 0$$ I also can not explain, why this parametrization is good: but for this particular example it is pretty clear, that f(x, y, z) looks much better in this coordinate substitution: $f(\alpha) = \alpha^2$ and we dont loos any information.
I think, that by knowing the structure of equipotential surfaces (let's say, a family of parametrizations of this surfaces), I want to get parametrization of the rest of the space.
I think that what you are looking for is explicit relationships between variables. I will go through an example:
\begin{equation} f = x+y = 2 \end{equation} obviously, no parameterization is needed here. Partially, because we can say that $ dx/dy = -1 $ is defined everywhere.
Next, for the following: \begin{equation} f = x^2+y^2 = 1 \end{equation} the parameterization is known: $x = \cos(t)$ and $y = \sin(t)$, for example. However, can there exist an explicit relationship $ x = f(y)$ to trace the circle? No, since there are points on the circle where $ dx/dy = -2y/2x $ is impossible to define (i.e. (0;1)).
So, for some specific domains, you may find an explicit relationship (i.e. x = sqrt(1-y^2) ). So you need a domain and to make sure that the derivatives are defined within it.