Finding a cutting plane that splits a polyhedron evenly

Question

Say we have a polyhedron in standard form:

\begin{equation*} \begin{array}{rl} \mathbf{A}\mathbf{x} = \mathbf{b} \\ \mathbf{x} \ge 0 \end{array} \end{equation*}

Are there any known methods for finding a hyperplane $\mathbf{d} \mathbf{x} +d_0= 0$ that splits the polyhedron in a way that the number of vertices on each side of the hyperplane is approximately the same? (i.e. an algorithm that minimizes the absolute difference of vertex cardinalities on the two sides of the split).

Also, are there any known results regarding the complexity of this problem?

Addendum: Restricting the types of cuts:

Here is a variation of the original problem with the hope that it is easier to solve than the original one:

Is there a way to efficiently compute or estimate for which coordinate $i$ a hyperplane of the form $d_ix_i + d_0 = 0$ would yield the lowest absolute difference of vertex cardinalities on both sides of the split? By efficient I mean anything more efficient than the exhaustive enumeration of vertex cardinalities for all possible such splits.

Note: After a few days of little progress, I posted this question at MathOverflow too.

Answer 1

I can't remember the analytic way to do this!

But this a classical problem for Genetic Programming! If you are familiar with it you can use a normalized vectors in the center of the polyhedron that describes the cutting plane.

So your population is a set of [x,y,z,...] normalized vectors and as fitting function you use the difference between the 2 split volumes!

So, if the difference tends to zero more "fit" is your vector/plane!