TITLE: Finding a hyperplane that splits a convex polytope evenly
QUESTION [5 upvotes]: Say we have a convex polytope 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. a hyperplane that minimizes the absolute difference of vertex cardinalities on the two sides of the split).
Also, are there any known results regarding the computational 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 such possible splits.
Note:
I first asked this question in CSTheory.stackexchange.com last week. Since the question has not seen any significant progress since then, I thought I could try here.

REPLY [2 votes]: This is not an answer, it's a warning of a dead end.
I considered using a Markov Chain to randomly sample the vertices.  Just start at an arbitrary vertex, and at each time step, pick an edge-connected neighbor of the current vertex and step to that vertex.  The distribution of points hit by this chain converges to the uniform distribution on the vertices.  So we could use it to take a random sample of vertices and pick a splitting plane for that sample.  This would be an approximation to the real splitting plane.  Unfortunately I was able to construct a polytope where the convergence takes time exponential in the number of constraints.  So as long as you're taking exponential time, you might as well just enumerate all the vertices.