TITLE: Polytopes with few vertices.
QUESTION [10 upvotes]: Suppose I have a convex polytope in $\mathbb{R}^d$ which I know has few vertices (in the case which prompted this question, I seem to have a polytope in $\mathbb{R}^9$ which has sixteen vertices). Is there some constructive way to enumerate the possibilities? If the polytope has $k$ vertices, is there some not-too-horrible upper bound on the number of possibilities?

REPLY [4 votes]: I've got some information about a lower bound.
We know how many binary codes there are of length 5 on n words - http://oeis.org/A034190.  Each corresponds to a subset of the vertices of a cube and thus a convex polytope.
For length 5 with 16 words we get 169112, and a total of 1,226,525 up to symmetry.
I've computed the flag vectors for all such polytopes in dimension 5 - http://arxiv.org/abs/1011.4269.  Including polars I get 688,298 flag vectors.  (I did this for a good reason.)
From this calculation I can conclude that in dimension 5 with up to 32 vertices there are at least 344,149 such polytopes that are combinatorially distinct.