Let $P$ be a $d$-dimensional convex polytope.
$P$ is contained in $[0,1]^d$ and all vertices have only integral coefficients.

Given a set of facets of $P$, how to check that this set is maximal. i.e. that it is the set of all facets of $P$?

**[update]** I don't know yet if this is valid, but here's a simplification. Let's say, every facet contains the same number of vertices and this number is strictly bigger than d. Is this task easier?**[/update]**

**[update 2]** Does the general form of Eulers formula for polytopes help? It is $f\_0 - f\_1 + f\_2 - ... + (-1)^{d-1}\cdot f\_{d-1} = 1 - (-1)^d$ with $f\_0$ beeing vertices, $f\_1$ beeing the edges ... and $f\_{d-1}$ beeing the facets. **[/update 2]**

As a side note: For my specific polytope, it is virtually impossible to calculate all facets by combining vertices computationally. It would take a million modern dual-core processors since the big bang to have the numbers crunched. Roughly though, could be only 10% done by now. Believe me, I've tried and my code is good. If specific information of the polytope could help you, here it is: if $n$ is an arbitrary integer, there are $n!$ vertices and the dimension is $3\cdot n\cdot(n-1)/2$.