Faces and Facets in a convex polyhedron

Question

I am willing to find the number of faces and facets in a convex polyhedron. Suppose, in the cube polyhedron there exists $8$ vertices, $12$ edges, and $6$ faces. It satisfies the Euler equation as follows: $$ V - E + F = 2 ; \quad \text{where}\{v:\text{vertices}, e:\text{edges}, f: \text{faces}\}$$

Now, I would calculate this for a simple polytope: $P=\{x_1-x_2\leq0,\ -x_1+x_2\leq1,\ 2x_2\geq5,\ 8x_1-x_2\leq16,\ x_1+x_2\geq4,\ x\in\text{R}^2 \}$.

In this case, there are $3$ vertices as $\{(3/2, 5/2), (17/7, 24/7), (37/16,5/2) \}$, $3$ edges, and based on the mentioned Euler formulation it would have $2$ faces. Also, there are $3$ facets. What I am looking for is, how the number of faces is $2$?

Answer 1

The Euler equation was originally for polyhedra in three dimensions. When applied in two dimensions, it is for planar graphs (graphs with no edge crossings). Your example qualifies as a planar graph. A planar graph partitions the plane into regions (including the exterior of the graph), with each region termed a "face". See, for instance, http://discrete.openmathbooks.org/dmoi3/sec_planar.html. In your example, there are indeed two faces: the interior of the triangle and the exterior of the triangle.

Answer 2

For software for characterizing facets and/or extreme points of a polytope you might look at: http://porta.zib.de/