Can I uniquely identify a convex, closed polyhedron given the number of sides in each face?

by Justin   Last Updated April 25, 2018 14:20 PM

FYI: I'm coming from a Computer Science/Computer Graphics background, so my terminology might not be 100% precise.

I have a collection of polyhedra that I would like to decompose into tetrahedra. Instead of using a general-purpose algorithm for any polyhedron, I'd like to create an algorithm that is more efficient because it knows something about the shape.

For simplicity's sake, assume that each polyhedron is closed and convex. If I list out the number and type of each of a polyhedron's face (for example, a tetrahedron has 4 3-sided faces and a hexagonal prism has 2 6-sided faces and 6 4-sided faces), is that enough to uniquely identify it?

(If there is some special word for "uniquely identifying it", please let me know)

Answers 1

The term would be: combinatorially equivalent polyhedra ( if they are structured in a similar way).

Just the information on the faces is not enough. If we know that all its faces are triangles ( $f$ of them), then the number of edges is $e=3f/2$, so $$f + v = 3f/2+ 2$$ so that determines the number of edges and the number of vertices. But the structure may be different. As an example, start with a tetrahedron, and raise each face with three other triangular faces. We get a polyhedron with $f=12$ faces, all triangular. We have $v=8$ and $e=18$.

Another polyhedron with $12$ triangular faces can be obtained from joining at the base two pyramids with a hexagonal base. Again, $v=8$, $e=18$.

However, if we look at the number of edges coming out of each vertex, they are for the first polyhedron $$3,3,3,3,6,6,6,6$$ while for the second one $$4,4,4,4,4,4,6,6$$ so the polyhedrons are not combinatorially equivalent.

A necessarily and sufficient condition for two polyhedra to be comtinatorially equivalent is that the grahs formed with the vertices and the edges are isomorphic. ( clearly necessary). See also this post.

April 25, 2018 14:19 PM

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