Is there a formular, and if yes, what is it, to describe the relation of faces, edges and vertices in three-dimensional convex bodies.
for regular shapes:
A tetrahedron has 4 faces, 6 edges and 4 vertices
Cube: 6 faces, 12 edges, 8 vertices
Octahedron: 8 faces, 12 edges, 6 vertices
Pentagonal dodecahedron: 12 faces, 30 edges, 20 vertices
What about a n-faced polyhedron? n faces, but how many edges and vertices? Is there a formula to calculate the number of vertices and edges, given a specific number of faces? Or a range of possible numbers of vertices and edges?
Add-on: What happens under the assumption of irregular shapes with that formula?
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$\begingroup$Yes, there is such a formula. It is called Euler's characteristic formula, and it states that if $V$ is the number of vertices, $E$ the number of edges, and $F$ the number of faces of a polyhedron, then $$V-E+F=2$$
For example, the cube has $8$ vertices, $6$ faces and $12$ edges, and$8-12+6=2$. The octahedron has $6$ vertices, $8$ faces and $12$ edges, and again $6-12+8=2$.
$\endgroup$ $\begingroup$Euler's formula even allows dimensional regression. Let $F=\{f_0, f_1, f_2, …\}$ be the facet vector of your polytope $P$, i.e. $P$ has $f_0$ $0$-facets (vertices), $f_1$ $1$-facets (edges), $f_2$ $2$-facets (faces), etc. Then you have$$\sum_{k=0}^{d-1} (-1)^k f_k = 1-(-1)^d$$
That formula also holds for non-regular polytopes, provided you do not encounter holes and other odd stuff. Esp. it is valid for all convex polytopes. You just have to sum on the various types of different vertices, on the different types of edges, on the different faces, etc. And thus it not only holds for the polytope $P$ itself, but also for all its sub-elements individually too! That is for its edges (all edges are incident to 2 vertices), its (possibly different types of) faces, etc. But also for its (possibly different types of) vertex figures, its edge figures, etc.
Consider the incidence matrix of a square pyramid
1 * | 4 0 | 4 0 tip vertex
* 4 | 1 2 | 2 1 base vertices
----+-----+----
1 1 | 4 * | 2 0 lacing edges
0 2 | * 4 | 1 1 base edges
----+-----+----
1 2 | 2 1 | 4 * lacing triangles
0 4 | 0 4 | * 1 base squareThere you clearly have $(1+4)-(4+4)+(4+1)=2$ on the diagonal (the absolute counts of the respective elements), but it also applies to every sub-diagonal and every super-diagonal part of each row, eg. $(1+1) = 2$ (lacing edges connect the tip to a base vertex each) or $(1+2)-(2+1)=0$ (triangular faces have 1 tip vertex and 2 base ones, as well as 2 lacing edges and 1 base one).
And finally there is even more to this incidence matrix. If you consider such matrices as $M=(m_{i,j})_{i,j}$ then you will have everywhere$$m_{i,i}\cdot m_{i,j}=m_{j,i}\cdot m_{j,j}\ \ (i<j)$$Cf. for details to and lots of individual incidence matrices for various polytopes are provided on that site too.
--- rk
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