# Converse of Euler's formula for polyhedra

Is it true that for any positive integers $$V, E, F$$ with $$V - E + F = 2$$ there exists a polyhedron with $$V$$ vertices, $$E$$ edges and $$F$$ faces?

In case there is a silly counterexample (say, with $$F=1$$), then what about large $$V,E,F$$ - say, all greater than or equal to $$10$$?

Any help appreciated!

• Euler's formula applies not only to (certain) polyhedra, but to planar graphs. In that context, $F=1$ is not a silly counterexample: a graph with two vertices connected by a single edge works; the single face is the "unbounded region" of the plane in which the graph lies. (More generally, a "tree" graph always has $V-E=1$.) By Steinitz' Theorem, graphs corresponding to polyhedra have a property called "three-connectedness" (removing fewer than three vertices and their adjacent edges will not disconnect the graph). Thus, simply satisfying $V-E+F=2$ is not enough to guarantee a polyhedron.
– Blue
Jun 30, 2019 at 12:14
• There are large counterexamples. For any large $n$, set $V=n$, $E=\frac{n(n+1)}{2}$, and $F=\frac{n(n+1)}{2}-n+2$. It's obvious that $V-E+F=2$, but a graph with $n$ vertices can have no more than $\frac{n(n-1)}{2}$ edges, so these values cannot correspond in any meaningful way to a graph/polyhedra. Jun 30, 2019 at 13:20

For any $$F$$ there exist $$V$$ and $$E$$ with $$V-E+F=2$$ but no polyhedron with $$V$$ vertices, $$E$$ edges, and $$F$$ faces. Indeed, the number of edges which can fit on the $$F$$ polygons is finite and hence we can take $$E$$ much larger and define $$V:=2-F+E$$.
There is no polyhedron $$P\subset{\mathbb R}^3$$ with $$7$$ edges. This is well known, and is proven, e.g., here as an exercise: https://people.math.ethz.ch/~blatter/Mathesis.pdf