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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!

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    $\begingroup$ 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. $\endgroup$
    – Blue
    Jun 30, 2019 at 12:14
  • $\begingroup$ 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. $\endgroup$
    – CardioidAss22
    Jun 30, 2019 at 13:20

2 Answers 2

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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 $V:=2-F+E$.

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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

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  • $\begingroup$ nor is one with fewer than 6. $\endgroup$
    – Dr. Richard Klitzing
    Jun 30, 2019 at 20:18

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