Permalink Submitted by Philip Mallinson on June 19, 2020

This is a very lucid exposition BUT... I have a question. It is clear that you can remove a face of a cube and then flatten it to get a planar graph. But suppose your polyhedron had every face a legitimate polygon but is very spiky and complicated with all sorts of ins and outs but no holes or tunnels. How can you guarantee that your algorithm for flattening the polyhedron to an equivalent planar graph still works?

## Proof of Euler's formula

This is a very lucid exposition BUT... I have a question. It is clear that you can remove a face of a cube and then flatten it to get a planar graph. But suppose your polyhedron had every face a legitimate polygon but is very spiky and complicated with all sorts of ins and outs but no holes or tunnels. How can you guarantee that your algorithm for flattening the polyhedron to an equivalent planar graph still works?