On Jan 17, 6:28 pm, Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote:
> Surely it depends on the polyhedron? E.g. for an octahedron there > are 4 pairwise vertex-independent paths between non-adjacent > vertices.
4 paths for adjacent vertices too.
Indeed the number of pairwise vertex-independent paths for any pair of non-identical vertices from any Platonic polyhedron is equal to the number of edges meeting at each vertex (so 3 for a tetrahedron, cube or dodecahedron, 4 for an octahedron and 5 for an icosahedron).