
Re: A New Graph Coloring Conjecture.
Posted:
Jun 26, 2013 9:20 PM


as in, "a complete graph on four vertices "is" a tetrhaedron, which is autodual; the dual to the graph is the map that is apparently 4colorable, but the tetrahedron (resp., tetraasteron) isn't 3colorable
> As I previously noted, if a graph with 4 vertices is not > 3colorable, then it must be complete (that is, it must be

