Re:Decades-Old Graph Problem Yields to Amateur Mathematician
by jcranmer on Wednesday April 18, 2018 @12:38AM
(#16864664)
You're horribly mistaken, but rather than just say that you're an idiot because there's a proof that every planar graph is 4-colorable, let me explain your confusion.
You are trying to construct a complete graph on N vertices in the plane, giving an example for N=5. This cannot be done. Indeed, every graph is either planar xor contains either the complete graph on 5 vertices or the bipartite graph of 3 vertices in each group as an implied subgraph.
You can embed a complete graph on N vertices in 3 dimensions easily, with thin rods. But this doesn't work for 2 dimensions because that connecting line divides 2-d space, so you can't cross it with another line. This property imposes some sharp constraints on what planar graphs have to look like, with implications for its colorability. The Four-Color theorem amounts to an exhaustive enumeration of the possible situations arising from these constraints and then showing that all of them can be colored with only 4 colors.
You are trying to construct a complete graph on N vertices in the plane, giving an example for N=5. This cannot be done. Indeed, every graph is either planar xor contains either the complete graph on 5 vertices or the bipartite graph of 3 vertices in each group as an implied subgraph.
You can embed a complete graph on N vertices in 3 dimensions easily, with thin rods. But this doesn't work for 2 dimensions because that connecting line divides 2-d space, so you can't cross it with another line. This property imposes some sharp constraints on what planar graphs have to look like, with implications for its colorability. The Four-Color theorem amounts to an exhaustive enumeration of the possible situations arising from these constraints and then showing that all of them can be colored with only 4 colors.