Talk:Famous Theorems of Mathematics/Four color theorem

Minumum Counter Example
Is it really appropriate for the "amateur" to consider the mininum counter-example approach as a format for proving the 4CT? How can we compete with the experience and sophisticated computing power of the professionals?

Jillbones 18:52, 5 May 2006 (UTC)

Simple math tells us that the fully triangulated planar graph for which Grinberg's graph is the dual would have only 24 vertices. Does such a graph exist?

Jillbones 01:26, 12 March 2007 (UTC)

"The dual of a fully-triangulated planar graph is a cubic planar graph. Cubic means that every vertex has 3 edges. However, not every cubic planar graph has a Hamiltonian cycle -- if it did it would prove the Four Color Theorem."

Can anyone explain why this is true? In your explanation, kindly state how each graph was created and which feature of the graph is being colored; ie (face, edge or vertex). For instance, in the map, it is the faces that are being colored. If, in the graph equivalent, the colorability of the faces is also being considered, describe how such a graph equivalent is achieved. If, in the graph equivalent; the colorability of the vertices is being considered; describe how that graph equivalent was generated.

Jillbones 22:33, 15 March 2007 (UTC)


 * Ok. We have a fully-triangulated planar graph G, and are trying to four-colour its vertices. This is equivalent to trying to four-colour the faces of the dual graph, which is cubic. A standard trick involving the Klein four group means that four-colouring the faces of a cubic planar graph is equivalent to three-colouring the edges. But if any cubic graph had a hamiltonian cycle, then we could easily three-colour the edges: just alternate red, blue along the hamiltonian cycle, and colour the remaining edges green. 131.111.8.96 10:47, 7 June 2007 (UTC)

This works iff the faces of the dual have a one-to-one correspondence with the vertics of G. Fortunately, this is always true whenever G is fully triangulated. By design, G itself is the dual of a map or graph. The generating map is 3-regular and the generating graph is \ a cubic planar graph. If M is the generating original map or graph, then G is the dual of M and M is the dual of G.

This means that if M is a 3-regular map or a cubic planar graph and M is 3 edge colorable. then M is 4 face colorable and the 4CT is true. You do not need the "triangulation"!

Of course, not all cubics have a "triangulation" for a dual!