Visual Proof for Four colour theorem

1.1k Views Asked by At

I found a seemingly elegant, visual argument that shows why the four colour theorem (4CT) is true. The argument is as follows (I drew some pictures as well so I hope that helps):

A summary of the proof.

Once again, I have a strong feeling that this proof is incorrect given that I cannot find a peer-reviewed mathematics paper that details its derivation.

Question: Is it a fake proof?

Source of Proof: https://www.youtube.com/watch?v=X8nA7x6Db1I The proof is from 2:15 - 3:45

2

There are 2 best solutions below

0
On BEST ANSWER

You can triangulate a planar graph by adding edges to it. If the resulting triangulated graph is 4-colourable, then that same colouring will also be a valid 4-colouring of the original graph. Adding edges only imposes more restrictions, making colouring it harder. So if you can prove that all triangular planar graphs are 4-colourable, then all planar graphs are. But nowhere in that quoted text is there anything resembling a proof that all triangular planar graphs are 4-colourable - it just shows a few triangular planar graph examples and no more. That is not a proof.

Looking at the video, it is not much better. At a certain point it claims without proof that you only need a fourth colour for the border area. At that point it seems to confuse the colouring of the regions of a map and the colourings of its dual graph.

It is fairly easy to construct a graph (or its dual map) that needs 4 colours and for which all four colours must occur on some of the internal vertices (or regions on the dual map).

Here is one such graph:
enter image description here

If you colour it starting from the three corners (which must have distinct colours because they connect as a triangle), it becomes clear that there is only one way to colour the four internal vertices and that they use all 4 colours.

enter image description here

So there are graphs/maps that need at least 4 colours on internal vertices/regions. The video assumes that this is not possible, and so certainly does not prove that there is no graph/map that needs a fifth colour on some vertex/region.

2
On

Two other places where the video needs to be fact-checked, aside from the part where the proof is completely fake:

(0:40) "In 2004, a man named G. [incomprehensible] verified that you only need four colors to illustrate any map."

This is probably referring to G. Gonthier who, along with B. Werner, verified the 1996 proof Robertson, Sanders, Seymour, Thomas proof of the theorem in Coq (see MathWorld on the 4-color theorem). This is certainly an important contribution, but it's not like it's the first proof of the theorem. (Even the 1996 proof is a simplification of the 1977 proof by Appel and Haken.)

(6:04) If one of your borders is a point, you may need more than four colors. Don't believe me? Pause the video and go back.

I didn't believe her, so I paused the video and went back. Here is a 4-coloring of the map shown:

4-coloring

In general, the interpretation that the video seems to be giving to the point border, based on the partial coloring around 6:09, is that the region colored in red is adjacent to all the regions that meet at that point. Such situations are not a problem for the 4-color theorem. Just delete a small disk around that point and join it to the red region. Now no three regions meet at a point, but adjacencies are the same, so the 4-color theorem applies.