I need relax. Let’s go to the beach!

Personal note: I don’t know where I am headed, so don’t follow me. I am just wandering around and taking notes.

Suppose that there exist maps that are not four-colorable, perhaps only one, perhaps a few, or even infinitely many.

Now, remove from this set all maps that have regions with 2, 3, or 4 borders. For the scope of the four color theorem and without lack of generality, we can skip them. 🤞🏻 I am confident it is true, but not completely sure I can do this passage. Or at least it should be proved.

Now select a pentagonal region and regard it as the outer boundary: the ocean that surrounds all other lands.

And now walk along the shoreline.

What can we say about the five regions that face the ocean?

Can one of them be always an F5?

The answer is no, not always. Fullerene graphs show that it is possible for all F5 faces to be isolated, with no two pentagons sharing an edge. The standard example is the Buckminsterfullerene. Or, more commonly a soccer ball.

Can one of them be always an F6?

Euler’s formula neither guarantees nor rules out this possibility.

1 F5 – 0 F6 – 1 F7 – 2 F8 – 3 F9 – … = 12

👉 I don’t know if there are studies done to analyze these maps. F5 always surrounded by F7 or higher.

To be continued …

• Remove an edge between F5-F5, or F5-F6 or 😦 F5-F7+
• Analyze all combinations of colors along the new shoreline of the reduced maps
• And search for a contradiction to the hypothesis that there exist maps that cannot be colored with four colors