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.
C60
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
Fullerenes are a perfect playground to test my four color algorithm.
I used the C6000 fullerene. 6000 atoms! A fullerene is an allotrope of carbon. Its molecules consist of carbon atoms connected by single and double bonds. A perfect 3-regular planar graphs ready to be tested.
And this is the four colored version you can play with, created using:
But a reinforcement learning agent can learn to make better selections over time to reduce color conflicts, I hope :-), and shed light on some properties of the graph that can be used in the proof.
Remove one edge at a time from F2, F3, F4, or F5 faces. These faces represent the basic unavoidable set. This process will reduce the map to three faces, including the ocean as the external area. Afterward, I will restore the edges in reverse order, consistently working with 3-regular planar graphs. I will consider Tait coloring of the edges and apply, when necessary, Kempe’s chain color switching to the edges.
Remove edges a then restore them
For the Agent:
Environment initialization = The starting uncolored 3-regular planar graphs
Action = Select the edge to remove from the 3-regular planar graphs
State = The updated 3-regular planar graphs
Reward = +1 if Random Kempe’s chain color switching was not necessary, -1 otherwise
Transform it into a 3-regular graph, by adding a frame and modifying the vertices with more than 3 edges, as shown below
Consider the surrounding area (out of the frame) as part of the map (the ocean) and color also that
It is known that we can also suppose that exactly three edges meet at each vertex since maps that have vertexes with more than three edges can be easily simplified without affecting the search of the coloring. In other words, if you can find a coloring for the map on the right, you can use the same coloring for the original map on the left.
Here is an example of these steps applied to the following Voronoi diagram. Three vertices have a degree greater than three.
If you can prove that it is possible to decompose a 3-regular graph into cycles, all with even length, you have also proved the four color theorem.
I think this fact is already known, but I never paid too much attention to it before.
Naturally, also the opposite is true: since the 4CT is true, it means that it is possible to color it, and also get the Tait coloring. And for any two colors you choose (for example Red and Green), the related Kempe cycles form a partition of the graph and each cycle has an even number of edges.
Once the decomposition is found, since the cycles have an even number of edges, it is possible to color the edges of a cycle with two alternate colors (for example Red and Green), forming this way Kempe cycles. When all cycles are colored (with R and G), it is possible to use a third color (for example Blue) to color the remaining edges.
The resulting graph will have a Tait coloring for all edges, which is equivalent to coloring all faces with four different colors.
But, how difficult is it to find all cycles?
This one has 6 Red-Green cycles (all of even length), that form a division on the 3-regular planar graph into cycles.
I realized that for some posts I may have used “Kempe loops” and “Kempe cycles” to describe closed “Kempe chains”, that in other words form closed paths (cycles). I think it would have been better always to use “Kempe cycles”.
Considering the smallest known non-Hamiltonian 3-regular planar graphs, discovered by Barnette-Bosák-Lederberg, I computed the Tait coloring and then deformed it to put in evidence the R-G cycles that in this case are four.
Q2) Verify if at least one of the cycles (R-G, R-B, or G-B) form an Hamiltonian circuit
False.
From the Tait’s conjecture we now know (now = 1946 by William Thomas Tutte) that 3-regular graphs exist also without an Hamiltonian circuit. It means that, no matter what, it is not possible to find a cycle that touches all vertices, and therefore no single Kempe cycles of any two color selection, can exist.
From Wikipedia, the free encyclopedia
This article is about graph theory. For the conjectures in knot theory, see Tait conjectures.
In mathematics, Tait’s conjecture states that “Every 3-connectedplanarcubic graph has a Hamiltonian cycle (along the edges) through all its vertices“. It was proposed by P. G. Tait (1884) and disproved by W. T. Tutte (1946), who constructed a counterexample with 25 faces, 69 edges and 46 vertices. Several smaller counterexamples, with 21 faces, 57 edges and 38 vertices, were later proved minimal by Holton & McKay (1988). The condition that the graph be 3-regular is necessary due to polyhedra such as the rhombic dodecahedron, which forms a bipartite graph with six degree-four vertices on one side and eight degree-three vertices on the other side; because any Hamiltonian cycle would have to alternate between the two sides of the bipartition, but they have unequal numbers of vertices, the rhombic dodecahedron is not Hamiltonian. The conjecture was significant, because if true, it would have implied the four color theorem: as Tait described, the four-color problem is equivalent to the problem of finding 3-edge-colorings of bridgeless cubic planar graphs. In a Hamiltonian cubic planar graph, such an edge coloring is easy to find: use two colors alternately on the cycle, and a third color for all remaining edges. Alternatively, a 4-coloring of the faces of a Hamiltonian cubic planar graph may be constructed directly, using two colors for the faces inside the cycle and two more colors for the faces outside.
Next post will be on the four coloring of the barnette-bosák-lederberg graph, which is the smallest known non-Hamiltonian graph, and the analysis of its Kempe cycles.
Q3) What about if I start with finding cycles, of even length (# of edges), that cross all vertices?
Some initial considerations: If cycles have to fit with Kempe cycles (closed paths of two alternating colors), they have to be of even length, and that would also mean that the total number of vertices crossed by all the cycles of the same two colors should be even too. Is it like this, or am I missing something? What about graphs with an odd number of vertices? Am I missing something?
Four Color Theorem 