It all started with this drawing, an idea and an algorithm.
And with the help of AI and computers, the very enemies I am trying to defeat here :-), see the creation of maps without F2, F3, or F4.
Watch a planar map grow, live, on a sphere, visualizing the combinatorics behind the Four Color Theorem.
This is an interactive visualization of a growing cubic planar map, rendered on a fixed-size sphere (“Waterworld colonization”). It starts from the simplest possible thing: a single face floating in an empty ocean, no vertices, not yet a graph at all. The first move already sets the pattern for every move after it: a new border is drawn across the current coastline, adding one region, like a coastline advancing across the ocean.
Each new region follows simple combinatorial rules:
A sealing rule guarantees no region with fewer than 5 sides ever gets fully enclosed (“buried”) by its neighbors, a key structural invariant used in Four Color Theorem style reduction arguments.
The running quantity Σ(6−n) * Fₙ − C = 6 is tracked live in the “Ship’s log” panel and stays mathematically invariant at every single step, no matter how the coastline grows, a live visual proof of this invariant.
The camera is fully interactive: drag to orbit the globe, scroll or pinch to zoom, and use the Start/Stop and speed controls to watch the island grow at your own pace. Growth halts automatically once the island has expanded to fill the sphere.
PS: Once again, I came across many incredible stories of people chasing their dreams.
I was convinced that cubic maps on the sphere without F2, F3, and F4 faces were forced to have large clusters of F5 faces. But I’ve actually found an infinite family of counterexamples showing that it is possible to build maps of arbitrary size, with faces of arbitrary size, with no face smaller than a pentagon, and with every pentagon kept strictly isolated from every other.
Anyway, I don’t mind chasing wild geese, quite the contrary actually, because this wonderful theorem keeps showing me things I didn’t know, bringing me to the work of people I didn’t know.
The basic act of surgery is brutal and simple: delete one edge together with its two endpoints, and smooth the two resulting degree-two corners. Four faces feel it. The two faces that shared the edge merge into one face of size a+b−4 (6+6−4=8 sides); the two flanking hexagons drop to pentagons. This is the famous 5-8-5 defect of graphene physics (a reconstructed divacancy: two missing atoms) Volterra studied. Net charge: +1−2+1=0. Crucially, the surrounding ring of hexagons is combinatorially untouched. The defect is invisible from one ring away.
From there, the whole game is choosing the right edge so that the F5 face glides over the ocean to be kept as isolated as wanted.
I was already familiar with Paul Wernicke and Philip Franklin, who hunted the Four Color Theorem by proving unavoidability results: in any cubic spherical map with all faces of size at least five, some pentagon must sit next to a face of size at most six (Wernicke), in fact, next to two such faces (Franklin). Their method, discharging, the art of letting positive charge flow along edges until a contradiction surfaces, is the ancestor of the 1976 computer proof of the Four Color Theorem.
Victor Eberhard, a German geometer who lost his sight as a young man and did all of this work blind, proved in 1891 that any face-count sequence satisfying the law of twelve can be realized by an actual polyhedron, provided you are allowed to throw in enough neutral hexagons. Eberhard’s theorem is the existential backbone of everything here: the hexagons are the silent majority that makes any charge distribution geometrically possible
Vito Volterra, the great Italian analyst, supplied the deepest tool without knowing it. His theory of distorsioni in elastic bodies, cut the material, shift one lip of the cut by a lattice step, re-glue, is exactly the theory of dislocations that crystallographers and graphene physicists use today. On a hexagonal map, a Volterra cut-and-shift leaves a pentagon at one end of the seam and a heptagon at the other, with unharmed hexagons everywhere in between: a +1+1 and a -1-1, created in pairs, separable at will.
Michael Goldberg classified the highly symmetric cubic maps made of twelve pentagons and any number of hexagons, the Goldberg polyhedra, rediscovered by chemists as fullerenes and by virologists in virus capsids. A giant Goldberg map is the perfect blank canvas: a near-infinite sea of neutral hexagons with twelve pentagonal hills parked at the corners of an icosahedron, as far from each other as you please.
Stay with me 👀, because this is going to like you.
I came up with an idea that feels so simple and obvious that I can’t help but wonder why I hadn’t thought of it earlier. Maybe difficult to implement 😭 but the idea it is good and simple.
I’m curious if anyone has explored something similar, or if there are known related results 🤔
Would love to hear your thoughts!
I’ve been working on the Four Color Theorem 🎨, specifically on reducing planar 3-regular graphs down to a minimal configuration and then back reconstructing them step by step using, when necessary, different color Kempe chain switches techniques.
The challenge is that while my reduction strategy works in most cases ✅, there are configurations where the reconstruction gets stuck ❌ and Kempe switches don’t resolve the conflict. Only by shuffling the order of face selection 🔀 I’ve always been able to eventually color the map.
Here’s the idea 🚀: instead of trying to design the perfect reduction strategy upfront, why not start from the opposite direction?
Since I can already color arbitrarily complex graphs (even if it requires trial, backtracking, and random Kempe switches 🎲), I can treat successful colorings as “ground truth”.
From there, I can analyze the sequence of reductions on an already colored graph and observe which choices preserve a valid path back to a full coloring 🔍
In other words, rather than guessing the best reduction strategy, I can learn it from examples of graphs that have already been four-colored.
By removing an edge between two F5 faces, you obtain an F6 face. The two adjacent faces that shared the removed vertices each lose one edge. If the two adjacent faces are actually the same face, the number of edges decreases by two.
General case F5-Fn
By removing an edge between an F5 face and a Fn face, you obtain an F(n+1) face. The two adjacent faces that shared the removed vertices each lose one edge. If the two adjacent faces are actually the same face, the number of edges decreases by two.
The problem is that maps on the sphere can exist:
for sure, without adjacent F5-F5 faces, as in C60 Fullerene and other structures
Personal note: I don’t know where I am headed, so don’t follow me. I am just wandering around 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 step of removing these maps with no further considerations. 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.
Four Color Theorem 