Four color theorem: simplified maps and fullerenes


Analyzing all 3-regular graphs that have only faces with 5 edges or more (simplified), I empirically found (using a computer program) that many hypothetically possible graphs, that by Euler’s identity may exist (F5=12+F7+2F8+3F9+...), do not actually exist. Using a VF2 algorithm to filter out isomorphic maps, I also noticed that not so many graphs as I expected existed. And that one general category of graphs, that always represents a simplified 3-regular graph, is that of Fullerenes (with 12 faces F5 and an arbitrary number of F6). Here is a list of what I found, so far, for each class of graphs, from 12 faces to 20 faces (surrounding area included).

The question is: Since the computation of maps with 17, 18, 19, 20 faces (simplified and not containing isomorphic graphs) it is taking me very long time (days of CPU time on a PC), is this sequence already known?

I did post this question on: math.stackexchange.com

  • 12 faces: 1 (only 1 graph exists)
    • On 3 dimentional space (sphere) it is a dodecahedron
    • It is a fullerene: 20-fullerene Dodecahedral graph
  • 13 faces: 0 (no simplified graphs exist with 13 faces)
    • The hypothetical (by Euler’s identity) map of 12 F5 and 1 F6 does not exist
  • 14 faces: 1 (12 F5 + 2 F6)
    • The hypothetical (by Euler’s identity) map of 13 F5 and 1 F7 does not exist
    • It is a fullerene: GP (12,2) Generalized Petersen graph
  • 15 faces: 1 (12 F5 + 3 F6)
    • The hypothetical (by Euler’s identity) map of 14 F5 and 1 F8 does not exist
    • It is a fullerene: 26-Fullerene
  • 16 faces: 3 (Two graphs are 12 F5 + 4 F6. The other has 14 F5 + 2 F7)
    • The hypothetical (by Euler’s identity) map of 14 F5 and 2 F7 does exists
    • The other two are Fullerenes
  • 17 faces: ???
  • 18 faces: ???
  • 19 faces: ???
  • 20 faces: ???

Many fullerenes are here: http://hog.grinvin.org/Fullerenes.

Last things implemented in the Java program (github):

  • Save and restore all lists (maps and todoList) to disk
    • Done: 19/Gen/2013
  • Verify the various filters when generating maps
    • Done: 19/Gen/2013
  • Browse the todoList
    • Done: 20/Gen/2013

Here are the maps … four colored … up to 16 faces (ocean included):

12 14 15
16 16 16

And the same graphs elaborated using yWorks:

12 14 15
16 16 16

Four color theorem: work in progress


😦

Too many more things to do and little time:

  • Filter out duplicates. I finally found a java library to efficiently filter out all isomorphic graphs. It is a library part if the sspace project. Using it I will be able to test more complex simplified maps with lots of vertices … not risking to spend CPU time on duplicates
  • Finish the implementation of the Cahit algorithm to color the edges of a bridgeless cubic planar graph. I still need to well understand why Kempe’s chain color switching works using Cahit spiral chain method … especially when there is more than one spiral chain
  • Convert the swing application to javafx. This way I’ll be able to eliminate many third party dependencies: G, swixml, …

Bye

Four color theorem: Cahit spiral chains (step two)


Now the application is able to find all spiral chains of a graph.

I still need to:

  • Implement the Cahit coloring algorithm using the spiral chains
  • Add some additional features to the Java application
    • Modify the settings to color the edges
    • Manual selection of the starting vertex
    • Manual coloring of the Edges

The application can be downloaded from the github site.

About the Cahit coloring algorithm there are some things that I need to undestrand before implementing the Java code.

Four color theorem: Cahit spiral chains


Hi,

I’ve found some time to implement the first version of Cahit Spiral Chains algorithm.

I still need to:

  • Find all spiral chains of a given graph and not only one (changing the starting point)
  • I need to implement the concept of “nearest unused vertex to the last vertex of the last spiral chain”. This is needed to find the starting point of the next spiral <– DONE

Here is the video on youtube:

Four color theorem: other representations of maps


Here are some new representation of graphs:

Thanks to:

Four color theorem: representations of maps


For the scope of the four color problem and without lack of generality, maps can be represented in different ways. This is generally done to have a different perspective on the problem.

For example, the graph-theoretic representation of maps has become so common and important that generally the four color problem is stated and analyzed directly in terms of graph theory: http://en.wikipedia.org/wiki/Four_color_theorem.

I am trying to collect other representations that may in some way help to get a different point of view on the problem. If you know one of these representations that is not listed and wish to share, report it here. If you also have a web reference that explains or shows the representation, it would be great.

The representations have to be general and applicable to all maps with the simplification that only regular maps (no exclaves or enclaves, 3 edges meeting at each vertex, etc.) can be considered.

These are some classic representations:

  • Natural: As a 3-regular planar graph (boundaries = edges)
  • Canonical: As the dual graph of the “natural” representation (region = vertex, neighbors = edges)
  • As a straight line drawing graph (Fáry’s theorem)
  • As a graph with vertices arranged on a grid
  • As a rectilinear cartogram

Plus, I found these:

  • As a circular map
  • As a rectangular map
  • As clefs (derived from rectangular maps)
  • As pipes map (derived from the clefs representation)

New representations (answers from mathoverflow):

Here is an example of some of these representations for the original map shown:

See next post for other representations.

Counting maps


I’ve posted this question on mathoverflow.

Is there a formula to count how many different topological regular maps can be created with n faces (on a sphere)?

  • For “regular” I intend maps in which the boundaries form a 3-regular planar graph
  • For “different” I intend maps that cannot be topologically transformed one into another (faces have to be considered unnamed)

I’ve been looking for a formula, but it is too difficult for me. Maybe it has a simple solution but I don’t see it.

This was my best guess, but I already know that it is not correct because full of symmetries, as it can be verified manually.

General formula:

2\sum _{s_{(f-3)}=2f-5}^{2f-5+2} \text{...}\sum _{s_3=7}^{s_4} \sum _{s_2=5}^{s_3} \sum _{s_1=3}^{s_2} s_1\left(s_1-1\right)\left(s_2-3\right)\left(s_3-5\right)\text{...}\left(s_{(f-3)}-((2f-5)-2)\right)

4 faces = 2\sum _{s_1=3}^5 s_1\left(s_1-1\right)
5 faces = 2\sum _{s_2=5}^7 \sum _{s_1=3}^{s_2} s_1\left(s_1-1\right)\left(s_2-3\right)

Here are the first results that can be found manually (excluding simmetrical maps):

  • 2 faces = 0 possible regular map (an island and the ocean) (not to be counted, because not regular)
  • 3 faces = 1 possible regular map (an island with two regions and the ocean) (two islands and the ocean wouldn’t be regular)
  • 4 faces = 3 possible regular maps (can be verified adding a face from the previous map)
  • 5 faces = 16 possible regular maps (with homeomorphics eliminated)

These are all maps up to 5 faces:

Four color theorem: recap


Recap (some facts about maps and coloring):

  • All regular maps (3-regular planar graphs) can be topologically transformed (represented) as circular or rectangular maps
  • In searching for a solution of the four color problem, it is possible to exclude maps with faces that have less then five neighbours
  • Simplified maps (no faces with less than five neighbours) of 13 faces do not exist!
  • The nunber of proper colorings (not considering permulations of colors) can be count using the “Chromatic polynomial” and dividing the result by 4! (factorial that counts the permutations). Chromatic polynomial is only known for few types of graphs (Triangle K3, Complete graph Kn, Petersen graph, …)
  • Dual graphs derived from maps can be arranged with vertices positioned on a grid (see “Rectangular grid drawings of plane graphs” by Md. Saidur Rahman, Shin-ichi Nakano, Takao Nishizeki)

To implement on the Java application (https://github.com/stefanutti/maps-coloring-java):

Are these different colorings?


UPDATE (18/Apr/2011)

The nunber of proper colorings (not considering permutations of colors) can be count using the “Chromatic polynomial” and dividing the result by 4! (factorial that counts the permutations). But, the chromatic polynomial is only known for few types of graphs (Triangle K3, Complete graph Kn, Petersen graph, …). I haven’t found papers on this problem. Highly propably, since the difficulty to find the Chromatic polynomial of complex graphs, there cannot be a direct formula to calculate how many different colorings exist.

I’ve posted this question here on mathoverflow.

END UPDATE

Take a look at these three graphs:

Are these really different colorings?

Before to dismiss this, do the following on the first graph:

  • Change Blue to Yellow (Blue was just an arbitrary color among infinite others)
  • Change Green to Blue
  • Change Yellow to Green

Now you have the second graph. Continue with this:

  • Change Green to Yellow
  • Change Red to Green
  • Change Yellow to Red

Now you have the third graph. Since the arbitrary nature of the choice of colors, the first graph could have been colored as the third one since the beginning.

The same result, can be obtained by changing all three colors of the first graph to three other colors. If applying the same method to the third graph, you can get the same configuration for the two graphs, it means that they can be considered, for some areas of investigation, as the same configuration.

There are more complex cases in which swapping colors won’t help to get from one coloring to the other.

In the following picture taken from http://en.wikipedia.org/wiki/Graph_coloring, the graphs named (A) and (B) are the only ones that cannot be converted into one another by swapping colors.

In particular I’m interested in regular maps, excluding all maps that can be colored with 2 or 3 colors. For what I need to analyze, maps have to be regarded as differently colored, if the same coloring cannot be obtained by subsequent exchanges of colors.

In other words, for example, once a map has been properly colored, I don’t want to count all other configurations that derive from subsequent exchanges of colors.

Since the arbitrary nature of choosing colors, these derived configurations are equivalent (for what I’m analyzing) to the first one, since they could have been obtained just choosing different colors in the first place. Instead, there are colorings that differ in such a way that exchanging colors won’t help to transform one configuration into the other.

My question is: how many “different” colorings (in the meaning I explained) exist for a given map?

I’ve only found an article on http://en.wikipedia.org/wiki/Graph_coloring that count all possible colorings including swaps.

Is there a paper that can help me on this?

T1 was already known


The theorem I proved in T1 was already known. It was found by Kempe back in 1879 in terms of graph theory (see http://en.wikipedia.org/wiki/Four_color_theorem: “Kempe also showed correctly that G can have no vertex of degree 4″). Only 132 years later … not bad.

People from http://cstheory.stackexchange.com helped me on this (to find that it was already known). See here: http://cstheory.stackexchange.com/questions/5822.

I still think my proof it is of some value, since it is not expressed in terms of graph theory (of the dual graph derived from the map).