335 reputation
26
bio website 4coloring.wordpress.com
location Rome, Italy
age 45
visits member for 3 years
seen Apr 16 at 10:36

Jun
15
asked Four color theorem, 3-regular planar graph, Hamiltonian path and spiral chains
May
30
answered Can the borders of a map be deformed to give arbitrary area to any region?
Mar
17
accepted How many different four coloring exist for a given regular map?
Mar
16
accepted Left or right edge in cubic planar graph
Mar
16
comment Left or right edge in cubic planar graph
Your argument about mirroring the planar image of the graph convinced me. What a pity! I think I have to go the other way and elaborate the image by pattern recognition. In my case, since I have a particular kind of maps, it is not going to be so difficult. Thanks!
Mar
16
comment Left or right edge in cubic planar graph
I am trying to identify some particular chains of a cubic planar graph (see 4coloring.wordpress.com/spiral-chains). To get these chains I was considering to convert the graph that I have into a graph theory model in memory (adjacency list) and then apply the algorithm to find the chain, but is it possible algorithmically to identify left and right edges? For example I know it is possible to use an algorithm to test planarity (en.wikipedia.org/wiki/Planarity_testing), but is it possible to do something similar to identify left and right edges?
Mar
16
asked Left or right edge in cubic planar graph
Mar
16
comment How to find all proper colorings (four coloring) of a graph with a brute force algorithm
Yes it does, but that is exactly why I'm looking for an algorithm. I'm working with graphs that have 30-50 vertices and that makes impossible to find all possible colorings and then filter them out. I'l like to write something similar to the function colorIt() inside this java code: - maps-coloring.git.sourceforge.net/git/… implementing for example a new method called findAllColorings()
Mar
15
awarded  Commentator
Mar
15
comment How to find all proper colorings (four coloring) of a graph with a brute force algorithm
How do I find all functions f:V→{1,2,3,4}? All possible functions, without filtering them first (for not valid coloring), seem to be a very high number of solutions.
Mar
15
comment How to find all proper colorings (four coloring) of a graph with a brute force algorithm
Wow, how fast! I'll take a look into it. For my level it seems too difficult to understand and implement, but i'll try to see what I can do. I was hoping for a simpler approach to the problem! Thanks for the info
Mar
15
asked How to find all proper colorings (four coloring) of a graph with a brute force algorithm
Feb
11
comment planar embedding a graph
Draw your graph on the sphere. Your edge $e$ divides two faces. Choose one of these two faces. Now imagine the sphere made of very elastic rubben, make a small hole to the chosen face and blow into it. Enlarge the hole and flatten the balloon (the sphere) to adhere to the plane. The face you chosed is now the outer face of the graph.
Dec
3
comment Help with distribution and disposition of faces in a map (four color theorem)
Do you know if there is an online version of this paper or an equivalent study on the subject?
Nov
25
comment Help with distribution and disposition of faces in a map (four color theorem)
Thanks. I'll try to get a copy!
Nov
25
comment Help with distribution and disposition of faces in a map (four color theorem)
I used a computer program I am building to create simplified maps (sourceforge.net/projects/maps-coloring). The program did not return any map of 13 faces. At the beginning I thought to a bug in the software, then I sketched a proof I thing is right (4coloring.wordpress.com/open-points-and-notes)
Nov
25
revised Help with distribution and disposition of faces in a map (four color theorem)
added 38 characters in body
Nov
25
comment Help with distribution and disposition of faces in a map (four color theorem)
$$F = F_2+F_3+F_4+\cdots$$ $$2E = 3V = 2F_2+3F_3+4F_4+\cdots$$ Since a region bounded by n edges has n vertices and each vertex belongs to three regions, by Euler's formula V-E+F = 2 we have: $$6V-6E+6F = 12$$ $$4E-6E+6F = 12$$ $$6F-2E = 12$$ $$6(F_2+F_3+F_4+\cdots)-(2F_2+3F_3+4F_4+\cdots) = 12$$ $$4F_2+3F_3+2F_4+F_5+0F_6-F_7-2F_8-3F_9-\cdots = 12$$ That becomes: $$F_5=12+F_7+2F_8+3F_9+\cdots$$
Nov
25
comment Help with distribution and disposition of faces in a map (four color theorem)
I am pretty sure the identity is: $$F_5=12+F_7+2F_8+\cdots$$ Why the ">"? About the simplified map with 13 faces, I verified manually (computer program) that it does not exist. I agree that 13 pentagons is not possible because it would lead to the identity to be 13 = 12. Any other face with more than 6 edges would not work either. If we try with an $F_7$, the equilibrium would bend on the right of the equality and we would need 13 faces of type $F_5$ to balance it. The only possible solutions is to have 12 faces of type $F_5$ + 1 face of type $F_6$ = 13 faces. But these maps do not exist.
Nov
24
asked Help with distribution and disposition of faces in a map (four color theorem)