Mario Stefanutti
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 Mar15 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. Mar15 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 Mar15 asked How to find all proper colorings (four coloring) of a graph with a brute force algorithm Feb11 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. Dec3 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? Nov25 comment Help with distribution and disposition of faces in a map (four color theorem) Thanks. I'll try to get a copy! Nov25 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) Nov25 revised Help with distribution and disposition of faces in a map (four color theorem) added 38 characters in body Nov25 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: $$F_5=12+F_7+2F_8+3F_9+\cdots$$ Nov25 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. Nov24 asked Help with distribution and disposition of faces in a map (four color theorem) Jun1 awarded Teacher Jun1 answered Publishing elementary proofs of theorems May30 revised Is value of $\pi = 4$? added 358 characters in body May30 answered Is value of $\pi = 4$? May21 answered Interesting properties for numbers 1-20? May19 awarded Scholar May19 accepted Circle packing representation of a given graph May19 revised Circle packing representation of a given graph added 264 characters in body May17 awarded Supporter