Mario Stefanutti
Reputation
445
17/100 score
 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) Jun 1 awarded Teacher Jun 1 answered Publishing elementary proofs of theorems May 30 revised Is value of $\pi = 4$? added 358 characters in body May 30 answered Is value of $\pi = 4$? May 19 awarded Scholar May 19 accepted Circle packing representation of a given graph May 19 revised Circle packing representation of a given graph added 264 characters in body May 17 awarded Supporter May 16 asked Circle packing representation of a given graph Apr 13 awarded Editor Apr 13 revised How many different four coloring exist for a given regular map? added 235 characters in body Apr 7 comment How many different four coloring exist for a given regular map? In other words, for example, for maps with hundreds of faces, once a map has been properly four colored, I don't want to count all other colorings that derive from subsequent exchanges of colors. My doubt is how many "different" (in the meaning I explained) coloring exist for a given map? Is there a study on this that can help me?