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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?