Coloring triangles in a Delaunay triangulation on the surface of a 3d sphere.












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Suppose a delaunay triangulation over the surface of a 3d sphere (or generally some 3d surface of something topologically equivalent to the sphere). How many colors do I need to color its triangles so that triangles sharing an edge have different colors?



My idea: For the delaunay triangulation of a set of points on the plane, 3 colors are always enough.



Proof: taking the 1-ring (all the triangles that touch an epsilon small circle around a vertex) 3 colors are always enough to color it.



enter image description here



This is taken from wikipedia. The 4rth point from the bottom, (4rth in the sense of y coordinate) has a 1-ring of size 5, thus I need three colors to color it.



I think the argument still holds for the surface of a sphere. Am I correct?



Will it still hold even if there is genus?










share|cite|improve this question






















  • The small-ring argument would work for general maps, won't it? Even for a tertrahedron ...
    – Hagen von Eitzen
    Nov 23 at 20:10










  • I think yes. Even genus seems irrelevant. Seems a little suspicious, that is why I asked
    – Paramar
    Nov 23 at 20:12










  • The point being if it works for a tetrahedron, it must be wrong. You need four colors to color a tetrahedron.
    – WhatToDo
    Nov 23 at 20:48
















0














Suppose a delaunay triangulation over the surface of a 3d sphere (or generally some 3d surface of something topologically equivalent to the sphere). How many colors do I need to color its triangles so that triangles sharing an edge have different colors?



My idea: For the delaunay triangulation of a set of points on the plane, 3 colors are always enough.



Proof: taking the 1-ring (all the triangles that touch an epsilon small circle around a vertex) 3 colors are always enough to color it.



enter image description here



This is taken from wikipedia. The 4rth point from the bottom, (4rth in the sense of y coordinate) has a 1-ring of size 5, thus I need three colors to color it.



I think the argument still holds for the surface of a sphere. Am I correct?



Will it still hold even if there is genus?










share|cite|improve this question






















  • The small-ring argument would work for general maps, won't it? Even for a tertrahedron ...
    – Hagen von Eitzen
    Nov 23 at 20:10










  • I think yes. Even genus seems irrelevant. Seems a little suspicious, that is why I asked
    – Paramar
    Nov 23 at 20:12










  • The point being if it works for a tetrahedron, it must be wrong. You need four colors to color a tetrahedron.
    – WhatToDo
    Nov 23 at 20:48














0












0








0







Suppose a delaunay triangulation over the surface of a 3d sphere (or generally some 3d surface of something topologically equivalent to the sphere). How many colors do I need to color its triangles so that triangles sharing an edge have different colors?



My idea: For the delaunay triangulation of a set of points on the plane, 3 colors are always enough.



Proof: taking the 1-ring (all the triangles that touch an epsilon small circle around a vertex) 3 colors are always enough to color it.



enter image description here



This is taken from wikipedia. The 4rth point from the bottom, (4rth in the sense of y coordinate) has a 1-ring of size 5, thus I need three colors to color it.



I think the argument still holds for the surface of a sphere. Am I correct?



Will it still hold even if there is genus?










share|cite|improve this question













Suppose a delaunay triangulation over the surface of a 3d sphere (or generally some 3d surface of something topologically equivalent to the sphere). How many colors do I need to color its triangles so that triangles sharing an edge have different colors?



My idea: For the delaunay triangulation of a set of points on the plane, 3 colors are always enough.



Proof: taking the 1-ring (all the triangles that touch an epsilon small circle around a vertex) 3 colors are always enough to color it.



enter image description here



This is taken from wikipedia. The 4rth point from the bottom, (4rth in the sense of y coordinate) has a 1-ring of size 5, thus I need three colors to color it.



I think the argument still holds for the surface of a sphere. Am I correct?



Will it still hold even if there is genus?







general-topology graph-theory coloring triangulation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 23 at 20:07









Paramar

257111




257111












  • The small-ring argument would work for general maps, won't it? Even for a tertrahedron ...
    – Hagen von Eitzen
    Nov 23 at 20:10










  • I think yes. Even genus seems irrelevant. Seems a little suspicious, that is why I asked
    – Paramar
    Nov 23 at 20:12










  • The point being if it works for a tetrahedron, it must be wrong. You need four colors to color a tetrahedron.
    – WhatToDo
    Nov 23 at 20:48


















  • The small-ring argument would work for general maps, won't it? Even for a tertrahedron ...
    – Hagen von Eitzen
    Nov 23 at 20:10










  • I think yes. Even genus seems irrelevant. Seems a little suspicious, that is why I asked
    – Paramar
    Nov 23 at 20:12










  • The point being if it works for a tetrahedron, it must be wrong. You need four colors to color a tetrahedron.
    – WhatToDo
    Nov 23 at 20:48
















The small-ring argument would work for general maps, won't it? Even for a tertrahedron ...
– Hagen von Eitzen
Nov 23 at 20:10




The small-ring argument would work for general maps, won't it? Even for a tertrahedron ...
– Hagen von Eitzen
Nov 23 at 20:10












I think yes. Even genus seems irrelevant. Seems a little suspicious, that is why I asked
– Paramar
Nov 23 at 20:12




I think yes. Even genus seems irrelevant. Seems a little suspicious, that is why I asked
– Paramar
Nov 23 at 20:12












The point being if it works for a tetrahedron, it must be wrong. You need four colors to color a tetrahedron.
– WhatToDo
Nov 23 at 20:48




The point being if it works for a tetrahedron, it must be wrong. You need four colors to color a tetrahedron.
– WhatToDo
Nov 23 at 20:48















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