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.
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
asked Nov 23 at 20:07
Paramar
257111
add a comment |
up vote
0
down vote
favorite
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.
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
asked Nov 23 at 20:07
Paramar
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
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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.
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
asked Nov 23 at 20:07
Paramar
257111
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.
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
general-topology graph-theory coloring triangulation
asked Nov 23 at 20:07
Paramar
257111
asked Nov 23 at 20:07
Paramar
257111
asked Nov 23 at 20:07
Paramar
257111
asked Nov 23 at 20:07
Paramar
257111
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
add a comment |
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
add a comment |
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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