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The four-color theorem is typically formulated for planar figures. It also applies to the surface of anything homeomorphic to a sphere, because a stereographic
by dietrichepp 14y ago
The four-color theorem is typically formulated for planar figures. It also applies to the surface of anything homeomorphic to a sphere, because a stereographic projection will preserve the four-color property. It does not apply to other shapes -- a simple torus may need seven colors. Adding holes lets you increase the number of required colors without bound.
- deleted 14y ago[deleted]
- xentronium 14y agoIs there a statement for polyhedra similar to four-color theorem? I could only google a book called "Map Coloring Polyhedra and the Four Color Problem". Unfortunately, it's not available to read online. From what I've found, it seems to me that there is only a handful of proven facts about colorings of some polyhedra. [ADD]: four color theorem works for every planar graph and it looks like one can make such graph corresponding to any convex polyhedron, so it seems that original OP statement holds.
- deleted 14y ago[deleted]
- Vvector 14y agoJust make a hole in the center of one side of the polyhedra. Then stretch out the hole, flattening the polyhedra in the process. Eventually you get a planar map, while retaining the same topology as the original polyhedra.
- dietrichepp 14y agoOr, in math-speak, the polyhedron is homeomorphic to the sphere, and a stereographic projection of the sphere will preserve the four-color property while projecting the sphere to a plane. I first thought, "That's exactly what I said," and then I realized that "homeomorphic" isn't exactly a household word.