# Four colour theorem in $3$ dimensions?

There is, of course the 4 colour theorem, which has been proven - every map can be coloured in just 4 colours.

However, has anything been examined in $3$ dimensions? By that, I mean how many different colours would I need to be able to colour in the parts of any cut up $3$D object?

Is it infinite, because any shape can have infinite sides?

• Imagine a bowl of spaghetti where each noodle touches every other noodle. There's no upper bound. Mar 14, 2015 at 11:55

If you take a set of square columns in the z=0 plane, such that there is a set for y=0, y=1, etc. Now take a second set, turned at right angles and at z=1, so you have x=0, x=1, &c. Now join these by pairs at x=n, y=n.

You now have an infinite number of X-shaped figures, each pair neighbouring another exactly twice (ie at i,j and j,i).

So there exists a construction fot infinite colours.

• Surely you could colour that with less than infinite colours, just repeating them? Also, missing ).
– Tim
Mar 14, 2015 at 11:33
• Each piece touches every other piece, so no two can be the same color. Mar 14, 2015 at 11:36
• Even simpler construction: n spheres where every sphere is expanding by a pipe, making connection to the other. So we have construction with n elements in 3-d world connected to each other. Jun 2, 2016 at 8:57

There are meaningful generalizations, if you consider surfaces like sphere, torus, Möbius band etc. as "3D objects". The minimum number of required colors for the mentioned surfaces is 4, 7 and 6, respectively.