# Thread: N-dimensional hypercubes

1. ## N-dimensional hypercubes

I'm not sure where I should put the question, it's not for school or university, but just out of curiosity.

2D space hypercubes are squares. Let's paint the sides of a square with red and blue by the special way: its blue side must be opposite its red side. We can match any two squares painted according to this rule, only rotating them without taking them out of the plane (2D space). It seems we can do similar things with 3D cubes, painting their faces by the similar rule (a blue face opposite to a red face) and then matching them without taking out from 3D space.

So, the question is: is that valid for any N-dimensional hypercubes in any N-Dimensional space?

2. ## Re: N-dimensional hypercubes

What do you mean "match any two squares"? Match them in what way? So that the orientation of red and blue is the same? Are all sides painted?

3. ## Re: N-dimensional hypercubes

Originally Posted by SlipEternal
So that the orientation of red and blue is the same? Are all sides painted?
Both yes.

4. ## Re: N-dimensional hypercubes

I highly doubt that this would work. In general, an n-cube can be "unfolded" into 2n (n-1)-cubes. Those 2n (n-1)-cubes have too many different coloring options. I doubt they would be unique up to rotations.

My guess would be that this painting scheme would work for up to 4-d cubes, but at 5+ dimensions, these painting schemes would be non-unique. This is just a guess, but it is based on the fact that quartic polynomials are solvable, but higher dimensional polynomials are not. These colorations seem similar to solving polynomials in a sense, but I have not thought at all of how that might be so or how to explain it.

5. ## Re: N-dimensional hypercubes

Originally Posted by SlipEternal
an n-cube can be "unfolded" into 2n (n-1)-cubes.
How so? Let's say, 3D cube must have according to this formula 2*3*(3-1) faces, therefore 12 faces. Actually, it has 6 faces. Do you mean something different?

6. ## Re: N-dimensional hypercubes

A 3-cube has 2*3 faces of dimension 3-1

Squares have 2*2 edges of dimension 2-1
4-cubes have 2*4 cubes of dimension 4-1

An n-cube is a cube of dimension n. An (n-1)-cube is a cube of dimension n-1.

7. ## Re: N-dimensional hypercubes

Here's a good one. A 5-cube has 10 4-cube faces. Every 4-cube is adjacent to 8 other 4-cubes and opposite exactly 1. Consider one pair of opposite faces. They are both adjacent to four pairs of opposite 4-cubes. Let's paint those 4 pairs. We will consider only two possible paintings. RRRRBBBB, RBRBBRBR. Now, this is a circular ordering. You can test this out yourself. No matter how you paint the two remaining 4-cubes, there is no rotation that will allow these two painting schemes to match.

8. ## Re: N-dimensional hypercubes

I am not sure I am being clear with the term "circular ordering". What I am picturing is something like this:

$\begin{matrix} & & C & & \\ & B & & D & \\ A & & Z & & A \\ & D & & B & \\ & & C & & \end{matrix}$

Each letter represents a 4-cube face. Face Z is adjacent to all of the other 8 4-cubes. The A cubes are left and right of Z. The C cubes are up and down from Z. The B and D cubes are in the third and fourth spacial directions from Z (left and right prime and up and down prime). What you cannot see is the opposite Z because it is at the "bottom" of this diagram.

Now, you can color each of the letters A through D and their opposite side gets the opposite color. Hopefully this visualization will make it a little clearer what I meant.

9. ## Re: N-dimensional hypercubes

Thanks a lot for the explanation! Now I need to think and sort the things out.
Is there any way to transfer these geometric ideas into algebra and find a solution in which N-dimensional spaces the rule works and in which it doesn't work?

10. ## Re: N-dimensional hypercubes

Btw, for 1D space the rule doesn't work either. We can't match two segments with blue and red ends without taking them out 1D space, if they are a mirror reflection of each other. Probably, that's because rotation doesn't available in 1D space.
I just thought, maybe, there are some forms of rotation in higher dimensional spaces which aren't available in 3D space and which can do the job (I mean matching)?

11. ## Re: N-dimensional hypercubes

Originally Posted by Fox333
Thanks a lot for the explanation! Now I need to think and sort the things out.
Is there any way to transfer these geometric ideas into algebra and find a solution in which N-dimensional spaces the rule works and in which it doesn't work?
Probably. I do not really have the time to work on it, but it is likely going to be solutions to certain polynomials of degree equal to the dimension of the cube you are considering. So, as stated in post #4, you can probably make it work up to 4-d cubes, and any dimension higher than that will not work. I have not thought about which polynomials would be used to represent specific painting schemes. This problem can be converted to a graph theory problem by considering isomorphisms of graph colorings. The idea would be that you have 2n vertices, each vertex is incident to every other vertex except its "opposite" face. Then, you create two modified graphs generated by coloring the vertices where opposite vertices are colored opposite colors. The question becomes given two arbitrary colorings, does there exist an isomorphism between the two. The scholarly writings about this likely encompass this exact topic, but I have not researched it specifically and would not have a recommendation where to look. Perhaps another graph theorist on the forums has seen something like this? Based on my own studies, these problems tend to break down into polynomials of degree equal to the degree of the object I am viewing.

Here is some good reading about solving higher degree polynomials:
https://math.stackexchange.com/quest...are-unsolvable

Here is the article explaining graph coloring:
https://en.wikipedia.org/wiki/Graph_coloring

This may be over your head, but this is an attempt to look at isomorphisms between colorings bounded by color multiplicity of 3 (a different problem than the one you are looking at):
http://pages.cs.wisc.edu/~dieter/Papers/3gi.pdf

12. ## Re: N-dimensional hypercubes

Btw, I applied the method you suggested for a 5D hypercube to a 4D hypercube. I found out then it doesn't meet the rule either. Am I right?

The circles represent the tesseract's cubic faces, the curves between them are squares, shared by those faces. A and B letters represent opposite faces.

13. ## Re: N-dimensional hypercubes

That is correct.

14. ## Re: N-dimensional hypercubes

In general, for an n-cube, there are $2^n$ different painting schemes. This comes from if you choose n of the faces (none of them opposites), you can paint each face red or blue, so there are two choices for each face.
For the symmetries of the n-cube, look at the hyperoctohedral group. It has order $2^n(n!)$. Although the hyperoctohedral group includes symmetries for the n-cube as though it could be inverted (like picture a square, then flip it over and you still have a square). So, you probably need a different group that the hyperoctohedral group, but I'm not quite sure which one would work.

Once you find the correct group of symmetries, there may be a way to mod out the group of symmetries by the coloring schemes to determine the number of possible coloring schemes that are distinct after symmetries.

Edit: I found this about the group of rotational symmetries of the n-cube.
https://math.stackexchange.com/quest...ymmetric-group

15. ## Re: N-dimensional hypercubes

Originally Posted by Fox333
Btw, for 1D space the rule doesn't work either. We can't match two segments with blue and red ends without taking them out 1D space, if they are a mirror reflection of each other. Probably, that's because rotation doesn't available in 1D space.
I just thought, maybe, there are some forms of rotation in higher dimensional spaces which aren't available in 3D space and which can do the job (I mean matching)?
A "cube" in 1D space is a line segment. The two ends are the 0D endpoints. You color one of them blue and one of them red.

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