1
$\begingroup$

Imagine a $4\times 4\times 4$ cube, consisting of $64$ cubelets. Each cubelet must be labeled either $1,2,3,$ or $4$ such that if you divide the cube into four parallel $4$x$4$ layers in any dimension, each row and column of each layer has one cubelet with each label.

For example, a layer of the cube could be
$$ \begin{array} & 1 & 2& 3& 4 \\ 2 &3 &4& 1 \\ 3& 4 &1& 2 \\ 4& 1& 2& 3\end{array}$$

You can think of this as a generalization of a sudoku (I apologize for the pun in the title).

My question is: suppose we have constructed such a cube. What is the group of transformations that sends our cube to another such cube? Some subgroups include the group of rotations of the cube (isomorphic to $S_4$) and the group of permutations of parallel layers; each group of such permutations with respect to a single axis is of course $S_4$.

My conjecture is that the answer is in fact $S_4 \times S_4 \times S_4$, since not only have I convinced myself that all permutations of layers (not necessarily with respect to the same axis) in fact commute (though I may be incorrect), I suspect that you can generate any rotation by a composition of permutations of layers.

Also, can anyone determine the total number of distinct ways to label the cube?

$\endgroup$
3
$\begingroup$

You're describing a Latin cube of order $4$, and there are more transformations than $S_4 \times S_4 \times S_4$, such as the rotations you mention (and replacing parallel line with its inverse).

Yes, we can permute the parallel layers to obtain another Latin cube of order $n$, giving $S_n \times S_n \times S_n$.

If $M=m_{ijk}$ is a Latin cube of order $n$, we can write set of entries of $M$ as $$E(M)=\{(i,j,k,m_{ijk}):i,j,k \in [n]\}$$ where $[n]=\{1,2,\ldots,n\}$. We can form the entry sets of other Latin cubes of order $n$ by permuting the coordinates of $E(M)$, e.g., $$\{(k,j,m_{ijk},i):i,j,k \in [n]\}$$ will be the entry set of a conjugate Latin cube.

This gives the group of transformations $S_n \times S_n \times S_n \rtimes S_4$ acting on the set of Latin cubes of order $n$ (generalizing the notion of the paratopy group acting on the set of Latin squares).

There are five inequivalent Latin cubes of order $4$, listed on Prof. Brendan McKay's webpage: Latin cubes and hypercubes.

If transformations other than permuting parallel layers were equivalent to permuting parallel layers, then we'd have an autoparatopy. If they were all equivalent, this would require |Par(A)|/|Is(A)| (in McKay's notation) to be equal to $|S_4|=24$ when $n=4$. This doesn't happen for two of the five Latin cubes of order $4$:

0123 1032 2301 3210 1032 0123 3210 2301 2301 3210 1032 0123 3210 2301 0123 1032  128 4
0123 1032 2310 3201 1032 2310 3201 0123 2310 3201 0123 1032 3201 0123 1032 2310  64 8

In fact, since neither of these |Par(A)|/|Is(A)| values are divisible by $3$, we can conclude that no rotations about a diagonal are equivalent to permuting parallel layers.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I am unfamiliar with the notion of paratopy but this seems like an excellent, thorough answer. Will accept when I fully understand it $\endgroup$ – mikefallopian Jan 13 '16 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.