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?


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.

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  • $\begingroup$ 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

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