[1]organization=School of Mathematics and Physics, ARC Centre of Excellence,
Plant Success in Nature and Agriculture, addressline=The University of
Queensland, city=Brisbane, postcode=4072, country=Australia

# Latin hypercubes realizing integer partitions

Diane Donovan<EMAIL_ADDRESS>Tara Kemp<EMAIL_ADDRESS>James Lefevre
<EMAIL_ADDRESS>
###### Abstract

For an integer partition $h_{1}+\dots+h_{n}=N$, a 2-realization of this
partition is a latin square of order $N$ with disjoint subsquares of orders
$h_{1},\dots,h_{n}$. The existence of 2-realizations is a partially solved
problem posed by Fuchs. In this paper, we extend Fuchs’ problem to $m$-ary
quasigroups, or, equivalently, latin hypercubes. We construct latin cubes for
some partitions with at most two distinct parts and highlight how the new
problem is related to the original.

###### keywords:

latin square , permutation cube , realization

##  1 Introduction

L. Fuchs asked the following question [7]: If $n$ is any positive integer and
$n=n_{1}+n_{2}+\dots+n_{k}$ any fixed partition of $n$, is it possible to find
a quasigroup $(Q,\circ)$ of order $n$ which contains subquasigroups on the
sets $Q_{1},Q_{2},\dots,Q_{k}$ of orders $n_{1},n_{2},\dots,n_{k}$
respectively, whose set theoretical union is $Q$? The problem of finding such
a quasigroup is equivalent to finding a latin square of order $n$ with
disjoint subsquares of orders $n_{1},n_{2},\dots,n_{k}$ (these subsquares must
be row, column and symbol disjoint).

It is easy to find partitions for which such a quasigroup does not exist,
however the problem of existence is far from complete. Much work has been done
by considering latin squares, which are equivalent structures to quasigroups,
and such a latin square is known as a realization or partitioned incomplete
latin square (PILS).

For partitions with at most four parts, existence has been completely
determined. Many cases have been studied for partitions with 5 or 6 parts,
however the problem is not yet solved. For any partition, existence has been
determined when the parts are of at most two distinct sizes.

The same problem can be posed for $m$-ary quasigroups, which are equivalent to
$m$-dimensional latin hypercubes. In this paper, we investigate this problem
by first considering latin cubes and also by examining the relationships
between latin hypercubes of different dimenions.

######  Definition 1.1.

A _latin square_ of order $N$ is an $N\times N$ array, $L$, on a set of $N$
symbols such that each symbol occurs exactly once in each row and column. An
entry of $L$ is defined by its row and column position as $L(i,j)$.

An $m\times m$ sub-array of $L$ that is itself a latin square of order $m$ is
a _subsquare_ of $L$. Subsquares are disjoint if they share no row, column or
symbol of $L$.

For any positive integer $a$, we define $[a]$ to be the set of integers
$\\{1,2,\dots,a\\}$. Given another integer $b$, $b+[a]$ is the set
$\\{b+1,b+2,\dots,b+a\\}$.

We will assume that all latin squares of order $N$ use the symbols of $[N]$.

######  Ex. 1.2.

The array in Figure 1 is a latin square of order 5, and the highlighted cells
indicate a subquare of order 2 and three subsquares of order 1.

$\begin{array}[]{|c|c|c|c|c|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&4&5&3\\\
\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}1&5&3&4\\\ \hline\cr
4&5&\pagecolor{lightgray}3&2&1\\\ \hline\cr 5&3&1&\pagecolor{lightgray}4&2\\\
\hline\cr 3&4&2&1&\pagecolor{lightgray}5\\\ \hline\cr\end{array}$

Figure 1:

######  Definition 1.3.

An _integer partition_ of $N$ is a sequence of non-increasing integers
$P=(h_{1},h_{2},h_{3},\dots)$ where $\sum_{i=1}^{\infty}h_{i}=N$.

An integer partition $P=(h_{1},h_{2},\dots,h_{n})$ will be represented by
$(h_{1}h_{2}\dots h_{n})$ or
$(h_{i_{1}}^{\alpha_{1}}h_{i_{2}}^{\alpha_{2}}\dots h_{i_{m}}^{\alpha_{m}})$,
where there are $\alpha_{j}$ copies of $h_{i_{j}}$ in the partition.

######  Definition 1.4.

For an integer partition $P=(h_{1},h_{2},\dots,h_{n})$ of $N$ with $h_{1}\geq
h_{2}\geq\dots\geq h_{n}>0$, a _2-realization_ of $P$, denoted
$\operatorname{2-RP}(h_{1}h_{2}\dots h_{n})$, is a latin square of order $N$
with pairwise disjoint subsquares of order $h_{i}$ for each $i\in[n]$. We say
that a 2-realization is in _normal form_ if the subsquares are along the main
diagonal and the $i^{th}$ subsquare, $H_{i}$, is of order $h_{i}$.

The latin square in Example 1.2 is a 2-realization of the partition
$(2^{1}1^{3})$.

The existence of 2-realizations for $n=1,2,3,4$ has been completely determined
in [4].

######  Theorem 1.5.

Take a partition $(h_{1},h_{2},\dots,h_{n})$ of $N$ with $h_{1}\geq
h_{2}\geq\dots\geq h_{n}>0$. Then a $\operatorname{2-RP}(h_{1}h_{2}\dots
h_{n})$

  * 1.

always exists when $n=1$;

  * 2.

never exists when $n=2$;

  * 3.

exists when $n=3$ if and only if $h_{1}=h_{2}=h_{3}$;

  * 4.

exists when $n=4$ if and only if $h_{1}=h_{2}=h_{3}$, or $h_{2}=h_{3}=h_{4}$
with $h_{1}\leq 2h_{4}$.

The problem has also been solved for partitions with subsquares of at most two
distinct orders. Dénes and Pásztor [2] gave the result for subsquares all of
the same order and the result for two different orders was considered by
Heinrich [3] and finished by Kuhl and Schroeder [9].

######  Theorem 1.6.

For $n\geq 1$ and $a\geq 1$, a $\operatorname{2-RP}(a^{n})$ exists if and only
if $n\neq 2$.

The following theorem completes the problem for all partitions with at most
two distinct orders, considering only the cases which haven’t already been
covered in the previous theorems.

######  Theorem 1.7.

For $a>b>0$ and $n>4$, a $\operatorname{2-RP}(a^{u}b^{n-u})$ exists if and
only if $u\geq 3$, or $0<u<3$ and $a\leq(n-2)b$.

Colbourn [1] has determined two necessary, but not sufficient, conditions for
the existence of a 2-realization.

######  Theorem 1.8.

If a $\operatorname{2-RP}(h_{1}h_{2}\dots h_{n})$ exists, then
$h_{1}\leq\sum_{i=3}^{n}h_{i}$.

######  Theorem 1.9.

If a $\operatorname{2-RP}(h_{1}h_{2}\dots h_{n})$ exists, then

$N^{2}-\sum_{i=1}^{n}h_{i}^{2}\geq 3\Bigg{(}\sum_{i\in
D}h_{i}\Bigg{)}\Bigg{(}\sum_{j\in\overline{D}}h_{j}\Bigg{)}$

for any $D\subseteq\\{1,2,\dots,n\\}$ and
$\overline{D}=\\{1,2,\dots,n\\}\setminus D$.

Fuchs’ problem considers subgroups of quasigroups with binary operations which
are equivalent to latin squares, and so this problem can be extended to
quasigroups with ternary operations by looking at latin cubes. These are also
known as permutation cubes.

######  Definition 1.10.

A _latin cube_ $C$ of order $N$ is an $N\times N\times N$ array on a symbol
set of size $N$, with entries defined by $C(j_{1},j_{2},j_{3})$, such that if
$(j_{1},j_{2},j_{3})$ and $(j^{\prime}_{1},j^{\prime}_{2},j^{\prime}_{3})$
differ in exactly one position, then $C(j_{1},j_{2},j_{3})\neq
C(j^{\prime}_{1},j^{\prime}_{2},j^{\prime}_{3})$.

If $C$ is instead an $r\times s\times t$ array with $r,s,t\leq N$, then we
call $C$ a _latin box_ of order $r$ by $s$ by $t$.

We will consider a _line_ of $C$ to be a set of $N$ cells with two fixed
coordinates. A _column_ , _row_ or _file_ is thus a line represented by
$([N],j_{2},j_{3})$, $(j_{1},[N],j_{3})$ or $(j_{1},j_{2},[N])$ respectively.
A _layer_ of $C$ is the set of cells with only one fixed coordinate. Layers
can thus be represented as $(j_{1},[N],[N])$, $([N],j_{2},[N])$ or
$([N],[N],j_{3})$.

Similar to latin squares, a _subcube_ of $C$ is an $m\times m\times m$ sub-
array which is itself a latin cube of order $m$. Subcubes are disjoint if they
share no row, column, file or symbol.

As with latin squares, we will assume that all latin cubes use the symbols of
$[N]$.

######  Definition 1.11.

For an integer partition $P=(h_{1},h_{2},\dots,h_{n})$ of $N$ with $h_{1}\geq
h_{2}\geq\dots\geq h_{n}>0$, a _3-realization_ of $P$, denoted
$\operatorname{3-RP}(h_{1}h_{2}\dots h_{n})$, is a latin cube of order $N$
with pairwise disjoint subcubes of order $h_{i}$ for each $i\in[n]$.

As with 2-realizations, a 3-realization is in _normal form_ if the subcubes
are along the main diagonal (cells of the form $(i,i,i)$ for all $i\in[N]$ and
the additional cells needed to form the subsquares) and the $i^{th}$ subcube,
$H_{i}$, is of order $h_{i}$.

######  Ex. 1.12.

The arrays in Figure 2 are the layers of a $\operatorname{3-RP}(2^{2}1^{1})$.
The subcubes are highlighted.

$\begin{array}[]{|c|c|c|c|c|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&3&5&4\\\
\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}1&4&3&5\\\ \hline\cr
3&4&5&1&2\\\ \hline\cr 4&5&1&2&3\\\ \hline\cr 5&3&2&4&1\\\
\hline\cr\end{array}$
$\begin{array}[]{|c|c|c|c|c|}\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}1&4&3&5\\\
\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&5&4&3\\\ \hline\cr
5&3&1&2&4\\\ \hline\cr 3&4&2&5&1\\\ \hline\cr 4&5&3&1&2\\\
\hline\cr\end{array}$ $\begin{array}[]{|c|c|c|c|c|}\hline\cr 4&3&5&2&1\\\
\hline\cr 3&5&2&1&4\\\ \hline\cr
1&2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&5\\\ \hline\cr
5&1&\pagecolor{lightgray}4&\pagecolor{lightgray}3&2\\\ \hline\cr 2&4&1&5&3\\\
\hline\cr\end{array}$ $\begin{array}[]{|c|c|c|c|c|}\hline\cr 5&4&2&1&3\\\
\hline\cr 4&3&1&5&2\\\ \hline\cr
2&5&\pagecolor{lightgray}4&\pagecolor{lightgray}3&1\\\ \hline\cr
1&2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&5\\\ \hline\cr 3&1&5&2&4\\\
\hline\cr\end{array}$ $\begin{array}[]{|c|c|c|c|c|}\hline\cr 3&5&1&4&2\\\
\hline\cr 5&4&3&2&1\\\ \hline\cr 4&1&2&5&3\\\ \hline\cr 2&3&5&1&4\\\ \hline\cr
1&2&4&3&\pagecolor{lightgray}5\\\ \hline\cr\end{array}$

Figure 2:

The problem of existence of a $\operatorname{3-RP}(h_{1}h_{2}\dots h_{n})$ is
different to that of a $\operatorname{2-RP}(h_{1}h_{2}\dots h_{n})$. To see
this, consider the latin cube in Figure 3. This latin cube is a 3-realization
of the partition $P=(1,1)$, so a $\operatorname{3-RP}(1^{2})$. From Theorem
1.5, we know that a $\operatorname{2-RP}(1^{2})$ does not exist. Thus, there
are partitions which can be realized in latin cubes but not in latin squares.

1221

2112

Figure 3:

##  2 Constructions

We now give some construction methods for latin cubes which will be used later
to find 3-realizations. Construction 2.1 is quite general, but this allows it
to be used to construct more specific latin cubes in later sections.
Construction 2.2 is a generalisation of a construction method used for latin
squares.

######  Construction 2.1.

Given any order $N$ latin square $L$, we can construct an order $N$ latin cube
$C$ as follows.

Let the entries of $L$ be determined by $L(r,c)$. Construct $C$ as
$C(r,c,l)=L(L(r,l),c)$. Observe that if $C(r,c,l)=C(r^{\prime},c,l)$ then
$L(L(r,l),c)=L(L(r^{\prime},l),c)$, and since $L$ is a latin square,
$L(r,l)=L(r^{\prime},l)$. Again, this implies that $r=r^{\prime}$. If
$C(r,c,l)=C(r,c^{\prime},l)$, then $L(L(r,l),c)=L(L(r,l),c^{\prime})$, and
since $L$ is a latin square, $c=c^{\prime}$. Also, if
$C(r,c,l)=C(r,c,l^{\prime})$, then $L(L(r,l),c)=L(L(r,l^{\prime}),c)$. As
before, this implies that $L(r,l)=L(r,l^{\prime})$ and so $l=l^{\prime}$.

Thus, $C$ is a latin cube.

######  Construction 2.2.

Given any latin cube, $C$, of order $N$, we can construct a latin cube of
order $tN$, for any $t\in\mathbb{Z}^{+}$.

To do this, take a latin cube $T$ of order $t$ on the symbol set
$\\{1,2,\dots,t\\}$ and an empty $tN\times tN$ array $C^{\prime}$. For every
entry $C(i,j,k)$ of $C$, place a copy of $T$ with the symbols
$(C(i,j,k)-1)t+[t]$ across the cells
$C^{\prime}(i^{\prime},j^{\prime},k^{\prime})$ of $C^{\prime}$ where
$i^{\prime}\in(i-1)t+[t]$, $j^{\prime}\in(j-1)t+[t]$ and
$k^{\prime}\in(k-1)t+[t]$.

In this way, every entry of $C$ corresponds to a copy of $T$ in $C^{\prime}$
with a set of symbols determined by the symbol in $C$.

This is the analogue of a construction method for latin squares known as
_inflation_.

This method of construction gives some immediate results for 3-realizations.
Similarly to what can be done for latin squares, if a latin cube has a
transversal of $N$ symbols on the main diagonal, then an inflation of this by
$a$ will be a $\operatorname{3-RP}(a^{N})$, since the subsquares correspond to
copies of $T$ with distinct symbols. Also, if a
$\operatorname{3-RP}(h_{1}\dots h_{n})$ exists, then an inflation of this is a
$\operatorname{3-RP}((ah_{1})^{1}\dots(ah_{n})^{1})$.

_Prolongation_ is another method which has been used to find $2-$realizations.
The article [5] defines a generalisation of this method for latin cubes but it
is unknown whether it can be used for latin cubes of any order.

###  2.1 Outline Rectangles and Boxes

######  Definition 2.3.

Given partitions $P,Q,R$ of $N$, where $P=(p_{1},\dots,p_{u})$,
$Q=(q_{1},\dots,q_{v})$, $R=(r_{1},\dots,r_{t})$, let $O$ be a $u\times v$
array of multisets, with elements from $[t]$. For $i\in[u]$ and $j\in[v]$, let
$O(i,j)$ be the multiset of symbols in cell $(i,j)$ and let $|O(i,j)|$ be the
number of symbols in the cell, including repetition.

Then $O$ is an _outline rectangle_ associated to $(P,Q,R)$ if

  1. 1.

$|O(i,j)|=p_{i}q_{j}$, for all $(i,j)\in[u]\times[v]$;

  2. 2.

symbol $l\in[t]$ occurs $p_{i}r_{l}$ times in the row $(i,[v])$;

  3. 3.

symbol $l\in[t]$ occurs $q_{j}r_{l}$ times in the column $([u],j)$.

######  Ex. 2.4.

For partitions $P=(4^{1}1^{2})$, $Q=(3^{1}2^{1}1^{1})$ and $R=(2^{3})$, the
array in Figure 4 is an outline rectangle associated to $(P,Q,R)$.

$\begin{array}[]{|ccc|cc|c|}\hline\cr 1&1&1&2&2&1\\\ 1&1&1&2&2&1\\\
2&2&3&3&3&2\\\ 3&3&3&3&3&2\\\ \hline\cr 2&2&3&1&1&3\\\ \hline\cr
2&2&3&1&1&3\\\ \hline\cr\end{array}$

Figure 4:

######  Definition 2.5.

Given partitions $P,Q,R$ of $N$, where $P=(p_{1},\dots,p_{u})$,
$Q=(q_{1},\dots,q_{v})$ and $R=(r_{1},\dots,r_{w})$, and a latin square $L$ of
order $N$, the _reduction modulo $(P,Q,R)$_ of $L$, denoted $O$, is the
$u\times v$ array of multisets obtained by amalgamating rows
$(p_{1}+\dots+p_{i-1})+[p_{i}]$ for all $i\in[u]$, columns
$(q_{1}+\dots+q_{j-1})+[q_{j}]$ for all $j\in[v]$, and symbols
$(r_{1}+\dots+r_{k-1})+[r_{k}]$ for all $k\in[w]$.

When amalgamating symbols, for $k\in[w]$ we will map all symbols in
$(r_{1}+\dots+r_{k-1})+[r_{k}]$ to symbol $k$.

######  Ex. 2.6.

The second array in Figure 5 is a reduction modulo $(P,Q,R)$ of the latin
square in Example 1.2 (the left array) with $P=(2^{2}1^{1})$, $Q=(4^{1}1^{1})$
and $R=(2^{1}1^{3})$. In this reduction of $L$, symbols 1 and 2 have been
amalgamated into symbol 1, and for the remaining symbols, symbol $i$ became
$i-1$.

$\begin{array}[]{|c|c|c|c|c|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&4&5&3\\\
\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}1&5&3&4\\\ \hline\cr
4&5&\pagecolor{lightgray}3&2&1\\\ \hline\cr 5&3&1&\pagecolor{lightgray}4&2\\\
\hline\cr 3&4&2&1&\pagecolor{lightgray}5\\\ \hline\cr\end{array}$
$\begin{array}[]{|cccc|c|}\hline\cr 1&1&1&1&2\\\ 2&3&4&4&3\\\ \hline\cr
1&1&2&2&1\\\ 3&3&4&4&1\\\ \hline\cr 1&1&2&3&4\\\ \hline\cr\end{array}$

Figure 5:

The following theorem, and its proof, can be found in Hilton [6].

######  Theorem 2.7.

For every outline rectangle $O$, there is a latin square $L$ of order $N$ and
partitions $P$, $Q$ and $R$ such that $O$ is the reduction modulo $(P,Q,R)$ of
$L$.

This result provides a construction method which has been used to prove the
existence of many 2-realizations. To construct a latin square $L$, we do not
need to find $L$ itself. It is sufficient, by Theorem 2.7, to find an outline
rectangle $O$ which is a reduction of $L$. For a realization of partition
$P=(h_{1},\dots,h_{n})$, a reduction modulo $(P,P,P)$ with each cell $(i,i)$
on the main diagonal filled with $h_{i}^{2}$ copies of symbol $i$ will force
the resulting latin square $L$ to have the required subsquares.

Given the usefulness of outline rectangles in finding 2-realizations, we will
define the analogous concept for latin cubes.

######  Definition 2.8.

Given partitions $P,Q,R,S$ of $N$, where $P=(p_{1},\dots,p_{u})$,
$Q=(q_{1},\dots,q_{v})$, $R=(r_{1},\dots,r_{w})$, $S=(s_{1},\dots,s_{t})$, let
$O$ be a $u\times v\times w$ array of multisets, with elements from $[t]$. For
$i\in[u]$, $j\in[v]$ and $k\in[w]$, let $O(i,j,k)$ be the multiset of symbols
in cell $(i,j,k)$ and let $|O(i,j,k)|$ be the number of symbols in the cell,
including repetition.

Then $O$ is an _outline box_ associated to $(P,Q,R,S)$ if

  1. 1.

symbol $l\in[t]$ occurs $N^{2}s_{l}$ times in $O$;

  2. 2.

$|O(i,j,k)|=p_{i}q_{j}r_{k}$, for all $(i,j,k)\in[u]\times[v]\times[w]$;

  3. 3.

symbol $l\in[t]$ occurs $p_{i}q_{j}s_{l}$ times in the line $(i,j,[w])$;

  4. 4.

symbol $l\in[t]$ occurs $p_{i}r_{k}s_{l}$ times in the line $(i,[v],k)$;

  5. 5.

symbol $l\in[t]$ occurs $q_{j}r_{k}s_{l}$ times in the line $([u],j,k)$.

######  Definition 2.9.

For partitions $P,Q,R,S$ of $N$, where $P=(p_{1},\dots,p_{u})$,
$Q=(q_{1},\dots,q_{v})$ $R=(r_{1},\dots,r_{w})$ and $S=(s_{1},\dots,s_{t})$,
and a latin cube $C$ of order $N$, the _reduction modulo $(P,Q,R,S)$_ of $C$,
denoted $O$, is the $u\times v\times w$ array of multisets obtained by
amalgamating layers $(j_{1},[N],[N])$, $([N],j_{2},[N])$ and $([N],[N],j_{3})$
for all $j_{1}\in(p_{1}+\dots+p_{i-1})+[p_{i}]$ for all $i\in[u]$,
$j_{2}\in(q_{1}+\dots+q_{j-1})+[q_{j}]$ for all $j\in[v]$ and
$j_{3}\in(r_{1}+\dots+r_{k-1})+[r_{k}]$ for all $k\in[w]$, and symbols
$(s_{1}+\dots+s_{l-1})+[s_{l}]$ for all $l\in[t]$.

Unlike with outline rectangles, these outline boxes are not always the
reduction of a latin cube. Kochol [8] has shown that not all $k\times n\times
n$ latin boxes can be extended to a latin cube and we will use this result in
the following proof.

######  Theorem 2.10.

Not every outline box $O$ associated to $(P,Q,R,S)$ is the reduction of a
latin cube $C$.

###### Proof.

Suppose that $O$ is the reduction of a latin cube $C$ modulo $(P,Q,R,S)$.

Consider the outline box where $Q=R=S=(1^{N})$ and $P=(N-k,1^{k})$. Then $O$
is obtained by amalgamating $N-k$ layers of $C$. Thus, $O$ has $k$ layers
which form a $k\times N\times N$ latin box. Since $C$ only has those layers
amalgamated, $C$ can be obtained by completing this latin box to a full latin
cube. However, from above, this is not always possible.

Thus, there are outline boxes which are not the reduction of a latin cube. ∎

######  Ex. 2.11.

The outline box associated to $(P,Q,R,S)$ given in Figure 6, with
$P=R=S=(1^{4})$ and $Q=(2^{2})$ and each array representing a layer, is not
the reduction of a latin cube.

$\begin{array}[]{|cc|cc|}\hline\cr 3&4&1&2\\\ \hline\cr 3&4&1&2\\\ \hline\cr
1&2&3&4\\\ \hline\cr 1&2&3&4\\\ \hline\cr\end{array}$
$\begin{array}[]{|cc|cc|}\hline\cr 2&4&1&3\\\ \hline\cr 2&3&1&4\\\ \hline\cr
1&3&2&4\\\ \hline\cr 1&4&2&3\\\ \hline\cr\end{array}$
$\begin{array}[]{|cc|cc|}\hline\cr 1&3&2&4\\\ \hline\cr 1&2&3&4\\\ \hline\cr
2&4&1&3\\\ \hline\cr 3&4&1&2\\\ \hline\cr\end{array}$
$\begin{array}[]{|cc|cc|}\hline\cr 1&2&3&4\\\ \hline\cr 1&4&2&3\\\ \hline\cr
3&4&1&2\\\ \hline\cr 2&3&1&4\\\ \hline\cr\end{array}$

Figure 6:

##  3 Some conditions on 3-realizations

In this section, we give some conditions on the existence of a
$\operatorname{3-RP}(h_{1}h_{2}\dots h_{n})$. We begin with a theorem that
shows an important relationship between 2-realizations and 3-realizations.

######  Theorem 3.1.

If there exists a $\operatorname{2-RP}(h_{1}\dots h_{n})$, then there exists a
$\operatorname{3-RP}(h_{1}\dots h_{n})$.

###### Proof.

Suppose that $L$ is a $\operatorname{2-RP}(h_{1}\dots h_{n})$. First, we
assume that $L$ is in normal form and that the entries of subsquare $H_{i}$,
corresponding to $h_{i}$, are in $S_{i}=(h_{1}+\dots+h_{i-1})+[h_{i}]$. Thus,
for all $(r,c)\in S_{i}\times S_{i}$, we have $L(r,c)\in S_{i}$. If this is
not the case, then the rows, columns and symbols can be permuted to satisfy
this condition.

We construct a latin cube $C$ using the method given in Construction 2.1.

We now check that the subcubes are maintained in $C$. For any part $h_{i}$, if
$(r,c,l)\in S_{i}\times S_{i}\times S_{i}$ then $L(r,l)=a\in S_{i}$ and
$L(a,c)\in S_{i}$ by our earlier assumption. Thus, $C(r,c,l)=L(L(r,l),c)\in
S_{i}$. Therefore, the cubes formed by the columns, rows and layers in $S_{i}$
are filled with only the elements of $S_{i}$. So $C$ has the required
subcubes. ∎

This theorem means that we can use all results on the existence of
$\operatorname{2-RP}(h_{1}h_{2}\dots h_{n})$ when solving the problem in three
dimensions.

We now provide analogues to the two necessary conditions given for
$\operatorname{2-RP}(h_{1}h_{2}\dots h_{n})$.

The first condition is similar to that of Theorem 1.8, by placing a bound on
the size of the largest subcube.

######  Theorem 3.2.

If a $\operatorname{3-RP}(h_{1}\dots h_{n})$ exists and $n\geq 2$, then
$h_{i}\leq\frac{N}{2}$ for all $1\leq i\leq n$.

###### Proof.

Suppose that there is a latin cube $\operatorname{3-RP}(h_{1}\dots h_{n})$.
Rearrange the layers of the cube so that the subcube of order $h_{i}$ is in
the cells $(i,j,k)$ with $i,j,k\in[h_{i}]$, making it the first subcube on the
main diagonal. Let this subcube be $H_{i}$. Then the cells $(i,j,1)$ form the
latin square shown in Figure 7, where the subarray $A_{1}$ is a $h_{i}\times
h_{i}$ array and $A_{4}$ is an $(N-h_{i})\times(N-h_{i})$ array.

$A_{1}$$A_{2}$$A_{3}$$A_{4}$ Figure 7:

Observe that the cells of $A_{1}$ are within the subcube $H_{i}$ of order
$h_{i}$. Thus, the symbols of $H_{i}$ cannot appear in $A_{3}$. These symbols
need to appear in every row and column, so each row of the subarray $A_{4}$
must contain all $h_{i}$ symbols in $H_{i}$. This gives that $h_{i}\leq
N-h_{i}$. ∎

The next condition is similar to Theorem 1.9.

######  Theorem 3.3.

If a $\operatorname{3-RP}(h_{1}\dots h_{n})$ exists then
$N^{3}-\sum_{i=1}^{n}h_{i}^{3}\geq N^{2}\sum_{i\in D}h_{i}+3(\sum_{i\in
D}h_{i}^{2})(N-\sum_{i\in D}h_{i})-\sum_{i\in D}h_{i}^{3}$ for any subset $D$
of $\\{1,\dots,n\\}$.

###### Proof.

Suppose that a $\operatorname{3-RP}(h_{1}\dots h_{n})$, $C$, exists. We assume
that $C$ is in normal form and that subcube $H_{i}$ contains the symbols of
$S_{i}=(h_{1}+\dots+h_{i-1})+[h_{i}]$. For some fixed subset $D$ of $[n]$, let
$S=\cup_{i\in D}S_{i}$.

Fix $i\in D$ and let $L_{i}$ be the set of lines defined by
$(s,s^{\prime},[N])$, $(s,[N],s^{\prime})$ and $([N],s,s^{\prime})$ for all
$s,s^{\prime}\in S_{i}$. The only cells appearing in more than one of these
lines are the cells of $H_{i}$, which appear three times. Also, every symbol
of $S$ appears within every line, with the symbols of $S_{i}$ being only
within $H_{i}$.

Consider the union of these sets $L_{i}$ for all $i\in D$. Clearly the cells
in the $L_{i}$ are disjoint. Thus, the number of times that the symbols of $S$
occur in the union is

$\sum_{i\in D}h_{i}^{3}+3\sum_{\begin{subarray}{c}i,j\in D\\\ i\neq
j\end{subarray}}h_{i}^{2}h_{j}.$

Therefore, the number of times they occur not within these lines is

$N^{2}|S|-\sum_{i\in D}h_{i}^{3}-3\sum_{\begin{subarray}{c}i,j\in D\\\ i\neq
j\end{subarray}}h_{i}^{2}h_{j}=N^{2}\sum_{i\in D}h_{i}-\sum_{i\in
D}h_{i}^{3}-3\sum_{\begin{subarray}{c}i,j\in D\\\ i\neq
j\end{subarray}}h_{i}^{2}h_{j}=N^{2}\sum_{i\in D}h_{i}-(\sum_{i\in
D}h_{i})^{3}+6\sum_{\begin{subarray}{c}i,j,k\in D\\\
i<j<k\end{subarray}}h_{i}h_{j}h_{k}.$

We now count the number of cells not within these lines which can contain the
symbols of $S$. This is all of the cells not within the lines, and not within
the subcubes $H_{j}$ for $j\in\overline{D}$. The number of cells is

$N^{3}-\sum_{i=1}^{n}h_{i}^{3}-\sum_{i\in
D}3h_{i}^{2}(N-h_{i})=N^{3}-\sum_{i=1}^{n}h_{i}^{3}-3N\sum_{i\in
D}h_{i}^{2}+3(\sum_{i\in D}h_{i}^{2})(\sum_{i\in D}h_{i})+\sum_{i\in
D}h_{i}^{3}-(\sum_{i\in D}h_{i})^{3}+6\sum_{\begin{subarray}{c}i,j,k\in D\\\
i<j<k\end{subarray}}h_{i}h_{j}h_{k}.$

Therefore, combining these two numbers,

$N^{3}-\sum_{i=1}^{n}h_{i}^{3}\geq N^{2}\sum_{i\in D}h_{i}+3(\sum_{i\in
D}h_{i}^{2})(N-\sum_{i\in D}h_{i})-\sum_{i\in D}h_{i}^{3}.$

∎

We now begin to update the established existence results of 2-realizations for
3-realizations.

######  Theorem 3.4.

A $\operatorname{3-RP}(a^{n})$ exists for all $a,n\in\mathbb{Z}^{+}$.

###### Proof.

$\operatorname{3-RP}(a^{n})$ exists for all $n\neq 2$ by Theorem 1.6 and
Theorem 3.1.

For $\operatorname{3-RP}(a^{2})$, take the latin cube $C$ of order 2 given in
Figure 3.

Now take a latin cube $A$ of order $a$ and inflate $C$ with $A$ to get a latin
cube of order $2a$ with 2 disjoint subcubes of order $a$ on the main diagonal.
∎

The following theorem provides both a condition for existence and also a
method for constructing 3-realizations of partitions.

######  Theorem 3.5.

A $\operatorname{3-RP}(h_{1}\dots h_{n})$ with $h_{1}=\frac{N}{2}$ exists if
and only if a $\operatorname{3-RP}(h_{2}\dots h_{n})$ exists.

###### Proof.

If $C_{1}$ is a $\operatorname{3-RP}(h_{2}\dots h_{n})$, then let
$h_{1}=\sum_{i=2}^{n}h_{i}$. We construct $C_{2}$ by taking a
$\operatorname{3-RP}(h_{1}^{2})$, which we know exists from Theorem 3.4, and
replacing one of the subcubes of order $h_{1}$ with $C_{1}$. Thus, $C_{2}$ is
a $\operatorname{3-RP}(h_{1}\dots h_{n})$ with $h_{1}=\frac{N}{2}$.

If $C_{1}$ is a $\operatorname{3-RP}(h_{1}\dots h_{n})$ with
$h_{1}=\frac{N}{2}=\sum_{i=2}^{n}h_{i}$, then it can be represented as in
Figure 8, where all cells are $h_{1}\times h_{1}\times h_{1}$ arrays, $A$ is
the order $h_{1}$ subcube $H_{1}$ and $H$ contains all other subcubes.

ABCD

EFGH

Figure 8:

Since $C_{1}$ is a latin cube, all symbols from $H_{1}$ must appear in the
subarrays $A$, $D$, $F$ and $G$, with the remaining cells only containing
symbols from the remaining subcubes. Thus, $H$ is a latin cube on the symbols
from subcubes $H_{2},\dots,H_{n}$. Therefore $H$ is a
$\operatorname{3-RP}(h_{2}\dots h_{n})$. ∎

######  Ex. 3.6.

The partition $(3^{2}1)$ satisfies the condition from Theorem 3.2 but not the
condition from Theorem 3.3.

The partition $(10^{1}2^{1}1^{7})$ satisfies the condition from Theorem 3.3
but not the condition from Theorem 3.2. This shows that both of these
conditions are needed as necessary conditions for the existence of a
3-realization.

The partition $(19^{1}10^{1}2^{1}1^{7})$ satisfies both conditions, but the
existence of a $\operatorname{3-RP}(19^{1}10^{1}2^{1}1^{7})$ would imply the
existence of a $\operatorname{3-RP}(10^{1}2^{1}1^{7})$ by Theorem 3.5. The
partition $(10^{1}2^{1}1^{7})$ cannot be realized due to Theorem 3.2. Thus, we
can conclude that the first two conditions are necessary but not sufficient to
show existence.

##  4 Partitions with at most two distinct parts

This section will focus on partitions with two distinct parts, given by
$(a^{u}b^{n-u})$, where $a>b$. Theorems 1.5 and 1.7 give the existence without
bounds of a 2-realization for many such partitions, so we need only consider
the cases where $1<n<5$ or $u<3$.

######  Lemma 4.1.

A $\operatorname{3-RP}(h_{1}h_{2})$ exists if and only if $h_{1}=h_{2}$.

###### Proof.

If $h_{1}=h_{2}$, then we are considering the partition $(h_{1}^{2})$. By
Theorem 3.4, a $\operatorname{3-RP}(h_{1}^{2})$ exists.

Suppose that there is a latin cube $\operatorname{3-RP}(h_{1}h_{2})$. From
Theorem 3.2, $h_{1}\leq\frac{N}{2}$ and $h_{2}\leq\frac{N}{2}$. Since
$h_{1}+h_{2}=N$, $h_{1}=h_{2}=\frac{N}{2}$. ∎

We now consider the case where $u=1$. For latin squares, $a$ is limited by the
bound $a\leq(n-1)b$ given in Theorem 1.7, and for $n<4$ the 2-realization does
not exist. We begin with the case where $n=3$.

######  Lemma 4.2.

A $\operatorname{3-RP}(a^{1}b^{2})$ with $a>b$ exists if and only if $a\leq
2b$.

###### Proof.

Suppose that a $\operatorname{3-RP}(a^{1}b^{2})$ exists, then by Theorem 3.2
we have that $a\leq 2b$. Now suppose that this is satisfied.

If $a=2b$ then a $\operatorname{3-RP}(a^{1}b^{2})$ can be constructed by
Theorem 3.5 with a $\operatorname{3-RP}(a^{2})$.

If $a<2b$, then we begin by constructing an outline rectangle $O$, with the
structure shown in Figure 9a. We take the partitions
$P=Q=R=(b^{1}(a-b)^{1}b^{2})$ for the outline rectangle, so we are working
with the symbols $\\{1,2,3,4\\}$. In $O$, we want the cells $A_{1}$, $A_{2}$,
$A_{3}$ and $A_{4}$ to contain a total of $a^{2}$ copies of only symbols 1 and
2, and the cells $B^{\prime}$, $C^{\prime}$ and $D^{\prime}$ to be filled with
$b^{2}$ copies of 3, 4 and 1 respectively.

A1A2A3A4B′C′D′D′ a The outline rectangle $O$

ABCDD b The subsquare structure of $L$

Figure 9:

Fill the array $O$ as in Figure 10, where $x:y$ indicates that there are $y$
copies of symbol $x$ in that cell.

$1:b(2b-a)$ $2:b(a-b)$ | $1:b(a-b)$ | $3:b^{2}$ | $4:b^{2}$
---|---|---|---
$1:b(a-b)$ | $2:(a-b)^{2}$ | $4:b(a-b)$ | $3:b(a-b)$
$4:b^{2}$ | $3:b(a-b)$ | $1:b^{2}$ |  $2:b(a-b)$ $3:b(2b-a)$
$3:b^{2}$ | $4:b(a-b)$ |  $2:b(a-b)$ $4:b(2b-a)$ | $1:b^{2}$
Figure 10: The outline rectangle $O$

The array $O$ satisfies all of the conditions of an outline rectangle when
$N=a+2b$, so this array is the reduction of a latin square $L$ by Theorem 2.7.
Clearly this array also has the cells $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$,
$B^{\prime}$, $C^{\prime}$ and $D^{\prime}$ filled as described above.

Symbols 1 and 2 will become the symbols of $[a]$ in $L$, and since there are
$a^{2}$ of them in $A_{1}\cup A_{2}\cup A_{3}\cup A_{4}$, $L$ will have a
subsquare of order $a$, shown by $A$ in Figure 9b. Similarly, the entries of
$B^{\prime}$ force $B$ to be a subsquare of order $b$ on the symbols of
$a+[b]$, $C^{\prime}$ makes $C$ a subsquare of order $b$ on the symbols of
$(a+b)+[b]$, and $D^{\prime}$ causes $D$ to be a subsquare of order $b$ on the
symbols of $[b]$. Therefore, due to the structure of $O$, any latin square
$L$, constructed from $O$, will have the subsquare structure shown in Figure
9b.

The latin square $L$ has subsquares $H_{1},H_{2},H_{3}$ on the main diagonal
with $h_{1}=a$, $h_{2}=h_{3}=b$ and symbols $1$ to $h_{i}$ in each subsquare
$H_{i}$. So if $S_{i}=(h_{1}+\dots+h_{i-1})+[h_{i}]$, then for all $(r,c)\in
S_{i}\times S_{i}$ we have that the symbols in that subsquare are in
$[h_{i}]$. Also, the same subsquares, but with disjoint symbols between
subsquares, appear in the first $a$ or $b$ rows respectively. So if
$(r,c)\in[h_{i}]\times S_{i}$ for some $i\in[3]$, then the symbols in that
subsquare are in $S_{i}$.

We construct a latin cube $C$ using Construction 2.1, similar to the method
used in Theorem 3.1. To accomodate for the fact that a realization does not
exist, the symbol-disjoint subsquares are now at the top of $L$, instead of on
the main diagonal, but they are still column-disjoint. We know that $C$ must
be a latin cube from our construction, so we now check that it has the
required subcubes on the main diagonal.

Note that for any part $h_{i}$, if $(r,c)\in S_{i}\times S_{i}$ then
$L(r,c)\in[h_{i}]$ by the subsquare structure of $L$. Also, if
$(r,c)\in[h_{i}]\times S_{i}$ then $L(r,c)\in S_{i}$ by the earlier
assumption. So if $l,c,r\in S_{i}$ then $L(r,l)=a\in[h_{i}]$ and $L(a,c)\in
S_{i}$. Thus, $C(r,c,l)=L(L(r,l),c)\in S_{i}$. Therefore, the cubes formed by
the columns, rows and layers of $S_{i}$ are filled with only the elements
$S_{i}$. So $C$ has the required subcubes. ∎

######  Ex. 4.3.

The latin squares in Figures 11 and 12 can be used to construct a
$\operatorname{3-RP}(a^{1}b^{2})$ with Construction 2.1.

$\begin{array}[]{|cc|c|cc|cc|}\hline\cr 1&1&1&3&3&4&4\\\ 2&2&1&3&3&4&4\\\
\hline\cr 1&1&2&4&4&3&3\\\ \hline\cr 4&4&3&1&1&2&2\\\ 4&4&3&1&1&3&3\\\
\hline\cr 3&3&4&2&2&1&1\\\ 3&3&4&4&4&1&1\\\ \hline\cr\end{array}$

a The outline rectangle $O$

$\begin{array}[]{|ccc|cc|cc|}\hline\cr 1&3&2&4&5&6&7\\\ 3&2&1&5&4&7&6\\\
\hline\cr 2&1&3&6&7&4&5\\\ \hline\cr 6&\lx@intercol\hfil
7\hfil\lx@intercol\vrule\lx@intercol&4&1&2&5&3\\\ 7&\lx@intercol\hfil
6\hfil\lx@intercol\vrule\lx@intercol&5&2&1&3&4\\\ \hline\cr
4&\lx@intercol\hfil 5\hfil\lx@intercol\vrule\lx@intercol&6&7&3&1&2\\\
5&\lx@intercol\hfil 4\hfil\lx@intercol\vrule\lx@intercol&7&3&6&2&1\\\
\hline\cr\end{array}$

b The latin square $L$

Figure 11: A latin square $L$ for a $\operatorname{3-RP}(3^{1}2^{2})$ and the
outline rectangle used to find $L$.

$\begin{array}[]{|cccc|ccc|ccc|}\hline\cr 1&2&4&3&5&6&7&8&9&10\\\
2&4&3&1&6&7&5&9&10&8\\\ 4&3&1&2&7&5&6&10&8&9\\\ \hline\cr
3&1&2&4&8&9&10&5&6&7\\\ \hline\cr 8&9&\lx@intercol\hfil
10\hfil\lx@intercol\vrule\lx@intercol&5&1&2&3&4&7&6\\\ 9&10&\lx@intercol\hfil
8\hfil\lx@intercol\vrule\lx@intercol&6&2&3&1&7&4&5\\\ 10&8&\lx@intercol\hfil
9\hfil\lx@intercol\vrule\lx@intercol&7&3&1&2&6&5&4\\\ \hline\cr
5&6&\lx@intercol\hfil 7\hfil\lx@intercol\vrule\lx@intercol&8&4&10&9&1&2&3\\\
6&7&\lx@intercol\hfil 5\hfil\lx@intercol\vrule\lx@intercol&9&10&4&8&2&3&1\\\
7&5&\lx@intercol\hfil 6\hfil\lx@intercol\vrule\lx@intercol&10&9&8&4&3&1&2\\\
\hline\cr\end{array}$ a A latin square $L$ for a
$\operatorname{3-RP}(4^{1}3^{2})$.

$\begin{array}[]{|ccccc|ccc|ccc|}\hline\cr 1&4&5&2&3&6&7&8&9&10&11\\\
5&2&4&3&1&7&8&6&10&11&9\\\ 4&5&3&1&2&8&6&7&11&9&10\\\ \hline\cr
3&1&2&4&5&9&10&11&6&7&8\\\ 2&3&1&5&4&10&11&9&7&8&6\\\ \hline\cr
9&10&\lx@intercol\hfil
11\hfil\lx@intercol\vrule\lx@intercol&6&7&1&2&3&8&4&5\\\
10&11&\lx@intercol\hfil
9\hfil\lx@intercol\vrule\lx@intercol&7&8&2&3&1&5&6&4\\\ 11&9&\lx@intercol\hfil
10\hfil\lx@intercol\vrule\lx@intercol&8&6&3&1&2&4&5&7\\\ \hline\cr
6&7&\lx@intercol\hfil
8\hfil\lx@intercol\vrule\lx@intercol&9&10&11&4&5&1&2&3\\\
7&8&\lx@intercol\hfil
6\hfil\lx@intercol\vrule\lx@intercol&10&11&5&9&4&2&3&1\\\
8&6&\lx@intercol\hfil
7\hfil\lx@intercol\vrule\lx@intercol&11&9&4&5&10&3&1&2\\\
\hline\cr\end{array}$ b A latin square $L$ for a
$\operatorname{3-RP}(5^{1}3^{2})$.

Figure 12: A latin square $L$ which can be used to construct a
$\operatorname{3-RP}(a^{1}b^{2})$.

We will now extend the previous lemma to any value of $n\geq 3$.

######  Theorem 4.4.

For $n\geq 3$, a $\operatorname{3-RP}(ab^{n-1})$ exists if and only if
$a\leq(n-1)b$.

###### Proof.

Suppose that a $\operatorname{3-RP}(ab^{n-1})$ exists with $a>b$. From Theorem
3.2 we may deduce that $a\leq(n-1)b$.

Now suppose that this condition is satisfied, and we will show that a
$\operatorname{3-RP}(ab^{n-1})$ exists.

  * 1.

If $n=3$ then we have already shown existence in Lemma 4.2.

  * 2.

We know that a $\operatorname{2-RP}(ab^{n-1})$ exists for all $b<a\leq(n-2)b$
with $n\geq 4$ from Theorems 1.5 and 1.7. Thus, from Theorem 3.1 we have
existence of the 3-realization.

  * 3.

If $a=(n-1)b$ then existence has already been shown in Theorem 3.5.

So suppose that $(n-2)b<a<(n-1)b$.

We prove existence by contructing a similar outline rectangle $O$ to the one
in Lemma 4.2 and then likewise constructing a latin square and cube from this.
We will use the partitions

$P=Q=R=(b^{n-2}(a-b(n-2))^{1}b^{n-1}),$

which means that our outline rectangle will be an $2(n-1)\times 2(n-1)$ array
on $2(n-1)$ symbols, which we take to be $[2(n-1)]$. We now separate our
outline rectangle into four sections, as shown in Figure 13, each of which is
an $(n-1)\times(n-1)$ array which can have multiple entries per cell.

ABCD Figure 13:

We now construct each of these arrays. For the array $D$, let entries of cell
$(i,j)\in[n-1]\times[n-1]$ be defined as follows. All values modulo $n-1$ are
considered with residues in $[n-1]$; that is, we consider $k(n-1)\mod n-1$ to
be $n-1$ instead of $0$.

  * 1.

$b^{2}$ copies of symbol $-i+j+1\mod n-1$ if $-i+j\not\equiv n-2\mod n-1$;

  * 2.

$b(a-(n-2)b)$ copies of symbol $n-1$ and $b((n-1)b-a)$ copies of $n-1+i$ if
$-i+j\equiv n-2\mod n-1$.

The array $D$ will be filled as in Figure 14, where $c=b(a-(n-2)b)$ and
$d=b((n-1)b-a)$.

$1:b^{2}$ | $2:b^{2}$ | $\dots$ | $n-2:b^{2}$ |  $n-1:c$ $n:d$
---|---|---|---|---
$n-1:c$ $n+1:d$ | $1:b^{2}$ | $\dots$ | $n-3:b^{2}$ | $n-2:b^{2}$
$\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$
$3:b^{2}$ | $4:b(a-b)$ | $\dots$ | $1:b^{2}$ | $2:b^{2}$
$2:b^{2}$ | $3:b(a-b)$ | $\dots$ |  $n-1:c$ $2n-2:d$ | $1:b^{2}$
Figure 14: The array $D$

For array $B$ we use a similar construction. The entries of cell
$(i,j)\in[n-1]\times[n-1]$ are

  * 1.

$b^{2}$ copies of symbol $(-i+j+1\mod n-1)+n-1$ if $i\neq n-1$;

  * 2.

$b(a-(n-2)b)$ copies of symbol $(-i+j+1\mod n-1)+n-1$ if $i=n-1$.

For array $C$, the entries of cell $(i,j)\in[n-1]\times[n-1]$ are

  * 1.

$b^{2}$ copies of symbol $(i+j\mod n-1)+n-1$ if $j\neq n-1$;

  * 2.

$b(a-(n-2)b)$ copies of symbol $(i+j\mod n-1)+n-1$ if $j=n-1$.

For array $A$, the entries of cell $(i,j)\in[n-1]\times[n-1]$ are

  * 1.

$b(a-(n-2)b)$ copies of symbol $i+j\mod n-1$ if $i=n-1$ or $j=n-1$;

  * 2.

$b^{2}$ copies of symbol $i+j\mod n-1$ if $i,j<n-1$ and $i+j>n-1$;

  * 3.

$b(a-(n-2)b)$ copies of symbol $i+j$ and $b((n-1)b-a)$ copies of $i+j-1$ if
$i+j\leq n-1$.

Combining these and ignoring repetitions, the outline rectangle will look like
the array in Figure 15.

2,1 | 3,2 | $\dots$ | $n-1,n-2$ | 1 | $n$ | $n+1$ | $\dots$ | $2n-3$ | $2n-2$
---|---|---|---|---|---|---|---|---|---
3,2 | 4,3 | $\dots$ | 1 | 2 | $2n-2$ | $n$ | $\dots$ | $2n-4$ | $2n-3$
$\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$
$n-1,n-2$ | 1 | $\dots$ | $n-3$ | $n-2$ | $n+2$ | $n+3$ | $\dots$ | $n$ | $n+1$
1 | 2 | $\dots$ | $n-2$ | $n-1$ | $n+1$ | $n+2$ | $\dots$ | $2n-2$ | $n$
$n+1$ | $n+2$ | $\dots$ | $2n-2$ | $n$ | 1 | 2 | $\dots$ | $n-2$ | $n-1,n$
$n+2$ | $n+3$ | $\dots$ | $n$ | $n+1$ | $n-1,n+1$ | 1 | $\dots$ | $n-3$ | $n-2$
$\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$
$2n-2$ | $n$ | $\dots$ | $2n-4$ | $2n-3$ | 3 | 4 | $\dots$ | $1$ | 2
$n$ | $n+1$ | $\dots$ | $2n-3$ | $2n-2$ | 2 | 3 | $\dots$ | $n-1,2n-2$ | 1
Figure 15: The outline rectangle $O$

The array $O$ satisfies all of the requirements for an outline rectangle, thus
we can obtain a latin square $L$ from this array using Theorem 2.7. The
subarray $A$ only contains symbols from $[n-1]$, and this symbol set becomes
$[a]$ in $L$. Since $A$ contains $a^{2}$ of these symbols, the subarray $A$
becomes a subsquare of order $a$ in $L$.

Considering the subarray $B$, the first row of cells each contain $b^{2}$
copies of a specified symbol. Each of these symbols is unique and represents a
disjoint set of $b$ symbols in $L$. Thus, there are $n-1$ symbol-disjoint
subsquares of order $b$ in the first $b$ rows of $B$, and they contain
disjoint symbols to the subsquare of order $a$.

In subarray $D$, the main diagonal cells contain only $b^{2}$ copies of symbol
1. In $L$, each of these becomes a subsquare of order $b$ on the symbols of
$[b]$.

The structure of $L$ is thus similar to a 2-realization of $(a^{1}b^{n-1})$,
except the symbol-disjoint subsquares of order $b$ are moved to the top of the
array and are replaced by subsquares with the symbols of $[b]$. For $i\in[n]$,
where $h_{1}=a$ and $h_{i}=b$ for $i\geq 2$, let
$S_{i}=(h_{1}+\dots+h_{i-1})+[h_{i}]$. Then formally, for all $(r,c)\in
S_{i}\times S_{i}$, we have $L(r,c)\in[h_{i}]$, and for all
$(r,c)\in[h_{i}]\times S_{i}$, we have $L(r,c)\in S_{i}$. This is the same
structure as the latin square $L$ in Lemma 4.2.

Thus, we can use 2.1 to build a latin cube $C$, and $C$ will have the required
subcubes for a 3-realization by the same proof as given in Lemma 4.2.

∎

This completes the existence problem where $u=1$. We now consider the case
when $u=2$. For 2-realizations, this case has the same bounds as for $u=1$.

######  Lemma 4.5.

A $\operatorname{3-RP}((2b)^{2}b^{1})$, a
$\operatorname{3-RP}((3b)^{2}(2b)^{1})$ and a
$\operatorname{3-RP}((4b)^{2}(3b)^{1})$ exist for all $b\geq 1$.

###### Proof.

Observe that a $\operatorname{3-RP}(2^{2}1^{1})$ is given in Example 1.12. To
construct a $\operatorname{3-RP}((2b)^{2}b^{1})$, inflate this latin cube with
a latin cube of order $b$ using 2.2.

A $\operatorname{3-RP}(3^{2}2^{1})$ is given in A.1. Use inflation to get a
$\operatorname{3-RP}((3b)^{2}(2b)^{1})$. The same process can be used for the
$\operatorname{3-RP}(4^{2}3^{1})$ in A.2 ∎

######  Lemma 4.6.

A $\operatorname{3-RP}(a^{2}b^{2})$ exists for $a\leq 2b$.

###### Proof.

If $a=2b$ then take a $\operatorname{3-RP}(a^{3})$ and replace one of the
subcubes with a $\operatorname{3-RP}(b^{2})$.

Suppose that $a<2b$. Since we have that $a<2b<2a$, we know from Theorem 4.4
that a $\operatorname{3-RP}((2b)^{1}a^{2})$ exists. Take this latin cube and
replace the subcube of order $2b$ with a $\operatorname{3-RP}(b^{2})$. This is
a $\operatorname{3-RP}(a^{2}b^{2})$. ∎

This does not provide a complete solution, since there are partitions for
which the 3-realization exists outside of this bound. For example, a
$\operatorname{3-RP}(4^{2}1^{2})$ exists, since this can be constructed using
a $\operatorname{3-RP}(4^{2}2^{1})$.

The following lemma provides a new upper bound for the case when $u=2$.

######  Lemma 4.7.

If a $\operatorname{3-RP}(a^{2}b^{n-2})$ exists for $n\geq 3$, then $a\leq
2(n-2)b$.

###### Proof.

Firstly, for $n=3$, using the condition from Theorem 3.3 and $D=\\{1,2\\}$ for
$h_{1}=h_{2}=a$, the condition simplifies to $b\geq a/2$. For $n=4$, taking
$D=\\{1,2,3\\}$ in the condition from Theorem 3.3 gives that $a\leq
4b=2(n-2)b$. Now consider the case where $n\geq 5$.

We know that a latin square $\operatorname{2-RP}(a^{2}b^{n-2})$ exists for
$a\leq(n-2)b$ where $n\geq 5$, so the corresponding latin cube also exists. So
now we consider $a>(n-2)b$.

Consider the outline rectangle in Figure 16, which is an amalgamation of the
layer $([N],[N],1)$ of a latin cube $\operatorname{3-RP}(a^{2}b^{n-2})$.
$A$,$B$,$D$ and $G$ represent $a\times a$ subarrays, while $C$ and $F$
represent $(n-2)b\times a$ subarrays, and $E$ is an $a\times(n-2)b$ subarray.
$A$ is within the subcube $H_{1}$ and $G$ is in the same files as the subcube
$H_{2}$. We have amalgamated the symbols using the partition
$(a^{2}((n-2)b)^{1})$.

ADBGCFE Figure 16:

Let $x=(n-2)b$. Observe that there must be $ax$ of symbol 3 across $B$ and
$C$, but there can be at most $x^{2}$ of this symbol in $C$. So there must be
at least $(a-x)x$ of 3 in $B$ since $a>x$. We can repeat this for cells $D$
and $E$.

Since there are at least $(a-x)x$ copies of 3 in $D$, there are at most
$a^{2}-(a-x)x$ of symbol 2 in $D$. Also, since $G$ is in the same files as the
subcube $H_{2}$, $G$ cannot contain any copies of symbol 2. Thus, there must
be at least $(a-x)x$ copies of symbol 2 in $F$. This is the same for cells $B$
and $C$.

Therefore, there are at least $2(a-x)x$ copies of symbol 2 across $C$ and $F$.
There can be at most $ax$ of this symbol across those cells, so $2(a-x)x\leq
ax$. This simplifies to $a\leq 2x=2(n-2)b$. ∎

For $n=3$ and $n=4$, this is the same bound that is achieved using the
condition from Theorem 3.3, by taking $D=\\{1,2\\}$ and $D=\\{1,2,3\\}$
respectively. For all greater values of $n$ this is a stricter bound than that
condition. For example, the partition $(43^{2}7^{3})$ satisfies the conditions
from Theorems 3.2 and 3.3, but not Lemma 4.7. This partition is also not
excluded by Theorem 3.5. Therefore, we can conclude that the three conditions
together are still not sufficient for existence.

The existence of a $\operatorname{3-RP}(a^{2}b^{n-2})$ is incomplete. Latin
squares provide existence for all $a\leq(n-2)b$, where $n\geq 5$, and Lemma
4.6 gives the same result for $n=4$. The following lemma shows that if a
$\operatorname{3-RP}(a^{2}b^{1})$ exists for the entire range given in Lemma
4.7, then the remaining existence cases for a
$\operatorname{3-RP}(a^{2}b^{n-2})$ follow immediately.

######  Lemma 4.8.

If a $\operatorname{3-RP}(a^{2}b^{1})$ exists for all $a\leq 2b$ then a
$\operatorname{3-RP}(a^{2}b^{n-2})$ exists for all $(n-2)b<a\leq 2(n-2)b$,
where $n\geq 4$.

###### Proof.

Take any $a$ such that $(n-2)b<a\leq 2(n-2)b$. Then there exists a
$\operatorname{3-RP}(a^{2}((n-2)b)^{1})$. Replacing the subsquare of order
$(n-2)b$ with a $\operatorname{3-RP}(b^{n-2})$ gives a
$\operatorname{3-RP}(a^{2}b^{n-2})$. ∎

##  5 $m$-dimensional hypercubes

This idea of a realization of a partition can be extended to higher
dimensions.

######  Definition 5.1.

An _m-dimensional latin hypercube_ $L=(L_{j_{1},\dots,j_{m}})$ of order $N$ is
an $m$-dimensional array of side $N$ on a symbol set of size $N$, such that
for all $(j_{1},\dots,j_{m}),(j^{\prime}_{1},\dots,j^{\prime}_{m})\in S^{m}$,
and all $\alpha\in\\{1,\dots,m\\}$, if $j_{i}=j^{\prime}_{i}$ for all
$i\neq\alpha$ and $j_{\alpha}\neq j^{\prime}_{\alpha}$ then
$L_{j_{1},\dots,j_{m}}\neq L_{j^{\prime}_{1},\dots,j^{\prime}_{m}}$.

As before, a _subhypercube_ of $L$ is an $m$-dimensional sub-array which is
itself a latin hypercube.

######  Definition 5.2.

For an integer partition $P=(h_{1},h_{2},\dots,h_{n})$ of $N$ with $h_{1}\geq
h_{2}\geq\dots\geq h_{n}>0$, an _m-realization_ of $P$, denoted
$m\operatorname{-RP}(h_{1}h_{2}\dots h_{n})$, is an $m$-dimensional latin
hypercube of order $N$ with pairwise disjoint subhypercubes of order $h_{i}$
for each $i\in[n]$.

######  Theorem 5.3.

If there exists an $m\operatorname{-RP}(h_{1}\dots h_{n})$ and a
$k\operatorname{-RP}(h_{1}\dots h_{n})$, then there exists an
$(m+k-1)\operatorname{-RP}(h_{1}\dots h_{n})$.

###### Proof.

Let $F$ be a latin hypercube which gives an $m\operatorname{-RP}(h_{1}\dots
h_{n})$ and let $G$ be a latin hypercube which gives a
$k\operatorname{-RP}(h_{1}\dots h_{n})$. We will assume that these are in
normal form and with symbols in increasing order across the subhypercubes.

Let $f(x_{1},\dots,x_{m})$ be an entry of $F$ and $g(x_{1},\dots,x_{k})$ be an
entry of $G$. For $l=k+m-1$, we will construct an $l$-dimensional hypercube,
$H$, by

$h(x_{1},\dots,x_{l})=f(g(x_{1},\dots,x_{k}),x_{k+1},\dots,x_{l}).$

Suppose that
$h(x_{1},\dots,x_{i},\dots,x_{l})=h(x_{1},\dots,x_{i}^{\prime},\dots,x_{l})$
for some $i\in[l]$. If $i>k$ then

$f(g(x_{1},\dots,x_{k}),x_{k+1},\dots,x_{i},\dots,x_{l})=f(g(x_{1},\dots,x_{k}),x_{k+1},\dots,x_{i}^{\prime},\dots,x_{l}).$

Since $F$ is a latin hypercube, we must have that $x_{i}=x_{i}^{\prime}$.

If $i\leq k$, then

$f(g(x_{1},\dots,x_{i},\dots,x_{k}),x_{k+1},\dots,x_{l})=f(g(x_{1},\dots,x_{i}^{\prime},\dots,x_{k}),x_{k+1},\dots,x_{l}).$

This gives that
$g(x_{1},\dots,x_{i},\dots,x_{k})=g(x_{1},\dots,x_{i}^{\prime},\dots,x_{k})$,
and since $G$ is a latin hypercube, $x_{i}=x_{i}^{\prime}$. Therefore $H$ is a
latin hypercube.

We now show that this $H$ contains the required subhypercubes. Let
$S_{i}=(h_{1}+\dots+h_{i-1})+[h_{i}]$ for some part $h_{i}$. Given our earlier
assumptions about $F$ and $G$, we know that if $(x_{1},\dots,x_{m})\in
S_{i}\times S_{i}$ then $f(x_{1},\dots,x_{m})\in S_{i}$, and if
$x_{1},\dots,x_{k}\in S_{i}$ then $g(x_{1},\dots,x_{k})\in S_{i}$.

So if $x_{1},\dots,x_{l}\in S_{i}$ for some $i$, then
$g(x_{1},\dots,x_{k})=a\in S_{i}$. Thus,
$h(x_{1},\dots,x_{l})=f(a,x_{k+1},\dots,x_{l})\in S_{i}$. Therefore there are
subhypercubes found along the diagonal with the symbols in increasing order.
Thus, $H$ is an $(m+k-1)\operatorname{-RP}(h_{1}\dots h_{n})$. ∎

We now prove an analogue of Theorem 3.1 for a hypercube of any dimension.

######  Corollary 5.4.

If there exists a $\operatorname{2-RP}(h_{1}\dots h_{n})$, then there exists
an $m\operatorname{-RP}(h_{1}\dots h_{n})$ for all $m\geq 3$.

###### Proof.

The proof for $m=3$ is given in Theorem 3.1. Now suppose that the statement is
true for some $m$.

So we have a latin square $L$ which gives a $\operatorname{2-RP}(h_{1}\dots
h_{n})$ and a latin hypercube, $G$, which gives an
$m\operatorname{-RP}(h_{1}\dots h_{n})$. By Theorem 5.3, there exists an
$(m+1)\operatorname{-RP}(h_{1}\dots h_{n})$ since $2+m-1=m+1$.

Therefore, we have proven the result by induction. ∎

The question that naturally follows from this is whether the existence of an
$m\operatorname{-RP}(h_{1}\dots h_{n})$ implies the existence of an
$(m+1)\operatorname{-RP}(h_{1}\dots h_{n})$. This can be easily disproven in
the general case (for $m\geq 3$) with the partition $(1^{2})$. There is no
2-realization $\operatorname{2-RP}(1^{2})$ but there is a
$\operatorname{3-RP}(1^{2})$, and it can be easily seen that there is no
$4\operatorname{-RP}(1^{2})$. In fact, there is no
$m\operatorname{-RP}(1^{2})$ for any even $m$.

The following theorem is a further generalisation, which starts with an
$m$-realization.

######  Corollary 5.5.

If there exists an $m\operatorname{-RP}(h_{1}\dots h_{n})$, then there exists
a $k\operatorname{-RP}(h_{1}\dots h_{n})$ for all $k\geq m\geq 2$ with
$k\equiv 1\pmod{m-1}$.

###### Proof.

If $k\equiv 1\pmod{m-1}$ then $k=k^{\prime}(m-1)+1$ for some
$k^{\prime}\in\mathbb{Z}^{+}$.

When $k^{\prime}=1$, $k=m$ and the statement is true.

Suppose that it is true for some $k^{\prime}\geq 1$. Then there is a latin
hypercube $F$ which gives an $m\operatorname{-RP}(h_{1}\dots h_{n})$ and a
latin hypercube $G$ which gives a $k\operatorname{-RP}(h_{1}\dots h_{n})$.
Since $l=(k^{\prime}+1)(m-1)+1=k+m-1$, there exists an
$l\operatorname{-RP}(h_{1}\dots h_{n})$ by Theorem 5.3. ∎

######  Lemma 5.6.

An $m\operatorname{-RP}(1^{2})$ exists if and only if $m$ is even.

###### Proof.

Suppose such a hypercube exists and the entries are represented by
$f(x_{1},\dots,x_{m})$. Without loss of generality, the subhypercubes must be
filled as $f(1,\dots,1)=1$ and $f(2,\dots,2)=2$. For every cell with exactly
one $x_{i}=2$, we know that $f(x_{1},\dots,x_{m})=2$ since it is in a line
with the first subhypercube. Repeating this, for every cell of the hypercube,
we get that

$f(x_{1},\dots,x_{m})=\begin{cases}1,&\text{if $x_{i}=2$ for an even number of
$i$,}\\\ 2,&\text{otherwise.}\end{cases}$

If $m$ is even, then the second subhypercube must contain $1$ by this process.
Thus, $m$ must be odd.

Since a $\operatorname{3-RP}(1^{2})$ exists, we can use Corollary 5.5 to find
an $m\operatorname{-RP}(1^{2})$ for any odd $m$. ∎

We have shown that many partitions can be realized in latin cubes which cannot
in latin squares. We note here that although these realizations cannot all be
extended to a $4$-realization, there are some partitions which exist in
$4$-dimensional hypercubes but not latin squares. One such hypercube is given
in A.3.

## Appendix A

######  A.1.

The following arrays represent the layers of a
$\operatorname{3-RP}(3^{2}2^{1})$.

$\begin{matrix}\scalebox{0.9}{
$\begin{array}[]{|cccccccc|}\hline\cr\pagecolor{lightgray}3&\pagecolor{lightgray}1&\pagecolor{lightgray}2&5&6&4&8&7\\\
\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&8&4&5&7&6\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}1&4&7&6&5&8\\\
4&8&6&2&3&7&1&5\\\ 5&6&4&7&8&2&3&1\\\ 6&5&7&3&1&8&2&4\\\ 8&7&5&6&2&1&4&3\\\
7&4&8&1&5&3&6&2\\\
\hline\cr\end{array}$\quad$\begin{array}[]{|cccccccc|}\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}1&6&4&8&7&5\\\
\pagecolor{lightgray}3&\pagecolor{lightgray}1&\pagecolor{lightgray}2&7&5&6&4&8\\\
\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&5&6&4&8&7\\\
7&4&5&3&8&1&6&2\\\ 4&5&8&1&7&3&2&6\\\ 5&6&4&8&2&7&1&3\\\ 6&8&7&2&1&5&3&4\\\
8&7&6&4&3&2&5&1\\\
\hline\cr\end{array}$\quad$\begin{array}[]{|cccccccc|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&4&5&7&6&8\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}1&5&6&4&8&7\\\
\pagecolor{lightgray}3&\pagecolor{lightgray}1&\pagecolor{lightgray}2&6&8&5&7&4\\\
5&6&4&8&7&2&3&1\\\ 6&7&5&2&1&8&4&3\\\ 8&4&6&7&3&1&5&2\\\ 7&5&8&1&4&3&2&6\\\
4&8&7&3&2&6&1&5\\\
\hline\cr\end{array}$\quad$\begin{array}[]{|cccccccc|}\hline\cr
4&7&5&8&3&2&1&6\\\ 5&8&4&2&1&7&6&3\\\ 8&6&7&1&2&3&4&5\\\
1&3&8&\pagecolor{lightgray}6&\pagecolor{lightgray}5&\pagecolor{lightgray}4&2&7\\\
7&1&3&\pagecolor{lightgray}5&\pagecolor{lightgray}4&\pagecolor{lightgray}6&8&2\\\
3&2&1&\pagecolor{lightgray}4&\pagecolor{lightgray}6&\pagecolor{lightgray}5&7&8\\\
2&4&6&3&7&8&5&1\\\ 6&5&2&7&8&1&3&4\\\ \hline\cr\end{array}$}\\\ \\\
\scalebox{0.9}{\;$\begin{array}[]{|cccccccc|}\hline\cr 7&8&6&3&2&1&5&4\\\
6&4&7&1&8&3&2&5\\\ 4&5&8&7&1&2&3&6\\\
3&2&1&\pagecolor{lightgray}5&\pagecolor{lightgray}4&\pagecolor{lightgray}6&7&8\\\
8&3&2&\pagecolor{lightgray}4&\pagecolor{lightgray}6&\pagecolor{lightgray}5&1&7\\\
2&7&3&\pagecolor{lightgray}6&\pagecolor{lightgray}5&\pagecolor{lightgray}4&8&1\\\
5&1&4&8&3&7&6&2\\\ 1&6&5&2&7&8&4&3\\\
\hline\cr\end{array}$\quad$\begin{array}[]{|cccccccc|}\hline\cr
6&5&8&1&7&3&4&2\\\ 8&7&6&3&2&1&5&4\\\ 7&4&5&2&3&8&6&1\\\
2&1&7&\pagecolor{lightgray}4&\pagecolor{lightgray}6&\pagecolor{lightgray}5&8&3\\\
3&2&1&\pagecolor{lightgray}6&\pagecolor{lightgray}5&\pagecolor{lightgray}4&7&8\\\
1&8&2&\pagecolor{lightgray}5&\pagecolor{lightgray}4&\pagecolor{lightgray}6&3&7\\\
4&6&3&7&8&2&1&5\\\ 5&3&4&8&1&7&2&6\\\
\hline\cr\end{array}$\quad$\begin{array}[]{|cccccccc|}\hline\cr
8&6&4&7&1&5&2&3\\\ 4&5&8&6&7&2&3&1\\\ 5&8&6&3&4&7&1&2\\\ 6&7&3&1&2&8&5&4\\\
2&4&7&8&3&1&6&5\\\ 7&1&5&2&8&3&4&6\\\
1&3&2&4&5&6&\pagecolor{lightgray}7&\pagecolor{lightgray}8\\\
3&2&1&5&6&4&\pagecolor{lightgray}8&\pagecolor{lightgray}7\\\
\hline\cr\end{array}$\quad$\begin{array}[]{|cccccccc|}\hline\cr
5&4&7&2&8&6&3&1\\\ 7&6&5&4&3&8&1&2\\\ 6&7&4&8&5&1&2&3\\\ 8&5&2&7&1&3&4&6\\\
1&8&6&3&2&7&5&4\\\ 4&3&8&1&7&2&6&5\\\
3&2&1&5&6&4&\pagecolor{lightgray}8&\pagecolor{lightgray}7\\\
2&1&3&6&4&5&\pagecolor{lightgray}7&\pagecolor{lightgray}8\\\
\hline\cr\end{array}$}\end{matrix}$

######  A.2.

The following arrays represent the layers of a
$\operatorname{3-RP}(4^{2}3^{1})$.

$\begin{array}[]{|ccccccccccc|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&5&6&7&8&11&9&10\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&6&7&8&5&9&10&11\\\
\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&7&8&5&6&10&11&9\\\
\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&10&11&9&7&5&8&6\\\
5&6&7&10&11&4&3&9&8&2&1\\\ 6&7&8&5&2&9&10&11&1&4&3\\\
7&8&5&11&9&10&1&4&3&6&2\\\ 8&5&6&9&4&1&11&10&2&3&7\\\
11&9&10&8&3&5&2&1&6&7&4\\\ 9&10&11&6&8&2&4&3&7&1&5\\\
10&11&9&7&1&3&6&2&4&5&8\\\ \hline\cr\end{array}$
$\begin{array}[]{|ccccccccccc|}\hline\cr\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&8&10&11&9&6&5&7\\\
\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&5&6&7&8&11&9&10\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&6&7&8&5&9&10&11\\\
\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&7&8&5&6&10&11&9\\\
9&5&6&7&10&1&2&11&3&4&8\\\ 10&6&7&8&9&11&1&4&5&3&2\\\
6&7&8&5&11&3&9&10&2&1&4\\\ 11&8&5&6&1&9&10&2&4&7&3\\\
5&11&9&10&2&4&6&3&7&8&1\\\ 7&9&10&11&4&5&3&1&8&2&6\\\
8&10&11&9&3&2&4&7&1&6&5\\\ \hline\cr\end{array}$
$\begin{array}[]{|ccccccccccc|}\hline\cr\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&7&8&5&6&10&11&9\\\
\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&9&5&10&11&7&6&8\\\
\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&5&6&7&8&11&9&10\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&6&7&8&5&9&10&11\\\
7&11&5&6&3&2&9&10&1&8&4\\\ 8&9&6&7&11&10&2&3&4&1&5\\\
5&10&7&8&1&9&11&2&6&4&3\\\ 6&7&8&5&10&11&4&9&3&2&1\\\
10&6&11&9&4&3&1&7&8&5&2\\\ 11&8&9&10&2&1&6&4&5&3&7\\\
9&5&10&11&8&4&3&1&2&7&6\\\ \hline\cr\end{array}$
$\begin{array}[]{|ccccccccccc|}\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&6&7&8&5&9&10&11\\\
\pagecolor{lightgray}3&\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&7&8&5&6&10&11&9\\\
\pagecolor{lightgray}4&\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&11&9&6&10&8&7&5\\\
\pagecolor{lightgray}1&\pagecolor{lightgray}2&\pagecolor{lightgray}3&\pagecolor{lightgray}4&5&6&7&8&11&9&10\\\
6&7&8&5&9&10&11&1&4&3&2\\\ 7&8&11&6&10&4&3&9&2&5&1\\\
8&5&9&7&4&11&10&3&1&2&6\\\ 5&6&10&8&3&2&9&11&7&1&4\\\
9&10&7&11&8&1&4&2&5&6&3\\\ 10&11&5&9&1&3&2&7&6&4&8\\\
11&9&6&10&2&5&1&4&3&8&7\\\ \hline\cr\end{array}$

$\begin{array}[]{|ccccccccccc|}\hline\cr 11&8&7&9&1&2&3&10&4&6&5\\\
6&9&10&11&2&3&4&1&5&8&7\\\ 9&10&5&8&3&4&1&11&7&2&6\\\
8&5&11&10&4&1&2&9&6&7&3\\\
1&2&3&4&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&11&9&10\\\
2&3&4&1&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&9&10&11\\\
3&4&1&2&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&10&11&9\\\
10&11&9&3&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&1&4&2\\\
7&1&6&5&11&9&10&4&2&3&8\\\ 4&6&8&7&9&10&11&2&3&5&1\\\
5&7&2&6&10&11&9&3&8&1&4\\\ \hline\cr\end{array}$
$\begin{array}[]{|ccccccccccc|}\hline\cr 10&5&6&11&9&1&2&3&7&8&4\\\
9&11&5&8&10&2&3&4&1&7&6\\\ 11&7&9&10&2&3&4&1&6&5&8\\\
5&9&10&6&11&4&1&2&8&3&7\\\
4&10&11&9&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&2&1&3\\\
1&2&3&4&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&11&9&10\\\
2&3&4&1&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&9&10&11\\\
3&4&1&2&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&10&11&9\\\
6&8&2&7&1&11&9&10&3&4&5\\\ 8&1&7&5&3&9&10&11&4&6&2\\\
7&6&8&3&4&10&11&9&5&2&1\\\ \hline\cr\end{array}$
$\begin{array}[]{|ccccccccccc|}\hline\cr 7&6&9&10&3&11&1&2&5&4&8\\\
11&10&6&7&4&9&2&3&8&5&1\\\ 5&9&11&6&1&10&3&4&2&8&7\\\
10&11&8&9&2&3&4&1&7&6&5\\\
3&4&1&2&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&10&11&9\\\
9&1&10&11&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&3&2&4\\\
1&2&3&4&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&11&9&10\\\
2&3&4&1&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&9&10&11\\\
8&7&5&3&10&2&11&9&4&1&6\\\ 6&5&2&8&11&4&9&10&1&7&3\\\
4&8&7&5&9&1&10&11&6&3&2\\\ \hline\cr\end{array}$
$\begin{array}[]{|ccccccccccc|}\hline\cr 9&10&11&5&2&3&4&1&8&7&6\\\
10&8&7&9&3&4&11&2&6&1&5\\\ 8&11&10&7&4&1&9&3&5&6&2\\\
7&6&9&11&1&2&10&4&3&5&8\\\
2&3&4&1&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&9&10&11\\\
3&4&1&2&\pagecolor{lightgray}7&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&10&11&9\\\
11&9&2&10&\pagecolor{lightgray}8&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&4&3&1\\\
1&2&3&4&\pagecolor{lightgray}5&\pagecolor{lightgray}6&\pagecolor{lightgray}7&\pagecolor{lightgray}8&11&9&10\\\
4&5&8&6&9&10&3&11&1&2&7\\\ 5&7&6&3&10&11&1&9&2&8&4\\\
6&1&5&8&11&9&2&10&7&4&3\\\ \hline\cr\end{array}$

$\begin{array}[]{|ccccccccccc|}\hline\cr 5&9&8&7&10&4&6&11&1&3&2\\\
8&6&9&5&11&10&1&7&4&2&3\\\ 6&5&7&9&8&11&10&2&3&4&1\\\
9&7&6&8&3&5&11&10&2&1&4\\\ 10&8&2&11&1&9&4&3&7&6&5\\\
11&10&5&3&4&2&9&1&6&7&8\\\ 4&11&10&6&2&1&3&9&8&5&7\\\
7&1&11&10&9&3&2&4&5&8&6\\\
3&4&1&2&7&6&5&8&\pagecolor{lightgray}10&\pagecolor{lightgray}11&\pagecolor{lightgray}9\\\
2&3&4&1&6&7&8&5&\pagecolor{lightgray}9&\pagecolor{lightgray}10&\pagecolor{lightgray}11\\\
1&2&3&4&5&8&7&6&\pagecolor{lightgray}11&\pagecolor{lightgray}9&\pagecolor{lightgray}10\\\
\hline\cr\end{array}$ $\begin{array}[]{|ccccccccccc|}\hline\cr
6&11&5&8&4&9&10&7&2&1&3\\\ 5&7&11&6&8&1&9&10&3&4&2\\\
7&6&8&11&10&5&2&9&1&3&4\\\ 11&8&7&5&9&10&6&3&4&2&1\\\
8&9&10&3&2&11&1&4&5&7&6\\\ 4&5&9&10&1&3&11&2&8&6&7\\\
10&1&6&9&3&2&4&11&7&8&5\\\ 9&10&2&7&11&4&3&1&6&5&8\\\
2&3&4&1&6&7&8&5&\pagecolor{lightgray}11&\pagecolor{lightgray}9&\pagecolor{lightgray}10\\\
1&2&3&4&5&8&7&6&\pagecolor{lightgray}10&\pagecolor{lightgray}11&\pagecolor{lightgray}9\\\
3&4&1&2&7&6&5&8&\pagecolor{lightgray}9&\pagecolor{lightgray}10&\pagecolor{lightgray}11\\\
\hline\cr\end{array}$ $\begin{array}[]{|ccccccccccc|}\hline\cr
8&7&10&6&11&5&9&4&3&2&1\\\ 7&5&8&10&1&11&6&9&2&3&4\\\
10&8&6&5&9&2&11&7&4&1&3\\\ 6&10&5&7&8&9&3&11&1&4&2\\\
11&1&9&8&4&3&10&2&6&5&7\\\ 5&11&2&9&3&1&4&10&7&8&6\\\
9&6&11&3&10&4&2&1&5&7&8\\\ 4&9&7&11&2&10&1&3&8&6&5\\\
1&2&3&4&5&8&7&6&\pagecolor{lightgray}9&\pagecolor{lightgray}10&\pagecolor{lightgray}11\\\
3&4&1&2&7&6&5&8&\pagecolor{lightgray}11&\pagecolor{lightgray}9&\pagecolor{lightgray}10\\\
2&3&4&1&6&7&8&5&\pagecolor{lightgray}10&\pagecolor{lightgray}11&\pagecolor{lightgray}9\\\
\hline\cr\end{array}$

######  A.3.

The six arrays in each row form a latin cube, and when all six latin cubes are
combined gives a latin hypercube with a realization of $(2^{2}1^{2})$.

$\begin{array}[]{|cccccc|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&5&6&3&4\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}1&6&5&4&3\\\ 3&6&2&4&5&1\\\
4&5&1&3&6&2\\\ 6&3&4&2&1&5\\\ 5&4&3&1&2&6\\\ \hline\cr\end{array}$
$\begin{array}[]{|cccccc|}\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}1&6&5&4&3\\\
\pagecolor{lightgray}1&\pagecolor{lightgray}2&5&6&3&4\\\ 5&4&3&1&2&6\\\
6&3&4&2&1&5\\\ 4&5&1&3&6&2\\\ 3&6&2&4&5&1\\\ \hline\cr\end{array}$
$\begin{array}[]{|cccccc|}\hline\cr 6&5&4&3&1&2\\\ 3&4&2&1&5&6\\\
4&3&1&2&6&5\\\ 1&2&6&5&3&4\\\ 5&6&3&4&2&1\\\ 2&1&5&6&4&3\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 5&6&3&4&2&1\\\
4&3&1&2&6&5\\\ 2&1&5&6&3&4\\\ 3&4&2&1&5&6\\\ 1&2&6&5&4&3\\\ 6&5&4&3&1&2\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 3&4&2&1&5&6\\\
6&5&4&3&1&2\\\ 1&2&6&5&4&3\\\ 5&6&3&4&2&1\\\ 2&1&5&6&3&4\\\ 4&3&1&2&6&5\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 4&3&1&2&6&5\\\
5&6&3&4&2&1\\\ 6&5&4&3&1&2\\\ 2&1&5&6&4&3\\\ 3&4&2&1&5&6\\\ 1&2&6&5&3&4\\\
\hline\cr\end{array}$

$\begin{array}[]{|cccccc|}\hline\cr\pagecolor{lightgray}2&\pagecolor{lightgray}1&6&5&4&3\\\
\pagecolor{lightgray}\pagecolor{lightgray}1&\pagecolor{lightgray}2&5&6&3&4\\\
4&5&1&3&6&2\\\ 3&4&2&1&5&6\\\ 5&6&3&4&2&1\\\ 6&3&4&2&1&5\\\
\hline\cr\end{array}$
$\begin{array}[]{|cccccc|}\hline\cr\pagecolor{lightgray}1&\pagecolor{lightgray}2&5&6&3&4\\\
\pagecolor{lightgray}2&\pagecolor{lightgray}1&6&5&4&3\\\ 6&3&4&2&1&5\\\
5&6&3&4&2&1\\\ 3&4&2&1&5&6\\\ 4&5&1&3&6&2\\\ \hline\cr\end{array}$
$\begin{array}[]{|cccccc|}\hline\cr 4&3&1&2&6&5\\\ 5&6&3&4&2&1\\\
3&4&2&1&5&6\\\ 2&1&5&6&4&3\\\ 6&5&4&3&1&2\\\ 1&2&6&5&3&4\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 3&4&2&1&5&6\\\
6&5&4&3&1&2\\\ 1&2&6&5&4&3\\\ 4&3&1&2&6&5\\\ 2&1&5&6&3&4\\\ 5&6&3&4&2&1\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 5&6&3&4&2&1\\\
4&3&1&2&6&5\\\ 2&1&5&6&3&4\\\ 6&5&4&3&1&2\\\ 1&2&6&5&4&3\\\ 3&4&2&1&5&6\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 6&5&4&3&1&2\\\
3&4&2&1&5&6\\\ 5&6&3&4&2&1\\\ 1&2&6&5&3&4\\\ 4&3&1&2&6&5\\\ 2&1&5&6&4&3\\\
\hline\cr\end{array}$

$\begin{array}[]{|cccccc|}\hline\cr 5&6&4&3&1&2\\\ 6&5&3&4&2&1\\\
1&3&6&2&4&5\\\ 2&1&5&6&3&4\\\ 3&4&2&1&5&6\\\ 4&2&1&5&6&3\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 6&5&3&4&2&1\\\
5&6&4&3&1&2\\\ 4&2&1&5&6&3\\\ 3&4&2&1&5&6\\\ 2&1&5&6&3&4\\\ 1&3&6&2&4&5\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 2&1&5&6&3&4\\\
4&3&1&2&6&5\\\ 6&5&\pagecolor{lightgray}3&\pagecolor{lightgray}4&1&2\\\
5&6&\pagecolor{lightgray}4&\pagecolor{lightgray}3&2&1\\\ 1&2&6&5&4&3\\\
3&4&2&1&5&6\\\ \hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr
1&2&6&5&4&3\\\ 3&4&2&1&5&6\\\
5&6&\pagecolor{lightgray}4&\pagecolor{lightgray}3&2&1\\\
6&5&\pagecolor{lightgray}3&\pagecolor{lightgray}4&1&2\\\ 4&3&1&2&6&5\\\
2&1&5&6&3&4\\\ \hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr
4&3&1&2&6&5\\\ 2&1&5&6&3&4\\\ 3&4&2&1&5&6\\\ 1&2&6&5&4&3\\\ 5&6&3&4&1&2\\\
6&5&4&3&2&1\\\ \hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr
3&4&2&1&5&6\\\ 1&2&6&5&4&3\\\ 2&1&5&6&3&4\\\ 4&3&1&2&6&5\\\ 6&5&4&3&2&1\\\
5&6&3&4&1&2\\\ \hline\cr\end{array}$

$\begin{array}[]{|cccccc|}\hline\cr 6&5&3&4&2&1\\\ 5&6&4&3&1&2\\\
2&4&5&1&3&6\\\ 1&3&6&2&4&5\\\ 4&2&1&5&6&3\\\ 3&1&2&6&5&4\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 5&6&4&3&1&2\\\
6&5&3&4&2&1\\\ 3&1&2&6&5&4\\\ 4&2&1&5&6&3\\\ 1&3&6&2&4&5\\\ 2&4&5&1&3&6\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 3&4&2&1&5&6\\\
1&2&6&5&4&3\\\ 5&6&\pagecolor{lightgray}4&\pagecolor{lightgray}3&2&1\\\
6&5&\pagecolor{lightgray}3&\pagecolor{lightgray}4&1&2\\\ 2&1&5&6&3&4\\\
4&3&1&2&6&5\\\ \hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr
4&3&1&2&6&5\\\ 2&1&5&6&3&4\\\
6&5&\pagecolor{lightgray}3&\pagecolor{lightgray}4&1&2\\\
5&6&\pagecolor{lightgray}4&\pagecolor{lightgray}3&2&1\\\ 3&4&2&1&5&6\\\
1&2&6&5&4&3\\\ \hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr
1&2&6&5&4&3\\\ 3&4&2&1&5&6\\\ 4&3&1&2&6&5\\\ 2&1&5&6&3&4\\\ 6&5&4&3&2&1\\\
5&6&3&4&1&2\\\ \hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr
2&1&5&6&3&4\\\ 4&3&1&2&6&5\\\ 1&2&6&5&4&3\\\ 3&4&2&1&5&6\\\ 5&6&3&4&1&2\\\
6&5&4&3&2&1\\\ \hline\cr\end{array}$

$\begin{array}[]{|cccccc|}\hline\cr 3&4&1&2&5&6\\\ 4&3&2&1&6&5\\\
6&2&4&5&1&3\\\ 5&6&3&4&2&1\\\ 2&1&5&6&3&4\\\ 1&5&6&3&4&2\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 4&3&2&1&6&5\\\
3&4&1&2&5&6\\\ 1&5&6&3&4&2\\\ 2&1&5&6&3&4\\\ 5&6&3&4&2&1\\\ 6&2&4&5&1&3\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 1&2&6&5&4&3\\\
6&5&4&3&1&2\\\ 2&1&5&6&3&4\\\ 4&3&1&2&5&6\\\ 3&4&2&1&6&5\\\ 5&6&3&4&2&1\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 2&1&5&6&3&4\\\
5&6&3&4&2&1\\\ 3&4&1&2&6&5\\\ 1&2&6&5&4&3\\\ 6&5&4&3&1&2\\\ 4&3&2&1&5&6\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 6&5&4&3&1&2\\\
1&2&6&5&4&3\\\ 5&6&3&4&2&1\\\ 3&4&2&1&6&5\\\
4&3&1&2&\pagecolor{lightgray}5&6\\\ 2&1&5&6&3&4\\\ \hline\cr\end{array}$
$\begin{array}[]{|cccccc|}\hline\cr 5&6&3&4&2&1\\\ 2&1&5&6&3&4\\\
4&3&2&1&5&6\\\ 6&5&4&3&1&2\\\ 1&2&6&5&4&3\\\ 3&4&1&2&6&5\\\
\hline\cr\end{array}$

$\begin{array}[]{|cccccc|}\hline\cr 4&3&2&1&6&5\\\ 3&4&1&2&5&6\\\
5&1&3&6&2&4\\\ 6&2&4&5&1&3\\\ 1&5&6&3&4&2\\\ 2&6&5&4&3&1\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 3&4&1&2&5&6\\\
4&3&2&1&6&5\\\ 2&6&5&4&3&1\\\ 1&5&6&3&4&2\\\ 6&2&4&5&1&3\\\ 5&1&3&6&2&4\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 5&6&3&4&2&1\\\
2&1&5&6&3&4\\\ 1&2&6&5&4&3\\\ 3&4&2&1&6&5\\\ 4&3&1&2&5&6\\\ 6&5&4&3&1&2\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 6&5&4&3&1&2\\\
1&2&6&5&4&3\\\ 4&3&2&1&5&6\\\ 2&1&5&6&3&4\\\ 5&6&3&4&2&1\\\ 3&4&1&2&6&5\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 2&1&5&6&3&4\\\
5&6&3&4&2&1\\\ 6&5&4&3&1&2\\\ 4&3&1&2&5&6\\\ 3&4&2&1&6&5\\\ 1&2&6&5&4&3\\\
\hline\cr\end{array}$ $\begin{array}[]{|cccccc|}\hline\cr 1&2&6&5&4&3\\\
6&5&4&3&1&2\\\ 3&4&1&2&6&5\\\ 5&6&3&4&2&1\\\ 2&1&5&6&3&4\\\
4&3&2&1&5&\pagecolor{lightgray}6\\\ \hline\cr\end{array}$

## Acknowledgement

Funding: This work was supported by The Australian Research Council, through
the Centre of Excellence for Plant Success in Nature and Agriculture
(CE200100015) and the second author would like to acknowledge the support of
the Australian Government through a Research Training Program (RTP)
Scholarship.

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