Balanced equi-n-squares

We define a d-balanced equi-n-square L = (lij), for some divisor d of n, as an n×n matrix containing symbols from Zn in which any symbol that occurs in a row or column, occurs exactly d times in that row or column. We show how to construct a d-balanced equi-n-square from a partition of a Latin square of order n into d×(n/d) subrectangles. In design theory, L is equivalent to a decomposition of Kn,n into dregular spanning subgraphs of Kn/d,n/d. We also study when L is diagonally cyclic, defined as when l(i+1)(j+1) = lij + 1 for all i, j ∈ Zn, which corresponds to cyclic such decompositions of Kn,n (and thus α-labellings). We identify necessary conditions for the existence of (a) d-balanced equi-nsquares, (b) diagonally cyclic d-balanced equi-n-squares, and (c) Latin squares of order n which partition into d× (n/d) subrectangles. We prove the necessary conditions are sufficient for arbitrary fixed d > 1 when n is sufficiently large, and we resolve the existence problem completely when d ∈ {1, 2, 3}. Along the way, we identify a bijection between α-labellings of d-regular bipartite graphs and what we call d-starters: matrices with exactly one filled cell in each topleft-to-bottom-right unbroken diagonal, and either d or 0 filled cells in each row and column. We use d-starters to construct diagonally cyclic d-balanced equi-n-squares, but this also gives new constructions of α-labellings. Mathematics Subject Classifications: 05B15, 05C51, 05C78 the electronic journal of combinatorics 27(4) (2020), #P4.8 https://doi.org/10.37236/9118


Introduction
An n × n matrix containing exactly n copies of each symbol from a set of size n is called an equi-n-square (or gerechte framework [9,11], or gerechte skeleton [33]). We typically use Z n as the symbol set. We call an equi-n-square d-balanced if any symbol that occurs in a row or column, occurs exactly d times in that row or column. The question we focus on in this paper is: For what parameters d and n do there exist d-balanced equi-n-squares?
Examples of balanced equi-n-squares are given in Figure 1. Balanced equi-n-squares are a generalization of Latin squares: Latin squares are precisely 1-balanced equi-n-squares.  A square matrix L = (l ij ) on the symbol set Z n , with rows and columns indexed by Z n , which satisfies the property l (i+1)(j+1) = l ij + 1 (1) is called diagonally cyclic. The 3-balanced equi-9-square in Figure 1 is a diagonally cyclic matrix. Diagonally cyclic equi-n-squares are a generalization of diagonally cyclic Latin squares [34] (which are diagonally cyclic equi-1-squares). We observe that: (a) diagonally cyclic n × n matrices on the symbol set Z n are equin-squares (since each of the n broken diagonals contains each symbol exactly once), and (b) diagonally cyclic matrices L = (l ij ) are uniquely determined by their first row.
A subrectangle (resp. subsquare) of a Latin square is a rectangular (resp. square) submatrix in which the number of distinct symbols equals the number of columns. The following lemma gives necessary conditions for the existence of d-balanced equi-n-squares, and describes a relationship between d-balanced equi-n-squares and Latin squares of order n which decompose into d × (n/d) subrectangles.

Lemma 1 (Necessary conditions).
1. For a d-balanced equi-n-square to exist, d must be a divisor of n, and d 2 n.
2. If d is even, for a diagonally cyclic d-balanced equi-n-square to exist, d must be a divisor of n/2.
3. For a diagonally cyclic 1-balanced equi-n-square (i.e., a diagonally cyclic Latin square) to exist, n must be odd. 4. For d 1, a Latin square of order n that decomposes into d × (n/d) subrectangles exists only if a d-balanced equi-n-square exists.
Proof. The symbols in the first row of a d-balanced matrix induce a partition of the n column indices into parts of size d, so d must divide n. A d-balanced equi-n-square L = (l ij ) has d rows in which the symbol 0 occurs, and in each of those rows the symbol 0 occurs d times. Thus, since the symbol 0 occurs n times, we must have d 2 n. This proves the first claim. Now suppose L is diagonally cyclic and d is even. The sum of the symbols in any row or column of L must be divisible by d (as each symbol which occurs, occurs exactly d times). In particular, the first column of L contains the multiset of symbols {l 0j − j} n−1 j=0 , from which we obtain If d = 1 the first row and first column have the same sum, implying n(n − 1)/2 ≡ 0 (mod n) which is satisfied only if n is odd, which implies the third claim; now assume d 2. As d divides n and d 2, we know d is coprime to n − 1, so the above equation implies that d divides n/2, proving the second claim. We defer the proof of the fourth claim to Lemma 8.
This paper proves, for d ∈ {1, 2, 3} the necessary conditions in Lemma 1 are sufficient (with some small exceptions), and for all d 1 the necessary conditions in Lemma 1 are sufficient except for finitely many n. Specifically, we prove the following two theorems. Theorem 1. For n 1,

1.
• a 1-balanced equi-n-square exists for all n, • a diagonally cyclic 1-balanced equi-n-square exists if and only if n is odd, and • there exists a Latin square of order n which decomposes into 1×n subrectangles for all n; 2.
• a 2-balanced equi-n-square exists if and only if n is even and n = 2, • a diagonally cyclic 2-balanced equi-n-square exists if and only if 4 divides n, and • there exists a Latin square of order n which decomposes into 2 × (n/2) subrectangles for all even n ∈ {2, 6}; and 3.
• a 3-balanced equi-n-square exists if and only if 3 divides n and n 9, • a diagonally cyclic 3-balanced equi-n-square exists if and only if 3 divides n and n 9, and • there exists a Latin square of order n which decomposes into 3 × (n/3) subrectangles if and only if 3 divides n and n 9. When d = 1, we are working with Latin squares of order n, which exist for all n 1 (e.g. the Cayley table of Z n ); the rows of a Latin square of order n partition it into 1 × n subrectangles. It is well known that diagonally cyclic Latin squares exist for odd n 1 and do not exist for even n (an early proof was given by Euler [16]; the d = 1 case of Lemma 1 is essentially the same proof). This proves Theorem 1 for d = 1. The rest of this paper is primarily devoted to proving the remaining cases.
Except for 2-balanced equi-6-squares (which exist; see Figure 1), constructing the d-balanced equi-n-squares required to prove Theorems 1 and 2 is achieved through constructing Latin squares of order n that decompose into d × (n/d) subrectangles, then applying the construction in Section 3.
An equi-n-square that is orthogonal to a Latin square (i.e., like symbols in the equin-square correspond to n distinct symbols in the Latin square) are together called a gerechte design (attributed to [5] by [2]). When re-using a field after an agricultural experiment, a gerechte design can be used to balance the carry-over effects from the previous experiment [2]. Vaughan [33] showed the NP-completeness of deciding if an equi-n-square is orthogonal to some Latin square. A Latin square with a decomposition into d × (n/d) subrectangles differs from gerechte designs: in a d-balanced equi-n-square (a) each symbol in the equi-n-square corresponds to a set of n cells in the Latin square that contains exactly d copies of n/d distinct symbols, and (b) we insist on the subrectangles being d × (n/d) rectangular matrices.

Diagonally cyclic balanced equi-n-squares and α-labellings
A d-balanced equi-n-square is equivalent to a decomposition of K n,n into d-regular n-edge spanning subgraphs of K n/d,n/d (the k-th component has a biadjacency matrix corresponding to the occurrences of symbol k in the d-balanced equi-n-square). The decomposition equivalent to the 2-balanced equi-6-square in Figure 1 is given in Figure 2. A decomposition G of K n,n with vertex partition {1, . . . , n} ∪ {n + 1, . . . , 2n} is cyclic if mapping each vertex to its next highest integer (except where n is mapped to 1 and 2n is mapped to n + 1) is a permutation of the decomposition G. Diagonally cyclic dbalanced equi-n-squares L = (l ij ) are equivalent to cyclic decompositions of K n,n into d-regular spanning subgraphs of K n/d,n/d as follows. We assume K n,n has the vertex bipartition {u i } i∈Zn ∪ {v i } i∈Zn . We construct a d-regular spanning subgraph from L for each k by taking the set of edges u i v j when l ij = k. In particular, the starter is this dregular spanning subgraph when k = 0. Note that by construction, each of the d-regular spanning subgraph can be obtained by cyclically rotating the starter.
We identify a particular type of starter (in matrix form) which is helpful in constructing diagonally cyclic d-balanced equi-n-squares. For d 1, we call an r × s matrix A a d-starter if: • every row either contains 0 or d filled cells, • every column either contains 0 or d filled cells, • every top-left to bottom-right (unbroken) diagonal contains exactly one filled cell, and • the number of filled cells in A is r + s − 1.
Actually, the third condition above implies the fourth, but it is useful to make it explicit. Figure 3 depicts some 3-starters. A d-starter A with n filled cells gives rise to a starter for K n,n with an edge u i v j whenever cell (i, j) is filled. A starter for K n,n arising from a d-starter has three special properties: it has n edges, vertices have degree d or 0, and each value of i − j (mod n) is used exactly once. (Actually, this last property holds for all starters.) A d-starter with n filled cells also describes the placement of zeroes in a diagonally cyclic d-balanced equi-n-square, with an example depicted in Figure 4.
Lemma 2. If a d-starter with n filled cells exists, then a diagonally cyclic d-balanced equi-n-square exists.
Proof. We embed the d-starter in an n × n diagonally cyclic matrix L over Z n . The filled cells in the d-starter form the 0s in L. The requirement that each unbroken diagonal in a d-starter has exactly one filled cell implies this process indeed generates a diagonally cyclic matrix. The diagonally cyclic property ensures L is an equi-n-square.
Suppose L is not d-balanced, and some row (resp. column) i contains x ∈ {0, d} copies of some symbol k. The diagonally cyclic property implies row (resp. column) i−k contains x zeroes, contradicting the d-starter property.
We make use of d-starters because of their simplicity, and because they admit the following direct product construction. We illustrate how this construction works in Figure 5. Also, we feel d-starters are interesting combinatorial matrices in their own right. Proof. We embed B in the top-right corner of an n 2 ×n 2 matrix B * . There are n 2 unbroken diagonals of B * which contain a cell in the top row, and each of them contains exactly one filled cell.
We blow up A, replacing each filled cell with a copy of B * , and each empty cell with an n 2 × n 2 all-empty matrix (this is the direct product of A and B * , also known as the Kronecker product). We then delete any boundary rows and columns that do not contain a zero. The result is what we call A ⊗ B.
Suppose A is an r 1 × s 1 matrix, and B is an r 2 × s 2 matrix. In this process, we delete the bottom n 2 − r 2 empty rows and left-most n 2 − s 2 empty columns. Thus we end up with an (r 1 n 2 − n 2 + r 2 ) × (s 1 n 2 − n 2 + s 2 ) matrix. The number of unbroken diagonals is (r 1 n 2 − n 2 + r 2 ) + (s 1 n 2 − n 2 + s 2 ) − 1 = (r 1 + s 1 − 1)n 2 − n 2 + r 2 + s 2 − 1 = n 1 n 2 since n 1 = r 1 + s 1 − 1 and n 2 = r 2 + s 2 − 1. Since the number of filled cells is n 1 n 2 , to prove that A ⊗ B is a d 1 d 2 -starter, we check there are no empty unbroken diagonals. Prior to deleting empty boundary rows and columns, the direct product is composed of n 2 ×n 2 blocks: some empty, and some containing the d 2 -starter. Shrinking the blocks to a single cell regains the d 1 -starter. Moreover, any unbroken diagonal of the direct product (after deleting the empty boundary rows and columns) maps to an unbroken diagonal of the d 1 -starter corresponding to the n 2 × n 2 blocks it intersects in their top rows. This unbroken diagonal of the d 1 -starter contains exactly one filled cell, which corresponds to the only non-empty n 2 ×n 2 block that the original unbroken diagonal of the direct product intersects in its top row. Thus, any unbroken diagonal in the direct product intersects the top row of some non-empty n 2 × n 2 block. Since each non-empty n 2 × n 2 block contains a copy of the d 2 -starter in the top-right corner, this unbroken diagonal has a filled cell.
It is clear from the construction that each row and each column contain either 0 or d 1 d 2 filled cells.  Lemma 1 implies that for a diagonally cyclic d-balanced equi-n-square to exist when d 2, we must have either • d is odd, and d divides n, or • d is even, and 2d divides n.
Thus, for the purposes of proving Theorem 2 in the diagonally cyclic case, we need only consider prime d, then take direct products according to Lemma 4. Since diagonally cyclic d-balanced equi-n-squares are equivalent to cyclic decompositions of K n,n into isomorphic copies of a d-regular graph with n edges (ignoring isolated vertices), we begin by looking at constructions in the graph-decomposition literature.
For the d = 2 case, we require n ∈ {4, 8, 12, . . .} for a diagonally cyclic d-balanced equin-square to exist (by Lemma 1). Rosa [27] and Huang and Rosa [22] showed that when n is a multiple of 4, the complete bipartite graph K n,n can be cyclically decomposed into copies of C n (see also [14]), which proves the d = 2 diagonally cyclic case of Theorem 1. The constructions of [22,27] are effectively the 2-starters depicted in Figure 6, in the present paper's terminology.
For the d = 3 case, we require n ∈ {3, 6, 9, . . .}. Constructions of d-balanced equi-nsquares arise from α-labellings (attributed to Rosa [27] in Gallian's dynamic survey [19]). A graceful labelling of an n-edge graph G is an injection f from V (G) to {0, 1, . . . , n} such that when each edge xy is assigned the label |f (x) − f (y)|, the resulting edge labels are distinct. An α-labelling is a graceful labelling with the additional property that there exists an integer k so that for each edge xy either f (x) k < f (y) or f (y) k < f (x). An example of an α-labelling is given in Figure 7.  Figure 7: The α-labelling (blue) of the 14-vertex 21-edge Möbius ladder given in [24], with parameter k = 6.
El-Zanati and Vanden Eynden [12] showed that if an n-edge graph G admits an αlabelling, then K n,n cyclically decomposes into subgraphs isomorphic to G as follows. Suppose K n,n has the vertex bipartition {u i } i∈Zn ∪ {v i } i∈Zn . Given an α-labelling f of G, we map edges as follows: The distinct values of |f (x) − f (y)| ensure edges are used exactly once when cyclically rotated. If G is a d-regular graph with n edges, then this decomposition is equivalent to a diagonally cyclic d-balanced equi-n-square. Figure 8 depicts the biadjacency matrix of the Möbius ladder in Figure 7 arising from its given α-labelling. The filled cells form a 3-starter. We number the filled cells from bottom-left to top-right using the numbers 1 to 21 sequentially; they correspond to the edge labels arising from the α-labelling in Figure 7.
A technicality for α-labellings is that we treat n and 0 distinctly in the α-labelling (since we require |n − 0| = n), but when constructing the bipartite graph or diagonally cyclic d-balanced equi-n-square, these represent the same index (since indices are in Z n ).
Theorem 3. For d 1, a d-starter A is equivalent to an α-labelling of the d-regular bipartite graph with biadjacency matrix corresponding to A. Thus, a d-starter exists if and only if some d-regular bipartite graph admits an α-labelling.
We construct a bipartite graph with biadjacency matrix B = (b ij ) with b ij = 1 if and only if a ij is filled. The rows of B correspond to vertices labeled {0, 1, . . . , r − 1} and the columns of B correspond to vertices labeled {r, r + 1, . . . , r + s − 1}. We delete vertices corresponding to an empty row or column. We observe the labelling is a graceful labelling: (a) if there is a filled cell in row i and column j, then by definition we have an edge from vertex j to vertex i labelled j − i, and this label is unique since A is a d-starter, and (b) the number of edges is r + s − 1. It is an α-labelling with parameter k = r − 1. Now suppose we have an α-labelling f of a d-regular bipartite graph G, with parameter k and vertex labels belonging to {0, 1, . . . , h} (where a vertex labelled h exists). We construct a k × (h − k + 1) d-starter A = (a ij ), with rows {0, 1, . . . , k − 1} and columns {k, k + 1, . . . , h} with a filled cell a ij whenever there is an edge between vertex i and vertex j. Since G is d-regular, we have either d or 0 filled cells per row and column. By definition of a graceful labelling, each value of |f (x) − f (y)| occurs exactly once; this implies each unbroken diagonal in A contains exactly one filled cell. This verifies that A is a d-starter.
The zeroes in Figure 4 (right) form the biadjacency matrix of the graph 3C 4 (i.e., ), which does not admit an α-labelling, so Theorem 3 implies it does not come from a 2-starter (even after cyclically permuting its rows and columns). Thus not all diagonally cyclic d-balanced equi-n-squares arise from d-starters, or equivalently from α-labellings.
In light of Theorem 3, Lemma 4 implies that if bipartite graphs with biadjacency matrices A and B both admit α-labellings, then so does the bipartite graph which has the biadjacency matrix A ⊗ B, where ⊗ denotes the Kronecker product of matrices. In general, this graph product is not the same as the Cartesian product for which there are many known constructions of α-labellings: it is essentially a "half-Kronecker product" of the graphs.
We turn our attention to proving the existence of d-starters with n filled cells when d and n satisfy the necessary conditions (Lemma 1) for sufficiently large n. Theorem 3 implies Lemma 5 is equivalent to K d,d having an α-labelling, thus Lemma 5 also follows from work in [3,27], which [19] claims contain α-labellings of K d,d .
We color the filled cells blue, green, and red. This matrix is depicted for d ∈ {2, 3, 4, 5} in Figure 10. We claim this is a d-starter. We first note that (2d 2 − 4d + 3) + (2d − 2) − 1 = 2d 2 − 2d, i.e., the required number of filled cells in a d-starter with these dimensions. Check I : no monochromatic clashes. Cells in the same row cannot clash (i.e., belong to the same unbroken diagonal). If two cells (a, b) and (a , b ) with a > a clash, then a − a = b − b . Thus, since indices of distinct non-empty rows differ by at least d − 1, Thus, monochromatic clashes are not possible due to the restrictions on column indices j. For example, red-red clashes are excluded since if a =r i Check II : no blue-red and green-red clashes. For blue cells (a, b), the maximum difference a − b is d 2 − 3d + 2, and for green cells Then the difference a − b for a red cell is always larger than the difference for a blue cell or a green cell. As each cell in an unbroken diagonal has the same difference between its row and column, this means that any unbroken diagonal containing a red cell cannot also contain a blue or green cell. Thus there are no blue-red and green-red clashes.
Check III : no blue-green clashes. Since indices of rows containing blue and green cells differ by a multiple of d − 1, if blue cell (r i , j) and green cell (r i , j ) clash, then The blue cells in row r i belong to columns {0, . . . , i} j and the green cells in row r i = r i+1 belong to columns {d + i, . . . , 2d − 3} j + d − 1, which is impossible.
Check IV : non-empty rows and columns contain d filled cells. By design, each row contains either 0 or d filled cells. Also by design, each non-empty row is a cyclic shift of the other non-empty rows in the array, where all possible cyclic shifts occur in some row; this suffices to show each column contains d filled cells.
The graph G corresponding to the d-starter in Lemma 6 is the d-regular bipartite graph with vertex bipartition {x i } i∈Zn ∪ {y i } i∈Zn , where n = 2d − 2, and edges between each x i with each vertex in {y i , y i+1 , . . . , y i+d−1 }. Theorem 3 therefore implies G admits an α-labelling.
The following lemma describes a way to "adjoin" special types of d-starters. It is equivalent to the d-regular case of [13, Th. 1] by El-Zanati and Vanden Eynden which adjoins "left-free" and "right-free" α-labellings of bipartite graphs.

Lemma 7.
If there exists a d-starter with n 1 filled cells with an empty second row and a d-starter with n 2 filled cells and an empty second-last row, then there exists a d-starter with n 1 + n 2 filled cells.
The basic idea behind adjoining d-starters in Lemma 7 is depicted in Figure 11. We can use Lemma 7 recursively: if where each A i has at least 4 rows, and both the second row and second-last row of each A i is empty, then there exists a d-starter with n 1 + · · · + n k filled cells, for all k 1. We use this idea to prove the following theorem.  If d is even, a d-starter with n filled cells exists for all sufficiently large n divisible by 2d; it does not exist if 2d does not divide n.
Proof. When d 3, Lemma 5 gives a d-starter with n 1 := d 2 filled cells, and Lemma 6 gives a d-starter with n 2 := 2d 2 − 2d filled cells: these both have more than 4 rows and both have empty second rows and second-last rows. Thus, we recursively use Lemma 7 on these two d-starters. Since d-starters exist for all sufficiently large n that satisfy the necessary conditions in Lemma 1.
Theorem 4 implies diagonally cyclic d-balanced equi-n-squares exist for all sufficiently large admissible n.

Decompositions into subrectangles
The following lemma gives a method for using a Latin square of order n which has been partitioned into d × (n/d) subrectangles, for some divisor d of n, to obtain a d-balanced equi-n-square. Figure 12 illustrates Lemma 8 where n = 10 and d = 2.
Lemma 8. Suppose for some divisor d of n there exists a Latin square L = (l ij ) of order n whose entries can be partitioned into d×(n/d) subrectangles. If we index the subrectangles 0, 1, . . . , n − 1, and define an n × n matrix M where cell (l ij , j) contains the index of the subrectangle of L containing the cell (i, j), then M is a d-balanced equi-n-square.
Proof. For each of the d copies of symbol k occurring in the t-th subrectangle of L, the symbol t occurs in row k of M . Hence any row of M containing t contains exactly d copies of t.
Similarly, for each of the entries in the j-th column of the t-th subrectangle of L, the symbol t occurs in column j of M . Hence any column of M containing t contains exactly d copies of t.
We require some modifications to this general approach in individual cases. For brevity, we use standard partition notation such as to represent (2).

Case d = 2
When d = 2, for a 2-balanced equi-n-square to exist (and for a Latin square of order n that can be partitioned into 2 × (n/2) subrectangles to exist), Lemma 1 implies n must be even. When n ≡ 0 (mod 4) there is an easy construction: a direct product gives a Latin square of order n which decomposes into four (n/2) × (n/2) subsquares, each of which decomposes into 2 × (n/2) subrectangles. So we consider n ≡ 2 (mod 4). An exhaustive computer search reveals that no 6×6 Latin square decomposes into 2×3 subrectangles. Figure 12 gives an example of order 10 and Figure 16 gives an example of order 14. The next smallest case is 18.
We get close to decomposing a Latin square of order 6 into 2 × 3 subrectangles with the following: Here, the unhighlighted pairs of rows decompose into three 2 × 2 subsquares.
Then there exists a 2m×2m Latin square that can be partitioned into 2×m subrectangles.
Proof. We prove the theorem with the aid of a running example for m = 13, beginning with the matrix in Figure 17 (left). We highlight how to select 1 or 2 rows from the 3 × 3 submatrices U t . Lemma 9 implies that we can construct an m × m Latin square L, with symbols in {0, 1, 2} wherever the symbol 0 occurs in M , symbols in {3, 4, 5} wherever the symbol 1 occurs in M , symbols in {6, 7, 8} wherever the symbol 2 occurs in M , and symbols in {9, 10, . . . , m − 1} wherever an empty cell occurs in M . Moreover, the chosen rows of the 3 × 3 submatrices U t in M map to rows of 3 × 3 subsquares of L. This is depicted in Figure 17 (right); we highlight the subsquares we subsequently discuss in this proof.
We next take a direct product of L with a 2 × 2 Latin square, giving a 2m × 2m Latin square. Consequently, we also blow up the selected rows of the 3 × 3 subsquares of L to 2 × 6 subrectangles. Denote these subrectangles S i for i ∈ {0, 1, . . . , m − 1}.
This selection covers all but a few boundary rows. We still need to choose rows for indices • We choose row 2 from J 2, m/3 −2 .
Examples of this construction are given in Figure 20. We conclude that this construction satisfies the conditions of Theorem 5. Finally, combining the results in this section (i.e., a direct product for m ≡ 0 (mod 2); explicit examples for m ∈ {5, 7, 11}; a special case of Theorem 5 for m = 9; and this construction for odd m 13), we completely resolve the existence problem for Latin squares of order n which decompose into 2 × (n/2) subrectangles.

Case d = 3
The construction we give for decomposing Latin squares of order n into 3 × (n/3) subrectangles is similar to the 2 × (n/2) construction, so we only explain the differences. In fact, the construction is made easier in this case because there exist Latin squares of order 12 which decompose into 3 × 4 subrectangles (as opposed to Latin squares of order 6 which do not decompose into 2 × 3 rectangles); we include one example in Figure 21    We have designed M 4,m,3 (which exists when m ∈ {12, 16, 17, 18} and m 20) so that it satisfies the two matrix conditions in Theorem 6. (We omit the details of a formal check of these properties; they are straightforward to prove, and are apparent from Figure 15). Thus the conditions of Theorem 6 are satisfied when m ∈ {12, 16, 18} ∪ {20, 21, . . .}, in which case there exists a 3m × 3m Latin square that can be partitioned into 3 × m subrectangles.
We still need to resolve the m ∈ {3, 4, . . . , 11} ∪ {13, 14, 15, 17, 19} cases. If m is divisible by 3, we take a direct product of a Latin square of order 3 and a Latin square of order m to obtain a Latin square of order 3m which can be partitioned into 3 × 3 subsquares, which we group together to give a decomposition into 3 × m subrectangles. When m ∈ {4, 5, 7, 11} the existence problem is settled by examples in Figure 21 and Case m = 17. This is similar to the m = 13 case, so we just highlight the differences. We blow up the matrix on the right in Figure 23. We make the replacements tabulated below: after blowing up. . . we replace it with. . . R, S Figure 22 (top left) U , V , X, Y Figure 21 (left) . . .after adjusting the symbols In the resulting Latin square of order 51, sets of 3 consecutive rows decompose into one of the following: • exactly three 3 × 5 subrectangles, and exactly twelve 3 × 3 subsquares; • exactly three 3 × 5 subrectangles, exactly three 3 × 4 subrectangles, and exactly eight 3 × 3 subsquares; or • exactly six 3 × 4 subrectangles, and exactly nine 3 × 3 subsquares. describes partitions of 51 into three partitions of 17, in every case, it is possible to group these subrectangles and subsquares together to obtain three 3 × 17 subrectangles.

Sufficiently large n
We begin with the following Latin square where sets of d consecutive rows can be partitioned into one d × x subrectangle and one d × y subrectangle, where x ≡ 1 (mod d) and y ≡ −1 (mod d). Proof. We define an n × n matrix containing symbols 0 and 1 (and some empty cells) in which every row and every column has exactly d + 1 occurrences of symbol 0 and exactly d + 1 occurrences of symbol 1. The construction is illustrated in Figure 24 for d ∈ {3, 4}. We add all-i blocks according to the table below (all other cells are empty; we "mark" the submatrices to match Figure 24): dimensions top-left corner symbol mark The d × (d + 1) and 1 × (d + 1) all-1 blocks combine to form (d + 1) × (d + 1) all-1 submatrices. Applying Lemma 9 (König's Theorem) to this matrix gives a Latin square L, which has (d + 1) × (d + 1) subsquares wherever an all-i submatrix occurs.
By taking a direct product of the Latin square of order d(d + 1) in Theorem 7 with a Latin square of order (d − 1) 2 , we let S d be a Latin square of order d(d We define the m × m matrix P = M p,m,2 where p := (d + 1)(d − 1) 2 and observe that every row of P intersects a p × p all-0 block and/or a p × p all-1 block. We use Lemma 9 (König's Theorem) to show that there exists an m×m Latin square L with p×p subsquares occurring whenever a p × p all-1 block occurs in P . We take a direct product of L with a Latin square of order d and replace the dp × dp subsquares which arise due to all-i blocks in P with copies of S d after relabelling the symbols appropriately. This gives an n × n  1), which is enough when m is large. This gives a construction for Latin squares of order n which decompose into d × (n/d) subrectangles for all sufficiently large n divisible by d.
We note that "sufficiently large n" in this construction is at least asymptotically d 4 , which likely far exceeds the actual minimum n.
The smallest unresolved case is the existence of a 4-balanced equi-20-square, and a 20 × 20 Latin square that decomposes into 4 × 5 subrectangles. An exhaustive computer search rules out the existence of a 5-starter with 30 filled cells (and thus K 6,6 − I, where I is a 1-factor, does not admit an α-labelling). The proof of Theorem 3 also works for α-labellings of bipartite graphs in general (after dropping the condition that there are "d filled cells in each row and column"), which we illustrate in Figure 25. While we do not need this generality in this paper, it would be interesting to explore α-labellings from this perspective in future research. Moreover, the proof of the product construction in Lemma 4 holds for this generalization of d-starters, which implies the following lemma.
Lemma 10. If two bipartite graphs with the biadjacency matrices A and B admit αlabellings, then the graph with biadjacency matrix A⊗B, where ⊗ is the matrix Kronecker product, also admits an α-labelling.
Brankovic and Wanless [8] found a relation between α-labellings of a path and partial transversals of the Cayley table of the cyclic group, which has a similar style to this construction. Graceful labelings of paths give rise to cyclic oriented triangular embeddings of complete graphs [20]. It is straightforward to construct α-labellings (and hence graceful labelings) of all caterpillars via these matrices: if the main path has degree sequence (d i ) n i=1 , we start at the bottom right, fill d 1 = 1 cells vertically, then d 2 cells horizontally, then d 3 cells vertically, and so on. This is depicted in Figure 26. Graceful labellings of caterpillars in particular are studied in connection with multi-protocol label switching in IP networks [1,4]. We observe that if we take a d-starter with n zeroes, rotate it by 90 • , then superimpose it on the multiplication table of Z n , we obtain a kind of generalized transversal. In this way, 1-starters correspond to transversals of the multiplication table of Z n . For example, a 24-element 4-starter embeds in the multiplication table of Z 24 as in Figure 27. This selection of entries has a unique copy of each symbol in Z n , and each row and each column is either unrepresented, or is represented exactly d times.
In this paper, we also describe decomposing a Latin square into subrectangles. This leads to a range of interesting problems, where many problems related to graph decomposition have analogues with Latin squares. The subrectangles we consider are not necessarily structurally equivalent (isotopic), so we have not resolved the problem of decomposing Latin squares into isotopic copies of a subrectangle. In fact, we might consider dropping the constraint that the subrectangles be horizontally aligned. Moreover, we need not limit ourselves to rectangular submatrices: it is interesting to ask when there exists a Latin square that decomposes into isotopic copies of a given partial Latin square.
Along these lines, the existence of a Latin square that decomposes into d × (n/d) submatrices containing all n symbols was resolved in [11]: it is possible for all divisors d of n. In [6], the authors observe that all Latin squares of order n have an O(n 1/2+ε ) × O(n 1/2+ε ) submatrix containing n−O(n 1/2+ε ) distinct symbols, which raised the existence problem for Latin squares of order n 2 which cannot decompose into n × n submatrices which contain all n symbols.
Diagonally cyclic equi-n-squares are equivalent to equi-n-squares that admit an n-cycle automorphism. Latin squares that admit automorphisms are used in secret sharing [32], and the Latin square could easily be replaced by an equi-n-square, and interpreted as a graph decomposition [7] (see [30] for a broad and detailed treatment). In abstract algebra, diagonally cyclic equi-n-squares correspond to n-element magmas (sometimes called groupoids) that admit n-cycle automorphisms.
Diagonally cyclic equitable rectangles [17] are another kind of generalization of diago-nally cyclic Latin squares. Equitable rectangles are studied in connection with generalized mix functions [26,31]. Like diagonally cyclic d-balanced equi-n-squares, they admit a compact description, where we need only store one entry from each diagonal.