On Completing Partial Latin Squares with Prescribed Diagonals

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Introduction
Historically, a (partial) latin square L of order n is an n × n array in which each cell contains (at most) one symbol in S(n) = {1, 2, . . . , n} and each of the symbols in S(n) occurs (at most) once in each row and (at most) once in each column. Let L(i, j) denote the symbol in cell (i, j) of L, and let N L (i) (or simply N (i) if L is clear) be the number of cells that contain symbol i in L. A (partial) incomplete latin square of order n (also referred to as a (partial) latin array of order n) on the symbols in S(t) is an n × n array in which each cell contains (at most) one symbol in S(t) and each of the symbols in S(t) occurs at most once in each row and at most once in each column. A partial or incomplete latin square L of order n is said to be embedded in the latin square T of order t if for each cell (i, j) of L that contains a symbol, L(i, j) = T (i, j). The cells (i, i) for n+1 i t are said to be the diagonal of T outside L. A latin square of order n is said to be idempotent if L(i, i) = i for 1 i n, and is said to be symmetric if L(i, j) = L(j, i) for 1 i j n.
There is a rich history of papers that consider the embedding of partial and incomplete latin squares; the following is a sample of such results. The classic result of Ryser [12] shows that an incomplete latin square L of order n on the symbols in S(t) can be embedded in a latin square of order t if and only if N L (i) 2n − t for 1 i t. This condition is known as the Ryser condition. Evans [7] obtained a related result for partial latin squares, proving that any partial latin square of order n can be embedded in a latin square of order t for any t 2n. This result is best possible in that there are partial latin squares of order n that cannot be embedded in a latin square of order t if t < 2n. Cruse [6] then found necessary and sufficient conditions for a partial latin square of order n to be embedded in a symmetric latin square of order t, and also to be embedded in an idempotent symmetric latin square of order t, where in both cases t > n is arbitrary. It turns out that embedding partial and incomplete latin squares in an idempotent latin square is a very difficult problem. The Ryser conditions can naturally be extended to provide a necessary condition for an incomplete idempotent latin square L of order n with symbol set S(t) to be embedded in an idempotent latin square of order t with symbol set S(t), namely that N L (i) 2n − t + f (i) for 1 i t, where f (i) = 0 for 1 i n and f (i) = 1 for n + 1 i t. It was shown by Andersen et al. [3,4] that for all t < 2n these Ryser-type conditions are not sufficient: there exists an incomplete idempotent latin square of order n which cannot be embedded in an idempotent latin square of order t. In some cases, just swapping the placement of symbols in two cells results in one which does have an idempotent embedding. So, for the first time in these sorts of embedding problems, the arrangement of the symbols in L can determine its embeddibility, thus making the idempotent setting quite special. The case where t 2n was finally settled after various results reduced the bound on t. Treash [13] showed that a finite embedding of a partial idempotent latin square was always possible, Lindner [8] reduced the bound to around 6n, conjecturing that 2n + 1 was the right lower bound (the Ryser-type conditions come into play when t 2n), Andersen [5] further reduced it to t 4n and t = 4n + 1, and finally Andersen et al. [2] settled the Lindner conjecture which states that any partial idempotent latin square can be embedded in an idempotent latin square of order t, for any t 2n + 1. The idempotent embedding for incomplete idempotent latin squares was then settled for all t 2n by Rodger [10].
A natural generalization to embedding an incomplete latin square L of order n with symbol set S(t) into an idempotent latin square T of order t is to more generally prescribe what is to occur on the diagonal: suppose it is required that for 1 i t symbol i should occur f (i) times in the diagonal cells of T outside L. Then the Ryser-type conditions are again necessary, and if t 2n + 1 then Rodger [11] proved they, along with two other necessary conditions, are also sufficient. It is the case that if f (i) = 1 for some symbol, i, then Andersen et al. [3] again showed that when t < 2n the arrangement of symbols in L can determine if L can be embedded in T with the given prescribed diagonal of T outside L. Rodger conjectured that if f (i) = 1 for 1 i t then, even when t 2n, the Ryser-type conditions are sufficient. It is this 30 year old conjecture that we prove in this paper.

Previous Results
Before proving the main result, Theorem 4, we note the following three results.
Andersen et al. [1] proved Theorem 1, which completely settles the embedding problem providing not all of the diagonal is prescribed.
Then L can be embedded in a latin square T of order t on the same symbols in which each symbol i appears at least f (i) times on diagonal of T outside L if and only if N L (i) 2n − t + f (i) for 1 i t.
The following classic theorem, proven by Ryser [12], will be used in Step 1 of the proof of Theorem 4. Theorem 2 ( [12]). An incomplete latin square L of order n on the symbols in S(t) can be embedded in a latin square of order t on the same symbols if and only if N L (i) 2n − t for 1 i t.
A family L of sets is said to be a laminar set if X, Y ∈ L implies that X ⊆ Y , Y ⊆ X, or X ∩ Y = ∅. Nash-Williams [9] proved the following result which will play a critical role in Step 3 of the proof of Theorem 4.

Main Result
In the following proof, f (i) will be modified in various ways. With this in mind, the incomplete latin square L of order n is said to be (f, t)-satisfied if N L (i) 2n − t + f (i) for 1 i t. We say a symbol i satisfies Ryser's condition if N L (i) 2n − t + f (i).
Theorem 4. Let t n > 0. Let L be an incomplete latin square of order n on the symbols in S(t). Let f : S(t) → N such that n i=1 f (i) = t − n and f (i) = 1 for 1 i t. Then L can be embedded in a latin square T of order t on the same symbols in which each symbol i appears f (i) times on the diagonal of T outside L if and only if N L (i) 2n − t + f (i) for 1 i t.
Proof. The necessity is well known, so assume that N L (i) 2n − t + f (i) for 1 i t.
Suppose there exists a symbol α for which f (α) Then by Theorem 1, L can be embedded in a latin square T of order t in which for 1 i t, symbol i occurs at least f (i) times on the diagonal of T outside L. By a permutation of rows and columns if needed, assume T (n + 1, n + 1) = α. Define the incomplete latin square L of order n + 1 by L (a, b) = T (a, b) for 1 a, b n + 1. We now show that L is (f , t)-satisfied. Because T (n + 1, n + 1) = α, Also, since L is embedded in T , by the necessary condition in Theorem 1, N L (i) 2(n + 1) − t + f (i) for 1 i t, i = α. Thus L is an incomplete latin square of order n + 1 satisfying the conditions of the theorem. Therefore, by repeating this process, we can assume that f (i) ∈ {0, 2} for 1 i t; so t − n is even.
The remainder of the proof is completed in three steps. In each step, two rows and columns are added so that the resulting incomplete latin square satisfies the necessary condition after appropriately modifying f to allow for the symbol placed in both the added diagonal cells.
Step 1. Suppose t − n = 2. Then f (α) = 2 for exactly one symbol α, and f (i) = 0 for all symbols i = α. By assumption, N L (α) 2n − t + f (α) = 2n − (n + 2) + 2 = n. Because L is of order n, N L (α) = n. Use Theorem 2 to embed L in a latin square T of order t. Because N L (α) = n, symbol α must appear twice in the 2 × 2 square formed with rows and columns t − 1 and t of T . If α is on the diagonal, we are done. If not, then permute columns t − 1 and t to obtain the required embedding. Thus we can assume t − n 4.
Step 2. Suppose t − n 8. Let s = (t − n)/2. By renaming symbols, we can assume that f (i) = 2 for 1 i s and f (i) = 0 for s + 1 i t. We wish to extend L by 2 rows and 2 columns embedding L in a latin square of order n + 2 that satisfies the conditions of the theorem. Define Thus by Theorem 1 and  Figure 1. We now have s − 1 candidates for extending L by two rows and two columns, namely L 1 , . . . , L s−1 . We now show that at least one of them must satisfy the necessary conditions of the theorem. (It is only symbol s that is potentially problematic because f (s) = f (s). However, we show for at least one value of i, 1 i s − 1, s appears the necessary number of times in A i ∪ B i ∪ D i , so L i meets the necessary conditions of the theorem.) Suppose 1 i s − 1. Permute the rows and columns of T to produce a latin square T i such that L i is embedded in T i and for 1 j t, j = i, symbol j appears in at least . Because i appears 2 more times on the diagonal of L i than it did in L, Since L i is embedded in T i , by the necessity of Theorem 1, for 1 j t, j / ∈ {i, s}, Also, by the necessity of Theorem 1, We claim that for some i, 1 i s − 1, s satisfies Ryser's condition in L i , so in actuality .
Symbol s appears n times in the first n rows of T i (by the definition of a latin square), but does not appear in the (t − 1) th column of the first n rows because symbol s appears on the diagonal in that column. Symbol s could possibly appear in the t th column of the first n rows. Thus So, because s 4, N L (s) (2n − t + 1), contradicting our original assumption. Therefore, for some value of i, Thus L α is an incomplete latin square of order n + 2 that is (f α , t)-satisfied and thus satisfies the conditions of the theorem. By repeating this process, we may now assume t − n 6.
Step 3. Suppose t − n ∈ {4, 6}. Form a bipartite multigraph G * c with bipartition C = {c 1 , c 2 , . . . , c n , c * } and S = {σ 1 , σ 2 , . . . , σ t } of the vertex set as follows. For 1 i n and 1 j t, join c i to σ j if and only if symbol j is missing from column i of L and join c * to σ j with f (j) edges. Similarly, form a bipartite multigraph G * ρ with bipartition R = {ρ 1 , ρ 2 , . . . , ρ n , ρ * } and S = {σ 1 , σ 2 , . . . , σ t } of the vertex set as follows. For 1 i n and 1 j t, join ρ i to σ j if and only if symbol j is missing from row i of L, and join ρ * to σ j with f (j) edges. Because each column and row of L contains n symbols, for Because For 1 j t, symbol j is missing from n − N L (j) rows of L and n − N L (j) columns of . Thus, by (3), So, ∆(G c * ) = ∆(G ρ * ) = t−n and z(j) measures how far σ j and σ j are from this maximum degree. Define a laminar set L 1 of subsets of E(G * c ) ∪ E(G * ρ ) as follows. For 1 i n, let C i ∈ L 1 be the set of edges incident to c i . Let C * ∈ L 1 be the set of edges incident to c * . For 1 j t such that f (j) > 0, let C * j ∈ L 1 be the subset of C * given by the two element set of the pair of edges joining c * and σ j . Similarly, for 1 i n, let R i ∈ L 1 be the set of edges incident to ρ i . Let R * ∈ L 1 be the set of edges incident to ρ * . For 1 j t such that f (j) > 0, let R * j ∈ L 1 be the subset of R * given by the two element set of the pair of edges joining ρ * and σ j . Define a second laminar set L 2 of subsets of E(G * c ) ∪ E(G * ρ ) as follows. For 1 j t, let S j ∈ L 2 be the set of edges incident to σ j , S j ∈ L 2 be the set of edges incident to σ j , and Σ j ∈ L 2 be the set of all edges incident to either σ j or σ j . By Theorem 3, there exists a set J ⊆ (E(G * c ) ∪ E(G * ρ )) for which Let G J be the graph induced by the edges of G * c and G * ρ in J. Later, a modified version of G J will be colored with 2 colors and be used to fill rows and columns n + 1 and n + 2 to embed L in an incomplete latin square of order n + 2. But first we explore G J to see what modifications are needed.
Recall, deg G J (c * ) = 2. Because C * j ∈ L 1 at most one edge {c * , σ j } ∈ C * j is in J. So, the two edges in J incident to c * are incident to two different vertices in S. Similarly, there are exactly two edges in J incident to ρ * , each of which is incident to two different vertices in S . Because c * and ρ * are incident to σ j and σ j respectively for the same two (if t − n = 4) or three (if t − n = 6) values of j, there must exist an α such that 1 α t, {c * , σ α } ∈ J and {ρ * , σ α } ∈ J.
In what follows we construct another set of edges J through a modest modification of J so both edges in C * α and both edges in R * α will be in J . The graph G J is a bipartite graph with maximum degree 2. Thus, the edges of G J can be properly colored with 2 colors, say 1 and 2. Consider the graphs G * c − J and G * ρ − J. They are bipartite graphs with maximum degree t − n − 2. Thus, the edges of G * c − J and G * ρ − J can be properly colored with t − n − 2 colors, say 3, . . . , t − n. These two edge-colorings naturally induce a proper (t In what follows we construct an edge-coloring X : E(G * c ∪ G * ρ ) → {1, 2, . . . , t − n} by interchanging colors on two 2-colored trails, T 1 and T 2 , in X. In X the edges in C α will be colored 1 and 2 and the edges in R α will be colored 1 and 2. Suppose the edges in C α are colored 1 and 3 by X. Consider the maximal trail, T 1 , containing the edge {c * , σ α } colored 3, in which the edges are alternately colored 2 and 3 by X. Because the edge-coloring is proper, T 1 is either a cycle or a path. Interchange the colors on T 1 and let this new edge-coloring be X on the edges in G * c . The edges in C α are now colored 1 and 2 by X . If T 1 is a cycle, interchanging colors did not impact the number of edges of each color incident to each vertex. Suppose T 1 is a path. Interchanging colors did not impact the number of edges of each color incident to each vertex in the interior of T 1 , but did impact the endpoints. For each c ∈ C, by (1) and (2), deg(c) = t − n. So there is exactly one edge colored 2 and one edge colored 3 by X incident to vertex c. Thus c is not an endpoint of T 1 , so the endpoints of T 1 must be in S. Because G * c is bipartite and both ends of T 1 are in S, exactly one of the ends was incident to an edge colored 2 by X. This end cannot be σ α because σ α was incident to an edge colored 3 by X. So one end of T 1 is a vertex σ u ∈ S \ {σ α } that now does not have an edge colored 2 by X incident to it. The other end of T 1 was incident to an edge colored 3 by X. So this vertex now is incident to an edge colored 2 by X . Similarly, we can use a trail T 2 to modify the proper edge-coloring, X, of G * ρ and define X on the edges of G * ρ so the edges in R α are now colored 1 and 2 in X . After recoloring, at most one vertex in S \ {σ α } has lost an edge colored 2 incident to it in X . If such a vertex exists, name it σ v . All other vertices in G * c ∪ G * ρ have an equal or greater number of edges colored 2 incident to them. Thus, we have the revised edge-coloring X : E(G * c ∪ G * ρ ) → {1, 2, . . . , t − n}. Define J to be the set of edges colored 1 and 2 by X , and let G J be the graph induced by the edges in J .
It is important to note a property that will be used later in the proof if σ u and/or σ v have been defined. In X , σ u and σ v do not have an edge colored 2 incident to them, To finally arrive at the desired edge-coloring, X : Case 3 below). To do this we construct the edge-coloring X by interchanging colors on up to two trails, T 3 and T 4 , whose edges are colored 1 and 2 in X as follows. We Let e u and e v be the edges incident to σ u and σ v in G J respectively. In X , we make sure that one of e u and e v is colored 1 and the other is colored 2. If X (e u ) = X (e v ), then we already have the desired property, so define X (e) = X (e) for all e ∈ E(G * c ). Otherwise, X (e u ) = X (e v ). Take a maximal trail, T 3 , in G J that begins with e u . Because ∆(G J ) = 2, the trail T 3 is necessarily a path. For each c ∈ C, deg G J (c) = 2. Thus c is not an endpoint of T 3 , so the endpoint of T 3 must be in S. The path T 3 cannot end with e v because X (e u ) = X (e v ) and G J is a bipartite graph. So, because deg G J (σ v ) = 1, T 3 does not include e v . Interchange colors along T 3 . All interior vertices of T 3 still have exactly one incident edge colored 1 and exactly one incident edge colored 2. The endpoints of T 3 now have the opposite color incident to them. Define X on the edges in G * c to be this new edge-coloring. Thus, in any case, if e u and e v exists, we can assume For convenience, if deg G J (σ u ) = 1 or deg G J (σ v ) = 1, so we are not in the above case, then define X (e) = X (e) for all e ∈ E(G * c ). Similarly, we can define X on the edges in G * ρ by interchanging colors on a trail T 4 in X if needed. So, if deg G J (σ u ) = 1 and deg G J (σ v ) = 1 and we let e u and e v be the edges incident to σ u and σ v in G J respectively, then For convenience, if deg G J (σ u ) = 1 or deg G J (σ v ) = 1, so we are not in the above case, then define X (e) = X (e) for all e ∈ E(G * ρ ). Thus, we have the revised edge-coloring X : Define J to be the set of edges colored 1 and 2 by X , and let G J be the graph induced by the edges in J .
We will use G J to fill in rows and columns n + 1 and n + 2 and extend L to an incomplete latin square of order n + 2. For 1 i t, deg G J (c i ) = deg G J (ρ i ) = 2. For 1 j t, deg G J (σ j ) 2 and deg G J (σ j ) 2. Also, for 1 j t, by (5), where j = 0 if j / ∈ {u, v}, j = 1 if j ∈ {u, v} and u = v, and j = 2 if j = u = v. Form a partial incomplete latin square L α of order n + 2 by adding two new rows and columns to L using J as follows. Let L α (a, b) = L(a, b) for 1 a n and 1 b n. Define the sets of cells A, B and D as follows: {(a, b) : 1 a n, n + 1 b n + 2}, B : = {(a, b) : n + 1 a n + 2, 1 b n}, and D : = {(a, b) : n + 1 a, b n + 2}.
Let A, B, and D be the n × 2, 2 × n, and 2 × 2 latin subrectangles of L α formed by the cells in A, B, and D respectively. Similarly, let A ∪ B be the array formed by the cells in A and B. We now fill A and B using G J . For 1 k 2 and 1 i n, let L α (n + k, i) = j if and only if {c i , σ j } is colored k in G J . Similarly, for 1 k 2 and In a later modification of L α we will place α in the two new diagonal cells, so define f (α) = f (α) − 2 and f (j) = f (j) for 1 j t and j = α. For 1 j t and j / ∈ {u, v, α}, by (10), Thus, L α is a partial incomplete latin square of order n + 2 with all cells except those in D filled and all symbols satisfying Ryser's condition except possibly α (which will satisfy Ryser's condition once placed twice on the diagonal in D) and possibly u and v if they exist. The aim is to construct L α through a modest modification of L α to form a partial incomplete latin square of order n + 2 on S(t) which is (f , t)-satisfied. By (10), N Lα (u) 2(n + 2) − t + f (u) − u and N Lα (v) 2(n + 2) − t + f (v) − v . We will now modify L α to form L α so that if N Lα (j) < 2(n + 2) − t + f (j) for any j ∈ {u, v}, u and/or v will be placed in cells (n + 1, n + 2) and/or (n + 2, n + 1) of L α as needed to ensure that N L α (u) 2(n + 2) − t + f (u) and Let j ∈ {u, v}. If N Lα (j) < 2(n + 2) − t + f (j), by (10), N Lα (j) = 2n − t + f (j) + 2 or N Lα (j) = 2n − t + f (j) + 3. These two cases of N Lα (j) correspond to L α needing 2 or 1 more occurrence of j respectively. To reveal more about these potentially problematic cases, consider the following properties. Recall, by (6), N L (j) 2n − t + f (j) + 1.
In the edge-coloring X , σ u is missing exactly one of the colors 1 or 2 (because T 1 ended on this vertex). So, since σ u is incident to an edge of every color in G * c except one, σ u must be incident to the other color (1 or 2). Therefore, deg G J (σ u ) = 1, and so because u appears at most 2 times If N Lα (j) 2(n + 2) − t + f (j) for j ∈ {u, v}, then let L α (a, b) = L α (a, b) for 1 a, b n + 2. Otherwise we will make use of (i-iv) to modify L α to define L α and place u and/or v in cells (n + 1, n + 2) and/or (n + 2, n + 1) of L α as needed to ensure that, for j ∈ {u, v}, N L α (j) 2(n + 2) − t + f (j). The following three cases are considered. The first two cases consider if exactly one of u or v, say u, does not meet Ryser's condition. So, by (10), N Lα (u) = 2(n + 2) − t + f (u) − 2 or N Lα (u) = 2(n + 2) − t + f (u) − 1. The third case considers when both u and v do not meet Ryser's Condition, so, by (10), Then u does not appear in A nor in B. Define L α (a, b) = L α (a, b) for (a, b) ∈ A ∪ B and for 1 a, b n. Also, define L α (n + 1, n + 2) = L α (n + 2, n + 1) = u. Thus N L α (u) = N Lα (u) + 2 = 2(n + 2) − t + f (u).
Case 2: Suppose that for exactly one of u or v, say u, N Lα (u) = 2(n + 2) − t + f (u) − 1 and for the other, say v, N Lα (v) 2(n + 2) − t + f (v), u = v, or v does not exist. By (i,iii,iv), u is in at most one row of B, say n + 2, and at most one column of A, say n + 1 (permuting the columns and/or rows of A and/or B respectively if need be). Define L α (a, b) = L α (a, b) for (a, b) ∈ A ∪ B or 1 a, b n. Also, define L α (n + 1, n + 2) = u. Thus N L α (u) = N Lα (u) + 1 = 2(n + 2) − t + f (u).
By (i-ii, iv), u and v each appear at most once in A and at most once in B. By (7) and (8), we can assume u and v appear in different rows of B and different columns of A. Thus, permuting rows and/or columns if necessary we can assume u does not appear in row n + 1 nor in column n + 2 of L α and v does not appear in row n + 2 nor in column n + 1 of L α . Define L α (a, b) = L α (a, b) for (a, b) ∈ A ∪ B or 1 a, b n. Also, define L α (n + 1, n + 2) = u and L α (n + 2, n + 1) = v. So N L α (u) = N Lα (u) + 1 = 2(n + 2) − t + f (u) and Thus, in any case, we can place u and/or v in cells (n + 1, n + 2) and/or (n + 2, n + 1) if needed so that for j ∈ {u, v}, Define L α (n + 1, n + 1) = L α (n + 2, n + 2) = α. So N Lα (α) = N L (α) + 2 2(n + 2) − t + f (j).
Also for 1 j t and j / ∈ {u, v, α}, N L α (j) = N Lα (j), so by (11), Thus, L α is a partial incomplete latin square of order n + 2 with all symbols satisfying Ryser's condition and all cells filled except possibly cells (n + 1, n + 2) and (n + 2, n + 1). We now define L α through a modest modification of L α to fill cells (n+1, n+2) and/or (n + 2, n + 1) if needed to form an (f , t)-satisfied incomplete latin square. Suppose cell (n + 1, n + 2) of L α is empty. Form the bipartite graph B with bipartition C = {c i | 1 i n + 2} and S = {σ j | 1 j t} of the vertex set as follows. For 1 i n + 2 and 1 j t, join c i to σ j if and only if symbol j is missing from column i of L α or L α (n + 1, i) = j. For c i ∈ C \ {c n+1 }, deg B (c i ) = t − n − 1. Because j appears at most once in row n + 1, for σ j ∈ S, deg B (σ j ) n + 2 − (N L α (j) − 1) n + 2 − (2(n + 2) − t + f (j) − 1) = t − n − f (j) − 1 t − n − 1.
Define the matching M by letting {c i , σ j } ∈ E(B) be in M if and only if L α (n + 1, i) = j.
Because symbol α appears in cell (n + 1, n + 1), {c n+1 , σ α } ∈ M . Let B be the induced subgraph of B formed by removing vertices c n+1 and σ α . We wish to find an M -augmenting path in B starting at c n+2 . For a contradiction, suppose there does not exist an Maugmenting path in B starting at c n+2 . Let W be the subgraph of B induced by the set of vertices that can be reached by an M -alternating path starting at c n+2 . All maximal Malternating paths starting at c n+2 end at an M -saturated vertex in C \ {c n+1 }. So V (W ) contains say x vertices from S \ {σ α } and V (W ) contains x + 1 vertices from C \ {c n+1 }, namely c n+2 and the M -neighbors of the x vertices from S \ {σ α }. Let C W = C ∩ V (W ) denote the set of these x + 1 vertices. By the definition of W , every edge in B incident to a vertex in C W must be an edge in W (which implies the equality in the relations below). Because deg B (σ α ) t − n − 1 (by (12)) and σ α is adjacent to c n+1 in B, at most t − n − 2 vertices in C W have degree t − n − 2 in B and all other vertices in C W have degree t − n − 1 in B . So, This is a contradiction. Thus, there exists an M -augmenting path in B starting at c n+2 . Form the matching M from M by interchanging edges in M with edges not in M along this path. Now replace row n + 1 of L α to form L α using M by letting L α (n + 1, i) = j if and only if {c i , σ j } is in M for 1 i n + 2 and 1 j t and letting L α (a, b) = L α (a, b) for 1 a n or a = n + 2 and 1 b n + 2. Thus, L α contains the same symbols as L α with the addition of one more symbol in row n + 1. Now, all cells in row n + 1 are filled. Similarly, if cell (n + 2, n + 1) is empty, we can modify L α using the same approach. So we can assume all cells of L α are filled and all symbols satisfy Ryser's condition. Thus, L α is an incompete latin square of order n + 2 that is (f , t)-satisfied and thus satisfies the conditions of the theorem.