Completing partial Latin squares with one filled row, column and symbol

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Completing partial Latin squares with one filled

row, column and symbol

Carl Johan Casselgren and Roland Häggkvist

Linköping University Post Print

N.B.: When citing this work, cite the original article.

Original Publication:

Carl Johan Casselgren and Roland Häggkvist, Completing partial Latin squares with one filled row, column and symbol, 2013, Discrete Mathematics, (313), 9, 1011-1017.

http://dx.doi.org/10.1016/j.disc.2013.01.019

Copyright: Elsevier

http://www.elsevier.com/

Postprint available at: Linköping University Electronic Press

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Completing partial Latin squares with one filled row,

column and symbol

Carl Johan Casselgren

∗

Department of Mathematics

Link¨

oping University

SE-581 83 Link¨

oping, Sweden

Roland H¨

aggkvist

†

Department of Mathematics

Ume˚

a University

SE-901 87 Ume˚

a, Sweden

Abstract. Let P be an n × n partial Latin square every non-empty cell of which lies in

a fixed row r, a fixed column c or contains a fixed symbol s. Assume further that s is the symbol of cell (r, c) in P . We prove that P is completable to a Latin square if n ≥ 8 and n is divisible by 4, or n ≤ 7 and n /∈ {3, 4, 5}. Moreover, we present a polynomial algorithm for the completion of such a partial Latin square.

1 Introduction

Throughout this paper, n is assumed to be a positive integer. Consider an n × n array P where each cell contains at most one symbol from {1, . . . , n}. P is called a partial Latin square if each symbol occurs at most once in every row and column. If no cell inP is empty, then it is a Latin square.

The cell in position (i, j) in an array A is denoted by (i, j)_A, and the symbol in cell (i, j)_A is denoted by A(i, j). If cell (i, j)_A is empty, then we write A(i, j) = ∅; and if A(i, j) = k, then we say that k is an entry of cell (i, j)_A. An r × s array with entries from {1, . . . , n} is called a Latin rectangle if each symbol occurs at most once in every row and column, and each cell has precisely one entry.

An n × n Latin square L is a completion of an n × n partial Latin square P if L(i, j) = P (i, j) for each non-empty cell (i, j)P of P . P is completable if there is such a Latin square.

Otherwise, P is non-completable. The problem of completing partial Latin squares is a classic within combinatorics and there is a wealth of results in the literature. Let us here just mention a few.

In general, it is an NP -complete problem to determine if a partial Latin square is com-pletable [3]. Thus it is natural to ask if particular families of partial Latin squares are completable. A classic result due to Ryser [8] states that if n ≥ r, s, then every n × n partial Latin square whose non-empty cells lie in an r × s Latin rectangle Q is completable if and

∗_{E-mail address: carl.johan.casselgren@liu.se} †_{E-mail address: roland.haggkvist@math.umu.se}

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only if each of the symbols 1, . . . , n occurs at least r + s − n times in Q. Another classic result is Smetaniuk’s proof [9] of Evans’ conjecture [7] that every n × n partial Latin square with at most n − 1 entries is completable. This was also independently proved by Andersen and Hilton [1]. Let us also mention the following conjecture of H¨aggkvist.

Conjecture 1.1. Any nr × nr partial Latin square whose non-empty cells lie in (n − 1)

disjoint r × r squares can be completed.

In [5] Conjecture 1.1 was proved for the caser = 3. Some other cases were resolved in [4] and [6].

In this paper we are interested in completions of a particular family of partial Latin squares. We propose the following conjecture.

Conjecture 1.2. Let r, c, s ∈ {1, . . . , n}, and let P be an n × n partial Latin square where

each non-empty cell lies in row r or column c, or has entry s, and assume further that P (r, c) = s. If n /∈ {3, 4, 5}, then P is completable.

Note that the condition P (r, c) = s is necessary, since removing it yields a conjecture with obvious counterexamples such as partial Latin squares of the form as in Figure 1.

1 1

1 2

Figure 1: A non-completable partial Latin square with entries only on the main diagonal. The non-completable partial Latin squares in Figure 2 show that the condition n /∈ {3, 4, 5} in Conjecture 1.2 is necessary. 1 2 3 2 1 3 1 1 3 4 2 2 1 3 1 4 1 1 3 2 4 5 2 1 3 1 4 1 5 1

Figure 2: Non-completable partial Latin squares of order 3, 4 and 5.

In this paper we present the following theorem which provides some evidence for Conjec-ture 1.2.

Theorem 1.3. Let r, c, s ∈ {1, . . . , n}, and let P be an n × n partial Latin square every

non-empty cell of which is in row r or column c, or has entry s, and assume further that P (r, c) = s. If n /∈ {3, 4, 5} and n ≤ 7, or n = 4k for some integer k ≥ 2, then P is completable.

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In the remaining part of the paper we prove this theorem. We also note the following slightly more general result that follows from Theorem 1.3

Corollary 1.4. Let r, c, s ∈ {1, . . . , n}, and let P be an n × n partial Latin square every

non-empty cell of which is in row r or column c, or has entry s. Assume further that P (r, c) is non-empty and that symbol s appears in row r and column c. If n /∈ {3, 4, 5} and n ≤ 7, or n = 4k for some integer k ≥ 2, then P is completable.

Proof. Suppose thatP (r, c) = s, since otherwise the result follows immediately from Theorem 1.3. Note also that by a simple application of Hall’s theorem, we may assume that each cell in row r and column c is non-empty, and that there are n cells with entry s in P . The case when n ∈ {1, 2} is trivial, so suppose that n ≥ 6. Let r, c be integers such that P (r, c) = s and that P (r, c) =s. Deﬁne a new n × n partial Latin square R from P by for each integer i ∈ {1, . . . , n} \ {c, c_{}, setting R(r, i) = ∅ and R(r}_{, i) = P (r, i), R(r}_{, c}_{) =} _{P (r, c), and}

retaining the entry (or non-entry) of every other cell in P . It is easy to see that R satisﬁes the hypothesis of Theorem 1.3, and thus is completable to a Latin square L. From L we deﬁne the Latin square L by for each integeri ∈ {1, . . . , n} \ {c, c}, setting L(r, i) = L(r, i) and L(r, i) = L(r, i), and retaining the entry of every other cell in L. L is a completion of P .

2 Preliminaries

Two partial Latin squares L and L are isotopic ifL can be obtained fromL by permuting rows, permuting columns and/or permuting symbols inL. Note that if L and L are isotopic, then L is completable if and only if L is completable.

A 2-square in a partial Latin squareL is a set

S = {(i1, j1)L, (i1, j2)L, (i2, j1)L, (i2, j2)L}

of cells in L such that

L(i1, j1) =L(i2, j2) and L(i1, j2) = L(i2, j1)

(where possibly L(i₁, j₁) =∅ or L(i₁, j₂) =∅, i.e. (i₁, j₁)_Land (i₂, j₂)_Lor (i₁, j₂)_Land (i₂, j₁)_L might be empty). Note that a 2-square is uniquely determined by two cells in L, if at least one of those cells is non-empty. An (s₁, s₂)-factor inL is a non-empty set S of 2k cells, where k is a positive integer, satisfying the following conditions:

(i) each row and column inL contain either two or no cells from S, (ii) each cell in S has entry s₁ or s₂.

An (s₁, s₂)-factor S in L is called an (s₁, s₂)-cycle if there is no subset S ⊆ S such that S _{= S and S} _{is a (}_s

1, s2)-factor. Note that an (s1, s2)-cycle in L is uniquely determined by

the two symbols s₁, s₂ along with a speciﬁed row or column. If

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is a 2-square in a partial Latin square L with L(i₁, j₁) =s₁ and L(i₁, j₂) =s₂ (where possibly s1 = ∅ or s2 = ∅), then a swap on S denotes the operation L → L, where L is a partial

Latin square with L₍_i

1, j1) =L(i2, j2) =s2, L(i1, j2) =L(i2, j1) =s1

and L(i, j) = L(i, j) for all other 1 ≤ i, j ≤ n.

A generalized diagonal (or just a diagonal ) in an n × n array A is a set of n cells in A no two of which lie in the same column or row. The main diagonal in A is the set {(i, i)A : i = 1, . . . , n}. The lower main diagonal in A is the set {(i, i − 1)A : i = 2, . . . , n},

and the upper main diagonal in A is the set {(i − 1, i)_A:i = 2, . . . , n}.1

In the following, n is an even positive integer. Given an n × n array A and an integer j ∈ {1, . . . , n}, the jth upper back diagonal in A is the set

{(j, 1)A, (j − 1, 2)A, . . . , (1, j)A},

and for j ∈ {2, . . . , n − 2} the jth small lower back diagonal in A is the set {(n − 1, j + 1)A, (n − 2, j + 2)A, . . . , (j + 1, n − 1)A}.

For odd j, the jth broken upper back diagonal in A is the set

{(j, 1)A, (j − 1, 2)A, . . . , (1, j)A} \ {(j+1₂ ,j+1₂ )A} ∪ {(n,j+1₂ )A, (j+1₂ , n)A},

and for even j, the jth broken small lower back diagonal in A is the set

{(n − 1, j + 1)A, (n − 2, j + 2)A, . . . , (j + 1, n − 1)A} \ {(n+j₂ ,n+j₂ )A} ∪ {(n,n+j₂ )A, (n+j₂ , n)A}.

We now deﬁne the symmetric n × n Latin square A_n as follows: • assign symbol 1 to every cell on the main diagonal,

• for even j ≤ n, assign symbol j to every cell on the jth upper back diagonal in An,

• for even j < n, assign symbol j to every cell on the jth broken small lower back diagonal,

• for odd 1 < j < n, assign symbol j to every cell on the jth broken upper back diagonal in A_n,

• for odd 1 < j < n − 1, assign symbol j to every cell on the jth small lower back diagonal.

It is readily veriﬁed that for each even positive integer n, A_n is a symmetric Latin square of order n.

1_{The last two definitions involve a slight abuse of terminology, because these sets are not diagonals.}

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1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 5 6 2 1 4 5 6 3 3 4 1 6 2 5 4 5 6 1 3 2 5 6 2 3 1 4 6 3 5 2 4 1 1 2 3 4 5 6 7 8 2 1 4 5 6 7 8 3 3 4 1 6 7 8 2 5 4 5 6 1 8 2 3 7 5 6 7 8 1 3 4 2 6 7 8 2 3 1 5 4 7 8 2 3 4 5 1 6 8 3 5 7 2 4 6 1 Figure 3: The Latin squares A₄, A₆ and A₈.

Next, for each positive integer n that is divisible by 4, we deﬁne a Latin square B_n of order n by, for each i, j ∈ {1, . . . , n/2}, setting

Bn(2i − 1, 2j − 1) = Bn(2i, 2j) = 2An/2(i, j) − 1 and Bn(2i − 1, 2j) = Bn(2i, 2j − 1) = 2An/2(i, j). 1 2 3 4 5 6 7 8 2 1 4 3 6 5 8 7 3 4 1 2 7 8 5 6 4 3 2 1 8 7 6 5 5 6 7 8 1 2 3 4 6 5 8 7 2 1 4 3 7 8 5 6 3 4 1 2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 2 1 4 3 6 5 8 7 10 9 12 11 3 4 1 2 7 8 9 10 11 12 5 6 4 3 2 1 8 7 10 9 12 11 6 5 5 6 7 8 1 2 11 12 3 4 9 10 6 5 8 7 2 1 12 11 4 3 10 9 7 8 9 10 11 12 1 2 5 6 3 4 8 7 10 9 12 11 2 1 6 5 4 3 9 10 11 12 3 4 5 6 1 2 7 8 10 9 12 11 4 3 6 5 2 1 8 7 11 12 5 6 9 10 3 4 7 8 1 2 12 11 6 5 10 9 4 3 8 7 2 1 Figure 4: The Latin squares B₈ and B₁₂.

3 Proof of Theorem 1.3

In this section we prove Theorem 1.3. Let P be an n × n partial Latin square where every non-empty cell lies in row r or column c, or has entry s, and assume further that P (r, c) = s. Without loss of generality, we assume thats = 1, r = 1 and c = 1. Note further that without loss of generality we may assume that all cells in row 1 and column 1 are non-empty, and that there are n cells in P with entry 1. (Otherwise, we can just ﬁll in the missing symbols in row 1 and column 1; and, similarly, by Hall’s theorem it is always possible to extend the set of cells with entry 1 in P to a generalized diagonal.) Finally, we assume that P is in standard form, that is, P (1, j) = j and P (j, 1) = j, for each j ∈ {1, . . . , n}.

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Deﬁne a permutation ϕ on {1, . . . , n} by setting ϕ(i) = j if P (i, j) = 1. Note that the partial Latin squareP is uniquely determined (up to isotopy) by the permutation ϕ. We say that any partial Latin square which is isotopic to P is associated with ϕ.

The proof of Theorem 1.3 consist of two parts: the ﬁrst part deals with the case n ≥ 8, and in the second part we prove that the theorem holds when n ∈ {1, 2, 6, 7}.

So suppose thatn = 4k, for some positive integer k ≥ 2. We use the algorithm described below for constructing a partial Latin square R that is isotopic to P . The algorithm proceeds by steps, and step i (i ≥ 2) consist of two parts. In the ﬁrst part of step i, the algorithm determines the entries of (1, 2i − 1)_R, (2i − 1, 1)_R, (2i − 2, 2i − 1)_R, (2i − 1, 2i − 2)_R and (2i − 1, 2i − 1)_R, depending on the entry of (2i − 2, 2i − 2)_R determined at the previous step and on the permutation ϕ.

In the second part of step i, the algorithm determines the entries of the cells (1, 2i)_R, (2i, 1)_Rand (2i, 2i)_R. The entries of these cells will depend on the entries of (2i − 1, 1)_Rand (1, 2i − 1)_R which were determined in the ﬁrst part of step i, and on the permutation ϕ.

Algorithm

Step 1. Set R(1, 1) = 1. Choose an arbitrary symbol c₁ not present in R and set R(2, 1) = R(1, 2) = c1. If ϕ(c1) =c1, set R(2, 2) = 1, otherwise set R(2, 2) = ∅. Step i. (i ≥ 2)

Part 1.

Case 1a. R(2i − 2, 2i − 2) = 1:

Choose an arbitrary symbol s_i not already present inR and set

R(2i − 1, 1) = si, R(1, 2i − 1) = ϕ(si) and R(2i − 1, 2i − 1) = 1.

Case 1b. R(2i − 2, 2i − 2) = ∅: Set

R(1, 2i − 1) = ϕ(R(2i − 2, 1)), R(2i − 2, 2i − 1) = 1, R(2i − 1, 1) = ϕ−1₍_{R(1, 2i − 2)) and R(2i − 1, 2i − 2) = 1.}

Part 2.

Case 2a. R(2i − 1, 1) = R(1, 2i − 1): Choose an arbitrary symbol c_i not already present in R and set R(2i, 1) = c_i and R(1, 2i) = c_i. If ϕ(c_i) =c_i, then set

R(2i, 2i) = 1, otherwise set R(2i, 2i) = ∅. If 2i < n, then go to step (i + 1), otherwise Stop.

Case 2b. R(2i − 1, 1) = R(1, 2i − 1): Set R(2i, 1) = R(1, 2i − 1) and

R(1, 2i) = R(2i − 1, 1). If ϕ (R(2i, 1)) = R(1, 2i), then set R(2i, 2i) = 1, otherwise set R(2i, 2i) = ∅. If 2i < n, then go to step (i + 1), otherwise Stop.

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It follows from the description of the algorithm above that it will stop after stepn/2 and that R is a partial Latin square where each non-empty cell is in row 1 or column 1, or has entry 1. Moreover, it is not hard to verify that each symbol in {1, . . . , n} is present in row 1 and column 1. In particular, we have that {R(2i − 1, 1), R(2i, 1)} = {R(1, 2i − 1), R(1, 2i)} for each i ∈ {1, . . . , n/2}.

Furthermore, each row and column in R contains exactly one cell with entry 1. The permutation ϕ along with previously determined entries decides which cell in each row and column is assigned symbol 1; each cell with entry 1 lies on the main diagonal, or on the upper or lower main diagonal ofR. Additionally, if R(j, j−1) = 1, then j is odd and R(j−1, j) = 1. (This is Case 1b in the algorithm.) Since P is uniquely determined by ϕ up to isotopy, R is isotopic to P . Hence, we have the following:

Claim 3.1. The algorithm above produces a partial Latin square R isotopic to P , such that

(i) each entry distinct from 1 lies in row 1 or column 1, and for each i = 1, . . . , n/2, {R(2i − 1, 1), R(2i, 1)} = {R(1, 2i − 1), R(1, 2i)},

(ii) each cell with entry 1 in R lies on the main diagonal, or on the upper or lower main diagonal in R. Moreover, for each j ∈ {1, . . . , n}, if R(j, j − 1) = 1, then j is odd and R(j − 1, j) = 1.

Without loss of generality we will assume that {R(2i − 1, 1), R(2i, 1)} = {2i − 1, 2i}. (A suitable permutation of the symbols in R yields such a partial Latin square.) It follows from Claim 3.1 (ii) that by performing a sequence of swaps on 2-squares in R deﬁned by pair of cells of the form{(j, j −1)_R, (j −1, j)_R}, we obtain a partial Latin square R with entries only in row 1, column 1 and on the main diagonal, and such that all cells on the main diagonal have entry 1. We will now show that R is completable to a Latin squareT, and then show how to modify T to obtain a Latin square T , which is a completion of R.

Consider the symmetric Latin squareB_nof ordern deﬁned in the preceding section. Note that each cell on the main diagonal ofB_nhas entry 1. Furthermore, for eachi ∈ {2, . . . , n/2}, the sets

Ii ={(2i − 1, 1)Bn, (2i − 1, 2)Bn, (2i, 1)Bn, (2i, 2)Bn}

and

Ji ={(1, 2i − 1)Bn, (2, 2i − 1)Bn, (1, 2i)Bn, (2, 2i)Bn}

are 2-squares in B_n. It follows from Claim 3.1 (i) and the construction ofR that performing swaps on some of the 2-squares in {I₂, . . . , I_n/2, J₂, . . . , J_n/2} yields a Latin square T which is a completion of R. We will now show how a completionT of R can be obtained from T. We have to consider two diﬀerent cases.

Case 1. R(2, 2) = 1 or R(3, 1) = R(1, 3):

By the construction of T, we have T(1, j) = R(1, j) and T(j, 1) = R(j, 1), for each j ∈ {1, . . . , n}. Moreover, for each i ∈ {3, . . . , n − 1}, the set

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is a 2-square in T. Suppose ﬁrst that R(2, 2) = 1. It follows from Claim 3.1 (ii) that by performing swaps on a subset of pairwise disjoint 2-squares from {U₃, . . . , U_n−1}, we obtain a Latin square T that is a completion of R.

Now suppose instead that R(3, 1) = R(1, 3) and R(2, 2) = 1. Then R(3, 1) = R(1, 3) and thus T(3, 2) = T(2, 3). Hence, if R(3, 1) = R(1, 3), then

U2 ={(2, 2)T, (3, 2)_T, (3, 2)_T, (3, 3)_T}

is a 2-square in T. It now follows from Claim 3.1 (ii) that by performing swaps on some of the 2-squares U₂, . . . , U_n−1 we obtain a Latin square T which is a completion of R.

Case 2. R(2, 2) = 1 and R(3, 1) = R(1, 3):

Consider the Latin square A_n/2 defined in the preceding section. LetC₁ be the (2, n/2 − 1)-cycle with cells in the first row of A_n/2, and let C₂ be the (2, n/2 − 1)-cycle with cells in the first column of A_n/2.

Claim 3.2. C₁∩ C₂ =∅.

Proof. Clearly, if C₁ ∩ C₂ = ∅, then C₁ = C₂. However, for each cell (i, j)_A_n/2 of C₁, if i ≤ n/2, then i is an odd number, and for each cell (i, j)An/2 of C2, if i ≤ n/2, then i is an

even integer.

From C₁ and C₂ we deﬁne two sets D₁ and D₂ of cells in T: for each cell (i, j)_A_n/2 ∈ C₁ we include the cells (2i, 2j)_T and (2i, 2j − 1)_T in D₁, and for each cell (i, j)_A_n/2 ∈ C₂ we

include the cells (2i, 2j)_T and (2i − 1, 2j)_T in D₂. Since C₁∩ C₂ = ∅, D₁ ∩ D₂ =∅. Note

that by the construction of T, we have

{T₍₂_{i, 2j), T}₍₂_{i, 2j − 1)} = {3, 4} or {T}₍₂_{i, 2j), T}₍₂_{i, 2j − 1)} = {n − 3, n − 2}}

for each pair of cells {(2i, 2j)_T, (2i, 2j − 1)_T} in D₁, and

{T₍₂_{i, 2j), T}₍₂_{i − 1, 2j)} = {3, 4} or {T}₍₂_{i, 2j), T}₍₂_{i − 1, 2j)} = {n − 3, n − 2}}

for each pair of cells {(2i, 2j)_T, (2i − 1, 2j)_T} in D₂. Additionally, no cell in D₁∪ D₂ lies in

the ﬁrst row or ﬁrst column of T.

Next, we deﬁne a new Latin square S from T in the following way:

• for each pair of cells (i1, j1)T and (i₂, j₁)_T in D₁ which lie in the same column of T,

we set S(i₁, j₁) =T(i₂, j₁) and S(i₂, j₁) =T(i₁, j₁);

• for each pair of cells (i3, j3)T and (i₃, j₄)_T in D₂ which lie in the same row of T, we

set S(i₃, j₃) =T(i₃, j₄) and S(i₃, j₄) =T(i₃, j₃); • we retain the symbol of every other cell of T_.

Now, consider the 2-square X in S containing the cells (3, 2)_S and (4, 2)_S. Each cell in X contain symbol n − 3 or n − 2 and does not lie in the ﬁrst row or column of S. If S(3, 2) = S(2, 3), then we perform a swap on X to obtain the Latin square S_{. Otherwise,}

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Since T(1, j) = R(1, j) and T(j, 1) = R(j, 1), for each j ∈ {1, . . . , n}, we have that S₍₁_{, j) = R(1, j) and S}₍_{j, 1) = R(j, 1) for each j ∈ {1, . . . , n}. Moreover, for each i ∈}

{2, . . . , n − 1}, the set

Wi ={(i, i)S, (i + 1, i + 1)_S, (i + 1, i)_S, (i, i + 1)_S}

is a 2-square in S. Similarly as in Case 1, it now follows that by performing swaps on some of the 2-squares W₂, . . . , W_n−1 we obtain a Latin square T which is a completion of R. This completes the proof of Theorem 1.3 in the case when n = 4k for some integer k ≥ 2.

We now prove that Theorem 1.3 holds in the case when n ∈ {1, 2, 6, 7}. The cases n = 1 and n = 2 are trivial. Since P is uniquely determined by the permutation ϕ up to isotopy, when n ∈ {6, 7} it suﬃces to verify that for each permutation γ of {1, . . . , n} with γ(1) = 1, some partial Latin square associated with γ is completable. If γ and γ are permutations on{1, . . . , n}, satisfying γ(1) = 1 andγ(1) = 1, respectively, and with the same number of cycles of length i, for each i = 1, . . . , n − 1, then each partial Latin square associated with γ _{is isotopic to each partial Latin square associated with} _γ_{. Hence, it suﬃces to check that}

for all permutations γ on n, satisfying γ(1) = 1 and with a diﬀerent number of cycles of length j for some j ∈ {1, . . . , n − 1}, at least one partial Latin square associated with γ is completable. For n = 6, there are 7 such permutations, and for n = 7, there are 11 such permutations (the number of partitions of 5 and 6, respectively.) We deal with each case separately, and an explicit completion of a partial Latin square associated with each of these permutations is given in the Appendix. This completes the proof of Theorem 1.3.

Finally, let us also mention that we have veriﬁed by a computer search that Theorem 1.3 holds in the case when n ∈ {9, 10}.

Acknowledgement

The authors thank Jonas H¨agglund for helping them with some computer simulations at an early stage of the research. They also thank the referees for helpful suggestions.

References

[1] L. D. Andersen, A. J. W. Hilton, Thank Evans!, Proc. London Math. Soc. 47 (1983), pp. 507-522.

[2] A. S. Asratian, T. M. J. Denley, R. H¨aggkvist, Bipartite graphs and their applications, Cambridge University Pres, Cambridge, 1998.

[3] C. J. Colbourn, The complexity of completing partial Latin squares, Discrete Applied Mathematics 8 (1984), pp. 25–30.

[4] T. Denley, On a conjecture of H¨aggkvist on partial Latin squares, Proceedings of the Thirty-second Southeastern International Conference on Combinatorics, Graph Theory and Computing, vol. 150, Baton Rouge, LA (2001), 73-78.

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[5] T. Denley, R. H¨aggkvist, Completing some Partial Latin squares, European Journal of Combinatorics 21 (2000), 877–880.

[6] T. Denley, J. S. Kuhl, On a generalization of the Evans conjecture, Discrete Mathematics 20 (2007), 4763–4767.

[7] T. Evans, Embedding incomplete latin squares, American Mathematical Monthly 67 (1960), 958–961.

[8] H. J. Ryser, A combinatorial theorem with an application to Latin squares, Proc. Amer. Math. Soc. 2 (1951), 550–552.

[9] B. Smetaniuk, A new construction for Latin squares I. Proof of the Evans conjecture, Ars Combinatoria 11 (1981), pp. 155–172.

Appendix

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