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

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1 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, Copyright: Elsevier Postprint available at: Linköping University Electronic Press

2 Completing partial Latin squares with one filled row, column and symbol Carl Johan Casselgren Department of Mathematics Linköping University SE Linköping, Sweden Roland Häggkvist Department of Mathematics Umeå University SE Umeå, Sweden Abstract. Let P be an n n partial Latin square every non-empty cell of which lies in afixedrowr, a fixed column c or contains a fixed symbol s. Assume further that s is the symbol of cell (r, c) inp. We prove that P is completable to a Latin square if n 8andn is divisible by 4, or n 7andn/ {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 in P is empty, then it is a Latin square. The cell in position (i, j) in an array A is denoted by (i, j) A,andthesymbolincell(i, j) A is denoted by A(i, j). If cell (i, j) A is empty, then we write A(i, j) = ; andifa(i, j) =k, then we say that k is an entry of cell (i, j) A.Anr 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 completable [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, theneveryn n partial Latin square whose non-empty cells lie in an r s Latin rectangle Q is completable if and address: carl.johan.casselgren@liu.se address: roland.haggkvist@math.umu.se 1

3 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 Häggkvist. 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 case r = 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}, andletp 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. Ifn/ {3, 4, 5}, thenp 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 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 Figure 2: Non-completable partial Latin squares of order 3, 4and5. In this paper we present the following theorem which provides some evidence for Conjecture 1.2. Theorem 1.3. Let r, c, s {1,...,n}, andletp 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, orn =4k for some integer k 2, thenp is completable. 2

4 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}, andletp 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, thenp is completable. Proof. Suppose that P (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.Thecase 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 integer i {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 if L can be obtained from L by permuting rows, permuting columns and/or permuting symbols in L. NotethatifL and L are isotopic, then L is completable if and only if L is completable. A 2-square in a partial Latin square L is a set of cells in L such that S = {(i 1,j 1 ) L, (i 1,j 2 ) L, (i 2,j 1 ) L, (i 2,j 2 ) L } L(i 1,j 1 )=L(i 2,j 2 )andl(i 1,j 2 )=L(i 2,j 1 ) (where possibly L(i 1,j 1 )= or L(i 1,j 2 )=, i.e. (i 1,j 1 ) L and (i 2,j 2 ) L or (i 1,j 2 ) L and (i 2,j 1 ) L might be empty). Note that a 2-square is uniquely determined by two cells in L, ifatleast one of those cells is non-empty. An (s 1,s 2 )-factor in L is a non-empty set S of 2k cells, where k is a positive integer, satisfying the following conditions: (i) each row and column in L contain either two or no cells from S, (ii) each cell in S has entry s 1 or s 2. An (s 1,s 2 )-factor S in L is called an (s 1,s 2 )-cycle if there is no subset S S such that S S and S is a (s 1,s 2 )-factor. Note that an (s 1,s 2 )-cycle in L is uniquely determined by the two symbols s 1,s 2 along with a specified row or column. If S = {(i 1,j 1 ) L, (i 1,j 2 ) L, (i 2,j 1 ) L, (i 2,j 2 ) L } 3

5 is a 2-square in a partial Latin square L with L(i 1,j 1 )=s 1 and L(i 1,j 2 )=s 2 (where possibly s 1 = or s 2 = ), then a swap on S denotes the operation L L,whereL is a partial Latin square with L (i 1,j 1 )=L (i 2,j 2 )=s 2, L (i 1,j 2 )=L (i 2,j 1 )=s 1 and L (i, j) =L(i, j) for all other 1 i, j n. A generalized diagonal (orjustadiagonal) inann 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}, thejth 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, thejth broken upper back diagonal in A is the set {(j, 1) A, (j 1, 2) A,...,(1,j) A }\{( j+1 2, j+1 2 ) A} {(n, j+1 2 ) A, ( j+1 2,n) A}, and for even j, thejth 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+j 2, n+j 2 ) A} {(n, n+j 2 ) A, ( n+j 2,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 A n, 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. However, for simplicity of presentation, we have chosen to use this terminology. 4

6 Figure 3: The Latin squares A 4,A 6 and A 8. 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 B n (2i 1, 2j 1) = B n (2i, 2j) =2A n/2 (i, j) 1and B n (2i 1, 2j) =B n (2i, 2j 1) = 2A n/2 (i, j) Figure 4: The Latin squares B 8 and B 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 that s =1,r =1andc = 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, foreachj {1,...,n}. 5

7 Define a permutation ϕ on {1,...,n} by setting ϕ(i) =j if P (i, j) = 1. Note that the partial Latin square P 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 that n =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) R and (2i, 2i) R. The entries of these cells will depend on the entries of (2i 1, 1) R and (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 c 1 not present in R and set R(2, 1) = R(1, 2) = c 1.Ifϕ(c 1 )=c 1,setR(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 in R and set R(2i 1, 1) = s i,r(1, 2i 1) = ϕ(s i )andr(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.ifϕ(c i )=c i,thenset R(2i, 2i) = 1, otherwise set R(2i, 2i) =. If 2i <n,thengotostep(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) =. If2i<n,thengotostep(i + 1), otherwise Stop. 6

8 It follows from the description of the algorithm above that it will stop after step n/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 orlower main diagonal of R. 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, andforeachi =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}, ifr(j, j 1) = 1, thenj 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 square T,andthenshow how to modify T to obtain a Latin square T, which is a completion of R. Consider the symmetric Latin square B n of order n defined in the preceding section. Note that each cell on the main diagonal of B n has entry 1. Furthermore, for each i {2,...,n/2}, the sets I i = {(2i 1, 1) Bn, (2i 1, 2) Bn, (2i, 1) Bn, (2i, 2) Bn } and J i = {(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 of R that performing swaps on some of the 2-squares in {I 2,...,I n/2,j 2,...,J n/2 } yields a Latin square T which is a completion of R. We will now show how a completion T 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,wehaveT (1,j)=R(1,j)andT (j, 1) = R(j, 1), for each j {1,...,n}. Moreover, for each i {3,...,n 1}, theset U i = {(i, i) T, (i +1,i+1) T, (i +1,i) T, (i, i +1) T } 7

9 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 {U 3,...,U n 1 },weobtain alatinsquaret that is a completion of R. Now suppose instead that R(3, 1) = R(1, 3) and R(2, 2) 1. ThenR (3, 1) = R (1, 3) and thus T (3, 2) = T (2, 3). Hence, if R(3, 1) = R(1, 3), then U 2 = {(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 U 2,...,U n 1 we obtain a Latin square T which is a completion of R. Case 2. R(2, 2) 1andR(3, 1) R(1, 3): Consider the Latin square A n/2 defined in the preceding section. Let C 1 be the (2,n/2 1)- cycle with cells in the first row of A n/2, and let C 2 be the (2,n/2 1)-cycle with cells in the first column of A n/2. Claim 3.2. C 1 C 2 =. Proof. Clearly, if C 1 C 2, thenc 1 = C 2. However, for each cell (i, j) An/2 of C 1,if i n/2, then i is an odd number, and for each cell (i, j) An/2 of C 2,ifi n/2, then i is an even integer. From C 1 and C 2 we define two sets D 1 and D 2 of cells in T : for each cell (i, j) An/2 C 1 we include the cells (2i, 2j) T and (2i, 2j 1) T in D 1, and for each cell (i, j) An/2 C 2 we include the cells (2i, 2j) T and (2i 1, 2j) T in D 2. Since C 1 C 2 =, D 1 D 2 =. Note that by the construction of T,wehave {T (2i, 2j),T (2i, 2j 1)} = {3, 4} or {T (2i, 2j),T (2i, 2j 1)} = {n 3,n 2} for each pair of cells {(2i, 2j) T, (2i, 2j 1) T } in D 1,and {T (2i, 2j),T (2i 1, 2j)} = {3, 4} or {T (2i, 2j),T (2i 1, 2j)} = {n 3,n 2} for each pair of cells {(2i, 2j) T, (2i 1, 2j) T } in D 2.Additionally,nocellinD 1 D 2 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 (i 1,j 1 ) T and (i 2,j 1 ) T in D 1 which lie in the same column of T, we set S(i 1,j 1 )=T (i 2,j 1 )ands(i 2,j 1 )=T (i 1,j 1 ); for each pair of cells (i 3,j 3 ) T and (i 3,j 4 ) T in D 2 which lie in the same row of T,we set S(i 3,j 3 )=T (i 3,j 4 )ands(i 3,j 4 )=T (i 3,j 3 ); 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 3orn 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, we set S = S. NotethatS (3, 2) = S (2, 3) 8

10 Since T (1,j)=R(1,j)andT (j, 1) = R(j, 1), for each j {1,...,n}, wehavethat S (1,j)=R(1,j)andS (j, 1) = R(j, 1) for each j {1,...,n}. Moreover, for each i {2,...,n 1}, theset W i = {(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 W 2,...,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}. Thecasesn =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 and γ (1) = 1, respectively, and with the same number of cycles of length i, foreachi =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 Hägglund 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 [2] A.S.Asratian,T.M.J.Denley,R.Häggkvist, Bipartite graphs and their applications, Cambridge University Pres, Cambridge, [3] C. J. Colbourn, The complexity of completing partial Latin squares, Discrete Applied Mathematics 8 (1984), pp [4] T. Denley, On a conjecture of Häggkvist on partial Latin squares, Proceedings of the Thirty-second Southeastern International Conference on Combinatorics, Graph Theory and Computing, vol. 150, Baton Rouge, LA (2001),

11 [5] T. Denley, R. Häggkvist, Completing some Partial Latin squares, European Journal of Combinatorics 21 (2000), [6] T. Denley, J. S. Kuhl, On a generalization of the Evans conjecture, Discrete Mathematics 20 (2007), [7] T. Evans, Embedding incomplete latin squares, American Mathematical Monthly 67 (1960), [8] H. J. Ryser, A combinatorial theorem with an application to Latin squares, Proc. Amer. Math. Soc. 2 (1951), [9] B. Smetaniuk, A new construction for Latin squares I. Proof of the Evans conjecture, Ars Combinatoria 11 (1981), pp Appendix Figure 5: Completable partial Latin squares of order 6. Prescribed entries in bold. 10

12 Figure 6: Completable partial Latin squares of order 7. Prescribed entries in bold. 11