Completing Partial Latin Squares with Blocks of
Non-empty Cells
Jaromy Kuhl
Michael William Schroeder
Department of Mathematics and Statistics
University of West Florida
Pensacola, FL 32514
Department of Mathematics
Marshall University
Huntington, WV 25755
[email protected]
[email protected]
December 24, 2014
Abstract
In this paper we develop two methods for completing partial latin squares and
prove the following. Let A be a partial latin square of order nr in which all non-empty
cells occur in at most n − 1 r × r squares. If t1 , . . . , tm are positive integers for which
n > t21 + t22 + · · · + t2m + 1 and if A is the union of m subsquares each with order rti ,
then A can be completed. We additionally show that if n > r + 1 and A is the union of
n identical r × r squares with disjoint rows and columns, then A can be completed. For
smaller values of n we show that a completion does not always exist.
1
Introduction
Let n be a positive integer and S be a finite symbol set. If A is an n × n array in which each
cell contains a subset of S, then we call A an array over S. We use [n] = {1, 2, . . . , n} to
index the rows and columns of A and A(i, j) to denote the subset of symbols in the (i, j) cell
of A. If the symbols are arranged so that each occurs at most once in each column, then A is
called column-latin. Row-latin is defined analogously.
Let r be a positive integer and S have cardinality nr. We say that A is a r-semi partial
latin square of order n, or A ∈ PLSr (n), if |A(i, j)| 6 r for each {i, j} ⊆ [n] and A is
both column- and row-latin. If |A(i, j)| = r for each {i, j} ⊆ [n], A is called a r-semi latin
square, or A ∈ LSr (n). The sets of all partial latin squares and latin squares of order n
are PLS(n) = PLS1 (n) and LS(n) = LS1 (n) respectively. For an integer i and set S, let
i + S = {i + s | s ∈ S}. For an nr × nr array A and {i, j} ⊆ [n], let Aij denote the r × r
sub-array of A at the intersection of rows (i − 1)r + [r] and columns (j − 1)r + [r].
1
In this paper we develop two methods of completing partial latin squares, and consequently,
prove two cases of the following conjecture [3].
Conjecture 1. All partial latin squares of order nr with symbols occurring in at most n − 1
disjoint r × r squares are completable.
Conjecture 1 is a possible generalization of a theorem famously known as the Evans Conjecture
[4].
Theorem 2. If A ∈ PLS(n) with at most n − 1 non-empty cells, then A can be completed to
a latin square of order n.
Proofs of Theorem 2 were given independently by Häggkvist for n > 1111 [5], and by Anderson
and Hilton [1] and Smetaniuk [11] for all n. The upper bound on the number of non-empty
cells is sharp. In [1], the authors determine all incompletable partial latin squares of order
n with exactly n non-empty cells. One such square has symbol 1 in the first k diagonal
cells and symbols 2, 3, . . . , n − k + 1 in the last n − k cells of column k + 1. We name this
incompletable partial latin square Bk,n , where k < n.
Let P ∈ PLS(n) and Sn be the symmetric group acting on [n]. For θ = (α, β, γ) ∈ Sn ×Sn ×Sn ,
we use θ(P ) ∈ PLS(n) to denote the array in which the rows, columns, and symbols of P
are permuted according to α, β, and γ respectively. The mapping θ is called an isotopism,
and P and θ(P ) are said to be isotopic. Each partial latin square can be written as a subset
of ordered triples of [n] × [n] × [n], where (i, j, k) ∈ P if k ∈ P (i, j). A conjugate of P is an
array in which the coordinates of each triple of P are uniformly permuted. There are six,
not necessarily distinct, conjugates of P . The main class of P is the set of all partial latin
squares that are isotopic to some conjugate of P . The following theorem is Anderson and
Hilton’s answer to the Evans conjecture [1].
Theorem 3. If P ∈ PLS(n) with exactly n non-empty cells, then P is completable if and
only if P is not in the main class of Bk,n for each k, 1 6 k < n.
In [9], Theorem 2 is generalized to elements of PLSr (n).
Theorem 4. If A ∈ PLSr (n) with at most n − 1 non-empty cells, then A can be completed
to an element of LSr (n).
The next two theorems are applications of Theorem 4 and special cases of Conjecture 1. By
uniform we mean that each fixed r × r square is a filled Aij for some {i, j} ⊆ [n].
Theorem 5. If A ∈ PLS(nr) is uniform with symbols appearing in at most (n − 1) r × r
squares filled as latin squares, then A can be completed.
If we remove the uniformity and latin conditions of Theorem 5 and insist that the fixed r × r
squares are column-disjoint, it has been shown that some of these partial latin squares are
completable [9].
2
Theorem 6. Let A ∈ PLS(nr) with symbols appearing in at most (n − 1) r × r squares filled
in distinct columns. There is a permutation of the symbols appearing in each row of each
fixed r × r square such that the resulting element of PLS(nr) is completable.
It is also known that Conjecture 1 is true for n = 3 [3] (n = 1 is trivially true and n = 2 is
an application of Ryser’s Theorem [10]). Furthermore, if we assume that n is odd and only
the sub-arrays Aii contain non-empty cells, then the following cases are known [8].
Theorem 7. If A ∈ PLS(nr) with
can be completed.
n+1
2
filled r × r squares along the main diagonal, then A
Theorem 8. If A ∈ PLS(nr) with n − 1 filled r × r squares along the main diagonal and
r > n − 2, then A can be completed.
In Section 2 we introduce lifts and contractions to show that if symbols occur in at most n − 1
disjoint sub-arrays Aij and if the sub-arrays together form latin squares, then a completion
exists. This result generalizes Theorem 4. In Section 3 we introduce permissible sets to show
that when all of the filled cells occur in the block diagonal {Aii | i ∈ [n]} and the blocks are
identical, then a completion exists when n > r + 1. This confirms Conjecture 1 when the
n − 1 r × r squares are identical, and column and row disjoint.
2
Lifts and Contractions
Let a, r, and n be positive integers. A balanced partition of a set is a partition in which
each part has the same cardinality. Let S be a symbol set of cardinality anr2 and let S =
{S1 , S2 , . . . , Sanr } be a balanced partition of S. Let L ∈ PLSar2 (n) over S and A ∈ PLSa (nr)
over [anr]. The array A is a lift of L (or L is a contraction of A) if for all {i, j} ⊆ [n] and
σ ∈ [anr], |L(i, j) ∩ Sσ | = t if and only if symbol σ appears t times in the sub-array Aij .
Arrays A and L are called a lift/contraction pair. See Figure 1. We use the word contraction
because in the bipartite graph representation of A, L is obtained from A by a series of vertex
contractions (or identifications), and substituting each occurrence of edge color σ for a color
in Sσ . Lifts and contractions are similar to the disentanglements and amalgamations found in
[7] respectively, with the primary difference being the symbol sets on which each are formed.
The existence of lifts of r-semi latin squares for each integer r follows from a result of Hilton
[7], which we discuss below.
Let P = (p1 , . . . , pu ) and Q = (q1 , . . . , qv ) be partitions of n and S = {S1 , S2 , . . . , Sw } be
a partition of a symbol set S of order n. The reduction modulo (P, Q, S) of a latin square
L ∈ LS(n) over S is obtained from L by amalgamating rows p1 + · · · + pi−1 + [pi ], columns
q1 + · · · + qj−1 + [qj ], and symbols of Sk , where 1 6 i 6 u, 1 6 j 6 v, and 1 6 k 6 w. Thus,
the reduction modulo (P, Q, S) is a u × v array over [w] such that k ∈ [w] appears t times in
cell (i, j) if and only if symbols from the set Sk appear exactly t times in the corresponding
3
1
3
2
5
4
6
1123 4456 2356
2345 1236 1456
4566 1325 1234
L
2
1
3
4
6
5
4
6
1
2
5
3
5
4
6
3
2
1
3
2
5
6
1
4
6
5
4
1
3
2
A
Figure 1: Members of LS4 (3) and LS(6) which are a lift/contraction
pair, where r = 4, n = 3, and Sσ = {σ, σ} for σ ∈ [6].
sub-array of P . In Figure 1, L (with bars removed) is a reduction modulo (P, Q, S) of A in
which P = Q = (2, 2, 2) and S is the partition of S = [6] into singletons.
Let C be a u × v array over [w] in which symbols may be repeated in a cell. Let ρλ and cµ
denote the number of symbols which occur in row λ and column µ, respectively, and let τν
be the number of occurrences of ν ∈ [w] in C. Then C is called an outline rectangle if for
some integer n the following properties are met for each λ ∈ [u], µ ∈ [v], and ν ∈ [w]:
1. n divides each ρλ , cµ , and τν ;
2. cell (λ, µ) contains
ρλ cµ
n2
symbols (including repetitions); and
3. the number of times ν appears in row λ (column µ) is
ρλ τν
n2
( cµnτ2ν );
Again, by removing the bars in Figure 1, L is an outline rectangle with ρλ = cµ = 12 and
τν = 6 for each {λ, µ} ⊆ [3] and ν ∈ [6]. Hilton proved the result below in [7].
Theorem 9. To each outline rectangle C there is a latin square L and partitions P , Q, and
S such that C is the reduction of L modulo (P, Q, S).
The existence of lifts of semi-latin squares which are latin squares follows from Theorem 9.
Let L ∈ LSr2 (n) over S and S = {S1 , . . . , Snr } be a balanced partition of S. Construct L0
from L by replacing each occurrence of a symbol in Si with i for each i ∈ [nr]. Then L0 is an
outline rectangle. So there exists M ∈ LS(nr) for which L0 is its reduction modulo (P, Q, S),
where P = Q = (r, r, . . . , r). It follows that M is a lift of L. Thus, we have part (a) of the
following result when a = 1.
Theorem 10.
(a) For each L ∈ LSar2 (n), there exists A ∈ LSa (nr) such that A is a lift of L.
(b) For each A ∈ LSa (nr), there exists L ∈ LSar2 (n) such that L is a contraction of A.
4
The proof of Theorem 9 can be extended to the case a > 1 in Theorem 10(a). We prove
Theorem 10(b) formally as a special case of a more general result since it is not readily
implied from [7]. For a given A ∈ PLSa (nr), we wish to find a contraction L ∈ PLSar2 (n) of
A. See Figure 2.
2
14
5z
38
9x
67
46
1 58
2
67 3z
1v
5z
78 1z 26 35 4v
39 48 5v 2z 1x
122446
135678z
13558vzz
3677899x 124568vz 12345vxz
A
L
Figure 2: Members of PLS2 (6) and PLS8 (3) which are a lift/contraction pair,
where a = 2, r = 2, n = 3, and Sσ = {σ, σ} for σ ∈ [9] ∪ {v, x, z}.
Lemma 11. Every array of PLSa (nr) is the lift of some array of PLSar2 (n).
Proof. Let A ∈ PLSa (nr) over [anr] and let S = {S1 , . . . , Sanr } be a balanced partition of
a symbol set S of cardinality anr2 . Define G as the bipartite graph with vertex partition
R = {r1 , . . . , rn } and C = {c1 , . . . , cn }, and (ri , cj ) ∈ E(G) with multiplicity matching the
number of filled cells in Aij . For each σ ∈ [anr], let Mσ be the edge set such that for all
{i, j} ⊆ [n], (ri , cj ) ∈ Mσ with multiplicity t if and only if σ appears exactly t times in Aij .
Let Bσ denote the bipartite graph induced on Mσ for each σ ∈ [anr]. See Figure 3. Since
A ∈ PLSa (nr), the degree of each vertex in each Bσ is at most |Sσ | = r. Hence for each
σ ∈ [anr], there exists a proper edge-coloring fσ : Mσ → Sσ . Let L be an n × n array such
that for each {i, j} ⊆ [n],
L(i, j) = {fσ (e) | σ ∈ [nr], e ∈ Mσ , and e = (i, j)}.
By this construction, L ∈ PLSar2 (n) is a contraction of A.
Let A ∈ PLS(n) over S. We write A as a subset of [n] × [n] × S, where (i, j, k) ∈ A if
k ∈ A(i, j). We prove the following case of Conjecture 1, which is a generalization of Theorem
5.
Theorem 12. Let A ∈ PLS(nr) with symbols appearing in at most n − 1 disjoint r × r Aij
sub-arrays. Furthermore, symbols are arranged such that if symbol z occurs in a cell of Aij ,
then z occurs r times in the rows and columns occupied by Aij . Then A can be completed.
Proof. Without loss of generality, we assume that A has n − 1 disjoint r × r sub-arrays
B1 , . . . , Bn−1 which satisfy the hypotheses of Theorem 12. See Figure 4(a) for an example.
5
r1
c1
r1
c1
r1
c1
r1
c1
r1
c1
r1
c1
r2
c2
r2
c2
r2
c2
r2
c2
r2
c2
r2
c2
r3
c3
r3
c3
r3
c3
r3
c3
r3
c3
r3
c3
B1
B2
B3
B4
B5
B6
r1
c1
r1
c1
r1
c1
r1
c1
r1
c1
r1
c1
r2
c2
r2
c2
r2
c2
r2
c2
r2
c2
r2
c2
r3
c3
r3
c3
r3
c3
r3
c3
r3
c3
r3
c3
B7
B8
B9
Bv
Bx
Bz
Figure 3: The graphs of Bσ from Lemma 11 applied to A in Figure 2.
Black (gray) edges of Bσ correspond to σ ∈ Sσ (σ ∈ Sσ ).
Let S be a symbol set of size nr2 and let {S1 , . . . , Snr } be a balanced partition of S. From
Lemma 11, there is a contraction L0 ∈ PLSr2 (n) of A in which |L0 (i, j) ∩ Sσ | = t if and only
if symbol σ occurs t times in Aij . Since there are exactly n − 1 non-empty cells in L0 , by
Theorem 4, L0 has a completion. Let L denote a completion of L0 .
0
By Theorem 10(a), L can be lifted to a latin square A0 of order nr. Let B10 , . . . , Bn−1
⊆ A0
denote the r × r sub-arrays corresponding to B1 , . . . , Bn−1 in A. Note that if Bi = Bi0 for each
0
i, then A0 is a completion of A. Let A00 = A0 \ {B10 , . . . , Bn−1
}. If (x, y, z) ∈ Bi , then symbol z
00
does not occur in the rows and columns of A occupied by Bi . Therefore A00 ∪ {B1 , . . . , Bn−1 }
is a completion of A.
1 2 3
3 4 1
4
2
4 3
2 1
4
2
3 2 1
1 4 3
1 2
3 4
1
2
3
4
2
3
4
1
3
4
1
2
4
1
2
3
5 6
6 5
1 2
2 1
7 8
8 7
1 2
2 1
9 10
10 9
(a)
Figure 4: (a)
(b)
(b)
A completable PLS by Theorem 12 for which Ryser’s Theorem
does not apply [r = 2, n = 9, padded with 8 empty rows/columns]
A completable PLS by Corollary 13
[t1 = 2, t2 = 1, padded by at least 6 empty rows/columns.]
6
We show a case of Theorem 12 not covered by Theorem 5. Let L ∈ LS(n). We say that
M ⊂ L is a subsquare of order q if M ∈ LS(q) after permuting rows, columns, and symbols
of L. Two subsquares are disjoint if they have no rows, columns, or symbols in common.
Corollary 13. Let t1 , . . . , tm be positive integers for which n > t21 + t22 + · · · + t2m + 1 and let
A ∈ PLS(nr) be the union of m subsquares, each with order rti . Then A can be completed.
Proof. Each subsquare can be partitioned into t2i filled r × r squares, making A the union of
at most n − 1 filled r × r squares satisfying the hypotheses of Theorem 25.
It is worth noting that Corollary 13 includes the case of m disjoint subsquares. This adds to
the literature on the existence of latin squares with disjoint subsquares. See Chapter 4 of [2].
3
Identical Blocks
Let n and r be positive integers and Σ be a symbol set of cardinality nr. Let LS(r; Σ) denote
the set of column- and row-latin arrays of order r in which each cell is filled with an element
of Σ. Let A ∈ LS(r; Σ). Denote nA ∈ PLS(nr) as the partial latin square in which the
sub-array at rows and columns (i − 1)r + [r] is A for each i ∈ [n], and all other cells are
empty. See Figure 5. In this section we examine when nA can and cannot be completed. A
completion clearly exists when r = 1, and so we assume that r > 2.
1 2 4
2 5 6
4 3 5
1 2 4
2 5 6
4 3 5
A
1 2 4
2 5 6
4 3 5
2A
1
2
4
5
3
6
2
5
3
6
4
1
4
6
5
3
1
2
5
3
6
1
2
4
6
4
1
2
5
3
3
1
2
4
6
5
Completion of 2A
Figure 5: Arrays A ∈ LS(3; [6]) and 2A ∈ PLS(6), and a completion of 2A.
In the construction below we show that if n 6 r − 1, there exists A ∈ LS(r; Σ) for which
nA is not completable. When then introduce some definitions and lemmas to show that if
n > r + 1, then nA is always completable. Lastly, we comment on the case n = r.
Construction 14. Let A0 ∈ LS(r − 1) and A ∈ LS(r; [2r − 1]) such that A = A0 ∪ {(i, r, 2r −
i), (r, i, 2r − i) | i ∈ [r]}. See Figure 6. In nA, each symbol from [r − 1] occurs n(r − 1) times.
Let B denote the n × n sub-array of nA induced by rows and columns {r, 2r, . . . , nr}. To
complete nA, each symbol from [r − 1] must occur n times in B. Observe that symbol r
already occurs n times in B. Therefore, n2 > r2 and so n > r. Thus, if n 6 r − 1, nA can
not be completed.
7
1 2 5
2 1 4
5 4 3
1
2
3
7
2
3
1
6
3
1
2
5
7
6
5
4
1
2
3
4
9
2
3
4
1
8
3
4
1
2
7
4
1
2
3
6
9
8
7
6
5
Figure 6: Arrays A as constructed in Construction 14 with r = 3, 4, 5.
3.1
Permissible Sets
Let I ⊆ Σ with |I| 6 r. Define A(I) ∈ PLS(r) over I and C(I) ⊆ [r] × [r] as
A(I) = {(i, j, s) ∈ A | s ∈ I} and
C(I) = {(i, j) ∈ A | (i, j, s) ∈ A and s ∈ I}.
(1)
We say that A(I) is the partial latin square from A induced by I, and C(I) is the set of
locations of the filled cells in A(I). Furthermore, I is A-permissible (or simply permissible) if
A(I) is completable over some r-subset of Σ containing I. A partition S of Σ is A-permissible
if each part of S is permissible. Note that a permissible set has cardinality at most r. For
a symbol α ∈ Σ, let σα denote the number of occurrences of α in A, and similarly let σ(I)
denote the number of occurrences of symbols from I in A.
Lemma 15. Let A ∈ LS(r; Σ) and |Σ| = nr. If there exists a permissible, balanced partition
{S1 , S2 , . . . , Sn } of Σ, then nA is completable over Σ.
Proof. For each k ∈ [n], let Bk be the completion of A(Sk ). For each k ∈ [n] and l ∈ [r],
define Bkl ∈ PLS(r) over Sk and Rl ∈ LS(r; Σ) as
Bkl = {(i, j, s) ∈ Bk | (i, j) ∈ C(Sl )}
and
Rl =
n
[
Bmm0 ,
m=1
where m + l = m0 mod n. Let B be the nr × nr array in which for every {i, j} ⊆ [n], the
sub-array at rows (i − 1)r + [r] and columns (j − 1)r + [r] is Rk , where i + k = j mod n.
Observe that each Rn is a copy of A. Therefore B ∈ LS(nr) and B is a completion of nA.
Example 16. Let n = 2, r = 3, and A be the 3 × 3 array over [6] given in Figure 5. Let
S1 = {1, 2, 4} and S2 = {3, 5, 6}. Note A(S1 ) and A(S2 ) are both completable, say to B1 and
B2 . From B1 and B2 we derive four partial latin squares B11 , B12 , B21 , B22 and construct R1
and R2 (and R2 = A). See Figure 7. This gives the completion of 2A seen in Figure 5.
Let ΣA ⊆ Σ be the set of symbols which occur in A.
Observation 17. Suppose that ΣA has a partition S = {S1 , S2 , . . . , St } into A-permissible
sets. We say that ΣA has a A-permissible partition with t parts. If n > t, there exists a
8
1 2 4
2
4
A(S1 ) = B11
5 6
3 5
A(S2 ) = B22
1 2 4
2 4 1
4 1 2
5 6 3
3 5 6
6 3 5
B1
B2
4 1
1 2
B12
5 6 3
3
6
5 6 3
3 4 1
6 1 2
B21
R1
Figure 7: Arrays used in Example 16.
partition S 0 = {S10 , S20 , . . . , Sn0 } of Σ such that |Si | = r for each i ∈ [n], and Si ⊆ Si0 for
each i ∈ [t]. By its construction, any such partition S 0 is A-permissible. Therefore, nA is
completable if ΣA has a partition into at most n A-permissible sets.
Note that ΣA , with A as defined in Construction 14, does have a A-permissible partition
{{x} | x ∈ [r − 1]} ∪ {[2r − 1]\[r − 1]}
into r parts. So if n > r, then nA is completable. In Section 4 we conjecture that the
condition n > r is sufficient. In the meantime, we assume that n > r + 1 and begin with
observations and lemmas on completing partial latin squares.
3.2
Completion Lemmas and Observations
Unless otherwise specified, we assume that n > r + 1.
Observation 18.
(a) A partial latin square over one distinct symbol is completable.
(b) Let P ∈ PLS(r) over [2]. Suppose that σ1 = r − 1 and σ2 6= r − 1. Suppose further that
row i ∈ [r] and column j ∈ [r] do not contain symbol 1. Then P is completable if and
only if (i, j, 2) ∈
/ P.
(c) Let P ∈ PLS(r) over [2]. Suppose that σ1 6= r − 1 and σ2 6= r − 1. Then P is completable.
Lemma 19. Let A ∈ LS(r; Σ) with |Σ| = nr and T = {i ∈ ΣA | σi = r − 1}. Then either
(a) there exist Y ⊆ ΣA \T and a bijection τ : Y → T such that {y, τ (y)} is A-permissible for
every y ∈ Y , or
(b) nA is completable.
Proof. Assume that (a) does not hold. We will show that (b) follows. Observe that |T | 6 r+1,
and we first consider the case |T | = r + 1. Then σ(T ) = r2 − 1, so there exists α ∈ ΣA with
9
σα = 1, and ΣA = T ∪ {α}. Let β be any symbol of T which shares a row or column with α.
By Observations 18 (a) and (b), {α, β} is permissible and
{{α, β}} ∪ {{x} | x ∈ T and x 6= β}
is an A-permissible partition of ΣA with r + 1 parts. So nA is completable by Observation 17.
Now suppose that |T | 6 r. Let Y be the set of subsets of ΣA \T such that for each Y 0 ∈ Y,
there exists an injection τ 0 : Y 0 → T such that {y, τ 0 (y)} is A-permissible for each y ∈ Y 0 .
Let Y ∈ Y be a set of maximal cardinality with injection τ . By our assumption, τ is not a
bijection. Let x ∈ T \τ (Y ).
Without loss of generality, assume that x appears in the first r − 1 rows and columns of
A, and let (r, r, α) ∈ A. Let β ∈ ΣA and β 6= α. If β ∈
/ T , then {x, β} is A-permissible by
Observation 18 (b). Then {β} ∪ Y ∈ Y and hence β ∈ Y by the maximality of Y . Therefore
ΣA = T ∪ Y ∪ {α}. If α ∈
/ T ∪ Y , then
/ τ (Y )}
{{α}} ∪ {{y, τ (y)} | y ∈ Y } ∪ {{x} | x ∈ T and x ∈
is an A-permissible partition of ΣA with at most r + 1 parts. Otherwise α ∈ T ∪ Y and
/ τ (Y )}
{{y, τ (y)} | y ∈ Y } ∪ {{x} | x ∈ T and x ∈
is an A-permissible partition of ΣA with at most r parts. Thus nA is completable by
Observation 17.
Lemma 20. Let I ⊆ {i ∈ ΣA | σi = 2 or 3}. Then either
(a) σ(I) 6 r − 1,
(b) there exists I 0 ⊆ I such that σ(I 0 ) ∈ {r, r + 1}, or
(c) there exists I 0 ⊆ I such that σ(I 0 ) = r + 2 and σα = 3 for all α ∈ I 0 .
Proof. Assume all three statements are false; that is assume that σ(I) > r − 1, and for every
I 0 ⊆ I, σ(I 0 ) ∈
/ {r, r + 1} and either σ(I 0 ) 6= r + 2 or σα = 2 for some α ∈ I 0 . Observe that
this implies σ(I) > r + 1. Let J ⊆ I be a subset of minimum occurrence in A such that
σ(J) > r + 1.
Suppose that σ(J) = r + 2. Then σα = 2 for some α ∈ J. Then σ(J\{α}) = σ(J) − 2 = r,
which is a contradiction.
Suppose now that σ(J) > r + 3. Let α ∈ J and J 0 = J\{α}. Then σ(J 0 ) > r. If σ(J 0 ) > r + 2,
then J is not minimal, which is a contradiction. So σ(J 0 ) ∈ {r, r + 1}, which is also a
contradiction. The result follows.
Lemma 21. Let I ⊆ {α ∈ ΣA | σα = 1}. Then either |I| < 2r − 1 or there exists a
permissible r-subset I 0 ⊆ I.
10
Proof. Suppose that |I| > 2r − 1. Let I 00 be any r-subset of I. If I 00 is permissible, then we
are done. If not, by Theorem 3, A(I 00 ) has a row R and column C such that their intersection
is empty and all filled cells belong to R or C. Without loss of generality, let R contain at
least 2 symbols α and β. There are 2r − 2 cells in the disjoint union of R and C, and hence
there is some symbol γ ∈ I which occurs outside of the disjoint union. Again, by Theorem 3,
either I 00 ∪ {γ}\{α} or I 00 ∪ {γ}\{β} is permissible.
Let P ∈ PLS(r). Suppose that symbol α ∈ [r] occurs in P and that σα = k. By permuting
the rows and columns of P , we may assume that α occurs in the rows and columns indexed
by [k]. We use Gα to denote the bipartite graph with vertex parts R = C = {k + 1, . . . , r}
and edge set {(i, j) | P (i, j) = ∅}. If Gα contains a perfect matching M , we place symbol α
in cell (x, y) of P if and only if (x, y) ∈ M . We say that symbol α can be completed in P .
From Hall’s Theorem, if each vertex in Gα has degree at least r−k
, then M exists.
2
Lemma 22. Let t 6 r be a nonnegative integer and let P ∈ PLS(r) over [t]. Suppose that
σ1 > σ2 > . . . > σt .
(a) If t 6
r−σ1
2
+ 1, then P is completable.
(b) If r > 5, σ1 = σt = 2, and t 6
r+1
,
2
then P is completable.
(c) If r > 6, σ1 = 3, σ2 = σt = 2, and t 6 2r , then P is completable.
Proof. Since symbols t + 1, . . . , n do not occur in P , we need only to show that symbols from
[t] can be completed. For each k, 0 6 k 6 t, we wish to find Pk ∈ PLS(r) such that P ⊆ Pk
and symbols from [k] are completed in Pk . By convention, P0 = P .
1
Suppose that t 6 r−σ
+ 1, 1 6 k 6 t, and Pk−1 exists. We must show that Pk exists. The
2
degree of each vertex of Gk is
r − σ1
r − σk
d > r − σk − (t − 1) > r − σk −
>
2
2
Thus Gk has a perfect matching and hence Pk exists.
Suppose that r > 5 and σ1 = σt = 2. If t < r+1
, then the result follows from (a). Therefore,
2
we assume that t = r+1
.
The
degree
of
each
vertex
of Gk is
2
r−1
r−3
d > r − 2 − (t − 1) = r − 2 −
=
2
2
There is no perfect matching in Gk if there is a subset S ⊆ R in which |S| < |N (S)|. Suppose
that such a set S exists. Certainly |N (S)| > r−3
. If |S| > r−1
, there is a vertex i ∈ C\N (S)
2
2
r−3
with d(i) 6 |R\S| < 2 . Thus |S| = |C\N (S)| = t − 1 = r−1
and |N (S)| = r−3
.
2
2
11
Let M denote the sub-array of Pk−1 at the intersection of S and C\N (S). Since there are
no edges between S and C\N (S) in Gk , the cells in M are each filled. Since t > 3 and
(t − 1)2 6 (k − 1)(t − 1) + 2(t − k) (the number of filled cells in M is at most the number of
filled cells in the rows indexed by S), it follows that k = t.
Observe that M is a subsquare of Pt−1 over [t − 1]. Furthermore, the filled cells of Pt−1 in
rows from S or columns from C\N (S) are precisely M .
Let α ∈ [t − 1]. The fixed occurrences of α are the original two occurrences of α in P . Since
t > 3, there are unfixed occurrences of α both in M and outside of S and C\N (S), say
(i1 , j1 , α) ∈ M and (i2 , j2 , α) ∈ Pt−1 \M , respectively. Observe that cells (i1 , j2 ) and (i2 , j1 ) are
empty in Pt−1 . Thus P 0 = (Pt−1 \ {(i1 , j1 , α), (i2 , j2 , α)}) ∪ {(i1 , j2 , α), (i2 , j1 , α)} ∈ PLS(r)
has no subsquare of order t − 1. Therefore the graph Gt formed from P 0 now contains a
perfect matching. It follows that symbol t can be completed in P 0 , and such a completion
contains P .
Lastly, suppose that r > 6, σ1 = 3, σ2 = σt = 2, and t 6 2r . If t < 2r , then the result follows
from part (a). Therefore, we assume that t = 2r . We begin by first finding P1 . The degree of
each vertex of G1 is
r−4
r−2
d > r − 3 − (t − 1) = r − 3 −
>
2
2
There is no perfect matching in G1 if there is a subset S ⊆ R in which |S| = r−2
and
2
r−4
r−2
|N (S)| = 2 . This implies that there is a subsquare of order 2 in P . However, since
Pt
(r−2)2
6 r − 2 implies that r 6 5. Thus, no such subsquare exists and there
i=2 σi = r − 2,
4
is a perfect matching in G1 , which implies that symbol 1 can be completed and therefore P1
exists.
Consider now symbol k > 1. The degree of each vertex of Gk is
r−2
r−2
d > r − 2 − (t − 1) = r − 2 −
=
2
2
Thus Gk has a perfect matching and thus Pk exists, where 2 6 k 6 t.
Let L ∈ LS(2) and both P ∈ PLS(5) over [5] and Q ∈ PLS(6) over [6] be defined as
P = L ∪ {(3, 3, 3), (4, 4, 3)}
Q = L ∪ {(3, 3, 3), (4, 4, 3), (5, 5, 3)}
Observe that P and Q are both incompletable and satisfy the conditions in Lemma 22 (b)
and (c), respectively, excluding the condition on r. Therefore the inequalities for r in the
lemma are sharp.
Observation 23. Let P ∈ PLS(r) over [t].
12
r− r
(a) Suppose 4 6 σα 6 2r for each α ∈ [t] and t 6 b 4r + 1c. Then t 6 2 2 + 1, and hence by
Lemma 22 (a), P is completable. Furthermore, σ([t]) > 4b 4r + 1c > r + 1.
(b) Suppose 2 6 σα 6 3 for each α ∈ [t] and t 6 b r−1
c. Suppose that at least two symbols
2
have occurrence 3. By Lemma 22 (a), P is completable and σ([t]) > 2t + 2 > r + 1.
(c) Suppose 2 6 σα 6 3 for each α ∈ [t] and t 6 b 2r c. Suppose there is exactly one symbol
with occurrence 3. By Lemma 22 (b), P is completable and σ([t]) = 2t + 1 > r + 1.
(d) Suppose σα = 2 for each α ∈ [t] and t 6 b r+1
c. By Lemma 22 (c), P is completable and
2
σ([t]) = 2t > r + 1.
3.3
Main Result
Let r > 2, n > r + 1 and A ∈ LS(r; Σ), with |Σ| = nr. If 2 6 r 6 6, it can be shown through
tedious case-analysis that there exists an A-permissible partition of ΣA into at most r + 1
parts. So we proceed with r > 7.
Lemma 24. If r > 7, n > r + 1, and A ∈ LS(r; Σ) with |Σ| = nr, then nA is completable
over Σ.
Proof. Let T = {α ∈ Σ | σα = r − 1}. By Lemma 19, either nA is immediately completable,
or there exist Y ⊆ ΣA \T and τ for which |T | = |Y | and {y, τ (y)} is permissible for each
y ∈ Y . We assume the latter, and let T = {{y, τ (y)} | y ∈ Y }. We partition the remaining
symbols in ΣA \(T ∪ Y ) into the following sets:
S = {α ∈ ΣA \(T ∪ Y ) | σα = r},
Q = {α ∈ ΣA \(T ∪ Y ) |
r
2
< σα 6 r − 2},
P = {α ∈ ΣA \(T ∪ Y ) | 4 6 σα 6 2r },
Z = {α ∈ ΣA \(T ∪ Y ) | 2 6 σα 6 3}, and
K = {α ∈ ΣA \(T ∪ Y ) | σα = 1}.
We begin by partitioning each of the above sets into as many A-permissible sets as possible,
each with occurrence at least r. We then combine any remaining symbols into A-permissible
sets to achieve an A-permissible partition of ΣA into at most r + 1 parts.
Since A(S) is completable, S is A-permissible. Let S = {S} if S is nonempty, and let S be
empty otherwise. Thus (perhaps vacuously) every set in S has occurrence at least r.
Let q = |Q| mod 2. Let Q0 be a q-subset of Q and Q a partition of Q\Q0 into 2-subsets. By
Observation 18 (c), each part of Q is A-permissible with occurrence at least r + 1. If q = 1,
by Observation 18 (a), Q0 is A-permissible.
13
Let p = |P | mod (b 4r + 1c). Let P 0 be a p-subset of P and let P be a partition of P \P 0 into
(b 4r + 1c)-subsets. By Observation 23 (a), each part of P is A-permissible with occurrence at
least r + 1, and by Theorem 2, P 0 is A-permissible.
By an inductive application of Lemma 20, there exists Z 0 ⊆ Z and a partition Z of Z\Z 0 such
that σ(Z 0 ) 6 r − 1 and σ(I) ∈ {r, r + 1, r + 2} for each I ∈ Z, and either σ(I) ∈ {r, r + 1} or
σ(I) = r + 2 with σα = 3 for every α ∈ I. By Observations 23 (b), (c), and (d), each I ∈ Z
is A-permissible with occurrence at least r + 1, and by Theorem 2, Z 0 is A-permissible.
By an inductive application of Lemma 21, there exists K 0 ⊆ K and a partition K of K\K 0
such that each part of K is an A-permissible r-subset (and hence has occurrence r) and
|K 0 | < 2r − 1.
Define A = T ∪ S ∪ P ∪ Q ∪ Z ∪ K, which is a partition of ΣA \(Q0 ∪ P 0 ∪ Z 0 ∪ K 0 ). Observe
that σ(I) > r for each I ∈ A. Therefore σ(Q0 ∪ P 0 ∪ Z 0 ∪ K 0 ) 6 r2 − r|A| = (r − |A|)r. We
show that the remaining symbols in Q0 ∪ P 0 ∪ Z 0 ∪ K 0 can be partitioned into A-permissible
sets to produce an A-permissible partition of ΣA into at most r + 1 parts.
Suppose that |A| 6 r − 4. Partition K 0 into K10 and K20 so that |Ki0 | 6 r − 1 for i ∈ [2].
By Theorem 2, Ki0 is A-permissible for i ∈ [2]. Therefore A ∪ {Q0 , P 0 , Z 0 , K10 , K20 } is a
A-permissible partition with at most r + 1 parts.
Suppose that |A| = r − 3. Then σ(Q0 ∪ P 0 ∪ Z 0 ∪ K 0 ) 6 3r. Partition K 0 into KQ0 , KP 0 , KZ 0
and K 00 so that |KI | 6 r − 1 − σ(I) for each I ∈ {Q0 , P 0 , Z 0 } and |K 00 | 6 3 6 r − 1. Define
Q00 = Q0 ∪ KQ0 , P 00 = P 0 ∪ KP 0 , and Z 00 = Z 0 ∪ KZ 0 . By Theorem 2, Q00 , P 00 , Z 00 , and K 00 are
each A-permissible. Hence A ∪ {Q00 , P 00 , Z 00 , K 00 } is a A-permissible partition with r + 1 parts.
Suppose that |A| = r − 2. Then σ(Q0 ∪ P 0 ∪ Z 0 ∪ K 0 ) 6 2r. Partition K 0 into KQ0 , KP 0 ,
and KZ 0 with |KI | 6 r − 1 − σ(I) for each I ∈ {Q0 , P 0 , Z 0 }, which is possible since r > 3.
Define Q00 , P 00 , and Z 00 as before, and again by Theorem 2, each are A-permissible. Hence
A ∪ {Q00 , P 00 , Z 00 } is a A-permissible partition with r + 1 parts.
Suppose that |A| = r − 1. Then σ(Q0 ∪ P 0 ∪ Z 0 ∪ K 0 ) 6 r. If Q0 ∪ P 0 ∪ Z 0 ∪ K 0 is empty, then
A is an A-permissible partition of ΣA with r − 1 parts. Otherwise, let α ∈ Q0 ∪ P 0 ∪ Z 0 ∪ K 0 .
Then by Observation 18 (a) and Theorem 2, {α} and (Q0 ∪ P 0 ∪ Z 0 ∪ K 0 )\{α} are both
A-permissible. Hence A ∪ {{α}, (Q0 ∪ P 0 ∪ Z 0 ∪ K 0 )\{α}} is a A-permissible partition with
r + 1 parts.
Suppose that |A| > r. Since σ(I) > r for each I ∈ A, it follows that |A| = r and σ(I) = r
for each I ∈ A. Therefore A is a partition of ΣA which is A-permissible.
We summarize these results with the theorem below.
Theorem 25. Let n and r be positive integers. If n > r + 1, then for every A ∈ LS(r; [nr]),
nA is completable. If n 6 r − 1, then there exists A ∈ LS(r; [nr]) for which nA is not
completable.
14
4
Concluding Remarks
Theorem 25 does not include the case when n = r. For every example of A we investigated,
there did exist a A-permissible partition of ΣA into at most r parts. However, there are
several special cases that arise when forming the partition, some of which seem difficult to
reconcile with larger values of r. This leads us to the following conjecture.
Conjecture 26. Let r and n be positive integers. Then nA is completable for every A ∈
LS(r; [nr]) if and only if n > r.
We end with an example illustrating that the converse of Lemma 15 does not hold – that there
exists an array A ∈ LS(r; [nr]) for which nA is completable and [nr] has no A-permissible
partition.
Example 27. Let A ∈ LS(4; [8]) be the array in Figure 8. In a balanced partition {S1 , S2 }
of [8], at least two elements of {1, 2, 3} belong to the same part, say S1 . Hence at least one
of the arrays B1 , B2 , and B3 is contained in A(S1 ). Observe that B1 , B2 , and B3 are not
completable. Therefore no balanced bipartition of [8] will induce completable partial latin
squares. However, 2A does have a completion. See Figure 8.
1
5
4
3
4
2
5
1
6
4
3
2
A
3
1
2
4
1
1
2
1 2
B1
3
1
1
2
3
2
3
3 1
3
B2
3 2
2
B3
1
5
4
3
2
8
7
6
4
2
5
1
7
3
6
8
6
4
3
2
8
7
1
5
3
1
2
4
5
6
8
7
2
8
7
6
1
5
4
3
7
3
6
8
4
2
5
1
8
7
1
5
6
4
3
2
5
6
8
7
3
1
2
4
A completion of 2A
Figure 8: The arrays from Example 27.
We generalize the example above to an infinite family of completable partial latin squares of
the form 2A in which there is no permissible bipartition of [2r].
Construction 28. Let r > 4 be given, and let r = 3k + l, where l ∈ {4, 5, 6}. Let L3 ∈ LS(3)
over [3]. Define Tl and L ∈ PLS(r) over [r] as given in Figure 9. Observe that for all
A ∈ LS(r; [2r]), if L ⊆ A, then by a similar argument used in Example 27, no balanced
bipartition of [2r] will induce completable partial latin squares from A.
Let M ∈ PLS2 (r) over [2r] be M = L ∪ {(1, 1, 2), (2, 2, 3), (3, 3, 1)}. Observe that symbols
1, 2, and 3 each occur r times in M . Using Hall’s Theorem for bipartite multi-graphs [6],
M can be completed to a 2-multi latin square M 0 ∈ LS2 (r). Let A, B ∈ LS(r; [2r]) such
that A ∪ B = M 0 and L ⊆ A. Observe that 2A has a completion (put copies of B in the
off-diagonal squares), but A does not satisfy the hypotheses of Lemma 15.
15
1
1
1
2
3
2
Tl
3
3
2
2 3 1
3 1
2
T4
T5
L3
1
3
3
2
1
3 2
3 1 2
1
3
L3
2
L3
1 2
2
3
2
3 1
1
L3
T6
L
Figure 9: Arrays for Construction 28.
12
23
3
1
13
2
M
3
1
2
12
58
47
36
47
23
56
18
68
47
13
25
35
16
28
47
M0
1
5
4
3
4
2
5
1
6
4
3
2
A
3
1
2
4
2
8
7
6
7
3
6
8
8
7
1
5
5
6
8
7
B
Figure 10: The arrays from Construction 28 applied to the example in Figure 6.
References
[1] L.D. Anderson, A.J.W. Hilton, Thanks Evans!, Proc. London Math. Soc. 47 (1983)
507-522.
[2] J. Dénes, A. Keedwell, Latin Squares, Annals of Discrete Mathematics, New developments
in the theory and applications, With contributions by G.B. Belyavskaya, A.E. Brouwer,
T. Evans, K. Heinrich, C.C. Lindner and D.A. Preece, With a foreword by Paul Erdős.
North-Holland Publishing Co. Amsterdam 46 (1991).
[3] T. Denley, R. Häggkvist, Completing some partial Latin squares, Europ. J. Combin. 21
(1995) 877-880.
[4] T. Evans, Embedding incomplete Latin squares, Amer. Math. Monthly 67 (1960) 958-961.
[5] R. Häggkvist, A solution to the Evans conjecture for Latin squares of large size, Colloqu.
Math. Soc. Janos Bolyai 18 (1978), 405-513.
[6] P. Hall, On representatives of subsets, J. London. Math. Soc. 10 (1935) 26-30.
[7] A.J.W. Hilton, The reconstruction of latin squares with applications to school timetables
and experimental design, Math. Prog. Study. 13 (1980) 68-77.
[8] L-D. Öhman, A note on completing Latin squares, Australas. J. of Combin. 45 (2009)
117-123.
16
[9] J. Kuhl, T. Denley, On a generalization of the Evans Conjecture, Discrete Math. 308
(2008) 4763-4767.
[10] H.J. Ryser, A combinatorial theorem with an application to Latin squares, Proc. Amer.
Math. Soc. 2 (1951) 550-552.
[11] B. Smetaniuk, A new construction for latin squares I. Proof of the Evans conjecture,
Ars Combin. 11 (1981) 155-172.
17