Embedding a Latin Square in a Pairof Orthogonal Latin Squares

Peter JenkinsCentre for Discrete Mathematics and Computing, Department of Mathematics,The University of Queensland, Queensland 4072, Australia,E-mail: pdj@maths.uq.edu.au

Received January 9, 2005; revised July 18, 2005

Published online 21 October 2005 in Wiley InterScience (www.interscience.wiley.com).

DOI 10.1002/jcd.20087

Abstract: In this paper, it is shown that a latin square of order n with n � 3 and n 6¼ 6 can be

embedded in a latin square of order n2 which has an orthogonal mate. A similar result for

idempotent latin squares is also presented. # 2005 Wiley Periodicals, Inc. J Combin Designs 14: 270–

276, 2006

Keywords: orthogonal latin squares; embedding; orthogonal mate

1. INTRODUCTION

A partial latin square of order n is an n� n array, possibly with some empty cells,whose entries are chosen from a set of n symbols such that each symbol occurs atmost once in each row and in each column. A latin square of order n is a partial latinsquare of order n with no empty cells. Unless otherwise stated, it will be assumed thata (partial) latin square of order n is based on the symbols 0; 1; . . . ; n� 1, and that therows and columns are indexed by 0; 1; . . . ; n� 1.

A (partial) latin square of order n is said to be idempotent if cell ði; iÞ contains thesymbol i for each i 2 f0; 1; . . . ; n� 1g. The (partial) latin square L is said to beembedded in the latin square S provided that the top left corner of S agrees with L.Two (partial) latin squares of the same order n (and with precisely the same filledcells) are orthogonal if they produce a set of distinct ordered pairs when they aresuperimposed. Figure 1 shows a pair of orthogonal partial latin squares of order 4.Notice that the empty cells of these latin squares cannot be filled to form a pair oforthogonal latin square of order 4. Hence, the problem of embedding pairs of (partial)orthogonal latin squares arises.

# 2005 Wiley Periodicals, Inc.

270

It is known [4] that a pair of orthogonal latin squares of order n can be embeddedin a pair of orthogonal latin squares of order t if t � 3n, the bound of 3n being bestpossible. Obtaining an analogous result for pairs of partial orthogonal latin squareshas proved to be an extremely challenging problem. Although it is known [6] that apair of partial orthogonal latin squares can always be finitely embedded in a pair oforthogonal latin squares, there is no known method which obtains an embedding ofpolynomial order (with respect to the order of the partial arrays).

In [5], Hilton et al. formulate some necessary conditions for a pair of orthogonalpartial latin squares to be extended to a pair of orthogonal latin squares. Un-fortunately, proving the sufficiency of these conditions appears to be beyond thereach of current methods.

In this paper, the less difficult problem of embedding a single partial latin square in apair of orthogonal latin squares is investigated. Since a partial latin square of order n canalways be embedded in a latin square of order 2n (see [3]), we restrict our attention toproving that a latin square can always be embedded in a latin square of ‘‘small’’ orderwhich has an orthogonal mate. A similar result for idempotent latin squares is alsoobtained. It should be noted that latin squares exist which have no orthogonal mate.

2. PRELIMINARY RESULTS

Unless otherwise stated, it will be assumed that a (partial) latin square of order n isbased on the symbols 0; 1; . . . ; n� 1, and that the rows and columns are indexed by0; 1; . . . ; n� 1.

The back circulant latin square of order n, denoted Bn, is the latin square formedby filling each cell ði; jÞ; i; j 2 f0; 1; . . . ; n� 1g, with the symbol iþ jðmod nÞ. Notethat if n is odd, the diagonal elements of Bn are distinct.

If L is a latin square based on a set of integers S and i is an integer, let Lþ idenote the latin square based on the set fxþ i j x 2 Sg formed by adding i to everyentry in L.

Let n be a positive integer and x be an integer. For convenience, the notationx will be used to denote the unique integer in the set f0; 1; . . . ; n� 1g such thatx � x ðmod nÞ.

Let AðnÞ be the n� n array in which cell ði; jÞ is occupied by the symbol niþ j forall i; j 2 f0; 1; . . . ; n� 1g. Furthermore, for any r; c 2 f0; 1; . . . ; n� 1g, let Ac

rðnÞdenote the array formed by filling each cell ði; jÞ; i; j 2 f0; 1; . . . ; n� 1g, with thesymbol in cell ði� r; j� cÞ of AðnÞ. Thus, Ac

rðnÞ is the array formed by cyclicallyshifting the rows of AðnÞ forward (downward) by r, and the columns of AðnÞ forward(to the right) by c. It is easy to verify that the entry in cell ði; jÞ of Ac

rðnÞ is given bynði� rÞ þ j� c. See Figure 2 for an example.

FIGURE 1. A pair of orthogonal partial latin squares of order 4.

EMBEDDING A LATIN SQUARE 271

The following well-known theorem (which is a consequence of several results inSection II.2 of [2]) gives the possible orders of pairs of orthogonal latin squares, andpairs of orthogonal idempotent latin squares.

Theorem 2.1 [2]. There exists a pair of orthogonal latin squares of order n for allintegers n � 3; n 6¼ 6, and there exists a pair of idempotent orthogonal latin squaresof order n for all integers n � 4; n 6¼ 6.

3. MAIN RESULTS

We now present the main embedding results of this paper.

Theorem 3.1. Let L be a latin square of order n with n � 3 and n 6¼ 6. Then L canbe embedded in a latin square of order n2; which has an orthogonal mate.

Proof. Let L be a latin square of order n where n � 3 and n 6¼ 6. Let fS; Tg be a pairof orthogonal latin squares of order n such that cell ð0; 0Þ of S contains the symbol 0.We shall construct a pair of orthogonal latin squares fU;Vg such that L is embeddedin the top left hand corner of U.

For each i; j 2 f0; 1; . . . ; n� 1g, if cell ði; jÞ of S contains the symbol k, and cellði; jÞ of T contains the symbol l, then let

Ui; j ¼ L if k ¼ 0;Bn þ kn if k > 0;

�

and Vi; j ¼ AlkðnÞ:

Now, let U be the array formed by replacing the entry in cell ði; jÞ of S with Ui; j.Similarly, let V be the array formed by replacing the entry in cell ði; jÞ of T with Vi; j.(An example illustrating the construction of the arrays U and V follows this proof.)

It is clear that both U and V are latin squares of order n2 based on the symbolsf0; 1; . . . ; n2 � 1g, and that the latin square L lies in the top left corner of U. Itremains to show that U and V are orthogonal.

Suppose that cells ðw; xÞ and ðy; zÞ of U both contain the same symbol s. We claimthat cells ðw; xÞ and ðy; zÞ of V contain distinct symbols.

Let ~ww ¼ bw=nc; ~xx ¼ bx=nc; ~yy ¼ by=nc; and ~zz ¼ bz=nc. Thus, cell ðw; xÞ of U(respectively V) corresponds to cell ðw; xÞ of U ~ww;~xx (respectively V ~ww;~xx). Similarly, cellðy; zÞ of U (respectively V) corresponds to cell ðy; zÞ of U~yy;~zz (respectively V~yy;~zz).

Since cells ðw; xÞ and ðy; zÞ contain the same symbol s, it follows that U ~ww;~xx ¼ U~yy;~zz.First suppose that ~ww ¼ ~yy and ~xx ¼ ~zz. Then the symbol s must occur in cells ðw; xÞ and

FIGURE 2. The arrays Að4Þ (left) and A21ð4Þ (right).

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ðy; zÞ of U ~ww;~xx. This implies that w 6¼ y; (for otherwise, the symbol s would occurtwice in the same row of U ~ww;~xx.) Since the rows of V ~ww;~xx are disjoint, the cells ðw; xÞand ðy; zÞ of V ~ww;~xx must contain distinct symbols. The claim is thus true in this case.

Alternatively, suppose that ~ww 6¼ ~yy and ~xx 6¼ ~zz. (If precisely one of these statementswere true, then it would imply that a symbol occurs twice in the same row or columnof S.) Now, let cells ð~ww;~xxÞ and ð~yy;~zzÞ of S contain the symbol �. Since S and T areorthogonal, these cells must contain distinct symbols in T: let cell ð~ww;~xxÞ of T containsthe symbol � and cell ð~yy;~zzÞ of T contains the symbol �. It follows that V ~ww;~xx ¼ A�

�ðnÞand V~yy;~zz ¼ A�

�ðnÞ. If w ¼ y, then we must also have x ¼ z. Since � 6¼ �, cell ðw; xÞ ofA��ðnÞ contains a symbol different from that in cell ðw; xÞ of A�

�ðnÞ. If w 6¼ y, thensince the symbols in row w of A�

�ðnÞ are disjoint from the symbols in row y of A��ðnÞ,

the cells ðw; xÞ of A��ðnÞ and ðy; zÞ of A�

�ðnÞ must contain distinct symbols. Thisverifies the aforementioned claim; thus, U and V are indeed orthogonal. &

Figures 3, 4, and 5 illustrate the process by which a latin square L of order 4 isembedded in a pair fU;Vg of orthogonal latin squares of order 16.

The following theorem gives a similar result to Theorem 3.1 for idempotent latinsquares.

Theorem 3.2. Let L be an idempotent latin square of odd order n with n � 5. Then Lcan be embedded in an idempotent latin square of order n2, which has an idempotentorthogonal mate.

Proof. Let fS; T 0g be a pair of idempotent orthogonal latin squares of odd order n.(Such a pair exists by Theorem 2.1.) Let T be the latin square obtained from T 0 byreplacing each occurrence of the symbol x where x 2 f0; 1; . . . ; n� 1g, by the symbol2x. (Thus, T has diagonal entries 0; 2; 4; . . . ; n� 1; 1; 3; . . . ; n� 2.) It is clear that S andT are orthogonal. We claim that if fU;Vg is the pair of orthogonal latin squares obtainedfrom fS; Tg in the manner described in the proof of Theorem 3.1, then the diagonalelements of U are distinct, and the diagonal elements of V are distinct.

FIGURE 3. A latin square L of order 4, and a pair fS; Tg of orthogonal latin squares oforder 4.

EMBEDDING A LATIN SQUARE 273

Now, the set of diagonal elements of U is simply the union of the set of diagonalelements of the latin squares

L;Bn þ n;Bn þ 2n; . . . ;Bn þ ðn� 1Þn:

Since L is idempotent, and the diagonal elements of Bn are distinct, this set ofdiagonal elements is clearly f0; 1; . . . ; n2 � 1g.

FIGURE 4. The latin square U:

FIGURE 5. The latin square V:

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Similarly, the set of diagonal elements of V is the union of the diagonal elementsof the arrays

A00ðnÞ;A2

1ðnÞ;A42ðnÞ; . . . ;An�1

ðn�1Þ=2ðnÞ;A1ðnþ1Þ=2ðnÞ;A3

ðnþ3Þ=2ðnÞ; . . . ;An�2n�1ðnÞ:

Suppose that the entry in cell ði; iÞ of A2xx ðnÞ is equal to the entry in cell ðj; jÞ of

A2yy ðnÞ, where x; y; i; j 2 f0; 1; . . . ; n� 1g. We shall show that x ¼ y and i ¼ j. From

the definition of the array AcrðnÞ, we have

nði� xÞ þ i� 2x ¼ nðj� yÞ þ j� 2y:

It follows that

i� x � j� yðmod nÞ ð1Þ

and

i� 2x � j� 2yðmod nÞ: ð2Þ

Eliminating i and j in (1) and (2) gives

x � y ðmod nÞ:

Since x; y 2 f0; 1; . . . ; n� 1g, we must have x ¼ y. Furthermore, since the arrayA2xx consists of n2 distinct symbols, we must have i ¼ j. Thus, the diagonal elements

of V are distinct.Let U� and V� be the latin squares formed by relabeling the symbols of U and V ;

respectively, so that U� and V� are idempotent. Then fU�;V�g is a pair of idempotentorthogonal latin squares. Furthermore, since the cells ð0; 0Þ; ð1; 1Þ; . . . ; ðn� 1; n� 1Þof U contain the symbols 0; 1; 2; . . . ; n� 1; respectively, the latin square L (whichoccurs in the top left corner of U) must occur in the top left corner of U�. &

4. CONCLUDING REMARKS

In [3], Evans proved that a partial latin square of order n � 1 can be embedded in alatin square of order 2n. Similarly, Andersen et al. [1] have shown that a partialidempotent latin square of order n � 1 can be embedded in an idempotent latinsquare of order 2nþ 1. Combining these results with Theorems 3.1 and 3.2 gives thefollowing two theorems.

Theorem 4.1. A partial latin square of order n � 4 can be embedded in a latinsquare of order 4n2 which has an orthogonal mate.

Theorem 4.2. A partial idempotent latin square of order n � 3 can be embedded inan idempotent latin square of order ð2nþ 1Þ2; which has an idempotent orthogonalmate.

EMBEDDING A LATIN SQUARE 275

While the results of this paper do not extend to embedding pairs of orthogonalpartial latin squares, they do have applications to enclosing certain types of partialK4-designs. For example, consider a partial K4-design with the property that everyblock contains a vertex not in any other block. Then it is possible to show that thereexists an enclosing K4-design of quadratic order and index 12 for such a partialdesign.

REFERENCES

[1] L. D. Andersen, A. J. W. Hilton, and C. A. Rodger, A solution to the embedding problemfor partial idempotent latin squares, J London Math Soc 26(2) (1982), 21–27.

[2] C. J. Colbourn and J. H. Dinitz (Editors), The CRC Handbook of Combinatorial Designs,CRC Press, Boca Raton, FL, 1996.

[3] T. Evans, Embedding incomplete latin squares, Amer Math Mon 67 (1960), 958–961.

[4] K. Heinrich and L. Zhu, Existence of orthogonal latin squares with aligned subsquares,Discrete Math 59 (1986), 69–78.

[5] A. J. W. Hilton, C. A. Rodger, and J. Wojciechowski, Prospects for good embeddings ofpairs of partial orthogonal latin squares and of partial Kirkman triple systems, J CombinMath Combin Comput 11 (1992), 83–91.

[6] C. C. Lindner, Embedding orthogonal partial latin squares, Proc Amer Math Soc 59(1)(1976), 184–186.

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