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/ Maklumat dan Sains Kuantitatif

II

UNIVERSITI TEKNOLOGI MARA

ISSN:1823-0822

Jilid 6, Bil. 1,2004

JURNAL TEKNOLOGI MAKLUMAT DAN SAINS KUANTITATIF Kandungan

Muka Surat

Measuring Outlier Effects in Bilinear Time Series Processes Mohamad Said Zainol, Ibrahim Mohamed, Azami Zaharim

IX

Orthogonal Latin Square Constructions Haslinda Ibrahim

U) OC

Is

T

Pemeringkatan Kriteria Pemilihan Pelajar Matrikulasi ke Universiti Menggunakan Model Set Kabur Wan Rosmanira Ismail, Vmmul Khair Salma Hi.Din, Norngainy Mohd Tawil

19

A Case Study of Intrusion Signature Identifications Featuring SNORT Saadiah Yahaya, Abdul Hamid Osman, Nik Mariza

31

Optimization of Overtime in a Manufacturing System Ruzzakiah Jenal, Amelia Natasya Abdul Wahab, Suhaila Mohamed Yusof

39

Adopting a Common Mathematics Curriculum in a Diverse Population: Is it a Wise Strategy? Mohd Sahar Sauian, Shukri Shamsudin

49

Universiti Teknoloei MARA

ISSi urnal Tek. Maklumat & Sains Kuantitatif Jilid 6, Bil. 1, 2004 (9-17)

Orthogonal latin square constructions Haslinda Ibrahim Faculty of Quantitative Sciences, UUM, 06010 Sintok, Kedah

Abstract The 18th-century mathematician Euler credited with the development of the latin square combinatorial design, was responsible for numerous theorems and constructions of latin squares. MacNeish, another prominent mathematical scholar, further developed the latin square conjecture in the 1920s, but more recent work by Bose, Shrikhande and Parker led to the collapse of Euler and MacNeish conjectures. This paper illustrates the sequential unfolding of our understanding of orthogonal latin square constructions.

Keywords: Latin squares, mutually orthogonal latin squares, orthogonal latin squares.

1.

Introduction

A latin square of order n is an n x n array based on set S of n symbols (or treatments) with the following property: every row and every column contains every symbol exactly once. The symbols can be letters, integers or pictures. In our discussion we will use integers to represent symbols. Here are four examples of latin squares: Order 2 12 21

Order 3 123 23 1 312

Order 4 1234 1234 2341 2143 3412 3412 4 123 432 1

You can see that the first three examples follow the same pattern: each row is determined by cyclically rotating the previous row to the right. It is obvious that this procedure will work for any size n to produce a latin square. The fourth example illustrates that we can have more than one latin square of a given order. This raises some interesting questions. Does there exist a latin square of any order? How many latin squares of a given order are there? These questions will be discussed below.

e-mail: [email protected]

10

Haslinda Ibrahin

Latin squares were introduced by the great mathematician Leonhard Euler. In 1779, Euler posed the following problem: "Is it possible to arrange 36 officers, each having one of six different ranks and belonging to one of six different regiments, in a square formation 6 by 6, so that each row and each file shall contain just one officer of each rank and just one from each regiment? The solution requires two latin squares of order 6; in one the symbols represent ranks, and in the other the symbols represent regiments. In addition to these requirements, when these two latin squares are superimposed each ordered pair must occur exactly once. Two latin squares having the characteristics mentioned are called orthogonal latin squares. Definition: Two latin squares L] = | a | and L, = \ b..\ on n symbols 1, 2,...,n are said to be orthogonal if every ordered pair of symbols occurs exactly once among the n2 pairs (a.., bf) where i= 1, 2,...,n; /'= 1, 2,...,n. Consider for example a pair of orthogonal latin squares of order 3:

L;=

123 23 1 3 12

L, = "

123 3 12 23 1

When we superimpose L; onto L, the resultant square has nine distinct ordered pairs.

Lp=

1.1 2,2 3,3 2,3 3,1 1,2 3.2 1,3 2,1

Thus L; is orthogonal to Lr Definition: A set of latin squares {L p L2,...,Lk} is called a set of k mutually orthogonal latin squares (MOLS) if L. and L. are orthogonal for all i, j e {1, 2,...,k} and i * j . For example consider these latin squares of order 4:

L, =

1234 2143 34 1 2 4321

L.

1342 42 13 243 1 3 124

L

,=

1423 3241 4132 2 3 14

Lr L, and L, are orthogonal to each other. Thus the set {L^L^} is a set of three MOLS of order 4. Now, the question that arises is: How many MOLS exist for every order?

Orthogonal latin square construction 11

2.

Some results

Theorem 1. For any positive integer n, there is a latin square of order n. Proof. Place the integers 0, 1,.. ,,n-l in the first row of an nxn array, as shown. 1 2

2 3

n-1 0

1

0 1

n-1 0

...

n-2

first row second row

nth row

Then to build the second row, we simply shift each integer one position to the left and move 0 to the far right, as shown above. We continue this process by shifting the symbols of a given row one position to the left to form the next row, until we have constructed an n x n square. Now, we can see that each row and column has n distinct elements. Theorem 2. No more than n - 1 MOLS of order n can exists. Proof. Let N{n) = the maximum positive number of MOLS of order n. Suppose L} and L, are two MOLS of order n in a standard form: 0 1

1 2

2 3

.. ..

n-1 0

L,=

0 1 L

2

n-1 0

1

...

n-2

1 2

2 3

. .. .. .

n-1 0

n-1 0

1

...

n-2

=

Neither symbol x in L; nor y in L, can be 0 if L, and L, are latin. In addition x *• y, because if x = y = i the ordered pair ((',/), which already occurs in the first row when L; is superimposed on L2, will violate the orthogonality condition. Thus, there are at most n - 1 possible symbols that can be in the first position of the second row of the squares in an orthogonal set with no repetitions. Hence N(n) < n - 1. Theorem 3. For q, a prime power the set of polynomials of the form f (x,y) = ax + y, with a ^ 0 e F represents a complete set of q - 1 MOLS of order q. (The proof can be found in various text books (Laywine, and Mullen 1998, Lindner and Roger 1997, and Wallis 1988). 3.

Conjecture about MOLS

Euler failed to solve his 36 officer problem, that is, to find a pair of MOLS of order 6. Since he was also unable to construct a pair of MOLS of order 10, he was led to what has become known as Euler's conjecture.

12 Haslinda Ibrahin

Conjecture (Euler): A pair of orthogonal latin squares of order n exists if and only if n- 0, 1 or 3 (mod 4). For n\= 2, it is easy to see that there is no pair of orthogonal latin squares of order 2. In 1900 Tarry proved, that there did not exist a pair of orthogonal latin squares of order 6 by brute force. For a non-brute-force method, see Stinson (1984). Conjecture (H.FMacNeish, 1921): Let n = pf'.pf...p™ where each of p,, p2, P.,,---,PX is a distinct prime. Then the maximum number of MOLS(n) is N(n) = min (p/', p,'-,...,p/*) - 1. Note that Euler's conjecture is a special case of MacNeish's conjecture, because if'n'=2 (mod 4), then n is twice an odd number, whence its factorization has the form l.pf'.pf.. .p™. Thus N(«) = min (2, pf'.pf.. .,p/ x ) - 1 = 2 - 1= 1 ' In 1957, Parker constructed three MOLS(21), thus disproving the MacNeish's conjecture. Since Euler's conjecture is a special case of MacNeish's conjecture, it may not be suprising to learn that in 1958 Bose and Shrikhande were able to construct a pair of orthogonal latin squares of order 22 = 4(5) + 2. This disproved Euler's conjecture. A year later Parker (1959) produced a pair of MOLS of order 10. Then, joining forces Bose, Shrikhande, and Parker proved in 1960 that a pair of orthogonal latin squares of order n exists for all n other than 2 and 6. 4.

Latin squares constructions

To begin our own construction of latin squares, let's start with two examples involving prime powers. We shall use the idea of finite field construction. Let (F,+,.) be a finite field of order n. Assume that F= {l,2,...,n} with the proviso that n represents the zero field element. For each k *• n,k e F, we define a binary operation •(&) on F by x • (k) y = x.k + y. Note that the arithmetic is computed in the finite field (F,+,»). Theorem 4 . Let (F, +, •) be a finite field of order n, where F = {l,2,3,...,n=zero}. Then the quasigroup (F, °(1)), (F, °(2)),..., (F,°(n-1)) are complete set of MOL of order n. Example 1. To construct a complete set of three MOLS of order 4 requires the finite field F4= {0, 1, x, 1 + x -x2 }. Here we use the irreducible polynomial 1 +x +x2 of order 2.

+

0

X

1 1 0 l +x

0 1 x l+x

0 1 1+x

X

X

l +x

X

l+x

l+x 0 1

X

1 0

0 1 x 1 +x

0

1

X

l+x

0 0 0 0

0 1

0 X

0 l +x 1

X

1 +x

l +x 1

X

Orthogonal latin square construction 13

Renaming the elements 0, 1, x and 1 + X with 4, 1,2 and 3, respectively, gives the following tables: Table A + 4 1 2 3

4 4 1 2 3

1 1 4 3 2

Table B 2 2 3 4 1

3 3 2 1 4

4 1 2 3

4 4 4 4 4

1 4 1 2 3

2 4 2 3 1

3 4 3 1 2

Now, we can construct three MOLS of order 4 since there are three nonzero field elements. Case 1: x.l + y

°(D 4 1 2 3

4 4 1 2 3

1 1 2 3 4

2 2 3 4 1

3 3 4 1 2

Calculation for the first column: 4.1 = 4 , 4 + 4 = 4 1.1 = 1 , 1 + 4 = 1 2.1 = 2 , 2 + 4 = 2 3.1=3,3 + 4 = 3

The calculation for the rest of the columns is similar, just use tables A and B as a reference. Do the same for the following cases. Case 2: x.2 + y °(2) 4 1 2 3

4 4 2 3 1

1 1 3 4 2

2 2 4 1 3

3 3 1 2 4

Calculation for the first column: 4.2 = 4, 4 + 4 = 4 1.2 = 2,2 + 4 = 2 2.2 = 3,3 + 4 = 3 3.2=1,1+4=1

1 1 4 2 3

2 2 1 3 4

3 3 2 4 1

Calculation for the first column: 4.3 = 4,4 + 4 = 4 1.3 = 3,3 + 4 = 3 2.3 = 1, 1 + 4 = 1 3.3 = 2,2 + 4 = 2

Case 3: x.3 + y °(2) 4 1 2 3

4 4 3 1 2

Example 2. We use the finite field construction to construct four MOLS(5). The finite field of order 5 is, of course, the integer mod 5. Thus we can easily obtain the four MOLS of order 5 by defining x ° (k)y = xk + y, and renaming zero as 5

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Haslinda Ibrahin

X.l+y °(D 1 2 1 3 2 4 3 5 4 1 5

2 3 4 5 1 2

3 4 5 1 2 3

4 5 1 2 3 4

5 1 2 3 4 5

°(2) 1 2 3 4 5

x.2 + 1 3 5 2 4 1

y 2 4 1 3 5 2

x.3 + y °(3) 1 2 1 4 5 2 2 3 3 5 1 3 4 4 1 2 5

3 1 4 2 5 3

4 2 5 3 1 4

5 3 1 4 2 5

°(4) 1 2' 3 4 5

x.4 + 1 5 4 3 2 1

2 l 5 4 3 2

3 5 2 4 1 3

3 2 1 5 4 3

4 1 3 5 2 4

5 2 4 1 3 5

4 3 2 1 5 4

5 4 3 2 1 5

Let's now consider how to construct latin squares with nonprime powers. Suppose A is a latin square of order m and B a latin square of order n. We shall denote the entry at row i and column / of A by a... Similarly, the (;,/')th entry of B is b... The Kronecker product of A and B is the mn x mn square A ® B: (a„,B) (a21,B)

(ap,B) (fl„,B)

K„B)

(a,;,B) (a,2,B) Here each entry a of A (a, B) is the n x n matrix

(a.*„) (a,bv) (a,B) = (a,b ,)

(a,b ,)

(a,6 )

Example 3. Let us construct a latin square of order 6 by using the Kronecker product, A= 12 21

B =123 23 1 3 12

A ® B gives the following 6 by 6 square whose elements are ordered pairs:

Orthogonal latin square construction 15

11 12 12 13 13 11 21 22 22 23 23 21

13 11 12 23 21 22

21 22 23 11 12 13

22 23 21 12 13 11

23 21 22 13 11 12

Rename the ordered pairs 11, 12, 13, 21, 22, 23 with the integers 0, 1,2, 3, 4, 5 to obtain a latin square of order 6. This yields the latin square below:

0 1 2 3 4 5

1 2 0 4 5 3

2 0 1 5 3 4

34 45 ' 53 01 12 20

5 3 4 2 0 1

Here is another application of the Kronecker product. We shall use it to construct a set of MOLS. Theorem 5. If there is a pair of MOLS of order n and a pair of MOLS of order m, then there is a pair of MOLS of order mn. Example 4. Construct a pair of MOLS of order 12. We can apply theorem 4 by using a pair of MOLS of order 3 and a pair of MOLS of order 4. Since there are only two MOLS of order 3, we will use both of them:

A=

1 2 3 2 3 1 3 1 2

A=

1 2 3 1 2 3

3 2 1

There are three MOLS of order 4, so we can use any two of them. Let BJ and B, be any two of them: 1 2 3 4 1 3 4 2 B,= 2 1 4 3 4 2 1 3

3 4

4 3

1 2

2 1

V

2 3

4 1

3 2

1 4

16

Haslinda Ibrahin

Now we use the Kronecker product LI-AI®BI theorem 4. 11 12 13 14 21 22 23 24 31 32 33 34

12 11 14 13 22 21 24 23 32 31 34 33

13 14 11 12 23 24 21 22 33 34 31 32

14 13 12 11 24 23 22 21 34 33 32 31

21 22 23 24 31 32 33 34 11 12 13 14

22 21 24 23 32 31 34 33 12 11 14 13

11 13 14 12 12 14 13 11 31 33 34 32 32 34 33 34 21 23 24 22 22 24 23 21

14 11 13 12 34 31 33 32 24 21 23 22

12 13 11 14 32 33 31 34 22 23 21 24

21 23 24 22 22 24 23 21 11 13 14 12 12 14 13 11 31 33 34 32 32 34 33 31

and L2 = A2® Br Both L, and L2 are orthogonal by

23 24 21 22 33 34 31 32 13 14' 11 12

24 23 22 21 34 33 32 31 14 13 12 11

31 32 32 31 33 34 34 33 11 12 12 11 13 14 14 13 21 22 22 21 23 24 24 23

33 34 34 33 31 32 32 31 13 14 14 13 11 12 12 11 23 24 24 23 21 22 22 21

24 21 23 22 14 11 13 12 34 31 33 32

22 23 21 24 12 13 11 14 32 33 31 34

31 34 32 33 21 24 22 23 11 14 12 13

34 31 33 32 24 21 23 22 14 11 13 12

33 32 34 31 23 22 24 21 13 12 14 11

32 33 31 34 22 23 21 24 12 13 11 14

We can find other ways to construct latin squares and orthogonal latin squares in various text books, for example Laywine, and Mullen (1998), Lindner and Roger (1997), and Wallis (1988). For examples of constructions showing the falsity of Euler's and MacNeish's conjectures, see Lindner and Roger (1997). 5.

Conclusion

This paper has illuminated the development of latin squares and its sequential steps of unfolding Euler's and MacNeish's conjecture. In addition to that, we discussed the construction of orthogonal and mutually orthogonal latin squares from finite field constructions for prime number. For a nonprime number we used Kronecker product to illustrate the construction.

Orthogonal latin square construction 17

References Bose, R. C , & Shrikhande, S. S. 1959. On the falsity of Euler's conjecture about the nonexistence of two orthogonal latin squares of order 4t+2, Proc. Nat. Acad. Sci.USA, 45, 734-737. Bose, R. C , Shrikhande S.S & E. T. Parker. 1960 Further results on the construction of mutually orthogonal latin squares and the falsity of Euler's conjecture, Can. J. Math. 12, 189-203. Laywine, C.F. & Mullen, G.L. 1998 Discrete Mathematics Using Latin Squares, Wiley, New York. Lindner, C.C & C. A. Rodger, C.A. 1997 Design Theory, CRC Press, New York. Parker, E.T. 1959 Orthogonal latin squares, Proc. Nat. Acad. Set USA, 45, 859-862. Stinson, D.R. 1984 A short proof of the Nonexistence of a pair of orthogonal Latin Squares of order six, J. Combinatorial Theory, Series A, 36, 373-376. Tarry, G. 1900 Le probleme des 36 officers, C.R. Assoc. France Av. Sci. 29(2), 170-203. Wallis, W.D. 1988. Combinatorial Designs, Marcel Dekker, New York.