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A note on Smarandache mukti-squares.

[section] 1. Introduction

Smarandache theory is one of the new branches in mathematics that were first defined and studied by Prof. F. Smarandache and W. B. Kandasamy. Study of algebraic structers and Smarandache algebraic structers is one of the interesting problmes in Smarandache theory. In this note, we will introduced Smarandache Mukti-Squares (SMS) which are non-associative algebraic structures. Arun S. Muktibodh in [2] define and study Smarandache Mukti-Squares (SMS) and gave some open problems. We will extend some SMS to Latin squares that satisfies in problems propounded by Muktibodh [2]. First we recall some definitions and theorems.

Definition 1.1. An ra x n array containing symbols from some alphabet of size m with m [greater than or equal to] n is called a square of order n.

Definition 1.2. A Latin Square of order n is an n by n array containing symbols from some alphabet of size n, arrenged so that each symbol appears exactly once in each row and exactly once in each column.

Definition 1.3. If a Latin square L contains a Latin square S properly, then S is called a Sub Latin square.

Definition 1.4. An square in which

1. No element in the first row is repeated, 2. No element in the first column is repeated, 3. Elements in first row and first column have similar arrange, is called a Mukti-Square.

Definition 1.5. If a square contains a Latin Square properly the square is called a Smarandache Mukti-Square or SMS.

Example 1. The following are examples of Mukti-Squares of order 3 with alphabets {0, 1, 2, 3, 4}.
0 1 2
1 2 0
2 0 1

and

0 1 2
1 2 3
2 3 4

and

0 1 2
1 3 4
2 4 3


Also maybe in other rows or columns we have elements that repeated. for example:
0 1 2 3
1 2 0 3
2 0 2 3


is an Mukti-Square with element 3 that repeated in third column.

In this note we consider Mukti-Squares that have no repeated elements in rows and columns, and therefore they have Latin squares structures. Hence we see that because of a Latin square is equivalent to an quasigroup and vise versa, in quasigroup language. we have only 6 Mukti Square for quasigroups with alphabet {0, 1, 2}. Also for quasigroups constructed by useing all elements of alphabet {0, l, 2, 3} we have 96 such Mukti-Square.

[section] 2. Orthogonal Smarandache Mukti-Squares

Two SMS are said to be orthogonal if the Latin squares contained in them are orthogonal. Therefore for two SN1S
1 2 3 4
2 1 4 3
3 4 2 1
4 3 1 2

and

2 1 4 3
1 2 4 3
4 3 1 2
3 4 2 1


are orthogonal and we have
(1,2) (2,1) (3,4) (4,3)
(2,1) (1,2) (4,3) (3,4)
(3,4) (4,3) (2,1) (1,2)
(4,3) (3,4) (1,2) (2,1)


that is an SMS too.

[section] 3. Mukti-Squares of order 2

We investigate some SMS that have one element, without repeatetion, outside of his alphabet.

For SMS with alphabet A = {0, 1, 2}, let we have
0 1
1 a


with a [??] A, and only in one place. Then we can extend this SMS to:
0 1 2 a
1 a 0 2
2 0 a 1
a 2 1 0


For SMS with two element outside the alphabet of SMS, for example:
a 2
2 b


with alphabet A, we can contruct the following SMS:
b 2 1 a
2 a b 1
1 b a 2
a 1 2 b.


For SMS with three element outside the alphabet of SMS:
0 a
b c


we have:
a 1 0 b c 2
1 0 a c 2 b
0 b c 2 a 1
b c 2 a 1 0
c 2 b 1 0 a
2 a 1 0 b c


Finally for SMS with four element outside of SMS alphabet:
a b
c d


we can extend to the following form:
0 1 a b
1 0 c d
a b 0 1
c d 1 0


It is obvious that for SKIS with one element outside of alphabet that repeated in SMS, for example
1 a
a 1,


construction of Latin square is simpler. Also this construction is simpler for two and more element with repeatation.

[section] 4. SMS of order 3

In this section at first we recall Theorem 3.3 of [2] .

Theorem 4.1. A Latin square of order 3 does not posses an SMS. proof. (See [2], Theorem 3.3).

Now let
2 1 0
1 0 2
0 2 1


be a Latin square. Therefore we see that
2 1
1 0

and

1 0
0 2

and

0 2
2 1


SMS's that are in Latin squre.

Theorem 4.2. It is not possible to extend an SMS of order 3 to a Latin square if the Latin square contained in the SMS has two elements outside the alphabet of the SKIS. proof. (See [2], Theorem 4.3).

It has been practically tried out but could not construct the Latin square. Contrexample:

Let in alphabet {0,1, 2}
0 1 2
1 3 4
2 4 3


be an SKIS with {3, 4} outside of alphabet A. Then
4 0 1 2 3
0 1 3 4 2
1 2 4 3 0
2 3 0 1 4
3 4 2 0 1


is a Latin square. Also
3 0 1 2 4
2 1 3 4 0
0 2 4 3 1
4 3 0 1 2
1 4 2 0 3


can be considered for this SMS.

[section] 5. Answer to Problems and some open problems

In [2] we have some open problems, that we answer to them in this section. 1. Can we extend an SMS of order 3 when all the elements of the Latin square contained in the SMS are outside the alphabet of the SMS?

Example 5.1. Let by alphabet A = {0, 1, 2} we have an SMS, such that {3, 4} [??] A. Then we have the following SMS:
0 1 2
1 3 4
2 4 3


Therefore we can construct the Latin square as:
4 0 1 2 3
0 1 3 4 2
1 2 4 3 0
2 3 0 1 4
3 4 2 0 1


Also we have the following Latin square:
3 0 1 2 4
2 1 3 4 0
0 2 4 3 1
4 3 0 1 2
1 4 2 0 3


that can be considered for this SMS.

Now, let we have an SMS of order 3 such that one element of the Latin square contained in SMS is outside of alphabet A = {0,1, 2} of SMS. For example:
0 1 2
1 2 0
2 0 3


with 3 [??] A. Then this SMS can be extended to the following Latin square of order 6:
0 1 2 3 4 5
1 2 0 5 3 4
2 0 3 4 5 1
3 5 4 0 1 2
4 3 5 1 2 0
5 4 1 2 0 3.


In other case if 3 [??] A appear in array [a.sub.22] of an SMS. That is
0 1 2
1 3 0
2 0 1.


Then this SMS can be extended to the following Latin square of order 6:
0 1 2 3 4 5
1 3 0 5 2 4
2 0 1 4 5 3
3 5 4 0 1 2
4 2 5 1 3 0
5 4 3 2 0 1.


Next, let we consider an SMS of the form
0 1 2
1 2 0


where 3, 4 [??] {0,1, 2}. Then this SMS can be extended to the following Latin square:
0 1 2 3 4 5
1 2 0 5 4 3
2 3 4 5 0 1
3 5 4 0 1 2
4 5 3 1 2 0
5 0 1 2 3 4.


2. Can we extend an SMS of order 4 when one element of the Latin square contained in the SMS is outside the alphabet of the SMS?

Let us consider the following SMS of order 4, with one element outside of alphabet {l, 2, 3, 4}:
1 2 3 4
2 3 4 1
3 4 1 2
4 1 2 a.


Then we can extend this SKIS to the following Latin square:
1 2 3 4 a 5 6 7
2 3 4 1 6 7 a 5
3 4 1 2 7 a 5 6
4 1 2 a 5 6 7 3
a 6 7 5 1 2 3 4
5 7 a 6 2 3 4 1
6 a 5 7 3 4 1 2
7 5 6 3 4 1 2 a


3. Can we extend an SMS of order 4 when two element of the latin square contained in the SMS are out side the alphabet of SMS?

To answer this quastion we refree the reader to our last construction. It is easily constructable.

At the end we consider the following quastions for next works:

1. Can we construct an algorithm for caculate all SKIS of order 3 with alphabet containing n element?

2. Can we construct an algorithm for caculate all SKIS of order 4 with alphabet containing n element?

3. Can we write an computer programme for calculating all SMS of order 3?

4. Can we write an computer programme for calculating all SN1S of order 4?

5. Can we found applications of SNIS in cryptography?

6. Can we found application of SMS in Matroid Theory?

7. what is the interpretation of SMS in geometry, probablity, combinatorial theory or other sciences?

References

[1] C. F. Laywine and G. L. Mullen, Discrete Mathematics Using Latin Squares, John Wiley Ans Sons, Inc., 1998, 305.

[2] Arun S. Muktibodh, Smarandache Mukti-Squares, Scientia Magna, 3(2007), No. 1, 10-107.

Khalil Shahbazpour

Department of Mathematics, University of Urmia, Urimia-Iran

E-mail: shahbazpour2003@hotmail.com
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Author:Shahbazpour, Khalil
Publication:Scientia Magna
Article Type:Report
Geographic Code:7IRAN
Date:Jun 1, 2009
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