This document summarizes research on generating new sets of sequences (called M-sequences) that can be used in communication channels. M-sequences are linear recurring sequences that form additive groups and have applications in mixing and extracting information in communication channels. The research aims to compose existing M-sequences to generate new sets with longer lengths and larger minimum distances, which can increase security and error correction. It provides definitions and theorems related to M-sequences, linear recurring sequences, orthogonal sequences, and codes. Examples are given to illustrate the composition of M-sequences of order 3 and 7 to generate new orthogonal sets and codes.

1. INTERNATIONAL JOURNAL OF COMPUTER ENGINEERING &
International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 5, Issue 9, September (2014), pp. 73-89 © IAEME
= l + l + + l + l l
Î = −
... , , 0,1,..., 1
a a a a F i k
n k k n k k n k n i q
73
COMPOSED SHORT M-SEQUENCES
Dr. Ahmad Hamza Al Cheikha
Department of Mathematical Science, College of Arts-science and Education/
Ahlia University, Exhibition Street, Manama, Bahrain
ABSTRACT
M - Sequences (which formed a closed sets under the addition and with the corresponding
null sequence formed additive groups and generated by feedback registers) are used widely at the
forward links of communication channels to mix the information on connecting to and at the
backward links of these channels to sift through this information is transmitted to reach the receivers
this information in correct form, specially in the pilot channels, the Sync channels, and the Traffics
channel.
This research is useful to generate new sets of sequences (which are also with the
corresponding null sequence additive groups) by compose m-sequences with the bigger lengths and
the bigger minimum distance that assists to increase secrecy of these information and increase the
possibility of correcting mistakes resulting in the channels of communication.
Index Terms: M-sequences, Additive group, Orthogonal sequences. Coefficient of Correlation,
Code.
1. INTRODUCTION
M- Linear Recurring Sequences
Let k be a positive integer and l ,l0,l1,...,lk −1 are elements in the field Fq , then the
sequence a0,a1,...
is called non homogeneous linear recurring sequence of order k iff :
(1)
+ − + − − + −
1 1 2 2 0
−
1
= +
l l
or a a
+ +
n k i n i
=
1
k
i
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( ) = +l 1 +...+l +l −
k (2)
x . [1]-[ 4]
Î Î − £
£ − Î ¹ Î −
X Y G X Y xi yi or R
74
The elements a0 ,a1,...,ak −1 are called the initial values (or the vector (a0 ,a1,...,ak −1) is
called the initial vector). If l = 0 then the sequence a0 ,a1,...
is called homogeneous linear recurring sequence (H. L. R. S. ), except the zero initial vector, and
the polynomial
1 0
1
f x x k
k
− x x is called the characteristic polynomial. In this study, we are limited tol0 =1 . [1]-[3]
II. RESEARCH METHODS AND MATERIALS
Definition 1. The Ultimately Periodic Sequence a0, a1,....with the smallest period r is called a
periodic iff: an+r = an ; n = 0,1, ... [1]-[ 4]
Definition 2. The complement of the binary vector X = (x1, x2 ,..., xn ) is the vector
X = (x1, x2 ,..., xn )when
=
1 0
=
=
if x
i
0 1
i
i
if x
Definition 3. Any Periodic Sequence a0,a1,....over F2 with prime characteristic polynomial is an
orthogonal cyclic code and ideal auto correlation [1]-[ 10].
Definition 4. Suppose G is a set of binary vectors of length n
G = {X ; X = ( x0 , x1,..., xn −1 ), xi Î F2 = {0,1}, i = {0,..., n − 1}}
Let 1*= -1 and 0* =1. The set G is said to be orthogonal if the following two conditions are satisfied
1
, { 1 , 0 ,1}, , 0 1 .
X G x i or R
0
−
=
*
x
n
i
1
, ( ), { 1,0,1}, , 1.
0
=
* *
x y
n
i
That is, the absolute value of the number of agreements minus the number of disagreements is
equal to or less than 1. [1]
Definition 5. Hamming weight: The Hamming weight of binary vector x = ( x0 , x1 ,..., xn −1 ) is
the number of non zero components of x. [9]
Definition 6. Hamming distance d(x, y) : The Hamming distance between the binary vectors
x = ( x0 , x1 ,..., xn −1 ) and y = ( y0 , y1,..., yn−1) is the number of the disagreements of the
corresponding components of x and y.[9]
Definition 7. Minimum distance d: The minimum distance d of a set C of binary vectors is:
= . [9]
min ( , )
,
d d x y
x yÎC
Definition 8. The code C of the form [n, k, d] if each element (Codeword) has the: length n, the rank
k is the number of information components (Message), minimum distance d.[9]

3. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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n i
R i i (3)
75
Theorem 1.
i. If a0,a1,.... is a homogeneous linear recurring sequence of order k in F2 , satisfies (1) then
this sequence is periodic
ii. If this sequence is homogeneous linear recurring sequence, periodic with the period r, and its
characteristic polynomial f (x) then r ord f (x) . [6]
iii. If the polynomial f (x) is primitive then the period of the sequence is 2 −1 k , and this
sequence is called M – sequence.
Theorem 2.
The number of irreducible polynomials in Fq (x) of degree m and order e is j (e) /m, if e ³ 2,
When j is the Euler function and m is the order of q by mod e, and equal to 2 if m = e = 1, and
equal to zero elsewhere. [6]-[9]
Theorem 3.
If g(x) is a characteristic prime polynomial of the (H. L. R. S.)
a0,a1,....of degree k, and a is a root of g(x) in any splitting field of F2
then the general bound of the sequence is:
k n
i
i
C a
=
−
=
1 2
1
a . [11], [12].
Definition 9. Suppose x = ( x 0 , x1 ,..., x n −1 ) and y = ( y0 , y1,..., yn−1) are vectors of length n on
GF(2) ={0,1}. The coefficient of correlations function of x and y , denoted by Rx,y, is:
−
+ = −
, ( 1)
=
1
0
n
i
x y
x y
Where xi +yi is computed mod 2. It is equal to the number of agreements components minus the
number of disagreements corresponding to components. [13]
Definition 10. If M is a M - sequences and w is any binary vector then:
M(w) = {xi (w) : wi ÎM}, we replace each “ 1” in xi by w and each “ 0 “ in xi by w .[14]
Example 1: If a is a root of the prime polynomial ( ) 1 2 f x = x + x + and generates (2 ) 2 GF and
Suppose the Linear Binary Recurring Sequence be
n n n a +2 = a +1 + a or an+2 + an+1 + an = 0 (4)
With the characteristic equation 1 0 2 x + x + = and the characteristic polynomial ( ) 1 2 f x = x + x + ,
which is a prime then the general solution of equation (1) For the initial position: a1 =1 , a2 = 0 is
given by: n n
an 2 2 =a ×a +a ×a , and the sequence is periodic with the period 2 1 3 2 − =
and x1= (101) , by the cyclic permutations on x1 we have { } M3 = x1, x2, x3
where:
x1= (101), x2 = (110), x3 =(011), The first two digits in each sequence are the initial position of the
feedback register, and the set M3 is an orthogonal set and a cyclic code of the form [n =3, k=2,
d = 2].

4. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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Extend each sequence addition of 0 to the beginning Mr (or to the end Mr ) of each, we have
s
= (0101), x2 =
r
=(0110)) . The sets M3
an 2 4 2 4 =a ×a +a ×a +a ×a , and the sequence is periodic with the period 2 1 7 3 − = and
y1 = (1001011) , by the cyclic permutations on y1 we have { } M7 = y1, y2, y3, y4, y5, y6, y7 where:
y2 =(1100101), y3 =(1110010), y4 =(0111001),
y5 =(1011100), y6 =(0101110), y7 =(0010111), and the first three digits in each sequence are the
initial position of the feedback register, and the set M7 is an orthogonal set and a cyclic code of the
form [n =7, k=3, d = 4].
Extend each sequence addition of 0 to the beginning Mr (or to the end Mr ) of each,
we have: { } M7 y1, y2 , y3, y4 , y5 , y6 , y7
r r r r r r r r
76
{ } M3 x1, x2 , x3
s s s s
=
where: x1
s
s
=(0011), ( { } M3 x1, x2 , x3
(0110), x3
r r r r
= where:
r
= (1010), x2 =
x1
r
(1100), x3
s
r
and M3
are orthogonal sets
and a codes of the form [n =4, k=2, d = 2].
There is only one prime polynomial of order 2 that is ( ) 1 2 f x = x + x +
.
Example 2: If a is a root of the prime polynomial ( ) 1 3 f x = x + x + and generates (2 ) 3 GF and
Suppose the Linear Recurring Sequence be :
n n n a +3 = a +1 + a or 3 1 0 an+ + an+ + an = (5)
With the characteristic equation 1 0 3 x + x + = and the characteristic polynomial ( ) 1 3 f x = x + x + ,
which is a prime and generates 3 2
F and if ( 3 ) x =a ÎGF 2 is a root of f (x) and For the initial
position: a1 = 1 , a2 = 0 , a3 = 0 then the general solutions of equation (2) is given by
n n n
s s s s s s s s
=
where:
s
y1
= (01001011), 2 y s
=(01100101) , 3 y s
=(01110010),
4 y s
=(00111001), 5 y s
=(01011100),
6 y s
=(00101110), 7 y s
=(00010111), and ) M7 =
{ y1, y2 , y3, y4 , y5 , y6 , y7
} where:
1 y r
= (10010110), 2 y r
r
=(11001010), y3
=(11100100),
4 y r
=(01110010), 5 y r
=(10111000),
6 y r
=(01011100), 7 y r
=(00101110)).
s
Each of the sets M7
r
and M7
is an orthogonal set and a code of the form [n =8, k=3, d = 4].
There are only two prime polynomial of order 3 that are ( ) 1 3 f x = x + x + and its conjugate
( ) 1 3 2 g x = x + x + .
III. RESULTS AND DISCUSSION
First Step
A1. M3(M3 ) : We suppose the M-sequences of order 3 that are:
1 x = (101), x2 = (110), x3 = (011) . We will determine M3(x1) . we can see that:
R1 = x1(x1) = (101 010 101), R2 = x2 (x1) = (101 101 010), R3 = x3(x1) =(010 101 101)

5. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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1 0 0 1 1 0
1 0 1 0 0 1
1, 2 Rr r = − , and the rows of the matrix 3G3 also not orthogonal set for
0 1 1 0 0 1
1 0 0 1 0 1
77
If we write the matrix 3G3 such that their rows are: Ri = xi (x1), i =1,2,3 and theirs columns
are: Ci , j = 1,2,...,9 , after that we remove: all zero columns and all equals columns except one, we
have the matrix:

6. =
0 1 1 0 1 0
3G3
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j = 1,2,3,...,6we can denote the
rank of the matrix 3G3 is 3, Since the rank of the matrix 3G3 is 3 then the ranks of the matrix 3G3
is also 3.
• The rows of 3G3 are not orthogonal set, for example: r1 + r2 = (001111) and the coefficient of
correlation of them is 2
example: R1 + R2 = (000111111) and the coefficient of correlation of them is 3
1, 2 RR R = − .
* For M3(M3 ) Span {r1, r2, r3} and Span {R1, R2, R3} are not orthogonal sets.
s
B1. M3(M3)
: We suppose the M-sequences of order 3.
s
= (0101), We will determine M3 (x1)
That are: x1= (101), x2 = (110), x3 = (011), and x1
s
,
3
we can see that: R1 = x1(x1) = (0 1 0 1 1 0 1 0 0 1 0 1) , R2 = x2 (x1) =(0 1 0 1 0 1 0 1 1 0 1 0) ,
R3 = x3(x1) = (1 0 1 0 01 01 0 1 0 1) .
s
If we write the matrix 3G such that their rows are: Ri = xi (x1), i =1,2,3
and theirs columns are:
3
Ci , j = 1,2,...,12 , after that we remove: all zero columns and all equals columns except one, we have
s
the matrix 3G
:

7. =
0 1 0 1 1 0
3
s
3G
3
3
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j = 1,2,...,6we can denote
s
s
That the rank of the matrix 3G
is 3, and the rows of 3G
are not orthogonal set, for
Example: r1 + r2 =(001111) and the coefficient of correlation of them is 2
1, 2 Rr r = − , and the rows
of the matrix 3
3G also not orthogonal set for example:
R1 + R2 = (000011111111) and the coefficient of correlation of them is 4
1, 2 RR R = − .
s
* For M3(M3 )
Span {r1, r2, r3} and Span {R1, R2, R3} are not orthogonal sets.

8. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 5, Issue 9, September (2014), pp. 73-89 © IAEME
s
-sequences of order 3, That are:
3G such that their rows are: Ri = xi (x1), i = 1,2,3
1 1 0 0 1 1 0
1 1 0 1 0 0 1
1 0 1 1 0 1 0
s
3G x
s
-sequences of order 3 , That are:
s s
= = (1010 0101 1010 0101), R2 x2 (x1)
78
C1. M3(M3)
s
: We suppose x1= (101), M
s
= (0101), x2
x1
s
s
= (0110), x3
s
= (0011), We will determine M3 (x1)
, we can see that:
s
= = (010 101 010 101), R2 x2 (x1)
R1 x1(x1)
s
= = (010 101 101 010),
s
= = (010 010 101 101).
R3 x3(x1)
If we write the matrix 3
s
and theirs columns are:
Ci , j = 1,2,...,12 , after that we remove: all zero columns and all equals columns except one, we have
s
the matrix 3G
3
:

9. =
3G
3
s
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j = 1,2,3,...,7 we can denote
s
That the rank of the matrix 3G
3
is 3, and the ranks of the matrix 3G
3
is also 3.
We determine the other non zero elements of { } Span r1, r2, r3 we get:
r4 = r1 + r2 = (0001111), r5 = r1 + r3 = (0111100)
r6 = r2 + r3 =(0110011), r7 = r1 + r2 + r3 = (1010101)
Thus, Span { } r1, r2 , r3 is closed under the addition, orthogonal sequences and determine code of the
form [ n = 7, k = 3, d = 4].
We can see that R1 + R2 + R3 =(010 010 010 010) and Span { } R1, R2 , R3 is not orthogonal sequences
set.
s
• For M3(M3 )
Span {r1, r2, r3} is orthogonal set but Span {R1, R2, R3} is not orthogonal
sets.
s
3G x
By the same way each of 3 ( 2 )
s
3G x
, 3 ( 3 )
, 3 ( 1)
s
3G x
, 3( 2 )
s
3G x
, and 3 ( 3 )
a closed sets
under the addition and each of them is an orthogonal sequences set of the form [n = 7, k = 3, d = 4].
r
* By the same way M3 (M3 )
generates other 6 sets each of them contains 7 orthogonal sequences
with the: length n = 7, rank k = 3, and distance d = 4.
s s
D1. M3(M3)
: We suppose the M
s
= (0101), x2
x1
s
s
= (0110), x3
s s
= (0011). We will determine M3 (x1)
, we can see that:
R1 x1(x1)
s s
= = (1010 0101 0101 1010),
s s
= = ( 1010 1010 0101 0101)
R3 x3(x1)
If we write the matrix 3
s s
3G such that their rows are: Ri = xi (x1), i =1,2,3
and theirs columns
are: Ci , j =1,2,...,16 , after that we remove: all zero columns and all equals columns except one, we
have the matrix:

10. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 5, Issue 9, September (2014), pp. 73-89 © IAEME
1 0 1 1 0 0 1
1 0 1 0 1 1 0
1 1 0 0 1 0 1
s s
3G x
s
3G x
79

11. =
3G
3
s
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j =1,2,3,...,7we can denote that
s
3
the rank of the matrix 3G
is 3, and the ranks of the matrix 3
3G is also 3.
We determine the other non zero elements of { } Span r1, r2, r3 we get:
r4 = r1 + r2 = (0001111), r5 = r1 + r3 = (0111100)
r6 = r2 + r3 = (0110011), r7 = r1 + r2 + r3 = (1101010)
Thus, Span { } r1, r2 , r3 is closed under the addition, orthogonal sequences and determine code of the
form [ n = 7, k = 3, d = 4],
We can see that:
R4 = R1 + R2 = (0000 000011111111) R5 = R1 + R3 = (0000 1111 1111 0000)
R6 = R2 + R3 = (0000 1111 0000 1111), R7 = R1 + R2 + R3 = (1010 1010 1010 1010)
Thus, Span { } R1, R2 , R3 is a closed set under the addition and an orthogonal sequences set.
•
s s
For M3(M3 )
Span {r1, r2, r3} and Span {R1, R2, R3} are orthogonal sets.
s s
3G x
3
a. By the same way each of ( 2 )
3
, ( 3 )
s s
3G x
3
, ( 1)
s s
3G x
3
, ( 2 )
s s
3G x
3
, and ( 3 )
a closed sets
under the addition and each of them is an orthogonal sequences set of the form
[n = 7, k = 3, d = 4].
s
3G x
3
b. By the same way each of ( 2 )
3
, ( 3 )
s
3G x
3
, ( 1)
s
3G x
3
, ( 2 )
s
3G x
3
, and ( 3 )
a closed sets
under the addition and each of them is an orthogonal sequences set of the form [n = 16, k = 3,
d = 8].
r r
• By the same way for the M3(M3 )
generate 6 orthogonal sets of the form
[n = 7, k = 3, d = 4] and other 6 orthogonal sets of the form [n = 16, k = 3, d = 8].
Second step
A2. M3 (M7 ) : We suppose the M-sequences of order 3, That are:
x1= (101), x2 = (110), x3 = (011), and y1 = (1001011). We will determineM3( y1)
.
we can see that:
R1 = x1( y1) =(1001011 0110100 1001011), R2 = x2 ( y1) = (1001011 1001011 0110100)
R3 = x3( y1) =(0110100 1001011 1001011)
If we write the matrix 3G7 such that their rows are: Ri = xi ( y1), i = 1,2,3 and theirs columns are:
Ci , j =1,2,...,21, after that we remove: all zero columns and all equals columns except one, we have
the matrix

12. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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1 0 0 1 1 0
1 0 1 0 0 1
0 1 1 0 0 1
1 0 0 1 0 1
s
-sequences of order 3 That are: x1
80
1, 2 Rr r = − , and the rows of the matrix 7

13. =
0 1 1 0 1 0
3G7
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j = 1,2,...,6we can denote
That the rank of the matrix 3G 3 is 3 and the rank of 3G7 is also 3, but the rows of 3G 7 is not
orthogonal set, for example: r1 + r2 =(001111) and 2
1, 2 Rr r = −
, and the rows of the matrix 3G7
also not orthogonal set for example: R1 + R2 = (0000000 1111111 1111111), and the coefficient of
correlation of them is 7
1, 2 RR R = − .
• For M3(M7 ) Span {r1, r2, r3} and Span {R1, R2, R3} are not orthogonal sets.
s
B2 . M3 (M7 )
: We suppose the M-sequences of order 3, That are:
1 x = (101), 2 x = (110), 3 x = (011), and 1 y s
= (01001011),
s
We will determine M3 ( y1)
. We can see that:
s
= = (01001 011 10110100 010 01011) ,
R1 x1( y1)
s
= = (01001011 01001011 10110100)
R2 x2 ( y1)
s
= = (101 10100 01001011 01001011)
R3 x3( y1)
s
If we write the matrix 7
3G such that their rows are: Ri = xi ( y1), i = 1,2,3
and theirs columns
are: Ci , j = 1,2,...,24 , after that we remove: all zero columns and all equals columns except one, we
have the matrix:

14. =
0 1 0 1 1 0
7
s
3G
7
7
In this matrix if we consider rows ri : i =1,2,3 and columns, c j : j = 1,2,...,6we can denote
s
That the rank of the matrix 3G
is 3, and then the ranks of the matrix 3G is also 3.
s
But the rows of 7
3G
are not orthogonal set, for example: r1 + r2 = (001111) and
2
3G also not orthogonal set for example:
R1 + R2 = (00000000 11111111 11111111) and 8
1, 2 RR R = − .
s
• For M3(M7 )
Span {r1, r2, r3} and Span {R1, R2, R3} are not orthogonal sets.
s
C2. M3(M7 )
: We suppose the M
s
s
= (0101), x2
= (0110),
s
x3 = (0011), and y1 = (1001011). We will determine M3 ( y1)
, we can see that:

15. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 5, Issue 9, September (2014), pp. 73-89 © IAEME
s
= = (0110100 1001011 1001011 0110100),
s
= = (0110100 0110100 1001011 1001011),
3G such that their rows are: Ri = xi ( y1), i = 1,2,3
1 1 0 0 1 1 0
1 1 0 1 0 0 1
1 0 1 1 0 1 0
s
3G y
s
-sequences of order 3 That are:
s s
= = (10110100 01001011 10110100 01001011),
s s
= = (10110100 01001011 01001011 10110100),
s s
= = (10110100 10110100 01001011 01001011)
81
R1 x1( y1)
s
= = (0110100 1001011 0110100 1001011),
R2 x2 (y1)
R3 x3( y1)
If we write the matrix 7
s
and theirs columns are:
Ci , j =1,2,...,28, after that we remove: all zero columns and all equals columns except one, we have
the matrix:

16. =
3G
7
s
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j =1,2,...,7we can denote that
s
3G
the rank of the matrix 7
3G is also 3.
is 3, and rank of the matrix 7
We determine the other non zero elements of { } Span r1, r2, r3 we get:
r4 = r1 + r2 = (0001111), r5 = r1 + r3 = (0111100)
r6 = r2 + r3 = (0110011), r7 = r1 + r2 + r3 = (1010101)
Thus, Span { } r1, r2 , r3 is closed under the addition, orthogonal sequences and determine code of the
form [ n = 7, k = 3, d = 4].
We can see that:
R1 + R2 + R3 = (0110100 0110100 0110100 0110100) and 4
RR + R + R =
1 2 3
and the set
Span { } R1, R2 , R3 is not an orthogonal sequences set.
s
• For M3(M7 )
Span {r1, r2, r3} is orthogonal set but Span {R1, R2, R3} is not orthogonal
sets.
s
3G y
* By the same way each of 7( 2 )
,…, 7( 7 )
s
3G y
, 7( 1)
s
3G y
, 7( 2 )
s
3G y
, …,and 7( 7 )
a closed sets under the addition and each of them is an orthogonal sequences set of the form
[n = 7, k = 3, d = 4].
* The result is 14 sets closed under the addition and each of them contains 7 orthogonal sequences
with the: length n = 7, rank k = 3, and distance d = 4.
s s
D2. M3(M7 )
: We suppose the M
s
= (0101), x2
x1
s
s
= (0110), x3
= (0011), and
1 y = (1001011), 1 y s
= (01001011), We will determine
s s
M3 ( y1)
. We can see that:
R1 x1( y1)
R2 x2 (y1)
R3 x3( y1)

17. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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1 0 1 1 0 1 0
1 0 1 0 1 0 1
1 1 0 0 1 1 0
s s
3G y
7
s
3G y
7
82
If we write the matrix 7
s s
3G such that their rows are: Ri = xi ( y1), i = 1,2,3
and theirs columns
are: Ci , j =1,2,...,32 , after that we remove: all zero columns and all equals columns except one, we
have the matrix:

18. =
3G
7
In this matrix if we consider rows ri : i =1,2,3 and columns c j : j =1,2,...,7we can denote
s
3
That the rank of the matrix 3G
is 3 and rank of the matrix 7
3G are also 3, We determine the other
non zero elements of { } Span r1, r2, r3 we get:
r4 = r1 + r2 = (0001111), r5 = r1 + r3 = (0111100), r6 = r2 + r3 = (0110011)
r7 = r1 + r2 + r3 = (1101001), Thus, Span { } r1, r2 , r3 is closed under the addition, orthogonal
sequences and determine code of the form [ n = 7, k = 3, d = 4].
We can see that:
R4 = R1 + R2 = (0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 11 1 1 1 1 1 1 1 1 11)
R5 = R1 + R3 = (0 0 0 0 0 0 0 0 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0)
R6 = R2 + R3 = (0 0 0 0 0 0 0 0 1 11 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 )
R7 = R1 + R2 + R3 = (1 0 1 1 0 1 0 0 1 0 11 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 )
Thus, Span { } R1, R2 , R3 is an orthogonal set.
s s
• For M3(M7 )
Span {r1, r2, r3} and Span {R1, R2, R3} are orthogonal sets.
s s
3G y
7
a. By the same way each of ( 2 )
,…, ( 7 )
s s
3G y
7
, ( 1)
s
3G y
7
, ( 2 )
s s
3G y
7
, …,and ( 7 )
a
closed sets under the addition and each of them is an orthogonal sequences set of the form
[n = 7, k = 3, d = 8].
s
3G y
7
b. By the same way each of ( 2 )
,…, ( 7 )
s
3G y
7
, ( 1)
s
3G y
7
, ( 2 )
s
3G y
7
, …, and ( 7 )
a
closed sets under the addition and each of them is an orthogonal sequences set of the form
[n = 32, k = 3, d = 4].
Third step
A3 . M7 (M3 ) : We suppose the M-sequences of order 7 that are:
y1 = (1001011), y2 =(1100101) ; y3 =(1110010) ; y4 =(0111001)
y5 =(1011100) ; y6 =(0101110) ; y7 =(0010111) and x1= (101) in M3
We will determine M7 (x1) , we can see that:
R1 = y1(x1) = (101 010 010 101 010 101 101),
R2 = y2 (x1) = (101 101 010 010 101 010 101)
R3 = y3(x1) = (101 101 101 010 010 101 010)
R4 = y4 (x1)
= (010 101 101 101 010 010 101)

19. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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1 0 0 1 0 1 1 0 0 1 1 0 1 0
1 0 1 0 0 1 0 1 1 0 0 1 1 0
1 0 1 0 1 0 0 1 0 1 1 0 0 1
RR +R = − , then { } Span r1, r2 , r3, r4
= = (0101 1010 1010 0101 1010 0101 0101)
= = (0101 0101 1010 1010 01011010 0101)
= = (0101 0101 0101 1010 1010 0101 1010)
0 1 1 0 1 0 0 1 1 0 0 1 0 1
0 1 0 1 1 0 1 0 0 1 1 0 0 1
0 1 0 1 0 1 1 0 1 0 0 1 1 0
83
If we write the matrix 7G3 such that their rows are: Ri = yi (x1), i = 1,2,3,4 and theirs
columns are: Ci , j =1,2,...,21. after that we remove: all zero columns and all equals columns except
one, we have the matrix:

20. =
0 1 1 0 1 0 1 0 0 1 0 1 1 0
7G3
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,14 .
we can denote that r1 + r2 = (00110011111100) and 2
Rr +r = − and
1 2
R1 + R2 =(000 111 000 111 111 111 000) and 3
1 2
and Span { R1,R2,R3,R4} are not orthogonal sets.
• For M7 (M3) Span {r1, r2, r3, r4} and Span {R1, R2, R3, R4} are not orthogonal sets.
s
B3 . M7 (M3)
: We suppose the M-sequences of order 7 that are:
y1 = (1001011), y2 =(1100101) ; y3 =(1110010) ; y4 =(0111001)
s
y5 =(1011100) ; y6 =(0101110) ; y7 =(0010111) and x1
= (0101)
s
We will determine M7 (x1)
, we can see that:
s
R1 y1(x1)
s
R2 y2 (x1)
s
R3 y3(x1)
R4 = y4 (x1)
= (1010 0101 0101 0101 1010 1010 0101)
s
If we write the matrix 3
7G such that their rows are: Ri = yi (x1), i = 1,2,3,4
and theirs columns are:
Ci , j = 1,2,...,28 . after that we remove: all zero columns and all equals columns except one, we have
the matrix
:

21. =
1 0 0 1 0 1 0 1 1 0 1 0 0 1
3
s
7G
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,14we can denote:
r1 + r2 = (0011 00111111 00) and 2
Rr +r = −
1 2
, then { } Span r1,r2, r3,r4 is not orthogonal sequences,
RR +R = − , then:
and We can see that: R1 + R2 = (00001111 0000 1111 1111 1111 0000) and 4
1 2
Span { R1,R2,R3,R4} is not orthogonal set.

22. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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s
= = (010 101 010 010 101 010 101 101)
s
= = (010 101 101 010 010 101 010 101)
1 1 0 0 1 0 1 1 0 0 1 1 0 1 0
1 1 0 1 0 0 1 0 1 1 0 0 1 1 0
1 1 0 1 0 1 0 0 1 0 1 1 0 0 1
84
s
• For M7 (M3)
Span {r1, r2, r3, r4} and Span {R1, R2, R3, R4} are not orthogonal sets.
s
C3 . M7 (M3)
s
: We suppose the M7
-sequences That are:
1 y s
= (01001011), 2 y s
s
=(01100101) ; y3
=(01110010) ; 4 y s
=(00111001)
s
y5
=(01011100) ; 6 y s
=(00101110) ; 7 y s
=(00010111) and x1= (101).
s
We will determine M7 (x1)
, we can see that:
R1 y1(x1)
R2 y2 (x1)
R3 = y3(x1) = (010 101 101 101 010 010 101 010)
s
R4 =
y4 (x1)
= (010 010 101 101 101 010 010 101)
s
7G
If we write the matrix 3
s
such that their rows are: Ri = yi (x1), i = 1,2,3,4
and theirs
columns are: Ci , j =1,2,...,24 . after that we remove: all zero columns and all equals columns except
one, we have the matrix:

23. =
1 0 1 1 0 1 0 1 0 0 1 0 1 1 0
7G
3
s
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,15.
s
The rank of the matrix 7G
3
is 4, we can denote that:
r1 + r2 =(000110011111100), r1 + r3 = (000111111000011)
r1 + r4 =(011111100001100), r2 + r3 = (000001100111111)
r2 + r4 =(001001111110000), r3 + r4 = (011000011001111)
r1 + r2 + r3 = (1100110 10100101), r1 + r2 + r4 = (101011001101010)
r1 + r3 + r4 = (101010101010101), r2 + r3 + r4 = (101100110101001)
r1 + r2 + r3 + r4 = (01111000011 0 011)
s
The Span { r1, r2 , r3, r4 } is an orthogonal set (and the same result for M7 (x1)
).
We can see that: R1 + R3 + R4 = (010 010 010 010 010 010 010 010) and RR +R +R =
8
1 3 4
then Span { R1,R2,R3,R4} is not orthogonal set.
s
• For M7 (M3)
Span {r1, r2, r3, r4} is orthogonal set but Span {R1, R2, R3, R4} is not
orthogonal set.
s s
D3 . M7 (M3)
: We suppose the m-sequences of order 7 that are:
1 y s
= (01001011), 2 y s
s
=(01100101), y3
=(01110010), 4 y s
=(00111001)
s
y5
=(01011100) ; 6 y s
=(00101110) ; 7 y s
s
= (0101)
=(00010111) and x1

24. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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s s
= = (1010 0101 1010 1010 0101 1010 0101 0101)
s s
= = (1010 0101 0101 1010 1010 0101 1010 0101)
s s
= = (1010 0101 0101 0101 1010 1010 0101 1010)
s s
= = (1010 1010 0101 0101 0101 1010 1010 0101)
1 0 1 1 0 1 0 0 1 1 0 0 1 0 1
1 0 1 0 1 1 0 1 0 0 1 1 0 0 1
1 0 1 0 1 0 1 1 0 1 0 0 1 1 0
85
s s
We will determine M7 (x1)
, we can see that:
R1 y1(x1)
R2 y2 (x1)
R3 y3(x1)
R4 y4 (x1)
If we write the matrix 3
s s
7G such that their rows are: Ri = yi (x1), i = 1,2,3,4
and theirs columns are:
Ci , j =1,2,...,32 . after that we remove: all zero columns and all equals columns except one, we have
the matrix:

25. =
1 1 0 0 1 0 1 0 1 1 0 1 0 0 1
7G
3
s
3
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,15.
s
The rank of the matrix 7G
is 4 , we can denote that:
r1 + r2 = (000110011111100), r1 + r3= (000111111000011)
r1 + r4 = (011111100001100), r2 + r3= (000001100111111)
r2 + r4 = (011001111110000), r3 + r4= (011000011001111)
r1 + r2 + r3= (101100101011010), r1 + r2 + r3= (110101110000101)
r1 + r3 + r4= (110101010101010), r2 + r3 + r4= (110011001010110)
r1 + r2 + r3 + r4 = (01111000011 0 011)
w s
Thus, Span { } r1, r2 , r3, r4 is an orthogonal set also M7 (x1)
is an orthogonal set.
We can see that:
R1 + R2 = (0000 0000 1111 0000 1111 1111 1111 0000),
R1 + R3= (0000 0000 1111 1111 1111 0000 0000 1111),
R1 + R4 = (0000 1111 1111 1111 0000 0000 1111 0000),
R2 + R3 = (0000 0000 0000 1111 0000 1111 1111 1111),
R2 + R4= (0000 1111 0000 1111 1111 1111 0000 0000),
R3 + R4 = (0000 1111 0000 0000 1111 0000 1111 1111),
R1 + R2 + R3 = (1010 0101 1010 0101 0101 0101 1010 1010) = R5 ,
R1 + R2 + R4= (1010 1010 1010 0101 1010 0101 0101 0101) = R7 ,
R1 + R3 + R4 = (1010 0101 1010 1010 1010 1010 1010 0101),
R2 + R3 + R4 = (1010 1010 0101 1010 0101 0101 0101 1010) = R6 ,
R1 + R2 + R3 + R4 = (0000 1111 1111 0000 0000 1111 0000 1111)
Thus, Span { R1,R2,R3,R4} is an orthogonal set.

26. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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= = (01001011 10110100 10110100 01001011 10110100 01001011 01001011),
= = (0100101 01001011 10110100 10110100 01001011 10110100 01001011),
= = (01001011 01001011 01001011 10110100 10110100 01001011 10110100),
= = (10110100 01001011 01001011 01001011 10110100 10110100 01001011),
0 1 1 0 1 0 0 1 1 0 0 1 0 1
0 1 0 1 1 0 1 0 0 1 1 0 0 1
0 1 0 1 0 1 1 0 1 0 0 1 1 0
RR +R = − Then Span { } r1, r2 , r3, r4 and Span { R1,R2,R3,R4} are not orthogonal sets.
86
s s
• For M7 (M3)
Span {r1, r2, r3, r4} and Span {R1, R2, R3, R4} are orthogonal sets.
Forth step
A4 . M7 (M7 ) : The study showed that the result was suchM3(M3) that:
*Span{ } r1, r2 , r3, r4
and Span { R1,R2 ,R3,R4} are not orthogonal sets.
s
B4 . M7 (M7 )
: We suppose the M-sequences of order 7 That are:
y1 = (1001011), y2 =(1100101) ; y3 =(1110010) ; y4 =(0111001)
s
y5 =(1011100) ; y6 =(0101110) ; y7 =(0010111) , We will determine M7 ( y1)
.
s
We can see that y1 = (01001011)
and:
s
R1 y1( y1)
s
R2 y2 (y1)
s
R3 y3( y1)
s
R4 y4 (y1)
s
If we write the matrix 7
7G such that their rows are: Ri = yi ( y1), i = 1,2,3,4
and theirs
columns are: Ci , j =1,2,...,56 . after that we remove: all zero columns and all equals columns except
one, we have the matrix:

27. =
1 0 0 1 0 1 0 1 1 0 1 0 0 1
7
s
7G
3
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,14 , and the rank of the
s
matrix 7G
Rr +r = − and:
is 4 ,we can denote that: r1 + r2 =(0 0 1 1 0 0 1 1 1 1 1 1 0 0) , 2
1 2
R1 + R2 = (00000000 11111111 00000000 11111111 11111111 11111111 00000000),
8
1 2
s
* For M7 (M7 )
Span{ } r1, r2 , r3, r4
and Span { R1,R2 ,R3,R4} are not orthogonal sets.
s
C4 . M7 (M7 )
s
: We suppose the M
-sequences of order 7 that are:
y1 = (01001011), y2 =(01100101) ; y3=(01110010) ; y4 =(00111001)
y5=(01011100) ; y6 =(00101110) ; y7 =(00010111) , and y1 = (1001011)

28. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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7G such that their rows are: Ri = yi ( y1), i = 1,2,3,4
1 1 0 0 1 0 1 1 0 0 1 1 0 1 0
1 1 0 1 0 0 1 0 1 1 0 0 1 1 0
1 1 0 1 0 1 0 0 1 0 1 1 0 0 1
s
-sequences of order 7 that are:
87
We will determine M7 ( y1)
s
s
, when Ri = yi ( y1); i = 1,2,3,4
, we can see that:
R1= (0110100 1001011 0110100 0110100 1001011 0110100 1001011 1001011),
R2 = (0110100 1001011 1001011 0110100 0110100 1001011 0110100 1001011),
R3= (0110100 1001011 1001011 1001011 0110100 0110100 1001011 0110100),
R4 = (0110100 0110100 1001011 1001011 1001011 0110100 0110100 1001011),
If we write the matrix 7
s
and theirs columns are:
Ci , j =1,2,...,56 . after that we remove: all zero columns and all equals columns except one, we have
the matrix:

29. =
1 0 1 1 0 1 0 1 0 0 1 0 1 1 0
7
s
7G
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,15.
s
The rank of the matrix 7G
7
is 4, we can denote that:
r1 + r2 = (000110011111100), r1 + r3= (000111111000011)
r1 + r4 = (011111100001100), r2 + r3= (000001100111111)
r2 + r4 = (011001111110000), r3 + r4= (011000011001111)
r1 + r2 + r3= (101100101011010), r1 + r3 + r4= (110100110010101)
r1 + r3 + r4= (110101010101010), r2 + r3 + r4= (110011001010110)
r1 + r2 + r3 + r4 = (011110000110011)
s
Thus, Span { } r1, r2 , r3, r4 is an orthogonal set (the same result forM7 ( y1)
), and we can see that:
R1 + R3 + R4 = (0110100 0110100 0110100 0110100 0110100 0110100 0110100 0110100)
And the 8
1, 3 4 RR R + R = then Span { R1,R2,R3,R4} is not orthogonal set.
s
* For M7 (M7 )
Span{ } r1, r2 , r3, r4
is an orthogonal set but Span { R1,R2 ,R3,R4} is
not orthogonal sets.
s s
D4 . M7 (M7 )
: We suppose the M
s
y1
= (01001011), 2 y s
=(01100101) ; 3 y s
=(01110010) ; 4 y s
=(00111001)
5 y s
=(01011100) ; 6 y s
=(00101110) ; 7 y s
s s
=(00010111) . We will determine M7 ( y1)
, when

30. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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1 0 1 1 0 1 0 0 1 1 0 0 1 0 1
1 0 1 0 1 1 0 1 0 0 1 1 0 0 1
1 0 1 0 1 0 1 1 0 1 0 0 1 1 0
88
Ri = yi ( y1), i = 1,2,...,7
s s
. We can see that:
R1= (10110100 01001011 10110100 10110100 01001011 10110100 01001011 01001011)
R2 = (10110100 01001011 01001011 10110100 10110100 01001011 10110100 01001011)
R3= (10110100 01001011 01001011 01001011 10110100 10110100 01001011 10110100)
R4 = (10110100 10110100 01001011 01001011 01001011 10110100 10110100 01001011)
If we write the matrix 7
s s
7G such that their rows are: Ri = yi ( y1), i = 1,2,3,4
and theirs columns are:
Ci , j =1,2,...,64. after that we remove: all zero columns and all equals columns except one, we have
the matrix:

31. =
1 1 0 0 1 0 1 0 1 1 0 1 0 0 1
7
s
7G
7
In this matrix if we consider rows ri : i =1,2,3,4 and columns c j : j =1,2,...,15.
s
The rank of the matrix 7G
is 4, we can denote that:
r1 + r2 = (000110011111100), r1 + r3= (000111111000011)
r1 + r4 = (011111100001100), r2 + r3= (000001100111111)
r2 + r4 = (011001111110000), r3 + r4= (011000011001111)
r1 + r2 + r3= (101100101011010), r1 + r2 + r4 = (110100110010101)
r1 + r3 + r4= (110101010101010), r2 + r3 + r4= (110011001010110)
r1 + r2 + r3 + r4 = (011110000110011)
s Thus, Span { r1, r2 , r3, r4 } s
is an orthogonal set (the same result for M7 (yi ); i = 1,2,...,7
We can see that:
R1 + R2 = (00000000 00000000 11111111 00000000 11111111 11111111 11111111 00000000),
R1 + R3= (00000000 00000000 11111111 11111111 11111111 00000000 00000000 11111111),
R1 + R4 = (00000000 11111111 11111111 11111111 00000000 00000000 11111111 00000000),
R2 + R3 = (00000000 00000000 00000000 11111111 00000000 11111111 11111111 11111111),
R2 + R4 = (00000000 11111111 00000000 11111111 11111111 11111111 00000000 00000000),
R3 + R4 = (00000000 11111111 00000000 00000000 11111111 00000000 11111111 11111111),
R1 + R2 + R3 = (10110100 01001011 10110100 01001011 01001011 01001011 10110100 10110111) = R5 ,
R1 + R2 + R4= (10110100 10110100 10110100 01001011 10110100 01001011 01001011 01001011) = R7 ,
R1 + R3 + R4 = (10110100 10110100 10110100 10110100 10110100 10110100 10110100 10110100),
R2 + R3 + R4 = (10110100 10110100 01001011 10110100 01001011 01001011 01001011 10110100) = R6,
R1 + R2 + R3 + R4 = (00000000 11111111 11111111 00000000 00000000 11111111 00000000 11111111)
Thus, Span { R1,R2,R3,R4} is an orthogonal set.
s s
* For M7 (M7 )
Span{ } r1, r2 , r3, r4
and Span { R1,R2 ,R3,R4} are orthogonal sets

32. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
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89
IV. CONCLUSION
s s
1. Each of M3(M3 )
r r
and M3(M3 )
generates 6 sets each of them closed under the addition and
contains 7 orthogonal sequences of the form [n = 16, k = 3, d = 8].
s s
2. Each of M3(M7 )
r r
and M3(M7 )
generates 14 sets closed under the addition and each of
them contains 7 orthogonal sequences of the form [ n = 32, k = 3, d = 16].
s s
3. Each of M7 (M3)
r r
and M7 (M3)
generates 6 sets closed under the addition and each of them
contains 15 orthogonal sequences of the form [n = 32, k = 4, d = 16].
s s
4. Each of M7 (M7 )
r r
and M7 (M7 )
generates 14 sets closed under the addition and each of
them contains 15 orthogonal sequences of the form [ n = 64, k = 4, d = 32].
V. ACKNOWLEDGMENT
The author express his gratitude to Prof. Abdulla Y Al Hawaj, President of Ahlia University
for all the support provided.
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