Compose Binary Matrices

*Correspondence to Author:

Dr. Ahmad Hamza Al Cheikha Dep. of Mathematical Science, Col-lege of Arts-science and Education Ahlia Uni., Manama, Bahrain, al-cheikhaa @yahoo.com

How to cite this article:

Ahmad Hamza Al Cheikha. Com-pose Binary Matrices. American Journal of Computer Sciences and Applications, 2017; 1:10.

eSciPub LLC, Houston, TX USA. Website: http://escipub.com/

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

American Journal of Computer Sciences and Applications

(ISS

N

:2

5

7

5-

77

5X

)

Research Article AJCSA, 2017, 1:10

Compose Binary Matrices

Hadamard Matrices and M-Sequences (which formed a closed sets under the addition and with the corresponding null se-quence formed additive groups and generated by feedback registers) are used widely at the forward links of communica-tion channels to mix the informacommunica-tion 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

Traffic channel.

This research is useful to generate new sets of sequences (which are also with the corresponding null sequence additive groups) by compose Hadamard matrices and M-sequences with the big-ger lengths and the bigbig-ger minimum distance that assists to in-crease secrecy of these information and inin-crease the possibility of correcting mistakes resulting in the channels of communica-tion.

Keywords: hadamard matrices, Walsh Sequences, M-sequenc-es, Additive group, Coefficient of Correlation, Orthogonal se-quences.

Dr. Ahmad Hamza Al Cheikha

Dep. of Mathematical Science, College of Arts-science and Education, Ahlia Uni., Manama,

Bah-rain

ABSTRACT

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Introduction

▪ M- Sequences: Let k be a positive integer and  ,₀,₁,...,_k1 are elements in the fieldF₂, then the sequence a₀,a₁,...is called non homogeneous linear recurring sequence of order k iff :

) 1 ( 1

,..., 1 , 0 ,

, ...

1

0 2

0 2

2 1

1















 

 



 

 







 



   

  

k i

i n i k

n i

n k

n k k

n k k n

a a

or

k i

F

a a

a a

The elements a₀,a₁,...,a_k1 are called the initial values (or the vector (a₀,a₁,...,a_k1) is called the initial vector). If 0 then the sequence

... , , ₁

0 a

a is called homogeneous linear recurring sequence (H. L. R. S.), except the zero-initial vector, and the polynomial

0 1 1

1 ...

)

(x x  _ x    x

f k _k k (2)

is called the characteristic polynomial. In this study, we are limited to₀1.

If the characteristic polynomial is prime then the sequence a₀,a₁,... is called M-Sequence and

this sequence is periodic with period n2k 1, and each period contains n₁ 2k11of “0.s”

and n₂ 2k1of “1.s”.

The set of all n2k 1cyclic permutations of one period is closed under the addition by mod 2 and form an orthogonal set and any two different permutations contain 2k1of disagreements and (2k11) of agreements,

) 1 2

( 2

1  



k

f of the agreements are “0.s” and

2

2 2



 k

f of the agreements are “1.s”.

We can collect the all permutations of one period in one square matrix as following:

   

 

   

 



  



0 1 1

2 0

1

1 1

0

a a

a

a a

a

a a

a

A

n n n

n



   

 

Example 1: If  is a root of the prime polynomial f(x)x2x1and generates

) 2 ( 2

GF and Suppose the Linear Binary Recurring Sequence be

n n

n

a

a

a

_2



_1



_or a_n2a_n1a_n 0

With the characteristic equation x2x10

and the characteristic polynomial

1 )

(x x2x

f , which is a prime then the general solution of equation For the initial position: a₁1,a₂ 0 is given by:

n n

n

a  22 , and the sequence is periodic with the period 22 13 _and

x

₁= (101), by the cyclic permutations on

x

₁we have



_{1 2 3}



3 x ,x ,x

M  where:

1

x

= (101),

x

₂



(110),

x

₃=(011), The first two digits in each sequence are the initial position of the feedback register, and the set

M

₃ is an orthogonal set. The matrix of the all cyclic permutations is:

  

 

  

  

1 1 0

0 1 1

1 0 1

A

2

2

,

1

1

2

,

2

2

,

1

1

2

.

3

1

2

1 2 2 1

2 1

1 2 2 1

2 1 2



























 

 

f

f

n

n

n

Example 2: The prime polynomial

1 )

(x x3x

f _{is prime and generates} GF(23)

Suppose the Linear Recurring Sequence be :

n n

n a a

a _3  _1 or a_n3a_n1a_n 0

The characteristic equation x3x10 and the characteristic polynomial f(x)x3x1, which is a prime and generates ₃

2

F and for the

initial position: a₁ 1 ,a₂ 0, a₃ 0, and the

sequence is periodic with the period2317

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0003



_{1 2 3 4 5 6 7}



7 y ,y ,y ,y ,y ,y ,y

M  where:

y

₂=

(1100101) ,

y

₃= (1110010),

4

y = (0111001),

y

₅=(1011100),

y

₆=(0101110), 7

y

=(0010111), the first three digits in each

sequence are the initial position of the feedback register, and the set

M

₇ is an orthogonal set. The matrix of the all cyclic permutations is:

        

 

        

 



1 1 1 0 1 0 0

0 1 1 1 0 1 0

0 0 1 1 1 0 1

1 0 0 1 1 1 0

0 1 0 0 1 1 1

1 0 1 0 0 1 1

1 1 0 1 0 0 1

B

4 2 ,

3 1 2

, 4 2 ,

3 1 2 .

7 1 2

1 3 2 1

3 1

1 3 2 1

3 1 3

  

 

  

  

 

 

 

f g

n m

m

[7], [10], [12], [15-18], [20], [27]

▪ Hadmard Matrices: Hadamard matrices seem such simple matrix structures: they are square, have entries +1 or −1 and have orthogonal row vectors and orthogonal column vectors.

A Hadamard matrix is invented by Sylvester (1867), 26 years before Hadamard (1893) considered them. The nnHadamard matrixH_n

must have n(n1)/2 of “-1.s” and n(n1)/2 of ”1.s”. The binary representation of Hadamard matrix gets replaced each”1” by “0” and replaced each “-1” by”1”

Some basic properties of Hadamard matrices are given by following theorem:

LetH_hbe an Hadamard matrix of order h then:

1. h

t h hH hI

H  , where I_his the identity matrix of order h;

2. H h2h

1

det  ;

3. h

t h t h

hH H H

H  ;

4. Hadamard matrices may be changed into other Hadamard matrices by permuting rows and columns and by multiplying rows and columns by −1.

5. Matrices which can be obtained from one another by these methods are referred to as H-equivalent (not all Hadamard matrices of the same order are H-equivalent).

6. Every Hadamard matrix is H-equivalent to an Hadamard matrix which has every element of its first row and column equal +1 – matrices of this latter form are called normalized.

7. If H_4nis a normalized Hadamard matrix of order 4n, then every row (column), except the first, has 2n minus ones and 2n plus ones in each row (column), and the set these rows is called Walsh’s sequences

8. The order of an Hadamad matrix is 1, 2, 4n, where n is a positive integer.

(Sylvester) Let H₁and H₂be two Hadamard matrices of orders h₁and h₂, then the

Kronecker product of H₁and H₂ is an Hadamard matrix of order h₁h₂.

9. Standard Hadamard matrixH_h have

orthogonal rows (columns) vectors and each row (column) except the first row and first column contains h/2 of “1.s” and h/2 of “-1” and h/2 of disagreements and h/2 of agreements h/4 of agreements are “1.s” and h/4 of agreements are “-1”. [1-6], [8,9], [13,14], [19-26], [28,32].

The smallest examples are:

  ,

1 1

1 1

1 1

1 1 1 1 ,

1 1 1 ,

1 1 2

0 _{2 2}

2

   

 

   

 

 

 

        

 

 H H

H

                

 

                

 

   



    

  

 

 

  

 

  

 

  

   



   



  

 

  

  

  



          

 

          

 

   

   

   

   

   

   

   



1 1 1 1 1

1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1

1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

,

1 1 1 1

1 1 1

1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1 1 1 1 1

12

23 H

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0004

Research method and Material

Definition 1.The Ultimately Periodic Sequence ....

, , ₁

0 a

a with the smallest period r is called a periodic iff: a_nr a_{n ;} n0,1,... [1,2], [12], [15-18], [20-27]

Definition 2. The complement of the binary vector X (x₁,x₂,...,x_n),x_iF₂{0,1}is the vector

) ,..., ,

(x₁ x₂ x_n

X  , where:

  

  

1 0

0 1

i i i

x if

x if

x . [1,2] (3)

Definition 3. Supposex (x₀,x₁,...,x_n1)

and y(y₀,y₁,...,y_n1)are binary vectors of

length n on F2={0,1}. The coefficient of correlations function of x and y , denoted by Rx,y, is:

) 4 ( )

1 ( 1

0

,



 



  n

i

y x y

x

i i

R

Where xi + yi is computed mod 2. It is equal to the number of agreements components minus the number of disagreements corresponding to components or if x_i,y_i{1,1}(usually, replacing in binary vectors x and y each “1” by “-1” and each “0” by “1”) then







 1

0 ,

n

i

i i y

x x y

R , [1,2],[13-18] (5)

Definition 4. Any Periodic Sequence a₀,a₁,....

over F₂ with prime characteristic polynomial is an orthogonal cyclic code and ideal auto correlation [1-2], [12], [18].

Definition 5. Supposex (x₀,x₁,...,x_n1)

and y(y₀,y₁,...,y_n1)are binary vectors of

length n on GF(2)={0,1}, or components belong to {1, -1}, is said strictly orthogonal if R_x,y 0

and orthogonal if R_x,y {1,0.1} ). [1,2],[13-18]

Definition 6. Suppose G is a set of binary vectors of length n:

   

 

 

 



 

1 ,..., 1 , 0 }, 1 , 0 {

), ,...,

, ( ;

2

1 1

0

n i

F x

x x

x X

X G

i

n

Let’s 1*_{= -1 and 0}* _{=1, The set G is said to be} strictly orthogonal if the following two conditions are satisfied:

.

0

,

0

,

.

1

_,0

1

0









_



 

x n

i

i

or

R

x

G

X

(6)

)

7

(

.

0

0

),

(

,

.

2

_,

1

0















  

y x n

i i

i

y

or

R

x

Y

X

G

Y

X

That is, the absolute value of "the number of agreements minus the number of disagreements" is equal to zero, and orthogonal if the following two conditions are satisfied:

. 1 ,

1 ,

.

1 _,0

1

0

 









 

x n

i

i or R

x G

X

(8)

) 9 ( . 1 1

), (

, .

2 _,

1

0

 

 







  

y x n

i i

i y or R

x Y X G Y X

That is, the absolute value of "the number of agreements minus the number of disagreements" is equal to zero. [1,2],[13-18] Definition 7. The matrix A[a_ij] is called Hadamard matrix if it is a square matrix and each entry is equals 1 or ”“ (where denotes -1) with the property that if the size of A is then

h T T

I h AA A

A   , in the decimal counting system, and the distinct rows vectors are mutually orthogonal. [20-27]

Definition 8. If all entries of the first row and the first column in the Hadamard matrix are equal to “1” then the matrix is called standard Hadamard matrix.

Such matrices were first invented by Sylvester (1867) who observed that if H is an Hadamard matrix, then:

) 10 (

   

 

 

H H

H H H

Is also an Hadamard matrix. [3-12]

Theorem 1.

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0005 ii. (1) then this sequence is periodic

iii. If this sequence is homogeneous linear recurring sequence, periodic with the period r, and its characteristic polynomial

) (x

f then r ord f(x).

iv. If the polynomial f(x) is primitive then the period of the sequence is 2k 1, and this sequence is called M – sequence. [7], [16], [17], [27]

Lema 2. (Sylvester (1867)): There is an Hadamard matrix of order t

2 for all nonnegative t. The matrices of order 2tconstructed using Sylvester’s construction are usually Referred to as Sylvester-Hadamard matrices. [3-6], [13,15], Hadamard (1983) gave examples for a few small orders. [5-6],[8],[13-14],[22-23],[31-32]

Results and Discussion

First. Suppose A and B are tow square binary matrices of orders n, m respectively, compose the matrix A with the matrix B or A(B) is a the result of replacing each entry “0” in the matrix A by the full matrix B and replacing each “1” in the matrix A by full Bcomplement of the matrix B and A(B) is a square matrix of order n.m .

The study is similar when A or B is rectangular matrix.

Example 3. If :

      

0 1

1 1

A and

  

 

  

  

1 0 1

1 1 0

0 1 1

B

then:

  

 

  

  

0 1 0

0 0 1

1 0 0

B

and:

      

 

      

 



1 0 1

1 1 0

0 1 1

0 1 0

0 0 1

1 0 0

0 1 0

0 0 1

1 0 0

0 1 0

0 0 1

1 0 0

) (B A

If:

 

   

 

   

 

 

n n

ij

A A A

a A



2 1

,







   

 

   

 

 

m m

kl

B B B

b B



2 1

m n m n n n

m m

B A

B A

B A

B A

B A

B A

B A

B A

B A

B A

. 2 1 2

2 2

1 2 1

2 1

1 1

) (

) (

) (

) (

) (

) (

) (

) (

) (

) (

                  

 

                  

 



   

(11)

And A_ij  A_i(B_j)the result of compose the ith row in the matrix A with the jth _{row of the matrix} B.

If A_icontains n1 of “0.s” and n2 of ”1.s” then A_i contains n1 of “1.s” and n2 of ”0.s” and if B_j contains m1 of “0.s” and m2 of ”1.s” then B_j contains m1 of “1.s” and m2 of ”0.s” and

) ( _j

i j

i A B

A  contains n₁m₁n₂m₂of ‘0.s’ and

1 2 2

1m n m

n  of “1.s” and A_ij  A_i(B_j)contains The same number of ‘0.s’ and of “1.s”, from example 1.:

] 1 0 1 0 1 0 [ ]) 1 0 1 ]([ 0 1 [ ) ( ₃

2

23  A B  

A .

Our purpose now to study the following: From the information about two rowsA_i1,A_i2from the matrix A and two rows B_j1,B_j2from the matrix B then what is the number of agreements and the number of disagreements between

) ( )

( _{1 2 2}

1 j i j

i B and A B

A .

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0006

2

1 n

n

n  , where n1 is the number of “0.s” in each of them and n2 is the number of “1.s” in each of them and the number of agreements between them is f₁ r₁r₂where r1 of theme are “0.s” and r2 of theme are”1.s” such the number of disagreements is

) ( [ ) (

)

( 1 1 2 2 1 2

1

2 n f n r n r n r r

f          , by the

same way :

If there are two rows from the matrix B, for simplicityB₁, B₂have: the same length

2

1 m

m

m  , where m1 is the number of “0.s” in each of them and m2 is the number of “1.s” in each of them and the number of agreements between them isg₁ h₁h₂where h1 of theme are “0.s” and h2 of

theme are”1.s”such the number of

disagreements is

] 2 ) 1 ( [ ) 2 2 ( ) 1 1 ( 1

2 m g m h m h m h h

g          .

Such the representation of the information as following:

Representation 1: Compare A₁(B₁) and A₂(B₂)



) ( ₁

1 B

A 0 0 … 0 _𝐵

1 𝐵1 . .. 𝐵1

⏞

𝑟1

0 0 … 0 𝐵1 𝐵1 . .. 𝐵1

⏞

(𝑛1−𝑟1)

⏞

𝑛1 𝑜𝑓 "0.𝑠"

1 1 … 1 𝐵̅1 𝐵̅1 . .. 𝐵̅1

⏞

(𝑛2−𝑟2)

1 1 … 1 𝐵̅1 𝐵̅1 . .. 𝐵̅1

⏞

𝑟2

⏞

𝑛2 𝑜𝑓 "1.𝑠""



) ( ₂

2 B

A

𝐵₂ 𝐵₂ … 𝐵₂ 0 0 … 0 ⏟

𝑟1

𝐵̅₂ 𝐵̅₂ … 𝐵̅₂ 1 1 … 1 ⏟

(𝑛2−𝑟2)

⏟

𝑛1𝑜𝑓"0.𝑠"

𝐵₂ 𝐵₂ … 𝐵₂ 0 0 … 0 ⏟

(𝑛1−𝑟1)

𝐵̅₂ 𝐵̅₂ … 𝐵̅₂ 1 1 … 1 ⏟

𝑟2

⏟

𝑛2𝑜𝑓"1.𝑠"

• The number of agreements between A1 and A2 is: f₁ r₁r₂ .

• The number of disagreements between A1 and A2 is: f₂ n f₁ [n₁(r₁r₂)].

This representation showing a need to compare:B₁ and B₂, B₁ and B₂, B₁ and B₂and

2 1 and B

B ,

a. Compare B1 and B2 : The Representation of the rows B1 and B2 of matrix B is as following:

Representation 2: Compare B₁ and B₂



1

B

0 0 … 0 ⏞

ℎ1

0 0 … 0 ⏞

(𝑚1−ℎ1)

⏞

𝑚1 𝑜𝑓 "0.𝑠"

1 1 … 1 ⏞

(𝑚2−ℎ2)

1 1 … 1 ⏞

ℎ2

⏞

𝑚2 𝑜𝑓 "1.𝑠"



2

B

0 0 … 0 ⏟

ℎ1

1 1 … 1 ⏟

(𝑚2−ℎ2)

⏟

𝑚1𝑜𝑓"0.𝑠"

0 0 … 0 ⏟

(𝑚1−ℎ1)

1 1 … 1 ⏟

ℎ2

⏟

𝑚2𝑜𝑓"1.𝑠"

• The number of agreements between B1 and B2 is: g₁ h₁h_2.

• The number of disagreements between B1 and B2 is: g₂ mg₁ [m(h₁h₂)].

b. Compare B1 and B₂ : The Representation of the rows B1 and B₂ of matrix B is as following:

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0007 

1

B

0 0 … 0 ⏞

ℎ1

0 0 … 0 ⏞

(𝑚1−ℎ1)

⏞

𝑚1 𝑜𝑓 "0.𝑠"

1 1 … 1 ⏞

(𝑚2−ℎ2)

1 1 … 1 ⏞

ℎ2

⏞

𝑚2 𝑜𝑓 "1.𝑠"



2

B 1 1 … 1⏟

ℎ1

0 0 … 0 ⏟

(𝑚2−ℎ2)

⏟

𝑚1𝑜𝑓"1.𝑠"

1 1 … 1 ⏟

(𝑚1−ℎ1)

0 0 … 0 ⏟

ℎ2

⏟

𝑚2𝑜𝑓"0.𝑠"

• The number of agreements between B1 and B₂ is: g₂ mg₁ [m(h₁h₂)]_.

• the number of Disagreements between B1 and B2 is: g1 (h1h2).

c. Compare B₁ and B₂ : The Representation of the rows B1 and B₂ of matrix B is as following:

Representation 4: Compare B₁ and B₂



1

B

1 1 … 1 ⏞

ℎ1

1 1 … 1 ⏞

(𝑚1−ℎ1)

⏞

𝑚1 𝑜𝑓 "1.𝑠"

0 0 … 0 ⏞

(𝑚2−ℎ2)

0 0 … 0 ⏞

ℎ2

⏞

𝑚2 𝑜𝑓 "0.𝑠"



2

B

0 0 … 0 ⏟

ℎ1

1 1 … 1 ⏟

(𝑚2−ℎ2)

⏟

𝑚1𝑜𝑓"0.𝑠"

0 0 … 0 ⏟

(𝑚1−ℎ1)

1 1 … 1 ⏟

ℎ2

⏟

𝑚2𝑜𝑓"1.𝑠"

• The number of agreements between B₁ and B2is: g2 mg1 [m(h1h2)]_.

• The number of Disagreements between B1 and B2is: g1 h1h2.

d. Compare B₁ and B_{2 : The Representation of the rows} B₁ and B₂ of matrix B is as following:

Representation 5: Compare B₁ and B₂



1

B

1 1 … 1 ⏞

ℎ1

1 1 … 1 ⏞

(𝑚1−ℎ1)

⏞

𝑚1 𝑜𝑓 "1.𝑠"

0 0 … 0 ⏞

(𝑚2−ℎ2)

0 0 … 0 ⏞

ℎ2

⏞

𝑚2 𝑜𝑓 "0.𝑠"



2

B

1 1 … 1 ⏟

ℎ1

0 0 … 0 ⏟

(𝑚2−ℎ2)

⏟

𝑚1𝑜𝑓"1.𝑠"

1 1 … 1 ⏟

(𝑚1−ℎ1)

0 0 … 0 ⏟

ℎ2

⏟

𝑚2𝑜𝑓"0.𝑠"

• The number of agreements between B₁and B₂ is; g₁ h₁h₂.

• The number of disagreements between B₁and B₂ is: g₂ [m(h₁h₂)]_.

First Step: For one row Aiof the matrix A and two different rows Bj,Bkof the matrix B then

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0008

Representation 6: Compare A_i(B_j) and A_i(B_k)

 ) ( _j

i B

A 0 0 … 0 _𝐵

𝑗 𝐵𝑗 . .. 𝐵𝑗

⏞

𝑟1

0 0 … 0 𝐵𝑗 𝐵𝑗 . .. 𝐵𝑗

⏞

(𝑛1−𝑟1)

⏞

𝑛1 𝑜𝑓 "0.𝑠"

1 1 … 1 𝐵̅𝑗 𝐵̅𝑗 . . . 𝐵̅𝑗

⏞

(𝑛2−𝑟2)

1 1 … 1 𝐵̅𝑗 𝐵̅𝑗 . .. 𝐵̅𝑗

⏞

𝑟2

⏞

𝑛2 𝑜𝑓 "1.𝑠""

 ) ( _k

i B

A

𝐵_𝑘 𝐵_𝑘 … 𝐵_𝑘 0 0 … 0 ⏟

𝑟1

𝐵_𝑘 𝐵_𝑘 … 𝐵_𝑘 0 0 … 0 ⏟

(𝑛1−𝑟1)

⏟

𝑛1𝑜𝑓"0.𝑠"

𝐵̅_𝑘 𝐵̅_𝑘 … 𝐵̅_𝑘 1 1 … 1 ⏟

(𝑛2−𝑟2)

𝐵̅_𝑘 𝐵̅_𝑘 … 𝐵̅_𝑘 1 1 … 1 ⏟

𝑟2

⏟

𝑛2𝑜𝑓"1.𝑠"

We have :

• The length of A_i(B_j) and A_i(B_k)is: n.m .

• The number of “0.s” in each of them is:

2 2 1

1m n m

n  .

• The number of “1.s” in each of them is:

1 2 2

1m n m

n  .

• The number of agreements is:

)

( _{1 2}

1 n h h

g

n   .

• The number of disagreements is:

)] (

[ _{1 2}

2 n m h h

g

n    .

• The difference d between the disagreements and agreements is:

) (

)] (

[ _{1 2 1 2}

1

2 ng n m h h n h h

g n

d       

)] (

2

[m h₁ h₂

n

d   

Second Step: For two different rows A_i,A_jof the matrix A and one row B_kof the matrix B and from the representations (2-5) and from the following representation of A_i,A_j,B_k:

Representation 7: Compare A_i(B_k) and A_j(B_k)

 ) ( _k

i B

A 0 0 … 0 _{𝐵𝑘 𝐵𝑘 . .. 𝐵𝑘}

⏞

𝑟1

0 0 … 0 𝐵𝑘 𝐵𝑘 . .. 𝐵𝑘

⏞

(𝑛1−𝑟1)

⏞

𝑛1 𝑜𝑓 "0.𝑠"

1 1 … 1 𝐵̅𝑘 𝐵̅𝑘 . . . 𝐵̅𝑘

⏞

(𝑛2−𝑟2)

1 1 … 1 𝐵̅𝑘𝐵̅𝑘 . .. 𝐵̅𝑘

⏞

𝑟2

⏞

𝑛2 𝑜𝑓 "1.𝑠""



) ( _k

j B A

𝐵_𝑘 𝐵_𝑘 … 𝐵_𝑘 0 0 … 0 ⏟

𝑟1

𝐵̅𝑘 𝐵̅𝑘 … 𝐵̅𝑘

1 1 … 1 ⏟

(𝑛2−𝑟2)

⏟

𝑛1𝑜𝑓"0.𝑠"

𝐵_𝑘 𝐵_𝑘 … 𝐵_𝑘 0 0 … 0 ⏟

(𝑛1−𝑟1)

𝐵̅𝑘 𝐵̅𝑘 … 𝐵̅𝑘

1 1 … 1 ⏟

𝑟2

⏟

𝑛2𝑜𝑓"1.𝑠"

We have :

• The length of A_i(B_k) and A_j(B_k)is: n.m .

• The number of “0.s” in each of them is:

2 2 1

1m n m

n  .

• The number of “1.s” in each of them is:

1 2 2

1m n m

n  .

• The number of agreements is:

)

(_{1 2}

1 m r r

f

m   .

• The number of disagreements is:

)] (

[ _{1 2}

2 m n r r

f

m    .

• The difference d between the disagreements and agreements is:

m r r m r r n m f m f m

d  ₂  ₁ [ (₁ ₂)] (₁ ₂)

m r r n

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0009 Third Step: For two different rows A_i,A_jof the matrix A and two different rows B_k,B_lof the

Matrix B and from the representations (1-5) and from the following representation of A_i,A_j , B_k,B_l :

Representation 8: Compare A_i(B_k) and A_j(B_l)

 ) ( _k

i B

A 0 0 … 0 _{𝐵𝑘 𝐵𝑘 . .. 𝐵𝑘}

⏞

𝑟1

0 0 … 0 𝐵𝑘 𝐵𝑘 . .. 𝐵𝑘

⏞

(𝑛1−𝑟1)

⏞

𝑛1 𝑜𝑓 "0.𝑠"

1 1 … 1 𝐵̅𝑘 𝐵̅𝑘 . . . 𝐵̅𝑘

⏞

(𝑛2−𝑟2)

1 1 … 1 𝐵̅𝑘𝐵̅𝑘 . .. 𝐵̅𝑘

⏞

𝑟2

⏞

𝑛2 𝑜𝑓 "1.𝑠""



) ( _l

j B A

𝐵_𝑙 𝐵_𝑙 … 𝐵_𝑙 0 0 … 0 ⏟

𝑟1

𝐵̅𝑙 𝐵̅𝑙 … 𝐵̅𝑙

1 1 … 1 ⏟

(𝑛2−𝑟2)

⏟

𝑛1𝑜𝑓"0.𝑠"

𝐵_𝑙 𝐵_𝑙 … 𝐵_𝑙 0 0 … 0 ⏟

(𝑛1−𝑟1)

𝐵̅𝑙 𝐵̅𝑙 … 𝐵̅𝑙

1 1 … 1 ⏟

𝑟2

⏟

𝑛2𝑜𝑓"1.𝑠"

We have:

• The number of agreements between Ai and Aj is: f₁ (r₁r₂).

• The number of disagreements between Ai and Aj is: f₂ [n(r₁r₂)]

• The number of agreements between Bk and Bl is: g₁ (h₁h₂).

• The number of disagreements between Bk and Bl is: g₂ [m(h₁h₂)]

• The length of A_i(B_k) and A_j(B_l)is: n.m .

• The number of “0.s” in each of them is:

2 2 1

1m n m

n  .

• The number of “1.s” in each of them is:

1 2 2

1m n m

n  .

• The number of agreements between

) ( _k

i B

A and A_j(B_l)is: f₁g₁ f₂g₂, or is:

)] (

)][ (

[ ) )(

(r₁r₂ h₁h₂  n r₁r₂ m h₁h₂

• The number of disagreements between

) ( _k

i B

A and A_j(B_l)is: f₁g₂  f₂g₁, or is:

) )](

( [ )] (

)[

(r₁r₂ m h₁h₂  n r₁r₂ h₁h₂

• The difference d between the agreements and disagreements of

) ( _k

i B

A and A_j(B_l)is:

) (

)

(f₁g₁ f₂g₂ f₁g₂ f₂g₁

d   

) )(

(f₁ f₂ g₁ g₂

d  

Or:

) )( ( 4 ) ( 2 ) (

2n h₁ h₂ m r₁ r₂ r₁ r₂ h₁ h₂ nm

d       

Example 4. Using binary representations of the standard Hadmard matrices:

,... , ,

1 1 0 1 0 1 0 1 0 1 0 0

1 1 0 0 1 1 1 0 0 0 1 0

0 0 1 1 1 1 0 0 0 1 1 0

0 1 0 1 1 0 1 0 1 1 0 0

1 0 0 1 1 0 0 1 1 0 1 0

0 0 0 0 0 1 1 1 1 1 1 0

1 0 1 0 1 0 1 1 0 1 0 0

0 1 1 1 0 0 1 1 0 0 1 0

1 1 1 0 0 0 0 0 1 1 1 0

1 0 1 1 0 1 1 0 1 0 0 0

0 1 1 0 1 1 0 1 1 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

,

1 0

0 0 0 1

1 1

0 1

1 1 0 1

1 1

0 1

1 1 1 0

0 0

1 0

0 0 1 0

0 0

0 1

1 1 1 0

0 0

1 0

0 0 1 0

0 0

0 1

1 1 1 0

0 0

1 0

0 0 1 0

0 0

,... , 0 1

1 1 1 0

0 0

1 0

0 0 1 0

0 0 , 1 0

0 0

1

4 12

8 4

2 H H H H h

H 

                

 

                

 



         

 

         

 

    

 

   

 

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0010 The set of row of a standard Hadamard matrix

form an additive group for addition by mod 2 and the set of this rows except the first row also closed under the addition (Walsh’s Sequences).

Except H2 , First row contains only “0.s” each of the other rows contains 2h₁of “0.s” , 2h₁of “1.s”, 2h₁of disagreements and 2h₁of agreements, h1 of the agreements are “0.s” and the other

h1 are ”1.s”.

i. Is very clear: H2 (H2) = H4 is an Hadamard matrix.

ii. All entries of the first row in H₂(H_4h)are “0”

and each other row contains 4h of “0.s” and 4h of “1.s” and orthogonal with the first row, the order of the matrix H₂(H_4h)is 8h.

       h h h h h H H H H H H 4 4 4 4 4

2( )

iii. Each row except the first row contains 4h of “0.s”, 4h of “1.s” and any two different rows except the first contain 4h of disagreements and 4h of agreements (from third. The difference between the agreements and

disagreements is:

0 ) 2 2 )( 1 1 ( ) )(

( ₁ _{2 1} ₂    

 f f g g h h

d ) 2h

of the agreements are “0.s” and the other 2h are “1.s” also because this set of rows is closed under the addition. Thus these two rows are orthogonal and the columns of

) ( ₄

2 H h

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0011 In this matrix, except the first row (column),

each row (column) contains 12 0f “0.s” , 12 of “1.s”, 12 of disagreements and 12 of agreements, 6 of the agreements are “0” and 6 of agreements are “1.s”

iv. All entries of the first row in ₄ ( _{4 2})

1 h

h H

H are “0”

and each other row contains 8h₁h₂ of “0.s”

and 8h₁h₂ of “1.s” and orthogonal with the first

row and the order of the matrix is:16h₁h₂. For any two different rows, except the first row

• The number of agreements between Ai and Aj is: f₁ (h₁h₁)2h₁.

• The number of disagreements between Ai and Aj is: f₂ [4h₁2h₁]2h₁

• The number of agreements between Bk and Bl is: g₁ (h₂ h₂)2h₂.

• The number of disagreements between Bk and Bl is: g₂ [4h₂ 2h₂]2h₂

❖ In the compositionA_i(B_j)and A_i(B_k):

• The number of agreements is:

2 1 2 1

1 4h (2h ) 8hh

g

n   .

• The number of disagreements is:

2 1 2 1

2 4h (2h ) 8hh

g

n   .

• The difference d between the disagreements and agreements is: d 0

Thus, second condition of the orthogonal is verified.

❖ In the compositionA_i(B_k)and A_j(B_k):

• The number of agreements is:

2 1 1 2

1 4h (2h ) 8hh

f

m   .

• The number of disagreements is:

) 2 (

4 _{2 1}

2 h h

f

m  .

• The difference d between the disagreements and agreements is: d 0

Thus, second condition of the orthogonal is verified.

❖ In the compositionA_i(B_k)and A_j(B_l):

• The number of agreements is:

2 1 2 1 2 1 2 2 1

1g f g 2h (2h ) 2h (2h ) 8hh

f     .

• The number of disagreements is:

2 1 2 1 2 1 1 2 2

1g f g 2h (2h ) 2h (2h ) 8hh

f     .

• The difference d between the disagreements and agreements is “0”, or :

0 ) 2 2 )( 2 2

( ₁ _{1 2}  ₂ 

 h h h h

d

• The columns of ₄ ( _{4 2})

1 h

h H

H have the same properties of the rows and H_4h1(H_4h2) is an Hadamard matrix. By the same way for

)

( ₁

2 4

4h H h

H is an Hadamard matrix.

Thus, second condition of the orthogonal is verified and rows of A(B) except the first row form an orthogonal set

Second. Compose binary matrices is associative: We will prove that

) ))( ( ( )) (

(B C A B C

A  for any A, B, C binary matrices.

Suppose:

    

 

    

 

 

    

 

    

 

 

    

 

    

 

 

3 3

2 2

1 1

2 1 2

1 2

1

] [ , ]

[ , ]

[

h h mn

h h kl

h h ij

C C C c

C B B B b

B A A A a

A

 



From the left side:

i.





₁

.

. , , 1,2,...,

)) ( ( )) ( (

3 2 1

h j

i C

B a C B A

h h h

ij 



• If a_ij 0 then:

From the left side:

    

 

    

 



    

 

    

 

 

) ( ....

) ( ) (

... ... ... ... ...

) ( ... ) ( ) (

) ( ... ) ( ) (

) (

) (

) (

)) ( ( )) ( (

2 2 2

2

2 2

2 1 2 1

2 22

21

1 12

11 2

1

C b C

b C b

C b C

b C b

C b C

b C b

C B

C B

C B C B C B a

h h h

h

h h

h

ij _

From the right side:

    

 

    

 



    

 

    

 

 

) ( .... ) ( ) (

... ... ... ... ...

) ( ... ) ( ) (

) ( ... ) ( ) (

) (

) (

) (

) )( ( ) ))( ( (

2 2 2

2

2 2

2 1 2 1

2 22

21

1 12

11 2

1

C b C b C b

C b C

b C b

C b C

b C b

C B

C B

C B

C B C B a

h h h

h

h h

h

ij _

Thus: A(B(C)) = (A(B)(C)

• If a_ij 1 then:

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0012                                   ) ( .... ) ( ) ( ... ... ... ... ... ) ( ... ) ( ) ( ) ( ... ) ( ) ( ) ( ) ( ) ( ) ) ( ( )) ( ( 2 2 2 2 2 2

2 1 2 1

2 22 21 1 12 11 2 1 C b C b C b C b C b C b C b C b C b C B C B C B C B C B a h h h h h h h ij 

From the right side:

                                 ) ( .... ) ( ) ( ... ... ... ... ... ) ( ... ) ( ) ( ) ( ... ) ( ) ( ) ( ) ( ) ( )) ( ( ) ))( ( ( 2 2 2 2 2 2

2 1 2 1

2 22 21 1 12 11 2 1 C b C b C b C b C b C b C b C b C b C B C B C B C B C B a h h h h h h h ij 

Is B(C)B(C)?

• If b_kl 0

        C C C b C C C b kl kl ) ( 0 ) ( ) ( 0 ) (

, if b_kl 1

          C C C C b C C C C b kl kl ) ( 1 ) ( ) ( 0 ) ( 1 ) (

Thus B(C)B(C)

Thus for any binary matrices A, B, C : A(B(C)) = (A(B)(C) and compose binary matrices is associative.

Result: If nn₁n₂n₃ and there are Hadmard matrices of orders n1 , n2 , n3 then there is Hadmard matrix of order nn₁n₂n₃.

Third. There is Identity for compose binary matrices that is Null matrix O = [0] and for any binary matrix A is A(O) = O(A) = A.

Forth. Compose binary matrices is not commutative: Example 6:                                              1 1 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 1 0 0 1 1 1 0 1 0 1 1 0                                          1 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 1

And clear that:



































































































0

1

1

0

1

1

0

0

1

1

1

0

1

1

1

0

0

1

1

1

0

1

0

1

1

0

Fifth. For the binary matrix A there is no inverse A’ such A(A’) = A’(A) = O = [0].

Example 7: Suppose (a)a₀,a₁,a₂,...,a_n1and

1 2

1 0, , ,...,

)

(b b b b b_m are one period of two binary M-Sequences with the lengths _n2k1 1

,_m2k2 1 _{respectively, and the matrix A is the}

all cyclic permutations of the sequence (a), the matrix B is the all cyclic permutation of the matrix (b).

Thus each row of the matrix A contains

) 1 2

( 1

1  1 



k

n of “0.s”and _n2 ₂k11of “1.s”.

The rows of the matrix A are closed under the addition mod 2 and any two different rows are orthogonal and contain _f2 ₂k11_{disagreements}

and _f1 (2k111)_{of agreements,} 2

1 2 1



 k

r of

agreements are “1.s”and _r2 (2k12 1) _of

agreements are “0.s”.

Thus each row of the matrix B contains

) 1 2

( 1

1  2 



k

m of “0.s” and _m2 ₂k21of “1.s”.

The rows of the matrix B are closed under the addition mod 2 and any two different rows are orthogonal and contain _g2 ₂k21

disagreements and_g1 (2k211)_of

agreements, _h1 ₂k22of agreements are “1.s”

and _h2 (2k22 1) of agreements are “0.s”.

Ahmad Hamza Al Cheikha, AJCSA, 2017; 1:10

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0013

• The length of each row is:

1 ) 2 2 ( 2

) 1 2 )( 1 2

( 1  2   1 2  1  2 

 k k kk k k

nm

• The number of “1.s” is:

1 1 1

1 2

2 1

1 (2 1 1)(2 2 1) 2 1 2 2

  

 _{  }



 k k k k

m n m n

1 ) 2 2

(

2 1 1 1

2 2 1

1   1 2  1  2 

 



k k k

k m n m n

• The number of “0.s” is:

) 1 2

( 2 2

) 1 2

( 1 1 1 1

1 2 2

1   1  2  1 2 

  

 k k k

k m n m

n

) 2 2

(

2 1 1 1

1 2 2

1 1 2 1 2

 



 _{ }



 k k k k

m n m n

• The difference between the number of “0.s” and the number of “1.s” is 1. Thus, the first condition of the orthogonal is exist.

❖ In the compositionA_i(B_j)and A_i(B_k):

• The number of agreements is:

1 ) 2 2 ( 2

) 1 2 )( 1 2

( 1 1 1

1 1  2   1 2  1  2 

 



 k k k k

k k

g n

• The number of disagreements is:

1 1

1

2 (21 1)(2 2 ) 21 2 2 2

 



 _{ }



 k k k k k

g

n .

• The difference d between the disagreements and agreements is:

1 2 1 

 k

d Thus, second condition of the orthogonal is unverified.

❖ In the compositionA_i(B_k)and A_j(B_k):

• The number of agreements is:

1 ) 2 2

( 2

) 1 2 )( 1 2

( 2 1 1 1 2 1 1 1 2

1       

 



 k k k k

k k

f m

.

• The number of disagreements is:

1 1

1

2 (2 2 1)(21 ) 2 1 2 2 1

 



 _{ }



 k k k k k

f

m .

• The difference d between the disagreements and agreements is:

1 2 2 

 k

d Thus, second condition of the orthogonal is unverified.

❖ In the compositionA_i(B_k)and A_j(B_l):

• The number of agreements is: f₁g₁ f₂g₂.

• The number of disagreements is:

1 2 2

1g f g

f  .

• The difference d between the disagreements and agreements is:

1 ] 2 ) 1 2 ][( 2 ) 1 2 [( ) )(

( ₁ _{2 1} ₂  11  1 1 2 1  2 1 

 k k k  k 

g g f f d

Thus, second condition of the orthogonal is verified

For remove the non-orthogonal situations need think about extend the rows of the matrix A and the rows of the matrix B by adding “0” at the beginning each row of them, thus we have:

1 2

1 1 2

1

1 1 1 1

1_{, 2 , 2 , 2}

2       

 k k k k

f f n

n n

1 2

1 1 2

1

1 2 2 2

2_{, 2 , 2 , 2}

2       

 k k k k

g g m

m m

❖ In the composition A~(B~): where the “~” is the

signal of extending, and A~(B~)of the size

2 1 2

1 _{1)(2 1) 2 2}

2

( k  k   k k .

• The length of each row is: 2

1 2 1_{2 2}

2k k k k

nm  

• The number of “1.s” is:

1 1

1 1

1 2

2 1

1 21 2 2 2 1 2 2 2 1 2

  

 

 _{ }



 k k k k k k

m n m n

• The number of “0.s” is:

1 1

1 1

1 1

2 2

1 21 2 2 2 1 2 2 2 1 2

  

 

 _{ }



 k k k k k k

m n m n

• The difference between the number of “0.s” and the number of “1.s” is 0.

Thus, the first condition of the orthogonal is verified.

❖ In the compositionA~_i(B~_j)and A~_i(B~_k):

• The number of agreements is:

1 1

1 2 12 2 2 1 2

   _

 k k k k

g

n .

• The number of disagreements is:

1 1

2 212 2 2 1 2

   _

 k k k k

g

n .

• The difference d between the disagreements and agreements is: d 0 Thus, second condition of the orthogonal is verified.

❖ In the compositionA~_i(B~_k)and A~_j(B~_k):

• The number of agreements is:

1 1

1 2 221 2 1 2

   _

 k k k k

f

m .

• The number of disagreements is:

1 1

2 2 221 21 2

   _

 k k k k

f

m .

• The difference d between the disagreements and agreements is: d 0

Http://escipub.com/american-journal-of-computer-sciences-and-applications/ 0014 ❖ In the compositionA~_i(B~_k)and A~_j(B~_l):

• The number of agreements is: f₁g₁ f₂g₂.

• The number of disagreements is:

1 2 2

1g f g

f  .

• The difference d between the disagreements and agreements is:

0 ) )(

( ₁ _{2 1} ₂ 

 f f g g d

Thus, second condition of the orthogonal is verified.

From example 1. And example 2.:

           1 1 0 0 1 1 1 0 1 A ,                                               0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 , 1 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 1 B B And                                                                    0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 ) (B A

❖ In the composition A(B):

• The length of each row is:

21 ) 7 ( 3 ) 1 2 )( 1 2

( 1  2   

 k k

nm

• The number of “1.s” is:

11 ) 4 ( 2 ) 3 ( 1 2 2 1

1m n m   

n

• The number of “0.s” is:

10 ) 3 ( 2 ) 4 ( 1 1 2 2

1m n m   

n

• The difference between the number of “0.s” and the number of “1.s” is 1.

Thus, the first condition of the orthogonal is exist.

❖ In the compositionA_i(B_j)and A_i(B_k):

• The number of agreements is: ng₁ 3(3)9 • The number of disagreements is:

12 ) 4 ( 3

2  

g

n .