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Page 1: Super Special Codes Using Super Matrices, by W. B. Vasantha Kandasamy, Florentin Smarandache, K. Ilanthenral

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W. B. Vasantha Kandasamy

Florentin Smarandache

K. Ilanthenral

SUPER SPECIAL CODES SUPER MATRICES

 

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 Svenska fysikarkivet (Swedish physics archive) is a publisregistered with the Royal National Library of Sweden (Ku biblioteket), Stockholm.

Postal address for correspondence: Näsbydalsvägen 4/11, 18331 Täby, Sweden

Peer-Reviewers:Prof. Mihàly Bencze, Department of MathematicsÁprily Lajos College, Braúov, RomaniaProf. Catalin Barbu, Vasile Alecsandri College, Bacau, RomaniDr. Fu Yuhua, 13-603, Liufangbeili

Liufang Street, Chaoyang district, Beijing, 100028 P. R. China

Copyright © Authors, 2010Copyright © Publication by Svenska fysikarkivet, 2010

Copyright note: Electronic copying, print copying and dthis book for non-commercial, academic or individual use  by any user without permission or charge. Any part of thicited or used howsoever in other publications must ackn  publication. No part of this book may be reproduced whatsoever (including storage in any media) for comwithout the prior permission of the copyright holder. R permission to reproduce any part of this book for commer be addressed to the Author. The Author retains his righ book as a whole or any part of it in any other publicationway he sees fit. This Copyright Agreement shall remain the Author transfers copyright of the book to another party

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CONTENTS

Preface

 Chapter One

INTRODUCTION TO SUPERMATRICESAND LINEAR CODES

1.1 Introduction to Supermatrices 1.2 Introduction to Linear Codes and their Prope

Chapter Two

SUPER SPECIAL VECTOR SPACES

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Chapter Four

APPLICATIONS OF THESE NEW CLASSES OFSUPER SPECIAL CODES

FURTHER READING

INDEX

ABOUT THE AUTHORS 

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PREFACE 

The new classes of super special codes arethis book using the specially constructed superspaces. These codes mainly use the super matric

can be realized as a special type of concatena book has four chapters.

In chapter one basic properties of codes andare given. A new type of super special vconstructed in chapter two of this book. Three

super special codes namely, super special rospecial column code and super special codes archapter three. Applications of these codes are gichapter.

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Our thanks are due to Dr. K. Kandasamy

reading this book. We also acknowledge our gKama and Meena for their help with correclayout.

W.B.VASANTHA

FLORENTIN SM

K.IL

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Chapter One

INTRODUCTION TO

SUPERMATRICES AND LINEAR C

 

This chapter has two sections. In section we on basic properties about supermatrices which are esuper special codes. Section two gives a brief

algebraic coding theory and the basic propertilinear codes.

1.1 Introduction to Supermatrices

The general rectangular or square array of numbe

A =2 3 1 4

5 0 7 8

ª º« »¬ ¼

, B =

1 2

4 5

ª « «

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We shall call them as simple matrices [10]. B

matrix we mean a matrix each of whose elements ordinary number or a letter that stands for a numbwords, the elements of a simple matrix are scalarquantities.

A supermatrix on the other hand is one whose ethemselves matrices with elements that can be eithe

other matrices. In general the kind of supermatricdeal with in this book, the matrix elements whichscalar for their elements. Suppose we have the four m

a11 =2 4

0 1

ª º« »¬ ¼

, a12 =0 40

21 12

ª º« »¬ ¼

 

a21 =

3 1

5 7

2 9

ª º« »« »« »¬ ¼

and a22 =

4 12

17 6

3 11

ª º« »« »« »¬ ¼

.

One can observe the change in notation aij denotes anot a scalar of a matrix (1 < i, j < 2).

Let

a = 11 12

21 22

a a

a a

ª º« »¬ ¼

;

we can write out the matrix a in terms of the origelements i.e.,

2 4 0 40ª º« »

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Thus far we have referred to the elements i

as matrices as elements. It is perhaps more uelements of a supermatrix as submatrices. Wsubmatrices within a supermatrix. Now we procethe order of a supermatrix.

The order of a supermatrix is defined in ththat of a simple matrix. The height of a sup

number of rows of submatrices in it. The width ois the number of columns of submatrices in it.All submatrices with in a given row must

number of rows. Likewise all submatrices wcolumn must have the same number of columns.

A diagrammatic representation is given by

figure:

In the first row of rectangles we have one rfor each rectangle; in the second row of rectanglrows of squares for each rectangle and in threctangles we have two rows of squares for

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  by partitioning. Furthermore, the order of superma

nothing about the orders of the submatrices wsupermatrix.  Now we illustrate the number of rows and co

supermatrix.

 Example 1.1.1: Let

a =

3 3 0 1 41 2 1 1 6

0 3 4 5 6

1 7 8 9 0

2 1 2 3 4

ª º« » « »« »« »

« »« »¬ ¼

.

a is a supermatrix with two rows and two columns.

 Now we proceed on to define the notion of partitioneIt is always possible to construct a supermatrix frommatrix that is not a scalar quantity.

The supermatrix can be constructed from a simthis process of constructing supermatrix is

 partitioning.A simple matrix can be partitioned by dividing o

the matrix between certain specified rows, or the pro be reversed. The division may be made first betweethen between columns.

We illustrate this by a simple example.

 Example 1.1.2: Let

ª º

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A1 =

3 0 1 1 2 01 0 0 3 5 2

5 1 6 7 8 4

0 9 1 2 0 1

2 5 2 3 4 6

1 6 1 2 3 9

ª º« »« »« »« »

« »« »

« »« »¬ ¼

.

 Now let us draw a thin line between the 2nd and 3This gives us the matrix A1. Actually A1 may

a supermatrix with two matrix elements formin

two columns. Now consider 

A2 =

3 0 1 1 2 0

1 0 0 3 5 2

5 1 6 7 8 4

0 9 1 2 0 12 5 2 3 4 6

1 6 1 2 3 9

ª º« »« »« »« »

« »« »« »« »¬ ¼

.

Draw a thin line between the rows 4 and 5 whinew matrix A2. A2 is a supermatrix with twocolumn.

 Now consider the matrix

3 0 1 1 2 0ª º« »

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11 12

21 22

a a

a a

ª º

« »¬ ¼ 

where

a11 =

3 0

1 0

5 1

0 9

ª º« »« »« »

« »¬ ¼

,

a12 =

1 1 2 0

0 3 5 2

6 7 8 4

1 2 0 1

ª º« »« »« »

« »¬ ¼

,

a21 =2 5

1 6

ª º« »¬ ¼

and a22 =2 3 4 6

1 2 3 9

ª º« »¬ ¼

.

The elements now are the submatrices defined as a11,a22 and therefore A3 is in terms of letters.

According to the methods we have illustratematrix can be partitioned to obtain a supermatrix that happens to suit our purposes.

The natural order of a supermatrix is usually detthe natural order of the corresponding simple matmore we are not usually concerned with natural osubmatrices within a supermatrix.

  Now we proceed on to recall the notion o

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symmetrically partitioned. We give an e

symmetrically partitioned matrix as, Example 1.1.3: Let

as =

2 3 4 1

5 6 9 2

0 6 1 95 1 1 5

ª º« »« »« »« »« »¬ ¼

.

Here we see that the matrix has been particolumns one and two and three and four. It

 partitioned between rows one and two and rows t

  Now we just recall from [10] the method partitioning of a symmetric simple matrix.

 Example 1.1.4: Let us take a fourth order symm

 partition it between the second and third rows anthe second and third columns.

a =

4 3 2 7

3 6 1 4

2 1 5 27 4 2 7

ª º« »« »

« »« »¬ ¼

.

We can represent this matrix as a superma

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a = 11 12

21 22

a a

a a

ª º

« »¬ ¼.

The diagonal elements of the supermatrix a are a11

also observe the matrices a11 and a22 are also matrices.

The non diagonal elements of this supermatri

matrices a12 and a21. Clearly a21 is the transpose of a12

The simple rule about the matrix elemsymmetrically partitioned symmetric simple matrix diagonal submatrices of the supermatrix are all matrices. (2) The matrix elements below the diagotransposes of the corresponding elements above the d

The forth order supermatrix obtained from a  partitioning of a symmetric simple matrix a is as follo

a =

11 12 13 14

12 22 23 24

13 23 33 34

14 24 34 44

a a a a

'a a a a

' 'a a a a' ' 'a a a a

ª º« »« »

« »« »¬ ¼

.

How to express that a symmetric matrix has been sym partitioned (i) a11 and at

11 are equal. (ii) atij (i z j); a

t jia = aij. Thus the general expression for a sym

 partitioned symmetric matrix;

11 12 1na a ... aª º

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D =

1

2

n

D 0 ... 0

0 D ... 0

0 0 ... D

ª º

« »c« »« »« »c c¬ ¼

 

0' only represents the order is reversed or tr

denote tija = a'ij just the ' means the transpose.D will be referred to as the super diagon

identity matrix

I =

s

t

I 0 0

0 I 00 0 I

ª º« »« »« »¬ ¼

 

s, t and r denote the number of rows and columsecond and third identity matrices respectivelymatrices with zero as all entries).

  Example 1.1.5: We just illustrate a general matrix d;

d =

3 1 2 0 0

5 6 0 0 00 0 0 2 5

0 0 0 1 3

0 0 0 9 10

ª º« »« »« »« »

« »« »¬ ¼

 

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 Example 1.1.6: Let

1 0 0 01 0 0 0

1 0 0 0

0 1 0 0

0 1 0 0

0 0 1 00 0 1 0

0 0 1 0

0 0 1 0

0 0 0 1

0 0 0 1

0 0 0 1

0 0 0 1

ª º« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »¬ ¼

.

Here the diagonal elements are only column unit

case of supermatrix [10] has defined the notiontriangular matrix as a supermatrix.

 Example 1.1.7: Let

u =

2 1 1 3 2

0 5 2 1 1

0 0 1 0 2

ª º« »« »« »¬ ¼

 

u is a partial upper triangular supermatrix.

 Example 1.1.8: Let

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L is partial upper triangular matrix partitioned as

Thus T =T

aª º« »c¬ ¼

where T is the lower triangular su

T =

5 0 0 0 0

7 2 0 0 0

1 2 3 0 0

4 5 6 7 0

1 2 5 2 6

ª º« »« »« »« »« »« »¬ ¼

and a' =1 2 3

0 1 0

ª « ¬

We proceed on to define the notion of superveccolumn supervector. A simple vector is a vectorelements is a scalar. It is nice to see the numbtypes of supervectors given by [10].

 Example 1.1.9: Let

v =

1

3

4

57

ª º« »« »« »« »

« »« »¬ ¼

    .

 This is a type I i.e., type one column supervector

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Type I row supervector is given by the following exa

  Example 1.1.10: v1 = [2 3 1 | 5 7 8 4] is a supervector. i.e., v' = [v'1, v'2, …, v'n] where each subvector; 1 d i d n.

 Next we recall the definition of type II supervectors.

Type II column supervectors.

DEFINITION 1.1.1: Let 

a =

11 12 1

21 22 2

1 2

...

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

...

ª º

« »« »« »« »¬ ¼

m

m

n n nm

a a a

a a a

a a a

a11 = [a11 … a1m ]

a21

= [a21 … a2m ]…

an1 = [an1 … anm ]

i.e., a =

111

2

1

ª º« »

« »« »« »« »¬ ¼

#

n m

a

a

a

is defined to be the type II column supervector.

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Clearly

a =

1

112

1

ª º« »« »« »« »« »¬ ¼

#

n m

aa

a

= [a1 a2 … am ]n

the equality of supermatrices.

 Example 1.1.11: Let

A =

3 6 0 4 5

2 1 6 3 0

1 1 1 2 10 1 0 1 0

2 0 1 2 1

ª º« »« »

« »« »« »« »¬ ¼

 

 be a simple matrix. Let a and b the supermatrix m

a =

3 6 0 4 5

2 1 6 3 0

1 1 1 2 1

0 1 0 1 0

2 0 1 2 1

ª º« »« »« »« »« »

« »¬ ¼

 

where

a11 =

3 6 0

2 1 6

ª º« »« » , a12 =

4 5

3 0

ª º« »« » ,

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 b =

3 6 0 4 52 1 6 3 0

1 1 1 2 1

0 1 0 1 0

2 0 1 2 1

ª º« »« »« »« »« »« »¬ ¼

= 11 12

21 22

 b b

 b b

ª º« »¬ ¼

 

where

 b11 =

3 6 0 4

2 1 6 3

1 1 1 2

0 1 0 1

ª º« »« »« »« »¬ ¼

, b12 =

5

0

1

0

ª º« »« »« »« »¬ ¼

,

 b21 = [2 0 1 2 ] and b22 = [1].

a =

3 6 0 4 5

2 1 6 3 0

1 1 1 2 10 1 0 1 0

2 0 1 2 1

ª º« »« »« »« »« »« »¬ ¼

 

and

 b =

3 6 0 4 5

2 1 6 3 0

1 1 1 2 1

0 1 0 1 0

ª º« »« »« »« »« »

.

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 Example 1.1.12:

a = 3 2 1 7 80 2 1 6 9

0 0 5 1 2

ª º« »« »« »¬ ¼

= [T' | a']

and

 b =

2 0 09 4 0

8 3 6

5 2 9

4 7 3

ª º« »« »« »« »« »« »¬ ¼

=T

 bª º« »c¬ ¼

 

are type III supervectors.

One interesting and common example of a typeis a prediction data matrix having both predictattributes.

The next interesting notion about supetranspose. First we illustrate this by an examplethe general case.

 Example 1.1.13: Let

a =

2 1 3 5 6

0 2 0 1 1

1 1 1 0 2

2 2 0 1 1

ª º« »« »« »« »« »« »

 

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 where

a11 =

2 1 3

0 2 0

1 1 1

ª º« »« »« »¬ ¼

, a12 =

5 6

1 1

0 2

ª º« »« »« »¬ ¼

,

a21 =2 2 0

5 6 1

ª º« »¬ ¼

, a22 =1 1

0 1

ª º« »¬ ¼

,

a31 =2 0 0

1 0 1

ª º« »

¬ ¼

and a32 =0 4

1 5

ª º« »

¬ ¼

.

The transpose of a

at = a' =

2 0 1 2 5 2 1

1 2 1 2 6 0 0

3 0 1 0 1 0 1

5 1 0 1 0 0 1

6 1 2 1 1 4 5

ª º« »

« »« »« »« »« »¬ ¼

.

Let us consider the transposes of a11

, a12

, a21

, a22

, a31

a

a'11 = t11

2 0 1

a 1 2 1

ª º« » « »« »

 

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a'31 = t31

2 1a 0 0

0 1

ª º« » « »« »¬ ¼

 

a'22 = t22

1 0a

1 1

ª º « »

¬ ¼

 

a'32 = t32

0 1a

4 5

ª º « »

¬ ¼.

a' = 11 21 31

12 22 32

a a aa a a

c c cª º« »c c c¬ ¼.

 Now we describe the general case. Let

a =

11 12 1m

21 22 2m

n1 n2 nm

a a aa a a

a a a

ª º« »« »« »« »¬ ¼

""

# # #

"

 

  be a n × m supermatrix. The transpose of thdenoted by

11 21 n1a a ac c cª º« »

"

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 Now we will find the transpose of a symmetricallysymmetric simple matrix. Let a be the symmetricallysymmetric simple matrix.

Let a be a m × m symmetric supermatrix i.e.,

a =

11 21 m1

12 22 m2

1m 2m mm

a a a

a a a

a a a

ª º« »« »« »« »¬ ¼

"

"# # #

"

 

the transpose of the supermatrix is given by a'

a' =

11 12 1m

12 22 2m

1m 2m mm

a (a ) (a )

a a ' (a )

a a a

c c c c cª º« »c c c« »« »« »c c c¬ ¼

"

"

# # #

"

 

The diagonal matrix a11 are symmetric matrices so ar by transposition. Hence

a'11 = a11, a'22 = a22, …, a'mm = amm.

Recall also the transpose of a transpose is the origina

Therefore(a'12)' = a12, (a'13)' = a13, …, (a'ij)' = aij.

Thus the transpose of supermatrix const i ll titi d t i i l t i

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If 

V =1

2

3

v

v

v

ª º« »« »« »¬ ¼

 

where

v1 =

3

12

ª º« »« »« »¬ ¼

, v2 =

4

57

ª º« »« »« »¬ ¼

and v3 =

5

1

ª º« »¬ ¼  

V' = [v'1 v'2 v'3].Thus if 

V =

1

2

n

vv

v

ª º« »« »« »« »¬ ¼

then

V' = [v'1 v'2 … v'n].

 Example 1.1.15: Let

t =

3 0 1 1 5 2

4 2 0 1 3 5

1 0 1 0 1 6

ª º« »

« »« »¬ ¼

 

= [T | a ]. The transpose of t

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SourceEncoder 

InformationSource

Figure 1.2: General Coding System

ChannelEncoder 

M(wr

C(SM

SourceDecoder 

DestinationChannelDecoder 

Dem(read

u v

r u

 1.2 Introduction of Linear Codes and thei

In this section we just recall the definition of enumerate a few important properties about themdescribing a simple model of a communicatiosystem given by the figure 1.2.

Messages go through the system starting fr(sender). We shall only consider senders with a fdiscrete signals (eg. Telegraph) in contrast sources (eg. Radio). In most systems the signals the source cannot be transmitted directly by thinstance, a binary channel cannot transmit worLatin alphabet. Therefore an encoder performtask of data reduction and suitably transforms thusable form. Accordingly one distinguishes bencoding the channel encoding. The former redu

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message. However this is time consuming, inefficienWe also note that the possibility of errors increasincrease in the length of messages. We want to fialgebraic methods (codes) to improve the reliabitransmission of messages. There are many types ocodes; here we give a few of them.

Throughout this book we assume that only frepresent the underlying alphabet for coding. Codingtransforming a block of k message symbols a1, a2, …into a code word x = x1 x2 … xn; xi  Fq, where n tfirst k i symbols are the message symbols i.e., xi = athe remaining n – k elements xk+1, xk+2, …, xn are cheor control symbols. Code words will be written informs x; x

1, x

2, …, x

nor (x

1x

2… x

n) or x

1x

2… x

nsymbols can be obtained from the message symbolway that the code words x satisfy a system of linear eHxT = (0) where H is the given (n – k) × n matrix wiin Fq = Z pn (q = pn). A standard form for H is (A, In–k

× k matrix and In–k , the n – k × n – k identity matrix.

We illustrate this by the following example:

 Example 1.2.1: Let us consider Z2 = {0, 1}. Take nThe message a1 a2 a3 is encoded as the code word  xx5 x6 x7. Here the check symbols x4 x5 x6 x7 are such

given matrix

4

0 1 0 1 0 0 0

1 0 1 0 1 0 0H = A;I

0 0 1 0 0 1 0

ª º« »« » « »

;

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code word x is 1 0 0 0 1 0 0. If the message a = 1x5 = 1, x6 = 1 = x7. Thus the code word x = 1 1 0

We will have altogether 23 code words given by

0 0 0 0 0 0 0 1 1 0 1 1 0 01 0 0 0 1 0 0 1 0 1 0 0 1 10 1 0 1 1 0 0 0 1 1 1 1 1 10 0 1 0 1 1 1 1 1 1 1 0 1 1

DEFINITION 1.2.1:  Let H be an n – k × n matriin Z q. The set of all n-dimensional vectors satisfover Z q is called a linear code(block code) C olength n. The matrix H is called the parity checcode C. C is also called a linear(n, k) code.

  If H is of the form(A, I n-k   ) then the k-symb

word  x is called massage(or information) symb

n – k symbols in are the check symbols. C is th

 systematic linear(n, k) code. If q = 2, then C isk/n is called transmission (or information) rate.

The set C of solutions of of HxT  = (0). i

  space of this system of equations, forms a s

  system of equations, forms a subspace of n

q Z  o

Since the code words form an additive group, C

  group code. C can also be regarded as the n

matrix H. 

  Example 1.2.2: (Repetition Code) If each codeconsists of only one message symbol a1  Z2 an

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[0 0 0], [1 0 0], [0 1 0], [0 0 1],[1 1 0], [1 0 1], [0 1 1] and [1 1 1]

x = > @

1 0 0 0 1 0 0

0 0 0 0 1 0 1 0 0 0

0 0 1 0 1 1 1

ª º« »« »« »¬ ¼

 

=> @0 0 0 0 0 0 0

x = > @

1 0 0 0 1 0 0

1 0 0 0 1 0 1 0 0 0

0 0 1 0 1 1 1

ª º« »« »« »¬ ¼

 

= > @1 0 0 0 1 0 0

x = > @

1 0 0 0 1 0 0

0 1 0 0 1 0 1 0 0 0

0 0 1 0 1 1 1

ª º« »« »« »¬ ¼

 

= > @0 1 0 1 0 0 0

x = > @

1 0 0 0 1 0 0

0 0 1 0 1 0 1 0 0 0

0 0 1 0 1 1 1

ª º« »« »« »¬ ¼

 

= > @0 0 1 0 1 1 1

1 0 0 0 1 0 0ª º« »

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=

0 0 0 0

0 0 0 00 0 0 0

ª º« »« »« »¬ ¼

.

We recall just the definition of HammingHamming weight between two vectors. This notcodes to find errors between the sent message amessage. As finding error in the received messagone of the difficult problems more so is the corrand retrieving the correct message from the recei

DEFINITION 1.2.3: The Hamming distance d(x,

vectors x = x1

x2

… xn

and y = y1

y2

… yn

inn

q

 F  i

coordinates in which x and y differ. The Hamm

of a vector x = x1 x2 … xn in n

q F is the number

ordinates in i x . In short Z (x) = d(x, 0).

We just illustrate this by a simple example.Suppose x = [1 0 1 1 1 1 0] and y [0 1 1 1 172F then D(x, y) = (x ~ y) = (1 0 1 1 1 1 0) ~ (0

(1~0, 0~1, 1~1, 1~1, 1~1, 1~0, 0~1) = (1 1 0 0 0Hamming weight Z of x is Z(x) = d(x, 0) = 5 an= 5.

DEFINITION 1.2.4: Let C be any linear code thedistance d min of a linear code C is given as

id min d( u,v ) .

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If H = (A, In–k ) be a parity check matrix in the stathen G = (Ik , –AT) is the canonical generator matrix o(n, k) code.

The check equations (A, In – k ) xT = (0) yield

1 1 1

2 2 2

n k k 

 x x a

 x a A A

 x x a

ª º ª º ª º« » « » « »« » « » « » « » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

# # #.

Thus we obtain

1 1

2 2k 

n k 

a

 x a I 

 A

 x a

ª º ª º« » « »ª º« » « »

« »« » « »¬ ¼« » « »¬ ¼ ¬ ¼

# # .

We transpose and denote this equation as

(x1 x2 … xn) = (a1 a2 … ak ) (Ik , –A7

)= (a1 a2 … ak ) G.

We have just seen that minimum distance

minu,v C

u v

d min d(u,v)

z

.

If d is the minimum distance of a linear code linear code of length n, dimension k and minimum dcalled an (n, k, d) code.

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Obviously before, this procedure is imposs but with the advent of computers one can easily

in few seconds and arrive at the result.We recall the definition of sphere of radius r. Th

nqF / d(x, y) d r} is called the sphere of radius r

In decoding we distinguish between the decorrection of error. We can say a code can corr

can detect t + s, s t 0 errors, if the structure of th possible to correct up to t errors and to detect errors which occurred during transmission over a

A mathematical criteria for this, given in theA linear code C with minimum distance dmin caerrors and can detect t + j, 0 < j d s, errors if and

dmin or equivalently we can say “A linear code C

distance d can correct t errors if and only if  t

real problem of coding theory is not merely to m  but to do so without reducing the tr

unnecessarily. Errors can be corrected by length  blocks, but this reduces the number of messagcan be sent per second. To maximize the transmwant code blocks which are numerous enough tomessage alphabet, but at the same time no necessary to achieve a given Hamming distan

main problems of coding theory is “Given blocHamming distance d, find the maximum num binary blocks of length n which are at distancesother”.

Let u = (u u u ) and v = (v v v

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vector spacen

q F  the orthogonal complement is of dim

k and an (n, n – k) code. It can be shown that if the c  generator matrix G and parity check matrix H t generator matrix H and parity check matrix G. 

Orthogonality of two codes can be expressed by GHT

 Example 1.1 .5: Let us consider the parity check m(7, 3) code where

1 0 0 1 0 0 0

0 0 1 0 1 0 0

1 1 0 0 0 1 0

1 0 1 0 0 0 1

 H 

ª º« »« »« »

« »¬ ¼

.

The code words got using H are as follows

0 0 0 0 0 0 0

1 0 0 1 0 1 10 1 0 0 0 1 0

0 0 1 0 1 0 1

1 1 0 1 0 0 1

 0 1 1 0 1 1 11 0 1 1 1 1 0

1 1 1 1 1 0 0

.

 Now for the orthogonal code the parity check matrcode happens to be generator matrix,

1 0 0 1 0 0 0ª º« »

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> @

1 0 0 1 0 0 0

0 0 1 0 1 0 01 0 0 1 1 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º« »

« » « »« »¬ ¼

= [0 0 1 1

> @

1 0 0 1 0 0 0

0 0 1 0 1 0 00 1 1 01 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º« »« »« »« »¬ ¼

= [1 1 1 0

> @

1 0 0 1 0 0 0

0 0 1 0 1 0 00 1 0 11 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º

« »« »« »« »¬ ¼

= [1 0 0 0

> @

1 0 0 1 0 0 0

0 0 1 0 1 0 00 0 1 1

1 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º« »« »« »« »¬ ¼

= [0 1 1 0

> @

1 0 0 1 0 0 0

0 0 1 0 1 0 01 1 1 0

1 1 0 0 0 1 0

1 0 1 0 0 0 1

 xª º« »« »« »« »¬ ¼

= [0 1 1 1

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> @

1 0 0 1 0 0 0

0 0 1 0 1 0 01 0 1 1 1 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º« »

« » « »« »¬ ¼

= [1 1

> @

1 0 0 1 0 0 0

0 0 1 0 1 0 00 1 1 11 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º« »« »« »« »¬ ¼

= [0 1

> @

1 0 0 1 0 0 0

0 0 1 0 1 0 01 1 1 11 1 0 0 0 1 0

1 0 1 0 0 0 1

 x

ª º

« »« »« »« »¬ ¼

= [1 1

The code words of C(7, 4) i.e., the orthogonal

are

^(0 0 0 0 0 0 0), (1 0 0 1 0 0 0), (0 0 1 0 1 0 0), (1 0 1 0 0 0 1), (1 0 1 1 1 0 0), (0 1 0 1 0 1 0), (01 1 0 1 1 0), (1 0 0 0 1 0 1), (0 1 1 0 0 1 1), (0 10 1 1 0 1), (1 1 1 1 0 1 1), (0 1 0 0 1 1 1), (1 1 0 1

Thus we have found the orthogonal code for t Now we just recall the definition of the cosets of

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 Now we give the decoding rule which is as follows:

 If a vector y is received then the possible error vthe vectors in the coset containing y. The most likely

vector  e with minimum weight in the coset of y.

decoded as  y e . [18-21]

 Now we show how to find the coset of y and d

above method. The vector of minimum weight incalled the coset leader.

 If there are several such vectors then we arbitra

one of them as coset leader. Let a(1) , a(2)  , …, a(t) bleaders. We first establish the following table

(1) (2) ( )

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

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

0 0 ½ °°¾°

°¿

!!

# # #

!

q

q

t t t q

 x x x code worda x a x a x

other cos

a x a x a xcoset leaders

 If a vector y is received then we have to find y i

 Let y = a(i) + x(j); then the decoder decides that the

the coset leader a(i). Thus y is decoded as the ( ) j x y e x . The code word  x occurs as the f

in the column of y. The coset of y can be found by eva so called syndrome.

 Let H be parity check matrix of a linear (n, k) the vector S(y) = HyT  of length n–k is called synd

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 Example 1.2.6: Let C be a (5, 3) code where tmatrix H is given by

H = 1 0 1 1 01 1 0 0 1

ª º« »¬ ¼

 

and

G =

1 0 0 1 1

0 1 0 0 1

0 0 1 1 0

ª º« »

« »« »¬ ¼

.

The code words of C are

{(0 0 0 0 0), (1 0 0 1 1), (0 1 0 0 1), (0 0 1 1 0),

1 0 1), (0 1 1 1 1), (1 1 1 0 0)}.

The corresponding coset table is

Message 000 100 010 001 110 101 code

words

00000 10011 01001 00110 11010 10101

10000 00011 11001 10110 01010 0010101000 11011 00001 01110 10010 11101

other cosets

00100 10111 01101 00010 11110 10001cosetleaders

If y = (1 1 1 1 0) is received, then y is found inthe coset leader (0 0 1 0 0)y + (0 0 1 0 0) = (1 1 1 1 0) + (0 0 1 0 0 ) = (1corresponding message

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 Example 1.2.7: Let

H =

1 0 1 1 1 1 0 0 1 0 1 1 01 1 0 1 1 1 1 0 0 1 0 0 1

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

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

ª « « « « ¬

which gives a C4(15, 11, 4) Hamming code.

Cyclic codes are codes which have been studied exteLet us consider the vector space n

qF over Fq. The

Z: nqF  o  n

qF  

where Z is a linear mapping called a “cyclic shift”…, an–1) = (an–1, a0, …, an–2)

A = (Fq[x], +, ., .) is a linear algebra in a vectorFq. We define a subspace Vn of this vector space by

Vn = {v Fq[x] / degree v < n}= {v0 + v1x + v2x2 + … + vn–1x

n–1 / vi  Fq; 0 d

We see that Vn #  nqF as both are vector spaces defin

same field Fq. Let * be an isomorphism

*(v0, v1, …, vn–1) o {v0 + v1x + v2x2 + … + vn–1xn–1}w: n

qF Fq[x] / xn – 1

i.e., w (v0, v1, …, vn–1) = v0 + v1x + … + vn–1xn–1.

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  Example 1.2.8: Let C 7

2F be defined by the gen

G =

1 1 1 0 1 0 0

0 1 1 1 0 1 0

0 0 1 1 1 0 1

ª º« » « »« »¬ ¼

The code words generated by G are ^(0 0 0 0 0 00), (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)`.

Clearly one can check the collection of all co

satisfies the rule if (a0 … a5) C then (a5 a0 … codes are cyclic. Thus we get a cyclic code.

 Now we see how the code words of the Hammlike.

 Example 1.2.9: Let 1 0 0 1 1 0 1

H 0 1 0 1 0 1 1

0 0 1 0 1 1 1

ª º« » « »« »¬ ¼

 

 be the parity check matrix of the Hamming (7, 4)

 Now we can obtain the elements of a HammWe proceed on to define parity check ma

d i b l i l t i ti i

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 Let g = g 0 + g 1 x + … + g m xm  V n , g / xn –1 and deg

  Let C be a linear (n, k) code, with k = n – m def

 generator matrix,

0 1

0 1

0 1

0 0

0 0 x=

0 0 x

m

m m

m

 g g g  

 g g g  G

 g g g  

ª º ª « » « « » « « » «

« » « ¬ ¬ ¼

! !

! !

# #

Then C is cyclic. The rows of G are linearly indeprank G = k, the dimension of C.

  Example 1.2.10: Let g = x3 + x2 + 1 be the  polynomial having a generator matrix of the cycliwith generator matrix

G =

1 0 1 1 0 0 0

0 1 0 1 1 0 0

0 0 1 0 1 1 00 0 0 1 0 1 1

ª º« »« »« »« »¬ ¼

.

The codes words associated with the generator matrix

0000000, 1011000, 0101100, 0010110, 00010111001110, 1010011, 0111010, 0100111, 00111011111111, 1000101, 0110001, 1101001.

Th i h k l i l i d fi d b

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the parity check matrix H related with the generag is given by

k 1

k k 1 0

k 1 0

0 0 h h h

0 h h h 0H

h h h 0

ª « « « « ¬

! !

!

# # # #

! !

For the generator polynomial g = x3 + x2 +1 tmatrix

0 0 1 1 1 0 1

H 0 1 1 1 0 1 01 1 1 0 1 0 0

ª º« »

« »« »¬ ¼

 

where the parity check polynomial is given by x4

7

3 2

x 1

x x 1

. It is left for the reader to verify that

matrix gives the same set of cyclic codes.

We now proceed on to give yet another ndecoding procedure using the method of best app

We just recall this definition given by >9,

give the basic concepts needed to define this nothat n

qF is a finite dimensional vector space ove

Z2 = (0, 1) the finite field of characteristic two.

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(b) ¢cD /E² = c¢D/E² (c) ¢E/D² = ¢D/E² 

(d) ¢D/D² > 0 if D z 0.

On V there is an inner product which we call the sta product. Let D = (x1, x2, …, xn) and E = (y1, y2, …, yn

¢D /E² = i ii

x y¦ .

This is called as the standard inner product. ¢D/D² isnorm and it is denoted by ||D||. We have the Graorthogonalization process which states that if V

space endowed with an inner product and if E1, E2, …set of linearly independent vectors in V; then one maa set of orthogonal vectors D1, D2, …, Dn in V such tk = 1, 2, …, n the set {D1, …, Dk } is a basis for thspanned by E1, E2, …, Ek where D1 = E1.

1 12 2 12

1

3 1 3 23 3 1 22 2

1 2

/  

/ / 

E DD E DD

E D E DD E D D

D D

& &

& & & &

 

and so on.Further it is left as an exercise for the reader to v

a vector E is a linear combination of an orthogonal t th E i th ti

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 Example 1.2.11: Let us consider the set of vectoE2 = (–1, 0, 5) and E3 = (1, 9, 2) in the space R

the standard inner product.Define D1 = (2, 0, 3)

2

( 1, 0, 5) /(2, 0, 3)( 1, 0, 5) (2

13

D

13( 1, 0, 5) 2, 0, 313 = (–3, 0,

3

( 1, 9, 2) /(2, 0, 3)(1,9,2) (2,

13

D

(1, 9, 2) /( 3, 0, 2)( 3,0,2)

13

 

=8 1

(1,9,2) (2,0,3) ( 3,0,213 13

16 24 3 2(1,9,2) , 0, , 0,

13 13 13 13§ · § ¨ ¸ ¨ © ¹ ©

16 3 24 2(1,9,2) , 0,13 13- ½ § · ® ¾¨ ¸© ¹¯ ¿

= (1, 9, 2) – (1, 0, 2)= (0, 9, 0).

Clearly the set ^(2, 0, 3), (–3, 0, 2), (0, 9, 0)` i

set of vectors.  Now we proceed on to define the no

approximation to a vector  E in V by vectors ofwhere E W Suppose W is a subspace of an

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to be a vector  D in W such that E –  D is pe(orthogonal) to W and that there ought to be exactly

These intuitive ideas are correct for some finite dsubspaces, but not for all infinite dimensional subspa

We just enumerate some of the properties relateapproximation.

Let W be a subspace of an inner product space  be a vector in V.

(i) The vector  D in W is a best approximatvectors in W if and only if  E – D is orevery vector in W.

(ii) If a best approximation to E by vectors iit is unique.

(iii) If W is finite-dimensional and {D1, D2any orthonormal basis for W, then the ve

k k 2

k  k 

/E DD D

D¦& &

, where D is the (

approximation to E by vectors in W. Now this notion of best approximation for the first t

in coding theory to find the best approximated senreceiving a message which is not in the set of cFurther we use for coding theory only finite fields Ff . If C is a code of length n; C is a vector space ov#  k 

qF    nqF , k the number of message symbols in th

C is a C(n, k) code. While defining the notion of inon vector spaces over finite fields we see all axio product defined over fields as reals or complex in getrue. The main property which is not true is if 0 z

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  be divided by the pseudo inner product in the sufinding the pseudo best approximation.

 Now first we illustrate the pseudo inner product by a

 Example 1.2.12: Let V = Z2 × Z2 × Z2 × Z2 be a vover Z2. Define ¢,² p to be the standard pseudo innerV; so if x = (1 0 1 1) and y = (1 1 1 1) are in V then

inner product of ¢x, y² p = ¢(1 0 1 1 ), (1 1 1 1)² p = 1 + 0 + 1 + 1

 Now consider ¢x, x² p = ¢(1 0 1 1), (1 0 1 1)² p = 1 + 0 + 1 + 1

 but¢y, y² p = ¢(1 1 1 1), (1 1 1 1)² p = 1 + 1 + 1 + 1

We see clearly y z 0, yet the pseudo inner product is

 Now having seen an example of the pseudo inner  proceed on to illustrate by an example the notion of approximation.

 Example 1.2.13: LetV = 8

2 2 2 2

8 times

Z Z Z Z u u u!

 

 be a vector space over Z2. Now

W = ^0 0 0 0 0 0 0 0), (1 0 0 0 1 0 11), (0 1 0 0 1 1 00 1 1 1), (0 0 0 1 1 1 0 1), (1 1 0 0 0 0 1 0), (0 1 1 00 1 1 1 0 1 0), (0 1 0 1 0 1 0 0), (1 0 1 0 1 1 0 0), (1 0), (1 1 1 0 0 1 0 1), (0 1 1 1 0 0 1 1), (1 1 0 1 1 1 1

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 Suppose E = (1 1 1 1 1 1 1 1) is a vector in V us

approximations find a vector in W close to E. Threlative to the basis B of W where

4

k k  pk 1

,

D E D D¦  

= ¢(1 1 1 1 1 1 1 1), (0 1 0 0 1 0 0 1)² p D1 +¢(1 1 1 1 1 1 1 1), (1 1 0 0 0 0 1 0)² p D2 +¢(1 1 1 1 1 1 1 1), (1 1 1 0 0 1 0 1)² p D3 +¢(1 1 1 1 1 1 1 1), (1 1 1 1 1 0 0 0)² p D4.

= 1.D1 + 1.D2 + 1.D3 + 1.D4.

= (0 1 0 0 1 0 0 1) + (1 1 0 0 0 0 1 0) + (1 1 1 01 1 1 0 0 0)= (1 0 0 1 0 1 1 0) W.

 Now having illustrated how the pseudo best appvector E in V relative to a subspace W of V is d

we illustrate how the approximately the nearesobtained.

  Example 1.2.14: Let C = C(4, 2) be a code ob parity check matrix

1 0 1 0H1 1 0 1

ª º « »¬ ¼

.

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2

k k  pk 1

,

D E D D¦ = ¢(1 1 1 1), (0 1 0 1)² pD1 

+ ¢(1 1 1 1), (1 0 1 1)² pD2.

= (1 0 1 1) .

Thus the approximated code word is (1 0 1 1).

  Now having seen some simple properties of cod  proceed on to define super special vector spacesalgebraic structure basically makes used of supermmore super linear algebra please refer [10, 21].

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Chapter Two

SUPER SPECIAL VECTOR SPACE

In this chapter we for the first time define a newspecial vector spaces. We describe mainly tessential for us to define super special codes and

like decoding, etc. Throughout this chapter V dspace over a field F. F may be a finite characterinfinite characteristic field.

Th l V

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…, n. Thus any element v s   V  s is a super row

entries from the field F. If v s , w s   V  s the sum v s + w

to be

! ! 1 1 1 1 1 1 2 2 2 2 2

1 1 2 2 m m 1 1 2 2 m[ v w v w v w v w v w v

!n n n n n n

1 1 2 2 m mv w v w v w ]

where

! ! ! !1 1 1 2 2 2 n n

 s 1 2 m 1 2 m 1 2w [ w w w w w w w w Also

! ! ! 1 1 1 2 2 2 n

 s 1 2 m 1 2 m 1av [ av av av av av av av av

a  F.

We illustrate this by a simple example.  Example 2.1: Let Vs = [V1 | V2 | V3] where eachdimensional vector space over Q, the field of ratielement Qs Vs would be of the form

1 1 1 2 2 2 3 3 3s 1 2 3 1 2 3 1 2 3a a a | a a a | a a aª ºQ ¬ ¼

is a super row vector; where i ja Q; 1 d i, j d 3. Cl

super special finite dimensional super vector space o

Bs = {[0 0 1 | 0 0 1 | 0 0 1], [0 0 1 | 0 0 1 | 0 1[0 0 1 | 0 0 1 | 1 0 0], [0 0 1 | 0 1 0 | 0 0 1]

[0 0 1 | 0 1 0 | 0 1 0], [0 0 1 | 0 1 0 | 1 0 0][0 0 1 | 1 0 0 | 0 0 1], [0 0 1 | 1 0 0 | 1 1 0][0 0 1 | 1 0 0 | 1 0 0], [0 1 0 | 1 0 0 | 0 0 1][0 1 0 | 0 0 1 | 0 0 1], [0 1 0 | 0 0 1 | 0 1 0]

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forms a super basis of Vs. Clearly this Bs will g

Q. It is easily verified Bs Vs and the elemlinearly independent set in Vs.

We have seen a super special finite dimensionover Q. Now we proceed on to define the super a super special vector space over F.

DEFINITION 2.2: Let V  s = [V 1 | V 2 | … | V n ] be vector space where each V i is of dimension m o

 Let  ; d!1 2 t n

 s s s s B { v ,v , ,v 0 t m } be the eleme

each t 

 sv is a super row vector 1 d  p d  t. We s

 forms a super linearly independent set if, D 1  , …in F such that 

!1 2 t 

1 s 2 s t sv v vD D D  = (0 …0 | 0 …0 | …|

then each D i = 0. If equation I is true for some n

D 1 , …, D t  in F then we say !1 2 t 

 s s sv ,v , , v forms a

dependent set.  If B s forms a super linearly independent selement in V  s can be expressed as a super line

 from the set B s in a unique way then we call B s

 super basis of V  s or super special basis of V  s. Ifelements in B s is finite then we call V  s to be a fin

  super special vector space otherwise an infin super special vector space.

  Example 2.2: Let Vs = [V1 | V2 | V3] where eadi i l t Q i 1 2 3 b t

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m z m for at least one i z j for i = 1 2 n 1

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mi z m j for at least one i z  j for i = 1, 2, …, n, 1is defined to be a super special mixed dimension

 Example 2.4: Let Vs = [M2 × 2 | Q × Q × Q × Q | Q[x] and Q[x] is the set of all polynomials of degequal to 3 with coefficients from Q. We see Mvector space of dimension 4 and V3 is also a dimension 4.

Thus we see M2 × 2 # Q × Q × Q × Q; V2 #and V3 # Q × Q × Q × Q. So any vs Vs will be o

1 1 1 1 2 2 2 2 3 3 31 2 3 4 1 2 3 4 1 2 3q q q q | q q q q | q q qª ¬

where i jq Q with i = 1, 2, 3, 4.

Let us consider Bs = 1 2 3 1 2 3 1 2 3

1 1 1 1 1 2 1 1 3{[v v v ],[v v v ],[v v v ], ,[v!

where

1 1 1 11 2 3 4

1 0 0 1 0 0v , v , v , v

0 0 0 0 0 1

ª º ª º ª º « » « » « »

¬ ¼ ¬ ¼ ¬ ¼

21v = [0 0 0 1], 2

2v = [0 0 1 0], 23v = [0 1 0 0], v

31v = 1, 3

2v = x, 33v = x2 and 3

4v = x3. Clearl

Bs Vs and Bs is a special super basis of Vs over

We may have any other basis for Vs but it is truvector spaces even in special super vector spaceelements in each and every superbasis is only f

Example 2 5: Let V = [V1 | V2 | V3 | V4] be a sp

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  Example 2.5: Let Vs [V1 | V2 | V3 | V4] be a spmixed dimension vector space over Q, where V1 = Q

a vector space of dimension 3 over Q, V2 = M3 × 2; the× 2 matrices with entries from Q is a vector space o6 over Q, V3 = Q[x] ; the set of all polynomials of than or equal to four and V4 = P2 × 2 the set of all 2 ×with entries from Q. V4 is a vector space of dimensioThus Vs is a super special mixed dimensional vector

Q. Any element vs Vs is a super mixed row vector g

1 1 1 2 2 2 2 2 2s 1 2 3 1 2 3 4 5 6v {[v v v | v v v v v v |

3 3 3 3 3 4 4 4 41 2 3 4 5 1 2 3 4v v v v v | v v v v ]}

where i jv Vi ; 1 d j d 3, 4, 5 or 6 and 1 d i d 4.A super special subspace of Vs can be either a su

subspace or it can also be a super special mixedsubspace.

Now it is important at this point to mention ev

special vector space can have a super special mixedsubspace also a super special mixed dimension vectohave a super special vector subspace.

We illustrate this situation now by the following

 Example 2.6: Let Vs = [V1 | V2 | V3] where V1 is the

× 3 matrices with entries from Q, V2 is the set of alvector with entries from Q and V3 is the collec  polynomials of degree less than or equal tocoefficients from Q. All the three vector spaces V1

f di i 6 Q V i i l

W2 = [[x1 0 x2 0 x3 0] | x1 x2 x3 Q

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W2 [[x1 0 x2 0 x3 0] | x1, x2 , x3 Q# Q u {0} u Q u {0} u Q u {0},

W2 is a subspace of V2 of dimension 3. Lcollection of all polynomials of even degree (i.e4) with coefficients from Q}. W3 is a subsdimension 3 over Q.

Clearly Ws is a special super subvector spadimension or Ws is a super special mixed dimensVs but Vs is not a super special mixed dimensionis only a super special vector space over Q.Consider Ts = [T1 | T2 | T3] a proper subset of Vs,

1 2 31 1 2 3

a a aT a , a , a Q

0 0 0- ª º° ® « »

¬ ¼° ¯

a proper subspace of V1 of dimension three.

T2 = {[a1 a2 a3 0 0 0] | a1, a2, a3 Q# Q × Q × Q × {0} × {0} × {0}

is a proper subspace of V2 of dimension 3 over  polynomials of degree 1 and 3 with coefficients+ a1x + a2x

3] | a0, a1, a2 Q}. T3 is a sub

dimension three over Q. Thus Ts is a super spspace of Vs over Q and the dimension of each T2, 3.

dimension four over Q and V3 = {M2×2 = (aij) | aij

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dimension four over Q and V3 {M2×2 (aij) | aij 2} a space of dimension four over Q.

 Now let Ws = [W1 | W2 | W3] where W1 is a diagonthe form

a 0 0

0 b 0 a, b, c Q

0 0 c

- ½ª º° °« » ® ¾« »° °

« »¬ ¼¯ ¿

which is proper subspace of V1 and of dimension 3,

W2 = {[a b c 0] | a, b, c, 0 Q} # Q u Q u Q u

is a subspace of V2 of dimension three over Q and

3

a bW a, b, c Q

0 c

- ½ª º° ° ® ¾« »

¬ ¼° °¯ ¿

is the proper subspace of dimension three of V3. Ws

W3] is a super special subspace of Vs. Clearly Ws isspecial mixed dimensional subspace of Vs.

Let R s = [R 1 | R 2 | R 3], a proper subset of Vs, wherof all 3 × 3 upper triangular matrices} i.e.,

1

a b c

R 0 d e a, b, c, d, e, f Q

0 0 f

- ½ª º° °« » ® ¾« »° °« »¬ ¼¯ ¿

.

a 0- ½ª º° °

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3

a 0R a, b Q

0 b

- ½ª º° ° ® ¾« »

° °¬ ¼¯ ¿

;

R 3 is a subspace of V3 of dimension 2 over Q. super special mixed dimension subspace of Vs.

We mention here only those factors about th

vector spaces which are essential for the study aof super special codes.

  Now we proceed on to define dot product ovector spaces.

DEFINITION 2.5:   Let V  s be a real super speciover the field of reals F. A super special inner pinner product on V  s is a function which assigns

 pair of super row vectors D  s , E  s in V  s a scalar in

that for D  s , E  s , J  s in V  s and for all scalars c in F w

1. ( D  s +  E  s / J  s ) = ( D  s / J  s ) + (  E  s / J  s ) where

( D  s / J  s ) = ª « ¬ 1 2

1 1 1 2 2 2 n n

1 2 m 1 2 m 1 2v v ...v v v ...v ... v v

ª ¬ 1 2

1 1 1 2 2 2 n n

1 2 m 1 2 m 1 2w w ...w w w ...w ... w w ...

1 1

1 1 1 1 1 1 2 2 2 2

1 1 2 2 m m 1 1 2 2v w v w ... v w v w v w .

n n n n n n

1 1 2 2 m mv w v w ... v w

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[0 1 0 | 0 1 0 | 0 1 0], [0 1 0 | 0 0 1 | 1 0 0]

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[0 1 0 | 0 0 1 | 0 1 0], [0 1 0 | 0 0 1 | 0 0 1]

[0 0 1 | 1 0 0 | 0 0 1], [0 0 1| 1 0 0 | 0 1 0],[0 0 1 | 1 0 0 | 1 0 0], [0 0 1 | 0 1 0 | 0 0 1][0 0 1| 0 1 0 | 0 1 0], [0 0 1 | 0 1 0 | 1 0 0],[0 0 1 | 0 0 1 | 1 0 0], [0 0 1 | 0 0 1 | 0 1 0]

and [0 0 1 | 0 0 1 | 0 0 1]}.

Bs is a super special basis of the super special vectoLetTs = {[1 0 0 | 0 1 0 | 1 0 0], [1 0 0 | 0 1 0 | 0 1

[1 0 0 | 0 0 1 | 1 0 0], [1 0 0 | 0 1 0 | 0 1 0][0 0 1 | 0 1 0 | 1 0 0], [0 0 1 | 0 1 0 | 0 1 0]

[0 0 1 | 0 0 1 | 1 0 0] and [0 0 1 | 0 0 1 | 0 1 0

Ts is a super special basis of Ws and the number of Ts is 8.

Let us take R s = [R 1 | R 2 | R 3] where R 1 = {0} u Z2

{0} × {0} × Z2 and R 3 = {0} × Z2 × Z2. Clearly R

special mixed dimension subspace of Vs. Let

Ms = {[0 0 1 | 0 0 1 | 0 0 1], [0 0 1 | 0 0 1| 0 1[0 1 0 | 0 0 1 | 0 0 1] and [0 1 0 | 0 0 1 | 0 1 0

Ms is a super special basis of R s. We see R s is a s

dimension 4. Further we see all the super special baare only super row vectors.

We give yet another example of a super spe

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 Now the following factors are easily verified to be t[V | V | | V ] i i l i d di

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[V1 | V2 | … | Vn] is a super special mixed dimen

space over a field F and if Vi is of dimension ni over…, n, then the dimension of Vs = dimension of V1 ×of V2 × … × dimension of Vn = n1 × n2 × … × nn.

In the next chapter we define the notion of super spwhich are built mainly using these super special vect

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  partitions are done vertically between the m andcolumn 2m and (2m + 1)th column and so on and las

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column, 2m and (2m + 1) column and so on and las

and {(r – 1)m +1}th

column.For example

1 2 3 1 4 5 6 7 1 0 0 1

4 5 6 2 8 9 0 1 2 1 0 3

1 0 1 3 1 1 0 0 1 1 0 2

ª « « « ¬

is a super row vector or matrix, here r = 3, n = 3 andA super mixed row vector or matrix V = [V1 | V2

a super matrix such that each Vi is a n × mi matrix mleast one i z j, 1 d i, j d s i.e., V is a n × (m1 + … +

where vertical partitions are made between m1 andcolumn of V, m2 and (m2 + 1)th column and so partitioned between the ms–1 and (ms–1 + 1)th column.

For example

1 0 1 1 0 1 1 7 2 1 2 3

2 1 2 0 5 0 3 5 1 0 1 0V

3 1 1 1 0 1 1 2 3 1 1 1

4 2 0 0 1 0 4 5 6 0 0 1

ª

« « « « ¬

where n = 4, m1 = 2, m2 = 4, m3 = 3 and m4 = 5

Clearly V is a super mixed row matrix or vector. Forchapter one of this book.

Now we proceed on to define the new class of super

code word is the same i.e., the number of messathe length of the code word may not be the sa

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the length of the code word may not be the sa

code word consisting of n code words can be r super row vector;

ª ¬ ! ! ! 1 2

1 1 1 2 2 2 n n

 s 1 2 n 1 2 n 1 2 x x x x x x x x x

ni > k i , 1 d  i d  n. In this super row vector 

i

 j x a ,1, 2, …, n and the remaining ni – k i elements

are check symbols or control symbols; i = 1, 2, …

These n code words denoted collectively by xs wthe super special row code word.

As in case of usual code, the check symbols in such a way that the super special code wosuper system of linear equations; T

s sH x 0

super mixed row matrix given by Hs = [H1 | H2

each Hi is a m × ni matrix with elements from Fq

i.e.,T

s sH x = [H1 | H2 | … | Hn]1 2s sx x xª ¬ !

= T T1 2 n

1 s 2 s 2 sH x H x H xª « ¬

!

= [|(0) | (0) | … | (0)]

i.e., each Hi is the partity check matrix of the co= 1, 2, …, n. Hs = [H1 | H2 | … | Hs] will be knospecial parity check super special matrix of th

d C C ill l b k th li

Cs is then also called a systematic linear ((n1 n2 … nkn)) super special code

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k n)) super special code.

If q = 2 then Cs is a super special binary row co+ k n) | (n1 + n2 + … + nn) is called the super transuper information) rate.

It is important and interesting to note the set Cs

xs of  Ts sH x = (0) i.e., known as the super solution s

super system of equations. Clearly this will form

special subspace of the super special vector space super special dimension (k 1 k 2 … k n).

Cs being a super special subspace can be realigroup under addition known as the super special gwhere Hs is represented in the form given in equatioknown as the standard form.

  Now we will illustrate this super special row codexamples.

  Example 3.1.1: Suppose we have a super special code given by the super special parity check matrix

H2 | H3] where

1

0 1 1 1 0 0

H 1 0 1 0 1 0 ,

1 1 0 0 0 1

ª º« » « »« »¬ ¼

2

0 0 0 1 1 0 0H 0 1 1 0 0 1 0

1 1 0 1 0 0 1

ª º« » « »« »¬ ¼

0 1 1 1 0 0 0 0 0 1

H 1 0 1 0 1 0 0 1 1 0

ª«

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sH 1 0 1 0 1 0 0 1 1 0

1 1 0 0 0 1 1 1 0 1

«

««¬1 1 0 0 0 1 0 0

0 0 1 1 0 0 1 0

1 0 1 0 1 0 0 1

º»»»¼

= [(A1, I3) | (A2, I3) | (A3, I3)] ;

i.e., we have given the super special code whichcode. The super special code words are given by

1 1 1 1 1 1 2 2 2 2 2s 1 2 3 4 5 6 1 2 3 4 5x a a a x x x | a a a a x ª¬

3 3 3 3 3 3 3 31 2 3 4 5 6 7 8a a a a a x x x º¼ = 1 2

s sx xª ¬ T

s sH x = (0) gives 3 sets of super linear equation

(0) is read as

[H1 | H2 | H3] TT1 2 3 1 2

s s s 1 s 2 sx | x | x H x H xª ª º ¬ ¼ « ¬ = [(0) | (0) | (0)];

i.e.,

T1

1 sH x = (0)

gives 1 1 12 3 4a a x 0

1 1 11 3 5a a x 0

1 1 1 0

2 24 5a x 0

2 2 2 0

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2 2 22 3 6a a x 0

2 2 2 21 2 4 7a a a x 0 .

The set of codewords of the super special row subco by

{0 0 0 0 0 0 0, 1 0 0 0 0 0 1, 0 1 0 0 0 1 1, 0 0 1

0 0 0 1 1 0 1, 1 1 0 0 0 1 0, 0 1 1 0 0 0 1, 0 0 1 11 0 1 0 0 1 1, 1 0 0 1 1 0 0, 0 1 0 1 1 1 0, 1 1 1 00 1 1 1 1 0 0, 1 1 0 1 1 1 1, 1 0 1 1 1 1 0, 1 1 1 1 1 0

 Now the sublinear equation T3

3 sH x = (0) gives

3 3 31 2 6a a x 0 3 3 33 4 7a a x 0

3 3 3 31 3 5 8a a a x 0 .

{0 0 0 0 0 0 0 0, 1 0 0 0 0 1 0 1, 0 1 0 0 0 1 0 1, 0 0

0 0 0 1 0 01 0, 0 0 0 0 1 0 0 1, 1 1 0 0 0 0 0 1, 0 1 10 0 1 1 0 0 0 1, 0 0 0 1 1 0 1 1, 1 0 1 0 0 1 1 0, 1 0 01 0 0 0 1 1 0 1, 0 1 0 1 0 1 1 1, 0 1 0 0 1 1 0 1, 0 0 11 1 1 0 0 0 1 0, 0 1 1 1 0 1 0 1, 0 0 1 1 1 0 0 0, 1 1 01 1 0 0 1 0 0 0, 1 0 1 1 0 1 0 0, 1 0 0 1 1 1 1 0, 0 1 10 1 0 1 1 1 1 1, 1 0 1 0 1 1 1 0, 1 1 1 1 0 0 0 0, 1 1 1

0 1 1 1 1 1 0 0, 1 1 0 1 1 0 1 0, 1 0 1 1 1 1 0 1, 1 1 1 3sC .

3

The super transmission rate is 12/21. Thus this cadvantages which will be enumerated in the last

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g

  book. We give yet another example of super which every super special code word is a super not a super mixed row vector.

 Example 3.1.2: Let Hs = [H1 | H2 | H3] be the supcheck super matrix associated with the super spe

Here

1

0 1 1 0 1 0 0 0

1 0 0 1 0 1 0 0H

1 1 1 0 0 0 1 0

1 0 0 0 0 0 0 1

ª º« »« »« »« »

¬ ¼

2

1 1 0 0 1 0 0 0

1 1 1 0 0 1 0 0H

0 1 1 0 0 0 1 0

0 1 0 1 0 0 0 1

ª º« « « « ¬ ¼

and

3

0 1 1 1 1 0 0 0

0 0 1 0 0 1 0 0H

0 0 1 1 0 0 1 0

1 0 1 0 0 0 0 1

ª º« »« »« »

« »¬ ¼

i.e.,

0 1 1 1 1 0 0 0

0 0 1 0 0 1 0 0

º»»

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0 0 1 0 0 1 0 0

0 0 1 1 0 0 1 0

1 0 1 0 0 0 0 1

»»»»¼

is the super special parity check matrix of the super sCs. Now the super special system of equations is give

Ts sH x = (0) i.e.,

T31 21 2 3 s s s

H H H x x xª ºª º ¬ ¼ ¬ ¼ [(0) | (0) | (0)

T T T1 2 31 s 2 s 3 sH x H x H xª º

« »¬ ¼.

We call the linear equations given by Ti

i sH x

subequations of the super linear equations. Now th

equations given by T1

1 sH x = (0) is

1 1 12 3 1a a x 0 1 1 11 4 2a a x 0

1 1 1 11 2 3 3a a a x 0

1 11 4a x 0 ;

where 1 1 1 11 2 3 4a a a a is the set of message sy

 Now we use the sublinear equation of the super

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got from Ts sH x = (0), we get T

22 sH x = (0). Thi

2 2 21 2 1a a x 0

2 2 2 21 2 3 2a a a x 0

2 2 22 3 3a a x 0 2 2 22 4 4a a x 0 .

The super special subcode 2sC associated with t

equations is given by :

2sC = {0 0 0 0 0 0 0 0, 1 0 0 0 1 1 0 0, 0 1 0

0 0 1 0 0 1 1 0, 0 0 0 1 0 0 0 1, 1 1 0 0 00 1 1 0 1 0 0 1, 0 0 1 1 0 1 1 1, 1 0 1 0 11 0 0 1 1 1 0 1, 0 1 0 1 1 1 1 0, 1 1 1 0 00 1 1 1 1 0 0 0, 1 1 0 1 0 0 1 1, 1 0 1 1 1

1 1 1 1 0 1 0 0}.

 Now using the sublinear equation T3

3 sH x = (0)

3 3 3 32 3 4 1a a a x 0

2 33 2a x 0

3 3 33 4 3a a x 0 3 3 31 3 4a a x 0 .

Thus the super special code word of the super spec

will be 1 2 3s s s sC C C Cª º ¬ ¼ i.e., it is formed by takin

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s s s s¬ ¼ y

word from each one of the isC ; i = 1, 2, 3. Thus if x

any xs = [1 1 1 1 0 0 1 1 | 1 1 1 1 0 1 0 0| 1 1 1 1 1 1 0

So we can realize the super special row code to b

which each and every subcode isC of Cs have the sa

of check symbols.

 Now we proceed on to define the notion of super repetition code.

DEFINITION

3.1.2:  Let ª º

¬ ¼!

1 2 n

 s s s sC C C C  special row code in which each of the

i

 sC  is a repet

= 1, 2, …, n, then we define C  s to be a super speciarow code. Here if H  s = [H 1|H 2| …|H n ] is the super sp

check matrix of C  s , then each H i is a t – 1 u  t matri

have

 H 1 = H 2 = … = H n

u

ª º« »« »« »« »¬ ¼

"

"

# # # #

!t 1

1 1 0 0

1 0 1 0

1 0 0 1

is the parity check matrix. The super special cassociated with C  s are just super row vectors only anmixed row vectors The number of super special cod

1 1 0 0 0 0 1 1 0 0ª

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1 0 1 0 0 0 1 0 1 01 0 0 1 0 0 1 0 0 1

1 0 0 0 1 0 1 0 0 0

1 0 0 0 0 1 1 0 0 0

««««««¬

1 1 0 0 0 0 1 1 0 0 0

1 0 1 0 0 0 1 0 1 0 0

1 0 0 1 0 0 1 0 0 1 0

1 0 0 0 1 0 1 0 0 0 1

1 0 0 0 0 1 1 0 0 0 0Thus

Cs = {[0 0 0 0 0 0 | 0 0 0 0 0 0 | 0 0 0 0 0 0 | 0[1 1 1 1 1 1 | 1 1 1 1 1 1 | 1 1 1 1 1 1 | 1 1 [0 0 0 0 0 0 | 1 1 1 1 1 1 | 0 0 0 0 0 0 | 1 1

[0 0 0 0 0 0 | 1 1 1 1 1 1 | 0 0 0 0 0 0 | 0 0 [0 0 0 0 0 0 | 0 0 0 0 0 0 | 0 0 0 0 0 0 | 1 1 [1 1 1 1 1 1 | 0 0 0 0 0 0 | 0 0 0 0 0 0 | 0 0 [0 0 0 0 0 0 | 0 0 0 0 0 0 | 1 1 1 1 1 1 | 0 0 [1 1 1 1 1 1 | 1 1 1 1 1 1 | 0 0 0 0 0 0 | 0 0 [1 1 1 1 1 1 | 0 0 0 0 0 0 | 1 1 1 1 1 1 | 0 0

[1 1 1 1 1 1 | 0 0 0 0 0 0 | 0 0 0 0 0 0 | 1 1 [0 0 0 0 0 0 | 1 1 1 1 1 1 | 1 1 1 1 1 1 | 0 0 [0 0 0 0 0 0 | 0 0 0 0 0 0 | 1 1 1 1 1 1 | 1 1 [1 1 1 1 1 1 | 0 0 0 0 0 0 | 1 1 1 1 1 1 | 0 0

DEFINITION 3.1.3: Let C  s be a super special parity c1 2

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row code i.e., ª º ¬ ¼!1 2 n

 s s s sC C C C where C s

using the super special mixed row matrix H  s = [H 1 | Hwhere each H i is a unit row vector having t i number

i.e.,

times times tim

ª « ¬

" " ! "

1 2 n

 st t t 

1 1 1 1 1 1 1 1 H 

where at least one t i z t  j for i z j. Any super special cC  s would be of the form

ª º ª « « » ¬ ¬ ¼1 2 n

1 1 1 2 2 2 n n n 1

 s 1 2 t 1 2 t 1 2 t s s x x ...x x x ... x ... x x ...x x x

with T  s s H x = (0); i.e., each i

 s would contain only eof ones and the rest are zeros.

Cs = [C1 | C2 | … | Cn] is defined to be super special prow code. Cs is obtained from the parity check rovector Hs = [H1 | H2 | … | Hn ] where H1 = H2

m times

1 1 1ª º

« »« »¬ ¼

"

. Here a super special codeword in C

a super row vector of the form 1 2 ns s sx x xª º¬ ¼!

i i i is 1 2 mx x x ... xª º ¬ ¼ where only even number of  i

 jx a

the rest zero, 1 d j d m and i = 1, 2, …, n.

  Now we will illustrate the two types of super sp

[0 0 0 | 0 0 0 | 1 0 1], [0 0 0 | 0 0 0 | 0 1 1], [0 0 0 [0 0 0 | 1 1 0 | 1 1 0], [0 0 0 | 1 1 0 | 1 0 1], [0 0 0

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[0 0 0 | 0 1 1 | 0 0 0], [0 0 0 | 0 1 1 | 1 0 1], [0 0 0 [0 0 0 | 0 1 1 | 0 1 1], [0 0 0 | 1 0 1 | 1 1 0], [0 0 0 [0 0 0 | 1 0 1 | 0 1 1], [0 0 0 | 1 0 1 | 1 0 1], [1 1 0 [1 1 0 | 0 0 0 | 1 0 1], [1 1 0 | 0 0 0 | 1 1 0], [1 1 0 [1 1 0 | 1 1 0 | 0 0 0], [1 1 0 | 1 1 0 | 0 1 1], [1 1 0 [1 1 0 | 1 1 0 | 1 1 0], [1 1 0 | 1 0 1 | 0 0 0], [1 1 0

[1 1 0 | 1 0 1 | 0 1 1], [1 1 0 | 1 0 1 | 1 1 0], [1 1 0 [1 1 0 | 0 1 1 | 1 1 0], [1 1 0 | 0 1 1 | 1 0 1], [1 1 0 [0 1 1 | 0 0 0 | 0 0 0], [0 1 1 | 0 0 0 | 0 1 1], [0 1 1 [0 1 1 | 0 0 0 | 1 0 1], [0 1 1 | 1 1 0 | 0 0 0], [0 1 1 [0 1 1 | 1 1 0 | 0 1 1], [0 1 1 | 1 1 0 | 1 0 1], [0 1 1 [0 1 1 | 0 1 1 | 1 0 1], [0 1 1 | 0 1 1 | 1 1 0], [0 1 1

[0 1 1 | 1 0 1 | 0 0 0], [0 1 1 | 1 0 1 | 0 1 1], [0 1 1 [0 1 1 | 1 0 1 | 1 1 0], [1 0 1 | 0 0 0 | 0 0 0], [1 0 1 [1 0 1 | 0 0 0 | 1 1 0], [1 0 1 | 0 0 0 | 1 0 1], [1 0 1 [1 0 1 | 0 1 1 | 1 1 0], [1 0 1 | 0 1 1 | 1 0 1], [1 0 1 [1 0 1 | 1 0 1 | 0 0 0], [1 0 1 | 1 0 1 | 1 0 1], [1 0 1 [1 0 1 | 1 0 1 | 1 1 0], [1 0 1 | 1 1 0 | 0 0 0], [1 0 1

[1 0 1 | 1 1 0 | 0 1 1],[1 0 1 | 1 1 0 | 1 0 1]}.Thus |Cs| = 43 = 64. We see in every super number of non zero ones is even.

  Next we give an example of a super specrow code Cs.

  Example 3.1.5: Let 1 2s s sC C C ª º¬ ¼ be a supe

check mixed code with the associated super speci d t H [H | H ] [1 1 1 1 | 1

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Let xs = 1 2 3 4s s s sx x x xª º¬ ¼ = [1 1 0 1 0 1 0 | 1

1 1 0 1 0 | 1 1 1 1 0 0 1 0 0] Cs be the sent sup

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Suppose ys = 1 2 3 4s s s sy y y yª º¬ ¼ = [1 0 1 0 1 1

0| 1 1 0 1 0 1| 1 1 1 0 1 0 1 0 0] be the receiword. The super error vector is given by

ys – xs = 1 2 3 4s s s sy y y yª º¬ ¼ –  1 2 3

s s sx x xª ¬

= 4 41 1 2 2 3 3s ss s s s s s y xy x y x y xª º ¬ ¼

= [0 1 1 1 1 0 0 | 0 0 0 0 1 1 0 0 | 0 0 1 0 0 0 0]

= 1 2 3 4s s s se e e eª º¬ ¼

= es.

Clearly ys + es = xs.

DEFINITION 3.1.5: The super Hamming dista

between two super row vectors of the super spec

V  s  , where x s =1 2 n

 s s s x xª º¬ ¼! and y s =1

 s yª ¬ is the number of coordinates in which i

 s x and  y

1, 2, …, n. The super Hamming weight w s(x s ) of t

 x s = 1 2 n

 s s s x x xª º¬ ¼! in V  s is the numbe

coordinates in each i

 s  x ; i = 1, 2, …, n. In sho

(0)).

As in case of usual linear codes we define s

smind =

s s su , Cmin

Q ds (us, Qs)

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s su zQ

s s s

s s

1 2 n 1 2 ns s s s s s su , C

u

min d ,u u u v v vQ

zQ

ª º ª º ¬ ¼ ¬ ¼! !

1 1 2 2 n ns s s s s smin d u , d u , ... d u ,ª Q Q Q¬

 Now smind of the super special row code given in ex

is 7 verified using the fact in 1 2 3s s s sC C C C ª º¬ ¼ ;

s 2 s 3min s min sd C 2 and d C 2 . Hence s

min sd C = 3 + 2

we will denote smin s s sd min d u , Q by s

min sd C ,

us z Qs.

S 1 2 nmin s s sd C C Cª º¬ ¼! S 1 2min s sd C C ...ª ¬

1 1 1 2 2 2 n n ns s s s s ss s s

1 1 2 2 n ns s s s s s

1 1 2 2s s s s

x ,y C x ,y C x ,y C

x y x y x y

min d x , y min d x , y ... min d

z z z

 Now we proceed on to define the dual of a super code.

DEFINITION 3.1.6:  Let  ª º¬ ¼!1 2 n s  s s s

C  C C C 

  special row [(n1  , …, nn  ), (k 1  , …, k n  )] binary code special dual row code of C  s denoted by

A A AA ª º ¬ ¼!1 2 n s  s s s

C  C C C 

where A

i

 sC  = { i

 su | i i

 s su 0Q  for all  i i

 s sC Q   }, i = 1

can say the super special dual code would be def

if ni = 2k i and such that n1 = n2 = … = nn.

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We can define the new notion of super specisuper code words of a super special row analogous to syndrome of the usual codes.

DEFINITION 3.1.7: Let C  s be a super special ro

be the associated super special parity check m  super special syndrome of any element y s   V 

  super special subspace of the super special vec

 given by S(y s ) =T 

 s s H y . S(y s ) = (0) if and only if y

Thus this gives us a condition to find oreceived super code word is a right message or

is the received super special code word, we findif S(y s ) = (0) then we accept y s as the correct me

 s s H y z (0) then we can declare the received wo

We can find the correct word by the following

we give this method we illustrate how thesyndrome is calculated.

 Example 3.1.7: Let 1 2 3 4s s s s sC C C C C ª º¬ ¼  be

row code. Let Hs = [H1 | H2 | H3 | H4] be the supcheck matrix of Cs.

Let1 0 0 0 1 0 0 0 1 0

H 1 0 0 1 0 1 0 1 0 1

ª« «

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[1 1 1 0 | 1 1 0 1 1], [1 0 1 1 | 1 1 0 1 1], [0 1 0 1 | 1[0 0 0 0 | 0 1 1 1 0], [1 0 1 1 | 0 1 1 1 0], [0 1 0 1 | 0[0 0 0 0 | 1 0 1 0 1], [1 0 1 1 | 1 0 1 0 1], [0 1 0 1 | 1

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[0 0 0 0 | 1 1 1 0 0], [1 1 1 0 | 1 1 1 0 0], [1 0 1 1 | 1[0 1 0 1 | 1 1 1 0 0] and so on}.

Clearly |Cs| = 32 . Now the coset table of  1sC is given

Message code

0 0 1 0 0 1

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

Other cosets1 0 0 0 0 0 1 1 1 1 0 1 00 1 0 0 1 1 1 1 0 0 0 1 1

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

coset leaders

 Now the coset table of  2sC is given by

message 0 0 0

codewords 0 0 0 0 0

1 0 0 0 0

other cosets 0 1 0 0 0

0 0 1 0 0

-°®°¯

1 0 0

1 0 0 1 0

0 0 0 1 0

1 1 0 1 0

message 1 1 0 0 1 1

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codewords 1 1 0 1 10 1 0 1 1

other cosets 1 0 0 1 1

1 1 1 1 1

-°®°¯

0 1 1 1 01 1 1 1 0

0 0 1 1 0

0 1 0 1 0

message 1 0 1codewords 1 0 1 0 1

0 0 1 0 1

other cosets 1 1 1 0 1

1 0 0 0 1

-°®°

¯

1 1 1

1 1 1 0

0 1 1 0

1 0 1 0

1 1 0 0

Suppose ys = [1 1 1 1 | 1 1 1 1 1] is the received = T

s sH y z [(0) | (0)]. es = [0 1 0 0 | 0 0 1 0 0]

coset leader. Thus xs = ys + es = [1 0 1 1 | 1 1 0 1

With the advent of computers calculating the supleaders is not a very tedious job. Appropriateyield the result in no time.

 Now we proceed on to describe/define the sucyclic code.

DEFINITION 3.1.8:   Let C  s = [C 1 | C 2 | … |

 special row code. If every i

 sC is a cyclic code in

be a super special cyclic row code H = [H

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Let 1 2s s sG G G ª º¬ ¼

1 1 0 1 0 0 0 1 0 0 0

0 1 1 0 1 0 0 0 1 0 0

ª

««

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0 1 1 0 1 0 0 0 1 0 0

0 0 1 1 0 1 0 0 0 1 0

0 0 0 1 1 0 1 0 0 0 1

« « « « « ¬

 be the super row generator matrix which generateThe code words of  1

sC generated by G1 is giv

1sC = {(0 0 0 0 0 0 0), (1 1 0 1 0 0 0), (0 1

(0 0 1 1 0 1 0), (0 0 0 1 1 0 1), (1 0 1 1 (0 1 0 1 1 1 0), (0 0 1 0 1 1 1), (1 1 1 0

(1 1 0 0 1 0 1), (0 1 1 1 0 0 1), (1 0 0 0 (0 1 0 0 0 1 1), (1 0 1 0 0 0 1), (1 1 1 1 1 0 1), (1

2sC = {(0 0 0 0 0 0 0), (1 0 0 0 1 0 1), (0 1

(0 0 1 0 1 1 0), (0 0 0 1 0 1 1), (1 1 0 0 (0 1 1 0 0 0 1), (0 0 1 1 1 0 1), (1 0 0 1 (1 0 1 0 0 1 1), (0 1 0 1 1 0 0), (1 1 1 0

(0 1 1 1 0 1 0), (1 1 0 1 0 0 1), (1 0 1 1 0 0 0), (1

If  1 2s s s

x x x ª º¬ ¼ by taking 1 1s sx C and 2

sx

Clearly elements in Cs are super row vectors.

 Now we proceed to define super special mixed and its super special generator mixed row matrix

We illustrate this by the following example.

E l 3 1 11 L t 1 2 3 4C C C C Cª º¬ ¼ b

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  Example 3.1.11: Let 1 2 3 4s s s s sC C C C C ª º¬ ¼ b

special mixed row code. Let 1 2 3s s s sG G G G ª ¬

associated super special mixed row generator matrix

1 2 3 4

s s s s sG G G G G ª º¬ ¼

1 0 0 0 1 1 0 0 1 1 0

0 1 0 1 0 0 1 0 0 1 1

0 0 1 1 1 0 0 1 1 0 1

ª« ««¬

1 0 0 0 0 1 0

0 1 0 1 0 0 0

0 0 1 0 1 0 1

1 0 0 1 0 0

0 1 0 0 1 1

0 0 1 1 0 1

Clearly Gs is a super mixed row matrix. All the code  by G1, G2, G3 and G4 have the same number symbols. The code 1

sC = {(0 0 0 0 0), (1 0 0 0 1), (00 1 1 1), (1 1 0 1 1), (0 1 1 0 1), (1 0 1 1 0), (1 1 1codewords given by 2

sC = {(0 0 0 0 0 0), (1 0 0 1 1

1 1), (0 0 1 1 0 1), (1 1 0 1 0 1), (0 1 1 1 1 0), (1 0 1 1 0 0 0)}. The codes associated with 3

sC = {1 0 0 0 00 0 0 0), (0 1 0 1 0 0 0), (0 0 1 0 1 0 1), (1 1 0 1 0 1

THEOREM 3.1.1: Let  1 2 ª º¬ ¼! n s  s s s

C  C C C  be

row code with H  s = [H 1 | H 2 | …| H n ] , the supe

check matrix If each Hi = (Ai In–k ) i = 1 2

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check matrix. If each H i (Ai , I n k ), i 1, 2, …, | G2 | …| Gn  ] with Gi = (I k   , – AT   ); 1 d  i d  n if

length of each code word ini

 sC  is the same for i

 Proof: Suppose we are given Hs = [H1 | H2 | …

super special parity check matrix of the super spCs = [C1 | C2 | … | Cn]. We know every subcodthe same number of check symbols. Suppose wsuper special row code Cs the super special matrix Gs with T

s sG H = [(0) | (0) | … | (0)] . Thek 1 = n2 – k 2 = … = nn – k n and k 1 = k 2 = k 3 =

 possible if and only if n1 = n2 = … = nn. Hence thThus we see in this situation we have both thgenerator row matrix of the code Cs as well as th

 parity check matrix of the code Cs are not superow matrices; we see both of them have the sameach row.

We illustrate this situation by an exam proceed on to define more concepts.

  Example 3.1.12: Let 1 2 3 4s s s s sC C C C C ª º¬ ¼ be

row code. Hs = [H1 | H2 | H3 | H4]

0 1 1 1 0 0 0 1 0 0 1

1 0 1 0 1 0 0 0 1 1 0

ª«««

the super special row parity check matrix; then the respecial row generator matrix

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Gs = [G1 | G2 | G3 | G4]

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

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

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

ª« «

«¬

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

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

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

Now we findT

s sG H = [G1 | G2 | G3 | G4] u [H1 | H2 | H3 | H

= [G1 | G2 | G3 | G4] u1 2 3 4

T T T TH H H Hª º¬ ¼

= T T T T1 1 2 2 3 3 4 4G H G H G H G Hª º¬ ¼

=

0 1 1 1

1 0 1 1

1 0 0 0 1 1 1 1 1 1 00 1 0 1 0 1 1 1 0 0 0

0 0 1 1 1 1 0 0 1 0 0

ª ª º« « »« « »

« « »ª º« « »« »« « »« »« « »« »¬ ¼« « »

1 0 1 1ª º ª« » «

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1 0 1 10 1 1 0

1 0 0 1 0 1 00 1 1 1

0 1 0 1 0 0 11 0 0 0

0 0 1 1 1 0 00 1 0 0

0 0 1 00 0 0 1

ª º ª« » «« » «« » «ª º« » «« » « » «« »« » «« »¬ ¼« » «« » «« » «¬ ¼ ¬

1 0 1

0 1 1

1 0 0 1 0 1 1 1 0 1

0 1 0 0 1 1 1 1 0 0

0 0 1 1 0 1 0 0 1 0

0 0 1

0 0 0

ª «

« « ª º« « »« « »« « »¬ ¼ « «

« ¬

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

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

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

ª « « « ¬

which is a super special zero row vector.

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special parity check mixed row matrix of Cs yet for this Cs. This is in keeping with the theorem.

Likewise if we have a super special row chave the super special row matrix which gene

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have the super special row matrix which genemay not exist.

This is the marked difference between the sucodes and usual linear codes.

 Example 3.1.14: Let Cs = [C1 | C2 | C3] be a supwith super special row generator matrix

s

1 0 0 0 1 1 1 0 0 1

G 0 1 0 1 0 1 0 1 0 0

0 0 1 1 1 0 0 0 1 1 0

ª« ««¬

1 0 0 1 1

0 1 0 0 1

0 0 1 1 0

º»»»¼

= [G1 | G2 | G3].

Clearly Gs is a super special row mixed matrix. N

1

0 1 1 1 0 0

H 1 0 1 0 1 01 1 0 0 0 1

ª º

« » « »« »¬ ¼

where T2 2G H = (0). Also

3

1 0 1 1 0H

1 1 0 0 1

ª º

« »¬ ¼

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1 1 0 0 1« »¬ ¼

and is such that T3 3G H = (0). We see H1, H2 and H

made into a super special row matrix. Hence the claim  Now having defined the new class of super

(mixed row) codes we will now define new classesuper classes of mixed super special row codes Cs ihave the super special row code to contain classical Hamming code or cyclic code or code and its complement and so on.

 D EFINITION  3.1.11:   Let C  s = [C 1 | C 2 | …| C n  ]  special row code. If some of the C i’s are Hamming c

C  j’s are cyclic codes i z  j, some C k ’s are repetition

 some C t ’s are codes and C  p’s are dual codes of C t ’s

i, p < n then we call C  s to be a mixed super special ro

It is important to mention here that even if two typescodes are present still we call Cs as a mixed super code.

We will illustrate them by the following examples.

  Example 3.1.15: Let Cs = [C1 | C2 | C3 | C4] be a mspecial row code. Here C1 is a Hamming code, C2 thcode, C3 a code of no specific type and C4 a cyclic co

L t th i d i l it h k

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Further the transmission rate can never be equal to then will the transmission rate be always greater than

We will just illustrate a super special Hamming row

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j p p gfollowing example.

  Example 3.1.17: Let Cs = 1 2 3s s sC C Cª º¬ ¼ be a su

Hamming row code where

Hs =

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

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

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

ª«««¬

1 0 0 1 1 0 10 1 0 1 0 1 1

0 0 1 0 1 1 1

º»»»¼

= [H1 | H2 | H3]

is the super special matrix associated with Cs. We (7, 4) Hamming code; i = 1, 2, 3. Here m = 3 and n =

The code words associated with 1sC is

{(0 0 0 0 0 0 0), (1 0 0 0 0 0 1 1), (0 1 0 0 1 0 1), (0 0(0 0 0 1 1 1 1), (1 1 0 0 1 1 0), (0 1 1 0 0 1 1), (0 0 (1 0 1 0 1 0 1), (0 1 0 1 1 1 1), (1 0 0 1 1 0 0), (1 1 (0 1 1 1 1 0 0) (1 1 0 1 0 0 1) (1 0 1 1 1 0 1) (1 1 1

(0 0 1 0 1 1 1), (0 0 0 1 1 1 0), (1 1 0 0 (0 1 1 0 0 1 0), (0 0 1 1 0 0 1), (1 0 0 1 (1 0 1 0 1 0 0), (0 1 0 1 0 1 1), (1 1 1 0

(0 1 1 1 1 0 0), (1 1 0 1 0 1 1), (1 0 1 1 0 1 0), (1

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By taking one code word from 1sC , one code wo

one from 3sC we form the super special Hamm

Any xs = 1 2 3s s s(x x x ) where 1

sx 1sC , 2 2

s sx C

We seeHs

Tsx = [H1 | H2 | H3]

1 2 3s s s(x x x )

= [H11 Ts(x ) | H2

2 Ts(x ) | H3

3 Ts(x ) ]

= [(0) | (0) | (0)]for every xs Cs.

3.2 New Classes of Super Special Column

Suppose we are interested in finding super spwhich the number of check symbols will be d

subcodes. In such a situation we see certainly wwith the super special row codes Cs, for in this calways the number of check symbols to be the subcode in Cs, so we are forced to define this classes of super special codes.

DEFINITION 3.2.1: Suppose we have to describe  same length say m but with varying sets of chec single matrix. Then we define it using super co

be a set of m codes, C 1 , C 2  , ..., C m where all of the

 same length n but have n – k 1 , n – k 2 , ..., n – k m to be of check symbols and k 1 , k 2  , ..., k m are the number

  symbols associated with each of the codes C 1 , i l

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respectively.

Let us consider 

Hs =

1

2

m

H

H

H

ª º« »« »« »« »« »¬ ¼

#

where each Hi is the n – k i u n parity check matrix

Ci; i = 1, 2, ..., m. We call Hs

to be the super special pmixed column matrix of Cs and Cs is defined aspecial mixed column code.

The main difference between the super speciaand the super special column code is that in super codes always the number of check symbols in every

is the same as the number of message symbols inlength of the code Ci can vary where as in the sucolumn code, we will always have the same lengtcode Ci in Cs but the number of message symbonumber check symbols for each and every code Ci

not be the same. In case if the number of check symb

and every code Ci is the same. Then we call Cs tospecial column code.In case when we have varying number of che

h ll h d b i l i

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0

0

ª º« »« »

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=

T1 1

T2 2

T3 3

T4 4

0

0

H x 0

0H x0H x

0H x

0

0

00

« »« »« »« »« »ª º« »« »« »« » « »« »« »« »« »« »¬ ¼« »« »« »

« »« »« »¬ ¼

.

As in case of super special row codes we getspecial column code, the main criteria being every c

= [C1 | C2 | ... |Cm] would have the same length. It canumber of message symbols and any arbitrary numbsymbols.

Just we saw in example 3.2.1 a super special coOne of the main reason for us to have this new classthat when we have super special row codes we see i

of the code words are the same then automatically it the number of message symbols become equal to thecheck symbols Thus the transmission rate becomes

0 0 1 1 0 0 0

0 1 0 0 1 0 0

1 1 1 0 0 1 01 0 0 0 0 0 1

ª º« »« »

« »« »« »

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Hs =

1 0 0 0 0 0 1

1 1 0 0 0 1 0

1 1 0 1 0 0 1

1 0 1 0 1 0 0

0 1 1 0 0 1 0

1 1 1 1 0 0 1

« »« »« »« »« »

« »« »« »¬ ¼

.

We see every code word in Ci are of lennumber of message symbols in C1 is 3 and the n

symbols is 4 i.e., C1 is a (7, 3) code where as C2

and C3 is a (7, 4) code. The number of super cocode Cs is 23 u 25 u 24 = 212.The code words in

C1 = {(0 0 0 0 0 0 0), (1 0 0 0 0 1 1), (0 1 0

(0 0 1 1 0 1 0), (1 1 0 0 1 0 1), (0 1 1 1 (1 0 1 1 0 0 1), (1 1 1 1 1 1 1)}.

The code

C2 = {(0 0 0 0 0 0 0), (1 0 0 0 0 1 1), (0 1 0(0 0 1 0 0 0 0), (0 0 0 1 0 0 1), (0 0 0 0

(1 1 0 0 0 0 0), (0 1 1 0 0 1 1), (0 0 1 1 (0 0 0 1 1 0 1), (1 0 1 0 0 1 1), (1 0 0 1 (1 0 0 0 1 1 1), (0 1 0 1 0 0 1), (0 0 1 0

(0 0 1 1 1 1 0), (1 0 1 0 0 1 0), (0 1 0 1 0 1 0), (1 0 0(1 1 1 0 0 0 1), (0 1 1 1 1 0 1), (1 1 0 1 1 1 1

(1 0 1 1 0 1 1), (1 1 1 1 0 0 0)}.

Thus by taking one code word from each of the

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Thus by taking one code word from each of the the super special code word which is super row vect7. Thus Cs has 8 u 32 u 16 = 4096 super code words.

Unlike in the case of super special row codes gispecial column code Cs with associated super spcheck column matrix Hs we will always be in a positthe super special generator matrix Gs. We see in t3.2.2 just given we do not have for that Hs angenerator matrix Gs though clearly each Hi in Hs is standard form.

However for the example 3.2.1 we have angenerator matrix Gs for the

Hs =

1

2

3

4

H

H

H

H

ª º« »« »« »« »

« »¬ ¼

1 0 1 0 1 0 0

0 1 0 1 0 1 0

1 1 1 0 0 0 1

0 1 1 0 1 0 01 0 0 1 0 1 0

1 0 1 0 0 0 1

ª º« »« »« »« »« »« »« »« »

1 0 0 0 1 0 1

0 1 0 0 0 1 1

0 0 1 0 1 0 10 0 0 1 0 1 0

ª º« »« »

« »« »« »

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5

0 0 0 1 0 1 0

1 0 0 0 0 1 1

0 1 0 0 1 0 0G0 0 1 0 1 0 1

G0 0 0 1 0 1 0G

G1 0 0 0 1 1 1G0 1 0 0 0 1 0

0 0 1 0 0 1 0

0 0 0 1 1 0 11 0 0 0 0 1 1

0 1 0 0 1 1 0

0 0 1 0 1 1 1

0 0 0 1 0 1 1

« »« »« »« »« » ª

« » « « » « « » « « » « « » « ¬ « »« »« »« »« »« »« »« »« »¬ ¼

Clearly

GsTsH =

1

2

3

4

G

G

GG

ª º« »« »

« »« »« »¬ ¼

T T T T1 2 3 4H H H Hª

¬

1 0 0 0 1 0 1

0 1 0 0 0 1 1

ª º« »« »« »

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0 0 1 0 1 0 1

0 0 0 1 0 1 0

1 0 0 0 0 1 1

0 1 0 0 1 0 00 0 1 0 1 0 1

0 0 0 1 0 1 0

1 0 0 0 1 1 1

0 1 0 0 0 1 0

0 0 1 0 0 1 00 0 0 1 1 0 1

1 0 0 0 0 1 1

0 1 0 0 1 1 0

0 0 1 0 1 1 10 0 0 1 0 1 1

« »« »« »« »« »

« »« »« »« »

« »« »« »

« »« »« »« »« »« »« »

« »« »¬ ¼

u

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

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

1 0 1 1 0 1 0 1 0 1 1 10 1 0 0 1 0 1 0 1 0 1 1

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

ª « «

« « « «

0 0 0

0 0 0

0 0 00 0 0

ª º« »« »

« »« »« »

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0 0 0

0 0 0

0 0 0

0 0 0.

0 0 0

0 0 0

0 0 0

0 0 00 0 0

0 0 0

0 0 0

0 0 0

« »« »« »« »« »

« »« » « »

« »« »« »« »« »« »« »« »« »« »« »¬ ¼

We find conditions under which we have a supercode Cs with an associated super special paritymatrix Hs in the standard form to have a super sp

column matrix Gs with GsTsH =

0

0

ª º« »« »

« »¬ ¼

# .

Before we prove a result of this nature wetypes of super special generator column matr

  generates the super special column code C  s. We

 super special generator column matrix which generaeach of the codes C i in C  s  , we have same number

 symbols then we call G s to be a super special generamatrix; i = 1, 2, …, n. If each of the codes C i’s

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f fdifferent number of message symbols then we call

 super special generator mixed column matrix.

We say Gs is in the standard form only if each G

standard form. Further only when Gs is in the standaGs is a super special column matrix which is not a mwe have Hs the super special parity check column msame Cs with

GsTsH =

0

0

0

ª º« »

« »« »« »« »¬ ¼

#.

 Now we illustrate both the situations by examples.

 Example 3.2.3: Let Cs = [C1 | C2 | C3 | C4]t

be a sucolumn code generated by

1 0 0 1 1 0 1

0 1 0 0 1 1 0

0 0 1 1 0 0 1

1 0 0 0 1 0 1

0 1 0 1 0 0 0

ª º« »« »« »« »« »« »« »

1Gª º« »

the super special column matrix. Now the relatedand is given by

1 0 1 1 0 0 0

1 1 0 0 1 0 0

0 1 0 0 0 1 0

ª « « «

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1

2

3

4

0 1 0 0 0 1 0

1 0 1 0 0 0 1

0 1 0 1 0 0 0

1 0 0 0 1 0 0H 0 0 1 0 0 1 0H 1 0 1 0 0 0 1

HH 1 0 1 1 0 0 0H 1 1 0 0 1 0 0

1 0 1 0 0 1 01 1 0 0 0 0 1

1 0 0 1 0 0 0

0 1 0 0 1 0 0

1 1 1 0 0 1 0

1 0 1 0 0 0 1

« « « « «

« « ª º« « »« « » « « »« « »« « »¬ ¼« « « « « « «

« « ¬  Now

1Gª ºT

1Hª º TG Hª º

0

0

00

0

ª « «

« « « «

 Example 3.2.4: Let Cs = [C1 | C2 | C3]t be a su

column code. Suppose

1 0 0 1 0 0 00 1 0 1 1 0 0

ª º« »« »

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Gs =

0 1 0 1 1 0 0

0 0 1 0 1 1 0

1 0 1 1 0 0 1

0 1 0 0 1 1 0

1 0 0 0 0 1 0

0 1 0 0 1 1 0

0 0 1 0 1 1 1

0 0 0 1 0 1 1

« »« »« »« »« »

« »« »« »« »« »« »¬ ¼

=1

2

3

G

G

G

ª º« »

« »« »¬ ¼

 be the super special column matrix which generates C

H1 =

1 1 0 1 0 0 0

0 1 1 0 1 0 0

0 0 1 0 0 1 00 0 0 0 0 0 1

ª º« »« »

« »« »¬ ¼

,

H2 =

1 0 1 0 0 0 0

1 0 0 1 0 0 0

0 1 0 0 1 0 00 1 0 0 0 1 0

1 0 0 0 0 0 1

ª º« »« »

« »« »« »« »¬ ¼

1 1 0 1 0 0 0

0 1 1 0 1 0 0

0 0 1 0 0 1 00 0 0 0 0 0 1

ª º« »« »

« »« »« »« »

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Hs =

1 0 1 0 0 0 0

1 0 0 1 0 0 0

0 1 0 0 1 0 0

0 1 0 0 0 1 01 0 0 0 0 0 1

0 1 1 0 1 0 0

1 1 1 1 0 1 0

0 0 1 1 0 0 1

« »« »« »« »

« »« »« »« »« »« »« »« »¬ ¼

is the super special column matrix which is tcheck column matrix of Cs. Clearly Gs

TsH = (0

zero matrix. As in case of usual codes given tmatrix H or the generator matrix G in the standa

always get G from H or H from G and we have Gwise in the case of super special (mixed columnCs if Gs is the super special column generatostandard form we can always get the super

 parity check matrix Hs from Hs and we have Gs

super special zero column matrix.As in case of super special row code we can super special column codes.

1 1 0 0 ... 0

1 0 1 0 ... 0

1 0 0 0 1

ª º« »

« »« »« »« »

# # # # #

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=

1 0 0 0 ... 1

1 1 0 0 ... 0

1 0 1 0 ... 0

1 0 0 0 ... 1

« »« »« »« »« »

« »« »« »« »¬ ¼

#

# # # # #

where

H =

1 1 0 0 ... 0

1 0 1 0 ... 0

1 0 0 0 ... 1

ª º

« »« »« »« »¬ ¼

# # # # #.

It is important to note that unlike the super speci

row code which can have different lengths the surepetition column code can have only a fixed lengsuper special code word

xs = 1 2 ns s sx x xª º¬ ¼"

where  jsx = (1 1 … 1), n-times or (0 0 … 0) n-times

d n.

1 1 0 0 0 0

1 0 1 0 0 0

1 0 0 1 0 01 0 0 0 1 0

ª º« »« »

« »« »« »« »

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s

1 0 0 0 0 1

1 1 0 0 0 0

1 0 1 0 0 0

H 1 0 0 1 0 01 0 0 0 1 0

1 0 0 0 0 1

1 1 0 0 0 0

1 0 1 0 0 01 0 0 1 0 0

1 0 0 0 1 0

1 0 0 0 0 1

« »« »« »« »

« » « »« »« »« »« »« »« »« »« »« »« »« »¬ ¼

.

ThusCs = {[0 0 0 0 0 0 | 0 0 0 0 0 0 | 0 0 0 0

[1 1 1 1 1 1 | 1 1 1 1 1 1 | 1 1 1 1 1 [0 0 0 0 0 0 | 0 0 0 0 0 0 | 1 1 1 1 1 [0 0 0 0 0 0 | 1 1 1 1 1 1 | 1 1 1 1 1

[1 1 1 1 1 1 | 0 0 0 0 0 0 | 0 0 0 0 0 0[1 1 1 1 1 1 | 1 1 1 1 1 1 | 0 0 0 0 0 0[1 1 1 1 1 1 | 0 0 0 0 0 0 | 1 1 1 1 1 [0 0 0 0 0 0 | 1 1 1 1 1 1 | 0 0 0 0 0 0

DEFINITION 3.2.4: Let C  s = [C 1 | C 2 | …| C n ]t be a su

 parity check column code . Let the super special p

column matrix associated with C  s be given by

ª ºH

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 H  s =

ª º« »« »« »

« »« »¬ ¼

#

1

2

n

 H 

 H 

 H 

where

 H 1 = H 2 = … = H n =

m times

1 1 ... 1 .

Thus we see we cannot get different lengths of pcodes using the super special column code. Howsuper special row code we can get super special pcodes of different lengths.

We illustrate super special parity check column c

  Example 3.2.6: Let Cs = [C1| C2 | C3 | C4]t

be a su parity check column code with super special parity ch

H =

1

2

3

4

H

H

HH

ª º« »« »

« »« »« »¬ ¼

=

1 1 1 1 1

1 1 1 1 1

1 1 1 1 11 1 1 1 1

ª º« »« »

« »« »« »¬ ¼

.

 Example 3.2.7: Let Cs = [C1 | C2 | C3]t be a super

 parity check code. The super special column pari

1H 1 1 1 1ª º ª º« » « »

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Hs = 2

3

H 1 1 1 1

H 1 1 1 1

« » « »« » « »

« » « »¬ ¼ ¬ ¼

.

The codes related with H1 is C1 = {(0 0 0 0), (1 1(0 0 1 1), (1 0 0 1), (1 0 1 0), (0 1 0 1), (1 1 1Thus Cs contains 8 u 8 u 8 = 512 number of supwords.

Thus only when the user wants to send mes

  by the same parity check matrix he can use main advantage of this special parity check col[C1 | C2 | … | Cn]; has each code word in Ci whicthen Cs has n2m–1 number of code words hence ochannels were one is not very much concetransmission rate; for the transmission rate

increase in the length of the code words in Cs. Now having seen this new class of codes

check column code. We proceed on to built anothcolumn codes using Hamming codes.

DEFINITION 3.2.5: Let C  s = [C 1 | C 2 | … | C special column code if each of the codes C i in C  s – 1 – m) Hamming code for i = 1, 2, … , n then w

i l l H i d It

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C1 = {(0 0 0 0 0 0 0), (1 0 0 0 0 1 1), (0 1 0(0 0 1 0 1 1 0), (0 0 0 1 1 0 1), (1 1 0 0 1 0 1), ((0 0 1 1 1 0 0), (1 0 1 0 1 0 1), (1 0 0 1 1 0 0), (

(1 1 1 0 0 0 0), (0 1 1 1 1 0 0), (1 1 0 1 (1 0 1 1 1 1 1), (1 1 1 1 1 1 1)}

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and

C2 = {(0 0 0 0 0 0 0), (1 0 0 0 1 0 1), (0 1 0

(0 0 1 0 1 1 0), (0 0 0 1 0 1 1), (1 1 0 0 0 1 0), ((0 0 1 1 1 0 1), (1 0 1 0 0 1 1), (1 0 0 1 1 1 1), (

(1 1 1 0 1 0 1), (0 1 1 1 0 1 0), (1 1 0 1 (1 0 1 1 0 0 0), (1 1 1 1 1 1 1)}.

Thus we see C1 and C2 are different Hammi

C2 z I as well as C1 z C2.So we can get different sets of codes and

elements in Cs is 256. Now as in case of super special row codes w

super special column codes have mixed super codes. The only criteria being is that each code

of same length. Now we proceed on to define them.

DEFINITION 3.2.6: C  s = [C 1 | C 2 | …| C n ]t  is

 special column code if some C i’s are repetition n some C  j’s are Hamming codes of length n, so

check codes of length n and others are arbitraryd  n.

 be the super special parity check column matrix assoCs. Here

1 1 1 1 1 1 11 1 0 0 0 0 0

ª º« »« »« »

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Hs =

1

2

3

4

H

H

H

H

ª º« »« »« »« »« »¬ ¼

=

1 0 1 0 0 0 0

1 0 0 1 0 0 0

1 0 0 0 1 0 0

1 0 0 0 0 1 0

1 0 0 0 0 0 1

0 0 0 1 1 1 1

0 1 1 0 0 1 1

1 0 1 0 1 0 11 0 0 1 0 0 0

0 1 1 0 1 0 0

1 0 1 0 0 1 0

1 1 1 0 0 0 1

« »« »« »« »

« »« »« »« »« »« »« »

« »« »« »« »« »« »« »¬ ¼

;

here H1 is a parity check code of length 7, H2 is code with 6 check symbols, H3 is the Hamming codeand H4 a code with 3 message symbols. The transmis

Cs =6 1 4 3 14 1

7 7 7 7 28 2

.

N d d fi h i l

 Now we illustrate this by an example.

 Example 3.2.11: Let Cs = [C1 | C2 | C3]t be a sup

column code where each Ci is of length six.

 Now the associated super special parity check co

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p p p y

Hs =

1

2

3

H

HH

ª º« »« »« »¬ ¼

=

0 0 1 0 0 1

0 1 0 0 1 0

1 0 0 1 0 0

1 1 0 0 0 0

1 0 1 0 0 0

1 0 0 1 0 0

1 0 0 0 1 01 0 0 0 0 1

0 0 0 0 1 1

0 0 0 1 1 0

0 0 1 1 0 0

0 1 1 0 0 0

1 1 0 0 0 0

ª º«

« « « « « « «

« « « « « « « « « « « ¬ ¼

The cyclic codes given by the parity check m

C1 = {(0 0 0 0 0 0), (1 0 0 1 0 0), (0 1 0 0 1 0),(1 1 0 1 1 0), (0 1 1 0 1 1), (1 0 1 1 0 1), (1 1

C2 = {(1 1 1 1 1 1), (0 0 0 0 0 0)

3.3 New Classes of Super Special Codes

We have given the basic definition and propertie

matrices in chapter one. In this section we proceed onew classes of supper special codes and discuss a fewabout them.

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DEFINITION 3.3.1: Let 

C(S) =

ª º

« »« »« »« »« »¬ ¼

"

"

# # " #

"

1 1 1

1 2 n

2 2 2

1 2 n

m m m

1 2 n

C C C 

C C C 

C C C 

wherei

 jC are codes 1 d  i d  m and 1 d  j d  n. Furth!1 2 m

1 1 1C ,C , ,C are of same length !1 2 m

2 2 2C ,C , , C a

length and  !1 2 m

n n 2C ,C , , C are of same length. C

have same number of check symbols, !2 2 2

1 2 nC ,C , ,C

number of check symbols and  m m m

1 2 nC ,C ,...,C   have sa

of check symbols.We call C(S) to be a super special code. We ca

 super parity check matrix

 H(S) =

ª º« »« »« »« »« »¬ ¼

"

"

# # " #

"

1 1 1

1 2 n

2 2 2

1 2 n

m m m

1 2 n

 H H H  

 H H H  

 H H H  

 Now this super special code has two types of me

1. Array of super row vector messages i.e.,

1 1 11 2 n

2 2 2

C C Cª º¬ ¼

ª º

!

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2 2 21 2 n

m m m1 2 n

C C C

 

;C C C

ª º¬ ¼

ª º¬ ¼

!

# #

!we have m rows of super row vectors and one bsent at the receiving end; one after decoding puform of the same array of rows.

2. Array of super column vector messages

1 1 11 2 n2 2 21 2 n

m m m1 2 n

C C C

C C C

C C C

ª º ª º ª º« » « » « »« » « » « »« » « » « »« » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

"# # #

.

 Now we have n columns of super column vectorstaking transposes of each column the messages one as

t112

1

m1

C

C

C

ª º« »

« »« »« »« »¬ ¼

# which is a super row vector,

t212

2

m2

C

C

C

ª º« »

« »« »« »« »¬ ¼

# a sup

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[1 0 0 0 0 1 0 1 0 0 1

[1 1 1 0 1 1 1 0 1 0 1 0

[1 1 1 1 0 0 1 1 1 1 1 0As an array of the super column vector we have

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0

1

1 00 0

0 1

0 0

0 11 1

1 0

1 1

, .1 00 1

1 0

1 1

1 1

1 1

1 1

ª º« »« »

« »ª º « »« »« »« »« »« »« »« »« »« »

« »« » « »« »« »« »« »« »« »« »« »« »« »« »

« »« » « »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »« »

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received message is correct or not. Now we fimethod using the general case before we procwith specific examples. Suppose

1 1 11 2 n2 2 21 2 n

C C CC C C

C S

ª º« »« » « »

""

# # " #

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m m m1 2 nC C C

« »« »« »¬ ¼

# # #

"

 be the super special super code. Any super speword of C(S) say x(S) C(S) would be of the fo

1 1 11 2 n2 2 21 2 n

m m m1 2 n

x x x

x x x

x S

x x x

ª º« »« »

« »« »« »¬ ¼

"

"

# # " #

"

where each  jix is a row vector; 1 d j d m and 1 d

Suppose

1 1 1

1 2 n2 2 21 2 n

m m m1 2 n

H H HH H H

H S

H H H

ª º« »« » « »« »« »¬ ¼

""

# # " #

"

  be the super special parity check matrix associ Now R(S) be the received super special super cby

We define super special syndrome of C(S) as

t

S x S H S x S ª º

¬ ¼

t t t1 1 1 1 1 1

1 1 2 2 n nH x H x H xª º« »

" 0 0ª

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t t t2 2 2 2 2 21 1 2 2 1 1

t t tm m m m m m1 1 2 2 n n

H x H x H x

H x H x H x

« »« »« »« »

« »« »« »¬ ¼

"

# # " #

"

=

0 0

0 0

0 0

ª « « « « « ¬

# #

then we take x (S) C(S); if 

H(S) [x(S)]t z

0 0 0

0 0 0

0 0 0

ª º« »« »« »« »« »¬ ¼

"

"

# # " #

"

then we declare x(S) C(S) so if y(S) is the receivone calculates

H(S) [y(S)]t =

0 0 0

0 0 0

0 0 0

ª º« »

« »« »« »« »¬ ¼

"

"# # " #

"

,

Chapter Four

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APPLICATIONS

OF THESE NEW CLASSES

OFS

UPER S

PECIALC

ODES

We enumerate in this chapter a few applications of using the super special codes. Now the followhy this super special super code is better thcode. These codes can also be realised as a concatenated codes. We enumerate them in the fo

1. Instead of using ARQ protocols we can useC, in the super special super code

C(S) =

C C C

C C C

C C C

ª º« »« »

« »« »« »¬ ¼

"

"

# # " #"

.

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We can take the correct message. This saves band time. We can also send it is as array ofvectors i.e., if  x is the message to be sent then,

x x ... x

x x ... x

x x ... x ;

ª º¬ ¼

ª º¬ ¼

ª º¬ ¼

#

As an array of row super vector or the array of super vectors as

x x xx x x

x x x

ª º ª º ª º« » « » « »« » « » « »« » « » « »« » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

!# # #

where

txx

ª º« »« »

C(S) =

C C C

C C C

C C C

ª º« »« »

« »« »« »¬ ¼

"

"

# # " #"

h if i h d h h i

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then if one wishes to study the changes in tersame message can be sent in all cells and the gstage transformation can be seen (observed) ancan be got.

Here it is not sending a message and receiv  but is a study of transformation from time to from satellite or pictures of heavenly bodies. Efield this will find an immense use. Also thesespecial super codes can be used in scientific expchanges can be recorded very minutely or with hWhat one needs is a proper calibration linking ththose experiments were one is interested inchanges from time to time were the graphical rimpossible due to more number of variables.

3. Another striking advantage of the super speC(S) is that if one has to be doubly certain aboutthe received message or one cannot request ftransmission in those cases the sender can sendrow codes and send the same codes as the super codes.

After receiving both the messages, if botcodes are identical and without any error one

th i fi d th i h ll d

C S

I I Iª º« »I I I« »

« »« »I I I« »¬ ¼

"

"

# # " #"

or  i i iª º

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i i i1 2 nC S C C C ª º¬ ¼!

itself has the empty code word I in each of the i jC ’si = 1, 2, …, n and 1 d j d m.

4. At times the super special code C(S) may be otype i.e., if 

C(S) =

1 1 1

1 2 n2 2 21 2 n

m m m1 2 n

C C CC C C

C C C

ª º« »« »« »« »« »¬ ¼

""

# # " #

"

is such that1 2 m 1 2 m 1 21 1 1 2 2 2 n nC C C ,C C C , ,C C ! ! !

i.e., each row in C(S) i.e.,1 1 11 2 nC C C ª º¬ ¼!

2 2 21 2 n

...C C C ª º¬ ¼!

m m m

1 2 n ;C C Cª º¬ ¼!i.e.,

1 1 1C C Cª º"

accepted as the approximately correct messageeven a single pair of coinciding super row vchoose a super row which has a minimum numb

if 

1 1 11 2 n1 1 11 2 n

H H H

H H HH S

ª º« »« » « »

"

"

# # #

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1 1 11 2 n

H S

H H H

« »« »« »¬ ¼

# # " #

"

then we say ith super row has least number of erroxi(s) = 1 1 1

1 2 nx x xª º¬ ¼!

we find out

t t t1 1 1 1 1 11 1 2 2 n nH x ,H x , , H x! ; 1 d i

we choose the super row which has the maximzeros i.e., that super row consequently has minimerror. We find the correct code word from thosrow and accept it as the approximately correct r

already we have a super row in which t

1 11 1H x n then we accept that as the correct messagetransformation helps the receiver to study the reand the cells in C(S) which misbehave or that wan error message. Thus one can know not only

sent message but also know more about the prmachine) while the message is transmittedcorrections can be made so as to guarantee an

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When we send only array super row transreceived code word in the ith row (say) 1 1

i iy yª ¬

 be the same. This is true for i = 1, 2, …, m. If thfor any row just by observation we can conclumessage has an error and choose the row wnumber of differences

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number of differences. Now when array of column transmission tak

if y(S) is the received message then

t1 1 11 2 m

t1 1 11 2 m

t1 1 11 2 m

y y y

y y yy S

y y y

ª º¬ ¼

ª º¬ ¼

ª º¬ ¼

!

!

#

!

.

We take that column which has least number of has no error as the received message. The maisimultaneous transmission is we can compare e

received array of super row vectors and array ovectors.

6. This type of code can be used when the Aimpossible so that the same message can be fillethe super special code C(S) i.e.,

x x x

x x x

ª º« »« »

"

C C C

C C C16 rows

C C C

17 columns

ª º-« »°

° « »® « »° « »° « »¯ ¬ ¼

"

"

# # " #

"

,

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thus if x is sent message then

x x x

x x xx S

x x x

ª º« »« » « »« »« »¬ ¼

"

"

# # " #

"

so when

1 2 17

1 2 17

16 16 16

y y y

y y yy S

y y y

ª º« »« »« »« »« »¬ ¼

"

"

# # " #

"

is the received super special code then we see ifelement in y(S) is the same or that element in repeats itself is taken as the approximate correct mARQ protocols can be avoided thus saving both

money. Also when the number of cells in C(S) is inis greater than that of the number of elements in thwhere C is the code then we can easily be guarante

how the super special code C(S) functions so tcan never guess the same or break the message.Suppose

1 1 11 2 32 2 21 2 3

C C C C

C C C C

ª « « «

"

"

# # # #

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1 1 11 2 3

m 1 m 1 m 1 m1 2 3 n

1 1 11 2 3

C C C CC S

C C C C

C C C C

« «

«

« « « « « ¬

# # # " #

"

"

# # # " #

"

The real message carrying code is say in 1st rowand so on and the last row. Very different codecarry the message will also be repeated so thfrequency of the repetition intruder even cannot concerned who are the reliable part and communication work knows the exact super rothe messages so they would only look into thaguarantee the error freeness during transmissionrows carry the same message. In our opinion thlarger number of rows and columns in C(S) impossible for the intruder to break it.

We give yet another super special code impossible for any intruder to break and whichmaximum security.

8 L t

that C(S) has some number of codes repeated a few p1) then another code repeating (say some p2) timeother code repeating p3 times and so on finally yet a

repeating some pr  times. This is done for each andOnly the concerned persons who are part and pagroup know at what stages and which codes carry tand such keys are present only with the group who exinformation so it is impossible for an intr der to b

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information so it is impossible for any intruder to bhigh percentage of confidentiality and security is mai

We give yet another type of super special code C(S).

9. Let C(S) be a super special super code. Supposecodes arranged in the column and m codes along thesuper matrix of the code C(S). Now the m × n

arranged in special super blocks where by a block wu q array of same code i.e., say if C is the code thensuper block has

q columns

C C C

C C C p-rows

C C C

-ª º°« »

°« »®« »°« »°« »¬ ¼¯

"

"

# # " #

"

.

Thus the code C(S) has1 1q q1 1 2 1

1 1 1 1 1 1r s , r s , , r s ,u u u!

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1 1 1

1 1 1

1 1 1

1 1 1

1 1

1 1 1

C C C C C C

C C C C C C

C C C C C C

C C C C C CC S

C C C C C C

C C C C C C

ª º« »« »« »« » « »« »« »« »

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1 1 1

1 1 1C C C C C C« »« »¬ ¼

where C(S) is a 7 u 6 super special super code. NowC1 are only (n, k 1) codes or (n, k 2) codes k 1 z k 2.which uses this super special codes can agree uponcode C to carry the messages and C1 are misleadinanyone in this group will analyse only the codes in c

and 6 ignore columns 1, 3 and 4.

We give now the example of a block and misleadingfor the cryptographist.

 Example 4.2: Let the super special super code

1 1 1 2 2 3 3 3

1 1 1 2 2 3 3 3

1 1 1 2 2 8 8 8

4 4 5 2 2 8 8 8

4 4 5 10 10 8 8 8

4 4 5 10 10 8 8 8

C C C C C C C C

C C C C C C C C

C C C C C C C C

C C C C C C C C

C C C C C C C CC S

C C C C C C C C

ª « « « «

« « « «

1 1 1 2 2 3 3

1 1 1 2 2 3 3

1 1 1 2 2 8 8

4 4 5 2 2 8 8

4 4 5 10 10 8 8

H H H H H H H

H H H H H H H

H H H H H H H

H H H H H H H

H H H H H H HH S

ª «

« « « « « «

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4 4 5 10 10 8 8

4 4 5 11 11 11 11

4 4 5 11 11 11 11

6 6 5 11 11 11 11

6 6 7 7 7 7

H SH H H H H H H

H H H H H H HH H H H H H H

H H H H H H H

H H H H H H H

7

« « « « « « « « ¬

where each Hi is the parity check matrix of a co…, 12. Further all the code C1, C4, C5, C6, C7, CC3, C12, C9 have the same length.

Further all codes given by H1, H2, H8, H3, HH6, H12, H7 and H10 have the same number of

 Now any super code word x(S) in C(S) would be

1 1 1 2 2 3 31 2 3 1 2 1 21 1 1 2 2 3 34 5 6 3 4 5 61 1 1 2 2 8 87 8 9 5 6 1 24 4 5 2 2 8 81 2 1 7 8 4 5

4 4 5 10 10 8 83 4 2 1 2 7 84 4 5 10 10 8 85 6 3 3 4 10 11

x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x x

x x x x x x xx S

x x x x x x x

ª « « « « « « « «

ti

i pH y = (0) will make the receiver accept it otherw

it or find the error using some techniques discussed e Now only few blocks are real blocks for which

work and other blocks are misleading blocks. Since ahave the same number of check symbols and the lethe 12 codes are the same the intruder will not be into make any form of guess and break the message

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y g gwill not even be in a position to find out which of themisleading block of C(S) and which of them reallymessage.

Thus this provides a very high percentage of conand it is very difficult to know or break the messagewhen properly used will be a boon to the cryptograph

We give yet another example of a super special which would be of use to the cryptographist.

 Example 4.3: Let C(S) be a super special super code

1 1 1 1 1 2 2 2 2

2 2 2 2

1 1 1 1 1 2 2

1 1 1 1 1 2 2 3

5 5 5 6 6 6 6

6 6 6 4

5 5 4 4

5 6 6 6 4

C C C C C C C C C

C C C C C C C C CC C C C C C C C C

C C C C C C C C CC S

C C C C C C C C C

C C C C C C C C C

C C C C C C C C C

C C C C C C C C C

ª

« « « « « « « « « « «¬

H(S) =

1 1 1 1 1 2 2 2

2 2 2

1 1 1 1 1 2

1 1 1 1 1 2 2

5 5 5 6 6 6

6 6 6 4

H H H H H H H H H

H H H H H H H H H

H H H H H H H H H

H H H H H H H H H

H H H H H H H H

H H H H H H H H

ª « « « « « « « « «

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5 5 4 4

5 6 6 6

H H H H H H H H

H H H H H H H H H

« «

« ¬

Here only the parity check matrix H gives the nall other parity check matrices H1, H2, H3, H4, Hnot be even known to the owner of this system o

  Now everyone in group will be given a trufollows:

F F F F F F F F F F

T T T T T F F F F T

F F F F F T F T F F

F F F F F F T F F F

F F F F F F T T T T

T T T F F F F T T F

F T F T T T F F T F

T F T F F F T T F T

ª « « « «

« « « « « « « ¬

or equivalently they can be supplied with a true c

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u u u u u u u u uª º« »u« »« »u« »u« »« »u« »

u« »« »

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« »u¬ ¼

i.e., the group is advised to read only the messthe first row and the 5th column of C(S) all othnot carry any sense to them for they are omessages or the messages can take place as the every member of the group.

For instance K is the first alphabet of somegroup then we need to decode the message from

u uª º« »u u« »

« »u u« »u u« »

« »u u« »u u« »

« »u u

« »u u« »« »u u« »¬ ¼

1 2 3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18

19 20 21 22 23 24

25 26 27 28 29 30

31 32 33 34 35 36

y y y y y y

y y y y y y

y y y y y y

y y y y y y

y y y y y yy Sy y y y y y

ª º« »« »« »« »« »« »« »« »« »

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37 38 39 40 41 42

43 44 45 46 47 48

49 50 51 52 53 54

y y y y y y

y y y y y yy y y y y y

« »« »

« »« »« »¬ ¼

He needs to decode only the messages y1, y7, y13, yy37, y43, y49, y26, y21, y16, y11, y6, y33, y40, y47 and y

easily seen to form the letter K.It can also be at times symbols like ‘cross’ orstar.

For instance if y(S) is the received message given

1 2 3 4 5 6

8 9 10 11 12 13

15 16 17 18 19 20

22 23 24 25 26 27

29 30 31 32 33 34 3

36 37 38 39 40 41 4

43 44 45 46 47 48 4

y y y y y y y

y y y y y y yy y y y y y y

y y y y y y yy S

y y y y y y y

y y y y y y y

y y y y y y y

ª

« « « «

« « « « « ¬

DEFINITION 4.1: Let C(S) be a super special sup

of the codes C i in C(S) is cyclic then we call C special super cyclic code.

We illustrate this by the following example.

 Example 4.5: Let C(S) be a super cyclic code. L

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1 11 2

2 21 2

C C

C S C C

ª º

« »« »¬ ¼

where each i jC is a cyclic code 1 d i d 2 and 1 d

1 11 22 21 2

H HH S H H

ª º « »« »¬ ¼

where

11

0 0 1 0 1 1 1

H 0 1 0 1 1 1 0 H

1 0 1 1 1 0 0

ª º« » « »

« »¬ ¼and

1 22 2

0 0 1 0 0 1

H 0 1 0 0 1 0 H

1 0 0 1 0 0

ª º« » « »« »¬ ¼

i.e.,0 0 1 0 1 1 1 0 0 1ª

«

Given a super special cyclic code

x(S) =1 11 2

2 21 2

x x

x x

ª º

« »« »¬ ¼

where each i jx is a cyclic code 1 d i, j d 2. Any supe

x(S) in C(S) is of the form

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x(S) in C(S) is of the form

1 21 11 22 2

x xx x

ª º« »« »¬ ¼

= 1 0 0 0 1 1 0 1 1 0 1 1 0

. H S1 1 1 1 1 1 1 1 1 1 1 1 1

ª º« »

« »¬ ¼

T1 1 1 11 2 1 22 2 2 21 2 1 2

H H x x

H H x x

ª º ª º « » « »

« » « »¬ ¼ ¬ ¼

t t1 11 11 21 2

2 2 t t2 21 2

1 2

x xH H

H H x x

ª ºª º « » « » « »« »¬ ¼ « »¬ ¼

t t1 1 1 1

1 1 2 2

t t2 2 2 21 1 2 2

H x H xH x H x

ª º« » « »« »¬ ¼

1 2 n1 1 11 2 n2 2 2

1 2 nm m m

x x x

x x x

x x x

ª º« »« »« »

« »« »¬ ¼

"

"

# # " #

"

,

where each  jix is a row vector 1 d i d n and 1 d j

of s per matri from no on ards ill be kno

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of super matrix from now on wards will be know

cell matrix i.e., each cell in x(S) is a row vector. Now we say two super row cell matrices x(equal if and only if each  j j

i ix y . We see twomatrices are of same order if and only if numberx(S) is equal to number of row cells in y(S) columns cells in x(S) is equal to number of colum

and further the number of elements in the rowsame as the number of elements in the row cell

and 1 d j d m where

1 2 n1 1 1

1 2 n2 2 2

1 2 nm m m

y y y

y y yy(S)

y y y

ª º

« »« » « »« »« »¬ ¼

"

"

# # " #

"

.

We will illustrate this situation by examples. Example 4.6: Let

 be a super cell matrix.1 2 31 1 11 2 32 2 2

1 2 33 3 3

y y y

y(S) y y y

y y y

ª º« »

« »« »¬ ¼

 be a super cell row matrix i.e., let

0 1 1 0 0 1 1 1 1 0 0 1 1 1 1 0 0ª

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

y S 0 1 0 0 1 1 0 1 0 0 1 0 1 0 1 11 1 1 1 0 0 1 0 1 1 0 1 0 1 0 1

ª «

« « ¬

we say y(S) and x(S) are of same order of same typy(S) z x(S). Suppose

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

x S 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 01

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

ª «

« « ¬

and

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

y S 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 01

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

ª « «

« ¬

x(S) = y(S) if each row cell in them are identical. Ty(S). Now x(S), y(S) can also be called as supevectors. Now we define the dot product of two

1 2 m1 1 11 2 m2 2 2

1 2 mn n n

y y y

y y yy S

y y y

ª º« »« » « »

« »« »¬ ¼

"

"

# # " #

"

.

Then we define the dot product of x(S) with y(S)

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1 1 2 2

1 1 1 11 1 2 22 2 2 2

1 1 2 2n n n n

x , y x , y x

x , y x , y x(x(S), y )S

x , y x , y x

ª « « « « « « ¬

"

"

# # "

"

where i i j jx , y is the usual inner (dot) product o

d i d m, 1 d j d n. Basically all these row vectosubspaces of the vector spaces.

 Now we illustrate this situation by the following  Example 4.7: Let

110 111101

111 011101x S

001 100010

010 011001

ª º« »« »« »

« »« »¬ ¼d

1 1 0 , 0 1 0 1 1 1 1 0 1 , 1 1

1 1 1 , 1 0 1 0 1 1 1 0 1 , 1 0x S , y S

0 0 1 , 0 1 1 1 0 0 0 1 0 , 1 00 1 0 , 1 1 0 0 1 1 0 0 1 , 0

ª « «

«

« « « ¬

1 1ª º« »

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0 1

1 0

1 0

« »

« » « »« »« »¬ ¼

is just a super matrix which can be called as a superas usual 4 u 2 matrix is divided as cells.

  Now having defined the notion of inner produrow cell vectors or matrices, we proceed to dorthogonal super code.

DEFINITION 4.2: Let C(S) be a super special super i.e., the code words are from the field of characterist

 from Z 2 = {0, 1}. Let 

ª º« »« » « »

« »« »¬ ¼

"

"

# # " #

"

1 2 m

1 1 1

1 2 m

2 2 2

1 2 m

n n n

 x x x

 x x x x S 

 x x x

is such that 

ª º« »« »

« »« »« »¬ ¼

"

"

# # " #

"

0 0 0

0 0 0

 x S , y S  

0 0 0

 ,

then we say the super code words are orthogona

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 Now suppose C(S) is a super special super code

all super codes {w(S) / (w(S), (x(S)) = (0)(S) for is defined to be the super special orthogonal co

 special code C(S).

  Here w(S)    {The collection of all n u  m vectors of same order as x(S) with entries from {

 fact this collection can be realised as super spec

over Z 2 = {0, 1}. We denote this super special orcode by (C(S))

A. We see 0(S) is the only code wo

the whole space V(S) is orthogonal to 0(S).

We will illustrate this situation by the follow

 Example 4.8: Let

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0,

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

-ª º ª º ª °®« » « » «

« » « » « °¬ ¼ ¬ ¼ ¬ ¯

1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 ;

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

ª º ª « » « « » «¬ ¼ ¬

ThusC C

C(S)C C

ª º « »

« »¬ ¼

where C is the linear binary (4, 2) code given by {(0 1 1), (0 1 0 1), (1 1 1 0)}. CA = {(0 0 0 0), (1 1 0 1), 0 1 0)} is the orthogonal code of C.

The super special orthogonal code of C(S) given

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The super special orthogonal code of C(S) given

given by(C(S)A) =

C C

C C

A A

A A

ª º« »« »¬ ¼

.

Clearly(C(S)), (C(S)A) = 0(S).

We give a general method of finding (C(S)A) given CLet

(C(S)) =

1 1 m1 2 n2 2 m1 2 2

1 2 mn n n

C C C

C C C

C C C

ª º« »« »« »« »« »¬ ¼

"

"

# # " #

"

;

i jC are linear codes 1 d i d m and 1 d j d n.

1 1 m1 2 1

2 2 m

C C C

C C C

A A A

A A A

ª º« »« »

"

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t t t1 1 2 2 m m1 1 1 1 1 1

t t t1 1 2 2 m m2 2 2 2 2 2

t t t1 1 2 2 m mn n n n n n

H x H x H x

H x H x H x0

H x H x H x

ª º« »« »« » « »« »« »« »¬ ¼

"

"

# # " #

"

x(S) C(S) otherwise we use the usual error

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x(S) C(S) otherwise we use the usual error

techniques to each cell i.e., to each  j

ti i jH x to

corrected code word; 1 d i d n; 1 d j d m.Thus we have indicated how these super specia

 be used when ARQ is impossible. Also these codes wuseful in cryptology. Further use of these codes can

time and economy. The applications of super speciaand super special column codes can be obtained special codes with appropriate modifications.

FURTHER R EADING

1. Berlekamp, E.R.,  Algebraic Coding TheoryInc, 1968.

2. Bruce, Schneier.,   Applied Cryptography, SJohn Wiley 1996

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John Wiley, 1996.

3. Gel'fand, I.M.,   Lectures on linear algebr New York, 1961.

4. Greub, W.H.,   Linear Algebra, Fourth EdVerlag, 1974.

5. Halmos, P.R.,   Finite dimensional vector  Nostrand Co, Princeton, 1958.

6. Hamming, R.W.,   Error Detecting and ecodes, Bell Systems Techical Journal, 29, 14

7. Herstein I.N., Abstract Algebra, John Wiley,8. Herstein, I.N., and David J. Winter, Matr

 Lienar Algebra, Maxwell Pub., 1989.

9. Hoffman, K. and Kunze, R.,  Linear algebrof India, 1991.

10. Horst, P., Matrix Algebra for Social S

13. Mac William, F.J., and Sloane N.J.A., The TheoCorrecting Codes, North Holland Pub., 1977.

14. Pless, V.S., and Huffman, W. C.,   Handbook

Theory, Elsevier Science B.V, 1998.

15. Shannol, C.E.,  A Mathematical Theory of ComBell Systems Technical Journal, 27, 379-423 an1948.

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16. Shilov, G.E.,   An Introduction to the TheorySpaces, Prentice-Hall, Englewood Cliffs, NJ, 196

17. Thrall, R.M., and Tornkheim, L., Vector smatrices, Wiley, New York, 1957.

18. VaN Lint, J.H.,   Introduction to Coding Theory

1999.

19. Vasantha Kandasamy and Rajkumar, R. Uapproximations in algebraic bicoding theory, Journal of Mathematical Sciences, 6, 509-516, 20

20. Vasantha Kandasamy and Thiruvegadam, N., of pseudo best approximation to coding theory,17, 139-144, 2005.

21. Vasantha Kandasamy, W.B., and SmarandacheSuper Linear Algebra, Infolearnquest, Ann Arbo

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INDEX

B

Best approximation, 45-8

C

Canonical basic matrix, 30Check symbols, 27-8Column super vector, 16-7Control symbols, 27-8Coset leader, 39-40Cyclic code, 42-4

G

Generator matrix, 30Group code, 29

H

Hamming code, 40-3Hamming distance, 32-3H i i ht 32 3

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Hamming weight, 32-3

I

Inner product, 45-7

L

Linear code, 29

M

Minimum distance, 32-4Mixed super special row code, 96

O

Orthogonal code, 35-6

P

S

Simple matrix, 8Standard inner product, 45-7Submatrices, 9-10Super diagonal matrix, 14-5Super error vector, 80Super error word, 80S H i di t 81

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Super Hamming distance, 81Super Hamming weight, 81Super inner product, 61-2Super linearly dependent, 55Super linearly independent, 55Super minimum distance, 81Super mixed row code, 89-90Super mixed special column code, 117Super row vector, 54-5Super special basis, 55Super special binary code, 150-1Super special codes, 120Super special column code, 99-101Super special column Hamming code, 115Super special column parity check matrix, 111-2Super special cyclic code, 145Super special cyclic column code, 118-9Super special cyclic row code, 87-8Super special dual row code, 82-3Super special generator mixed column matrix, 10Super special Hamming row code, 97-8

Super special row vector, 72-73Super special subspace, 56, 69-71Super special syndrome, 83Super special vector space of dimension m, 55Super special vector spaces, 53-4Super transmission, 69-71Supermatrices, 7-12Symmetric portioning of a matrix, 12-3Syndrome, 39-40

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