Algebriac Coding Theory

Basic DefinitionsWord : Let F be a field then the expression of

the form a1, a2, …, an where aiF i

is called a word of length n over the field F. We denote the set of all words

of length n over F by F(n).

Weight :

Let a=a1, a2, …, an F(n)

Then the number ai≠01 is called the weight of the word a & is denoted by

wt(a). i.e. weight of a word is the number of non zero entries of that

word.

Distance :

Let a=a1, a2, …, an & b=b1, b2, …, bn be two words of length n over the

field F. We define the distance between a & b (written d(a,b)) by

d(a,b)=i=1,…,n d(ai,bi)

Where d(ai,bi) = 0 if ai=bi

1 if ai≠bi

Code :

A block (m,n) code over a field F consist of an encoding function

E : F(m) F(n) (m < n) & a decoding function

D : F(n) F(m)

such that DoE is the identity function or almost near to identity function.

Elements of F(m) are called message words and the elements of image of E in F(n) are called code words.The collection of all the code words in F(n) is denoted by C. Thus C is a subset of the vector space F(n). If C becomes a subgroup of F(n) then we say that C is a Group Code.

Message

Word

Code Word

Received

Word

Code Word

Message Word

E D

Channel

Hamming code

Procedure for constructing Hamming code:

a) Let r be a positive integer greater then 2 (r>2). Then the message word in the code C are of length m = 2r-r-1 & the code words are of length n = 2r-1.

b) Let (b1,b2,…,bn) be the code word corresponding to the message word (a1,a2,…,an) in which b2^0, b2^1,…,

b2^(r-1) are check digits

And the remaining positions are filled by the digits of the message word in the order in which they are occuring in the message word.

c) Let M be a matrix of the type nr whose ith row is the binary representation of the number i. Consider the matrix equation bM=0. Let M1, M2,…,Mr be the r columns of M then the matrix equation bM=0 gives r linear equations bM1=0, bM2=0,…,bMr=0.

First We prove that each of these r equations contains exactly one check digit.

Suppose b2^i occurs in the equation which is obtained by multiplying b with the kth column of the matrix M. Then (2i)th entry of kth column is 1. This entry is in the (2i)th row of M & (2i)th row of M is the binary representation of the number 2i. Now the binary representation of 2i has 1 at (i +1)th place and 0 elsewhere i.e. 1 at the (r - i)th place from the left. k=r - i

Now consider the equation bMk=0Suppose this equation contains two check symbols b2^i & b2^j.Then as before k=r – i& k=r – ji.e. i = j. Thus there is atmost one check symbol present in the equation bMk=0 Again we have shown that each check symbol b2^i occurs in the linear equation bMk=0 for k=r-i. The number of check symbols & the number of

Linear equations is same and equal to r & none of the equation have more than one check symbol so there is exactly one check symbol in each of the r linear equations.

Solving these r linear equations we get the unique value of each check symbol. Hence E(a1,a2,…,am)=(b1,b2,…,bn)

is the encoding function for the code C.

Theorem: Prove that Hamming code is a group code.

Proof: Let C be a (m,n) hamming code over the field F. We know that m= 2r-r-1 & n=2r-1 for some positive integer r 2. The encoding function E of C is given by E: F(m) F(n)

E(a1,a2,…,am)=(b1,b2,…,bn)Where b2^0,b2^2,…,b2^(r-1) are check digits and the remaining positions are filled by the digits of the message word

in the order in which they appear in the message word. The check digits are given by the matrix equation bM=0 where M is the nr matrix in which ith row is the binary representation of the number i. Supppose E(a’)=E(a1’,a2’,…,am’) =b1’,b2’,…,bn’ Now a+a’=(a1+a1’),(a2+a2’),…,(am+am’)& b+b’=(b1+b1’),(b2+b2’),…,(bn+bn’)Also bM=0 & b’M=0

(b+b’)M=bM+b’M =0 + 0 = 0 Also in b+b’, b2^0+b’2^0, b2^1+b’2^1, …, b2^(r-1)+b’2^(r-1) are check digits & the remaining entries are a1+a’1, a2+a’2, …, am+a’m in their original order. Thus b+b’ corresponds to the message word a+a’. Hence E(a+a’) = b+b’ = E(a) + E(a’) Thus E is a homomorphism & so C is a group code .

Theorem: Prove that minimum distance of a hamming code is 3.

Proof: Let C be a (m,n) hamming code over the field F. We know that m= 2r-r-1 & n=2r-1 for some positive integer r 2. The encoding function E of C is given by E: F(m) F(n)

E(a1,a2,…,am)=(b1,b2,…,bn)Where b2^0,b2^2,…,b2^(r-1) are check digits and the remaining positions are filled by the digits of the message word

in the order in which they appear in the message word. The check digits are given by the matrix equation bM=0 where M is the nr matrix in which ith row is the binary representation of the number i. Again let M1, M2, …, Mr be the r columns of M then bM=0 is equivalent to r linear equations bM1=0, bM2=0, …, bMr=0. Also we know that each linear equation contains one & only one check symbol.

We have also shown that C is a group code. So to prove that minimum distance of C is 3 it is sufficient to prove that each non zero code word is of weight 3. Now by definition of C all the digits of the message word apppear in the code word so we must prove that the weight of a code word corresponding to the message word of weight 1 or 2 is always greater than or equal to 3.

Case (1) Let a=a1,a2,…,am be a message word of weight 1 & let b=b1,b2,…,bn be the corresponding code word. Let the non zero entry of a occurs at the ith position in b i.e. bi≠0. Since bi is a message digit so i is not a power of 2. Therefore the binary representation of i contains atleat two non zero entries i.e. ith row of M contains atleast two non zero entries. Suppose sth & tth entries of ith row are non zero.

So (i,s)th and (i,t)th entry of M is 1.Consider the equations bMt=0 & bMs=0

Let b2^k be the check symbol present in the equation bMt=0 & b2^l be the check symbol present in the equation bMs=0.

Since all the entries of the message word except bi are zero. So bMt=0 is of the form b2^k+bi=0

i.e. b2^k = -bi ≠ 0.

Similarly b2^l is also non zero.

Hence the code word b corresponding to the message word a of weight 1 contains atleast two non zero check digits.So wt(b)3 Case (2) Suppose the weight of the message word a=a1,a2,…,am is 2 then the corresponding code word b=b1,b2,…,bn

Contains two non zero entries corresponding to the message word a.

Let these entries be bi & bj. We may also suppose that i < j. Since bi & bj are the entries of a therefore i & j are not the power of 2. Again i ≠ j. The binary representation of i & j must differ atleast at one place. Let the binary representation of i & j differs at sth place (from the left). By interchanging i & j we may suppose that (i,s)th entry of M is 1 & (j,s)th entry of

M is zero. Consider the linear equation bMs=0.

This linear equation contain the unique check digit say b2^k. Since every digit of the message word except bi & bj is zero.

\bMs=0 is equivalent to b2^k+bi=0

i.e. b2^k= -bi ≠ 0.

Thus the code word b corresponding to the message word a contains atleast one non zero check symbol.

Since it already contains two non zero check symbols so wt(b) 3. Thus in all the cases wt(b) 3. Now to show minimum distancs of hamming code is 3 we are to find a code word of weight 3. Let b=b1, b2, …, bn be the code word corresponding to the message word a=a1, a2, …, am where a1=1 and ai=0 for 2 i m. So b3=1.

Consider the equation bMs=0 then the check symbol present in the equation takes the value 1 iff third entry of the sth column of M is 1. So the number of non zero check symbols in b is equal to the number of non zero entries in the third row of M. Third row of M is the binary representation of the number 3 which contains exactly 2 1’s. Thus b contains exactly two non zero check symbols and one non zero message symbol. So wt(b) = 3.

TEST

Do any two:

1) Construct a Hamming (4,7) code.

2) Prove that hamming code is a group code.

3) Prove that minimum distance of hamming code is 3.