The Error Correcting Codes

Coding Theory
R1 =(10101 110110 11100)
10101=1+x2 +x3 +x5
110110=1+x+x3 + x5 +
11100=1 + x + x2
R1 is a cyclic [3,5] over the polynomial g(x)=x3 + x4 + x5 +x6
R2 =( 11000 01110 10011)
11000=1 +x
01110=x +x2 + x3
10011=1 + x3 + x4
R2 is a cyclic [3,5]-order over polynomial g(x)=x3 + x4 + x5 +x6
R3 =(01000 00010 11111)
01000 =x
00010 =x4
11111=1 + x + x2 + x3 + x4
R3 is a cyclic [3,5] –order over polynomial (x)=x3 + x4 + x5 +x6
Problem 2
Part a
Let y=x
If g(y) is a generator polynomial of a cyclic code C,then you will have
F(y)=yn-1 =h(y)g(y) +p(y) where;
deg.p(ydegg(y)

Thus, p(y)=-g(y)h(y)mod(yn-1), hence p(y) ∈C
But;
P(x)∈is not possible unless p(y)=0
Assume that g(y) can divide xn-1 in regard g(y) is ideal and all multiples of
g(x) are reduced to modulo (yn-1)
Let there be a polynomial b(y) which is ideal and has got a smaller degree.
There is enough evidence to show that there is no such b(x) that would exist
and g(x) is the smallest degree polynomial, hence it is ideal and a generator.
For Instance;
Consider V[7,2] and g(y)h(y)=y7-1
To completely factorize g(x)h(x) over GF(2),you get;
y7-1=(y+1)(y3+y2+1)(y3 + y + 1)
And you can get the monic divisors to be ;
g 1(y) =1
g 2(y) =y+1
g 3(y)=y3 +y2 +1
g 4(y) = y3 +y +1
g 5(y)=(y +1)(y3 + y2 + 1)
g 6(y)=(y + 1)(y3 +y+1)
g 7(y)=(y3 +y2 +1)(y3 + y2 +1)
Assume that h(y) = F 2( y )
y n−1
You can see that C is a cyclic code and a polynomial g(y) is such that h(y) is
in a way that C=g(y)
Assume C1 is a subcode of the codomain hence deg(y) generates [n,k-1]code
Because C1 is cyclic,C1=g(y) for certain g(y)∈ h[y]

Therefore, degg(y) =(n-k)=n-k+1
As a result,we get;
C⊂C1 and f(y) divides g(y) in F 2( y )
y n−1 .This makes g(y)=(y-a)f(y)
When y-a ∈h(y) ,a=1 or a=0
As g(y) can divide yn-1 and y does not divide yn-1 in spite of being y-1 does,
you get a=1.
This may mean that y-1 is a divisor of g(h) and h(y).
However,g(y) is a generating polynomial of C1 and therefore, for all h(y) ∈ C
,you will realize that h(1) =0.
This is the roots of all codes in C1 and might be an indication that all
codewords in C1 are even in weight of vector C.
Because n is odd,yn-1 is a seperable polynomial and as g(h) divides yn-
1 ,g(y) is also seperable.
Thus,from the realization,you can note that y-1 is not a divisor of the
function.
As a result,it proves that f(1) ≠0,hence f(y) is odd in weight.
And C contains both odd and even vectors .in addition,a set of vect C is a
subcode of C1 with one as the dimension.
Consequently,C1 is precisely a set of all the even weight vectors in C.
You can confirm that :
g(y) would generate the full space of V[7,2]
g 8(y) would generate the trivial cyclic subspace {(0000000)}
g6(y) generates the cyclic code {(0000000),(1011100),(0101110),(0010111),
(1001011),(1100101),(1110010),(0111001)}
g 7(y) generate the cyclic code {(0000000),(1111111)}

V[7,2] precisely contains 8 cyclic codes
You can therefore conclude that code C with the polynomial g(y) is self-
orthogonal if and only if h*(y) divides g(y).
Part b
Consider a linear cyclic code over GF(4) to be of length n (4m-1) and assume
the polynomial generator g(x) to be self-orthogonal on condition that :
h (x)g(x) =0mod(xn-1)
Take h(x)= ∑
0
n −1
gr Xr
Thus g(x)=GCD(gn-rxr,xn-1)
Understand that the polynomial generator of the cyclic codes are expressed
in the form of their zeros in GF(n).
Assume that ʎ∈GF(4m) to be a primitive unity root.
Suppose g(x) has got a root of ʎ2
Then ,
g (ʎz)=g0+g1 ʎz +g2 ʎz +............+gr ʎ(n-1)z=0
=g0+g1 ʎ2z +g2 ʎ4z +............+gn-1 ʎ(n-1)2z=0
=
=g0+g1 ( ʎ-2z )+g2 (ʎ-2z)n-2 +............+gn-1 (ʎ-2z)=0
And therefore ,a polynomial g0+g1 xn-2 + +............+gn-1x=0 has got ʎ-2z as
the root.
Because the root xn-1 is ʎ-2z ,you can also confirm it to be the root of g(x)
It also follows that Z ⊂{0,1,2,3,..........,n-1} is a set of zeros of g(x) and this
may also mean that Z1={-2zmod n|z∈ Z} is a set of g(x)
Supposing that n=15,the cyclotomic consent modulo of n when multiplied
by 4 gives {0},{1,4},{2,8},{3,12},{5},{6,9},{7,13},{10},{11,14}.