Multiple-Step Word Problems -

MATHS
Problem 2
Part a
Hint:X=y
If g(y) is a polynomial generator of a cyclic code C,then you will have to say that F(y)=yn-1
=h(y)g(y) +q(y) where;
deg.q(ydegg(y)
Thus, q(y)=-g(y)h(y)mod(yn-1), hence q(y) ∈C i.e q of y is an element of C
But;
P(x)∈is not possible unless p(y) is equal to zero
Assume that g(y) can divide xn-1 in regard g(y) is true and exact, and all factors of g(x) are
decreased to modulo (yn-1)
Let there be a polynomial b(y) which is exact and has got a minimal level .
There is enough evidence to show that there is no such b(x) that would exist and g(x) is the
least level 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 as shown below;
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 sub code 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 the function 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(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 code words in C1 are even
in weight of vector C.
Because n is odd u number ,yn-1 is a separable and a polynomial generator and as g(h) divides
yn-1 ,g(y) is also separable.
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 vector C is a sub code 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(x) would generate the full space of V[7,2]
g 8(x) would produce the trivial circular minor space {(0000000)} from its vector quantity.
g6(y) produces the circular 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 precisely conclude that code C with the polynomial g(x) is self-orthogonal I
on the condition g(x) can be divided by h*(x)
Part b
Consider a straight line and circular code over GF(4) to be of magnitude 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}.
Thus (15,6) is a cyclic code over GF(4) where z is a zero set with z ={0,5,10,1,4,11,14,3,12}
You will realize that Zeros in GF(16) are self-orthogonal because, the non-zero elements like
S={7,13,2,8,6,9} are such that -2S and n⊂Z .
S has got only 4 consecutive integral hence we you can conclude that d=S is the minimum
distance of the dual code.
Consequently,[15,3,5] is a quantum code obtained from from self-orthogonal cyclic code.
You can also consider can the polynomial g(y)h(y)= y15- 1.
To completely factorize g(x)h(x) over GF(2) ,you get:
y7-1 =(y+1)(y2 + y+1 )(y4 +y + 1)(y4 + y3 + 1)(y4+y3 +1)(y4 +y3 +y2 +y + 1)
Assume C to be the [15,k] binary cycle code of length 15 generated by:
g(x)=(y+1)(y4 +y +1)
Hence
g 14(x) produces a cyclic code {(000000000000000),(010000000000000)}
V[15,2] precisely contains 16 cyclic codes.