FPGA Central

Hi,

Will the all e-codeword, e \in GF(2^m), i.e. the codeword e e e ... e,
always be present in a cyclic code. If this holds, how can I prove it?
If not, is this only true for MDS codes. (Again, how can I prove
it?)\

Your time, effort and suggestions will be greatly appreciated
Jaco

Jaco Versfeld asked:

> Will the all e-codeword, e \in GF(2^m), i.e. the codeword e e e ... e,
> always be present in a cyclic code. If this holds, how can I prove it?
> If not, is this only true for MDS codes. (Again, how can I prove
> it?)\


This looks a lot like homework but nonetheless here goes:

The codeword (e, e, ... , e) will not always be present in a cyclic
code (except, of course, for (0, 0, ... , 0)). The result is not
true for MDS codes either, unless some restricted specific type of
MDS code is being considered. The following is intended as a hint
towards figuring out when the claimed results are true.

Hint: (x^n - 1) = (x-1)(x^{n-1} + x^{n-2} + ... + x + 1) and
(x^n - 1) = h(x)g(x)

Dear Prof Sarwate and Prof Chapman.

> This looks a lot like homework but nonetheless here goes:

Wish it was . During undergrad I didn't spent much time on maths...
Now it seems to shoot me in foot.

I cross-posted the thread to sci.math/comp.dsp as well. Somehow
something didn't work. Prof Chapman responded there as well. (
http://groups.google.co.za/groups?hl...h%26start%3D50
)

> Hint: (x^n - 1) = (x-1)(x^{n-1} + x^{n-2} + ... + x + 1) and
> (x^n - 1) = h(x)g(x)

Prof Chapman's statement is: "The all-one vector is contained iff
(x^n-1)/(x-1) is not a multiple of the generator polynomial g(x)."

However, using my (limited) maths and the hints above, I get that it
should be:
The all-one vector is contained iff (x^n -1)/(x - 1) is a multiple of
the generator polynomial g(x).

From hint 1, we get an equation containing the all-one vector.

But if the all-one vector is a codeword/ code polynomial, it can be
expressed as d(x)g(x).

Thus (x^n - 1)/(x - 1) must be divisible by g(x), or (x^n -1)/(x - 1)
is a multiple of g(x).

Your comments will be highly appreciated
Jaco

Thanks to everyone that responded.

Jaco