Problem 1. Let a
be the primitive element of GF(24) which is the zero of the primitive polynomial
p(x) = x4 + x + 1 .
Let g(x) be the binary polynomial of smallest degree having
a and a5
as roots. Let V =
( g(x) ) be the cyclic code of
smallest length having g(x) as a generator polynomial. Use a
and a5 to
construct a parity check matrix H of V . (Do not explicitly compute g(x) . )
Hint: Let K
denote the matrix:
[ 1 a
a2
a3
a4
... a14 ]
K=
[
]
[ 1 (a5)1 (a5)2 (a5)3
(a5)4 ... (a5)]14
]
Then
f(x) = f0+f1x+f2x2+
... f14x14 e
V
iff
(f0,
f1, f2, ... , f14)KT = [ f(a), f(a5) ] = [ 0,
0 ]
Next let H' denote the
binary matrix formed by replacing each element of GF(24)
in K' with the corresponding binary 4-tuple written as a
column vector. Then the row space of H' is VPerp. Unfortunately, the rows of H' may be linearly dependent.
Form the parity check matrix H
by putting H' in row echelon canonical form and deleting the all
zero rows.
Problem 2.
Let x be the primitive element of GF(26)
which is the zero of the primitive polynomial:
p(x) = x6 + x + 1 .
Let g(x) be the polynomial of smallest degree having the
following zeros:
x, x2, x3, x4, x5, x6, x7, x8
Let V =
( g(x) ) be the corresponding cyclic
code of shortest length.
Note: If you have not installed the symbolic fonts on your web
browser, then the above ksi's will look like x's.