TOOLS FOR COSET WEIGHT ENUMERATORS

5

2.2. Cosets of codes of A. Let C be an [N, k)-code of A. We first recall the

MAc-WILLIAMS transform, which determines uniquely the weight polynomial of

the dual of C from the weight polynomial of the code C itself.

THEOREM

1. [16, p.146) Let C be a linear

[N,

k]-code. Define its weight poly-

nomial:

N

Wc(X, Y)

=

LAixN-iyi, Ai =card { c

E

C

I

w(c)

=

i},

i=O

where w(c) is the weight of c. Then the weight polynomial of the dual code Cj_

lS

(7)

1

W0 .L(X, Y)

=

kWc(X

+

(p -1)Y,X- Y).

p

Denote by Dx the

[N,

k

+

1)-code generated by C and a coset Cx = x

+

C

ofC:

Dx=

U

(ax+C).

aEK

As C

c

Dx , D;t

c

Cj_ ; then every weight of Cj_ can be a weight of D;t.

Moreover the weight distributions of x

+

C can be deduced from the weight

distribution of Cj_ and of D;t. Indeed suppose that the weight distribution of Cj_

is known and that we can compute the weight polynomial of D;t. The dimension

of Dx equals k

+

1 so that the dimension of D;t equals N- (k

+

1). Applying

(7) we obtain the weight polynomial of Dx:

1

Wvx(X, Y)

=

N-(k+l) Wvt(X

+

(p -1)Y,X- Y).

p

But, by definition, Wvx(X, Y) = (p-1)Qx(X, Y)

+

Wc(X,Y), where Qx(X, Y)

is the weight polynomial of the coset x

+

C . Then the expression of Qx(X, Y)

is as follows:

Qx(X, Y)

1

( ) N ( P W D.L (X

+

(p - 1) Y, X - Y)

p-1p

-k

X

(8)

- W0 .L(X

+

(p-1)Y,X-

Y)) .

In the following we always will assume that the polynomial

W

c

.L

(X, Y)

is

known and we want to have results on the polynomial Qx(X, Y), for every x.

Then the problem consists in the determination of the polynomials W D.L (X, Y).

Let. be a non zero weight of Cj_ and

A,\(x)

be the number of code'"words of

weight .in D;t. So A.x(x) is the number of elements of Cj_ of weight .which

are orthogonal to

x:

(9) A.x(x) =card { y

E

cj_

I

w(y) =.and y,x = 0}.

Indeed y

E

D;t if and only if y, z = 0 for all z in Dx. But z =ax+ c,

a

E

K and c

E

C; as y

E

Cj_, y,c = 0. Hence y

E

D;t is equivalent to

y,x = 0.