4G (Coding and Cryptography)

Solution

C \leq F_{2}^{n}

r

. Use the weight enumerator convention

W_{C} (s, t) = \sum_{c \in C} s^{w (c)} t^{n - w (c)},

where

w (c)

is the Hamming weight. This convention is essential here:

W_{C} (1, 0)

n

, rather than zero words. There are

2^{r}

W_{C} (1, 1) = 2^{r}

There is only one binary word of weight

n

, namely

1 = (1, \dots, 1)

. Thus

W_{C} (1, 0) = 1

precisely when

1 \in C

. If this holds, the linear map

c \mapsto c + 1

C

and replaces weight

w

n - w

; hence

W_{C} (s, t) = W_{C} (t, s)

. Conversely this symmetry gives

W_{C} (1, 0) = W_{C} (0, 1) = 1

, because the zero word is the unique word of weight zero.

The repetition code consists of

0, 1

W_{rep} (s, t) = t^{n} + s^{n} .

The single parity-check code consists of all words of even weight. Exactly

(j n)

j

W_{parity} (s, t) = j even \sum (j n) s^{j} t^{n - j} = \frac{( t + s ) ^{n} + ( t - s ) ^{n}}{2} .

There are no discussions about this article yet.

There are currently no matching articles.