**1 Answer**

written 2.4 years ago by |

Show that the (2,5) encoding function $e : B^2 \rightarrow B^5$ defined by

e(00) = 00000

e(01) = 01110

e(10) = 10101

e(11) = 11011

is a group code.

How many errors will it detect and correct?

**Solution:** Encoding function $e : B^2 \rightarrow B^5$defined as

e(00) = 00000 = $x_0$

e(01) = 01110 = $x_1$

e(10) = 10101 = $x_2$

e(11) = 11011 = $x_3$

$\therefore$ Range of encoding function is,

Range (e) = { $x_0 = 00000$, $x_1 = 01110$, $x_2 = 11011 $ , $x_3 =11011 $ }

Encoding function e : $B^2 \rightarrow B^5$ is said to be a group code if range of e is subgroup of $B^2$

$\therefore$ Prepare composition table for $(B_n, (+))$

$\because$ All the entries of composition table is closed.

$\therefore$ Encoding function $e : B^2 \rightarrow B^5$ is a group code.

$\because$ $e : B^2 \rightarrow B^5$ is a group codes

$\therefore$ mm distance of encoding function e is the min of weight of non-zero code words.

$\therefore$ |X_1 | = 3

| X_2 | = 3

| X_3 | = 4

$\therefore$ Minimum distance = 3

**Error detection:**

Encoding function $e : B^2 \rightarrow B^5$ can detect K or fewer errors if

min distance = k + 1

3 = k + 1

**$\therefore$ k = 2**

$\therefore$ Encoding function can detect 2 or fewer errors.

**Error correction:**

Encoding function $e : B^2 \rightarrow B^5$ can correct K or fewer errors if min distance = 2k + 1

3 = 2k + 1

**$\therefore$ k = 1.**

$\therefore$ Encoding function can correct 1 or fewer errors.