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Show that the (3,7) encoding function, How many errors will it detect and correct?
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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.