Question

Show that no $\left(6, 3\right)$code corrects double errors.

Short answer

It is proved that no $\left(6, 3\right)$code corrects double errors.

Step-by-step solution

Theorem 16.6 and Corollary 16.4

Theorem 16.6

• If G is a$k\times n$ standard generator matrix, then$\left\{uG|u\in B\left(k\right)\right\}$ is a systematic $\left(n,k\right)$code

• Corollary 16.4

A linear code detects2terrors and corrects terrors if and only if the Hamming weight of every nonzero codeword is at least$2t+1$ .

Showing that no 6, 3 code corrects double errors

Let’s assume there is a $\left(6, 3\right)$code that corrects double errors.

Then according to corollary 16.4, the humming weight of every nonzero word is at least $2\times 2+1=5$

Therefore, the possible code words can be:

$C=\left\{000000, 011111, 101111, 110111, 111011, 111101, 111110, 111111\right\}$

The message words of the code $\left(6,3\right)$are:

$000, 001, 010, 011, 100, 101, 110, 111$

If we combine message word $100, 010, 001$ and codeword, we are sure that the generator matrix will have two columns that consist of two 1 and one zero.

And if we combine the message word $111$ with that matrix thenthe generator matrix will have at least 2 zeros and 2 onesin code words. This contradicts because such codewords cannot exist.

Hence it is proved that no $\left(6, 3\right)$code corrects double errors.