Discussion
...................... is a received code word knowing that one-bit error occurred while transition of a group code word generated by G = \(\begin{vmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{vmatrix}\)
Answer: B
The received word is 0111.
The code is every sum (mod 2) of the rows of G. The three rows are 1001, 0101 and 0011, and their combinations give the eight code words:
0000, 1001, 0101, 0011, 1100, 1010, 0110, 1111
Every row has even weight, so every code word has even weight: the last column is a parity bit. A single-bit error changes the weight by one, so a word received after exactly one error must have odd weight.
0111 has weight 3, so it cannot be a code word and must be a corrupted one (it is one bit away from 0011, 0101, 0110 and 1111). 0011, 1111 and 1100 all have even weight and are themselves code words, so receiving them is consistent with no error, not with a one-bit error.
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