The code word ........................... sent by the sender using the group codes words generated by \(G=\begin{vmatrix}1 & 0 & 0 & 1 & 1 & 0\\0 & 1 & 0 & 0 & 1 & 1\\0 & 0 & 1 & 1 & 0 & 1\end{vmatrix}\), which is received as 001100 by the receiver.
001101
For a linear group code, every transmitted code word is a sum (modulo 2) of rows of the generator matrix G. The receiver decodes a received word to the code word at the smallest Hamming distance from it (maximum-likelihood decoding), which corrects any single-bit error because this code has minimum distance 3.
Step 1 — List the code words \(c = mG\) for the eight messages m: 000000, 100110, 010011, 001101, 110101, 101011, 011110, 111000.
Step 2 — Compare each with the received word 001100: the distances are 2, 3, 5, 1, 4, 4, 2 and 3 respectively.
Step 3 — The unique code word at distance 1 is 001101, so a single error occurred in the sixth bit. Syndrome decoding confirms this: the message part 001 should carry parity 101, the received parity is 100, the syndrome 001 matches the sixth column of the parity-check matrix \(H = [A^{T} \mid I_3]\), and flipping bit six gives 001101.
000000 is a valid code word but lies at distance 2, so two errors would be needed. 101011 is also a code word but lies at distance 4. 000111 cannot have been sent because it is not a combination of the rows of G.
001101 option is correct.
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