Determine the dual code $C ^\perp$ to the code C given by following generating matrix $G$
\begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 \end{bmatrix}
And determinate the coset leaders and syndromes. If $y = 01001$ is received, which word from $C^ \perp$ is most likely to have been transmitted?
What I've done so far:
$C ^\perp=<10010,11101>. $ Because we have $(C ^\perp)^\perp=C$, it means that the parity-check matrix of $C ^\perp$ is $G$. So I can calculate the syndromes and leaders with $S(y)=Gy^T$.
Let $e_1=(10000)$, $e_2=(01000)$, $\cdot \cdot \cdot , e_5=(00001)$
So we have
\begin{array}{|c|c|} \hline Syndromes & Coset\ leaders \\ \hline 000&0 \\ \hline 001&e_1=e_4 \\ \hline 100&e_2 \\ \hline 010&e_3 \\ \hline 111&e_5 \\ \hline 110&e_2+e_3 \\ \hline 011&e_1+e_3=e_4+e_3 \\ \hline 101&e_1+e_2=e_4+e_2 \\ \hline \end{array}
We know $y = c + e$, where $c$ is the codeword transmitted, and $e$ is the error vector. $eH^T = (y - c)H^T = yH^T - cH^T = yH^T - 0 = yH^T$
Calculating we get: $yH^T=(011)$
But $(011)$ can be from $e_1+e_3$ or $e_4+e_3$.
What do I do now?
If I choose $e_1+e_3$, then $c=01001-10100=11100$, if I choose $e_4+e_3$ , then $c=01001-00110=01111$.
Whats the real answer?