I have this question that I need to solve.
The Parity-Check matrix $H=\begin{bmatrix}1&1&1&0&0&0\\1&0&1&1&0&1\\0&1&1&0&1&1\end{bmatrix}$ solves 1 error in code $C$.
a) Find all element in $C$.
b) Correct the word \begin{bmatrix}0&1&1&1&1&1\end{bmatrix}
c) How many words can not be corrected?
Solutions:
a) $H=\begin{bmatrix}1&1&1&0&0&0\\1&0&1&1&0&1\\0&1&1&0&1&1\end{bmatrix}$~$\begin{bmatrix}1&0&0&0&1&1\\0&1&0&1&0&1\\0&0&1&1&1&0\end{bmatrix}$
This gives: $x_1=x_5+x_6$, $x_2=x_4+x_6$ and $x_3=x_4+x_5$.
Possible combinations of $x_4, x_5, x_6$:
$(0,0,0)$ $\implies$ $c1=(000000)$
$(1,0,0)$ $\implies$ $c2=(011100)$
$(0,1,0)$ $\implies$ $c3=(101010)$
$(0,0,1)$ $\implies$ $c4=(110001)$
$(1,1,0)$ $\implies$ $c5=(110110)$
$(0,1,1)$ $\implies$ $c6=(011011)$
$(1,0,1)$ $\implies$ $c7=(101101)$
$(1,1,1)$ $\implies$ $c8=(000111)$
b) Correcting the word \begin{bmatrix}0&1&1&1&1&1\end{bmatrix} gives $c6=(011011)$.
c) Can someone please help me here, I have no idea where to start.