[Ok people, before scolding me, let me tell you this thing is giving me headaches..]
I have a $C$ = $C(8,3)$ described by this matrix:
$$G=\begin {bmatrix} 1&0&0&1&0&1&1&0\\ 0&1&0&1&1&1&1&1\\ 0&0&1&1&0&0&1&0\\ \end{bmatrix}$$
I've been asked get a shortened code of $C$. I got:
$$G'=\begin {bmatrix} 1&0&1&0&1&1&0\\ 0&1&1&1&1&1&1\\ \end{bmatrix}$$
(Started from the very beginning in order to show you "where I'm coming from")
I should try to find a Reed-Muller code from $C'$.
As far as I know, $n = 2^m$ and $k = \sum_{i=1}^{r} 1 + \binom mi$ with hamming distance $d = 2^{m-r}$.
But if $d = 4$ and $n = 7$, $m$ (that should be an integer) in those formulas seems not matching at all. Am I doing something wrong?
Also, when calculating $k$, since $r$ is the upper bound of a sum, how big should it be?
but, my BIG question is, is it possible to get a Reed-Muller code from this matrix?