Let $C$ be the $[n,k,d]$-Code with the parity check matrix B in $\mathbb{F}_2$ with parity check matrix (sry, don't want to type that thing in Latex).
I already found out that $n=15, k=11$. For $d$ my idea was to calculate $C$ first, by solving $Bx=0$ and then calculate $d(C)$. My result is $d(c)=3=d$, is this true, or am I doing something wrong here?
Extra material
(Here you can see the solved system with matrixcalc)
Well, correct me if I'm wrong, but the parity check matrix contains each nonzero binary vector of length 4 as a column vector (15 vectors in total). This is a binary Hamming code with easy decoding.
In view of minimum distance, for each codeword $c\ne 0$, $Bc^t=0$ (transposition). But $Bc^t$ is a linear combination of the columns of $B$ (the columns at which $c$ has the nonzero entries). Such a nontrivial linear combination can only be zero if three or more columns are combined due to the structure of the columns of $B$. So the Hamming weight of $c$ is at least 3. Moreover, there exist three columns of $B$ that add up to zero and so the minimum Hamming weight of the code is indeed 3, i.e., the minimum distance is $d=3$,
In view of decoding, if $c$ is the vector sent, $y$ is the vector received, and $e=y-c$ is the error vector, then $Be^t = By^t-Bc^t = By^t$. If there is one error, then $By^t = Be^t\ne 0$ is a column of $B$. If its the $i$th column vector, then the error can be corrected by flipping the $i$th component in $y$.