$i$, i.e. out of the first $i-1$ columns of $\mathcal{H}_L$. After that, one proceeds like for $M_2$, i.e. next one has to choose $n-k$ columns out of $2n$ but not the first $i$, then $n-k$ out of $3n$ without the first $i$ and without the $n-k$ chosen in the preceding step and so on.

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It depends one the parameters of the code, which of the bounds is best. In the following, we give some examples:

1.  Case $L=0$: $M_1=M_2=M_3$ (in this case there are no trivially zero minors)

2.  Case $k=1$\
    $M_3=M_2\cdot\left(\frac{1}{n}+\frac{n-1}{\binom{n+L}{n-1}}\right)$

    -   $L\geq 1$: $M_3<M_2$

    -   $L=1\ (\Rightarrow \delta=1, n\geq 3)$: $M_3=(3n^2-n)(n-1)<M_1=(4n^2-2n)(n-1)<M_2=(n^3+n^2)(n-1)$

    -   $(2,1,1)$: $M_3=18\cdot 3<M_1=20\cdot 3<M_2=24\cdot 3$

    -   $(2,1,\delta)$ with $\delta\geq 2$: $M_1<M_3<M_2$\
        It holds $L\geq 4$, which implies $M_3<M_2$ according to (a), and for $L=4$, $M_1=252\cdot 5<M_3=480\cdot 5<M_2=720\cdot 5$. Moreover, $M_3$ is increasing more than $M_1$, when $L$ increases (to $L+1$). This is true since $M_1/(L+1)=\binom{2(L+1)}{L+1}$ increases with factor $\frac{(2L+3)(2L+4)}{(L+2)(L+2)}<4$ and $M_3/(L+1)=(L+2)!\cdot (\frac{1}{2}+\frac{1}{L+2})$ increases with factor $(L+2)\frac{L^2+7L+10}{L^2+7L+12}>5$ for $L\geq 4$.

Probability
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In this subsection, we want to compute the probability that a non-catastrophic convolutional code with arbitrary parameters is MDP. Therefore, we assign to each code the unique parity-check matrix from Remark \[kh\]. This is 