be a parity-check matrix of the convolutional code $\mathfrak{C}$ of rate $k/n$. Set $L:=\lfloor\frac{\delta}{n-k}\rfloor+\lfloor\frac{\delta}{k}\rfloor$. Then $$\begin{aligned}
\label{ppc}
\mathfrak{H}:=\left(\begin{array}{ccccc}
H_{\nu} & \cdots & H_0 &   & 0 \\ 
  & \ddots &   & \ddots &   \\ 
0 &   & H_{\nu} & \cdots & H_0
\end{array}\right)  \in\mathbb F^{(L+1)(n-k)\times (\nu+L+1)n}\end{aligned}$$ is called **partial parity-check matrix** of the code. Moreover, $\mathfrak{C}$ is called **complete MDP** convolutional code if for any of its parity-check matrices $H(z)$, every full size minor of $\mathfrak{H}$ which is not trivially zero is nonzero.

 \
(i) Every complete MDP convolutional code is a reverse MDP convolutional code. [@vp]\
(ii) A complete MDP convolutional code exists over a sufficiently large base field if and only if $(n-k)\mid\delta$. [@cmdp]

As for $\mathcal{H}_L$ - when considering MDP convolutional codes - and additionally for $\mathfrak{H}_L$ - when considering reverse MDP convolutional codes - one could describe the not trivially zero full size minors of the partial parity-check matrix $\mathfrak{H}$ by conditions on the indices of the columns one uses to form the corresponding minor.

[@vp]\[index\] \
A full size minor of $\mathfrak{H}$ formed by the columns $j_1,\hdots,j_{(L+1)(n-k)}$ is not trivially zero if and only if

-   $j_{(n-k)s+1}>sn$

-   $j_{(n-k)s}\leq sn+\nu n$

for $s=1,\hdots,L$.

This is 