all leaves in the network. This means that we made the wrong guess and that another set of guesses has to be checked. We say that a side $s$ is *lowest* if it does not yet have all its leaves, and there is no other such side $s'$ reachable from $s$. By reachable we mean that in the underlying generator $g$, there is a directed path from the head of side $s$ to the tail of side $s'$. Since $N$ is a directed acyclic graph, until all leaves in ${{\mathcal X}}$ have been added, there will always be a lowest side. For example, the side $F$ in Figure \[fig:exLemma1\](d) is lowest. The idea is to add leaves to the lowest side $s$, until all its leaves have been added. We then continue with remaining lowest sides until we have reconstructed $N$.

Given a lowest side $s$, with $s^{+}$ and $s^{-}$ fixed, it is possible to tell in polynomial time what the correct remaining leaves for $s$ are, as follows. Observe that a leaf $x$ that is on side $s$ in $N$ and which has not yet been added has the property $s^{+} \rightarrow x \rightarrow s^{-}$. Furthermore, there is at 