least one cluster $C \in {{\mathcal C}}$ such that $\{ x, s^{+}, s^{-} \} \cap C = \{ x, s^{-} \}$. There exists at least one such cluster because otherwise $\{ x, s^{+} \}$ would be not separated in ${{\mathcal C}}$, a contradiction since ${{\mathcal C}}$ is separating. We call such a cluster a *split cluster for side $s$*. Now, observe that for every split cluster $C$ for side $s$, and for every side $t \neq s$ that contains 2 or more leaves in $N$, either $\{t^{+}, t^{-}\} \cap C = \{t^{+}, t^{-}\}$ or $\{t^{+}, t^{-}\} \cap C = \emptyset$. This follows because the only edges in $N$ that represent $C$ lie on side $s$. If this is not the case, our set of guesses was incorrect and a new one has to be checked.

Now, consider any leaf $y$ that has not yet been added to the network. Assume that this leaf belongs to side $t$ for some $t$. We want to have a simple test to avoid wrongly placing it on side $s$, with $s\neq t$. Side $t$ will contain three or more leaves in $N$, so we can assume that $t^{+}$ and $t^{-}$ exist. If $s^{+} \rightarrow 