| Figure 9: A solution winding in a ring (top), and the “unwinding” or it (bottom). | |
| Having $P^*$ at hand, we replace $P'$ by $P^*$. This is done for every maximal degree-2 path, and thus we complete the construction of $R$. However, at this point, it is not clear why after we perform these replacements, the separators considered earlier remain separators, or that we even still have a tree. Roughly speaking, a scenario as depicted in Fig. 10 can potentially happen. To show that this is not the case, it suffices to prove that there cannot exist a vertex that belongs to two different rings. Towards that, we apply another preprocessing operation: we ensure that the radial completion of $G$ does not have $2^{ck}$ (for some constant $c$) concentric cycles that contain no vertex in $S \cup T$ by using another result by Adler et al. [2]. Informally, a sequence of concentric cycles is a sequence of vertex-disjoint cycles where each one of them is contained inside the next one in the sequence. Having no such sequences, we prove the following. | |
| **Lemma 2.3.** Let $R'$ be any Steiner tree. For every vertex $v$, there exists a vertex in $V(R')$ whose distance to $v$ (in the radial completion of $G$) is $2^{ck}$ for some constant $c$. | |
| To see why intuitively this lemma is correct, note that if $v$ was “far” from $R'$ in the radial completion of $G$, then in $G$ itself $v$ is surrounded by a large sequence of concentric cycles that contain no vertex in $S \cup T$. Having Lemma 2.3 at hand, we show that if a vertex belongs to a certain ring, then it is “close” to at least one vertex of the restriction of $R$ to that ring. In turn, that means that if a vertex belongs to two rings, it can be used to exhibit a “short” path between one vertex in the restriction of $R$ to one ring and another vertex in the restriction of $R$ to the second ring. By choosing constants properly, this path is shown to exhibit a detour in $R$, and hence we reach a contradiction. (In this argument, we use the fact that for every vertex $u$, towards the computation of the separator, we considered a vertex $u'$ of distance $2^{c_1k}$ from $u$—this subpath between $u$ and $u'$ is precisely that subpath that we will shortcut.) |