| A proof of this proposition can be obtained by first ordering the collection of inner and outer visitors by their 'distance' from the inner and outer interfaces, respectively, and the 'containment' relation between the cycles formed by them with the interfaces. This gives a partial order on the set of inner visitors and the set of outer visitors. Then if the proposition does not hold for $\mathcal{P}$, then the above ordering and containment relation can be used to reroute these paths along a suitable cycle. This will contradict the minimality of $\mathcal{P}$, since the rerouted linkage is aligned with it but uses strictly fewer edges outside of $\mathcal{C}$. The main result of this section can be now formulated as follows. (Its formulation and proof idea are based on Lemma 8.31 and Theorem 6.45 of [14].) | |
| **Lemma 7.1.** Let $G$ be a plane graph with radial completion $H$. Let $\text{Ring}(I_{\text{in}}, I_{\text{out}})$ be a ring in $H$. Suppose that $|I_{\text{in}}|, |I_{\text{out}}| \le l$ for some integer $l$, and that in $\text{Ring}(I_{\text{in}}, I_{\text{out}})$ there is an encircling tight concentric sequence of cycles $\mathcal{C}$ of size larger than $40l$. Let $\eta$ be a traversing path in the ring witnessing the tightness of $\mathcal{C}$, and fix $\eta$ as the reference path. Finally, let $\mathcal{P} = \mathcal{P}_{\text{traverse}} \uplus \mathcal{P}_{\text{visitor}}$ be a linkage in $G$, where $\mathcal{P}_{\text{traverse}}$ is a traversing linkage comprising the paths of $\mathcal{P}$ traversing $\text{Ring}(I_{\text{in}}, I_{\text{out}})$, while $\mathcal{P}_{\text{visitor}} = \mathcal{P} \setminus \mathcal{P}_{\text{traverse}}$ consists of the paths whose both endpoints lie in either $V(I_{\text{in}})$ or $V(I_{\text{out}})$. Further, suppose that $\mathcal{P}$ is minimal with respect to $\mathcal{C}$. Then, for every traversing linkage $\mathcal{Q}$ in $G$ that is minimal with respect to $\mathcal{C}$ such that every path in $\mathcal{Q}$ is disjoint from $\eta$ and $|\mathcal{Q}| \ge |\mathcal{P}_{\text{traverse}}|$, there is a traversing linkage $\mathcal{P}'_{\text{traverse}}$ in $G$ such that | |
| (a) $\mathcal{P}'_{\text{traverse}}$ is aligned with $\mathcal{P}_{\text{traverse}}$, | |
| (b) the paths of $\mathcal{P}'_{\text{traverse}}$ are disjoint from the paths of $\mathcal{P}_{\text{visitor}}$, and | |
| (c) $|\overline{\text{WindNum}}(\mathcal{P}'_{\text{traverse}}) - \overline{\text{WindNum}}(\mathcal{Q})| \le 60l + 6.$ | |
| *Proof.* Let $\mathcal{C} = (C_1, \dots, C_p)$, where $p > 40l$. Recall that $\mathcal{C}$ is a collection of cycles in $G$, and the path $\eta$ that witnesses the tightness of $\mathcal{C}$ contains $|\mathcal{C}|$ vertices of $V(G)$, one on each cycle of $\mathcal{C}$. Let $v_i$ denote the vertex where $\eta$ intersects the cycle $C_i \in \mathcal{C}$ for all $i \in \{1, 2, \dots, p\}$. Since $\mathcal{P}$ is minimal with respect to $\mathcal{C}$, Proposition 7.3 implies that the paths in $\mathcal{P}_{\text{visitor}}$ do not intersect any of the cycles $C_{10l}, C_{10l+1}, \dots, C_{p-10l+1}$ (note that since $p > 40l$, this sequence of cycles is non-empty). Call a vertex $x \in V(\text{Ring}(I_{\text{in}}, I_{\text{out}}))$ in the ring *non-separated* if there exists a path from $x$ to $C_{10l}$ whose set of internal vertices is disjoint from $\bigcup_{P \in \mathcal{P}_{\text{visitor}}} V(P)$. Otherwise, we say that the vertex $x$ is *separated*. Observe that every path in $\mathcal{P}_{\text{traverse}}$ is disjoint from the paths in $\mathcal{P}_{\text{visitor}}$ and intersects $C_{10l}$, hence all vertices on the paths of $\mathcal{P}_{\text{traverse}}$ are non-separated. Let $X$ denote the set of all non-separated vertices in the ring, and consider the graph $H[X]$. Observe that $H[X]$ is an induced subgraph of $\text{Ring}(I_{\text{in}}, I_{\text{out}})$, since it is obtained from $\text{Ring}(I_{\text{in}}, I_{\text{out}})$ by deleting the separated vertices. Further, observe that $H[X]$ is a ring of $H$. Indeed, the inner interface of $H[X]$ is the cycle $\hat{I}_{\text{in}}$ obtained as follows: let $\mathcal{P}_{\text{in}}$ be the set of inner visitors in $\mathcal{P}$; then, $\hat{I}_{\text{in}}$ is the outer face of the plane graph $H[V(I_{\text{in}}) \cup \bigcup_{P \in \mathcal{P}_{\text{in}}} V(P)]$. It is easy to verify that all vertices on $\hat{I}_{\text{in}}$ are non-separated, and any vertex of $\text{Ring}(I_{\text{in}}, I_{\text{out}})$ that lies in the strict interior of this cycle is separated. We then symmetrically obtain the outer interface $\hat{I}_{\text{out}}$ of $H[X]$ from the set $\mathcal{P}_{\text{out}}$ of outer visitors of $\mathcal{P}$. Then, $H[X] = \text{Ring}(\hat{I}_{\text{in}}, \hat{I}_{\text{out}})$. Here, $\hat{I}_{\text{in}}$ is composed alternately of subpaths of $I_{\text{in}}$ and inner visitors from $\mathcal{P}_{\text{visitor}}$, and symmetrically for $\hat{I}_{\text{out}}$. | |
| Note that the paths of $\mathcal{P}_{\text{traverse}}$ are completely contained in $\text{Ring}(\hat{I}_{\text{in}}, \hat{I}_{\text{out}})$ and they all traverse this ring. Thus, $\mathcal{P}_{\text{traverse}}$ can be regarded also as a traversing linkage in $\text{Ring}(\hat{I}_{\text{in}}, \hat{I}_{\text{out}})$. While $\mathcal{P}_{\text{traverse}}$ may have a different winding number in $\text{Ring}(\hat{I}_{\text{in}}, \hat{I}_{\text{out}})$ than in $\text{Ring}(I_{\text{in}}, I_{\text{out}})$, the difference is “small” as we show below. (Note that the two winding numbers in the following claim are computed in two different rings.) | |
| **Claim 7.1.** Let $P$ be a path in $G$ that is disjoint from all paths in $\mathcal{P}_{\text{visitor}}$, such that $P$ belongs to $\text{Ring}(I_{\text{in}}, I_{\text{out}})$ and traverses it. Then, $P$ also belongs to $\text{Ring}(\hat{I}_{\text{in}}, \hat{I}_{\text{out}})$, and $|\overline{\text{WindNum}}(P, \eta)| \le 20l$. ¹³ | |
| ¹³Here $\hat{\eta}$ is the reference path of $\text{Ring}(\hat{I}_{\text{in}}, \hat{I}_{\text{out}})$, which is a subpath of $\eta$ in this ring. |