The Utility of Untangling

Abstract

In this paper we show how techniques developed for untangling planar graphs by Bose et al. [Discrete & Computational Geometry 42(4): 570–585 (2009)] and Goaoc et al. [Discrete & Computational Geometry 42(4): 542–569 (2009)] imply new results about some recent graph drawing models. These include column planarity, universal point subsets, and partial simultaneous geometric embeddings (with or without mappings). Some of these results answer open problems posed in previous papers.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

1 Introduction

A geometric graph is a graph whose vertex set is a set of distinct points in the plane and each pair of adjacent vertices $\{v, w\}$ is connected by a line segment $\overline{vw}$ that intersects only the two vertices. A geometric graph is planar if its underlying combinatorial graph is planar. It is plane if no two edges cross other than in a common endpoint. A straight-line crossing-free drawing of a planar graph is a representation of that graph by a plane geometric graph.

Given a geometric planar graph, possibly with many crossings, to untangle it, means to move some of its vertices to new locations (that is, change their coordinates) such that the resulting geometric graph is plane. The goal is to do so by moving as few vertices as possible, or in other words, by keeping the locations of as many vertices as possible unchanged (that is, fixed). A series of papers have studied untangling of planar graphs or subclasses of planar graphs [9, 11, 14, 24, 26, 28, 30]. The best known (lower) bound for general planar graphs is due to Bose et al. [9] who proved that every n-vertex geometric planar graph can be untangled while keeping the locations of at least $\varOmega (n^{1/4})$ vertices fixed. On the other hand, Cano et al. [11] showed that for all large enough n, there exists an n-vertex geometric planar graph that cannot be untangled while keeping the locations of more than $\omega (n^{0.4948})$ vertices fixed.

The purpose of this paper is to highlight how the techniques developed by Bose et al. [9] and Goaoc et al. [24] can be used to establish new results on several recently studied graph drawing problems. Before presenting the new results we state the two key lemmas that are at the basis of all the results. The statements of these two lemmas are new, but their proofs are contained in and directly inferred by the work described in [9, 24].

Let G be a plane triangulation (that is, an embedded simple planar graph each of whose faces is bounded by a 3-cycle). Canonical orderings of plane triangulations were introduced by de Fraysseix et al. [22]. They proved that G has a vertex ordering $\sigma =(v_1:={x},v_2:={y}, v_3, \dots , v_n:={z})$, called a canonical ordering, with the following properties. Define $G_i$ to be the embedded subgraph of G induced by $\{v_1, v_2,\dots ,v_i\}$. Let $C_i$ be the subgraph of G induced by the edges on the boundary of the outer face of $G_i$. Then

• ${x}$, ${y}$ and ${z}$ are the vertices on the outer face of G.

• For each $i\in \{3,4,\dots ,n\}$, $C_i$ is a cycle containing ${x}{y}$.

• For each $i\in \{3,4,\dots ,n\}$, $G_i$ is biconnected and internally 3-connected; that is, removing any two interior vertices of $G_i$ does not disconnect it.

• For each $i\in \{3,4,\dots ,n\}$, $v_i$ is a vertex of $C_i$ with at least two neighbours in $C_{i-1}$, and these neighbours are consecutive on $C_{i-1}$.

The following structure was defined first in Bose et al. [9]. Using the above notation, a frame $\mathcal {F}$ of G is the oriented subgraph of G with vertex set $V(\mathcal {F}):=V(G)$, where:

• Edges ${x}$ ${y}$, ${x}v_1$ and $v_1{y}$ are in $E(\mathcal {F})$ where ${x}$ ${y}$ is oriented from ${x}$ to ${y}$, ${x}v_1$ is oriented from ${x}$ to $v_1$ and $v_1{y}$ is oriented from $v_1$ to ${y}$.

• For each $i\in \{4,5\dots ,n\}$ in the canonical ordering $\sigma$ of G, edges $pv_i$ and $v_ip'$ are in $E(\mathcal {F})$, where p and $p'$ are the first and the last neighbour, respectively, of $v_i$ along the path in $C_{i-1}$ from ${x}$ to ${y}$ not containing edge ${x}$ ${y}$. Edge $pv_i$ is oriented from p to $v_i$, and edge $v_ip'$ is oriented from $v_i$ to $p'$.

By definition, $\mathcal {F}$ is a directed acyclic graph with one source ${x}$  and one sink ${y}$. $\mathcal {F}$ defines a partial order $<_\mathcal {F}$ on $V(\mathcal {F})$, where $v<_\mathcal {F}w$ whenever there is a directed path from v to w in $\mathcal {F}$.

Subsequently, it has been observed that a frame of G can also be obtained by taking the union of any two trees in Schnyder 3-tree-decompositions where the orientation of the edges in one of the two trees is reversed. See, for example, page 13 in Di Giacomo et al. [16] for this alternative formulation.

Recall that a chain (antichain) in a partial order is a subset of its elements that are pairwise comparable (incomparable). Given a partial order $(V,\le )$ on a set of vertices V of some graph, we will often refer to a chain $V'\subseteq V$ (or antichain) and by that mean a subset of vertices of V that form a chain (antichain) in the given partial order $(V, \le )$. We also say that a chain $V'$ contains a chain $V''$ if $V'$ and $V''$ are both chains in $(V,\le )$ and $V''\subseteq V'$.

Consider an n-vertex planar graph G and a set P of $k\le n$ points in the plane together with a bijective mapping from a set $V_k$ of k vertices in G to P. Let D be a straight-line crossing-free drawing of G. We say that D respects the given mapping if each vertex of $V_k$ is represented in D by its image point as determined by the given mapping.

The following two lemmas are implicit in the work of Bose et al. [9] and Goaoc et al. [24]. Parts (b), (c) and consequently (d), in Lemma 1, are due to Goaoc et al. [24]. Note that, unlike here, the results of Goaoc et al. [24] are not expressed in terms of a chain in the frame of G but an equivalent structure: a simple path L in a plane triangulation, connecting two vertices x and y on the outer face x, y, z with the property that all chords of L lie on one side of L and z lies on the other.

Consider a graph G, a set $S\subseteq V(G)$ and a set P of |S| points in the plane together with a bijective mapping from S to P. For a vertex $v\in S$ mapped to a point $p\in P$, let $\textsf {x} (v)$ denote the $\textsf {x}$-coordinate of p.

Lemma 1

 [9, 24] Let G be an n-vertex plane triangulation with a partial order $<_\mathcal {F}$ associated with a frame $\mathcal {F}$ of G. Let $C\subseteq V$ be a chain in $<_\mathcal {F}$. Let H be the graph induced in G by a maximal chain that contains C in $<_\mathcal {F}$. The embedding of H is implied by the embedding of G. Then:

• (a) H is a 2-connected outerplane graph, i.e. a 2-connected embedded outerplanar graph all of whose vertices lie on the cycle bounding the infinite face.

• (b) Let $I\subseteq V(H)$ such that if $v, w\in I$ and $vw \in E(H)$ then vw lies on the outer face of H. Let P be any set of |I| points in the plane where no two points of P have the same $\textsf {x}$-coordinate. Given a bijective mapping from I to P such that, for every two vertices $v,w\in I$, $v<_\mathcal {F}w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$, there exists a straight-line crossing-free drawing of G that respects the given mapping.

• (c) There exists such a set I with at least $(V(H)+1)/2$ vertices.

• (d) There exists such a set I with at least |C| / 3 vertices of C.

While the lower bound in part (c) is stronger than the lower bound in part (d), part (d) ensures that a fraction of vertices of C are used. That will be critical for some applications (see Theorem 2 in Sect. 2 and Theorem 6 in Sect. 4.2). Part (d) follows from (b) as follows. Consider the graph $H'$ induced in H by the vertices of C. By part (a), $H'$ is outerplanar. Thus its vertices can be coloured with three colours such that adjacent vertices in $H'$ receive distinct colours. Thus there exists an independent set I in $H'$ that contains at least |C| / 3 vertices of C. The condition imposed on the vertex set I in part (b) are immediate since I is an independent set in $H'$ and H.

Note that, in an interesting recent development, Di Giacomo et al. [17] proved that every n-vertex plane triangulation has a frame where some chain has size at least $n^{1/3}$. Thus by part (a), $|V(H)|\ge n^{1/3}$ in that frame and consequently, every n-vertex plane triangulation has a 2-connected outerplane graph of size at least $n^{1/3}$ as an embedded induced subgraph.

The following is the second key lemma.

Lemma 2

 [9] Let G be an n-vertex plane triangulation with a partial order $<_\mathcal {F}$ associated with a frame $\mathcal {F}$ of G and the total order $<_\sigma$ associated with the corresponding canonical ordering. Let $A\subseteq V$ be an antichain in $<_\mathcal {F}$. Let P be any set of |A| points in the plane where no two points of P have the same $\textsf {x}$-coordinate. Given a bijective mapping from A to P such that, for every two vertices $v,w\in A$, $v<_\sigma w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$, then there exists a straight-line crossing-free drawing of G that respects the given mapping.

2 Column Planarity

Given a planar graph G, a set $R\subset V(G)$ is column planar in G if the vertices of R can be assigned $\textsf {x}$-coordinates such that given any arbitrary assignment of $\textsf {y}$-coordinates to R, there exists a straight-line crossing free drawing of G that respects the implied mapping of vertices of R to the plane.

The column planar sets were first defined by Evans et al. [21]. A slightly stronger notion¹ was used earlier (although not named) in [9] (see Lemma 1 and Lemma 6 in [9]) where such sets were studied and used to prove Lemma 2 in the previous section. In particular, define a set $R\subset V(G)$ as strongly column planar if the following holds: there exists a total order $\mu$ on R such that

• (a) given any set P of |R| points in the plane where no two points have the same $\textsf {x}$-coordinate; and,

• (b) given a bijective mapping from R to P such that, for every two vertices $v,w\in R$, $v<_\mu w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$,

then there exists a straight-line crossing-free drawing of G that respects the given mapping. Being strongly column planar implies being column planar but not the converse. We use this slightly stronger notion as it is needed in the later sections.

Notions similar to column planarity were studied by Estrella-Balderrama et al. [20] and Di Giacomo et al. [15].

It is implicit in the work of Bose et al. [9] (see the proof of Lemma 2 in [9]) that every tree has a strongly column planar set of size at least n / 2. For column planar sets, this result is improved to 14n / 17 by Evans et al. [21]. Having a bound greater than n / 2 is critical for an application of column planarity to partial simultaneous geometric embedding with mapping [21]. Barba et al. [6] prove that every n-vertex outerplanar graph has a column planar set of size at least n / 2.²

Evans et al. [21] pose as an open problem the question of developing any bound for column planar sets in general planar graphs. We provide here the first non-trivial (that is, better than constant) bound for this problem.

Theorem 1

For every n, every n-vertex planar graph G has a (strongly) column planar set of size at least $\sqrt{n/2}$.

Proof

If $|V(G)|\le 2$, the result is trivially true. Thus we may assume that G is a triangulated plane graph. Let $\mathcal {F}$ be a frame of G, let $<_\mathcal {F}$ be its associated partial order, and let $\sigma$ be the associated canonical ordering. Consider a chain in $<_\mathcal {F}$ of maximum size. (Hence, the chain starts with x and ends with y). Let H be the subgraph of G induced by that chain, as defined in Lemma 1. Let $I\subseteq V(H)$ be as defined in Lemma 1(b). Consider any set P of |I| points in the plane where no two points have the same $\textsf {x}$-coordinate and consider a bijective mapping from I to P such that, for every two vertices $v,w\in I$, it holds that $v<_\mathcal {F}w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$. By Lemma 1(b), there exists a straight-line crossing-free drawing of G that respects the given mapping and thus I, as ordered by $<_\mathcal {F}$, is a strongly column planar set. By Lemma 1(c), $|I|\ge |V(H)|/2$. Thus if the size of the maximum chain in $<_\mathcal {F}$ is at least $\sqrt{2n}$, and thus $|V(H)|\ge \sqrt{2n}$, we are done. Otherwise, by Dilworth’s theorem [18], $<_\mathcal {F}$ has a partition into at most $\sqrt{2n}$ antichains. By the pigeon-hole principle, there is an antichain in that partition with at least $n/\sqrt{2n}=\sqrt{n/2}$ vertices. Let $A\subseteq V(G)$ be the maximum antichain in $<_\mathcal {F}$. Consider any set P of |A| points in the plane where no two points have the same $\textsf {x}$-coordinate and consider a bijective mapping from A to P such that, for every two vertices $v,w\in A$, it holds that $v<_\sigma w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$. By Lemma 2, there exists a straight-line crossing-free drawing of G that respects the given mapping and thus A, as ordered by $<_\sigma$, is a strongly column planar set. This completes the proof since $|A|\ge \sqrt{n/2}$.    $\square$

We conclude this section by proving a slightly stronger statement (with a slightly weaker bound when $S=V$) than Theorem 1. This stronger statement relies on part (d) of Lemma 1, and is a critical strengthening for some applications, such as partial simultaneous geometric embeddings with mappings (see Theorem 6 in Sect. 4.2).

Theorem 2

Given any planar graph G and any subset $S\subseteq V$, there exists $R\subseteq S$ such that R is a strongly column planar set of G and $|R|\ge \sqrt{|S|/3}$.

Proof

If $|V(G)|\le 2$, the result is trivially true. Thus we may assume that G is a triangulated plane graph. Let $\mathcal {F}$ be a frame of G, let $<_\mathcal {F}$ be its associated partial order, and let $\sigma$ be the associated canonical ordering. Assume first that $<_\mathcal {F}$ has a chain C such that $C\subseteq S$ and $|C|\ge \sqrt{3|S|}$. Let H be the subgraph of G induced by a maximal chain that contains C in $<_\mathcal {F}$, as defined in Lemma 1. Let $I\subseteq S$ be as defined in Lemma 1(b) and (d). Consider any set P of |I| points in the plane where no two points have the same $\textsf {x}$-coordinate and consider a bijective mapping from I to P such that, for every two vertices $v,w\in I$, it holds that $v<_\mathcal {F}w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$. By Lemma 1(b), there exists a straight-line crossing-free drawing of G that respects the given mapping and thus I, as ordered by $<_\mathcal {F}$, is a strongly column planar set. By Lemma 1(d), $I\subseteq C$ and $|I|\ge |C|/3|$. Thus if $<_\mathcal {F}$ has a chain C such that $C\subseteq S$ and $|C|\ge \sqrt{3|S|}$, we are done. Otherwise, by Dilworth’s theorem [18], $<_\mathcal {F}$, when restricted to S, has a partition into at most $\sqrt{3|S|}$ antichains. By the pigeon-hole principle, there is an antichain $A\subseteq S$ in that partition that has at least $|S|/\sqrt{3|S|}=\sqrt{|S|/3}$ elements. Consider any set P of |A| points in the plane where no two points have the same $\textsf {x}$-coordinate and consider a bijective mapping from A to P, such that for every two vertices $v,w\in A$, $v<_\sigma w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$. By Lemma 2, there exists a straight-line crossing-free drawing of G that respects the given mapping and thus A, as ordered by $<_\sigma$, is a strongly column planar set. This completes the proof since $|A|\ge \sqrt{|S|/3}$ and $A\subseteq S$.    $\square$

3 Universal Point Subsets

A set of points P is universal for a set of planar graphs if every graph from the set has a straight-line crossing-free drawing where each of its vertices maps to a distinct point in P. It is known that, for all large enough n, universal pointsets of size n do not exist for all n-vertex planar graphs – as first proved by de Fraysseix et al. [22]. The authors also proved that the $O(n) \times O(n)$ integer grid is universal for all n-vertex planar graphs and thus a universal pointsets of size $O(n^2)$ exists. Currently the best known lower bound on the size of a smallest universal pointset for n-vertex planar graphs is $1.235n-o(n)$ [27] and the best known upper bound is $n^2/4 - O(n)$ [5]. Closing the gap between $\varOmega (n)$ and $O(n^2)$ is a major, and likely difficult, graph drawing problem, open since 1988 [22, 23].

This motivated the following notion introduced by Angelini et al. [2]. A set P of $k\le n$ points in the plane is a universal point subset for all n-vertex planar graphs if the following holds: every n-vertex planar graph G has a subset $S\subseteq V(G)$ of k vertices and a bijective mapping from S to P such that there exists a straight-line crossing-free drawing of G that respects that mapping.

Angelini et al. [2] proved that for every n there exists a set of points of size at least $\sqrt{n}$ that is a universal point subset for all n-vertex planar graphs. Di Giacomo et al. [16] continued this study and showed that for every n, every set P of at most $(\sqrt{\log _2n}-1)/4$ points in the plane is a universal point subset for all n-vertex planar graphs. They also showed that every one-sided convex point set P of at most $n^{1/3}$ points in the plane is a universal point subset for all n-vertex planar graphs. The following theorem improves all these results.

Theorem 3

Every set P of at most $\sqrt{n/2}$ points in the plane is a universal point subset for all n-vertex planar graphs.

The proof of this lemma can be derived directly from Lemmas 1 and 2, similarly to the proof of Theorem 1, but we will instead prove it using Theorem 1.

Proof

Rotate P to obtain a new pointset $P'$ where no two points of $P'$ have the same $\textsf {x}$-coordinate. By Theorem 1, every n-vertex planar graph has a strongly column planar set R of size |P|. Thus, by the definition of strongly column planar sets, there exists a total order $\mu$ on R such that given a bijective mapping from R to $P'$ where for every two vertices $v,w\in R$, $v<_\mu w$ if and only if $\textsf {x} (v)<\textsf {x} (w)$, there exists a straight-line crossing-free drawing of G that respects the given mapping. Such a mapping clearly exists since no two points of $P'$ have the same $\textsf {x}$-coordinate. Rotating $P'$ back to the original pointset completes the proof.    $\square$

It is not known if, for all n, there exist a universal point subset of size $n^{1/2+\epsilon }$ for some $\epsilon >0$. Better bounds are only known for outerplanar graphs. Namely, every pointset of size n in general position is universal for all n-vertex outerplanar graphs [8, 13, 25]. Should the results of Barba et al. [6] apply to strongly column planar sets, then arguments equivalent to those above would show that every pointset of size n / 2 is a universal point subset for all n-vertex outerplanar graphs.

4 (Partial) Simultaneous Geometric Embeddings

Simultaneous Geometric Embeddings were introduced by Braß et al. [10]. Initially there were two main variants of this problem, one in which the mapping between the vertices of the two graphs is given and another in which the mapping is not given. Since then there has been a plethora of work on the subject for various variants of the problem – see, for example a survey by Bläsius et al. [7].

4.1 Without Mapping

Whether the following statement, on simultaneous geometric embeddings, is true is an open question asked by Braß et al. [10] in 2003: For all n and for any two n-vertex planar graphs there exists a pointset P of size n such that each of the two graphs has a straight-line crossing-free drawing with its vertices mapped to distinct points of P. The statement is known not to be true when “two” is replaced by 7393 and $n=35$ [12].

This motivates a study of (partial) geometric simultaneous embeddings – various versions of which have been proposed and studied in the literature [7]. We start with the following version.

Two graphs $G_1$ and $G_2$, where $|V(G_1)|\ge |V(G_2)|$ are said to have a geometric simultaneous embedding with no mapping if there exists a pointset P of size $|V(G_1)|$ such that each of the two graphs has a straight-line crossing-free drawing where all of its vertices are mapped to distinct points in P. Angelini et al. [3] write: “What is the largest $k \le n$ such that every n-vertex planar graph and every k-vertex planar graph admit a geometric simultaneous embedding with no mapping? Surprisingly, we are not aware of any super-constant lower bound for the value of k.”

The following theorem answers their questions.

Theorem 4

For every n and every $k\le \sqrt{n/2}$, every n-vertex planar graph and every k-vertex planar graph admit a geometric simultaneous embedding with no mapping.

Proof

Let $G_1$ and $G_2$ be the two given planar graphs with $|V(G_1)|=n$ and $|V(G_2)|=k$. By Fáry’s theorem, $G_2$ has a straight-line crossing-free drawing on some set, $P_2$, of k points. By Theorem 3, $G_1$ has a straight-line crossing-free drawing where $|P_2|$ vertices of $G_1$ are mapped to distinct points in $P_2$. Consider now the set of points, P, defined by the vertices in the drawing of $G_1$. This set is our desired pointset as it is a set of n points such that each of $G_1$ and $G_2$ has a straight-line crossing-free drawing where all of its vertices are mapped to the points in P.    $\square$

Here is another variant of the (partial) geometric simultaneous embedding problem. For $k\le n$, two n-vertex planar graphs $G_1$ and $G_2$ are said to have a k-partial simultaneous geometric embedding with no mapping (k-PSGENM) if there exists a set P of at least k points in the plane such that each of the two graphs has a straight-line crossing-free drawing where |P| of its vertices are mapped to distinct points of P. Recall that Angelini et al. [2] proved that for every n there exists a set of points of size at least $\sqrt{n}$ that is a universal point subset for all n-vertex planar graphs. This implies that, for all n, any two n-vertex planar graphs have an $\sqrt{n}$-partial simultaneous geometric embedding with no mapping. Note however that this does not imply Theorem 4. Namely, if one starts with a straight-line crossing-free drawing of the smaller graph $G_2$ (say on $\sqrt{n}$ vertices), there is no guarantee with this result that the bigger, n-vertex graph, $G_1$ can be drawn while using all the points generated by the drawing of $G_2$.

4.2 With Mapping

The notion of k-partial simultaneous geometric embedding with mapping (k-PSGE) is the same as k-PSGENM except that a bijective mapping between $V(G_1)$ and $V(G_2)$ is given and the two drawings have a further restriction that if $v\in V(G_1)$ is mapped to a point in P then the vertex w in $V(G_2)$ that v maps to, has to be mapped to the same point in P. In other words, two n-vertex planar graphs $G_1$ and $G_2$ on the same vertex set, V, are said to have a k-partial simultaneous geometric embedding with mapping (k-PSGE) if there exists a straight-line crossing free drawing $D_1$ of $G_1$ and $D_2$ of $G_2$ such that there exists a subset $V'\subseteq V$ with $|V'|\ge k$ and each vertex $v\in V'$ is represented by the same point in $D_1$ and $D_2$.

It is known that, for every large enough n, there are pairs of n-vertex planar graphs that do not have an n-partial simultaneous geometric embedding with mapping, that is, an n-PSGE [10]. In fact the same is true for simpler families of planar graphs, for example for a tree and a path [4], for a planar graph and a matching [4] and for three paths [10].

k-PSGE was introduced by Evans et al. [21] who proved (using their column planarity result) that any two n-vertex trees have an 11n / 17-PSGE. Barba et al. [6] proved that any two n-vertex outerplanar graphs have an n / 4-PGSE. Evans et al. [21] also observed that the main untangling result by Bose et al. [9] implies that every pair of n-vertex planar graphs has an $\varOmega (n^{1/4})$-PSGE. Namely, start with a straight-line crossing-free drawing of $G_1$. Since the vertex sets of $G_1$ and $G_2$ are the same, the drawing of $G_1$ (or rather the drawing of its vertex set) defines a straight-line drawing of $G_2$. Untangling $G_2$ such that $\varOmega (n^{1/4})$ of its vertices remain fixed (which is possible by [9]) gives the result.

Theorem 5

[6] Every pair of n-vertex planar graphs has an $\varOmega (n^{1/4})$-partial simultaneous geometric embedding with mapping, that is, it has an $\varOmega (n^{1/4})$-PGSE.

However, the above untangling argument fails if we try to apply it one more time. Namely, consider the following generalization of the k-PGSE problem. Given any set $\{G_1, \dots , G_p\}$ of $p\ge 2$ n-vertex planar graphs on the same vertex set, V, we say that $G_1, \dots , G_p$ have a k-partial simultaneous geometric embedding with mapping (k-PSGE) if there exists a straight-line crossing-free drawing $D_i$ of each $G_i$, $i\in \{1,\dots ,p\}$ such that there exists a subset $V'\subseteq V$ with $|V'| \ge k$ and each vertex $v\in V'$ is represented by the same point in all drawings $D_i$, $i\in \{1,\dots ,p\}$.

If we try to mimic the earlier untangling argument that proves Theorem 5, it fails for $p=3$ already since we cannot guarantee that when $G_3$ is untangled the set of its vertices that stays fixed has a non-empty intersection with the set that remained fixed when untangling $G_2$. It is here that part (d) of Lemma 1 is needed, or rather the stronger result on column planarity from Theorem 2.

Theorem 6

Any set of $p\ge 2$ n-vertex planar graphs has an $\varOmega (n^{1/4^{(p-1)}})$-partial simultaneous geometric embedding with mapping, that is, it has an $\varOmega (n^{1/4^{(p-1)}})$-PGSE.

Proof

Let $\{G_1, \dots , G_p\}$ be the given set of p n-vertex planar graphs. The proof is by induction on p. The base case, $p=2$, is true by Theorem 5. Let $p\ge 3$ and assume by induction that the set $\{G_1,\dots , G_{p-1}\}$ has an $\varOmega (n^{1/4^{(p-2)}})$-PGSE. Let $V'\subseteq V$ be the set from the definition of k-PSGE and let $P'$ be the set of $|V'|$ points that $V'$ is mapped to in the drawings $D_1, \dots , D_{p-1}$. Thus $|V'|\in \varOmega (n^{1/4^{(p-2)}})$ by induction. We may assume that no pair of points in $P'$ has the same $\textsf {x}$-coordinate as otherwise we can just rotate the union of $D_1, \dots , D_{p-1}$. By Theorem 2, there exists $R\subseteq V'$ that is strongly column planar in $G_p$ and $|R|\ge \sqrt{|V'|/3}$. Since the vertices of $V'$ are bijectively mapped to $P'$, that mapping defines a bijective mapping from R to a subset $P_R$ of $P'$. Consider the total order $\mu$ of R (the total order from the definition of strongly column planar sets) and the total order $\phi$ of R as defined by the $\textsf {x}$-coordinates of $P_R$. By the Erdős–Szekeres theorem [19, 31], there exists a subset $R'$ of R of at least $\sqrt{|R|}\ge (|V'|/3)^{1/4}$ vertices such that the order of $R'$ in $\mu$ is the same or reverse as the order of $R'$ in $\phi$. In the second case the union of all the drawings of $D_1, \dots , D_p$ can be mirrored such that the order of $R'$ in $\mu$ is the same as the order of $R'$ in $\phi$. Thus in both cases, we can apply Theorem 2. Since the vertices of R are bijectively mapped to $P_R$, this defines a bijective mapping from $R'$ to a subset $P_R'$ of $P_R$. Since $R'$ is strongly column planar in $G_p$, we can apply Theorem 2 to conclude that $G_p$ has a straight-line crossing-free drawing $D_p$ that respects the mapping from $R'$ to $P_R'$ and thus each vertex $v\in R'$ is represented by the same point in all drawings $D_i$, $i\in \{1,\dots ,p\}$. Since $|V'|\in \varOmega (n^{1/4^{(p-2)}})$, and $|R'| \ge (|V'|/3)^{1/4}$, the lower bound holds.    $\square$

Note that the definition of k-PSGE, as introduced in Evans et al. [21], has one additional requirement, as compared with the definition used here. Namely, the additional requirement states that if $v, w\in V$ are mapped to a same point in $D_i$ and $D_j$, then $v=w$. However this additional requirement can always be met by the fact that it is possible to perturb any subset of vertices of a geometric plane graph without introducing crossings. More precisely, for any geometric plane graph there exists a value $\epsilon >0$ such that each vertex can be moved any distance of at most $\epsilon$, and the resulting geometric graph is also crossing-free.³

5 Conclusion

The main purpose of this paper is to draw attention to Lemmas 1 and 2 in the current form as they seem to have applications to numerous, some seemingly unrelated, graph drawing problems as evidenced by the results highlighted in the previous sections. The two lemmas appear in the current form for the first time here. Their original formulation was tailored towards specific application (untangling) and not directly applicable to any of the above mentioned problems.