The Clique Problem in Ray Intersection Graphs

Abstract

Ray intersection graphs are intersection graphs of rays, or halflines, in the plane. We show that any planar graph has an even subdivision whose complement is a ray intersection graph. The construction can be done in polynomial time and implies that finding a maximum clique in a segment intersection graph is NP-hard. This solves a 21-year old open problem posed by Kratochvíl and Nešetřil (Comment Math Univ Carolinae 31(1):85–93, 1990).

1 Introduction

The intersection graph of a collection of sets has one vertex for each set, and an edge between two vertices whenever the corresponding sets intersect. Of particular interest are families of intersection graphs corresponding to geometric sets in the plane. In this contribution, we will focus on segment intersection graphs, intersection graphs of line segments in the plane.

In a seminal paper, Kratochvíl and Nešetřil [12] proposed studying the complexity of two classical combinatorial optimization problems, the maximum independent set and the maximum clique, in geometric intersection graphs. While those problems are known to be hard to approximate in general graphs (see for instance [5, 9]), their restriction to geometric intersection graphs may be more tractable. They proved that the maximum independent set problem remains NP-hard for segment intersection graphs, even if those segments have only two distinct directions. It was also shown that in that case, the maximum clique problem can be solved in polynomial time. The complexity of the maximum clique problem in general segment intersection graphs was left as an open problem, and remained so until now. In their survey paper “On six problems posed by Jarik Nešetřil” [3], Bang-Jensen et al. describe this problem as being “among the most tantalizing unsolved problem in the area”.

Some progress has been made in the meanwhile. In 1992, Middendorf and Pfeiffer [13] showed, with a simple proof, that the maximum clique problem is NP-hard for intersection graphs of 1-intersecting curve segments that are either line segments or curves made of two orthogonal line segments. They also gave a polynomial time dynamic programming algorithm for the special case of line segments with endpoints of the form $(x,0),(y, i)$, with $i\in \{1,\ldots k\}$ for some fixed $k$. Another step was made by Ambühl and Wagner [2] in 2005, who showed that the maximum clique problem is NP-hard for intersection graphs of ellipses of fixed, arbitrary, aspect ratio. Unfortunately, this ratio must be bounded, which excludes the case of segments.

Our results. We prove that the maximum clique problem in segment intersection graphs is NP-hard, even if a representation as a set of segments is given as input. In fact, we prove the stronger result that the problem is NP-hard even in ray intersection graphs, defined as intersection graphs of rays, or halflines, in the plane. This complexity result is a consequence of the following structural lemma: every planar graph has an even subdivision whose complement is a ray intersection graph. Furthermore, the corresponding set of rays has a natural polynomial size representation. Hence solving the maximum clique problem in this graph allows us to recover the maximum independent set in the original planar graph, a task well known to be NP-hard [8]. The construction is detailed in Sect. 2.

Related work. We prove that the complement of some subdivision of any planar graph can be represented as a segment intersection graph. Whether the complement of every planar graph is a segment intersection graph remains an open question. In 1998, Kratochvíl and Kuběna [10] showed that the complement of any planar graph is the intersection graph of a set of convex polygons. More recently, Francis et al. [7] proved that the complement of any partial 2-tree is a segment intersection graph. Partial 2-trees are planar, and in particular every outerplanar graph is a partial 2-tree. The representability of planar graphs by segment intersection graphs, formerly known as Scheinerman’s conjecture, was proved recently by Chalopin and Gonçalves [4].

The maximum independent set problem in intersection graphs has been studied by Agarwal and Mustafa [1]. In particular, they proved that it can be approximated within a factor $n^{1/2+o(1)}$ in polynomial time for segment intersection graphs. This has been recently improved by Fox and Pach [6], who described, for any $\varepsilon >0$, an $n^{\varepsilon }$-approximation algorithm. In fact, their technique also applies to the maximum clique problem, and therefore $n^{\varepsilon }$ is the best known approximation factor for this problem too.

In 1994, Kratochvíl and Matoušek [11] proved that the recognition problem for segment intersection graphs is NP-hard, and in PSPACE.

Notation. For any natural number $m$ we use $[m]=\{1,\ldots , m\}$. In a graph $G$, a rotation system is a list $\pi =( \pi _v )_{v\in V(G)}$, where each $\pi _v$ fixes the clockwise order of the edges of $E(G)$ incident to $v$. When $G$ is an embedded planar graph, the embedding uniquely defines a rotation system, which is often called a combinatorial embedding. For the rest of the paper we use ray to refer to an open ray, that is, a ray does not contain its origin. Therefore, whenever two rays intersect they do so in the relative interior of both. Since our construction does not use degeneracies, we are not imposing any restriction by considering only open rays. A subdivision of a graph $G$ is said to be even if each edge of $G$ is subdivided an even number of times.

2 Construction

Let us start with an overview of the approach. We first construct a set of curves that will form the reference frame. This construction is quite generic and depends only on a parameter $k\in \mathbb N$. We then show that the complement of any tree has a special type of representation, called snooker representation, which is constructed iteratively over the levels of the tree. The number of levels of the tree is closely related to $k$, the parameter used for the reference frame. We then argue that if $G$ is a planar graph that consists of a tree $T$ and a few, special paths of length two and three, then the complement of $G$ can be represented as an intersection graph of rays by extending a snooker representation of $T$. Finally, we argue that any planar graph has an even subdivision that can be decomposed into a tree and a set of paths of length two and three with the required properties.

We first describe the construction using real coordinates. The construction does not rely on degeneracies, and thus we can slightly perturb the coordinates used in the description. This perturbation is enough to argue that a representation can be computed in polynomial time. Then, using a relation between the independence number of a graph $G$ and an even subdivision of $G$, we obtain that computing a maximum clique in a ray intersection graph is NP-hard, even if a representation is given as part of the input.

2.1 Reference Frame

Let $k\ge 3$ be an odd number to be chosen later. We set $\theta =\tfrac{k-1}{k}\pi$, and define for $i=0,\ldots ,k-1$ the points

$$\begin{aligned} p_i= ( \cos (i\cdot \theta ),\, \sin (i\cdot \theta )). \end{aligned}$$

The points $p_i$ lie on a unit circle centered at the origin; see Fig. 1a. In what follows, all indices are taken modulo $k$, hence in particular $p_k=p_0$. For each $i\in [k-1]$ we let $q_i$ be the point $p_i+(p_i-p_{i+1})$, symmetric to $p_{i+1}$ with respect to $p_i$, let $m_i$ be the midpoint between $p_i$ and $q_i$, and let $t_i$ be, among the two points along the line through $q_{i+1}$ and $m_i$ with the property $|m_i t_i|=|m_i p_i|$, the one that is furthest from $q_{i+1}$. We define $R_i$ to be the rectangle with vertices $p_i$, $t_i$, and $q_i$. The fourth vertex of $R_i$ is implicit and is denoted by $r_i$. Any two rectangles $R_i$ and $R_j$ are congruent with respect to a rotation around the origin. We have constructed the rectangles $R_i$ in such way that, for any $i\in [k-1]$, the line supporting the diagonal $p_i q_i$ of $R_i$ contains $p_{i+1}$ and the line supporting the diagonal $r_i t_i$ of $R_i$ contains $q_{i+1}$.

Fig. 1

The reference frame. a The points $p_0,\ldots , p_{k-1}$ and the rectangles $R_i$, for $k=11$, b The circular arcs $\alpha _i$

For each $i\in [k-1]$, let $\alpha _i$ be the circular arc tangent to both diagonals of $R_i$ with endpoints $q_i$ and $r_i$, and contained in $R_i$ (see Fig. 1b). Note that the curves $\alpha _i$ and the rectangles $R_i$ have been chosen so that any line that intersects $\alpha _i$ twice or is tangent to the curve $\alpha _i$ must intersect the curve $\alpha _{i+1}$. For any $i\in [k-2]$, let $\Gamma _i$ be the set of rays that intersect $\alpha _{i}$ twice or are tangent to $\alpha _{i}$ and have their origins on $\alpha _{i+1}$. We also define $\Gamma _0$ as the set of rays with origin on $\alpha _1$ and passing through $p_0$. The rays of $\Gamma _i$ that are tangent to $\alpha _i$ will play a special role. In fact, we will only use rays of $\Gamma _i$ that are “near-tangent” to $\alpha _i$.

Lemma 2.1

When $|j-i|>1$, any ray from $\Gamma _i$ intersects any ray from $\Gamma _j$.

Proof

Because of symmetry, it is enough to argue that all rays in $\Gamma _1$ intersect all rays of $\Gamma _j$ for $j\ge 3$. We provide a synthetic proof of this fact, illustrated on Fig. 2. For this, it is enough to show that the even points $p_4, p_6,\ldots$ are below the line through $t_1$ and $q_2$. By construction, this can be reduced to proving that $p_4$ is below the line through $t_1$ and $q_2$. Let $x$ be the point on the segment $q_1p_1$ at distance $|p_1 p_4|$ from $p_1$. Since $|p_1p_4|> \tfrac{1}{2} |p_1 p_2| = |p_1 m_1|$, the point $x$ lies on the segment $q_1m_1$. Because $\angle (p_2, p_1, p_4)=\tfrac{\pi }{k}$ we have $\angle (q_1, p_1, p_4)=\pi - \tfrac{\pi }{k}$. Since the triangle $\triangle (x,p_1,p_4)$ is isosceles because $|x p_1|=|p_1 p_4|$, and $\angle (q_1, p_1, p_4)=\pi - \tfrac{\pi }{k}$ we have $\angle (p_1, x , p_4)=\tfrac{\pi }{2k}$. Since $\angle (p_1, p_2, p_3)= \tfrac{\pi }{k}$ and $p_3,p_2,q_2$ are collinear, then $\angle (p_1, p_2, q_2)= \pi - \tfrac{\pi }{k}$ and, using that $\triangle (p_1, p_2, q_2)$ is isosceles, we have $\angle (p_2, p_1, q_2)= \tfrac{\pi }{2k}$. From $\angle (p_2, p_1, q_2)= \tfrac{\pi }{2k} = \angle (p_1, x , p_4)$ and the collinearity of $x,p_1,p_2$, we conclude that $p_1q_2$ and $xp_4$ are parallel. Since $m_1$ lies between $x$ and $p_1$, then the line through $m_1$ and $q_2$ passes above $p_4$. $\square$

Fig. 2

Illustration for the proof of Lemma 2.1

Lemma 2.2

Any ray of $\Gamma _{i+1}$ tangent to $\alpha _{i+1}$ at the point $x\in \alpha _{i+1}$ intersects any ray from $\Gamma _i$, except those having their origin at $x$.

Proof

Any ray of $\Gamma _{i+1}$ tangent to $\alpha _{i+1}$ intersects the boundary of $R_{i+1}$ in the segments $r_{i+1}p_{i+1}$ and $q_{i+1}t_{i+1}$, while any ray of $\Gamma _i$ intersects the segment $t_{i+1}p_{i+1}$. $\square$

Note that the whole construction depends only on the parameter $k$. We will refer to it as reference frame.

2.2 Complements of Trees

Let $T$ be a tree with a rotation system $\pi _T$, let $r$ be a vertex in $T$ and let $rs\in E(T)$ be an arbitrary edge incident to $r$. The triple $(\pi _T,r,rs)$ induces a natural linear order $\tau =\tau (\pi _T,r,rs)$ on the vertices of $T$. This order $\tau$ corresponds to the order followed by a breadth-first traversal of $T$ from $r$ with the following additional restrictions:

• (i) $s$ is the second vertex;

• (ii) the children of any vertex $v$ are visited according to the clockwise order $\pi _v$;

• (iii) if $v\not = r$ has parent $v'$, the first child of $u$ of $v$ is such that $vu$ is the successor of $vv'$ in the clockwise order $\pi _v$.

We say that vertices $v$ and $v'$ at the same level are consecutive when they are consecutive in $\tau$. See Fig. 3. The linear order will be fixed through our discussion, so we will generally drop it from the notation. Henceforth, whenever we talk about a tree $T$ and a linear order $\tau$ on $V(T)$, we assume that $\tau$ is the natural linear order induced by a triple $(\pi _T,r,rs)$. In fact, the triple $(\pi _T,r,rs)$ is implicit in $\tau$. For any vertex $v$ we use $v^+$ for its successor and $v^-$ for its predecessor.

Fig. 3

In this example, assuming that $\pi$ is as drawn, the linear order $\tau$ is $r,v_1^1,v_1^2,\ldots , v_{1}^5,v_2^1,\ldots$

A snooker representation of the complement of an embedded tree $T$ with linear order $\tau$ is a representation of $T$ with rays that satisfies the following properties:

• (a) Each vertex $v$ at level $i$ in $T$ is represented by a ray $\gamma _v$ from $\Gamma _i$. Thus, the origin of $\gamma _v$, denoted by $a_v$, is on $\alpha _{i+1}$. Note that this implies that $k$ is larger than the depth of $T$.

• (b) If a vertex $u$ has parent $v$, then $\gamma _u$ passes through the origin $a_v$ of $\gamma _v$. (Here it is relevant that we consider all rays to be open, as otherwise $\gamma _u$ and $\gamma _v$ would intersect.) In particular, all rays corresponding to the children of $v$ pass through the point $a_v$.

• (c) The origins of rays corresponding to consecutive vertices $u$ and $v$ of level $i$ are consecutive along $\alpha _{i+1}$. That is, no other ray in the representation has its origin on $\alpha _{i+1}$ and between the origins $\gamma _u$ and $\gamma _v$.

Lemma 2.3

The complement of any embedded tree with a linear order $\tau$ has a snooker representation.

Proof

Consider a reference frame with $k$ larger than the depth of $T$. The construction we provide is iterative over the levels of $T$. Note that, since we provide a snooker representation, it is enough to tell for each vertex $v\not = r$ the origin $a_v$ of the ray $\gamma _v$. Property (b) of the snooker representation provides another point on the ray $\gamma _v$, and thus $\gamma _v$ is uniquely defined. The ray $\gamma _r$ for the root $r$ is the ray of $\Gamma _0$ with origin $a_r$ at the center of $\alpha _1$.

Consider any level $i>1$ and assume that we have a representation of the vertices at level $i-1$. Consider a vertex $v$ at level $i-1$ and let $u_1,\ldots , u_d$ denote its $d$ children. If the successor $v^+$ of $v$ is also at level $i-1$, we take $a_v^+= a_{v^+}$, and else we take $a_v^+$ to be an endpoint of $\alpha _i$ such that no other origin is between the endpoint and $a_v$ (see Fig. 4). Similarly, if the predecessor $v^-$ of $v$ is at level $i-1$, we take $a_v^-= a_{v^-}$, and else we take $a_v^-$ to be an endpoint of $\alpha _i$ such that no other origin is between the endpoint and $a_v$. (If $v$ is the only one vertex at level $i$, we also make sure that $a_v^-\not = a_v^+$.) Let $\ell _v^+$ be the line through $a_v$ and $a_v^+$. Similarly, let $\ell _v^-$ be the line through $a_v$ and $a_v^-$. We then choose the points $a_{u_1},\ldots , a_{u_d}$ on the portion of $\alpha _{i+1}$ contained between $\ell _v^+$ and $\ell _v^-$ such that the $d+2$ points $\ell _v^-\cap \alpha _{i+1}, a_{u_1},\ldots , a_{u_d}, \ell _v^+\cap \alpha _{i+1}$ are regularly spaced. Since the ray $\gamma _{u_j}$ has origin $a_{u_j}$ and passes through $a_v$, this finishes the description of the procedure. Because $a_{u_j}$ lies between $\ell _v^+$ and $\ell _v^-$, the ray $\gamma _{u_j}$ either intersects $\alpha _i$ twice or is tangent to $\alpha _i$, and thus $\gamma _{u_j}\in \Gamma _{i}$.

Fig. 4

An example showing the construction of a snooker representation when $v$ has three children and $v^+$ and $v^-$ are at the same level as $v$

Recall that from Lemma 2.1, any ray from $\Gamma _i$ intersects any ray from $\Gamma _j$ when $|j-i|>1$. Therefore, vertices from levels $i$ and $j$, where $|i-j|>1$, intersect. For vertices $u$ and $v$ at levels $i-1$ and $i$, respectively, Lemma 2.2 and the choices for $a_v$ imply that $\gamma _v$ intersects $\gamma _u$ if and only if $u$ is not the parent of $v$ in $T$. For vertices $u$ and $v$ at the same level $i$, the rays $\gamma _u$ and $\gamma _v$ intersect: if they have the same parent $w$, then they intersect on $a_w$, and if they have different parents, the order of their origins $a_u$ and $a_v$ on $\alpha _{i+1}$ and the order of their intersections with $\alpha _{i}$ are reversed. $\square$

2.3 A Tree with a Few Short Paths

Let $T$ be an embedded tree with a linear order $\tau$. An admissible extension of $T$ is a graph $P$ with the following properties:

• $P$ is a union of vertex-disjoint paths;

• each maximal path in $P$ has 3 or 4 vertices;

• the endpoints of each maximal path in $P$ are leaves of $T$ that are consecutive and at the same level;

• the internal vertices of any path in $P$ are not vertices of $V(T)$.

Note that $T+P$ is a planar graph because we only add paths between consecutive leaves.

Lemma 2.4

Let $T$ be an embedded tree and let $P$ be an admissible extension of $T$. The complement of $T+P$ is a ray intersection graph.

Proof

We construct a snooker representation of $T$ using Lemma 2.3 where $k$, the number of levels, is the depth of $T$ plus 2 or 3, whichever is odd. We will use a local argument to represent each maximal path of $P$, and each maximal path of $P$ is treated independently. It will be obvious from the construction that rays corresponding to vertices in different paths intersect. We distinguish the case where the maximal path has one internal vertex or two.

Consider first the case of a maximal path in $P$ with one internal vertex. Thus, the path is $uwv$ where $u$ and $v$ are consecutive leaves in $T$ and $w\notin V(T)$ is not yet represented by a ray. The origins $a_u$ and $a_v$ of the rays $\gamma _u$ and $\gamma _v$, respectively, are distinct and consecutive along $\alpha _{i+1}$ because we have a snooker representation. We thus have the situation depicted in Fig. 5. We can then just take the ray $\gamma _w$ to be the line through $a_u$ and $a_v$. (This line can also be a ray with an origin sufficiently far away.) This line intersects the ray of any other vertex different from $\gamma _u$ and $\gamma _v$.

Fig. 5

Case 1 in the proof of Lemma 2.4: adding a path with one internal vertex

Consider now the case of a maximal path in $P$ with two internal vertices. Thus, the path is $uww'v$ where $u$ and $v$ are consecutive leaves in $T$ and $w,w'\notin V(T)$. In this case, the construction depends on the relative position of the origins $a_u$ and $a_v$, and we distinguish two scenarios: (i) shifting the origin $a_u$ of ray $\gamma _u$ towards $a_v$ while maintaining the slope introduces an intersection between $\gamma _u$ and the ray for the parent of $u$ or (ii) shifting the origin $a_v$ of ray $\gamma _v$ towards $a_u$ while maintaining the slope introduces an intersection between $\gamma _v$ and the ray for the parent of $v$. Note that exactly one of the two options must occur.

Let us consider only scenario (i), since the other one is symmetric; see Fig. 6. We choose a point $b_w$ on $\alpha _{i+1}$ between $a_u$ and $a_v$ very near $a_u$ and represent $w$ with a ray $\gamma _w$ parallel to $\gamma _u$ with origin $b_w$. Thus $\gamma _w$ does not intersect $\gamma _u$ but intersects any other ray because we are in scenario (i). Finally, we represent $w'$ with the line $\gamma _{w'}$ through points $b_w$ and $a_v$. With this, $\gamma _{w'}$ intersects the interior of any other ray but $\gamma _w$ and $\gamma _v$, as desired. $\square$

Fig. 6

Case 2(i) in the proof of Lemma 2.4: adding a path with two internal vertices

Lemma 2.5

Any embedded planar graph $G$ has an even subdivision $T+P$, where $T$ is an embedded tree and $P$ is an admissible extension of $T$. Furthermore, such $T$ and $P$ can be computed in polynomial time.

Proof

Let $r$ be an arbitrary vertex in the outer face $f$ of $G$. Let $B$ be the set of edges in a BFS tree of $G$ from $r$. With a slight abuse of notation, we also use $B$ for the BFS tree itself. Let $C=E(G)\setminus B$. In the graph $G^*$ dual to $G$, the edges $C^*=\{c^*\mid c\in C\}$ are a spanning tree of $G^*$, which with a slight abuse of notation we also denote by $C^*$. We root $C^*$ at the node $f^*$, corresponding to the outer face. This is illustrated in Fig. 7.

Fig. 7

Figure for the proof of Lemma 2.5. a A planar graph with a BFS from $r$ in bold. b The spanning tree $C^*$ of the dual graph. c The subdivision of the edges from $C$. d The resulting subdivided graph with $P$ marked in red and thinner lines. e The resulting graph $T$ after the removal of $P$, drawn so that the heights of the vertices correspond to their levels (Color figure online)

We define for each edge $e\in C$ the number $k_e$ of subdivisions it will undertake using a bottom-up approach. Any edge $e\in C$ that is incident to a leaf of $C^*$ gets assigned $k_e=4$. For any other edge $e\in C$, we define $k_e$ as $2$ plus the maximum $k_{e'}$ over all descendants $(e')^*\in C^*$ of $e^*$ in $C^*$. Let $H$ be the resulting subdivision. For any edge $e\in C$, let $Q_e$ be the path in $H$ that corresponds to the subdivision of $e$. We use in $H$ the combinatorial embedding induced by $G$.

We can now compute the tree $T$ and the paths $P$. Since $B$ is a BFS tree, every edge $e\in C$ connects two vertices that are either at the same level in $B$, or at two successive levels. For every edge $e\in C$ that connects two vertices at the same level in $B$, let $P_e$ be the length-three subpath in the middle of $Q_e$. For every edge $uv\in C$ that connects vertex $u$ at level $i$ to vertex $v$ at level $i+1$ in $B$, let $P_e$ be the length-two subpath in the middle of $Q_e-u$. We then take $P=\bigcup _{e\in C} P_e$ and take $T$ to be the graph $H-P$, after removing isolated vertices. In $T$ we use the rotation system inherited from $H$ and use the edge $rs$ to define the linear order, where $rs$ is an edge in the outer face $f$.

It is clear from the construction that $T+P=H$ is an even subdivision of $G$. We have to check that $P$ is indeed an admissible extension of $T$. The maximal paths of $P$ are vertex disjoint and connect leaves of $P$ because the paths $Q_e$, $e\in C$, are edge-disjoint and each $P_e$ is strictly contained in the interior of $Q_e$. Since in $H-P$ we removed isolated vertices, it is clear that no internal vertex of a path of $P$ is in $T$. The graph $T$ is indeed a tree because, for every edge $e\in C$, we have removed some edges from its subdivision $Q_e$. Since $B$ is a BFS tree and $P_{uv}$ is centered within $Q_{uv}$, when $u$ and $v$ are at the same level, or within $Q_{uv}-u$, when $u$ is one level below $v$, the maximal paths of $P$ connect vertices that are equidistant from $r$ in $H$.

It remains to show that the endpoints of any maximal path in $P$ are consecutive vertices in $T$. This is so because of the inductive definition of $k_e$. The base case is when $e\in C$ is incident to a leaf of $C^*$, and is subdivided $k_e=4$ times. The edge $e$ either connects two vertices at the same level in $B$, or at two successive levels. In both cases it can be checked that $P_e$ connects two consecutive vertices in $T$. The inductive case is as follows. We let $k_e$ be equal to $2$ plus the maximum $k_{e'}$ over all descendants $(e')^*\in C^*$ of $e^*$ in $C^*$. By induction, all the corresponding $P_{e'}$ connect two consecutive vertices of $T$, say at level $i$ in $T$. By definition, $P_e$ will connect two consecutive vertices of $T$ at level $i+1$.

The construction of $T$ and $P$ only involves computing the BFS tree $B$, the spanning tree $C^*$, and the values $k_e$, which can clearly be done in polynomial time. $\square$

Combining Lemmas 2.4 and 2.5 directly yields the following.

Theorem 2.6

Any planar graph has an even subdivision whose complement is a ray intersection graph. Furthermore, this subdivision can be computed in polynomial time.

2.4 Polynomial-Time Construction

So far, we used real coordinates in our construction. We next argue how to have a construction using only integers with a polynomial number of bits.

Lemma 2.7

In the construction of Lemma 2.4, if $n=|V(T+P)|$,

• any two origins of rays are at distance at least $n^{-O(n)}$ and at most $O(n)$;

• the distance between any ray and any other origin is at least $n^{-O(n)}$.

Proof

Recall that the circle containing points $p_0,p_1,\ldots$ has radius $1$. The rectangles $R_i$ are all congruent and have two diagonals of length $|p_0 p_1|=\Theta (1)$. The small side of rectangle $R_i$ has size $\Theta (1/n)$ and both diagonals of $R_i$ form an angle of $\Theta (1/n)$. It follows that the center of the circle supporting $\alpha _i$ has coordinates $\Theta (n)$. For points from different curves $\alpha _i$ there is at least a separation of $\Theta (1/n)$.

We first bound the distance between the origins for the rays representing vertices of $T$. Let us refer by features on the curve $\alpha _i$ the origins of rays lying on $\alpha _i$ and the extremes of $\alpha _i$. Let $\delta _i$ be the minimum separation between any two features on the curve $\alpha _i$. In the curve $\alpha _1$ there are three features: the two extremes of $\alpha _1$ and the origin $a_r$ of $\gamma _r$, which is in the middle of $\alpha _1$. Since $\alpha _1$ have length $\Omega (1)$, it follows that $\delta _1=\Omega (1)$.

We will bound the ratio $\delta _{i+1}/\delta _{i}$ for $i\ge 1$. Consider the construction of Lemma 2.3 to determine the features on $\alpha _{i+1}$. By induction, any two consecutive features along $\alpha _{i}$ are separated at least by $\delta _i$. Since $\alpha _{i}$ is supported by a circle of radius $\Theta (n)$, the lines $\ell _v^+$ and $\ell _v^-$ form an angle of at least $\Omega (\delta _i/n)$. This implies that the points $\ell _v^-\cap \alpha _{i+1}, a_{u_1},\ldots , a_{u_d}, \ell _v^+\cap \alpha _{i+1}$ have a separation of $\Omega (\delta _i/(nd))$. It follows that $\delta _{i+1}=\Omega (\delta _i/(nd))=\Omega (\delta _i/n^2)$, and thus $\delta _{i+1}/\delta _i=\Omega (1/n^2)$. Since $T$ has depth at most $n$, all features are at distance at least $n^{-O(n)}$.

We can now argue that the origins of the rays used in the construction of Lemma 2.4 also have a separation of $n^{-O(n)}$. In Case 1, we just add a line through previous features. In Case 2, it is enough to place $b_w$ at distance $|a_u a_v|/n$ from $a_u$, and thus the features keep a separation of at least $n^{-O(n)}$.

For the proof of the second item, consider a ray on level $i$ and the point $a_v$ at which it hits $\alpha _i$. Now consider another origin $b$ on $\alpha _i$. The angle $x$ made by the line $a_vb$ and the ray is bounded from below (like the angle between $\ell _v^-$ and $\ell _v^+$ in Fig. 4) by $n^{-cn}$ for some constant $c$, and so is the distance $d$ between $a_v$ and $b$. Now the distance between $b$ and the ray is $d \sin x \ge d 2 x / \pi \ge n^{-2cn} (2 / \pi ) = n^{-O(n)}$, as claimed, where the first inequality comes from the relation $\sin x \ge 2x/\pi$ for $x\in [0, \pi /2]$. $\square$

We can now give the following algorithmic version of Lemma 2.4.

Lemma 2.8

Let $T$ be an embedded tree and let $P$ be an admissible extension of $T$. We can find in polynomial time a family of rays described with integer coordinates whose intersection graph is isomorphic to the complement of $T+P$.

Proof

Recall that, after constructing the reference frame, each point in the construction is created either as (i) an intersection between a line through two existing points and a curve $\alpha _i$, or (ii) by equally spacing $O(n)$ points between two existing points on a curve $\alpha _i$. Consider that each point is moved by a distance $< \varepsilon$ after it is constructed. This causes any point further constructed from those to move as well. In case (ii), the new points would be moved by no more than $\varepsilon$ as well. In case (i) however, the error could be amplified. Let $a$ and $b$ be two previously constructed points, and suppose they are moved to $a'$ and $b'$, within a radius of $\varepsilon$. By the previous lemma, the distance between $a$ and $b$ is at least $n^{-O(n)}$ and so the angle between the line $ab$ and $a'b'$ is at most $\varepsilon n^{O(n)}$. Because the radius of the supporting circle of $\alpha _i$ is $\Theta (n)$, the distance between $ab \cap \alpha _i$ and $a'b' \cap \alpha _i$ is at most $O(n)\varepsilon n^{O(n)} = \varepsilon n^{O(n)}$.

Therefore, an error in one construction step expands by a factor $n^{O(n)}$. Now observe that each point of type (i) is constructed on the next level, and a point of type (ii) is always constructed from points of type (i). Therefore, as there are $O(n)$ levels, an error is propagated at most $O(n)$ times, and therefore, the total error propagated from moving each constructed point by a distance $< \varepsilon$ is at most $\varepsilon n^{O(n^2)}$.

By the previous lemma, there is a constant $c$ such that any origin in the construction of Lemma 2.4 is separated at least by a distance $A=1/n^{cn}$ from any other origin and any ray not incident to it. Therefore, by choosing $\varepsilon = n^{-c'n^3}$ for $c'$ large enough, the total propagation will never exceed $A$, and therefore perturbing the basic construction points of the reference frame and each further constructed point by a distance $<\varepsilon$ will not change the intersection graph of the modified rays.

Therefore, to construct the required set of rays in polynomial time, we multiply every coordinate by the smallest power of $2$ larger than $1/\varepsilon$ and snap every constructed point to the nearest integer while following the construction of Lemma 2.4. Each coordinate can then be represented by $O(n^3\log n)$ bits. $\square$

Let $\alpha (G)$ be the size of the largest independent set in a graph $G$. The following simple lemma can be deduced from the observation that subdividing an edge twice increases the independence number by exactly one.

Lemma 2.9

If $G'$ is an even subdivision of $G$ where each edge $e\in E(G)$ is subdivided $2k_e$ times, then $\alpha (G')=\alpha (G)+\sum _e k_e$. $\square$

By combining Lemmas 2.5, 2.8, and 2.9, we obtain a polynomial-time reduction from the maximum independent set problem in planar graphs to the maximum clique problem in ray intersection graphs.

Theorem 2.10

Finding a maximum clique in a ray intersection graph is NP-hard, even when the input is given by a geometric representation as a set of rays. $\square$