Triangles and Girth in Disk Graphs and Transmission Graphs

TL;DR

This paper presents efficient algorithms for detecting triangles and computing the girth in disk and transmission graphs, which are important in sensor network modeling, using new batched range searching techniques.

Contribution

The authors develop algorithms that find shortest triangles and girth in disk and transmission graphs in near-linear expected time, improving on previous methods.

Findings

Shortest triangle detection in O(n log n) expected time

Girth computation in O(n log n) expected time

New batched range searching tools introduced

Abstract

Let $S \subset \mathbb{R}^2$ be a set of $n$ sites, where each $s \in S$ has an associated radius $r_s > 0$. The disk graph $D(S)$ is the undirected graph with vertex set $S$ and an undirected edge between two sites $s, t \in S$ if and only if $|st| \leq r_s + r_t$, i.e., if the disks with centers $s$ and $t$ and respective radii $r_s$ and $r_t$ intersect. Disk graphs are used to model sensor networks. Similarly, the transmission graph $T(S)$ is the directed graph with vertex set $S$ and a directed edge from a site $s$ to a site $t$ if and only if $|st| \leq r_s$, i.e., if $t$ lies in the disk with center $s$ and radius $r_s$. We provide algorithms for detecting (directed) triangles and, more generally, computing the length of a shortest cycle (the girth) in $D(S)$ and in $T(S)$. These problems are notoriously hard in general, but better solutions exist for special graph classes such as planar graphs. We obtain similarly efficient results for disk graphs and for transmission graphs. More precisely, we show that a shortest (Euclidean) triangle in $D(S)$ and in $T(S)$ can be found in $O(n \log n)$ expected time, and that the (weighted) girth of $D(S)$ can be found in $O(n \log n)$ expected time. For this, we develop new tools for batched range searching that may be of independent interest.

Equations

$$w(P)=\min \{w(P_1),\dots ,w(P_r)\}.$$

$$|tu|\le |ts|+|su|\le r_s+r_u<r_t/2+r_t/2=r_t,$$

Full text

Triangles and Girth in Disk Graphs and

Transmission Graphs

Haim Kaplan

  

Katharina Klost

  

Wolfgang Mulzer

  

Liam Roditty

  

Paul Seiferth

  

Micha Sharir

We provide algorithms for detecting (directed) triangles and, more generally, computing the length of a shortest cycle (the girth) in $D(S)$ and in $T(S)$. These problems are notoriously hard in general, but better solutions exist for special graph classes such as planar graphs. We obtain similarly efficient results for disk graphs and for transmission graphs. More precisely, we show that a shortest (Euclidean) triangle in $D(S)$ and in $T(S)$ can be found in $O(n\log n)$ expected time, and that the (weighted) girth of $D(S)$ can be found in $O(n\log n)$ expected time. For this, we develop new tools for batched range searching that may be of independent interest.

1 Introduction

Given a graph $G$ with $n$ vertices and $m$ edges, does $G$ contain a triangle (a cycle with three vertices)? This is one of the most basic algorithmic questions in graph theory, and many other problems reduce to it [14, 23]. The best known algorithms use fast matrix multiplication and run in either $O(n^{\omega})$ time or in O\big{(}m^{2\omega/(\omega+1)}\big{)} time, where $\omega<2.37287$ is the matrix multiplication exponent [1, 12, 14]. Despite decades of research, the best available “combinatorial” algorithm111An algorithm is “combinatorial” if it does not need algebraic manipulations to achieve its goal. needs O\big{(}n^{3}\,\text{polyloglog}(n)/\log^{4}n\big{)} time [24], only slightly better than checking all vertex triples. This lack of progress can be explained by a connection to Boolean matrix multiplication (BMM): if there is a truly subcubic combinatorial algorithm for finding triangles, there is also a truly subcubic combinatorial algorithm for BMM [23]. Itai and Rodeh [14] reduced computing the girth (the length of a shortest cycle) of an unweighted undirected graph to triangle detection. For integer edge weights, Roditty and V. Williams [21] gave an equivalence between finding a minimum weight cycle (the weighted girth) and finding a minimum weight triangle.

For the special case of planar graphs, significantly better algorithms are known. Itai and Rodeh [14] and, independently, Papadimitriou and Yannakakis [19] showed that a triangle can be found in $O(n)$ time, if it exists. Chang and Lu [7] presented an $O(n)$ time algorithm for computing the girth. The weighted girth can be found in $O(n\log\log n)$ time both in an undirected and in a directed planar graph [17, 18].

In computational geometry, there are two noteworthy graph classes that generalize planar graphs: disk graphs and transmission graphs. We are given a set $S$ of $n$ planar point sites. Each $s\in S$ has an associated radius $r_{s}>0$ and an associated disk $D_{s}$ with center $s$ and radius $r_{s}$. The disk graph $D(S)$ is the undirected graph on $S$ where two sites $s,t\in S$ are adjacent if and only if $D_{s}$ and $D_{t}$ intersect, i.e., $|st|\leq r_{s}+r_{t}$, where $|\cdot|$ is the Euclidean distance. In a weighted disk graph, the edges are weighted according to the Euclidean distance between their endpoints. The transmission graph $T(S)$ is the directed graph on $S$ where there is an edge from $s$ to $t$ if and only if $t$ lies in $D_{s}$, i.e., $|st|\leq r_{s}$. Again, there is a weighted variant. Both graph classes have received a lot of attention, as they give simple and natural theoretical models for geometric sensor networks (see, e.g., [15, 16]).

Motivated by the vastly better algorithms for planar graphs, we investigate triangle detection and girth computation in disk graphs and transmission graphs. We will see that in a disk graph, a triangle can be found in $O(n\log n)$ time, using a simple geometric observation to relate disk graphs and planar graphs. By a reduction from $\varepsilon$-closeness [20], this is optimal in the algebraic decision tree model, a contrast to planar graphs, where $O(n)$ time is possible. Our method generalizes to finding a shortest triangle in a weighted disk graph in $O(n\log n)$ expected time. Moreover, we can compute the unweighted and weighted girth in a disk graph in $O(n\log n)$ time, with a deterministic algorithm for the unweighted case and a randomized algorithm for the weighted case. The latter result requires a method to find a shortest cycle that contains a given vertex. Finally, we provide an algorithm to detect a directed triangle in a transmission graph in $O(n\log n)$ expected time. For this, we study the geometric properties of such triangles in more detail, and we develop several new techniques for batched range searching that might be of independent interest, using linearized quadtrees and three-dimensional polytopes to test for containment in the union of planar disks. As before, this algorithm extends to the weighted version. We will assume general position, meaning that all edge lengths (and more generally shortest path distances) are pairwise distinct, that no site lies on a disk boundary, and that all radii are pairwise distinct.

2 Finding a (Shortest) Triangle in a Disk Graph

We would like to decide if a given disk graph contains a triangle. If so, we would also like to find a triangle of minimum Euclidean perimeter.

2.1 The Unweighted Case

The following property of disk graphs, due to Evans et al. [11], is the key to our algorithm. For completeness, we include a proof.

Lemma 2.1**.**

Let $D(S)$ be a disk graph that is not plane, i.e., the embedding that represents each edge by a line segment between its endpoints has two segments that cross in their relative interiors. Then, there are three sites whose associated disks intersect in a common point.

Proof 2.2**.**

Suppose the segments $st$ and $uv$ intersect in a point $a$. The sites $s$, $t$, $u$, and $v$ are pairwise distinct, and without loss of generality, we assume that: (i) $a\in D_{s}\cap D_{u}$; (ii) $r_{u}\leq r_{s}$; (iii) the point $s$ lies in the origin, the edge $st$ lies on the $x$-axis, with $t$ in the positive halfplane; and (iv) the site $u$ lies above the $x$-axis, the site $v$ lies below the $x$-axis; see Figure 1.

If $a\in D_{t}$, then $D_{s}\cap D_{t}\cap D_{u}\neq\emptyset$, and we are done. Thus, suppose that $a\not\in D_{t}$, and let $b$ be the first point on $st$ in $D_{t}$. If $b\in D_{u}$, then $D_{s}\cap D_{t}\cap D_{u}\neq\emptyset$, and we are done. Thus, suppose that $b\not\in D_{u}$. If $u$ lies in the northwest quadrant, then $v$ must be in the southeast quadrant. Furthermore, since $r_{u}\leq r_{s}$, it follows that in the southeast quadrant, $D_{u}$ is completely contained in $D_{s}$, so the first point on the segment $av$ that is in $D_{v}$ must also be in $D_{s}$ and $D_{u}$. Thus, $D_{s}\cap D_{u}\cap D_{v}\neq\emptyset$, and we are done. If $u$ lies in the northeast quadrant, since $r_{u}\leq r_{s}$ and since $b\not\in D_{u}$, it follows that below the $x$-axis, we have $D_{u}\subseteq D_{s}$, and the first point on the segment $av$ that is in $D_{v}$ must also be in $D_{s}$ and $D_{u}$, i.e. $D_{s}\cap D_{u}\cap D_{v}\neq\emptyset$.

If $D(S)$ is not plane, it contains a triangle by Lemma 2.1. If $D(S)$ is plane, we can construct it explicitly and then search for a triangle in $O(n)$ time [14, 19]. To check whether $D(S)$ is plane, we begin an explicit construction of $D(S)$ and abort if we discover too many edges.

Theorem 2.3**.**

Let $D(S)$ be a disk graph on $n$ sites. We can find a triangle in $D(S)$ in $O(n\log n)$ worst-case time, if it exists.

Proof 2.4**.**

For each $s\in S$ we split $\partial D_{s}$ into two $x$-monotone curves, namely the upper and the lower arc from the leftmost point of $D_{s}$ to the rightmost point. We use the Bentley-Ottmann sweepline algorithm to find the intersections between these boundary arcs. The intersections are reported one by one, and the total time to find the first $m$ intersections is $O(n\log n+m\log n)$ [2, Theorem 2.4].222The algorithm is presented for line segments, but it extends easily to continuous $x$-monotone curves. If the sweepline algorithm reports more than $6n-12$ intersection points, we can be sure that $D(S)$ is not plane, because an edge of $D(S)$ corresponds to at most two intersections. Then, $D(S)$ contains a triangle by Lemma 2.1, and we can find it in $O(n\log n)$ time with another plane sweep that gives an intersection between the edges we have generated so far.

If there are at most $6n-12$ intersections, we use another plane sweep to find the vertical decomposition of the arrangement of the disks $D_{s}$, $s\in S$, in $O(n\log n)$ time. We use the vertical decomposition to construct the remaining edges of $D(S)$ that are due to a disk being completely contained in another disk. For this, we walk through the pseudo-trapezoids of the vertical decomposition, keeping track of the disks that contain the current pseudo-trapezoid. When we enter a disk for the first time, we generate edges between this disk and the disks containing the current pseudo-trapezoid. If it turns out that $D(S)$ has more than $3n-6$ edges, we abort the generation of the edges, since then $D(S)$ is not plane and contains a triangle that can be found in $O(n\log n)$ time with a plane sweep over the edges generated so far. If $D(S)$ has at most $3n-6$ edges, we obtain an explicit representation of $D(S)$. We check if this representation is plane in $O(n\log n)$ time with a plane sweep, returning a triangle if this is not the case. Finally, if $D(S)$ is plane, we check for a triangle in $O(n)$ time [14, 19].

2.2 The Weighted Case

Suppose the edges in $D(S)$ are weighted by their Euclidean lengths. We would like to find a triangle of minimum perimeter, i.e., of minimum total edge length. For this, we solve the decision problem: given $W>0$, does $D(S)$ contain a triangle with perimeter at most $W$? Once a decision algorithm is available, the optimization problem can be solved with Chan’s randomized geometric optimization framework [5].

To decide if $D(S)$ contains a triangle with perimeter at most $W$, we use a grid with diameter $W/3$. We look for triangles whose vertices lie in a single grid cell, using the algorithm from Section 2.1. If no cell contains such a triangle, then $D(S)$ will be sparse and we will need to check only $O(n)$ further triples. Details follow.

Set $\ell=W/(3\sqrt{2})$. Let $G_{1}$ be the grid whose cells are pairwise disjoint, axis-parallel squares with side length $\ell$, aligned so that the origin $(0,0)$ is a vertex of $G_{1}$. The cells of $G_{1}$ have diameter $\sqrt{2}\cdot\ell=W/3$, so any triangle whose vertices lie in a single cell has perimeter at most $W$. We make three additional copies $G_{2}$, $G_{3}$, $G_{4}$ of $G_{1}$, and we shift them by $\ell/2$ in the $x$-direction, in the $y$-direction, and in both the $x$- and $y$-directions, respectively. In other words, $G_{2}$ has $(\ell/2,0)$ as a vertex, $G_{3}$ has $(0,\ell/2)$ as a vertex, and $G_{4}$ has $(\ell/2,\ell/2)$ as a vertex, see Figure 2. This ensures that if all edges in a triangle are “short”, the triangle lies in a single grid cell.

Lemma 2.5**.**

Let $\Delta$ be a triangle formed by three vertices $a,b,c\in\mathbb{R}^{2}$ such that each edge of $\Delta$ has length at most $\ell/2$. There is a cell $\sigma\in\bigcup_{i=1}^{4}G_{i}$ with $a,b,c\in\sigma$.

Proof 2.6**.**

We can enclose $\Delta$ with a square of side length $\ell/2$. This square must be completely contained in a cell of one of the four grids, see Figure 2.

We go through all nonempty grid cells $\sigma\in\bigcup_{i=1}^{4}G_{i}$, and we search for a triangle in the disk graph $D(S\cap\sigma)$ induced by the sites in $\sigma$, with Theorem 2.3. Since each site lies in $O(1)$ grid cells, and since we can compute the grid cells for a given site in $O(1)$ time (using the floor function), the total running time is $O(n\log n)$. If a triangle is found, we return YES, since the cells have diameter $W/3$ and thus such a triangle has perimeter at most $W$. If no triangle is found, Lemma 2.5 implies that any triangle in $D(S)$ has one side of length more than $\ell/2$ and hence at least one vertex with associated radius at least $\ell/4$. We call a site $s\in S$ large if $r_{s}>\ell/4$. A simple volume argument bounds the number of large sites in a grid cell.

Lemma 2.7**.**

Let $\sigma\in\bigcup_{i=1}^{4}G_{i}$ be a nonempty grid cell, and suppose that $D(S\cap\sigma)$ does not contain a triangle. Then $\sigma$ contains at most $18$ large sites.

Proof 2.8**.**

Suppose $\sigma$ contains at least $19$ large sites. We cover $\sigma$ with $3\times 3$ congruent squares of side length $\ell/3$. Then, at least one square $\tau$ contains at least $\lceil 19/9\rceil=3$ large sites. The associated radius of a large site is more than $\ell/4$ and each square has diameter $(\sqrt{2}/3)\ell<\ell/2$, so the large sites in $\tau$ form a triangle in $D(S\cap\sigma)$, a contradiction.

Let $\sigma\in G_{i}$, $i\in\{1,\dots,4\}$, be a grid cell. The neighborhood $N(\sigma)$ of $\sigma$ is the $5\times 5$ block of cells in $G_{i}$ centered at $\sigma$. Since the diameter of a grid cell is $W/3$, any two sites $u,v\in S$ that form a triangle of perimeter at most $W$ with a site $s\in S\cap\sigma$ must be in $N(\sigma)$. Let $S_{\ell}\subseteq S$ denote the large sites. At this stage, we know that any triangle in $D(S)$ has at least one vertex in $S_{\ell}$. By Lemma 2.7, for any $\sigma\in\bigcup_{i=1}^{4}G_{i}$, we have $|\cup_{\tau\in N(\sigma)}\tau\cap S_{\ell}|=O(1)$. Thus, to detect a triangle of perimeter at most $W$ with at least two large vertices, we proceed as follows: for each non-empty cell $\sigma\in G_{i}$, iterate over all large sites $s$ in $\sigma$, over all large sites $t$ in $N(\sigma)$, and over all (not necessary large) sites $u$ in $N(\sigma)$. Check whether $stu$ is a triangle of perimeter at most $W$. If so, return YES. Since the sites in each grid cell are examined $O(1)$ times for $O(1)$ pairs of large sites, the total time is $O(n)$.

It remains to detect triangles of perimeter at most $W$ with exactly one large vertex. We iterate over all grid cells $\sigma\in\bigcup_{i=1}^{4}G_{i}$, and we compute $D(S\cap\sigma)$. Since $D(S\cap\sigma)$ contains no triangle, Lemma 2.1 shows that $D(S\cap\sigma)$ is plane, has $O(|S\cap\sigma|)$ edges and can be constructed in time $O(|S\cap\sigma|\log|S\cap\sigma|)$. For every edge $st\in D(S\cap\sigma)$ with both endpoints in $S\setminus S_{\ell}$, we iterate over all large sites $u$ in $N(\sigma)$ and we test whether $stu$ makes a triangle in $D(S)$ with perimeter at most $W$. If so, we return YES. By Lemma 2.7, this takes $O(|S\cap\sigma|)$ time, so the total running time is $O(n\log n)$. If there is a triangle of perimeter at most $W$ with exactly one vertex in $S_{\ell}$, the edge with both endpoints in $S\setminus S_{\ell}$ has length at most $\ell/2$ and thus must lie in a single grid cell $\sigma\in\bigcup_{i=1}^{4}G_{i}$. To summarize:

Lemma 2.9**.**

Let $D(S)$ be a disk graph on $n$ sites, and let $W>0$. We can decide in $O(n\log n)$ worst-case time whether $D(S)$ contains a triangle of perimeter at most $W$.

We employ the following general lemma due to Chan [5]. Let $\Pi$ be a problem space, and for a problem $P\in\Pi$, let $w(P)\in\mathbb{R}$ be its optimum and $|P|\in{\mathbb{N}}$ be its size.

Lemma 2.10** (Lemma 2.1 in [5]).**

Let $\alpha<1$, $\varepsilon>0$, and $r\in{\mathbb{N}}$ be constants, and let $\delta(\cdot)$ be a function such that $\delta(n)/n^{\varepsilon}$ is monotone increasing in $n$. Given any optimization problem $P\in\Pi$ with optimum $w(P)$, suppose that within time $\delta(|P|)$, (i) we can decide whether $w(P)<t$, for any given $t\in\mathbb{R}$, and (ii) we can construct $r$ subproblems $P_{1},\dots,P_{r}$, each of size at most $\lceil\alpha|P|\rceil$, so that

[TABLE]

Then we can compute $w(P)$ in total expected time $O(\delta(|P|))$.

Now the following main theorem of this section is immediate.

Theorem 2.11**.**

Let $D(S)$ be a weighted disk graph on $n$ sites. We can compute a shortest triangle in $D(S)$ in $O(n\log n)$ expected time, if one exists.

Proof 2.12**.**

We apply Lemma 2.10. For Condition (i), we use Lemma 2.9. For Condition (ii), we construct four subsets $S_{0},\dots,S_{3}$ of $S$ as follows: enumerate the sites in $S$ as $S=\{s_{1},\dots,s_{n}\}$, and put the site $s_{i}$ into all sets $S_{j}$ with $i\not\equiv j\pmod{4}$. Then, for any three sites $a,b,c\in S$, there is at least one subset $S_{j}$ with $a,b,c\in S_{j}$. Now, Lemma 2.10 with $\alpha=3/4$, $\varepsilon=1$, $r=4$, and $\delta=O(n\log n)$ implies the theorem.

3 Computing the Girth of a Disk Graph

We extend the results from Section 2 to the girth. The unweighted case is easy: if $D(S)$ is not plane, the girth is $3$, by Lemma 2.1. If $D(S)$ is plane, we use the algorithm for planar graphs [7]. The weighted case is harder. If $D(S)$ is plane, we use the algorithm for planar graphs [17]. If not, Theorem 2.11 gives a shortest triangle $\Delta$ in $D(S)$. However, there could be cycles with at least four edges that are shorter than $\Delta$. To address this, we use $\Delta$ to split $D(S)$ into sparse pieces where a shortest cycle can be found efficiently.

3.1 The Unweighted Case

Chang and Lu [7, Theorem 1.1] showed how to find the girth of an unweighted planar graph with $n$ vertices in $O(n)$ time. Hence, we obtain a simple extension of Theorem 2.3.

Theorem 3.1**.**

Let $D(S)$ be a disk graph for a set $S$ of $n$ sites. We can compute the unweighted girth of $D(S)$ in $O(n\log n)$ worst-case time.

Proof 3.2**.**

We proceed as in Theorem 2.3. If $D(S)$ is not plane, the girth is $3$. If $D(S)$ is plane, we apply the algorithm of Chang and Lu [7, Theorem 1.1] to an explicit representation of $D(S)$.

3.2 The Weighted Case

We describe how to find the shortest cycle through a given vertex in a weighted graph with certain properties. This is then used to compute the weighted girth of a disk graph.

Let $G$ be a graph with nonnegative edge weights so that all shortest paths and cycles in $G$ have pairwise distinct lengths and so that for all edges $uv$, the shortest path from $u$ to $v$ is the edge $uv$. We present a deterministic algorithm that, given $G$ and a vertex $s$, computes the shortest cycle in $G$ containing $s$, if it exists.333Even though this seems to be a simple fact, we could not locate a previous reference for this. A simple randomized algorithm can also be found in Yuster [25, Section 2]. The next lemma states a structural property of the shortest cycle through $s$. It resembles Lemma 1 of Roditty and V. Williams [21] that deals with an overall shortest cycle in $G$.

Lemma 3.3**.**

The shortest cycle in $G$ that contains $s$ consists of two paths in the shortest path tree of $s$, and one additional edge.

Proof 3.4**.**

Let $C=v_{0},v_{1},v_{2},\dots,v_{\ell-1},v_{\ell}$ be the shortest cycle in $G$ containing $s$, where all vertices $v_{i}$, $0\leq i\leq\ell-1$ are pairwise distinct, $\ell\geq 3$, and $v_{0}=v_{\ell}=s$. For $v_{i}\in C$, let $d_{1}(v_{i})$ be the length of the path $s,v_{1},\dots,v_{i}$, and let $d_{2}(v_{i})$ be the length of the path $v_{i},v_{i+1},\dots,s$. Let $\pi(v_{i})$ denote the shortest path from $s$ to $v_{i}$, and let $|v_{i}v_{i+1}|$ be the length of the edge $v_{i}v_{i+1}$.

Suppose that $C$ is not of the desired form. Let $v_{k}v_{k+1}$ be the edge on $C$ with $d_{1}(v_{k})<|v_{k}v_{k+1}|+d_{2}(v_{k+1})$ and $d_{2}(v_{k+1})<d_{1}(v_{k})+|v_{k}v_{k+1}|$. By our assumptions on $G$, the edge $v_{k}v_{k+1}$ exists and $k\neq 0,\ell-1$. We distinguish two cases, illustrated in Figure 3.

First, suppose that $\pi(v_{k})\cap\pi(v_{k+1})=\{s\}$. Consider the cycle $C^{\prime}$ given by $\pi(v_{k})$, the edge $v_{k}v_{k+1}$, and $\pi(v_{k+1})$. Since $s\neq v_{k},v_{k+1}$ and since the edge $v_{k}v_{k+1}$ does not appear on $\pi(v_{k})$ nor on $\pi(v_{k+1})$, it follows that $C^{\prime}$ is a proper cycle. Furthermore, by assumption, $C^{\prime}$ is strictly shorter than $C$, because $\pi(v_{k})$ is shorter than $d_{1}(v_{k})$ or $\pi(v_{k+1})$ is shorter than $d_{2}(v_{k+1})$. This contradicts our assumption on $C$.

Second, suppose that $|\pi(v_{k})\cap\pi(v_{k+1})|\geq 2$. Since $\pi(v_{k})$ and $\pi(v_{k+1})$ are shortest paths, their intersection is a prefix of each path. By the assumption444Namely, that for all edges $uv$, the shortest path from $u$ to $v$ is the edge $uv$. on $G$ at least one of $v_{1},v_{\ell-1}$ is not in $\pi(v_{k})\cup\pi(v_{k+1})$. Without loss of generality, this vertex is $v_{1}$. Let $j\geq 1$ be the smallest index such that $v_{j}\in\pi(v_{k})\cup\pi(v_{k+1})$. We have $j\in\{2,\dots,k\}$. Consider the cycle $C^{\prime}$ that starts at $s$, follows $C$ along $v_{1},v_{2},\dots$ up to $v_{j}$, and then returns along $\pi(v_{k})$ or $\pi(v_{k+1})$ to $s$. By construction, $C^{\prime}$ is a proper cycle. Furthermore, $C^{\prime}\neq C$, because even if $j=k$, the path $\pi(v_{k})$ cannot contain the part of $C$ from $v_{k+1}$ to $s$, due to the choice of $k$. Finally, $C^{\prime}$ is strictly shorter than $C$, because the second part of $C^{\prime}$ from $v_{j}$ to $s$ follows a shortest path and is thus strictly shorter than $d_{2}(v_{j})$. Again, $C^{\prime}$ contradicts our choice of $C$.

Theorem 3.5**.**

Let $G=(V,E)$ be a weighted graph with $n$ vertices and $m$ edges that has the properties given at the beginning of this section. Let $s\in V$. We can compute the shortest cycle in $G$ that contains $s$ in $O(n\log n+m)$ time, if it exists.

Proof 3.6**.**

We find the shortest path tree $T$ for $s$ in $G$, and we traverse $T$ to find for each vertex $v$ in $T\setminus\{s\}$ the second vertex $b[v]$ on the shortest path from $s$ to $v$ (the vertex following $s$). Then, we iterate over all edges in $E$ that are not in $T$. For each such $e=uv$, we check if $b[u]\neq b[v]$. If so, $e$ closes a cycle in $T$ that contains $s$. We determine the length of this cycle (in $O(1)$ time). We return the shortest cycle found in this way.

The correctness follows from Lemma 3.3. As for the running time, it takes $O(n\log n+m)$ time to find the shortest path tree for $s$ with Dijkstra’s algorithm and Fibonacci heaps [10, Chapter 24.3]. After that, it takes $O(n)$ time to compute the nodes $b[v]$, for $v\in T\setminus\{s\}$, and $O(m)$ time to iterate over the edges not in $T$. The length of the cycle associated with such an edge $e$ can be computed in $O(1)$ time, using the shortest path distances in $T$ and the length of $e$.

Let $D(S)$ be a weighted disk graph on $n$ sites. A careful combination of the tools developed so far gives an algorithm for the weighted girth of $D(S)$.

Theorem 3.7**.**

Given a weighted disk graph $D(S)$ on $n$ sites, we can compute the weighted girth of $D(S)$ in $O(n\log n)$ expected time.

Proof 3.8**.**

We use Theorem 2.11 to find the shortest triangle in $D(S)$, if it exists, in $O(n\log n)$ expected time. If $D(S)$ has no triangle, it is plane by Lemma 2.1. As in the proof of Theorem 2.3, we can then explicitly construct $D(S)$ in $O(n\log n)$ time with a plane sweep. We determine the girth of $D(S)$ using the algorithm of Ła̧cki and Sankowski [17, Section 5], in additional $O(n\log\log n)$ time, and are done.

Now, suppose $D(S)$ contains a triangle, and let $W$ be the length of the shortest triangle in $D(S)$, an upper bound for the girth of $D(S)$. As in Section 2.2, we set $\ell=W/(3\sqrt{2})$, and we let let $G$ be the grid of side length $\ell$ that has the origin $(0,0)$ as a vertex. We call a site $s\in S$ large if $r_{s}\geq\ell/4$, and we let $S_{\ell}\subseteq S$ be the set of large sites.

We need to check whether $D(S)$ contains a cycle with more than three vertices and length less than $W$. If so, we must find the shortest such cycle. First, we consider cycles in the induced subgraph $D(S\setminus S_{\ell})$. The graph $D(S\setminus S_{\ell})$ has no triangle, as such a triangle would have length less than $3\cdot\ell/2<W$. Thus, by Lemma 2.1, $D(S\setminus S_{\ell})$ is plane. We can directly compute the weighted girth of $D(S\setminus S_{\ell})$ in $O(n\log n)$ time with a plane sweep and the algorithm of Ła̧cki and Sankowski [17, Section 5]. Let $\Delta_{1}$ be the weighted girth of $D(S\setminus S_{\ell})$.

Next, we consider cycles that have at least one large site. Let $\sigma$ be a cell of $G$. The induced subgraph $D(S\cap\sigma)$ has no triangle, since by the choice of $\ell$, such a triangle would have length less than $W$. Thus, Lemma 2.7 shows $|S_{\ell}\cap\sigma|=O(1)$. By the triangle inequality, the maximum distance between any two sites in a cycle of length less than $W$ is less than $W/2$. Thus, any such cycle containing a site $s\in S\cap\sigma$ completely lies in the $7\times 7$ neighborhood $N(\sigma)$ around $\sigma$. Since $N(\sigma)$ has $O(1)$ cells and since each cell contains $O(1)$ large sites, there are $O(1)$ large sites in $S_{\sigma}=S\cap(\bigcup_{\tau\in N(\sigma)}\tau)$.

Now, for each grid cell $\sigma$, we consider all large sites $s\in S_{\ell}\cap\sigma$. We must find the shortest cycle through $s$ in the subgraph $D(S_{\sigma})$ of $D(S)$ in $N(\sigma)$. Let $n_{\sigma}=|S_{\sigma}|$. Since the graph induced by $S_{\sigma}\setminus S_{\ell}$ is plane and since $|S_{\sigma}\cap S_{\ell}|=O(1)$, the graph $D(S_{\sigma})$ has $O(n_{\sigma})$ edges. Hence, we can construct $D(S_{\sigma})$ and apply Theorem 3.5 to compute the shortest cycle in $D(S_{\sigma})$ through $s$ in total time $O(n_{\sigma}\log n_{\sigma})$. Let $\Delta_{2}$ be the smallest length of such a cycle, over all grid cells $\sigma$ and all large sites $s\in S_{\ell}\cap\sigma$. Since each small site is involved in $O(1)$ neighborhoods, we get $\sum_{\sigma\in G}n_{\sigma}=O(n)$, and the overall running time of this step is $O(n\log n)$. Finally, we return $\min\{W,\Delta_{1},\Delta_{2}\}$. If we also want the shortest cycle itself, we simply maintain appropriate pointers in the algorithm. The total expected running time is $O(n\log n)$.

4 Finding a Triangle in a Transmission Graph

Given a transmission graph $T(S)$ on $n$ sites, we want to decide if $T(S)$ contains a directed triangle. We first describe an inefficient algorithm for this problem, and then we will explain how to implement it in $O(n\log n)$ expected time.

The algorithm iterates over each directed edge $e=st$ with $r_{t}\geq r_{s}$, and it performs two tests: first, for each directed edge $tu$ with $r_{u}\geq r_{t}/2$, it checks if $us$ is an edge in $T(S)$, i.e., if $s\in D_{u}$. If so, the algorithm reports the triangle $stu$. Second, the algorithm tests if there is a site $u$ such that $r_{u}\in[r_{s},r_{t}/2)$ and such that $us$ is an edge in $T(S)$, i.e., such that $s\in D_{u}$. If such a $u$ exists, it reports the triangle $stu$. If both tests fail for each edge $e$, the algorithm reports that $T(S)$ contains no triangle. The next lemma shows that the algorithm is correct.

Lemma 4.1**.**

A triple $stu$ reported by the algorithm is a triangle in $T(S)$. Furthermore, if $T(S)$ contains a triangle, the algorithm will find one.

Proof 4.2**.**

Let $stu$ be a triple reported by the algorithm. The algorithm explicitly checks that $st$ and $us$ are edges in $T(S)$. It remains to consider $tu$. If $r_{u}\geq r_{t}/2$, then $stu$ is reported by the first test, and the algorithm explicitly checks that $tu$ is an edge in $T(S)$. If $r_{u}<r_{t}/2$, then $stu$ is reported by the second test. We have $r_{s}<r_{t}/2$, since $s$ and $t$ are chosen so that $r_{s}\leq r_{u}$. Furthermore, $st$ and $us$ are edges of $T(S)$, so $t\in D_{s}$ and $s\in D_{u}$. Since the second test ensures that $r_{u}\leq r_{t}/2$, it follows from the triangle inequality that

[TABLE]

so $u\in D_{t}$, and $tu$ is an edge in $T(S)$. Thus, the reported triple $stu$ is a triangle in $T(S)$.

Now suppose that $T(S)$ contains a triangle $stu$, labeled such that $r_{s}\leq\min\{r_{t},r_{u}\}$. If $r_{u}\geq r_{t}/2$, then $stu$ is found by the first test for the edge $st$. If $r_{u}<r_{t}/2$, we have $s\in D_{u}$ and $r_{u}\in[r_{s},r_{t}/2)$. Thus, the second test will be successful for the edge $st$, and the algorithm will report a triple $stu^{\prime}$, such that $s\in D_{u^{\prime}}$ and $r_{u^{\prime}}\in[r_{s},r_{t}/2)$ (the site $u^{\prime}$ might be different from $u$). The first part of the proof shows that $stu^{\prime}$ is a triangle in $T(S)$.

There are several challenges for making the algorithm efficient. First of all, there might be many edges $st$ with $r_{t}\geq r_{s}$. However, the following lemma shows that if there are $\omega(n)$ such edges, the transmission graph $T(S)$ must contain a triangle.

Lemma 4.3**.**

There is an absolute constant $\alpha$ so that for any $r>0$, if there is an $r\times r$ square $\sigma$ that contains more than $\alpha$ sites $s\in S$ with $r_{s}\geq r/4$, then $T(S)$ has a directed triangle.

Proof 4.4**.**

We cover $\sigma$ with a $6\times 6$ grid of side length $r/6$; see Figure 5. There are $36$ grid cells. For every $s\in S\cap\sigma$ with $r_{s}\geq r/4$, the disk $D_{s}$ completely covers the grid cell containing $s$. If $\sigma$ contains more than $\alpha=72$ sites $s$ with $r_{s}\geq r/4$, then one grid cell contains at least three such sites. These sites form a directed triangle in $T(S)$.

Thus, to implement the algorithm, we must solve two range searching problems.

(R1)

EITHER determine that for every site $s\in S$, there are at most $\alpha$ outgoing edges $st$ with $r_{t}\geq r_{s}/2$ and report all these edges; OR find a square $\sigma$ of side length $r>0$ that contains more than $\alpha$ sites $s\in S$ with $r_{s}\geq r/4$.

(R2)

Given $O(n)$ query triples $(s,r_{1},r_{2})$ with $s\in S$ and $0<r_{1}<r_{2}$, find a site $u\in S$ such that there is a query triple $(s,r_{1},r_{2})$ with $u\neq s$, $r_{u}\in[r_{1},r_{2})$, and $s\in D_{u}$; or report that no such site exists.

The query (R1) indeed always has a valid outcome: suppose there is a site $s\in S$ with more than $\alpha$ outgoing edges $st$ with $r_{t}\geq r_{s}/2$. Then, all the endpoints $t$ lie in $D_{s}$, so the square $\sigma$ centered at $s$ with side length $r=2r_{s}$ contains more than $\alpha$ sites with associated radius at least $r/4$. The next theorem shows that we can detect a triangle in $T(S)$ with linear overhead in addition to the time needed for answering (R1) and (R2).

Theorem 4.5**.**

If (R1) and (R2) can be solved in time $R(n)$ for input size $n$, we can find a directed triangle in a transmission graph $T(S)$ on $n$ sites in time $R(n)+O(n)$, if it exists.

Proof 4.6**.**

First, we perform a range query (R1). If it reports a square $\sigma$ of side length $r$ that contains more than $\alpha$ sites $s\in S$ with $r_{s}\geq r/4$, we scan $S$ to find a set $S^{\prime}$ of $\alpha+1$ such sites. By Lemma 4.3, $T(S^{\prime})$ contains a triangle, and we find it in $O(1)$ time by testing all triples in $S^{\prime}$.

Otherwise, (R1) reports the set $E^{\prime}$ of all edges $st$ in $T(S)$ with $r_{t}\geq r_{s}/2$, where $|E^{\prime}|=O(n)$. We go through all edges $e=st$ in $E^{\prime}$ with $r_{t}\geq r_{s}$, and we check if there is an edge $tu$ in $E^{\prime}$ such that $us$ is an edge in $T(S)$, i.e., such that $s\in D_{u}$. If so, we report the triangle $stu$. This takes care of the first test in the algorithm, and we check only $O(n)$ triples, because for each site in $S$, there are at most $\alpha$ outgoing edges in $E^{\prime}$. If we have not been successful, we again go trough all edges $e=st$ in $E^{\prime}$, and if $r_{t}>2r_{s}$, we create the triple $(s,r_{s},r_{t}/2)$. We perform a range query (R2) on the resulting set of $O(n)$ triples. If (R2) finds a site $u\in S$ such that for a query triple $(s,r_{s},r_{t}/2)$, we have $u\neq s$, $r_{u}\in[r_{s},r_{t}/2)$, and $s\in D_{u}$, we report the triangle $stu$. Otherwise, we report that $T(S)$ does not contain a triangle.

By Lemma 4.1, we correctly report a triangle in $T(S)$, if it exists. The time for the additional steps is $O(n)$, so the total running time is $R(n)+O(n)$.

Using existing methods [22], it is easy to solve (R1) and (R2) in $O(n\log^{2}n)$ time. However, a better solution is possible. In the next section, we will implement (R1) and (R2) in $O(n\log n)$ expected time.

Theorem 4.7**.**

Let $T(S)$ be a transmission graph on $n$ sites. We can find a directed triangle in $T(S)$ in expected time $O(n\log n)$, if it exists.

5 Batched Range Searching

The range queries must handle subsets of sites whose associated radii lie in certain intervals: a query $s$ in (R1) concerns sites $t\in S$ such that $r_{t}\geq r_{s}/2$; and a query $(s,r_{1},r_{2})$ in (R2) concerns sites $t$ such that $r_{t}\in[r_{1},r_{2})$. Using a standard approach [2, 22], we subdivide each such query interval into $O(\log n)$ pieces from a set of canonical intervals. For this, we build a balanced binary tree $B$ whose leaves are the sites of $S$, sorted by increasing associated radius. For each vertex $v\in B$, let the canonical interval ${\mathcal{I}}_{v}$ be the sorted list of sites in the subtree rooted at $v$. There are $O(n)$ canonical intervals.

Next, we define canonical paths and canonical nodes. For a radius $r>0$, the (proper) predecessor of $r$ is the site $s\in S$ with the largest radius $r_{s}\leq r$ ($r_{s}<r$). The (proper) successor of $r$ is defined analogously. For a query $s$ in (R1), we consider the path $\pi$ in $B$ from the root to the leaf with the proper predecessor $t$ of $r_{s}/2$. If $t$ does not exist (i.e., if $r_{t}\geq r_{s}/2$, for all $t\in S$), we let $\pi$ be the left spine of $B$. We call $\pi$ the canonical path for $s$. The canonical nodes for $s$ are the right children of the nodes in $\pi$ that are not in $\pi$ themselves, plus possibly the last node of $\pi$, if $r_{t}\geq r_{s}/2$, for all $t\in S$, see Figure 6.

For a query $(s,r_{1},r_{2})$ in (R2), we consider the path $\pi_{1}$ in $B$ from the root to the leaf with the proper predecessor $t_{1}$ of $r_{1}$ and the path $\pi_{2}$ in $B$ from the root to the leaf for the successor $t_{2}$ of $r_{2}$. Again, if $t_{1}$ does not exist, we take $\pi_{1}$ as the left spine of $B$, and if $t_{2}$ does not exist, we take $\pi_{2}$ as the right spine of $B$. Then, $\pi_{1}$ and $\pi_{2}$ are the canonical paths for $(s,r_{1},r_{2})$. The canonical nodes for $(s,r_{1},r_{2})$ are defined as follows: for each vertex $v$ in $\pi_{1}\setminus\pi_{2}$, we take the right child of $v$ if it is not in $\pi_{1}$, and for each $v$ in $\pi_{2}\setminus\pi_{1}$, we take the left child of $v$ if it is not in $\pi_{1}$. Furthermore, we take the last node of $\pi_{1}$ if $t_{1}$ does not exist, and the last node of $\pi_{2}$ if $t_{2}$ does not exist. A standard argument bounds the number and total size of the canonical intervals.

Lemma 5.1**.**

The total size of the canonical intervals is $O(n\log n)$. The tree $B$ and the canonical intervals can be built in $O(n\log n)$ time. For any query $q$ in (R1) or (R2), there are $O(\log n)$ canonical nodes, and they can be found in $O(\log n)$ time. The canonical intervals for the canonical nodes of $q$ constitute a partition of the query interval for $q$.

Proof 5.2**.**

Since a site $s\in S$ appears in $O(\log n)$ canonical intervals, the total size of the canonical intervals is $O(n\log n)$. To construct $B$, we sort $S$ according to the radii $r_{s}$, and we build $B$ on top of the sorted list. To find the (sorted) canonical intervals, we perform a postorder traversal of $B$, copying and merging the child intervals for each parent node.

The bound on the canonical nodes for $q$ follows, since $B$ has height $O(\log n)$. To find them, we trace the canonical paths for $q$ in $B$. The partition property holds by construction.

5.1 Queries of Type (R1)

We build a compressed quadtree on $S$, and we perform the range searches the compressed quadtree. It is possible to compute a compressed quadtree for each canonical interval without logarithmic overhead. Since Lemma 4.3 gives us plenty of freedom in choosing the squares for our range queries, we take squares from the grid that underlies the quadtree. This allows us to reduce the range searching problem to predecessor search in a linear list, a task that can be accomplished by one top-down traversal of $B$. Details follow.

Hierarchical grids, Z-order, compressed

quadtrees.

We translate and scale $S$ (and the associated radii), so that $S$ lies in the interior of the unit square $U=[0,1]^{2}$ and so that all radii are at most $\sqrt{2}$. We define a sequence of hierarchical grids that subdivide $U$. The grid $G_{0}$ consists of the single cell $U$. The grid $G_{i}$, $i\geq 1$, consists of the $2^{2i}$ square cells with side length $2^{-i}$ and pairwise disjoint interiors that cover $U$. The hierarchical grids induce an infinite four-regular tree ${\mathcal{T}}$: the vertices are the cells of ${\mathcal{G}}=\bigcup_{i=0}^{\infty}G_{i}$. The unit square $U$ is the root, and for $i=1,\dots$, a cell $\sigma$ in $G_{i}$ is the child of the cell in $G_{i-1}$ that contains it. We make no explicit distinction between a vertex of ${\mathcal{T}}$ and its corresponding cell.

The $Z$-order $\leq_{Z}$ is a total order on the cells of ${\mathcal{G}}$; see [4] for more details. Let $\sigma,\tau\in{\mathcal{G}}$. If $\sigma\subseteq\tau$, then $\sigma\leq_{Z}\tau$: and if $\tau\subseteq\sigma$, then $\tau\leq_{Z}\sigma$, If $\sigma$ and $\tau$ are unrelated in ${\mathcal{T}}$, let $\rho$ be the lowest common ancestor of $\sigma$ and $\tau$ in ${\mathcal{T}}$, and let $\sigma^{\prime}$ and $\tau^{\prime}$ be the children of $\rho$ with $\sigma\subseteq\sigma^{\prime}$ and $\tau\subseteq\tau^{\prime}$. We set $\sigma\leq_{Z}\tau$ if $\sigma^{\prime}$ is before $\tau^{\prime}$ in the order shown in Figure 7; and $\tau\leq_{Z}\sigma$, otherwise. The next lemma shows that given $\sigma,\tau\in{\mathcal{G}}$, we can decide if $\sigma\leq_{Z}\tau$ in constant time.

Lemma 5.3** (Chapter 2 in Har-Peled [13]).**

Suppose the floor function and the first differing bit in the binary representations of two given real numbers can be computed in $O(1)$ time. Then, we can decide in $O(1)$ time for two given cells $\sigma,\tau\in{\mathcal{G}}$ whether $\sigma\leq_{Z}\tau$ or $\tau\leq_{Z}\sigma$.

For a site $s\in S$, let $\sigma_{s}$ be the largest cell in ${\mathcal{G}}$ that contains only $s$. The quadtree for $S$ is the smallest connected subtree of ${\mathcal{T}}$ that contains the root $U$ and all cells $\sigma_{s}$, for $s\in S$. The compressed quadtree ${\mathcal{C}}$ for $S$ is obtained from the quadtree by contracting any maximal path of vertices with only one child into a single edge. Vertices that were at the top of such a path are now called compressed vertices. The compressed quadtree for $S$ has $O(n)$ vertices, and it can be constructed in $O(n\log n)$ time (see, e.g., [3, Appendix A] and [13]).

The linearized compressed quadtree ${\mathcal{L}}$ for $S$ is the sorted sequence of cells obtained by listing the nodes of ${\mathcal{C}}$ according to a postorder traversal, were the children of a node $\sigma\in{\mathcal{C}}$ are visited according to the $Z$-order from Figure 7. The cells in ${\mathcal{L}}$ appear in increasing $Z$-order, and range searching for a given cell $\sigma\in{\mathcal{G}}$ reduces to a simple predecessor search in ${\mathcal{L}}$, as is made explicit in the following lemma.

Lemma 5.4**.**

Let $\sigma$ be a cell of ${\mathcal{G}}$, and let ${\mathcal{L}}$ be the linearized compressed quadtree on $S$. Let $\tau=\max_{Z}\{\rho\in{\mathcal{L}}\mid\rho\leq_{Z}\sigma\}$ be the $Z$-predecessor of $\sigma$ in ${\mathcal{L}}$ ($\tau=\emptyset$, if the predecessor does not exist). Then, if $\sigma\cap\tau=\emptyset$, then also $\sigma\cap S=\emptyset$, and if $\sigma\cap\tau\neq\emptyset$, then $\sigma\cap S=\tau\cap S$.

Proof 5.5**.**

Let ${\mathcal{C}}$ be the compressed quadtree on $S$, and let ${\mathcal{C}}_{\sigma}=\{\tau\in{\mathcal{C}}\mid\tau\subseteq\sigma\}$ be the cells in ${\mathcal{C}}$ that are contained in $\sigma$. If ${\mathcal{C}}_{\sigma}$ is non-empty, then ${\mathcal{C}}_{\sigma}$ is a connected subtree of ${\mathcal{C}}$. Let $\tau$ be the root of this subtree. Then, $\tau=\max_{Z}\{\rho\in{\mathcal{C}}_{\sigma}\}$, and $\tau\leq_{Z}\sigma$. Furthermore, all other cells in ${\mathcal{C}}\setminus{\mathcal{C}}_{\sigma}$ are either smaller than all cells in ${\mathcal{C}}_{\sigma}$ or larger than $\sigma$. Thus, $\tau$ is the $Z$-predecessor of $\sigma$ in ${\mathcal{L}}$, and $\sigma\cap S=\tau\cap S\neq\emptyset$. Otherwise, if ${\mathcal{C}}_{\sigma}=\emptyset$, the $Z$-predecessor of $\sigma$ in ${\mathcal{L}}$ either does not exist or is disjoint from $\sigma$. Thus, in this case, we have $\emptyset=\sigma\cap\tau=\sigma\cap S$.

The search algorithm.

For a site $s\in S$, we define the neighborhood $N(s)$ of $s$ as all cells in ${\mathcal{G}}$ with side length $2^{\lfloor\log_{2}{r_{s}}\rfloor}$ that intersect $D_{s}$. The neighborhood will be used to approximate $D_{s}$ for the range search in the quadtrees.

Lemma 5.6**.**

There is a constant $\beta$ such that $|N(s)|\leq\beta$ for all $s\in S$.

Proof 5.7**.**

We have $r_{s}/2<2^{\lfloor\log_{2}{r_{s}}\rfloor}$, and a $5\times 5$ grid with cells of side length $r_{s}/2$ covers $D_{s}$, no matter where $s$ lies; see Figure 8. Thus, the lemma holds with $\beta=25$.

We now show that a linearized compressed quadtree for each canonical interval can be found without logarithmic overhead.

Lemma 5.8**.**

We can compute for each $v\in B$ the linearized quadtree ${\mathcal{L}}_{v}$ for the sites in ${\mathcal{I}}_{v}$ in $O(n\log n)$ time.

Proof 5.9**.**

For each $v\in B$, we build the compressed quadtree ${\mathcal{C}}_{v}$ for ${\mathcal{I}}_{v}$, as follows: at the root, we compute the compressed quadtree ${\mathcal{C}}$ for $S$ in $O(n\log n)$ time [3, 13]. Then, we traverse $B$. Given the compressed quadtree ${\mathcal{C}}_{v}$ for a node $v\in B$, we compute ${\mathcal{C}}_{w}$ for a child $w$ of $v$ as follows. We do a postorder traversal of ${\mathcal{C}}_{v}$. In each leaf $\nu$ of ${\mathcal{C}}_{v}$, we check if the site $s$ in $\nu$ is in ${\mathcal{I}}_{w}$, by testing $r_{s}$. If not, we remove $\nu$; otherwise, we keep it. In each inner vertex $\nu$ of ${\mathcal{C}}_{v}$, we check if $\nu$ has any remaining children. If not, we remove $\nu$. If $\nu$ has exactly one remaining child that is not a compressed vertex, we mark $\nu$ as compressed and continue. If the only remaining child of $\nu$ is compressed, we remove this child, connect $\nu$ to its grandchild, and mark $\nu$ as compressed. This takes $O(|{\mathcal{C}}_{v}|)$ time and gives ${\mathcal{C}}_{w}$.

Once all the compressed quadtrees ${\mathcal{C}}_{v}$ are available, we traverse $B$ again to find the linearized compressed quadtrees ${\mathcal{L}}_{v}$ by a traversal of each ${\mathcal{C}}_{v}$. The total time to find the ${\mathcal{L}}_{v}$ is $O(n\log n+\sum_{v\in B}|{\mathcal{C}}_{v}|)=O(n\log n)$, since $|C_{v}|=O(|{\mathcal{I}}_{v}|)$, for all $v\in B$, and $\sum_{v\in B}|{\mathcal{I}}_{v}|=O(n\log n)$ by Lemma 5.1.

Using the linearized compressed quadtrees, the range searching problem can be solved by a batched predecessor search, using a single traversal of $B$.

Lemma 5.10**.**

The range searching problem (R1) can be solved in $O(n\log n)$ time.

Proof 5.11**.**

We apply Lemma 5.8 to find the linearized quadtree for every canonical interval in $B$. Remember that the queries in (R1) are the complete set $S$. We split each query $s\in S$ into subqueries, by considering the neighborhood $N(s)$ of $s$. Let {\mathcal{Q}}^{\prime}=\bigcup_{s\in S}\big{\{}(\sigma,s)\mid\sigma\in N(s)\big{\}} be the set of split queries. The purpose of the split queries is to approximate the associated disks for the query sites by cells from the hierarchical grid. By Lemma 5.6, $|{\mathcal{Q}}^{\prime}|=O(n)$.

We now perform range queries for the cells in the split queries. For this, we first sort the elements of ${\mathcal{Q}}^{\prime}$ in the $Z$-order of their first components, in $O(n\log n)$ time. Next, we distribute the split queries along their canonical paths in $B$. For each $v\in B$, let ${\mathcal{Q}}^{\prime}_{v}$ be the sorted sublist of queries in ${\mathcal{Q}}^{\prime}$ (in the $Z$-order of the first component) that have $v$ on their canonical path. By Lemmas 5.1 and 5.6, we have $\sum_{v\in B}|{\mathcal{Q}}^{\prime}_{v}|=O(n\log n)$. To find the lists ${\mathcal{Q}}^{\prime}_{v}$ for all $v\in B$ in $O(n\log n)$ time, we perform a pre-order traversal of $B$, computing the lists for the children from the lists of the parents. More precisely, given the sorted list ${\mathcal{Q}}^{\prime}_{v}$ for a node $v\in B$, we can find the sorted list ${\mathcal{Q}}^{\prime}_{w}$ for a child $w$ of $v$ in time $O(|{\mathcal{Q}}^{\prime}_{v}|)$ by scanning ${\mathcal{Q}}^{\prime}_{v}$ from left to right and by copying the elements that also appear in ${\mathcal{Q}}^{\prime}_{w}$. Finally, we distribute the split queries into their canonical nodes. The canonical nodes of a query are children of the nodes on its canonical path. Thus, we can find for each $v\in B$ the sorted list ${\mathcal{Q}}^{\prime\prime}_{v}$ of split queries with $v$ as a canonical node as follows: we iterate over all non-root nodes $v\in B$, and we scan the list ${\mathcal{Q}}^{\prime}_{w}$ of the parent node $w$ of $v$. We copy all queries that have $v$ as a canonical node from ${\mathcal{Q}}^{\prime}_{w}$ into ${\mathcal{Q}}^{\prime\prime}_{v}$. This takes $O(n\log n)$ time.

Next, we iterate over all $v\in B$, and we merge the lists ${\mathcal{Q}}^{\prime\prime}_{v}$ with the lists ${\mathcal{L}}_{v}$, in $Z$-order. This takes O\big{(}\sum_{v\in B}|{\mathcal{L}}_{v}|+|{\mathcal{Q}}^{\prime\prime}_{v}|\big{)}=O(n\log n) time. By Lemma 5.4, we obtain for each $(\sigma,s)\in{\mathcal{Q}}^{\prime\prime}_{v}$ a cell $\tau_{v}^{\sigma,s}$. If $\sigma\cap\tau_{v}^{\sigma,s}\neq\emptyset$ we know that $\sigma\cap{\mathcal{I}}_{v}=\tau_{v}^{\sigma,s}\cap{\mathcal{I}}_{v}$. Since these sites are all from ${\mathcal{I}}_{v}$ they all have radius at least $r_{s}/2$. We can find all these sites in $O(k)$ time, where $k$ is the output size. If $k>\alpha$, we stop and report $\sigma$ as a square with many sites of large radius.555Note that here the radii are $\geq r_{s}/2$ inside of the cells $\sigma$. This might be larger than the value $2^{\lceil\log_{2}r_{s}\rceil}/4$ needed by (R1). But still, if there are more than $\alpha$ sites in $\sigma$, we still have a triangle in a square. Otherwise we will later determine that each disk contains few sites of radius at least $r_{s}/2$.

Otherwise, we use the sites in $\sigma$ to accumulate the sites for the query disk $D_{s}$. This we can do by considering all canonical nodes of $s$ and for each cell $\sigma$ iterate over the sites contained in $\sigma$. In each such cell there are at most $\alpha$ sites. For each site $t\in\sigma$ we can check if $t\in D_{s}$. If we find a query disk $D_{s}$ with more than $\alpha$ sites of large radius, we stop and report its enclosing square with many sites of large radius.666$r=2r_{s}$ is the side length of the enclosing square, the radii are at least $r/4$ as desired. Otherwise, for each $s\in S$, we have found the at most $\alpha$ sites of radius at least $r_{s}/2$ in $D_{s}$. The whole algorithm takes $O(n\log n)$ time.

5.2 Queries of Type (R2)

We use the tree structure of the canonical intervals (i) to construct quickly the search structures for each canonical interval; and (ii) to solve all queries for a canonical interval in one batch. We exploit the connection between planar disks and three-dimensional polytopes. Let U=\big{\{}(x,y,z)\mid x^{2}+y^{2}=z\big{\}} be the three-dimensional unit paraboloid. For a site $s\in S$, the lifted site $\hat{s}$ is the vertical projection of $s$ onto $U$. Each disk $D_{s}$ is transformed into an upper halfspace $\widehat{D}_{s}$, so that the projection of $\widehat{D}_{s}\cap U$ onto the $xy$-plane is the set $\mathbb{R}^{2}\setminus D_{s}$;777This halfspace is bounded by the plane $z=2x_{s}x-x_{s}^{2}+2y_{s}y-y_{s}^{2}+r_{s}^{2}$, where $s=(x_{s},y_{s})$. see Figure 9. The union of a set of disks in $\mathbb{R}^{2}$ corresponds to the intersection of the lifted upper halfspaces in $\mathbb{R}^{3}$.

Lemma 5.12**.**

The range searching problem (R2) can be solved in $O(n\log n)$ expected time.

Proof 5.13**.**

For each $v\in B$, we construct a three-dimensional representation of the union of the disks in the canonical interval ${\mathcal{I}}_{v}$. As explained above, this is the intersection ${\mathcal{E}}_{v}$ of the lifted three-dimensional halfspaces $\widehat{D}_{s}$, for $s\in{\mathcal{I}}_{v}$. The intersection of two three-dimensional convex polyhedra with a total of $m$ vertices can be computed in $O(m)$ time [8, 6]. Therefore, we can construct all the polyhedra ${\mathcal{E}}_{v}$, for $v\in B$, in overall $O(n\log n)$ time, by a bottom-up traversal of $B$ (by Lemma 5.1, the total number of vertices of these polyhedra is $O(n\log n)$).

For the query processing, we compute a polytope $\widehat{Q}_{v}$ for each $v\in B$. The polytope $\widehat{Q}_{v}$ is obtained by determining all the points $p$ that appear in a query $(p,r_{1},r_{2})$ that has $v$ as a canonical node, lifting those points $p$ to their three-dimensional representations $\hat{p}$, and taking the convex hull of the resulting three-dimensional point set. The lifted query points all lie on the unit paraboloid $U$, so every lifted query point appears as a vertex on $\widehat{Q}_{v}$. To find all polytopes $\widehat{Q}_{v}$, for $v\in B$, efficiently, we proceed as follows: let $A$ be the three-dimensional point set obtained by taking all points that appear in a query and by lifting them onto the unit paraboloid. We compute the convex hull of $A$ in $O(n\log n)$ time. Then, for each $v\in B$, we find the convex hull of all lifted queries that have $v$ in their canonical path. This can be done in $O(n\log n)$ total expected time by a top-down traversal of $B$. We already have the polytope for the root of $B$. To compute the polytope for a child node, given that the polytope for the parent node is available, we use the fact that for any polytope ${\mathcal{E}}$ in $\mathbb{R}^{3}$ with $m$ vertices, we can compute the convex hull of any subset of the vertices of ${\mathcal{E}}$ in $O(m)$ expected time [9]. Once we have for each $v\in B$ the convex hull of the lifted query points that have $v$ on their canonical path, we can compute for each $v\in B$ the polytope $\widehat{Q}_{v}$ that is the convex hull of the lifted query points that have $v$ as a canonical node. For this, we consider the canonical path polytope stored at the parent node of $v$, and we again use the algorithm from [9] to extract the convex hull for the lifted query points that have $v$ as a canonical node.

Now that the polyhedra $\widehat{Q}_{v}$ and the polytopes ${\mathcal{E}}_{v}$ are available, for all $v\in B$, we can answer the query as follows: for each node $v\in B$, we must check for vertices of $\widehat{Q}_{v}$ that do not lie inside of ${\mathcal{E}}_{v}$. These are exactly the vertices of $\widehat{Q}_{v}$ that are not vertices of $\widehat{Q}_{v}\cap{\mathcal{E}}_{v}$. As mentioned, the intersections $\widehat{Q}_{v}\cap{\mathcal{E}}_{v}$ can be found in linear time for each node $v\in B$, for a total time $O(n\log n)$, and once the intersection is available, we can easily find all vertices $\hat{p}$ of $\widehat{Q}_{v}$ that are not vertices of $\widehat{Q}_{v}\cap{\mathcal{E}}_{v}$ (e.g., using radix sort). If for any such intersection $\widehat{Q}_{v}\cap{\mathcal{E}}_{v}$, there is a lifted site $\hat{s}\in\widehat{Q}_{v}$ that is not a vertex of $\widehat{Q}_{v}\cap{\mathcal{E}}_{v}$, we report $s$ as the result of the range search. Otherwise, we report that the range search is unsuccessful.

6 Finding the Shortest Triangle in a Transmission Graph

We extend Theorem 4.7 to find the shortest triangle in $T(S)$. As in Section 2.2, we solve the decision problem: given $W>0$, does $T(S)$ have a directed triangle of perimeter at most $W$? We set $\ell=W/\sqrt{27}$, and call a site $s\in S$ large if $r_{s}>\ell$. We let $S_{\ell}\subseteq S$ be the set of all large sites.

Lemma 6.1**.**

We can find a triangle in $T(S\setminus S_{\ell})$ of perimeter at most $W$ in $O(n\log n)$ time, if it exists.

Proof 6.2**.**

Any triangle in $T(S\setminus S_{\ell})$ has perimeter at most $W$: consider a directed triangle $stu$ in $T(S\setminus S_{\ell})$ with $r_{s}\geq\max\{r_{t},r_{u}\}$. Then we have $t,u\in D_{s}$, so the triangle $stu$ lies in $D_{s}$. Elementary calculus shows that a triangle of maximum perimeter in $D_{s}$ must be equilateral with its vertices on $\partial D_{s}$, so any triangle contained in $D_{s}$ has perimeter at most $3\cdot\sqrt{3}\cdot r_{s}\leq\sqrt{27}\cdot\ell=W$. We can find a triangle in $T^{\prime}$ in $O(n\log n)$ time by Theorem 4.7.

It remains to check for triangles of perimeter at most $W$ with at least one large vertex. Some such triangles have to be considered individually, while the others can be handled efficiently in batch mode. The following lemma shows that we may assume that there are few edges from $S\setminus S_{\ell}$ to $S_{\ell}$.

Lemma 6.3**.**

If $T(S)$ does not have a triangle of perimeter at most $W$, every site in $S_{\ell}$ has at most six incoming edges from $S\setminus S_{\ell}$. Furthermore, in $O(n\log n)$ time, we can either find a triangle of perimeter at most $W$ in $T(S)$ or determine for each site in $S_{\ell}$ all incoming edges from $S\setminus S_{\ell}$.

Proof 6.4**.**

Suppose there is a square $\sigma$ in the plane with side length $0<r<2\ell$ such that $\sigma$ contains more than $\alpha$ sites $s$ of radius $r_{s}\geq r/4$. Then, Lemma 4.3 shows that $T(S)$ contains a triangle that lies in $\sigma$. Specifically, since $\sigma$ has side length at most $r/6=2\ell/6$, the definition of $\ell$ implies that the triangle has perimeter at most $W$. If there is no such square, it follows that there is no site $s$ with $r_{s}\leq\ell$ such that $D_{s}$ contains more than $\alpha$ sites of radius at least $r_{s}/2$, as otherwise $s$ could be enclosed by a square of side length $2r_{s}\leq 2\ell$ that contains many sites of large radius. Thus, every small site $s$ has $O(1)$ outgoing edges to sites with radius at least $r_{s}/2$. In particular, there are $O(n)$ edges from small sites to large sites. We can use a suitable variant of (R1) so that in $O(n\log n)$ time we can EITHER find a square of side length $0<r<2\ell$ that contains more than $\alpha$ sites $s$ of radius $r_{s}\geq r/4$r OR determine that for every site $s$ with $r_{s}\leq\ell$, there are at most $\alpha$ sites in $D_{s}$ of radius at least $r_{s}/2$. Furthermore, in the second case, we explicitly get all sites of radius at least $r_{s}/2$ in each $D_{s}$ with $r_{s}\leq\ell$

Thus, if the second case applies, we obtain all edges from $S\setminus S_{\ell}$ to $S_{\ell}$. Suppose there is a large site $s$ of indegree at least $7$. Then, there must be two sites $t,u\in S\setminus S_{\ell}$ such that the angle between the edges $ts$ and $us$ is less than $\pi/3$. Thus, the distance $|tu|$ is less than the maximum of $|ts|$ and $|us|$, so there is a directed edge with endpoints $t$ and $u$ and the sites $s,t,u$ form a triangle of perimeter at most $3\ell\leq W$.

Next, we want to limit the number of relevant edges between large sites. For this, we subdivide the plane with a grid $G$ of side length $\ell/\sqrt{2}$. Then, we have the following:

Lemma 6.5**.**

A triangle contained in a cell $\sigma\in G$ has perimeter at most $W$. If there is no triangle in $\sigma$, then $\sigma$ contains $O(1)$ large sites. We can check for such triangles in $O(n\log n)$ overall expected time.

Proof 6.6**.**

The maximum perimeter of a triangle contained in $\sigma$ is $(1+\sqrt{2})\ell<W$. Furthermore, if there are at least three large sites in $\sigma$, these large sites form a triangle, since the disk of a large site covers $\sigma$. By applying Theorem 4.7 to the induced subgraph in each cell of $G$, we can find such a triangle in $O(n\log n)$ total expected time.

We define the neighborhood $N(\sigma)$ of a cell $\sigma\in G$ as the $5\times 5$ block of cells centered at $\sigma$. Let $t$ be a site and $\sigma$ the cell containing $t$, then the neighborhood $N(t)$ of $t$ are all sites contained in $N(\sigma)$. Since the side length of a grid cell is $W/3\sqrt{6}$, each triangle of perimeter at most $W$ is completely contained in the neighborhood of some cell.

Lemma 6.7**.**

We can check the remaining triangles in $O(n)$ overall time.

Proof 6.8**.**

Consider a remaining triangle $sut$ with $r_{t}\geq\max\{r_{u},t_{s}\}$. Then, $t\in S_{\ell}$, and $s,t,u$ all lie in $N(t)$. By Lemma 6.5, there are $O(1)$ large candidates for $u$ in $N(t)$, and by Lemma 6.3, there are $O(1)$ small candidates for $u$. Having fixed a $t$ and a possible candidate $u$, we iterate over all $s\in N(t)$ and check if $s$, $u$, and $t$ form a triangle with weight at most $W$. Every site $s$ is contained in $O(1)$ grid neighborhoods, and since there are $O(1)$ candidate pairs in each grid neighborhood, $s$ participates in $O(1)$ explicit checks. The result follows.

The following theorem summarizes the considerations in this section.

Theorem 6.9**.**

It takes $O(n\log n)$ expected time to find the shortest triangle in a transmission graph.

Proof 6.10**.**

We already saw that there is an $O(n\log n)$ time decision algorithm for the problem. As in Theorem 2.11, the result follows from an application of Chan’s randomized optimization technique [5] (restated in Lemma 2.10).

7 Conclusion

Once again, disk graphs and transmission graphs prove to be a simple yet powerful graph model where difficult algorithmic problems admit faster solutions. It would be interesting to find a deterministic $O(n\log n)$ time algorithm for finding a shortest triangle in a disk graph. Currently, we are working on extending our results to the girth problem in transmission graphs; can we find an equally simple and efficient algorithm as for disk graphs?