Routing in Histograms

TL;DR

This paper develops efficient routing schemes within specific geometric graph structures called histograms, achieving near-optimal path lengths with minimal label and table sizes.

Contribution

It introduces new routing algorithms for simple and double histograms, providing optimal or near-optimal path routing with small labels and tables.

Findings

Routing in double histograms achieves paths at most twice the shortest path length.

Routing in simple histograms achieves optimal path lengths.

Labels and routing tables are of size O(log n) bits.

Abstract

Let $P$ be an $x$-monotone orthogonal polygon with $n$ vertices. We call $P$ a simple histogram if its upper boundary is a single edge; and a double histogram if it has a horizontal chord from the left boundary to the right boundary. Two points $p$ and $q$ in $P$ are co-visible if and only if the (axis-parallel) rectangle spanned by $p$ and $q$ completely lies in $P$. In the $r$-visibility graph $G(P)$ of $P$, we connect two vertices of $P$ with an edge if and only if they are co-visible. We consider routing with preprocessing in $G(P)$. We may preprocess $P$ to obtain a label and a routing table for each vertex of $P$. Then, we must be able to route a packet between any two vertices $s$ and $t$ of $P$, where each step may use only the label of the target node $t$, the routing table and neighborhood of the current node, and the packet header. We present a routing scheme for double histograms that sends any data packet along a path whose length is at most twice the (unweighted) shortest path distance between the endpoints. In our scheme, the labels, routing tables, and headers need $O(\log n)$ bits. For the case of simple histograms, we obtain a routing scheme with optimal routing paths, $O(\log n)$-bit labels, one-bit routing tables, and no headers.

Figures

Equations

$$\text{Lab}(n)$$

$$\text{Tab}(n)$$

$$H(n)$$

$$\zeta (n)$$

$$I\big{(}\!\operatorname{bd}^{2}(s)\big{)}=I\big{(}\!\operatorname{bd}(\operatorname{td}(s))\big{)}\cup I\big{(}\!\operatorname{bd}(\operatorname{bd}(s))\big{)}.$$

$$I({\mathrm{td}}^2(s))=I(\mathrm{td}(\mathrm{td}(s)))\cup I(\mathrm{td}(\mathrm{bd}(s))).$$

$$I^{3}(s)=I\big{(}\!\operatorname{bd}^{2}(s)\big{)}\cup I\big{(}\!\operatorname{td}^{2}(s)\big{)}\\ =I\big{(}\!\operatorname{bd}(\operatorname{td}(s))\big{)}\cup I\big{(}\!\operatorname{bd}(\operatorname{bd}(s))\big{)}\cup I\big{(}\!\operatorname{td}(\operatorname{td}(s))\big{)}\cup I\big{(}\!\operatorname{td}(\operatorname{bd}(s))\big{)}=I^{2}\big{(}\!\operatorname{bd}(s)\big{)}\cup I^{2}\big{(}\!\operatorname{td}(s)\big{)},$$

Full text

We consider routing with preprocessing in $G(P)$. We may preprocess $P$ to obtain a label and a routing table for each vertex of $P$. Then, we must be able to route a packet between any two vertices $s$ and $t$ of $P$, where each step may use only the label of the target node $t$, the routing table and neighborhood of the current node, and the packet header.

We present a routing scheme for double histograms that sends any data packet along a path whose length is at most twice the (unweighted) shortest path distance between the endpoints. In our scheme, the labels, routing tables, and headers need $O(\log n)$ bits. For the case of simple histograms, we obtain a routing scheme with optimal routing paths, $O(\log n)$-bit labels, one-bit routing tables, and no headers.

1 Introduction

The routing problem is a classic question in distributed graph algorithms [16, 23]. We have a graph $G$ and would like to preprocess it for the following task: given a data packet located at some source vertex $s$ of $G$, route the packet to a target vertex $t$ of $G$, identified by its label. The routing should have the following properties: (A) locality: to determine the next step of the packet, it should use only information at the current vertex or in the packet header; (B) efficiency: the packet should travel along a path whose length is not much larger than the length of a shortest path between $s$ and $t$. The ratio between the length of this routing path and a shortest path is called the stretch factor; and (C) compactness: the space requirements for labels, routing tables, and packet headers should be small.

Obviously, we could store at each vertex $v$ of $G$ the complete shortest path tree of $v$. Then, the routing scheme is perfectly efficient: we can send the packet along a shortest path. However, the scheme lacks compactness. Thus, the challenge is to balance the (seemingly) conflicting goals of compactness and efficiency.

There are many compact routing schemes for general graphs [1, 2, 12, 13, 14, 24, 25]. For example, the scheme by Roditty and Tov [25] needs to store a poly-logarithmic number of bits in the packet header and it routes a packet from $s$ to $t$ on a path of length O\big{(}k\Delta+m^{1/k}\big{)}, where $\Delta$ is the shortest path distance between $s$ and $t$, $k>2$ is any fixed integer, $n$ is the number of nodes, and $m$ is the number of edges. The routing tables use $mn^{O(1/\sqrt{\log n})}$ space. In the late 1980’s, Peleg and Upfal [23] proved that in general graphs, any routing scheme with constant stretch factor must store $\Omega(n^{c})$ bits per vertex, for some constant $c>0$. Thus, it is natural to focus on special graph classes to obtain better routing schemes. For instance, trees admit routing schemes that always follow the shortest path and that store $O(\log n)$ bits at each node [15, 26, 28]. Moreover, in planar graphs, for any fixed $\varepsilon>0$, there is a routing scheme with a poly-logarithmic number of bits in each routing table that always finds a path that is within a factor of $1+\varepsilon$ from optimal [27]. Similar results are also available for unit disk graphs [19, 30] and for metric spaces with bounded doubling dimension [20].

Another approach is geometric routing: the graph resides in a geometric space, and the routing algorithm has to determine the next vertex for the packet based on the coordinates of the source and the target vertex, the current vertex, and its neighborhood, see for instance [10, 9] and the references therein. In contrast to compact routing schemes, there are no routing tables, and the routing happens purely based on the local geometric information (and possibly the packet header). For example, the routing algorithm for triangulations by Bose and Morin [11] uses the line segment between the source and the target for its routing decisions. In a recent result, Bose et al. [10] show that when vertices do not store any routing tables, no geometric routing scheme can achieve the stretch factor $o(\sqrt{n})$. This lower bound applies irrespective of the header size.

We consider routing in a particularly interesting class of geometric graphs, namely visibility graphs of polygons. Banyassady et al. [3] presented a routing scheme for polygonal domains with $n$ vertices and $h$ holes that uses $O(\log n)$ bits for the label, $O((\varepsilon^{-1}+h)\log n)$ bits for the routing tables, and achieves a stretch of $1+\varepsilon$, for any fixed $\varepsilon>0$. However, their approach is efficient only if the edges of the visibility graph are weighted with their Euclidean lengths. Banyassady et al. ask whether there is an efficient routing scheme for visibility graphs with unit weights (also called the hop-distance), arguably a more applied setting.

We address this open problem by combining the two approaches of geometric and compact routing: we use routing tables at the vertices to represent information about the structure of the graph, but we also assume that the labels of all adjacent vertices are directly visible at each node. This is reasonable from a practical point of view, because a node in a network must be aware of all its neighbors and their labels. The size of this list is not relevant for the compactness, since it depends purely on the graph and cannot be influenced during preprocessing. We focus our attention on $r$-visibility graphs of orthogonal simple and double histograms. Even this seemingly simple case turns out to be quite challenging and reveals the whole richness of the compact routing problem in unweighted, geometrically defined graphs. Furthermore, histograms constitute a natural starting point, since they are crucial building blocks in many visibility problems; see, for instance, [4, 5, 6, 8, 7, 18]. In addition, $r$-visibility is a popular concept in orthogonal polygons that enjoys many useful structural properties, see, e.g., [22, 18, 21, 17, 29].

A simple histogram is a monotone orthogonal polygon whose upper boundary consists of a single edge; a double histogram is a monotone orthogonal polygon that has a horizontal chord that touches the boundary of $P$ only at the left and the right boundary. Let $P$ be a (simple or double) histogram with $n$ vertices. Two vertices $v$ and $w$ in $P$ are connected in the visibility graph $G(P)$ by an unweighted edge if and only if the axis-parallel rectangle spanned by $v$ and $w$ is contained in the (closed) region $P$ (we say that $v$ and $w$ are co-visible). We present the first efficient and compact routing schemes for polygonal domains under the hop-distance. The following two theorems give the precise statements.

Theorem 1**.**

Let $P$ be a simple histogram with $n$ vertices. There is a routing scheme for $G(P)$ with a routing table with 1 bit, without headers, having label size $2\cdot\lceil\log n\rceil$, such that we can route between any two vertices on a shortest path.

Theorem 2**.**

Let $P$ be a double histogram with $n$ vertices. There is a routing scheme for $G(P)$ with routing table, label and header size $O(\log n)$, such that we can route between any two vertices with stretch at most $2$.

2 Preliminaries

Routing schemes.

Let $G=(V,E)$ be an undirected, unweighted, simple, connected graph. The (closed) neighborhood of a vertex $v\in V$, $N(v)$, is the set containing $v$ and its adjacent nodes. Let $v,w\in V$. A sequence $\pi:\langle v=p_{0},p_{1},\dots,p_{k}=w\rangle$ of vertices with $p_{i-1}p_{i}\in E$, for $i=1,\dots,k$, is called a path of length $k$ between $v$ and $w$. The length of $\pi$ is denoted by $|\pi|$. We define $d(v,w)=\min_{\pi}|\pi|$ as the length of a shortest path between $v$ and $w$, where $\pi$ goes over all paths with endpoints $v$ and $w$.

Next, we define a routing scheme. The algorithm that decides the next step of the packet is modeled by a routing function. Every node is assigned a (binary) label that identifies it in the network. The routing function uses local information at the current node, the label of the target node, and the header stored in the packet. The local information of a node $v$ has two parts: (i) the link table, a list of the labels of $N(v)$ and (ii) the routing table, a bitstring chosen during preprocessing to represent relevant topological properties of $G$. Formally, a routing scheme of a graph $G$ consists of:

• •

a label $\operatorname{lab}(v)\in\{0,1\}^{+}$ for each node $v\in V$;

• •

a routing table $\rho(v)\in\{0,1\}^{*}$ for each node $v\in V$; and

• •

a routing function f\colon\big{(}\{0,1\}^{*}\big{)}^{4}\rightarrow V\times\{0,1\}^{*}.

The routing function takes the link table and routing table of a current node $s\in V$, the label $\operatorname{lab}(t)$ of a target node $t$, and a header $h\in\{0,1\}^{*}$. Using these four inputs, it provides a next node $v$ adjacent to $s$ and a new header $h^{\prime}$. The local information in the packet is updated to $h^{\prime}$, and it is forwarded to $v$. The routing scheme is correct if the following holds: for any two sites $s,t\in V$, consider the sequence $(p_{0},h_{0})=(s,\varepsilon)$ and (p_{i+1},h_{i+1})=f\Big{(}\operatorname{lab}\big{(}N(p_{i})\big{)},\rho(p_{i}),\operatorname{lab}(t),h_{i}\Big{)}, for $i\geq 0$. Then, there is a $k=k(s,t)\geq 0$ with $p_{k}=t$ and $p_{i}\neq t$, for $i=0,\dots,k-1$. The routing scheme reaches $t$ in $k$ steps. Furthermore, $\pi:\langle p_{0},\dots,p_{k}\rangle$ is the routing path from $s$ to $t$. The routing distance is $d_{\rho}(s,t)=|\pi|$.

Now, let $\mathcal{R}$ be a family of correct routing schemes for a given graph class $\mathcal{G}$, i.e., $\mathcal{R}$ contains a correct routing scheme for every graph in $\mathcal{G}$ such that all routing functions use the same algorithm. There are several measures for the quality of $\mathcal{R}$. First, the various pieces of information used for the routing should be small. This is measured by the maximum label size $\operatorname{Lab}(n)$, the maximum routing table size $\operatorname{Tab}(n)$, and the maximum header size $H(n)$, over all graphs in $\mathcal{G}$ of a certain size. They are defined as

[TABLE]

Polygons.

Let $P$ be a simple orthogonal (axis-aligned) polygon in general position with vertex set $V(P)$, $|V(P)|=n$. No three vertices in $V(P)$ are on the same vertical or horizontal line. The vertices are indexed counterclockwise from [math] to $n-1$; the lexicographically largest vertex has index $n-1$. For $v\in V(P)$, we write $v_{x}$ for the $x$-coordinate, $v_{y}$ for the $y$-coordinate, and $v_{\operatorname{id}}$ for the index.

We consider $r$-visibility: two points $p,q\in P$ see each other (are co-visible) if and only if the axis-aligned rectangle spanned by $p$ and $q$ lies inside $P$ (we treat $P$ as a closed set). The visibility graph G(P)=\big{(}V(P),E(P)\big{)} of $P$ has an edge between two vertices $v,w\in V(P)$ if and only if $v$ and $w$ are co-visible. The distance $d(v,w)$ between two vertices $v,w\in V(P)$ is called the hop distance of $v$ and $w$ in $P$.

A histogram is an $x$-monotone orthogonal polygon where the upper boundary consists of exactly one horizontal edge, the base edge. Due to our numbering convention, the endpoints of the base edge are indexed [math] (left) and $n-1$ (right). They are called the base vertices. A double histogram is an $x$-monotone orthogonal polygon $P$ that has a base line, a horizontal line segment whose relative interior lies in the interior of $P$ and whose left and right endpoint are on the left and right boundary edge of $P$, respectively. We assume that the base line lies on the $x$-axis. Two vertices $v$, $w$ in $P$ lie on the same side if both are below or above the base line, i.e., if $v_{y}w_{y}>0$. Every histogram is also a double histogram. From now on, we let $P$ denote a (double) histogram.

Next, we classify the vertices of $P$. A vertex $v$ in $P$ is incident to exactly one horizontal edge $h$. We call $v$ a left vertex if it is the left endpoint of $h$; otherwise, $v$ is a right vertex. Furthermore, $v$ is convex if the interior angle at $v$ is $\pi/2$; otherwise, $v$ is reflex. Accordingly, every vertex of $P$ is either $\ell$-convex, r-convex, $\ell$-reflex, or r-reflex.

Visibility Landmarks.

To understand the structure of shortest paths in $P$, we associate with each $v\in V(P)$ three landmark points in $P$ (not necessarily vertices); Figure 1 gives an illustration.

The corresponding vertex of $v$, $\operatorname{cv}(v)$, is the unique vertex that shares a horizontal edge with $v$. To obtain the left point $\ell(v)$ of $v$, we shoot a leftward horizontal ray $r$ from $v$. Let $e$ be the vertical edge where $r$ first hits the boundary of $P$. If $e$ is the left boundary of $P$; then if $P$ is a simple histogram, we let $\ell(v)$ be the left base vertex; and otherwise $\ell(v)$ is the point where $r$ hits $e$. If $e$ is not the left boundary of $P$, we let $\ell(v)$ be the endpoint of $e$ closer to the base line. The right point $r(v)$ of $v$ is defined analogously, by shooting the horizontal ray to the right.

Let $p$ and $q$ be two points in $P$. We say that $p$ is to the left of $q$, if $p_{x}\leq q_{x}$. The point $p$ is strictly to the left of $q$, if $p_{x}<q_{x}$. The terms to the right of as well as strictly to the right of are defined analogously. The interval $[p,q]$ of $p$ and $q$ is the set of vertices in $P$ between $p$ and $q$, [p,q]=\big{\{}v\in V(P)\mid p_{x}\leq v_{x}\leq q_{x}\big{\}}. By general position, this corresponds to index intervals in simple histograms. More precisely, if $P$ is a simple histogram and $p$ is either an $r$-reflex vertex or the left base vertex and $q$ is either $\ell$-reflex or the right base vertex, then [p,q]=\big{\{}v\in V(P)\mid p_{\operatorname{id}}\leq v_{\operatorname{id}}\leq q_{\operatorname{id}}\big{\}}. The interval of a vertex $v$, $I(v)$, is the interval of the left and right point of $v$, $I(v)=[\ell(v),r(v)]$. Every vertex visible from $v$ is in $I(v)$, i.e., $N(v)\subseteq I(v)$. This interval plays a crucial role in our routing scheme and gives a very powerful characterization of visibility in double histograms.

Let $s$ and $t$ be two vertices with $t\in I(s)\setminus N(s)$. We define two more landmarks for $s$ and $t$. Assume that $t$ lies strictly to the right of $s$, the other case is symmetric. The near dominator $\operatorname{nd}(s,t)$ of $t$ with respect to $s$ is the rightmost vertex in $N(s)$ to the left of $t$. If there is more than one such vertex, $\operatorname{nd}(s,t)$ is the vertex closest to the base line. Since $t$ is not visible from $s$, the near dominator always exists. The far dominator $\operatorname{fd}(s,t)$ of $t$ with respect to $s$ is the leftmost vertex in $N(s)$ to the right of $t$. If there is more than one such vertex, $\operatorname{fd}(s,t)$ is the vertex closest to the base line. If there is no such vertex, we set $\operatorname{fd}(s,t)=r(s)$, the projection of $s$ on the right boundary. The interval I(s,t)=\big{[}\!\operatorname{nd}(s,t),\operatorname{fd}(s,t)\big{]} has all vertices between the near and far dominator; see Figure 2.

3 Simple Histograms

Let $P$ be a simple histogram with $n$ vertices. First, we give several intuitive characterizations of the visibility in $P$. Then, we analyze how the shortest paths between vertices behave. The idea for our routing scheme is as follows: as long as a target vertex $t$ is not contained in the interval $I(s)$ of a current vertex $s$, i.e., as long as there is a higher vertex that blocks visibility between $s$ and $t$, we have to leave the current pocket as far as possible. Once we have reached a high enough spike, we have to find the pocket containing $t$. Finding the right pocket is possible, but much harder than just going up. Details follow.

3.1 Visibility in Simple Histograms

We begin with some observations on the visibility in $P$. As $P$ is a simple histogram, we have that for all vertices $v\in V(P)$, the points $\ell(v)$ and $r(v)$ are vertices of $P$. Therefore, the far dominators also have to be vertices. The following observations are now immediate.

Observation 3.1**.**

Let $v\in V(P)$ be $r$-reflex or the left base vertex, and let $u\in[v,r(v)]$ be a vertex distinct from $v$ and $r(v)$. Then, $I(u)\subseteq[v,r(v)]$.

Proof.

Assume $\ell(u)$ or $r(u)$ is outside of $[v,r(v)]$. Then, $u$ has a larger $y$-coordinate than $v$. Thus, $v$ cannot see $r(v)$, a contradiction to the definition of $r(v)$. ∎

Observation 3.2**.**

Let $v\in V(P)$ be a left (right) vertex distinct from the base vertex. Then, $v$ can see exactly two vertices to its right (left): $\operatorname{cv}(v)$ and $r(v)$ ($\ell(v)$).

Proof.

Suppose that $v$ is a left vertex; the other case is symmetric. Any vertex visible from $v$ to the right of $v$ lies in in $[\operatorname{cv}(v),r(v)]$. If $\operatorname{cv}(v)$ is convex, the observation is immediate, since then $[\operatorname{cv}(v),r(v)]=\{\operatorname{cv}(v),r(v)\}$. Otherwise, $\operatorname{cv}(v)$ is $r$-reflex and $r(v)=r(\operatorname{cv}(v))$. By 3.1, we get that for all $u\in[\operatorname{cv}(v),r(v)]\setminus\{\operatorname{cv}(v),r(v)\}$, we have $I(u)\subseteq[\operatorname{cv}(v),r(v)]$. Thus, $v\notin I(u)$ for any such $u$, and since $N(u)\subseteq I(u)$, $v$ cannot see $u$. ∎

3.2 Paths in a Simple Histogram

We now analyze the structure of (shortest) paths in a simple histogram. The following lemma identifies certain “bottleneck” vertices that must appear on any path; see Figure 3.

Lemma 3.3**.**

Let $v,w\in V(P)$ be co-visible vertices such that $v$ is either $r$-reflex or the left base vertex and $w$ is either $\ell$-reflex or the right base vertex. Let $s$ and $t$ be two vertices with $s\in[v,w]$ and $t\notin[v,w]$. Then, any path between $s$ and $t$ includes $v$ or $w$.

Proof.

Let $I=[v,w]$. Since $t\notin I$, not both $v,w$ can be base vertices. Thus, suppose without loss of generality that $v_{y}<w_{y}$. Then, $\operatorname{cv}(v)$ is a left vertex and can see $w$. Hence, 3.2 implies that $r(v)=r(\operatorname{cv}(v))=w$. By 3.1, we get $N(u)\subseteq I(u)\subseteq I$, for every $u\in I\setminus\{v,w\}$. Thus, any path between $s$ and $t$ must include $v$ or $w$. ∎

An immediate consequence of Lemma 3.3 is that if $t\notin I(s)$, then any path from $s$ to $t$ uses $\ell(s)$ or $r(s)$. The next lemma shows that if $t\notin I(s)$, there is a shortest path from $s$ to $t$ that uses the higher vertex of $\ell(s)$ and $r(s)$, see Figure 3.

Lemma 3.4**.**

Let $s$ and $t$ be two vertices with $t\notin I(s)$. If $\ell(s)_{y}>r(s)_{y}$ ($\ell(s)_{y}<r(s)_{y}$), then there is a shortest path from $s$ to $t$ using $\ell(s)$ ($r(s)$).

Proof.

Assume $\ell(s)_{y}>r(s)_{y}$, the other case is symmetric. Let $\pi:\langle s=p_{0},\dots,p_{k}=t\rangle$ be a shortest path from $s$ to $t$. If $\pi$ contains $\ell(s)$, we are done. Otherwise, by Lemma 3.3, there is a $0<j<k$ with $p_{j}=r(s)$ and $p_{i}\neq r(s)$, for $i>j$. Thus, $p_{j+1}\notin I(s)$. Since we assumed $\ell(s)_{y}>r(s)_{y}$, it follows that $\ell(p_{j})=\ell(r(s))=\ell(s)$, so $p_{j+1}$ must be to the right of $p_{j}$. Therefore, by 3.2, we can conclude that $p_{j+1}\in\{\operatorname{cv}(p_{j}),r(p_{j})\}$. Now, since $\ell(s)$ is higher than $r(s)$, it can also see $\operatorname{cv}(p_{j})$ and $r(p_{j})$, in particular, it can see $p_{j+1}$. Hence, $\langle s,\ell(s),p_{j+1},\dots,p_{k}\rangle$ is a valid path of length at most $|\pi|$, so there exists a shortest path from $s$ to $t$ through $\ell(s)$. ∎

The next lemma considers the case where $t$ is in $I(s)$. Then, the near and far dominator are the potential vertices that lie on a shortest path from $s$ to $t$.

Lemma 3.5**.**

Let $s$ and $t$ be two vertices with $t\in I(s)\setminus N(s)$. Then, $\operatorname{nd}(s,t)$ is reflex and either $\operatorname{fd}(s,t)=\ell(\operatorname{nd}(s,t))$ or $\operatorname{fd}(s,t)=r(\operatorname{nd}(s,t))$.

Proof.

Without loss of generality, $t$ lies strictly to the right of $s$. First, assume that $\operatorname{nd}(s,t)$ is $\ell$-convex. Since $s$ can see $\operatorname{nd}(s,t)$ and since $\operatorname{nd}(s,t)$ is to the right of $s$, it follows that $s$ and $\operatorname{nd}(s,t)$ share the same vertical edge. Then, $\operatorname{cv}(\operatorname{nd}(s,t))$ is also visible from $s$ and its horizontal distance to $t$ is smaller. This contradicts the definition of $\operatorname{nd}(s,t)$.

Next, assume that $\operatorname{nd}(s,t)$ is $r$-convex. Let $v$ be the reflex vertex sharing a vertical edge with $\operatorname{nd}(s,t)$. Then, $N(\operatorname{nd}(s,t))\subseteq N(v)$ and $v\in N(s)$. Furthermore, since $t$ is strictly to the right of $v$ but still inside $I(s)$, the vertices $v$ and $r(s)$ must be distinct. Thus, $v_{y}<s_{y}$, so that $\operatorname{cv}(v)$ is also visible from $s$. Moreover, the horizontal distance of $\operatorname{cv}(v)$ and $t$ is smaller than the horizontal distance of $\operatorname{nd}(s,t)$ and $t$. This again contradicts the definition of $\operatorname{nd}(s,t)$. The first part of the lemma follows.

It remains to show that $\operatorname{fd}(s,t)=r(\operatorname{nd}(s,t))$. First of all, $\operatorname{fd}(s,t)$ is higher than $\operatorname{nd}(s,t)$, since otherwise $\operatorname{fd}(s,t)$ would not be visible from $s$. Moreover, if $\operatorname{nd}(s,t)$ and $\operatorname{fd}(s,t)$ are not co-visible, there must be a vertex $v$ strictly between $\operatorname{nd}(s,t)$ and $\operatorname{fd}(s,t)$ that is visible from $s$ and higher than $\operatorname{nd}(s,t)$. Now, either $t\in[\operatorname{nd}(s,t),v]$ or $t\in[v,\operatorname{fd}(s,t)]$. In the first case, the horizontal distance between $v$ and $t$ is smaller than between $t$ and $\operatorname{fd}(s,t)$, and in the second case, the horizontal distance between $v$ and $t$ is smaller than between $t$ and $\operatorname{nd}(s,t)$. Either case leads to a contradiction. Therefore, $\operatorname{fd}(s,t)$ is higher than $\operatorname{nd}(s,t)$, strictly to the right of $\operatorname{nd}(s,t)$ and visible from $\operatorname{nd}(s,t)$. Thus, 3.2 gives $\operatorname{fd}(s,t)=r(\operatorname{nd}(s,t))$. ∎

3.3 The Routing Scheme

We now describe our routing scheme and prove that it gives a shortest path.

Labels and routing tables.

Let $v$ be a vertex. If $v$ is convex and not a base vertex, it is labeled with its $\operatorname{id}$, i.e., $\operatorname{lab}(v)=v_{\operatorname{id}}$. Otherwise, suppose that $v$ is an $r$-reflex vertex or the left base vertex. The breakpoint of $v$, $\operatorname{br}(v)$, is defined as the left endpoint of the horizontal edge with the highest $y$-coordinate to the right of and below $v$ that is visible from $v$; analogous definitions apply to $\ell$-reflex vertices and the right base vertex; see Figure 4. Then, the label of $v$ consists of the $\operatorname{id}$s of $v$ and its breakpoint, i.e., $\operatorname{lab}(v)=(v_{\operatorname{id}},\operatorname{br}(v)_{\operatorname{id}})$. Therefore, $\operatorname{Lab}(n)=2\cdot\lceil\log n\rceil$. The routing table stores one bit, indicating whether $\ell(v)_{y}>r(v)_{y}$, or not. Hence, $\operatorname{Tab}(n)=1$.

The routing function.

We are given the current vertex $s$ and the label $\operatorname{lab}(t)$ of the target vertex $t$. The routing function does not use any information from the header, i.e., $H(n)=0$. If $t$ is visible from $s$, i.e., if $\operatorname{lab}(t)\in\operatorname{lab}(N(s))$, we directly go from $s$ to $t$ on a shortest path. Thus, assume that $t$ is not visible from $s$. First, we check whether $t\in I(s)$. This is done as follows: we determine the smallest and largest $\operatorname{id}$ in the link table $\operatorname{lab}(N(s))$ of $s$. The corresponding vertices are $\ell(s)$ and $r(s)$. Then, we can check whether $t_{\operatorname{id}}\in[\ell(s)_{\operatorname{id}},r(s)_{\operatorname{id}}]$, which is the case if and only if $t\in I(s)$. Now, there are two cases, illustrated in Figure 5.

First, suppose $t\notin I(s)$. If the bit in the routing table of $s$ indicates that $\ell(s)$ is higher than $r(s)$, we take the hop to $\ell(s)$; otherwise, we take the hop to $r(s)$. By Lemma 3.4, this hop lies on a shortest path from $s$ to $t$.

Second, suppose that $t\in I(s)\setminus N(s)$. This case is a bit more involved. We use the link table $\operatorname{lab}(N(s))$ of $s$ and the label $\operatorname{lab}(t)$ of $t$ to determine $\operatorname{fd}(s,t)$ and $\operatorname{nd}(s,t)$. Again, we can do this by comparing the $\operatorname{id}$s. Lemma 3.5 states that either $\operatorname{fd}(s,t)=\ell(\operatorname{nd}(s,t))$ or $\operatorname{fd}(s,t)=r(\operatorname{nd}(s,t))$. We discuss the case that $\operatorname{fd}(s,t)=r(\operatorname{nd}(s,t))$, the other case is symmetric. By Lemma 3.3, any shortest path from $s$ to $t$ includes $\operatorname{fd}(s,t)$ or $\operatorname{nd}(s,t)$. Moreover, due to Lemma 3.5, $\operatorname{nd}(s,t)$ is reflex, and we can use its label to access $b_{\operatorname{id}}=\operatorname{br}(\operatorname{nd}(s,t))_{\operatorname{id}}$. The vertex $b$ splits $I(s,t)=[\operatorname{nd}(s,t),\operatorname{fd}(s,t)]$ into two disjoint subintervals $[\operatorname{nd}(s,t),b]$ and $[\operatorname{cv}(b),\operatorname{fd}(s,t)]$. Also, $b$ and $\operatorname{cv}(b)$ are not visible from $s$, as they are located strictly between the far and the near dominator. Based on $b_{\operatorname{id}}$, we can now decide on the next hop.

If $t\in[\operatorname{nd}(s,t),b]$, we take the hop to $\operatorname{nd}(s,t)$. If $t=b$, our packet uses a shortest path of length $2$. Thus, assume that $t$ lies between $\operatorname{nd}(s,t)$ and $b$. This is only possible if $b$ is $\ell$-reflex, and we can apply Lemma 3.3 to see that any shortest path from $s$ to $t$ includes $\operatorname{nd}(s,t)$ or $b$. But since $d(s,b)=2$, our data packet routes along a shortest path.

If $t\in[\operatorname{cv}(b),\operatorname{fd}(s,t)]$, we take the hop to $\operatorname{fd}(s,t)$. If $t=\operatorname{cv}(b)$, our packet uses a shortest path of length $2$. Thus, assume that $t$ lies between $\operatorname{cv}(b)$ and $\operatorname{fd}(s,t)$. This is only possible if $\operatorname{cv}(b)$ is $r$-reflex, so we can apply Lemma 3.3 to see that any shortest path from $s$ to $t$ uses $\operatorname{fd}(s,t)$ or $\operatorname{cv}(b)$. Since $d(s,\operatorname{cv}(b))=2$, our packet routes along a shortest path. The following theorem summarizes our discussion.

Theorem 1 (restated).

Let $P$ be a simple histogram with $n$ vertices. There is a routing scheme for $G(P)$ with a routing table with 1 bit, without headers, having label size $2\cdot\lceil\log n\rceil$, such that we can route between any two vertices on a shortest path.

4 Double Histograms

Let $P$ be a double histogram with $n$ vertices. Similar to the simple histogram case, we first focus on the visibility and the structure of shortest paths in $P$. Again, if a target vertex $t$ is not in the interval $I(s)$ of a current vertex $s$, we should widen the interval as fast as possible. However, in contrast to simple histograms, we can now change sides arbitrarily often. Nevertheless, we can guarantee that in each step, the interval comes closer to $t$. Once we have reached the case that $t$ is in the interval of the current vertex, we again have to find the right pocket. Unlike in simple histograms, this case is now simpler to describe.

4.1 Visibility in Double Histograms

The structure of the shortest paths in double histograms can be much more involved than in simple histograms; in particular, Lemma 3.3 does not hold anymore. However, the following observations provide some structural insight that can be used for an efficient routing scheme.

Observation 4.1**.**

Two vertices $v,w$ are co-visible if and only if $v\in I(w)$ and $w\in I(v)$.

Proof.

The forward direction is immediate, since co-visibility implies $v\in N(w)\subseteq I(w)$ and $w\in N(v)\subseteq I(v)$. For the backward direction, let $Q$ be the rectangle spanned by $v$ and $w$. Since $v\in I(w)$ and $w\in I(v)$, the upper and lower boundary of $Q$ do not contain a point outside $P$. As $P$ is a double histogram, this implies that the left and right boundary of $Q$ also do not contain any point outside $P$. The claim follows since $P$ has no holes. ∎

Observation 4.2**.**

Let $a$, $b$, $c$, and $d$ be vertices in $P$ with $a_{x}\leq b_{x}\leq c_{x}\leq d_{x}$. If $a\in I(c)$ and $d\in I(b)$, then $b$ and $c$ are co-visible.

Proof.

This follows immediately from 4.1. ∎

Observation 4.3**.**

The intervals on one side of $P$ form a laminar family, i.e., for any two vertices $v$ and $w$ on the same side of the base line, we have (i) $I(v)\cap I(w)=\emptyset$, (ii) $I(v)\subseteq I(w)$, or (iii) $I(w)\subseteq I(v)$.

Proof.

Suppose there are two vertices $v$ and $w$ on the same side of $P$ with $\ell(v)_{x}<\ell(w)_{x}\leq r(v)_{x}<r(w)_{x}$. By 4.2, $\ell(w)$ and $r(v)$ are co-visible. Since $\ell(w)$ and $r(v)$ are on the same side of $P$, either $r(v)$ cannot see any vertex to the left of $\ell(w)$ or $\ell(w)$ cannot see any vertex to the right of $r(v)$. This contradicts the fact that the $\ell(v)$ and $r(v)$ as well as $\ell(w)$ and $r(w)$ must be co-visible. ∎

4.2 Paths in a Double Histogram

To understand shortest paths in double histograms, we distinguish three cases, depending on where $t$ lies relative to $s$. First, if $t$ is close, i.e., if $t\in I(s)$, we focus on the near and far dominators. Second, if $t\notin I(s)$ but there is a vertex $v$ visible from $s$ with $t\in I(v)$, then we can find a vertex on a shortest path from $s$ to $t$. Third, if there is no visible vertex $v$ from $s$ such that $t\in I(v)$, we can apply our intuition from simple histograms: go as fast as possible towards the base line. Details follow.

The target is close.

Let $s,t$ be two vertices with $t\in I(s)\setminus N(s)$. In contrast to simple histograms, $\operatorname{fd}(s,t)$ now might not be a vertex. Furthermore, $\operatorname{fd}(s,t)$ and $\operatorname{nd}(s,t)$ might be on different sides of the base line. In this case, Lemma 3.5 no longer holds. However, the next lemma establishes a visibility relation between them; see Figure 6.

Lemma 4.4**.**

Let $s,t\in V(P)$ with $t\in I(s)\setminus N(s)$. Then, $\operatorname{nd}(s,t)$ and $\operatorname{fd}(s,t)$ are co-visible.

Proof.

Without loss of generality, $t$ is strictly to the right of $s$. Suppose for a contradiction that $r(\operatorname{nd}(s,t))$ is strictly left of $\operatorname{fd}(s,t)$. Then, we get $r(\operatorname{nd}(s,t))\in I(s)$. Also, $s\in I(\operatorname{nd}(s,t))\subseteq I(r(\operatorname{nd}(s,t)))$. Hence, by 4.1, $s$ can see $r(\operatorname{nd}(s,t))$. But then $r(\operatorname{nd}(s,t))$ is a vertex strictly between the near and far dominator visible from $s$, contradicting the choice of the dominators. Thus, $s_{x}\leq\operatorname{nd}(s,t)_{x}\leq\operatorname{fd}(s,t)_{x}\leq r(\operatorname{nd}(s,t))_{x}$, and 4.2 gives the result. ∎

The proof of the next lemma uses Lemma 4.4 to find a shortest path vertex.

Lemma 4.5**.**

One of $\operatorname{nd}(s,t)$ or $\operatorname{fd}(s,t)$ is on a shortest path from $s$ to $t$. If $\operatorname{fd}(s,t)$ is not a vertex, then $\operatorname{nd}(s,t)$ is on a shortest path from $s$ to $t$.

Proof.

Without loss of generality, $t$ is to the right of $s$. Let $\pi:\langle s=p_{0},\dots,p_{k}=t\rangle$ be a shortest path from $s$ to $t$, and let $p_{j}$ be the last vertex outside of $I(s,t)$. If $j=0$, then $p_{j+1}$ must be one of the dominators, since by definition they are the only vertices in $I(s,t)$ visible from $s$. Now, assume $j\geq 1$. If $p_{j}$ is to the left of $\operatorname{nd}(s,t)$, we apply Lemma 4.4 and 4.2 on the four points $p_{j}$, $\operatorname{nd}(s,t)$, $p_{j+1}$, and $\operatorname{fd}(s,t)$ to conclude that $\operatorname{nd}(s,t)$ can see $p_{j+1}$. Symmetrically, if $p_{j}$ is to the right of $\operatorname{fd}(s,t)$, the same argument shows that the far dominator can see $p_{j+1}$. Thus, depending on the position of $p_{j}$ we can exchange the subpath $p_{1},\dots,p_{j}$ in $\pi$ by $\operatorname{nd}(s,t)$ or $\operatorname{fd}(s,t)$ and get a valid path of length $k-j+1\leq k$. The second part of the lemma holds because $p_{j}$ cannot be to the right of $\operatorname{fd}(s,t)$, if $\operatorname{fd}(s,t)$ is not a vertex but a point on the right boundary. ∎

Next, we consider the case where $\operatorname{fd}(s,t)$ is a vertex but not on a shortest path from $s$ to $t$. Then, $\operatorname{fd}(s,t)$ cannot see $t$, and we define $\operatorname{fd}^{2}(s,t)=\operatorname{fd}(\operatorname{fd}(s,t),t)$. By Lemma 4.4, $\operatorname{nd}(s,t)$ and $\operatorname{fd}(s,t)$ are co-visible, so $\operatorname{fd}^{2}(s,t)$ has to be in the interval $[\operatorname{nd}(s,t),t]$, and therefore it is a vertex. The following lemma states that $\operatorname{fd}^{2}(s,t)$ is strictly closer to $t$ than $s$; see Figure 7.

Lemma 4.6**.**

If $\operatorname{fd}(s,t)$ is a vertex but not on a shortest path from $s$ to $t$, then we have $d(\operatorname{fd}^{2}(s,t),t)=d(s,t)-1$.

Proof.

Without loss of generality, $t$ is to the right of $s$. By Lemma 4.5, $\operatorname{nd}(s,t)$ lies on a shortest path from $s$ to $t$. Let $\langle s=p_{0},\operatorname{nd}(s,t)=p_{1},p_{2},\dots,p_{k}=t\rangle$ be such a shortest path. We claim that $\operatorname{fd}^{2}(s,t)$ can see $p_{2}$. Then, $\langle\operatorname{fd}^{2}(s,t),p_{2},\dots,p_{k}=t\rangle$ is a valid path of length $k-1=d(s,t)-1$. To prove that $\operatorname{fd}^{2}(s,t)$ can indeed see $p_{2}$, we show that p_{2}\in I\big{(}\!\operatorname{fd}^{2}(s,t)\big{)} and $\operatorname{fd}^{2}(s,t)\in I(p_{2})$ and then apply 4.1.

First, we show $p_{2}\in I(\operatorname{fd}(s,t),t)$ by contradiction. Thus, suppose $p_{2}\notin I(\operatorname{fd}(s,t),t)$. Since $t\in I(\operatorname{fd}(s,t),t)$, there is a $j\geq 2$ with $p_{j+1}\in I(\operatorname{fd}(s,t),t)$ and $p_{j}\notin I(\operatorname{fd}(s,t),t)$. First, if $p_{j,x}<\operatorname{fd}^{2}(s,t)_{x}$, then $p_{j,x}<\operatorname{fd}^{2}(s,t)_{x}\leq p_{j+1,x}\leq\operatorname{nd}(\operatorname{fd}(s,t),t)_{x}$. By Lemma 4.4, $\operatorname{fd}^{2}(s,t)$ and \operatorname{nd}\big{(}\!\operatorname{fd}(s,t),t\big{)} are co-visible, so 4.2 implies that $\operatorname{fd}^{2}(s,t)$ and $p_{j+1}$ are co-visible. Then $\langle s,\operatorname{fd}(s,t),\operatorname{fd}^{2}(s,t),p_{j+1},\dots,p_{k}=t\rangle$ is a valid path of length $k-j+2\leq k$, contradicting the assumption that $\operatorname{fd}(s,t)$ is not on a shortest path. If $\operatorname{nd}(\operatorname{fd}(s,t),t)_{x}<p_{j,x}$, it follows with the same reasoning that $\operatorname{nd}(\operatorname{fd}(s,t),t)$ and $p_{j+1}$ are co-visible then $\operatorname{fd}(s,t)$ is on $\langle s,\operatorname{fd}(s,t),\operatorname{nd}(\operatorname{fd}(s,t),t),p_{j+1},\dots,p_{k}=t\rangle$ which is a valid path of length $k-j+2\leq k$. This again contradicts the assumption.

Now, since I(\operatorname{fd}(s,t),t)=\big{[}\!\operatorname{fd}^{2}(s,t),\operatorname{nd}(\operatorname{fd}(s,t),t)\big{]}\subseteq I(\operatorname{fd}^{2}(s,t)), we get $p_{2}\in I(\operatorname{fd}^{2}(s,t))$. Since $p_{2}$ sees $\operatorname{nd}(s,t)$ which is to the left of $\operatorname{fd}^{2}(s,t)$ and since $p_{2}$ is in $I(\operatorname{fd}(s,t),t)$, and thus to the right of $\operatorname{fd}^{2}(s,t)$, it follows that $\operatorname{fd}^{2}(s,t)\in I(p_{2})$. ∎

The target can be made close in one step.

Let $s,t$ be two vertices with $t\notin I(s)$ but there is a vertex $v\in N(s)$ with $t\in I(v)$. For clarity of presentation, we will always assume that $s$ is below the base line. The crux of this case is this: there might be many vertices visible from $s$ that have $t$ in their interval. However, we can find a best vertex as follows: once $t$ is in the interval of a vertex, the goal is to shrink the interval as fast as possible. Therefore, we must find a vertex $v\in N(s)$ whose left or right interval boundary is closest to $t$ among all vertices in $N(s)$. This leads to the following inductive definition of two sequences $a^{i}(s)$ and $b^{i}(s)$ of vertices in $N(s)$. For $i=0$, we let $a^{0}(s)=b^{0}(s)=s$. For $i>0$, if the set $A^{i}(s)=\{v\in N(s)\mid\ell(v)_{x}<\ell(a^{i-1}(s))_{x}\}$ is nonempty, we define $a^{i}(s)=\operatorname{argmin}\{v_{x}\mid v\in A^{i}(s)\}$; and $a^{i}(s)=a^{i-1}(s)$, otherwise. If the set $B^{i}(s)=\{v\in N(s)\mid r(v)_{x}>r(b^{i-1}(s))_{x}\}$ is nonempty, we define $b^{i}(s)=\operatorname{argmax}\{v_{x}\mid v\in B^{i}(s)\}$; and $b^{i}(s)=b^{i-1}(s)$, otherwise. We force unambiguity by choosing the vertex closer to the base line. Let $a^{*}(s)$ be the vertex with $a^{*}(s)=a^{i}(s)=a^{i-1}(s)$, for an $i>0$, and $b^{*}(s)$ the vertex with $b^{*}(s)=b^{i}(s)=b^{i-1}(s)$, for an $i>0$. If the context is clear, we write $a^{i}$ instead of $a^{i}(s)$ and $b^{i}$ instead of $b^{i}(s)$.

Let us try to understand this definition. For $i\geq 0$, we write $\ell^{i}$ for $\ell(a^{i})$; and we write $\ell^{*}$ for $\ell(a^{*})$. Then, we have $a^{0}=s$ and $\ell^{0}=\ell(s)$. Now, if $\ell(s)$ is not a vertex, then $a^{*}=s$, because there is no vertex whose left point is strictly to the left of the left boundary of $P$. On the other hand, if $\ell^{0}$ is a vertex in $P$, we have $a^{1}=\ell^{0}=\ell(s)$, and $[\ell^{1},a^{1}]$ is an interval between points on the lower side of $P$. Then comes a (possibly empty) sequence of intervals $[\ell^{2},a^{2}],[\ell^{3},a^{3}],\dots,[\ell^{k},a^{k}]$ between points on the upper side of $P$; possibly followed by the interval $[\ell(r(s)),r(s)]$. There are four possibilities for $a^{*}$: it could be $s$, $\ell(s)$, a vertex $a^{i}$ on the upper side of $P$, or $r(s)$. If $a^{*}\neq s$, then the intervals $[\ell^{1},a^{1}]\subset[\ell^{2},a^{2}]\subset\dots\subset[\ell^{*},a^{*}]$ are strictly increasing: $\ell^{i}$ is strictly to the left of $\ell^{i-1}$ and $a^{i}$ is strictly to the right of $a^{i-1}$; see Figure 8. Symmetric observations apply for the $b^{i}$; we write $r^{i}$ for $r(b^{i})$ and $r^{*}$ for $r(b^{*})$.

Lemma 4.7**.**

For $i\geq 1$, the vertices $\ell^{i-1}$ and $a^{i}$ as well as $r^{i-1}$ and $b^{i}$ are co-visible.

Proof.

We focus on $\ell^{i-1}$ and $a^{i}$. We show that $\ell^{i-1}\in I(a^{i})$ and $a^{i}\in I(\ell^{i-1})$; the lemma follows from 4.1. The claim $\ell^{i-1}\in I(a^{i})$ is due to the facts that $[\ell^{i},a^{i}]\subseteq I(a^{i})$ and $\ell^{i-1}\in[\ell^{i},a^{i}]$ (this holds also for $i=1$, as then $\ell^{i-1}=a^{i}$). Next, since $a^{i-1}\in I(\ell^{i-1})$, the vertex $a^{i-1}$ is to the left of $r(\ell^{i-1})$; and since $\ell^{i-1}\in I(r(\ell^{i-1}))$, the point $\ell(r(\ell^{i-1}))$ is to the left of $\ell^{i-1}$. Thus, if $r(\ell^{i-1})$ is visible from $s$, we have $a^{i}=r(\ell^{i-1})$, by the definition of $a^{i}$. On the other hand, if $r(\ell^{i-1})$ is not visible from $s$, the visibility must be blocked by $r(s)$, and then $a^{i}=r(s)$. In either case, we have $a^{i}\in[a^{i-1},r(\ell^{i-1})]\subseteq I(\ell^{i-1})$, as desired. ∎

Finally, the next lemma tells us the following: if $t\in[\ell^{*},r^{*}]$ we find a vertex $v\in N(s)$ with $t\in I(v)$. Its quite technical proof needs Lemma 4.7.

Lemma 4.8**.**

If $t\in[\ell^{i},\ell^{i-1}]$, for some $i\geq 1$, then $a^{i}$ is on a shortest path from $s$ to $t$. If $t\in[r^{i-1},r^{i}]$, for some $i\geq 1$, then $b^{i}$ is on a shortest path from $s$ to $t$.

Proof.

We focus on the first statement; see Figure 9. Let $\pi:\langle s=p_{0},\dots,p_{k}=t\rangle$ be a shortest path from $s$ to $t$, and let $p_{j}$ be the last vertex on $\pi$ outside of $[\ell^{i},\ell^{i-1}]$. If $j=0$, then $p_{j+1}$ must be $\ell^{0}=\ell(s)$, because this is the only vertex $\ell^{i-1}$ visible from $s$. Then, $i=1$ and $a^{i}=\ell(s)$ is on $\pi$. From now on, we assume that $j\geq 1$.

First, suppose that $p_{j+1}$ and $a^{i}$ are co-visible. Then $\langle s,a^{i},p_{j+1},\dots,p_{k}\rangle$ is a path from $s$ to $t$ that uses $a^{i}$ and has length $k-j+1\leq k$. Second, suppose that $p_{j+1}$ and $a^{i}$ are not co-visible. Then, the contrapositive of 4.2 applied to the four points $\ell^{i}$, $p_{j+1}$, $a^{i}$, and $p_{j}$ shows that $p_{j}$ is strictly to the left of $a^{i}$. There are two subcases, depending on whether $p_{j}$ is strictly to the left of $\ell^{i}$ or strictly to the right of $\ell^{i-1}$.

If $p_{j}$ is strictly to the left of $\ell^{i}$, then $j\geq 2$, since $\ell^{i}$ is to the left of $\ell^{0}=\ell(s)$ and we need at least two hops to reach a point strictly to the left of $\ell(s)$ from $s$. We apply 4.2 on the four points $p_{j}$, $\ell^{i}$, $p_{j+1}$, and $a^{i}$, and get that $\ell^{i}$ and $p_{j+1}$ are co-visible. Hence, $\langle s,a^{i},\ell^{i},p_{j+1},\dots,p_{k}\rangle$ is a path that uses $a^{i}$ and has length $k-j+2\leq k$.

Finally, assume that $p_{j}$ is strictly to the right of $\ell^{i-1}$. By Lemma 4.7, $a^{i}$ can see $\ell^{i-1}$. Thus, $p_{j+1}\neq\ell^{i-1}$ and there is no vertex strictly between $\ell^{i-1}$ and $a^{i}$ on the same side as $\ell^{i-1}$ that can see a vertex strictly to the left of $\ell^{i-1}$. Thus, $p_{j}$ and $a^{i-1}$ are on different sides of the base line. Let $b$ be the rightmost vertex that (i) lies on the same side of $P$ as $p_{j}$; (ii) is strictly between $\ell^{i-1}$ and $a^{i}$; (iii) is closest to the base line. The vertex $b$ exists (since $p^{j}$ is a candidate), is not visible from $s$ (because $b$ is strictly left of $a^{i}$ and can see strictly left of $\ell^{i-1}$); and thus strictly left of $\ell(s)$. The vertex $p_{j}$ cannot be strictly to the right of $b$, as otherwise $b$ would obstruct visiblity between $p_{j}$ and $p_{j+1}$. We conclude that $j\geq 2$, since we need at least two hops to reach a point strictly to the left of $\ell(s)$ from $s$. If $p_{j}\in\{b,\operatorname{cv}(b)\}$, $a^{i}$ can see $p_{j}$ and thus, $\langle s,a^{i},p_{j},p_{j+1},\dots,p_{k}\rangle$ is a path of length $k-j+2\leq k$ using $a^{i}$. If $p_{j}\not\in\{b,\operatorname{cv}(b)\}$, then $b$ is strictly closer to the base line than $p_{j}$. Then, we have $j\geq 3$, because we need two hops to cross the vertical line through $\ell(s)$ and one more hop to cross the horizontal line through $b$. We apply 4.2 on the four points $p_{j+1}$, $\ell^{i-1}$, $p_{j}$, and $b$ to conclude that $\ell^{i-1}$ can see $p_{j}$. Hence, $\langle s,a^{i},\ell^{i-1},p_{j},p_{j+1},\dots,p_{k}\rangle$ is a path of length $k-j+3\leq k$ using $a^{i}$. ∎

The target is far away.

Finally, we consider the case that there is no vertex $v\in N(s)$ with $t\in I(v)$, i.e., $t\notin[\ell^{*},r^{*}]$. The intuition now is as follows: to widen the interval, we should go to a vertex that is visible from $s$, but closest to the base line. In simple histograms, there was only one such vertex, but in double histograms there might be a second one on the other side. These two vertices are the dominators of $s$. These two dominators might have their own dominators, and so on. This leads to the following inductive definition.

For $k\geq 0$, we define the $k$-th bottom dominator $\operatorname{bd}^{k}(s)$, the $k$-th top dominator $\operatorname{td}^{k}(s)$, and the $k$-th interval $I^{k}(s)$ of $s$. For any set $Q\subset V(P)$, we write $Q^{-}$ (resp. $Q^{+}$) for all points in $Q$ below (resp. above) the base line. We set $\operatorname{bd}^{0}(s)=\operatorname{td}^{0}(s)=s$ and $I^{0}(s)=\{s\}$. For $k>0$, we set $I^{k}(s)=I(\operatorname{bd}^{k-1}(s))\cup I(\operatorname{td}^{k-1}(s))$. If $I^{k}(s)^{-}$ is nonempty, we let $\operatorname{bd}^{k}(s)$ be the leftmost vertex inside $I^{k}(s)^{-}$ that minimizes the distance to the base line. If $I^{k}(s)^{+}$ is nonempty, we let $\operatorname{td}^{k}(v)$ be the leftmost vertex inside $I^{k}(s)^{+}$ that minimizes the distance to the base line; see Figure 10. If one of the two sets is empty, the other one has to be nonempty, since $s\in I^{k}(s)$. In this case, we let $\operatorname{td}^{k}(s)=\operatorname{bd}^{k}(s)$. We write $\operatorname{bd}(s)$ for $\operatorname{bd}^{1}(s)$ and $\operatorname{td}(s)$ for $\operatorname{td}^{1}(s)$.

Observe, that $I^{1}(s)=I(s)$ and $I^{2}(s)=[\ell^{*},r^{*}]$. If $I(\operatorname{bd}^{k-1}(s))=V(P)$, we have $\operatorname{bd}^{k}(s)=\operatorname{bd}^{k-1}(s)$. The same holds for the top dominator. We provide a few technical properties concerning the $k$-th interval as well as the $k$-th dominators.

Lemma 4.9**.**

For any $s\in V(P)$ and $k\geq 0$, we have $I^{k}(s)\subseteq I(\operatorname{bd}^{k}(s))\cap I(\operatorname{td}^{k}(s))$.

Proof.

We have $I^{k}(s)\subseteq I(\operatorname{bd}^{k}(s))$, since by definition the interval $I^{k}(s)$ contains no vertex that is on the same side as $\operatorname{bd}^{k}(s)$ and strictly closer to the base line, so no vertex can obstruct horizontal visibility of $\operatorname{bd}^{k}(s)$ in $I^{k}(s)$. Analogously, we have $I^{k}(s)\subseteq I(\operatorname{td}^{k}(s))$. The claim follows. ∎

Lemma 4.10**.**

For any $s\in V(P)$ and $k\geq 0$, $\operatorname{bd}^{k}(s)$ and $\operatorname{td}^{k}(s)$ are co-visible.

Proof.

By definition and Lemma 4.9, $\operatorname{td}^{k}(s)\in I^{k}(s)\subseteq I(\operatorname{bd}^{k}(s))$ and $\operatorname{bd}^{k}(s)\in I^{k}(s)\subseteq I(\operatorname{td}^{k}(s))$. The claim now follows from 4.1. ∎

The following lemma seems rather specific but it will be needed later on to deal with short paths.

Lemma 4.11**.**

For any $s\in V(P)$, we have I^{3}(s)=I^{2}\big{(}\!\operatorname{bd}(s)\big{)}\cup I^{2}\big{(}\!\operatorname{td}(s)\big{)}.

Proof.

We begin by showing that

[TABLE]

If $\operatorname{bd}(s)$ is above the base line, then $\operatorname{bd}(s)=\operatorname{td}(s)$ and $I^{2}(s)=I(\operatorname{bd}(s))\cup I(\operatorname{td}(s))=I(\operatorname{td}(s))$. The definition of $\operatorname{bd}^{2}(s)$ then gives $\operatorname{bd}^{2}(s)=\operatorname{bd}(\operatorname{td}(s))$, and (1) follows.

If $\operatorname{bd}(s)$ is below the base line, the vertex $b_{1}=\operatorname{bd}(\operatorname{bd}(s))$ is below the base line. Let $b_{2}=\operatorname{bd}(\operatorname{td}(s))$. By Lemma 4.10, $\operatorname{bd}(s)$ and $\operatorname{td}(s)$ are co-visible, so $\operatorname{bd}(s)\in I(\operatorname{td}(s))^{-}$. Therefore, $b_{2}$ is below the base line. Since $I(b_{1})$ and $I(b_{2})$ are not disjoint (both contain $s$) and since $b_{1}$ and $b_{2}$ are on the same side of the base line, 4.3 gives $I(b_{1})\subseteq I(b_{2})$ or $I(b_{2})\subseteq I(b_{1})$. Because $\operatorname{bd}^{2}(s)$ is the highest vertex in \big{(}I(\operatorname{bd}(s))\cup I(\operatorname{td}(s))\big{)}^{-}, we get that $\operatorname{bd}^{2}(s)$ is $b_{1}$ or $b_{2}$, and (1) follows also in this case. Symmetrically, we have

[TABLE]

We use the definitions and (1,2) to get

[TABLE]

as desired. ∎

Intuitively, the meaning of $I^{k}(s)$ is as follows: let $\ell$ be the leftmost and $r$ be the rightmost vertex with hop distance exactly $k$ from $s$, then, $I^{k}(s)=[\ell,r]$. We do not really need this property. So we leave it as an exercise for the reader to find a proof for this. Instead, we prove the following weaker statement. For this, recall that due to its definition, $\operatorname{bd}^{k}(s)$ might not be on the lower side of the histogram (and $\operatorname{td}^{k}(s)$ might not be on the upper side).

Lemma 4.12**.**

Let $k\geq 0$ and let $s,t\in V(P)$ with $d(s,t)\leq k$. Then, $t\in I^{k}(s)$.

Proof.

We show that for any $j\geq 0$ and any vertex $v\in I^{j}(s)$, we have $N(v)\subseteq I^{j+1}(s)$. The lemma then follows by induction. If $v\in I^{j}(s)^{-}$, then $\operatorname{bd}^{j}(s)$ is on the lower side, and by definition, $I(v)\subseteq I(\operatorname{bd}^{j}(s))$. If $v\in I^{j}(s)^{+}$, by a similar argument $I(v)\subseteq I(\operatorname{td}^{j}(s))$. Thus, $N(v)\subseteq I(v)\subseteq I(\operatorname{bd}^{j}(s))\cup I(\operatorname{td}^{j}(s))=I^{j+1}(s)$, as desired. ∎

Let $k\geq 0$ and $s\in V(P)$. For $i=1,\dots,k$, by Lemma 4.9, $\operatorname{bd}^{i-1}(s),\operatorname{td}^{i-1}(s)\in I(\operatorname{bd}^{i}(s))\cap I(\operatorname{td}^{i}(s))$. Moreover, by definition, $\operatorname{bd}^{i}(s),\operatorname{td}^{i}(s)\in I(\operatorname{bd}^{i-1}(s))\cup I(\operatorname{td}^{i-1}(s))$. 4.1 now says that both $\operatorname{bd}^{i}(s)$ and $\operatorname{td}^{i}(s)$ can see at least one of $\operatorname{bd}^{i-1}(s)$ or $\operatorname{td}^{i-1}(s)$. Therefore, there is a path $\pi_{b}(s,k):\langle s=p_{0},\dots,p_{k}=\operatorname{bd}^{k}(s)\rangle$ from $s$ to $\operatorname{bd}^{k}(s)$ and a path $\pi_{t}(s,k):\langle s=q_{0},\dots,q_{k}=\operatorname{td}^{k}(s)\rangle$ from $s$ to $\operatorname{td}^{k}(s)$ with $p_{i},q_{i}\in\{\operatorname{bd}^{i}(s),\operatorname{td}^{i}(s)\}$, for $i=0,\dots,k$. We call $\pi_{b}(s,k)$ and $\pi_{t}(s,k)$ the canonical path from $s$ to $\operatorname{bd}^{k}(s)$ and from $s$ to $\operatorname{td}^{k}(s)$, respectively. The following two lemmas show that for every $t\notin I^{k+1}(s)$ one of the canonical paths is the prefix of a shortest path from $s$ to $t$. To show Lemma 4.14 we need Lemma 4.12 as well as Lemma 4.13.

Lemma 4.13**.**

Let $k\geq 1$ and $s\in V(P)$. If $I(\operatorname{bd}^{k-1}(s))\neq V(P)$ we have $d(s,\operatorname{bd}^{k}(s))=k$. If $I(\operatorname{td}^{k-1}(s))\neq V(P)$ we have $d(s,\operatorname{td}^{k}(s))=k$.

Proof.

On the one hand, $d(s,\operatorname{bd}^{k}(s))\leq|\pi_{b}(s,k)|=k$ and $d(s,\operatorname{td}^{k}(s))\leq|\pi_{t}(s,k)|=k$. On the other hand, we show that $\operatorname{bd}^{k}(s)\notin I^{k-1}(s)$ and $\operatorname{td}^{k}(s)\notin I^{k-1}(s)$. The claim then follows from the contrapositive of Lemma 4.12.

Case 1: First, assume that $\operatorname{bd}^{k-1}(s)\in I^{k-1}(s)^{-}$. Since $I(\operatorname{bd}^{k-1}(s))\neq P$, at least one of its bounding points is a vertex $v$ contained in $I^{k}(s)$. Then, $v$ is strictly closer to the base line than $\operatorname{bd}^{k-1}(s)$, and since $v$ is a candidate for $\operatorname{bd}^{k}(s)$, the same applies to $\operatorname{bd}^{k}(s)$. It follows that $\operatorname{bd}^{k}(s)\not\in I^{k-1}(s)$. Similarly, we get that if $\operatorname{td}^{k-1}(s)\in I^{k-1}(s)^{+}$, the vertex $\operatorname{td}^{k}(s)$ is not in $I^{k-1}(s)$.

Case 2: Second, assume that $\operatorname{bd}^{k-1}(s)\in I^{k-1}(s)^{+}$. Then, $\operatorname{bd}^{k}(s)\notin I^{k-1}(s)^{-}$, since this set is empty. Thus, suppose for a contradiction that $\operatorname{bd}^{k}(s)\in I^{k-1}(s)^{+}$. This can only be the case if $\operatorname{bd}^{k-1}(s)=\operatorname{td}^{k-1}(s)$ and $\operatorname{bd}^{k}(s)=\operatorname{td}^{k}(s)$. However, in Case 1 we showed that $\operatorname{td}^{k}(s)\not\in I^{k-1}(s)^{+}$ if $\operatorname{td}^{k-1}(s)\in I^{k-1}(s)^{+}$. Hence, $\operatorname{bd}^{k}(s)\notin I^{k-1}(s)^{+}$, as desired. ∎

Lemma 4.14**.**

Let $s$ and $t$ be vertices and $k\geq 1$ an integer such that $t\notin I^{k+1}(s)$. Then $\operatorname{bd}^{k}(s)$ or $\operatorname{td}^{k}(s)$ is on a shortest path from $s$ to $t$.

Proof.

First, observe that $I^{k+1}(s)=I(\operatorname{bd}^{k}(s))\cup I(\operatorname{td}^{k}(s))\neq P$, as $t\notin I^{k+1}(s)$. Let $\pi:\langle s=p_{0},\dots,p_{m}=t\rangle$ be a shortest path from $s$ to $t$, and $p_{j}$ the last vertex in $I^{k+1}(s)$. Without loss of generality, $p_{j+1}$ is strictly to the right of $s$. By Lemma 4.12, we get that $p_{j}$ is not in $I^{k}(s)$ and thus, again by Lemma 4.12, we have $j\geq d(s,p_{j})\geq k+1$.

First, suppose that $p_{j}$ and $\operatorname{bd}^{k}(s)$ (resp. $\operatorname{td}^{k}(s)$) are co-visible. Then, $\pi_{b}(s,k)\circ\langle p_{j},\dots,t\rangle$ (resp. $\pi_{t}(s,k)\circ\langle p_{j},\dots,t\rangle$) is a valid path of length $k+1+(m-j)\leq m$. Here, $\circ$ concatenates two paths. Second, suppose $p_{j}$ be visible from neither $\operatorname{bd}^{k}(s)$ nor $\operatorname{td}^{k}(s)$. First, we claim that $p_{j}$ is strictly to the right of $\operatorname{bd}^{k}(s)$ and $\operatorname{td}^{k}(s)$. Otherwise, since $p_{j+1}$ is strictly to the right of both dominators, we would get $\operatorname{bd}^{k}(s),\operatorname{td}^{k}(s)\in I(p_{j})$. Moreover, $p_{j}\in I^{k+1}(s)=I(\operatorname{bd}^{k}(s))\cup I(\operatorname{td}^{k}(s))$. 4.1 now would imply that $p_{j}$ can see $\operatorname{bd}^{k}(s)$ or $\operatorname{td}^{k}(s)$—a contradiction. The claim follows. Next, we claim $\operatorname{bd}^{k}(s)\neq\operatorname{td}^{k}(s)$. If not, $p_{j}\in I(\operatorname{bd}^{k}(s))=I^{k+1}(s)$, and since $p_{j}$ can see a point outside of $I^{k+1}(s)$, we would get $p_{j}\in\{\ell(\operatorname{bd}^{k}(s)),r(\operatorname{bd}^{k}(s))\}$, which again contradicts our assumption that $p_{j}$ cannot see $\operatorname{bd}^{k}(s)$. The claim follows. There are two cases, depending on which dominator sees further to the right.

Case 1: $r(\operatorname{bd}^{k}(s))_{x}<r(\operatorname{td}^{k}(s))_{x}$; see Figure 11. Let $b$ be the leftmost vertex in $[r(\operatorname{bd}^{k}(s)),r(\operatorname{td}^{k}(s))]^{-}$ closest to the base line. Observe that $b$ is strictly to the right of $I^{k}(s)$, because $r(\operatorname{bd}^{k}(s))$ is strictly to the right of $I^{k}(s)$. Since $p_{j}$ is not visible from $\operatorname{td}^{k}(s)$, it has to be strictly to the right of and strictly below $b$. Next, we claim that no vertex $v\in I^{k}(s)$ can see $p_{j}$. If one could, by 4.1, we would have $v\in I(p_{j})$. But since $p_{j}$ is strictly to the right of and strictly below $b$, then $v$ would be to the right of $b$, which is impossible. This shows the claim. Thus, by Lemma 4.12, $j\geq d(s,p_{j})\geq k+2$. We apply 4.2 to $\operatorname{td}^{k}(s)$, $p_{j}$, $r(\operatorname{td}^{k}(s))$ and $p_{j+1}$ and get that $r(\operatorname{td}^{k}(s))$ can see $p_{j}$. Therefore, $\pi_{t}(s,k)\circ\langle r(\operatorname{td}^{k}(s)),p_{j},\dots,t\rangle$ is a valid path of length $k+1+(1+m-j)\leq m$.

Case 2: $r(\operatorname{td}^{k}(s))_{x}<r(\operatorname{bd}^{k}(s))_{x}$. Let $b$ be the leftmost vertex in $[r(\operatorname{td}^{k}(s)),r(\operatorname{bd}^{k}(s))]^{+}$ closest to the base line. Observe that $b$ is strictly to the right of $I^{k}(s)$, because $r(\operatorname{td}^{k}(s))$ is strictly to the right of $I^{k}(s)$. Since $p_{j}$ is not visible from $\operatorname{bd}^{k}(s)$, it has to be strictly to the right of and strictly above $b$. Next, we claim that no vertex $v\in I^{k}(s)$ can see $p_{j}$. If one could, by 4.1, we would have $v\in I(p_{j})$. But since $p_{j}$ is strictly to the right of and strictly above $b$, then $v$ would be to the right of $b$, which is impossible. This shows the claim. Thus, by Lemma 4.12, $j\geq d(s,p_{j})\geq k+2$. We apply 4.2 to $\operatorname{bd}^{k}(s)$, $p_{j}$, $r(\operatorname{bd}^{k}(s))$ and $p_{j+1}$ and get that $r(\operatorname{bd}^{k}(s))$ can see $p_{j}$. Therefore, $\pi_{b}(s,k)\circ\langle r(\operatorname{bd}^{k}(s)),p_{j},\dots,t\rangle$ is a valid path of length $k+1+(1+m-j)\leq m$. ∎

4.3 Routing Scheme

Labels and routing tables

Let $v$ be a vertex. The label of $v$ consists of its $x$- and $y$-coordinate as well as the bounding $x$-coordinates of $I(v)$. We do not need $v_{\operatorname{id}}$ since $(v_{x},v_{y})$ identifies the vertex in the network. Thus, $\operatorname{Lab}(n)=4\cdot\lceil\log n\rceil$ since we can assume that $v_{x},v_{y}\in\{0,\dots,n-1\}$. In the routing table of $v$, we store the bounding $x$-coordinates of $I^{2}(\operatorname{bd}(v))$ as well as the bounding $x$-coordinates of $I^{2}(\operatorname{td}(v))$. Furthermore, we store $(\operatorname{bd}^{2}(v)_{x},\operatorname{bd}^{2}(v)_{y},bit)$ where $bit$ indicates whether $\operatorname{td}(v)$ or $\operatorname{bd}(v)$ is on the path $\pi_{b}(v,2)$. Thus, $\operatorname{Tab}(n)=6\cdot\lceil\log n\rceil+1$.

The routing function.

We are given a current vertex $s$ together with its routing table and link table, the label of a target vertex $t$, and a header. If $t\in N(s)$, then $\operatorname{lab}(t)$ is in the link table of $s$, and we send the data packet directly to $t$. If the header is non-empty, it will contain the coordinates of exactly one vertex visible from $s$. We clear the header and go to this respective vertex. The remaining discussion assumes that the header is empty and that $t\not\in N(s)$. The routing function now distinguishes four cases depending on whether $t\in I(s)$, $t\in I^{2}(s)$ or $t\in I^{3}(s)$. We can check the first and the second condition locally, using the link table of $s$ as well as the label of $t$ (note that from the link table of $s$, we can deduce $a^{*}(s)$ and $b^{*}(s)$, and their interval boundaries). To check the third condition locally, we use Lemma 4.11 which shows that $I^{3}(s)=I^{2}(\operatorname{bd}(s))\cup I^{2}(\operatorname{td}(s))$. Since we stored the bounding $x$-coordinates of these two intervals in the routing table of $s$, we can check $t\in I^{3}(s)$ easily.

Case 1 $(t\in I(s)\setminus N(s))$: if $\operatorname{fd}(s,t)$ is a vertex, we can determine it by using the link table and the label of $t$. The packet is sent to $\operatorname{fd}(s,t)$. If $\operatorname{fd}(s,t)$ is not a vertex, we determine $\operatorname{nd}(s,t)$ and send the packet there. The header remains empty.

Case 2 $(t\in I^{2}(s)\setminus I(s))$: there is an $i\geq 1$ with $t\in[\ell^{i},\ell^{i-1}]$ or $t\in[r^{i-1},r^{i}]$. We find $i$ using the link table and $\operatorname{lab}(t)$. The packet is sent to $a^{i}$ or $b^{i}$. The header remains empty.

Case 3 $(t\in I^{3}(s)\setminus I^{2}(s))$: if $t\in I^{2}(\operatorname{bd}(s))$, we send the packet to $\operatorname{bd}(s)$. Otherwise, $t\in I^{2}(\operatorname{td}(s))$, and we send the packet to $\operatorname{td}(s)$. In both cases, the header remains empty.

Case 4 $(t\notin I^{3}(s))$: in the routing table we find the entry $(\operatorname{bd}^{2}(s)_{x},\operatorname{bd}^{2}(s)_{y},bit)$. We store $(\operatorname{bd}^{2}(s)_{x},\operatorname{bd}^{2}(s)_{y})$ in the header and send the packet to $\operatorname{bd}(s)$ or $\operatorname{td}(s)$, whichever is indicated by $bit$.

Analysis.

Obviously, $H(n)=2\cdot\lceil\log n\rceil$. It remains to analyze the stretch factor. For this, we show that after one or two steps, the distance to the target vertex has decreased by at least one. This immediately gives a stretch factor of $2$.

Lemma 4.15**.**

Let $s,t\in V(P)$. After at most two steps of the routing scheme from $s$ with target label $\operatorname{lab}(t)$, we reach a vertex $v$ with $d(v,t)\leq d(s,t)-1$.

Proof.

First, if $t\in N(s)$, then we take one hop and decrease the distance to [math].

Second, suppose that $t\in I(s)\setminus N(s)$. If $\operatorname{fd}(s,t)$ is not a vertex, the next vertex is $\operatorname{nd}(s,t)$, which is on a shortest path from $s$ to $t$ due to Lemma 4.5. Otherwise, $\operatorname{fd}(s,t)$ is the next vertex. If $\operatorname{fd}(s,t)$ is on a shortest path from $s$ to $t$, we are done. Otherwise, $t$ is not visible from $\operatorname{fd}(s,t)$, so $\operatorname{fd}^{2}(s,t)$ has to be the second vertex on the routed path. By Lemma 4.6, we have $d(\operatorname{fd}^{2}(s,t),t)=d(s,t)-1$.

Third, if $t\in I^{2}(s)\setminus I(s)$, there is an $i\geq 1$ such that the next vertex is either $a^{i}$ or $b^{i}$. By Lemma 4.8, this vertex is on a shortest path.

Fourth, assume $t\in I^{3}(s)\setminus I^{2}(s)$. Let $v_{1}$ and $v_{2}$ be the next two vertices on the routing path. We use Lemma 4.10 and Lemma 4.14 to conclude $d(v_{1},t)\leq d(s,t)$, as $v_{1}$ is either $\operatorname{td}(s)$ or $\operatorname{bd}(s)$. Due to the construction of the routing function, we have $t\in I^{2}(v_{1})\setminus I(v_{1})$. Thus, there is an $i\geq 1$, such that $v_{2}=a^{i}(v_{1})$ or $v_{2}=b^{i}(v_{1})$. By Lemma 4.8, the vertex $v_{2}$ is on a shortest path from $v_{1}$ to $t$ and we can conclude $d(v_{2},t)=d(v_{1},t)-1\leq d(s,t)-1$.

Last, assume $t\notin I^{3}(s)$. Then, the packet is routed to a vertex $p\in\{\operatorname{bd}(s),\operatorname{td}(s)\}$, whichever is on a shortest path to $\operatorname{bd}^{2}(s)$, and then $\operatorname{bd}^{2}(s)$. Lemma 4.10 and Lemma 4.14 give $d(\operatorname{bd}^{2}(s),t)\leq d(s,t)-1$. ∎

A more detailed analysis gives that the label size can be reduced to $3\cdot\lceil\log n\rceil-1$, whereas the routing table size can be reduced to $5\cdot\lceil\log n\rceil-2$. However, this will not affect our second main result which follows from the discussion above.

Theorem 2 (restated).

Let $P$ be a double histogram with $n$ vertices. There is a routing scheme for $G(P)$ with routing table, label and header size $O(\log n)$, such that we can route between any two vertices with a stretch at most $2$.

5 Conclusion

We gave the first routing schemes for the hop-distance in simple polygons. In particular, we have a routing scheme for simple histograms with label size $2\cdot\lceil\log n\rceil$, routing table size 1, and stretch 1. We also presented a routing scheme for double histograms with label, routing table and header size $O(\log n)$ and stretch 2. This constitutes a first step towards an efficient routing scheme for the hop-distance in orthogonal polygons. The following open problems arise naturally.

First of all, the routing scheme for double histograms shows that it is possible to obtain a routing scheme for simple histograms with label size $\lceil\log n\rceil$. The stretch factor increases to $2$. The basic idea is as follows: if $t\in I(s)$, we determine the far dominator $\operatorname{fd}(s,t)$ and take the hop to $\operatorname{fd}(s,t)$, without looking at the breakpoint of the near dominator. Therefore, we save the $\lceil\log n\rceil$ bits that were necessary to store the $\operatorname{id}$ of the breakpoint. It remains open whether one can decrease the stretch simultaneously.

As a next step, it would be interesting to see how the routing scheme extends to monotone polygons as well as arbitrary orthogonal polygons, assuming $r$-visibility.

After that, it will be interesting to take a closer look at (orthogonal) polygons assuming $l$-visibility. Here, the structure of visibility – even in simple histograms – is much more complicated. Moreover, we can no longer assume integer coordinates.

Last but not least, it would be interesting to know, whether it is possible to decrease the stretch in double histograms to, say $1+\varepsilon$, for $\varepsilon>0$.