## Abstract

A data structure is presented for point location in connected planar subdivisions when the distribution of queries is known in advance. The data structure has an expected query time that is within a constant factor of optimal. More specifically, an algorithm is presented that preprocesses a connected planar subdivision of size and a query distribution to produce a point location data structure for . The expected number of point-line comparisons performed by this data structure, when the queries are distributed according to , is where is a lower bound on the expected number of point-line comparisons performed by any linear decision tree for point location in under the query distribution . The preprocessing algorithm runs in time and produces a data structure of size . These results are obtained by creating a Steiner triangulation of that has near-minimum entropy.

## 1Introduction

The planar point location problem is the classic search problem in computational geometry. Given a planar subdivision ,^{1}^{2}

The history of the planar point location problem parallels, in many ways, the history of binary search trees. After a few initial attempts [11], asymptotically optimal (and quite different) linear-space query time solutions to the planar point location problem were obtained by Kirkpatrick [17], Sarnak and Tarjan [24], and Edelsbrunner *et al.* [12] in the mid 1980s. These results were based on hierarchical simplification, data structural persistence, and fractional cascading, respectively. All three of these techniques have subsequently found many other applications. An elegant randomized solution, combining aspects of all three previous solutions, was later given by Mulmuley [20], and uses randomized incremental construction, a technique that has since become pervasive in computational geometry [10]. Preparata [22] gives a comprehensive survey of the results of this era.

In the 1990s, several authors became interested in determining the exact constants achievable in the query time. Goodrich *et al.* [13] gave a linear-size data structure that, for any query, requires at most point-line comparisons and conjectured that this query time was optimal for linear-space data structures.^{3}

Still not done with the problem, several authors considered the point location problem under various assumptions about the query distribution. All these solutions compare the expected query time to the *entropy bound*; in a planar subdivision with faces , if is the probability that is contained in , then no algorithm that makes only binary decisions can answer queries using an expected number of decisions that is fewer than

In the previous results on planar point location, none of the query times are affected significantly by the structure of ; they hold for arbitrary planar subdivisions. However, when studying point location under a distribution assumption the problem becomes more complicated and the results become more specific. A *connected subdivision* is a planar subdivision whose underlying (vertex and edge) graph is connected. A *convex subdivision* is a planar subdivision whose faces are all convex polygons, except the outer face, which is the complement of a convex polygon. A *triangulation* is a convex subdivision in which each face has at most 3 edges on its boundary.

Note that, if every face of has a constant number of sides, then can be augmented, by the addition of extra edges, so that it is a triangulation without increasing (Equation 1) by more than a constant. Similarly, in a convex subdivision where the query distribution is uniform within each face of , the faces of the subdivision can be triangulated without increasing the entropy by more than a constant [3]. Thus, in the following we will simply refer to results about triangulations where it is understood that these also imply the same result for planar subdivisions with faces of constant size or convex subdivisions when the query distribution is uniform within each face.

Arya *et al.* [2] gave two results for the case where the query point is chosen from a known distribution where the and coordinates of are chosen independently and is a convex subdivision. They gave an space data structure for which the expected number of point-line comparisons is at most and an space data structure for which the expected number of point-line comparisons is at most . The assumption about the independence of the and coordinates of is crucial to the these results.

For arbitrary distributions that are known in advance, several results exist. Iacono [15] showed that, for triangulations, a simple variant of Kirkpatrick’s original point location structure gives a linear space, expected query time data structure. Simultaneously, and independently, Arya *et al.* [5] showed that a variant of Mulmuley’s randomized data structure also achieves expected query time. A sequence of papers by Arya *et al.* [3] has recently culminated in an space data structure for point location in triangulations with query time [6].

In the current paper, we show that, for any connected planar subdivision, there exists a data structure of size that can answer point location queries using point/line comparisons. Here, is a lower bound on the expected cost of any linear decision tree that solves this problem. Note that is often greater than the quantity defined above and this is necessarily so. To see this, consider that the problem of testing whether a query point is contained in a simple polygon with vertices is a special case of planar point location in a connected planar subdivision. However, in this special case the subdivision only has 2 faces, so . It seems unlikely that, for any simple polygon and any probability measure over , it is always possible to test in expected time if a point drawn from is contained in . Indeed, it is not hard to design a convex polygon and distribution so that the expected cost of any algebraic decision tree for point location in , under query distribution , is .

Note that all known algorithms for planar point location that do not place special restrictions on the input subdivision can be described in the linear decision tree model of computation.^{4}

We achieve our results by showing how to compute a Steiner triangulation of that has nearly minimum entropy over all possible triangulations of and then proving that the entropy of a minimum-entropy Steiner triangulation of is a lower bound on the cost of any linear decision tree for point location in . By then applying the recent result of Arya *et al.* to the Steiner triangulation we obtain upper and lower bounds that match to within a lower-order term.

A preliminary version of this paper, which dealt only with convex subdivisions, has appeared in the Proceedings of the 19th ACM-SIAM Symposium on Discrete Algorithms (SODA 2008) [9].

The remainder of this paper is organized as follows: Section 2 presents definitions and notations used throughout the paper. Section 3 shows how to compute a near-minimum-entropy triangulation of a simple polygon. Finally, Section applies these tools to obtain our point location structure for connected planar subdivisions.

## 2Preliminaries

In this section we give definitions, notation, and background required in subsequent sections.

**Interiors and Boundaries.** For a set , we denote the boundary of by and the interior of by . The closure of is denoted by .

**Triangles and Convex Polygons.** For the purposes of this paper, a *triangle* is the common intersection of at most 3 closed halfplanes. This includes triangles with infinite area and triangles having 0, 1, 2, or 3, vertices. Similarly, a *convex -gon* is the common intersection of at most closed halfplanes.

For a closed region , a *triangulation* of is a set of triangles whose interiors are pairwise disjoint and whose union is . We use the convention that, unless is explicitly mentioned, the triangulation in question is a triangulation of . This definition of a triangulation is often referred to as a *Steiner triangulation* since it allows vertices of the triangles to be anywhere in , and not at some finite predefined set of locations.

**Simple Polygons, Pseudotriangles, and Geodesic Triangles.** A *(near-simple) polygon* is a closed subset of whose boundary is piecewise linear and such that is homeomorphic to an open disk. Note that this definition of a polygon implies that every bounded face of a connected planar subdivision is a polygon. Also, triangles, as defined above, are polygons. Note that near-simple polygons are slightly more general than simple polygons, for which is a simple closed curve. However, our definition is sufficiently close that algorithms designed for simple polygons continue to work with near-simple polygons.

A reflex chain in a polygon is a consecutive sequence of vertices of , where the internal angle at is at least , for all . A *pseudotriangle* is a polygon whose boundary consists of 3 reflex chains. An -convex pseudotriangle () is a pseudotriangle in which of the reflex chains consist of single line segments.

A *shortest path* between points , denoted is a curve of minimum length that is contained in and that has and as endpoints. For 3 points, , a *geodesic triangle* in , denoted is the union of all shortest paths of the form , where . Geodesic triangles are closely related to pseudotriangles. In particular, every geodesic triangle consists of a pseudotriangle and three paths joining the three convex vertices of to , , and .

**Classification Problems and Classification Trees.** A *classification problem* over a domain is a function . A -ary *classification tree* is a full -ary tree^{5}*search path* of an input in a classification tree starts at the root of and, at each internal node , evaluates and proceeds to the th child of . We denote by the label of the final (leaf) node in the search path for . We say that the classification tree *solves* the classification problem over the domain if, for every , .

In this paper, we are especially concerned with *linear decision trees*. These are binary classification trees for a problem over the domain . Each internal node of a linear decision tree contains a linear inequality , and the node evaluates to 1 or 0 depending on whether the query point satisfies the inequality or not, respectively. Geometrically, each internal node of is labelled with a directed line and the decision to go to the left or right child depends on whether is to the left or right (or on) this line. An immediate consequence of this is that, for each leaf of , the closure of is a convex polygon.

**Probability.** Throughout this paper is a probability measure over that represents the query distribution. The notation denotes the probability of event under the probability measure . The notation denotes the conditional probability of given , i.e., . For any set , we use the shorthand to denote .

For a set of subsets of , we define the *induced entropy of *, denoted by as . For two sets with , the *joint entropy* of and , is . It is well-known that (see, for example, Gray [14]).

We will sometimes abuse terminology slightly by referring to a triangulation of as a *partition* of into triangles, although strictly speaking this is not true since the triangles in are closed sets that overlap at their boundaries. We will then continue the abuse by computing the induced entropy of . This introduces a technical difficulty in that and inequality is possible if there exists sets such that the area of is 0 and . To avoid this technical difficulty, we will assume that is *nice* in the sense that, if the area of is 0 then . This implies that, for every in . This assumption will avoid lengthy technical but uninteresting cases in our analysis. In practice, this problem can be avoided by using a symbolic perturbation of the query point.

The probability measures used in this paper are usually defined over . We make no assumptions about how these measures are represented, but we assume that an algorithm can, in constant time, perform each of the following two operations:

given a triangle , compute , and

given a triangle and a point at the intersection of two of ’s supporting lines, compute a line that contains and that partitions into two open triangles and such that .

Requirement 2 is used only for convenience in describing our data structure. It is not strictly necessary, but its use greatly simplifies the exposition of our results. To eliminate requirement 2, one can use the same method described by Collette *et al.* [9].

For a classification tree that solves a problem and a probability measure over , the *expected search time* of is the expected length of the search path for when is drawn at random from according to . Note that, for each leaf of there is a maximal subset such that the search path for any ends at . Thus, the expected search time of (under distribution ) can be written as

where denotes the leaves of and denotes the length of the path from the root of to . For any tree we use to denote the vertices of .

The following theorem, which is a restatement of (half of) Shannon’s Fundamental Theorem for a Noiseless Channel [25], is what all previous results on distribution-sensitive planar point location use to establish their optimality:

Theorem ? is typically applied to the point location problem by treating point location as the problem of classifying the query point based on which face of contains it. In this way, we obtain the lower bound in (Equation 1).

## 3Minimum Entropy Triangulations

Let be a simple polygon with vertices, denoted as they occur, in counterclockwise order, on the boundary of . We will show how to find a triangulation of that has near-minimum entropy. That is, we will find a triangulation such that is near-minimum over all triangulations of . In order to shorten the formulas in this section, we will implicitly condition the distribution on . More precisely, throughout this section the notation should be treated as shorthand for .

### 3.1The Triangulation

Our triangulation algorithm is recursive and takes as input a polygon and a reflex chain on the boundary of . If is a triangle, then there is nothing to do, so the algorithm outputs and terminates. Otherwise, the algorithm first selects a point on the boundary of and adds all the edges of the geodesic triangle to the triangulation . Observe that removing from disconnects into components where is a polygon that shares a reflex chain with the pseudotriangle (see Figure ). The point is selected in such a way that, for all , .^{6}

To complete the triangulation all that remains is to partition into triangles. To do this, we first partition into at most one triangle and three 2-convex pseudotriangles as shown in Figure .a. Let be the connected component of that contains . To complete the triangulation we will partition into triangles, for each , using a recursive algorithm. This algorithm selects an edge of the reflex chain in and extends in both directions until it reaches the boundary of (see Figure .b). The resulting line segment partitions into a triangle , and two 2-convex pseudotriangles and that are triangulated recursively. At the same time, is partitioned into up to 4 pieces (see Figure .c):

the triangle , and

a subpolygon incident to ,

The two connected components and of that contain and , respectively.

The edge is selected so that for each .^{7}

(a) | (b) | (c) |

### 3.2The -Tree

In order to study the entropy of the triangulation defined above, we will impose a tree structure on the pieces of induced by the triangles in . The -tree for is a tree whose nodes are subpolygons of and which has the property that, for any node that is the child of a node , .

The tree has three different kinds of nodes, called P-nodes, T-nodes, and Q-nodes. The root of is the polygon and is a *P-node*. The root of has the following children (defined in terms of the construction algorithm in the previous section; see Figure ):

Each subpolygon whose boundary does not share a segment with is a child of and is a P-node.

The subpolygon is a child of and is called a

*T-node*.

The subpolygon has three children that are called *Q-nodes*. The subtree rooted at is a ternary tree corresponding to the recursive partitioning of and done by the algorithm. The leaves of this subtree are P-nodes and the internal nodes of this subtree are Q-nodes. Each internal node has up to 3 children, up to 1 of which may be a P-node corresponding to a subpolygon and up to two of which may be Q-nodes.

Note that the above definition yields a tree whose leaves are P-nodes that correspond to the subpolygons obtained by removing from . The subtree rooted at each such leaf is obtained recursively from the recursive triangulation of .

Now that we have defined the tree , we study some of its properties. Our first lemma says that does a good job of splitting based on its probabilities.

Assume , otherwise there is nothing to prove. Let be the fourth node on the path from the root, , of to . Let be the path in from to . If has at least two P-nodes, then . Otherwise, the second node in is a T-node followed by 2 Q-nodes. By construction, for any Q-node whose parent is a Q-node , . Therefore, . If , then and the proof is complete. Otherwise, apply the same argument inductively on the path from to , to obtain

and this completes the proof.

Our next lemma says that a single line segment does not intersect very many high probability triangles in .

There are 3 types of nodes in whose interiors intersect : Type 1 nodes contain one endpoint of in their interior, Type 2 nodes contain both endpoints of in their interior, and Type 0 nodes contain no endpoints of in their interior. Notice that each level of contains at most 1 Type 2 node or 2 Type 1 nodes, so the total number of Type 1 and Type 2 nodes at distance at most from the root is at most . Thus, all that remains is to bound the number of Type 0 nodes whose distance from the root is at most .

Let be any P-node such that does not contain either endpoint of . Since is a P-node, there is a reflex chain on the boundary of that is a shortest path between two points on the boundary of , and every path in from to intersects . Stated another way, the interior of a P-node does not intersect unless it contains at least one endpoint of . Therefore, all Type 0 nodes are either T-nodes or Q-nodes.

For every Type 0 node , there is a path in from to a T-node that is adjacent to a Type 1 or Type 2 node. Furthermore, the path consists of followed by 0 or more Q-nodes, and terminates with a T-node. Looking more closely at the definition of Q-nodes, we see that two sibling Q-nodes and are not mutually visible, i.e., there is no line segment that intersects both and .

All of this implies that each of the at most Type 1 or Type 2 nodes is adjacent to at most 1 Type 0 T-node, and this T-node is the endpoint of at most 3 paths of Type 0 Q-nodes. Each such path is of length at most . Therefore, the total number of nodes in that intersect is at most .

### 3.3Minimum-Entropy Triangulation

Next, we show that the triangulation defined above is nearly-minimum entropy over all possible triangulations of . We do this by developing a technique for lower-bounding the entropy of one triangulation in terms of the entropy of another triangulation. We then show how to apply this technique to lower bound the entropy of any triangulation in terms of the entropy of .

To obtain lower bounds on the entropy of a triangulation , consider the following easily proven observation: If each triangle in intersects at most triangles of some triangulation then .^{8}

Intuitively, Lemma ? can be thought of as follows: If we tell an observer which of the a point drawn according to occurs in then the amount of information we are giving the observer about the experiment is at most . However, after giving away this information, we are able to apply the simple observation in the previous paragraph, since each triangle in intersects at most elements of each . Thus, Lemma ? is really just applications of the simple observation. The following proof formalizes this:

and this completes the proof.

The remainder of our argument involves partitioning the triangles of into subsets and then showing that and are not too big. To help us, we will use the -tree . For a node in with children , let be the portion of not covered by ’s children. Note that is always either the empty set or is a triangle in (see Figure ). In fact, for every triangle , there is exactly one such that , and for every such that is non-empty there is exactly one such that . This implies that^{9}

For a node , we define is the probability that a point drawn from is contained in .

Next we apply Lemma ? to obtain a lower bound on the entropy of any triangulation .

Let be the -tree for . Partition the nodes of into *groups* where

In the following we will fix a value , , to be defined later. A group is *large* if it contains at least elements, otherwise is *small*. Let denote the index set of the large groups, i.e., . Let be the index set of the small groups.

Note that, for any group , Lemma ? ensures that all elements of have depth at most in . Therefore, Lemma ? ensures that any triangle of intersects at most triangles of . Therefore, applying Lemma ? with , we obtain:

Thus, all that remains is to bound the contribution of the last two terms on the right hand side of (Equation 2). First,

where the last equality follows from Jensen’s Inequality. Finally, we show that the contribution of is at most .

where the last equality is obtained using the Taylor series expansion for to obtain the inequality for close to 0. Continuing, we get

Where the last inequality is obtained by setting . Thus, we have shown that

which implies that . Applying this to the right hand side of (Equation 3) yields , completing the proof.

Lemma ? shows that the triangulation defined previously is nearly minimum-entropy over all triangulations of . The following theorem gives an algorithmic version of Lemma ?.

We show how the construction of the triangulation described in Section can be modified to run in time. When constructing the first step is to find the third vertex of the geodesic triangle . This can be accomplished in time by computing the shortest path trees from and to all other vertices of and using these to find . For an example of a similar computation, see Bose *et al.* [7].

Next, is split into three 2-convex pseudotriangles , which is easily accomplished in time. The last step, before recursing, is to triangulate each of . This step can be accomplished in time using a 2-sided exponential searching trick that was used by Mehlhorn [19] in the construction of biased binary search trees (see also, Collette *et al.* [9]).

Finally, the algorithm recurses on each of the pieces . In this way, we obtain a divide-and-conquer algorithm for constructing . Unfortunately, this algorithm may have running time since there is no bound significantly smaller than on the size of an individual subproblem . To overcome this, before recursing on a subproblem we check if it contains more than vertices. If so, then rather than recursing normally on we choose a geodesic triangle , one of whose sides is the reflex chain and such that removing from leaves a set of subpolygons each with at most vertices. This modification then yields an algorithm whose recursion tree has depth and at which the work done at each level is , so the total running time of this algorithm is .

Note that this algorithm yields a triangulation that is different from . In particular, there may exist one with . Despite this, all the proofs of Lemmas ?– ? continue to hold almost without modification. The only difference occurs in Lemma ?, which now only guarantees a bound of on the number of black edges, but this has almost no effect on subsequent computations.

Finally, to see that contains triangles, we count the different types of edges used in the triangulation . Some of these edges are edges of , of which there are at most . Some of these edges are edges of geodesic triangles, which always connect two vertices of and do not cross each other, so there are at most of these. The remaining edges are used to triangulate the interiors of pseudotriangles. A pseudotriangle that has vertices is triangulated using edges. Since the total number of vertices in all pseudotriangles is at most , this means that there are at most edges used to triangulate pseudotriangles. Therefore, the total number of edges used by triangles in , and hence the number of triangles in , is .

## 4Point Location in Simple Planar Subdivisions

Next we consider the problem of point location in simple subdivisions. The following theorem of Arya *et al.* [6] shows that a low entropy triangulation can be used to make a good point location structure.

The following lemma shows that the entropy of a minimum-entropy triangulation gives a lower bound on the cost of any point location structure.

Each leaf of has a region that is a convex polygon. If has sides then the depth of in is at least . To obtain the tree replace each such leaf of by a balanced binary tree of depth by repeatedly splitting the leaf into two children and whose regions have and vertices. For a leaf , let denote the set of leaves in in the subtree of . Then

where the last inequality is an application of Jensen’s Inequality.

Lemma ? says that for any linear decision tree for point location, there is an underlying triangulation. The entropy of this triangulation gives a lower bound on the cost of the decision tree. Thus, the entropy of a minimum entropy triangulation gives a lower bound on the expected cost of any linear decision tree for point location.

Keeping the above in mind, our point location structure is simple. Let be a connected planar subdivision whose faces are and let be a probability measure over . We assume, without loss of generality that the outer face of is the complement of a triangle, since otherwise we can add at most 3 vertices and 4 edges to to make this true. Adding these edges will not increase the entropy the minimum weight triangulation of by more than a constant. With this assumption, testing if the query point is in the outer face of can be done using 3 linear comparisons after which we may safely assume that the query point is contained in an internal face of .

We triangulate each internal face of (a near-simple polygon) using Theorem ? to obtain a triangulation . The union of all is a triangulation of , to which we apply Theorem ? to obtain a point location structure for point location in and hence also in . The following theorem shows that is nearly optimal:

The space and preprocessing requirements follow from Theorem ? and Theorem ?. To prove the bound on the expected query time, apply Lemma ? to the tree and consider the resulting tree , each of whose leaves have regions that are triangles and such that

Observe that each leaf of corresponds to a triangle in that is completely contained in one of the faces of . Let denote this set of triangles and let denote the subset of contained in . Consider the entropy of the distribution induced by the leaves of :

Similarly, the entropy of is given by

By Theorem ?, the triangles in form a nearly-minimum entropy triangulation of . More specifically,

Putting this all together, we have

Finally, since we preprocess using Theorem ?, the expected number of comparisons required to answer a query is

and this completes the proof, and the paper.

### Footnotes

- A
*planar subdivision*is a partitioning of the plane into points (called*vertices*), open line segments (call*edges*), and maximal connected 2-dimensional regions (called*faces*). - In the degenerate case where is a vertex or contained in an edge of any face incident on that vertex/edge may be returned as an answer.
- Here and throughout, logarithms are implicitly base 2 unless otherwise specified.
- Although significant breakthroughs have recently been made in this area [8], we deliberately do not survey algorithms that require the vertices of the subdivision to be on integer coordinates.
- A full -ary tree is a rooted ordered tree in which each non-leaf node has exactly children.
- The existence of such a point is readily established by a standard continuity argument; see Bose
*et al.*[7] for an example. - The existence of such an edge is assured by yet another continuity argument.
- Proof: Consider the set . Each triangle of contributes at most pieces to , so we have .
- Here, and throughout the remainder, we slightly abuse notation by using the convention that .

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