Lecture 4: Line Segment Intersection Reading: Chapter 2 in the 4M's. Geometric intersections: One of the most basic problems in computational geometry is that of computing intersections. Intersection computation in 2 and 3 space is basic to many different application areas. -- In solid modeling people often build up complex shapes by applying various boolean operations (intersection, union, and difference) to simple primitive shapes. The process in called constructive solid geometry (CSG). In order to perform these operations, the most basic step is determining the points where the boundaries of the two objects intersect. -- In robotics and motion planning it is important to know when two objects intersect for collision detection and collision avoidance. -- In geographic information systems it is often useful to overlay two subdivisions (e.g. a road network and county boundaries to determine where road maintenance responsibilities lie). Since these networks are formed from collections of line segments, this generates a problem of determining intersections of line segments. -- In computer graphics, ray shooting is an important method for rendering scenes. The computationally most intensive part of ray shooting is determining the intersection of the ray with other objects. Most complex intersection problems are broken down to successively simpler and simpler inter section problems. Today, we will discuss the most basic algorithm, upon which most complex algorithms are based. Line segment intersection: The problem that we will consider is, given n line segments in the plane, report all points where a pair of line segments intersect. We assume that each line segment is represented by giving the coordinates of its two endpoints. Observe that n line segments can intersect in as few as 0 and as many as (n choose 2) = O(n^2) different intersection points. We could settle for an O(n^2) algorithm, claiming that it is worst case asymptotically optimal, but it would not be very useful in practice, since in many instances of intersection problems intersections may be rare. Therefore it seems reasonable to look for an output sensitive algorithm, that is, one whose running time should be efficient both with respect to input and output size. Let I denote the number of intersections. We will assume that the line segments are in general position, and so we will not be concerned with the issue of whether three or more lines intersect in a single point. However, generalizing the algorithm to handle such degeneracies efficiently is an interesting exercise. See our book for more discussion of this. Complexity: We claim that best worst case running time that one might hope for is O(n log n+ I) time algorithm. Clearly we need O(I) time to output the intersection points. What is not so obvious is that O(n log n) time is needed. This results from the fact that the following problem is known to require O( n log n) time in the algebraic decision tree model of computation. Element uniqueness: Given a list of n real numbers, are all of these numbers distinct? (That is, are there no duplicates.) REDUCTION: Given a list of n numbers, (x1, x2, ..., xn ), in O(n) time we can construct a set of n vertical line segments, all having the same y coordinates. Observe that if the numbers are distinct, then there are no intersections and otherwise there is at least one intersection. Thus, if we could detect intersections in o(n log n) time (meaning strictly faster than Theta(n log n) time) then we could solve element uniqueness in faster than o(n log n) time. However, this would contradict the lower bound on element uniqueness. We will present a (not quite optimal) O(n log n + I log n) time algorithm for the line segment intersection problem. A natural question is whether this is optimal. Later in the semester we will discuss an optimal radomized O(n log n + I) time algorithm for this problem. * Note, this lower bound result assumes the algebraic decision tree model of computation, in which all decisions are made by comparisons made based on exact algebraic operations, (+,-,/,*) applied to numeric inputs. Althought this includes most "normal" geometric operations, there are alternative models... by taking mods, floors or ceilings, you can implement a hashing function which can check for element uniqueness in expected O(n) time. Even in this model, however, no-one knows an algorithm better than 0(n log n + I). Line segment intersection: In rather typical computational geometry fashion, our book does not discuss the issue of how to determine the intersection point of two line segments. Let ab and cd be two line segments, given by their endpoints. It is an easy exercise to determine whether these line segments intersect, simply by applying an appropriate combination of orientation tests. To determine the coordinates of the intersect point involves solving a small system of equations. The most natural way to set up this computation is to introduce the notion of a parametric representation of the line segment. Recall that any point on the line segment ab can be written as a convex combination involving a real parameter s: p(s) = (1 - s)a + sb for 0 <= s <= 1: Similarly for cd we may introduct a parameter t: q(t) = (1 - t)c + td for 0 <= s <= 1: An intersection occurs if and only if we can find s and t in the desired ranges such that p(s) = q(t). Thus we get the two equations: (1 - s)ax + s bx = (1 - t) cx + t dx (1 - s)ay + s by = (1 - t) cy + t dy The coordinates of the points are all known, so it is just a simple exercise in linear algebra to solve for s and t. The computation of s and t will involve a division. If the divisor is 0, this corresponds to the case where the line segments are parallel (and possibly collinear). These special cases should be dealt with carefully. If the divisor is nonzero, then we get values for s and t as rational numbers (the ratio of two integers). We can approximate them as floating point numbers, or if we want to perform exact computations it is possible to simulate rational number algebra exactly using high precision integers (and multiplying through by least common multiples). Once the values of s and t have been computed all that is needed is to check that both are in the interval [0; 1]. Plane Sweep Algorithm: Let S = (s1, s2, ..., sn) denote the line segments whose intersections we wish to compute. The method is called plane sweep. Here are the main elements of any plane sweep algorithm, and how we will apply them to this problem: Sweep line: We will simulate the sweeping of a vertical line L, called the sweep line from left to right. (Our text uses a horizontal line, but there is obviously no significant difference.) We will maintain the line segments that intersect the sweep line in sorted order (say from top to bottom). Events: Although we might think of the sweep line as moving continuously, we only need to update data structures at points of some significant change in the sweep line contents, called event points. Different applications of plane sweep will have different notions of what event points are. For this application, event points will correspond to the following: Endpoint events: where the sweep line encounters an endpoint of a line segment, and Intersection events: where the sweep line encounters an intersection point of two line segments. Note that endpoint events can be presorted before the sweep runs. In contrast, intersection events will be discovered as the sweep executes. Event updates: When an event is encountered, we must update the data structures associated with the event. It is a good idea to be careful in specifying exactly what invariants you intend to maintain. For example, when we encounter an intersection point, we must interchange the order of the intersecting line segments along the sweep line. There are a great number of nasty special cases that complicate the algorithm and obscure the main points. We will make a number of simplifying assumptions. They can be overcome through a more careful handling of these cases. (1) No line segment is vertical. (2) If two segments intersect, then they intersect in a single point (that is, they are not collinear). (3) No three lines intersect in a common point. Detecting intersections: We mentioned that endpoint events are all known in advance. But how do we detect intersection events. It is important that each event be detected before the actual intersection event occurs. Our strategy will be as follows. Whenever two line segments become adjacent along the sweep line, we will check whether they have an intersection occuring to the right of the sweep line. If so, we will add this new event (assuming that it has not already been added). A natural question is whether this is sufficient. In particular, if two line segments do intersect, is there necessarily some prior placement of the sweep line such that they are adjacent. Happily, this is the case, but it requires a proof. Figure 17: Plane sweep. Lemma: Given two segments s_{i}and s_{j}, which intersect in a single point p (and assuming no other line segment passes through this point) there is a placement of the sweep line prior to this event, such that s_{i}and s_{j}are adjacent along the sweep line (and hence will be tested for intersection). Proof: From our general position assumption it follows that no three lines intersect in a com mon point. Therefore if we consider a placement of the sweep line that is infinitessimally to the left of the intersection point, lines s_{i}and s_{j}will be adjacent along this sweepline. Consider the event point q with the largest x coordinate that is strictly less than p_{x}. The order of lines along the sweep line after processing q will be identical the the order of the lines along the sweep line just prior p, and hence s_{i}and s_{j}will be adjacent at this point. Figure 18: Correctness of plane sweep. Data structures: In order to perform the sweep we will need two data structures. Event queue: This holds the set of future events, sorted according to increasing x coordinate. Each event contains the auxiliary information of what type of event this is (segment endpoint or intersection) and which segment(s) are involved. The operations that this data structure should support are inserting an event (if it is not already present in the queue) and extracting the minimum event. It seems like a heap data structure would be ideal for this, since it supports insertions and extract min in O(log M ) time, where M is the number of entries in the queue. (See Cormen, Leiserson, and Rivest for details). However, a heap cannot support the operation of checking for duplicate events. There are two ways to handle this. One is to use a more sophisticated data structure, such as a balanced binary tree or skip list. This adds a small constant factor, but can check that there are no duplicates easily. The second is use the heap, but when an extraction is performed, you may have to perform many extractions to deal with multiple instances of the same event. Our book recommends the prior solution. If events have the same x coordinate, then we can handle this by sorting points lexico graphically by (x, y). (This results in events be processed from bottom to top along the sweep line, and has the same geometric effect as imagining that the sweep line is rotated infinitesimally counterclockwise.) Sweep line status: To store the sweep line status, we maintain a balanced binary tree or perhaps a skiplist whose entries are pointers to the line segments, stored in decreasing order of y coordinate along the current sweep line. Normally when storing items in a tree, the key values are constants. Since the sweep line varies, we need ``variable'' keys. To do this, let us assume that each line segment computes a line equation y = mx + b as part of its representation. The ``key'' value in each node of the tree is a pointer to a line segment. To compute the y coordinate of some segment at the location of the current sweep line, we simply take the current x coordinate of the sweep line and plug it into the line equation for this line. The operations that we need to support are those of deleting a line segment, inserting a line segment, swapping the position of two line segments, and determining the immediate predecessor and successor of any item. Assuming any balanced binary tree or a skiplist, these operations can be performed in O(log n) time each. The Complete Algorithm: We can now present the complete plane sweep algorithm. Plane Sweep Algorithm for Line Segment Intersection Intersect(S) : (1) Initially, we insert all of the endpoints of the line segments of S into the event queue. The initial sweep status is empty. (2) While the event queue is nonempty, extract the next event in the queue. There are three cases, depending on the type of event: Segment left endpoint: Insert this line segment into the sweep line status, based on the y coordinate of this endpoint and the y coordinates of the other segments currently along the sweep line. Test for intersections with the segment immediately above and below. Segment right endpoint: Delete this line segment from the sweep line status. For the entries immediately preceding and succeeding this entry, test them for intersections. Intersection point: Swap the two line segments in order along the sweep line. For the new upper segment, test it against its predecessor for an intersection. For the new lower segment, test it against its successor for an intersection. Analysis: The work done by the algorithm is dominated by the time spent updating the various data structures (since otherwise we spend only constant time per sweep event). We need to count two things: the number of operations applied to each data structure and the amount of time needed to process each operation. For the sweep line status, there are at most n elements intersecting the sweep line at any time, and therefore the time needed to perform any single operation is O(logn), from standard results on balanced binary trees. Since we do not allow duplicate events to exist in the event queue, the total number of elements in the queue at any time is at most 2n + I. Since we use a balanced binary tree to store the event queue, each operation takes time at most logarithmic in the size of the queue, which is O(log(2n + I)). Since I <= n^2 , this is at most O(log(n^2) ) = O(2 log n) = O(log n) time. Each event involves a constant number of accesses or operations to the sweep status or the event queue, and since each such operation takes O(log n) time from the previous paragraph, it follows that the total time spent processing all the events from the sweep line is O((2n + I) log n) = O((n + I) log n) = O(n log n + I log n): Thus, this is the total running time of the plane sweep algorithm.

This course is modeled on the Computational Geometry Course taught by Dave Mount, at the University of Maryland. These notes are modifications of his Lecture Notes, which are copyrighted as follows: Copyright, David M. Mount, 2000, Dept. of Computer Science, University of Maryland, College Park, MD, 20742. These lecture notes were prepared by David Mount for the course CMSC 754, Computational Geometry, at the University of Maryland, College Park. Permission to use, copy, modify, and distribute these notes for educational purposes and without fee is hereby granted, provided that this copyright notice appear in all copies.