International Journal of Computational Geometry & Applications c World Scientific Publishing Company

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1 International Journal of Computational Geometry & Applications c World Scientific Publishing Company FINDING SIMPLICES CONTAINING THE ORIGIN IN TWO AND THREE DIMENSIONS KHALED ELBASSIONI Max-Planck-Institut für Informatik, Saarbrücken, Germany elbassio@mpi-inf.mpg.de AMR ELMASRY Max-Planck-Institut für Informatik, Saarbrücken, Germany and Datalogisk Institut, University of Copenhagen, Denmark elmasry@mpi-inf.mpg.de KAZUHISA MAKINO Department of Mathematical Informatics, University of Tokyo, Japan makino@mist.i.u-tokyo.ac.jp Received (received date) Revised (revised date) Communicated by (Name) We show that finding the simplices containing a fixed given point among those defined on a set of n points can be done in O(n + k) time for the two-dimensional case, and in O(n 2 + k) time for the three-dimensional case, where k is the number of these simplices. As a byproduct, we give an alternative (to the algorithm in 4 ) O(n log r) algorithm that finds the red-blue boundary for n bichromatic points on the line, where r is the size of this boundary. Another byproduct is an O(n 2 +t) algorithm that finds the intersections of line segments having two red endpoints with those having two blue endpoints defined on a set of n bichromatic points in the plane, where t is the number of these intersections. Keywords: Enumeration; counting; simplices; output-sensitive algorithms; bichromatic problems; divide and conquer. 1. Introduction Let S R d be a given set of n points in the d-dimensional Euclidean space. Given a point Z R d, let us call a Z-simplex any minimal subset of points of S the convex hull of which contains Z. By Carathéodory theorem 5, the cardinality of any such set is at most d + 1 (under the general position assumption, this would be exactly d + 1). In the sequel, we may assume without loss of generality Z to be the origin. Parts of this paper appeared in References 1, 2, and 3. On leave from Alexandia University of Egypt. 1

2 2 K. Elbassioni, A. Elmasry and K. Makino It is a long-standing open problem, known in other polynomially-equivalent forms as the vertex enumeration or the convex hull problem, to find an algorithm for enumerating Z-simplices whose running time is polynomial in n, d and the number of Z-simplices 6,7. Since the convex hull problem has been extensively studied, assuming fixed dimension 8, it is natural to ask about the complexity of the enumeration of Z-simplices in fixed dimension. However, we only tackle the problem for d = 2 and 3, and leave out this challenging question about higher dimensions. In this paper, we give output-sensitive algorithms for finding Z-simplices: one runs in O(n+k) time for the case d = 2 and the other in O(n 2 +k) time for the case d = 3, where k is the number of such simplices. Throughout the paper, we assume that the points are in general positions, i.e., no two points are on the same line with Z and no three points are in the same plane with Z. Accordingly, we only assume simplices of cardinality 3 (triangles) for the two-dimensional case and of cardinality 4 for the three-dimensional case. To solve our Z-simplices problem in two dimensions, we map it to the problem of computing the decision boundary of a bichromatic set of points in one dimension. Bremner et al. 4 gave an optimal output-sensitive algorithm for computing the decision boundary that runs in O(n log r) time, where r is the number of points that contribute to the decision boundary. The algorithm in 4 relies on performing recursive calls to a median-finding algorithm. In Section 2.2, we give a more practical algorithm that also solves the problem in O(n log r) time by using the groupingand-querying paradigm (this paradigm was first used by Chan 13 to get an outputsensitive convex-hull algorithm, and later formalized by Nielsen 14 ). We also consider the problem of counting the Z-simplices in two dimensions. (The counting algorithm for d = 2 is a direct extension to the enumeration algorithm. This does not seem to apply for d = 3.) In 9,10,11 it was shown that, for a given set S of n points in two dimensions, counting the triangles with corners at S and containing a point Z can be done in O(n log n) time. It was furthermore shown in 9 that counting the triangles with corners at S for each of the points of S when treated as Z can be done in O(n 2 ) time. On the other hand, a lower bound of Ω(n log n) for the triangle-counting problem is known 12. In this paper, we sketch an output-sensitive counting algorithm for Z-simplices (triangles) in two dimensions that runs in O(n log (k/n)) time (which is obviously O(n log n) since k = O(n 3 )). It remains interesting to investigate whether there is a matching output-sensitive lower bound for this counting problem. One of the fundamental problems in computational geometry is the problem of reporting pairwise line-segment intersections. Balaban 15 gave an optimal algorithm that reports t intersecting pairs of n line segments in O(n log n + t) time and O(n) space. A variation of the general line-segment intersection problem is the bichromatic line-segment intersection reporting; Given a set of red segments and another of blue segments, the bichromatic intersection problem is the problem of reporting the intersections between red segments and blue segments. The case where there are no possible monochromatic intersections (intersections between segments having

3 Finding Simplices Containing the Origin 3 the same color) is a special case of the general line-segment intersection problem, and hence inherits the same O(n log n + t) time bound. Mairson and Stolfi 16 gave a simpler algorithm for such special case than that of Balaban. The problem becomes harder when monochromatic intersections exist. In this case, Agarwal 17 and Chazelle 18 showed how to report t bichromatic intersections in O(n 4/3 log O(1) n+t) time. A special case of the latter problem is when the union of each of the red and the blue segments is connected as a graph. In this case, Basch et al. 19 gave an O((n + t) log O(1) n) time algorithm that reports t such intersections. To solve our Z-simplices problem in three dimensions, we use the following bichromatic line-segment intersection problem as a subroutine: Consider a set of n points in the plane, each colored either red or blue; we need to find the intersections between line segments joining two red points and those joining two blue points. Note that the number of the line segments is Θ(n 2 ), and the number of the resulting bichromatic intersections is O(n 4 ). If we consider the possible Θ(n 2 ) red and blue line segments connecting the n-point set, our intersection problem is even a special case of that of Basch et al. 19. Applying their algorithm on these segments, we get an O((n 2 + t) log O(1) n) algorithm. In this paper, we introduce a simpler algorithm and get rid of the poly-logarithmic factor achieving an O(n 2 + t) time bound. 2. The Two-Dimensional Case In this section, we prove the following theorems: Theorem 1. Given a set S R 2 of n points, we can enumerate the Z-simplices from S in O(n + k) time, where k is the number of these simplices. Theorem 2. Given a set S R 2 of n points, we can count the Z-simplices from S in O(n log (k/n)) time, where k is the number of these simplices. In Section 2.1, we present a simple mapping from our two-dimensional problem to the problem of finding the bichromatic boundary of a set of bichromatic points in one dimension. (An alternative mapping was used in 20.) In Section 2.2, we give an O(n log r) algorithm that finds the bichromatic boundary points, whose size is r, for a set of n bichromatic points in one dimension. In Section 2.3, we show how to efficiently count the Z-simplices and as well prove Theorems 1 and A mapping to a sequence of zeros and ones Let l be an arbitrary line passing by Z but not by any other point of S. Let l be a line parallel to l and above all the points of S (all points lie on one side of l ). The points of S can be represented as a sequence S {0, 1} n of zeros and ones by projecting them on l as follows. The line segment joining every point p with Z is extended to meet l. Each point p below l is mapped to a zero, and each point p above l is mapped to a one. Together with every point is associated the angle whose extended line segment, joining it to Z, forms with l ; this angle indicates the

4 4 K. Elbassioni, A. Elmasry and K. Makino relative order of the point in S. This mapping is done in linear time; it is crucial to note that the sequence S is not in accordance given in sorted order (only the angles are given). It is straightforward to see that Z-simplices of S are in one-toone correspondence with subsequences of the sorted sequence of S, when sorted by angle, that have one of the two forms 010, 101. We use the following proposition in the proof of Lemma 1. Proposition 1. Let f(x, y) = c + c x x + c y y + c xy xy be a real-valued function of the two variables x and y, where c, c x, c y, c xy are real constants such that c xy 0. If x 1 and y 1, then min{f(x + y 1, 1), f(1, x + y 1)} f(x, y). Proof. Let δ 1 0 and δ 2 0 and note that f(x + δ 1, y δ 1 ) f(x, y) = δ 1 [c xy (δ 1 + x y) + c y c x ], (1) f(x δ 2, y + δ 2 ) f(x, y) = δ 2 [c xy (δ 2 + y x) + c x c y ]. (2) Since c xy 0, then it is not possible that both c xy (δ 1 + x y) + c y c x < 0 and c xy (δ 2 + y x) + c x c y < 0, for otherwise we get a contradiction c xy (δ 1 + δ 2 ) < 0. If c xy (δ 1 + x y) + c y c x 0, it follows from (1) that f(x + δ 1, y δ 1 ) f(x, y). Otherwise, it follows from (2) that f(x δ 2, y + δ 2 ) f(x, y). The proposition is concluded by taking δ 1 = y 1 and δ 2 = x 1. Consider the maximal blocks of consecutive zeros and consecutive ones within the sorted sequence of S. Let us denote by B1, 0 B1, 1 B2, 0 B2, 1..., Br 0, Br 1 the maximal blocks within this sorted sequence, where Bi 0 denotes a block of zeros, and Bi 1 denotes a block of ones. Also, let b 0 i = B0 i and b1 i = B1 i, for i = 1,..., r, be the sizes of these blocks, where b 0 1 0, b 1 r 0, and b j i > 0 for other values of i and j. The following lemma bounds the number of Z-simplices from below as a function of the number of maximal blocks in the sorted sequence of S. Lemma 1. Given a sequence S {0, 1} n corresponding to points in S, where the number of the maximal blocks of consecutive zeros (ones) in the sorted sequence of S is r, the number of Z-simplices defined by S is Ω(nr 2 ) when r 3. Proof. We claim that the number of Z-simplices is minimum when each block, but one, contains exactly one point. We start with a sequence S that has r maximal blocks of consecutive zeros, and transform it through several steps, that do not increase the number of Z-simplices, into another sequence that has the same number of blocks and with the claimed properties. Each step will involve moving points within one or two blocks, B j i Consider any two distinct blocks B j i and B j i simplices in terms of b j i and Bj i, where i, i {1,..., r} and j, j {0, 1}.. We can write the number of Z- and bj i as c 0+c 1 b j i +c 2 b j i +c 3 b j i bj i, where c 0, c 1, c 2 and c 3. Now, apply Proposition have non-negative values and depend on neither b j i nor bj i 1 with x b j i and y bj i to conclude that the minimum number of Z-simplices is

5 Finding Simplices Containing the Origin 5 achieved when b j i = 1 or bj i = 1. In other words, there is a choice of indices i and to Bj i, we do not increase the total j such that if we move all but one point of B j i number of Z-simplices. This establishes our claim. What remains is to count the Z-simplices in this case, which is a lower bound on their number for any sequence. Call the block that has n 2r+1 points the long block. First, consider the count of Z-simplices that have one point from this long block. There are Θ(r 2 ) possibilities for selecting two blocks out of the possible r indices other than the long block. Since r 3, this accounts for a total of (n 2r + 1)Θ(r 2 ) for these Z-simplices. Next, consider the Z-simplices that have none of the three points from the long block. There are Θ(r 3 ) possibilities for selecting these points from three blocks other than the long block. Hence, the total count of Z-simplices is (n 2r+1)Θ(r 2 )+Θ(r 3 ). If r = Θ(n), then the term Θ(r 3 ) implies that the number of Z-simplices is Θ(nr 2 ). If r = o(n), then the term (n 2r + 1)Θ(r 2 ) implies the same bound of Θ(nr 2 ). It follows that the number of Z-simplices is Ω(nr 2 ) Identifying the zero-one blocks Given the sequence S {0, 1} n corresponding to the points of S, we need to get the maximal blocks of consecutive zeros and consecutive ones, by identifying which points are in which blocks, within the sorted-by-angle sequence of S. To do that, we do not have to sort the points within the blocks, instead we need to get the points contributing to the boundary. This is exactly the problem of finding the bichromatic boundary for a set of bichromatic points on the line 4. We give next an O(n log r) time algorithm that does not involve median finding. (1) Partition S into a sequence of zeros S and a sequence of ones S. (2) Split each of S and S into at most n/m arbitrary groups each has at most m points (m is a parameter that will be determined later). (3) Sort each of these groups by angle (using O(m log m) per group). (4) Repeat at most m times alternating between the groups of S and S (4.1.) Find the point with the smallest angle among the groups of one sequence. (4.2.) Use binary search to find and remove all the points having a smaller angle within each of the groups of the other sequence. These points constitute the next block in our output. The above procedure is designed to run in O(n log m) time. If we set m to r, this procedure partitions the sequence correctly in O(n log r) time. Unfortunately, we do not know in advance the value of r. To overcome this problem, we call the above procedure repeatedly with m = 2 2i (i = 1, 2,... ), every time checking if the procedure terminates after partitioning the input sequence or not. It follows that (assuming r 2) the running time is O( log log r i=1 n2 i ) = O(n log r).

6 6 K. Elbassioni, A. Elmasry and K. Makino 2.3. Enumerating and counting Z-simplices Given the sequence S {0, 1} n corresponding to the points of S, we apply the above algorithm to identify the r blocks of consecutive zeros and consecutive ones in O(n log r) time. In accordance, enumerating the Z-simplices is straightforward and can be done in O(k) time. If r 2, we spend O(n) time to realize that there are no such Z-simplices. If r 3, the time spent by the algorithm is O(n log r + k), which is O(k) using Lemma 1. This establishes the proof of Theorem 1. For the counting problem, we need to count the number of subsequences of the form 010 and 101, which can be obtained in O(n) time once the zero-one block partitioning is available (see 9,11 for more details). Let us, without loss of generality, concentrate on counting the 010 subsequences. To do this, we scan the processed sequence of S (after applying the partitioning algorithm of Section 2.2) from left to right, computing the sizes of maximal blocks of consecutive zeros and ones b j i, for all i {1,..., r} and j {0, 1}. The number of 010 subsequences is given as 1 i 1 i 2<i 3 r b0 i 1 b 1 i 2 b 0 i 3. This formula can be computed in linear time as follows. Incrementally compute the prefix and postfix sums, for each i = 1,..., r 1: α 0 1 = b 0 1, α 0 i = α 0 i 1 + b 0 i, β 0 1 = r b 0 j, βi 0 = βi 1 0 b 0 i. j=2 Note that each of these values can be computed in constant time from the previously computed value. The number of 010 subsequences is precisely r 1 i=1 α0 i b1 i β0 i. Similarly, the values αi 1, β1 i can be defined for i = 1,..., r 1, from which the count of 101 subsequences can be computed. The running time of the counting algorithm is also asymptotically dominated by the bound of the partitioning. This establishes the proof of Theorem The Three-Dimensional Case In this section, we prove the following theorem: Theorem 3. Given a set S R 3 of n points, we can find the Z-simplices from S in O(n 2 + k) time, where k is the number of these simplices A mapping to two-dimensional problems Given a set S of n points and a point Z R 3, we perform the following mapping to reduce the dimensionality (as we did in the two-dimensional case). Let be an arbitrary plane passing by Z but not by any other point of S. Let be a plane parallel to and above all the points of S. The points of S can be represented as a bichromatic set S of points in two dimensions by projecting them on as follows.

7 Finding Simplices Containing the Origin 7 The line segment joining every point p with Z is extended to meet in an image point p. Each point below has its image point colored red, and each point above has its image point colored blue. This mapping can be done in linear time. Lemma 2. As a result of the above mapping, a set of four points forming a simplex that contains Z in the interior corresponds to either: (i) three points of the same color, the simplex of which contains the fourth point that has a different color, or (ii) two red points defining a red segment and two blue points defining a blue segment, and the two segments intersect. Proof. There are two possibilities for such a simplex: (i) three points p 1, p 2, p 3 lie on one side of, and one point p 4 lies on the other side; or (ii) two points p 1, p 2 lie on one side of and two points p 3, p 4 lie on the other side. We assume without loss of generality that Z = 0. First consider case (i). The point Z lies in the interior of the convex hull of p 1, p 2, p 3, p 4 if and only if there exist λ 1, λ 2, λ 3, λ 4 > 0 such that p 4 = λ1 λ 4 ( p 1 ) + λ 2 λ 4 ( p 2 ) + λ3 λ 4 ( p 3 ), or in other words, if and only if p 4 lies in the interior of the cone defined by the rays p 1, p 2, p 3. The latter condition holds if and only if the triangle formed by the intersection of the three rays p 1, p 2, p 3 with contains in its interior the point of intersection of the ray p 4 with. Next consider case (ii). The point Z lies in the interior of the convex hull of p 1, p 2, p 3, p 4 if and only if there exist λ 1, λ 2, λ 3, λ 4 > 0 such that λ 3 p 3 + λ 4 p 4 = λ 1 ( p 1 ) + λ 2 ( p 2 ), or in other words, if and only if the two cones formed by { p 1, p 2 } and { p 3, p 4 } intersect in the interior. We note that the two segments formed by the intersection of these two cones with cannot be collinear, for otherwise the given simplex is not full-dimensional. Furthermore, the two cones intersect in the interior if and only if these two segments also intersect in the interior. To enumerate simplices of the first type, we find for every point the twodimensional simplices that contain it among the points of the other color. This can be done by applying n calls to the two-dimensional algorithm given in Section 2, for a total of an O(n 2 ) time for the n calls in addition to the time to report the simplices. To find simplices of the second type, we need to apply an algorithm for reporting bichromatic intersections of line segments defined by bichromatic points in the plane. Next, we introduce such an algorithm Finding bichromatic line-segment intersections defined by bichromatic points in the plane Consider a set of n points in R 2, each colored either red or blue. A bichromatic intersection is an intersection between a segment with two red endpoints and a segment with two blue endpoints. In this section, we give an O(n 2 +t) time algorithm that reports the t bichromatic intersections defined by the n points.

8 8 K. Elbassioni, A. Elmasry and K. Makino A main result of the geometric duality is that the angular order of a given set of n points can be produced with respect to each of these n points, all in O(n 2 ) time and storage 21. We use this result, and assume that throughout the algorithm the angular orders are known and stored by a preprocessing phase. We use a divide-and-conquer approach to solve our problem. The set of n points is divided into two sets of almost equal sizes ( n/2 and n/2 ) via a line L. This can be done by finding the median of the points with respect to their Y -coordinates. Two categories of line segments are defined: A crossing segment is a segment whose endpoints are each on a different side of L. A non-crossing segment is a segment whose two endpoints are on the same side of L. Three types of intersections are thereby possible depending on the type of the two intersecting segments: crossing/non-crossing crossing/crossing non-crossing/non-crossing We start by finding the first two types of intersections (crossing/non-crossing and crossing/crossing), and then we recursively solve the problem on each of the two sides of L to find the third type of intersections (non-crossing/non-crossing). Being able to find the first two types of intersections for a single recursive step in O(n 2 ) time plus the time required to report these intersections, the claimed O(n 2 +t) time bound is concluded following the recursive relation: T (n) = T ( n/2 ) + T ( n/2 ) + O(n 2 ) whose solution is T (n) = O(n 2 ). For an efficient implementation, instead of finding the median of the Y - coordinates of the points within every recursive call, we sort the n points with respect to their Y-coordinates in a preprocessing phase. An iterative bottom-up (considering the recursion tree) implementation is possible Finding crossing/non-crossing intersections Fixing a point x, consider the line segments that cross L and join x to the points on the other side and have the same color as x. We call these segments the segments of the cone of x. The angular order of these segments around x has been computed in the preprocessing phase. We show how to efficiently find the intersections of the segments of the cone of x with the non-crossing line segments of the other color. The points that are on the same side as x and have a different color are identified and grouped such that the points between each two consecutive segments of the cone of x are in one group. Since the angular order of the points around x has been precomputed, these groups are identified in O(n) time. A special group is the group of the points that are outside the segments of the cone of x. Except for that special group, a line segment whose two endpoints are from the same group does not intersect with the segments of the cone of x. On the other hand, a line segment whose two endpoints are from two different groups intersects with all the segments of the cone of x that lie between (with respect to the angular order around x) these

9 Finding Simplices Containing the Origin 9 p 1 x p 6 p 2 p 4 p 3 p 5 Fig. 1. crossing/non-crossing intersections two groups. Finding these intersections is straightforward using an extra O(n) time. A line segment whose both endpoints are in the special group either intersects all the segments of the cone of x or none of them. To find these intersections, we start traversing the points of the special group in a clockwise increasing angular order around x. For each such point y, we start another traversal for the points of the special group in increasing anti-clockwise angular order. Once we reach a point y in the anti-clockwise traversal that together with y forms a line segment which does not intersect the cone of x, we proceed with the clockwise direction for another point y and restart the anti-clockwise traversal. The idea is that any line segment formed by the point y and a point after y in the anti-clockwise traversal does not intersect the cone of x. It follows that, for every point y, we spend a constant time in addition to the time to report the intersections. That way, we will be able to find the bichromatic intersections for the crossing line segments of the cone of x in O(n) time plus the time to report the intersections. Repeating the procedure for every point in S instead of the point x, we use an extra O(n 2 ) time to find all the crossing/non-crossing intersections. For the example in Figure 1, the points p 3 and p 4 are in the same group, while the points p 1, p 2 and p 6 are in the special group. The line segment p 2 p 6 intersects all the line segments of the cone of x, while the line segments p 1 p 2 and p 1 p 6 have no intersections with the segments of the cone of x Finding crossing/crossing intersections As stated in Section 3.2.1, the crossing line segments joining a point to the points with the same color but on the other side form a cone of segments. Also, the order of these segments around that point is determined in a preprocessing phase. Consider any two points x and y on the same side of L and having two different colors. There are two cases depending on the position of one point with respect to the cone of segments of the other. For Case 1, one of the two points lies inside the

10 10 K. Elbassioni, A. Elmasry and K. Makino x y c b a Fig. 2. crossing/crossing intersections cone of segments of the other. Assume without loss of generality that y lies in the cone of x. For every point y that lies inside the cone of x, the segment that precedes y and the segment that follows y in the sorted angular order around x can be found and stored all in O(n) time; this can be done by one angular traversal for the points around x. Hence, a total of O(n 2 ) time is required to perform such task for all the possible pairs x and y. Once we know these two segments for a pair of points x and y, we may consider the segments of the cone of x as two cones of segments and handle Case 1 as two problems of Case 2, where none of the points is inside the cone of the other point. Next, we show how to treat such case. As shown in Figure 2, we start from the segment with the largest angle around y among the segments of the cone of y. Let a be the intersection of this segment with L. Let c be the intersection of L with the segment of the cone of x forming the smallest angle with L. We traverse the segments of the cone of y in order until we reach the segment that intersects with L in b; this is the last line segment whose intersection with L is to the right of c. Note that each of the traversed segments of the cone of y intersects at least one segment of the cone of x, and hence the time spent in traversing these segments can be covered by the time spent in reporting such intersections. Once we reach the point b, we start traversing the segments of the cone of y again but now in the other direction. This can be done by maintaining a stack that holds these segments. Simultaneous to the reversed traversal of the segments of the cone of y, we traverse the segments of the cone of x in the same direction of increasing angular order. Whenever the intersection of the segment of the cone of y, which is at the top of the stack, with L is to the right of the intersection of the current segment of the cone of x with L, we report that the current segment of the cone of x intersects with all the segments of the cone of y that are still in the stack. We then proceed traversing the next segment of x, and in accordance perform as many pops to the stack of the segments of the cone of y as necessary. That way, we will be able to find the intersections of the segments of the cone of

11 Finding Simplices Containing the Origin 11 x with those of the cone of y in an O(1) time plus the time required for reporting these intersections. Repeating the procedure for all possible pairs of points x and y, we use an extra O(n 2 ) time for finding all the crossing/crossing intersections. 4. Conclusions We gave output-sensitive algorithms for the problem of enumerating simplices containing the origin in two and three dimensions. We also gave an output sensitive algorithm for the problem of counting simplices in two dimensions. The linear-time two-dimensional enumeration algorithm is obviously optimal. It remains to either prove lower bounds or develop better algorithms for the twodimensional counting problem as well as the three-dimensional problems. It is not difficult to modify our enumeration algorithms to output a prescribed number k of simplices in time O(n+k ) for the two-dimensional case, and O(n 2 +k ) for the three-dimensional case. For the the two-dimensional case, we have to replace the algorithm for identifying the zero-one blocks, that we use as a preprocessing step, with the straightforward incremental algorithm that works in O(nr ), where r is the required number of blocks needed to produce k simplices (which can be derived from k using a more precise bound than that in Lemma 1 including the constants). It does not seem that the methods used in the paper can be generalized to higher dimensions. So, it remains interesting to obtain non-trivial lower and upper bounds on counting and enumerating Z-simplices in any fixed dimension. The problem of finding bichromatic intersections for segments defined by bichromatic points is interesting in its own right. It is quite straightforward to modify the enumeration algorithm to perform the counting task in O(n 3 ) time. The decision problem about the existence of any bichromatic intersections is much easier though; one can answer this question in linear time by checking the separability of the set of red points from the set of blue points. References 1. A. Elmasry, Distribution-sensitive set multi-partitioning, Proc. 1st International Conference on Analysis of Algorithms, (2005) pp A. Elmasry and K. Elbassioni, Output-sensitive algorithms for counting and enumerating simplices containing a given point in the plane, Proc. 17th Canadian Conference on Computational Geometry, (2005) pp A. Elmasry and K. Makino, Finding intersections of bichromatic segments defined by points, Proc. 19th Canadian Conference on Computational Geometry, (2007) pp D. Bremner, E. Demaine, J. Erickson, J. Iacono, S. Langerman, P. Morin and G. Toussaint, Output-sensitive algorithms for computing nearest-neighbour decision boundaries, Discrete and Computational Geometry, 33(4) (2005) A. Schrijver, Theory of Linear and Integer Programming, Wiley-Interscience, (1986). 6. M. Bussieck and M. Lübbecke, The vertex set of a 0/1-polytope is strongly p- enumerable, Computational Geometry: Theory and Applications, 11(2) (1998) G. Swart, Finding the convex hull facet by facet, Journal of Algorithms, 6 (1985)

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