Divide-Conquer-Glue Algorithms

Transcription

1 Divide-Conquer-Glue Algorithms Closest Pair Tyler Moore CSE 3353, SMU, Dallas, TX Lecture DIVIDE AND CONQUER mergesort counting inversions closest pair of points randomized quicksort median and selection Some slides created by or adapted from Dr. Kevin Wayne. For more information see Some code reused from Python Algorithms by Magnus Lie Hetland. 2 / 19 Closest pair problem. Given points in the plane, find a pair of points with the smallest Euclidean distance between them. Closest pair problem. Given points in the plane, find a pair of points with the smallest Euclidean distance between them. Fundamental geometric primitive. Graphics, computer vision, geographic information systems, molecular modeling, air traffic control. Special case of nearest neighbor, Euclidean MST, Voronoi. fast closest pair inspired fast algorithms for these problems Brute force. Check all pairs with Θ distance calculations. 1d version. Easy algorithm if points are on a line. Nondegeneracy assumption. No two points have the same -coordinate. 3 / / 19

2 Closest-pair problem in one dimension Closest-pair problem has divide-and-conquer solution In the closest-pair problem, you are to select the pair of points (p 1, p 2 ) from a set S that are closest to each other. S l = {l 1, l 2, l 3 } S r = {r 1, r 2, r 3 } l 1 l 2 l 3 r 3 r 1 r 2 m A divide-and-conquer algorithm works as follows. 1 Base Case: If the list contains two points, then they must be the closest pair. 2 Divide: Divide the set into two halves (e.g., S l and S r in the figure above). Put all points less than the midpoint m in S l and all points greater than or equal to the midpoint in S r. 3 Conquer: Find the closest-pair for each half ((l 1, l 2 ) for S l and (r 1, r 2 ) for S r ). 4 Glue: To find the closest pair in the entire set, select from 3 options: 1 closest pair in the left half ((l 1, l 2 )); 2 closest pair in the right half ((r 1, r 2 )); 3 a pair with one point each from S l and S r. 5 / 19 6 / 19 1D Glue procedure A divide-and-conquer algorithm works as follows. Glue: To find the closest pair in the entire set, select from 3 options: 1 closest pair in the left half ((l 1, l 2 )); 2 closest pair in the right half ((r 1, r 2 )); 3 a pair with one point each from S l and S r. For the closest pair to take a point from both sets, each point must be within distance of the midpoint m between the two sets (here = min(distance(l 1, l 2 ), distance(r 1, r 2 ))). Only the largest point in the left set l max and the smallest point in the right set r min could be closer than. Compute distance(l max,r min ) and update the closest pair if less than. Sorting solution / 19 8 / 19

3 Sorting solution. Divide. Subdivide region into 4 quadrants / / 19 Divide. Subdivide region into 4 quadrants. Obstacle. Impossible to ensure points in each piece. Divide: draw vertical line so that points on each side. Conquer: find closest pair in each side recursively. Combine: find closest pair with one point in each side. Return best of 3 solutions. seems like Θ(N 2 ) L / / 19

4 Find closest pair with one point in each side, assuming that distance <. Observation: only need to consider points within of line. Find closest pair with one point in each side, assuming that distance <. Observation: only need to consider points within of line. Sort points in -strip by their -coordinate. Only check distances of those within positions in sorted list! why 11? L L / / 19 Def. Let be the point in the -strip, with the smallest -coordinate. Claim. If, then the distance between and is at least. Pf. No two points lie in same ½ -by-½ box. Two points at least rows apart have distance ½. Fact. Claim remains true if we replace with. 2 rows i j ½ ½ ½ / / 19

5 Theorem. The divide-and-conquer algorithm for finding the closest pair of points in the plane can be implemented in time. Θ Q. How to improve to? A. Yes. Don't sort points in strip from scratch each time. Each recursive returns two lists: all points sorted by -coordinate, and all points sorted by -coordinate. Sort by merging two pre-sorted lists. Lower bound. In quadratic decision tree model, any algorithm for closest pair (even in 1D) requires Ω quadratic tests. Theorem. [Shamos 1975] The divide-and-conquer algorithm for finding the closest pair of points in the plane can be implemented in time. Θ Pf. Θ Note. See SECTION 13.7 for a randomized time algorithm / 19 not subject to lower bound since it uses the floor function / 19 2D Divide-Conquer-Glue 2D Divide-Conquer-Glue (2, 6) Step 0: Get Sorted Lists (2, 6) Divide (4, 5) (6, 5) (4, 5) (6, 5) (9, 4) (9, 4) (3, 2) (3, 2) (1, 1) (6, 1) (1, 1) (6, 1) m = 5 seq = [(4,5),(1,1),(6,5),(3,2),(6,1),(9,4),(2,6)] lftx = [(1,1),(2,6),(3,2),(4,5)], rgtx=[(6,1),(6,5),(9,4)] seqx = [(1,1),(2,6),(3,2),(4,5),(6,1),(6,5),(9,4)] lfty = [(1,1),(3,2),(4,5),(2,6)], rgty=[(6,1),(9,4),(6,5)] seqy = [(1,1),(6,1),(3,2),(9,4),(6,5),(4,5),(2,6)] 19 / 19 Conquer: invoke cpp helper recursively with left and right halves 19 / 19

6 2D Divide-Conquer-Glue (2, 6) Glue (4, 5) (6, 5) d min < R( ) (9, 4) (3, 2) (1, 1) L( ) (6, 1) m = 5 lfty = [(1,1),(3,2),(4,5),(2,6)], rgty=[(6,1),(9,4),(6,5)] ymin = [(6,1),(3,2),(6,5),(4,5)] newy = [(1,1),(6,1),(3,2),(9,4),(6,5),(4,5),(2,6)] Check 6 neighbors of every point in ymin for distance < 19 / 19