Geometric Computation: Introduction. Piotr Indyk

Transcription

1 Geometric Computation: Introduction Piotr Indyk

2 Welcome to 6.850! Overview and goals Course Information Closest pair Signup sheet

3 Geometric Computation Geometric computation occurs everywhere: Robotics: motion planning, map construction and localization Geographic Information Systems (GIS): range search, nearest neighbor Simulation: collision detection Computer graphics: visibility tests for rendering Computer vision: pattern matching Computational drug design: spatial indexing

4 Computational Geometry Started in mid 70 s Focused on design and analysis of algorithms for geometric problems Many problems well-solved Many other problems remain open

5 Course Goals Introduction to Computational Geometry Classic results and techniques New directions

6 Syllabus Part I - Classic CG: Closest pair Segment intersection LP in low dimensions Polygon triangulation Range searching Point location Arrangements and duality Voronoi diagrams Delaunay triangulations Binary space partitions Motion planning and Minkowski sum Use Computational Geometry: Algorithms and Applications by de Berg, van Kreveld, Overmars, Schwarzkopf (2 nd edition). Part II - New directions in CG: Approximate nearest neighbor in low dimensions Approximate nearest neighbor in high dimensions: LSH Low-distortion embeddings Low-distortion embeddings II Geometric algorithms for external memory Geometric algorithms for streaming data Kinetic algorithms Pattern matching Combinatorial geometry Geometric optimization

7 Higher dimensions Best match for this?

8 Course Information H-level Graduate Credit Grading: 4 problem sets (see calendar): In each PSet: Core component (mandatory): style Two optional components:» More theoretical problems» Java programming assignments Can collaborate, but solutions written separately Midterm (but no final ) Prerequisites: understanding of algorithms and probability (6.046 level)

9 Closest pair (one algorithm)

10 Closest Pair Find a closest pair among p 1 p n R d Easy to do in O(dn 2 ) time For all p i p j, compute p i p j and choose the minimum We will aim for better time, as long as d is small For now, focus on d=2

11 Divide and conquer Divide: Compute the median of x- coordinates Split the points into P L and P R, each of size n/2 Conquer: compute the closest pairs for P L and P R Combine the results (the hard part)

12 Combine Let k=min(k 1,k 2 ) Observe: Need to check only pairs which cross the dividing line Only interested in pairs within distance < k Suffices to look at points in the 2k-width strip around the median line k 1 2k k 2

13 Scanning the strip Sort all points in the strip by their y- coordinates, forming q 1 q t, t n. Let y i be the y-coordinate of q i k min = k For i=1 to t j=i-1 While y i -y j < k If q i q j <k min then k min = q i q j j:=j-1 Report k min (and the corresponding pair) k k k k k

14 Analysis Correctness: easy Running time is more involved Can we have many q j s that are within distance k from q i? No Proof by packing argument k

15 Analysis, ctd. Theorem: there are at most 7 q j s, j<i, such that y i -y j k. Proof: Each such q j must lie either in the left or in the right k k square Within each square, all points have distance distance k from others We can pack at most 4 such points into one square, so we have 8 points total (incl. q i ) q i

16 At most 4 Split the square into 4 subsquares of size k/2 k/2 Diameter of each square is k/2 1/2 < k at most one point per sub-square

17 Running time Divide: O(n) Combine: O(n log n) because we sort by y However, we can: Sort all points by y at the beginning Divide preserves the y-order of points Then combine takes only O(n) We get T(n)=2T(n/2)+O(n), so T(n)=O(n log n)

18 Closest pair ctd. (two more algorithms for the curious; will try to cover this later in the term)

19 Closest Pair with Help Given: P={p 1 p n } of points from R d, such that the closest distance is in (t,c t] Goal: find the closest pair Will give an O((2c d 1/2 ) d n) time algorithm Note: by scaling we can assume t=1

20 Algorithm Impose a cubic grid onto R d, where each cell is a 1/d 1/2 1/d 1/2 cube Put each point into a bucket corresponding to the cell it belongs to Diameter of each cell is 1, so at most one point per cell For each p P, check all points in cells intersecting a ball B(p,c) How many cells are there? All are contained in a d-dimensional box of side 2(c+1/d 1/2 ) 2(c+1) At most (2 d 1/2 (c+1) ) d such cells Total time: O((2c d 1/2 ) d n) p

21 How to find good resolution t? Repeat: Choose a random point p in P Let t=nn(p)=min q P-{p} p-q Impose a grid with side t =t/(2d 1/2 )-δ Put the points into the grid cells Remove all points whose all adjacent cells are empty Until P is empty Observations: The diameter of two adjacent cells is 2t d 1/2 < t points p with NN(p ) t never survive Points p with NN(p )<t always survive t p t

22 Correctness Consider t computed in the final iteration There is a pair of points with distance t (it defines t) There is no pair of points with distance less than t (as per previous slide) We get c=t/t ~ 2 d 1/2

23 Running time Consider NN(p 1 ) NN(p m ) An iteration is lucky if NN(p i ) t for at last half of points p i The probability of being lucky is 1/2 Expected #iterations till a lucky one is 2 After we are lucky, the number of points is m/2 Total expected time = 3 d times O(n+n/2+n/4+ +1)

24 Conclusions Closest pair: O(n log n) for d=2 nd O(d) for general d

25 Higher dimensions (sketch) Divide: split P into P L and P R using the hyperplane x=t Conquer: as before Combine: Need to take care of points with x in [t-k,t+k] This is essentially the same problem, but in d-1 dimensions We get: T(n,d)=2T(n/2, d)+t(n,d-1) T(n,1)=O d (1) n Solves to: T(n,d)=n log d-1 n