COT5405: GEOMETRIC ALGORITHMS

Transcription

1 COT5405: GEOMETRIC ALGORITHMS Objects: Points in, Segments, Lines, Circles, Triangles Polygons, Polyhedra R n Alications Vision, Grahics, Visualizations, Databases, Data mining, Networks, GIS Scientific Comuting, Engineering, CAD Com Biology, Physics simulations, Oerations research, Prof. Aler Ungor - An Short Introduction to Comutational Geometry

2 SAMPLE PROBLEMS / CONCEPTS Convex Hulls or the unique convex olygon that contains P and whose vertices are al P might have interior oints that are not vertices of the convex hull. Triangulations? Voronoi diagram Motivation Voronoi diagrams Voronoi diagrams Satial interolation A set of oints and its convex hull. Convex hull vertices are black; interior oints are white. Nearest Neighborhood Diagrams Geometric decision Intersections to list vertices counterclockwise instead of clockwise is arbitrary Comutational Geometry Lecture 10: Voronoi diagrams To simlify the resentation of the convex hull algorithms, I will assu Range Queries Just to make things concrete, we will reresent the oints in P by th in two arrays X[1.. n] and Y [1.. n]. We will reresent the convex hull vertices in counterclockwise order. if the ith oint is a vertex of the conv the next vertex counterclockwise and red[i] is the index of the next ve next[i] = red[i] = 0. It doesn t matter which vertex we choose as th general osition, meaning (in this context) that no three oints lie on a like assuming that no two elements are equal when we talk about sorting to really imlement these algorithms, we would have to handle colinear t consistently. This is fairly easy, but definitely not trivial. E.2 Simle Cases

3 CONVEX HULLS Convex Hull, CH(P), of a set of oints P is the smallest convex olygon that contains P largest convex olygon whose vertices are all oints in P unique convex olygon that contains P and whose vertices are all oints in P

4 CONVEXITY A set C IR d is convex iff (, q) C 2 the line segment q is contained in C. q q Convex Non convex

5 CONVEX HULL PROBLEM Inut: the set P = { 1, 2,... n } IR 2 Outut: a sequence L = (c 1, c 2,... c h ) of vertices of CH(P ) in counterclockwise order Examle: L = ( 3, 4, 8, 6, 1, 5 ) CH(P )

6 CONVEX HULL - Characterization The directed edge (, q) is an edge of CH(P ) iff q CH(P )

7 CONVEX HULL - Characterization The directed edge (, q) is an edge of CH(P ) iff q r all r P \ {, q} lies to the left of line q (oriented by q).

8 CONVEX HULL PROBLEM The directed edge (, q) is an edge of CH(P ) iff q r r P \ {, q} the triangle (, q, r) is oriented counterclockwise.

9 CONVEX HULL We denote CCW (, q, r) = x x q x r y y q y r = (x q x )(y r y ) (x r x )(y q y ) Triangle (, q, r) is counterclockwise iff CCW (, q, r) > 0. How fast can we erform this test? 2 multilications and 5 subtractions takes O(1) time

10 CONVEX HULL - Naive Algorithm Algorithm SlowConvexHull(P) Inut: a set P of oints in IR 2 Outut: CH(P ) 1. E P 2 2. for all (, q, r) P 3 such that r / {, q} 3. if CCW (, q, r) 0 4. then remove (, q) from E 5. Write the remaining edges of E into L in counterclockwise order 6. Return L

11 CONVEX HULL - Divide & Conquer Comute median in x-coordinate and Slit P into two sets Comute Left and Right Hulls recursively Merge the two Hulls q q q q q q

12 CONVEX HULL - Gift Wraing Start with the leftmost oint and wra around =l l l l l l The execution of Jarvis s March.

13 CONVEX HULL - Gift Wraing JarvisMarch(X[1.. n], Y [1.. n]): l 1 for i 2 to n if X[i] < X[l] l i l reeat q + 1 Make sure q for i 2 to n if CCW(, i, q) q i next[] q; rev[q] q until = l

14 CONVEX HULL - Lower Bound let N = (x 1, x 2,... x n ) IR for all i let i = (x i, x 2 i ) comute CH(P ) P : y = x 2 CH(P )

15 Line-segment intersection Given n line segments, does any air intersect? a d e c b f

16 Line-segment intersection Given n line segments, does any air intersect? Obvious algorithm: O(n 2 ). e d a c b f

17 Swee-line algorithm Swee a vertical line from left to right (concetually relacing x-coordinate with time). Maintain dynamic set S of segments that intersect the swee line, ordered (tentatively) by y-coordinate of intersection. Order changes when new segment is encountered, existing segment finishes, or two segments cross segment endoints Key event oints are therefore segment endoints.

18 d e e e b a a a d d e d b e e c c c c d b d d d a b b b b b b f f f f a d e c b f

19 Swee-line algorithm Process event oints in order by sorting segment endoints by x-coordinate and looing through: For a left endoint of segment s: Add segment s to dynamic set S. Check for intersection between s and its neighbors in S. For a right endoint of segment s: Remove segment s from dynamic set S. Check for intersection between the neighbors of s in S.

20 Analysis Use red-black tree to store dynamic set S. Total running time: O(n lg n).

21 Correctness Theorem: If there is an intersection, the algorithm finds it.

22 Correctness Theorem: If there is an intersection, the algorithm finds it. Proof: Let X be the leftmost intersection oint. Assume for simlicity that only two segments s 1, s 2 ass through X, and no two oints have the same x-coordinate. At some oint before we reach X, s 1 and s 2 become consecutive in the order of S. Either initially consecutive when s 1 or s 2 inserted, or became consecutive when another deleted.

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

24 Divide & Conquer Divide: Comute the median of x- coordinates Slit the oints into P L and P R, each of size n/2 Conquer: comute the closest airs for P L and P R Combine the results (the hard art)

25 Divide & Conquer Combine 2k Let k=min(k 1,k 2 ) Observe: Need to check only airs which cross the dividing line Only interested in airs within distance < k Suffices to look at oints in the 2k-width stri around the median line k 1 k 2

26 Divide & Conquer Scanning the stri Sort all oints in the stri by their y-coordinates, forming q 1 q r, r n. Let y i be the y-coordinate of q i For i=1 to r j=i-1 While y i -y j < d Check the air q i,q j j:=j-1 d

27 Divide & Conquer - Analysis Correctness: easy Running time is more involved k Can we have many q j s that are within distance k from q i? No Proof by acking argument

28 Divide & Conquer - Analysis Theorem: there are at most 7 q j s 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 oints have distance distance k from others We can ack at most 4 such oints into one square, so we have 8 oints total (incl. q i ) q i

29 Divide & Conquer - Analysis Packing bound Proving 4 is not obvious Will rove 5 Draw a disk of radius k/2 around each oint Disks are disjoint The disk-square intersection has area (k/2)2/4 = /16 k2 The square has area k2 Can ack at most 16/ 5.1 oints

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