Point Location Problem in Computational Geometry

1. Point Location Problem in Computational Geometry I4B01 張繼文 Ref: Computational Geometry, Preparata and Shamos, Chap. 2

2. Topics Point Location Problem for • Polygon • Planar Subdivision

3. Point in Polygon Single-shot approach Division wedges for the convex inclusion

4. Problem (polygon inclusion) • Given a simple polygon P and a point Z, is Z in P?

5. P Z L The single-shot approach Assume that L intersects P and that L does not pass any vertex of P. • IN: odd number of crossings • OUT: even number of crossings • the result holds for concave polygons too

6. Degenerate Cases P 1,2 3 4 5 Z X X X L • If just one vertex of an edge belongs to L, then it must be counted if it is the vertex with larger ordinate, and ignored otherwise • If both vertices of an edge belong to L, this edge must be ignored

7. Theorem • Whether a point Z is internal to a simple N-gon P can be determined in O( N ) time, without preprocessing.

8. Division Wedges for Convex Inclusion • The method relies on the convexity of P. • The vertices of a polygon occur in angular order about any internal point, Q. • Treating Q as the origin of polar coordinates.

9. P2 P3 Z P P1 P4 Q P5 P8 P6 P7

10. Procedure CONVEX INCLUSION 1. Given a query point Z, determine by binary search the wedge in which it lies. 2. Determine that point Z lies between the rays defined by Pi and Pi+1

11. Theorem • The inclusion question for a convex N-gon can be answered in O( log N ) time, given O( N ) space and O( N ) preprocessing time.

12. A star-shaped polygon P Q

13. Theorem • The kernel Q can be founded in O( N ) time. • The inclusion question for an N-vertex star-shaped polygon can be answered in O( log N ) time and O( N ) storage, after O( N ) preprocessing time.

14. Point Location in a planar subdivision The slab method The chain method The triangulation refinement method

15. Divide the plane into easily manageable sections Divide the graph into slabs. Draw a vertical line through every vertex of the graph The slab method

16. The slab method All edges intersecting a slab • Have no endpoint in the slab • Don’t cross each other

17. Slab Method (cont) • Sort vertices in ascending x order. • For each interval ( Vi, Vi+1 ), store the segments crossing the interval ordered from top to bottom. • For a query point • Usebinary search to determine the slab 2.Use binary search to determine the face

18. Worst Case in Storage segment : O( N ) Total space : 2 2 2 4 6 8 10 10 8 6 4 2 O( N ) slab : O( N )

19. Theorem • Point-location in an N-vertex planar subdivision can be effected in O( log N )time using O( N2 ) storages.

20. Definition : A chain C = ( u1, … , up ) is a planar straight-line graph with vertex set { u1, … , up } and edge set { (ui, ui+1) : i = 1, …, p-1 } U1 U3 U2 U5 U6 U4 U8 U7 The Chain Method

21. Definition : A chain C = ( u1, …, up ) is said to be monotone with respect to a straight line L if a line orthogonal to L intersects C in exactly one point. U1 L U2 U3 U4 U5 The Monotone Chain

22. Where does the query point lie? U1 U2 P U3 U4 U5

23. The projection of P on L can be located with a binary search in a unique interval ( L(ui), L(ui+1) ) Determine on which side of the line containing uiui+1 the query point lies. U1 L U2 P L(U1) L(U2) U3 U4 L(P) L(U3) U5 L(U4) L(U5) A query point lies ?

24. Suppose there is a set C = { C1, …, Cr } of a polygon, we can apply the bisection to find the region query point lies. C1 C4 C2 C6 C2 C3 C6 C5 C4 C1 C3 C5 The Chain Method

25. The Chain Method (cont) • If there are rchains in C and the longest chain has pvertices, then the search worst-case time is O( log p * log r )

26. Steps to Construct the Chains … • First, regularize the PSLG • Second, assign weights on the graph using weight-balancing algorithm • Third, construct chains by traversing the graph

27. Definitions (for chains monotone w.r.t. y) • A vertex vj is said to be regular if there are vertices y(vi) < y(vj) < y(vk) such that ( vi, vj ) and ( vi, vk ) are edges of G. 2. Graph G is said to be regular if each yi is regular for 1 < j < N

28. Definition: cusp 2-pass sweep-line to remove cusps Connect to the upper end point of the lower neighbor e1 V e2 Regularize Nonregular Vertices (remove cusps)

29. Example: cusp removal

30. Example: cusp removal

31. Example: cusp removal

32. Example: cusp removal

33. Example: cusp removal

34. Example: cusp removal

35. Example: cusp removal

36. Example: cusp removal

37. Example: cusp removal

38. Example: cusp removal

39. Example: cusp removal

40. Example: cusp removal

41. Example: cusp removal

42. Example: cusp removal

43. Example: cusp removal

44. Weight Assignment of Edges • All edges satisfy : 1.Each edge has positive weight 2.For each Vj( j ≠1, N ), Win(Vi) = Wout(Vj) PS. Vin( Vi ) = | IN(v) | Vout( Vi ) = | OUT(v) |

45. WEIGHT-BALANCING REGULAR PSLG 1.Initialization Begin for each edge e do W(e) = 1 2. First pass for( i = 2; i <= N-1; i++ ) { Win(Vi)= sum of weight of incoming edges of Vi if (Win(Vi) > Vout(Vi)) { d1 = leftmost outgoing edge of Vi W(d1) = Win(Vi) – Vout(Vi) + 1 } }

46. WEIGHT-BALANCING REGULAR PSLG (cont) 3. Second pass For (i = N-1; i >= 2; i--){ Wout(Vi) = sum of weight of outgoing edges of Vi if (Wout(Vi) > Win(Vi)) { d2 = leftmost incoming edge of Vi W(d2) = Wout(Vi) – Win(Vi) + W(d2) } }

47. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Example: Initialization

48. 1 1 1 1 1 1 1 1 1 1 1 1 1=1 4>2 1<3 1 1 1=1 2=2 3>2 2>1 1 1 Example: 1st pass X 2 3 X X 2 if (Win(Vi) > Vout(Vi)) { d1 = leftmost outgoing edge(Vi) W(d1) = Win(Vi) – Vout(Vi) + 1 }

49. 1 2 4=4 1=1 3>1 3 1 1 2 1 1 1 1 1 1 1 1 2=2 1=1 2=2 3=3 1 1 Example: 2nd pass 3 X if (Wout(Vi) > Win(Vi)) { d2 = leftmost incoming edge (Vi) W(d2) = Wout(Vi) – Win(Vi) + W(d2) }

50. 1 2 3 1 1 2 1 1 1 1 1 1 1 1 3 1 Example: final result