Download presentation

Presentation is loading. Please wait.

Published byGordon Owen Modified over 2 years ago

1
One of the most important problems is Computational Geometry is to find an efficient way to decide, given a subdivision of E n and a point P, in which part of the subdivision lies P. A B C D E 4. Point location P

2
A natural way to start the search is to divide the plane into vertical stripes and to decide the position of P in two steps: - the stripe, (1) - the facet of the stripe (2). A B C D E P Point location (cont.)

3
But it turn out that the number of facets in this construction can be O (n 2 ) in terms of the number of edges! Point location (cont.) Example with 2(1 + 2 + 3 +...+ n) + 1 + n = ( n +1) 2 facets! If the vertical lines end at the fist intersection with the edges of the subdivision, the number of trapezoids is N 3 n + 1 (eq. 4.1).

4
A B C D E In this counting and further, we assume that, unlike the vertices of C, no two vertices of the subdivision have equal y coordinates, i.e. we assume that the edges of the subdivision are in general position. Point location (cont.)

5
This subdivision has the property that each trapezoid has unique upper bounding edge top ( ), unique lower bounding edge bottom ( ), unique left vertex leftp ( ), and unique right vertex rightp ( ). top ( ) bottom ( ) rightp ( ) leftp ( ) Each trapezoid has also 1-2 left and 1-2 right neighbors, e.g. lln ( ) is its left lower neighbor. Right upper neighbor Point location (cont.)

6
Finally, in this assignment, each of n input vertices can only once have either of these 3 roles, which proves eq. 4.1. Back to eq. 4.1. The number of trapezoids is L + 1 where L is the number of the vertices leftp ( ), since each trapezoid, with the exception of the leftmost one, has exactly one such vertex. Further, to each vertex leftp ( ) a unique edge may be assigned: bottom ( ), or top ( ), or top ( 1 ), 1 =lln ( ). Point location (cont.) 11

7
(1) Data structure (DS): For each edge e we save: -the right and the left endpoint of e, the facet whose top is e, and the facet whose bottom is e. For each facet we save: - leftp ( ), rightp ( ), top ( ), bottom ( ), and its neighbors. Point location ( Trapezoidal Map ) Algorithm 8 = Trapezoidal Map (Implementation = Exercise 20 ) The algorithm is incremental on the set of edges. So we may think about the structure as about a set S of edges. Adding edge at a time we update the data structure and the search structure. The edges may have only endpoints in common. They are not sorted.

8
Point location ( Trapezoidal Map ) (2) Search Structure (SS): Oriented graph without loops, whose - leafs are trapezoids of the Trapezoidal Map, - inner nodes are the edges from the initial set S, or their endpoints. - There is exactly one root. The out degree of inner nodes is 2. - If the inner node is an edge the deciding criterion is above or below*, if this is a point the criterion is left of right. SS is used not only to finally locate the given point, but also to build DS and SS itself inductively! More precisely, - in the step i we first locate the endpoints of the edge s i wrt already constructed SS i-1 after the step i-1, - then we again use SS i-1 to determine the trapezoids intersected by s i. *above will be depicted as „left“, below as „right“

9
Example1. A B C D E G F p1p1 q1q1 p2p2 q2q2 s1s1 s2s2 q1q1 p1p1 A s1s1 q2q2 G s2s2 B F p2p2 C s2s2 E D

10
Example2. A B C D E G F p1p1 q1q1 p2p2 q2q2 s1s1 s2s2 q1q1 p1p1 A s1s1 q2q2 G s2s2 B F p2p2 C s2s2 E D = p 3 q3q3 H I s3s3 I H q3q3 s3s3

11
00 A B C D 00 q i pipi A s i C D B q i pipi s i Example3.

12
11 22 33 00 s i pipi q i s i pipi q i A1A1 A2A2 B2B2 B1B1 D2D2 C2C2 C1C1 D1D1 E s i pipi q i A BD F C E Example4.1 The glueing process is necessary because the vertical lines intersected with s i need to be shortend to the first intersection!

13
s i B2B2 B1B1 q i D2D2 D1D1 E s i 11 22 00 33 C1C1 C2C2 A1A1 A2A2 B1B1 B2B2 q i s i D1D1 D2D2 E F A B C q i s i D E Example4.2

14
Theorem. A planar subdivision with n edges in general position admits an algorithm (1) whose running time has expected O (n log n) time, (2) the storage requires O (n) time, and (3) the location of a given point q takes expected O (log n) time. Proof of (2) : - 3n +1 leafs (trapezoids), out of which k i in step i (k i -1 new). - (k i - 1) inner nodes (easy to check) at each step i. ( Total number of inner nodes is equal to the number of leafs.) - Hence, the size of the search structure is 6n + 2 Point location ( T rapezoidal Map, expected behavior )

15
Proof of (3) : Let Xi be the number of nodes on the search path to q constructed in step i (in Example 1 (the path to q) X1=X2=3), and let E(X) be the expected value of X. Then: A B C D E G F p1p1 q1q1 p2p2 q2q2 s1s1 s2s2 q p 1 X1 q 1 s 1 p 2 X2 s 2 F - In the step i the path leading to q is increased by at most 3 nodes. - The increase happens only when a trapezoid of q is changed. -The change may only happen if s i is top ( ), bottom ( ), or defines leftp ( ) or rightp ( ), where is the new trapezoid! of q. Point location ( T rapezoidal Map, expected behavior )

16
Proof of (1): - O (log i) needed to locate the left endpoint of e i - O (k i ) needed to find the facets intersected by e i Hence, (O (log i) + O (E (k i )) ) = O (n log n) + O (n) = O (n log n) Proof of (3) – cont.: -The probability of each of these events is 1/i total = 34 1/i. Now E(X i) = 12/i = O (log n). Point location ( T rapezoidal Map, expected behavior )

Similar presentations

OK

Duality and Arrangements Computational Geometry, WS 2006/07 Lecture 7 Prof. Dr. Thomas Ottmann Algorithmen & Datenstrukturen, Institut für Informatik Fakultät.

Duality and Arrangements Computational Geometry, WS 2006/07 Lecture 7 Prof. Dr. Thomas Ottmann Algorithmen & Datenstrukturen, Institut für Informatik Fakultät.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on condition monitoring jobs Ppt on mineral and power resources for class 8 How to run ppt on ipad Ppt on tcp ip protocol chart Ppt on elections in india downloads Ppt on team building training module Raster scan display and random scan display ppt on tv Download ppt on trigonometric functions for class 11 Ppt on media and democracy Ppt on front office management in hospital