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

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.)

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( n) 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).

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.)

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.)

Finally, in this assignment, each of n input vertices can only once have either of these 3 roles, which proves eq Back to eq 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.) 11

(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.

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“

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

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

00 A B C D 00 q i pipi A s i C D B q i pipi s i Example3.

11 22 33 00 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!

s i B2B2 B1B1 q i D2D2 D1D1 E s i 11 22 00 33 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

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 )

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 )

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 )