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Point Location Strategies

Idit Haran13.4.2005

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Outline Problem Description Other strategies Our strategy - Landmarks Benchmark Working with conic arrangements

and some number theory Future work Summary

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Point Location – Problem Description

INPUT OUTPUT

Given A map (which divides space of interest into regions) A query point q (specified by coordinates)

Find the region of the map containing q

Requirement: (segments/conics) Fast query Reasonable preprocess time Good data structure (small space)

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Other Strategies Strategies implemented in CGAL:

Naïve Walk along line Random Incremental Algorithm

(Trapezoidal) Triangulation

Other strategies: Persistent search trees

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Naïve Approach (CGAL) Idea No need for any preprocess, just go over all

elements and find the closestPreprocess& memory

No preprocess.Storage space = 0.

Query Go over all vertices, and check if q equals a vertexGo over all halfedges and check if q is on edgeGo over all halfedges and find the closest edge upwards.O(n) query time

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Walk Along a Line Idea start from a known place in the arrangement and

walk from there towards the query point through a straight line

Preprocess& memory

No preprocess.Storage space = 0.

Query The implementation in CGAL: Start from the unbounded face Walk down to the point in through a vertical lineAsymptotically O(n) time In practice: quite good, easy to maintain

q

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Random Incremental Algorithm (Trapezoidal)

Idea Use hierarchy data structure that allows a logarithmic query time

Preprocess& memory

Add the segments one by one randomly. For each segment:

Update the trapezoidal map. Update the DAG.

O(n·logn) time, O(n) space

Query Search for the trapezoid in the DAG O(log n) time, if the segments were added randomly

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Vertical Decomposition Assume a bounding box. Extend the vertical line from each

vertex upward and downward, until it touches another segment.

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Trapezoidal Maps

Contains triangles and trapezoids. Each trapezoid or triangle is determined by two vertices

and two segments. A refinement of the original subdivision.

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Trapezoidal Maps Theorem: In a trapezoidal map of

n segments, there are at most 6n+4 vertices, and at most 3n+1 faces.

Proof:Vertices:

2n – original vertices 4n – 2 extensions for each original

vertex. 4 – Vertices of the bounding box. Total: 6n+4

Faces: Similar

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Trapezoidal Map Data Structure For each Trapezoid store:

The vertices that define its right and left sides.

The (up to two) neighboring trapezoids on right and left.

(Optional) The neighboring trapezoids from above and below. This number might be linear, so we store only the leftmost of these.

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The DAG Search StructureQ1

P1 Q2

P3

S3

B

C

S1

P2

A

S3 S2

D E F G D H

Q3

S2 S3

J K

BA

C

D

E

F

H

G

J K

P1

P2

P3

Q1

Q2

Q3

S1

S3

S2

DAG = Directed Acyclic Graph

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K

Q1

P1 Q2

P3

S3

B

C

S1

P2

A

S3 S2

D E F G D H

Q3

S2 S3

J

BA

C

D

E

F

H

G

J K

P1

P2

P3

Q1

Q2

Q3

S1

S3

S2

Using the DAG Search Structure

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TriangulationIdea Use a working algorithm on triangulation and test

it on the arrangement (only segments !)(triangles are much simpler than faces)

Preprocess& memory

Triangulate the planar map O(n·logn) time, O(n) space Keep relations between planar map vertices and triangulation

Query Find the triangle q is located in The locate is done by walk from an arbitrary vertex.They claim that it takes time O(n) in the worst case, but only O(sqrt(n)) on average if the vertices are distributed uniformly at random. Find the face in the arrangement that contains this triangle

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Persistent Search Trees (Sarnak-Tarjan)

Idea Divide the arrangement into slabs, like in sweep, and use logarithmic search

Preprocess& memory

Draw a vertical line through each vertex, splitting the plane into vertical slabs. Observations: sets of line segments intersecting - contiguous slabs are similar. Reduces the problem to storing a “persistent” sorted set. O(n · log n) time, O(n · log n) space

Query Find the slab Find the two segment above and below q O(log n) time.

AABB

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Other Strategies - SummaryNaïve Walk RIC Triangle Persiste

ntPreprocess & memory

none none O(n·logn) O(n)

O(n·logn) O(n)

O(n·logn) O(n)

Query O(n) < O(n) O(logn) O(sqrt(n)) – O(n)

O(logn)

Benefits Simple Simple and reasonable query time

Good locate time

Quite good locate time

Good locate time

Drawbacks

Bad query time

Query time not good enough

Complicated data structure, bugs, long preprocess, Very large memory needs

Long preprocess

Long preprocess, large memoryMore Algorithms on Delaunay Triangulations: The Delaunay Hierarchy, Jump &

Walk

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Point location using Landmarks

Idea Special points, “landmarks” are stored in a nearest neighbor search structure (e.g., kd-trees). During query time, “walk” from the landmarks towards the query point.

Preprocess Choose the landmarks, and locate them in the arrangement. Store the landmarks in a nearest neighbor search structure.

Query Given a query point q, find the closest landmark to the query point q, using the search structure.“walk” from to q.

q

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Choosing the Landmarks Number of landmarks Distribution of the landmarks in the arrangement:

Geometric entities of the arrangement: Vertices Edge Midpoints Points inside the faces:

Preprocess the arrangement into trapezoids and triangles Independent of the arrangement geometry:

Random

Grid

qq

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Quality of the landmarks – Arrangement Distance (AD)

The Arrangement Distance (AD) between two points is the number pf faces in which the straight line segment that connect these points passes.

Arrangement Distance ≠ Euclidean Distance

pv

v

p

AD = 0

AD = 4

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Nearest Neighbor Search Structure

Input: Landmarks Query point q

Question: Find nearest landmark to

the query point q Answer:

Voronoi? - Again, point location problem !

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Kd-trees

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6

5

1

3

2

9

8

10

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l5

l1 l9

l6

l3

l10 l7

l4

l8

l2

l1

l8

1

l2 l3

l4 l5 l7 l6

l9l10

3

2 5 4 11

9 10

8

6 7

The kd-tree is a powerful data structure that is based on recursively subdividing a set of points with alternating axis-aligned hyper-planes.

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Kd-trees - Construction

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6

5

1

3

2

9

8

10

11

l5

l1 l9

l6

l3

l10 l7

l4

l8

l2

l1

l8

1

l2 l3

l4 l5 l7 l6

l9l10

3

2 5 4 11

9 10

8

6 7

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6

5

1

3

2

9

8

10

11

l5

l1 l9

l6

l3

l10 l7

l4

l8

l2

l1

l8

1

l2 l3

l4 l5 l7 l6

l9l10

3

2 5 4 11

9 10

8

6 7

q

Kd-trees - Query

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Using Kd-trees with the landmarks

Quick: Approximate nearest neighbor Operations on points only Round all points to double –

Inexact computation. Final result is always exact !

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Landmarks point location – Query

Find Nearest

Landmark

Decide on Startup

Face

Cross to Next Face

Query PointLocated

Is query point

in face ?

Yes

No

Walk

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Landmarks point location – Query

Find Nearest

Landmark

Decide on Startup

Face

Cross to Next Face

Query PointLocated

Is query point

in face ?

Yes

No

Walk

Decide on startup face: From vertex:

Check all incident face to the vertex, and find the face in q’s direction.

From Edge: Choose between the two

incident face to the edge.

From Face: This is the startup face.

q

q

q

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Landmarks point location – Query

Find Nearest

Landmark

Decide on Startup

Face

Cross to Next Face

Query PointLocated

Is query point

in face ?

Yes

No

Walk

Is query point in face? Count the number of

edges on f’s boundary that are above q.

Odd – q is inside f – found !

Even –need to cross to the next face.

q

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Landmarks point location – Query

Find Nearest

Landmark

Decide on Startup

Face

Cross to Next Face

Query PointLocated

Is query point

in face ?

Yes

No

Walk

Cross to next face: Create a segment s from the

landmark to the query point q find the edge e in f

that intersects s flip e to get a new face f’. check if q is in f’…

ef

q

f’

s

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Benchmark Categories: Query & Preprocess time Algorithms: Naïve, Walk, Triangle, RIC,

2 variants of landmarks algorithm. Arrangements: random and degenerate. similar results.

Random Onebig

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Number of

Edges

Number of

Original Segment

s

Naïve Walk Triangle

RIC Landmarks

(vertices)

Landmarks

(random)

35 10 0.28 0.31 0.2 0.05 0.20 0.13

2301 100 8.9 3.9 1.7 0.20 0.31 0.23

9548 200 30.7 8.7 3.6 0.26 0.32 0.24

56070 500 163.4 24.3 8.6 0.33 0.35 0.26

114303 700 329.6 35.1 12.07 N/A 0.35 0.26

235553 1000 685.7 50.5 18.14 N/A 0.36 0.27

Benchmark: Query Arrangements of random line segments

Time in milliseconds

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Number of

Edges

Number of

Original Segments

Common Preprocessing: Constructing

the Arrangement

Algorithm Specific PreprocessTriangl

eRIC Landmark

s(vertices)

Landmarks

(random)

35 10 0.004 0.036 0.016 <0.001 0.03

2301 100 1.2 11 4.04 <0.001 2.3

9548 200 5.4 80 20 <0.001 10

56070 500 30 990 153 <0.001 68

114303 700 31 1463 N/A 1 166

235553 1000 65 7984 N/A 2 352

Benchmark: Preprocess Arrangements of random line segments

Time in seconds

Naïve and Walk algorithms do not require any specific preprocessing besides constructing the arrangement

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Number of Random

Landmarks

Total Preprocessing

Time [sec]

Net Preprocessing

Time [sec]

Query Time [msec]

% Queries with

AD = 0

10 72.8 0.1 3.95 2.9

100 72.9 0.2 1.41 14.2

1,000 73.3 0.5 0.55 30.9

10,000 76 3.5 0.33 53.0

50,000 90 17 0.26 73.9

100,000 106 34 0.22 82.0

500,000 242 169 0.21 87.7

Landmarks Algorithm Analysis

Results on an arrangement of 500 original segments, include ~56,000 edges and ~28,000 vertices

Algorithm performance for varying number of random landmarks

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Landmarks Type

Number of

Landmarks

Total Preprocessi

ng Time [sec]

Net Preprocessi

ng Time [sec]

Query Time

[msec]

% Queries

with AD = 0

Vertices (V)

28,000 72.8 0.6 0.32 92.7

Edge Midpoints (EM)

57,000 74.7 1.8 0.27 96.6

V + EM 85,000 74.5 1.9 0.27 97.6

Random 50,000 90.0 17 0.26 73.9

UniformGrid

50,000 89.3 16.7 0.24 77.5

Landmarks Algorithm Analysis Algorithm performance vs. the type of landmarks used

pv

q

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Landmarks Algorithm Analysis

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Conics Differences from segments Bench results Problems invoked, and their

solutions Cross to next face The chord method Number type theory

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Conics – differences from segments

More complex curves (every operator on the curves takes longer)

Working with irrational number types

Basically the same methods work (except triangle)

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Benchmark: Query Arrangements of random conic arcs

Time in seconds

Number of Edges

Number of Original Conics

Naïve Walk RIC Landmarks

(vertices)

Landmarks(random)

26 10 0.2 0.3 0.04 0.25 0.10

72 20 0.4 0.7 0.07 0.36 0.14

192 40 0.9 1.6 0.11 0.45 0.17

430 60 1.5 3.5 0.15 0.51 0.16

1000 100 3.4 8.5 0.2 0.62 0.15

7554 200 9.8 28.8 0.4 0.69 0.11

26160 400 29.6 71.4 0.35 0.72 0.12

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Conics - Cross to Next Face Find exact intersection point – a complex operation on conics ! Instead – use a simple method to decide what edge to flip. The idea: check vertical order on the left and right boundaries

of the common x-range.

q

s e

x-range

q

x-range

e

s

L

q

x-range

e

sL

q

x-range

e

s

q

x-range

e

s

(a) (b) (c)

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What’s the problem with the vertices in an arrangement of conic arcs?

v

Problem 1:vertex may be complex irrational number (algebraic with degree 4)

Problem 2:finding the intersection between vp and other curves

p

v’

v’ Properties: Rational On the conic Close to the v

The chord method:given a rationl point r on the conic, we can use the chord method to find v’

r

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2

2 2

tan2

2 1sin ,cos1 1

t

t tt t

2

2 2

1 2,1 1

t tt t

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The Chord MethodGiven: Unit circle equation:

x2 + y2 = 1 The point (-1,0) is on the circleFind: Rational point v’ close to vSolution: Every line passing through

this point is:y = t(x+1) where t is the slope of the line.

solve x2 + (t(x+1))2 = 1

-)1,0(

y=t(x+1)

x2+y2=1

v

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One Rational Point Apparently, not all conics with rational (or even integer)

coefficients have rational points on them

For example, the circle

x2 + y2 = 3

does not have any rational points on it.

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Change the equation into homogenic coordinates:

A: r x2 + s y2 + t xy + u x + v y + w = 0 ⇓ (x→x/z, y→y/z)

B: r x2 + s y2 + t xy + u xz + v yz + wz2 = 0

(Rational solution to A ⇔ Integer solution to B)

Theory 1: if there is a solution for every congruence modulo pk for every prime p and every k, then there is a solution.

Theory 2: Check only p=2 and p’s that divide the determinant of the equation coefficients. It is also sufficient to check only for k <= 3 + the largest exponent in the determinant factorization.

Does a conic have rational solution?

3 Variables

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How to find a rational solution(if there is one)

Theorem:if a conic equation has an integer solution,

then the size of the solution (3F),

where F is the sum of the absolute value of the coefficients.

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What to do in practice? Change the equation into homogenic coordinates Go over all small pk and check if there is a solution modulo

pk. One NO – no solution. All Yes –

search for pairs (x,y) so that |x|<3F, |y|<3F check whether z is also an integer.

YES – we found a solution. All NO – there is no solution.

Complexity: (3F)2

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If there are no rational points on the conic?

Euclid’s algorithm to approximate real number x into rational number r:

x = a0 + 1/x1, x1>1, a0 - integerx1 = a1 + 1/x2, x2>1, a1 - interger…Finally, r = a0 + 1/(a1 + (1/ (a2 + ….) ) ) = pn/qn.

If x is irrational, and the algorithm ends after n stages, then:

| x - pn/qn | < 1/qn2

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How to find ? = the approximation value.

should stay “close” to the vertex.1. Find the minimum distance between v and other vertices.

or 2. Find the minimum distance to all edges/vertices in the

adjacent faces to the vertex.

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Landmarks – future work Rationalization of the vertices More types of landmarks:

Combination of different landmarks types Memory usage of the algorithms

Halton sequence Hammersley Points

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Summary Point location using landmarks –

Fast query Efficient preprocess, using inexact computation Low memory Simple to understand and implement Many ways to improvements:

different search structure, other choice of points