a tutorial on computational geometry

A Tutorial on Computational Geometry

Pham Minh TriPh.D. Candidate and Project OfficerSchool of Computer Engineering

1 Mar 2008

presented by

What will be covered?

• Mainly popular static problems:– Point inside polygon– Convex hull– Line segment intersection– Delaunay triangulation– Closest pair of points

Point Inside Polygon Test

• Given a point, determine

if it lies inside a polygon

or not

Ray Test

• Fire ray from point• Count intersections

– Odd = inside polygon– Even = outside polygon

Problems With Rays



• Problems– Ray through vertex

Problems With Rays



• Problems– Ray through vertex– Ray parallel to edge

A Better Way

A Better Way

• One winding = inside

A Better Way

A Better Way

• zero winding = outside

Requirements

• Oriented edges• Edges can be processed

in any order

Computing Winding Number

• Given unit normal n• =0

• For each edge (p1, p2)

• for some integer k• k = winding number

– If k == 1, then inside– If k == 0, then outside

1p

2px

2k

xpxp

xpxp

xpxp

xpxpn

21

211

21

21 )()(cos

)()(

))()((

Advantages

• Extends to 3D!• Numerically stable • Even works on models with holes:

– Odd k: inside– Even k: outside

• No ray casting

Convex Hulls

Subset of S of the plane is convex, if for all pairs p,q in S the line segment pq is completely contained in S.

The Convex Hull CH(S) is the smallest convex set, which contains S.

p

q

pq

p

q

pq

Convex hull of a set of points in the plane

Rubber band experiment

The convex hull of a set P of points is the unique convex polygon whose vertices are points of P and which containsall points from P.

•

• •

•

• •

•

•

•

Convex Hull

• Input:– Set S = {s1, …, sn} of n points

• Output:– Find its convex hull

• Many algorithms:– Naïve – O(n3)– Insertion – O(n logn)– Divide and Conquer – O(n logn)– Gift Wrapping – O(nh), h = no of points on the hull– Graham Scan – O(n logn)

Graham Scan

• Polar sort the points around a point inside the hull

• Scan points in counter-clockwise (CCW) order– Discard any point that causes a

clockwise (CW) turn• If CCW, advance• If !CCW, discard current point

and back up

Graham-Scan : (1/11)

p1

p0

p3p4

p2

p5p6p7

p8p9

p10

p11

p12

Graham-Scan :(1/11)

p1

p0

p3p4

p2

p5p6p7

p8p9

p10

p11

p12

1.Calculate polar angle

2.Sorted by polar angle


p2p1p0

Stack S:

p1

p0

p3p4

p2

p5p6p7

p8p9

p10

p11

p12


Stack S:

p1

p0

p3p4

p2

p5p6p7

p8p9

p10

p11

p12

p3p1p0


p1

p0

p3p4

p2

p5p6p7

p8p9

p10

p11

p12

p4p3p1p0

Stack S:

Graham-Scan (5/11)

p1

p0

p3p4p2

p5p6p7

p8p9

p10

p11

p12

p5p3p1p0

Stack S:

Graham-Scan (6/11)

p1

p0

p3p4p2

p5p6p7

p8p9

p10

p11

p12

p8p7p6p5p3p1p0

Stack S:

Graham-Scan (7/11)

p1

p0

p3p4p2

p5p6

p7p8

p9

p10

p11

p12

p9p6p5p3p1p0

Stack S:

Graham-Scan (8/11)

p1

p0

p3p4p2

p5p6p7

p8

p9

p10

p11

p12

p10p3p1p0

Stack S:

Graham-Scan (9/11)

p1

p0

p3p4p2

p5p6p7

p8

p9

p10

p11

p12

p11p10p3p1p0

Stack S:

Graham-Scan (10/11)

p1

p0

p3p4p2

p5p6p7

p8

p9

p10

p11p12

p12p10p3p1p0

Stack S:

Time complexity Analysis

• Graham-Scan– Sorting in step 2 needs O(n log n).– Time complexity of stack operation is O(2n)– The overall time complexity in Graham-Scan is O(n log n).

• Demo:– http://www.cse.unsw.edu.au/~lambert/java/3d/hull.html

Line segment intersection

• Input:– Set S = {s1, …, sn} of n line segments, si = (xi, yi)

• Output:– k = All intersection points among the segments in S

Line segment intersection

• Worst case:– k = n(n –1)/2 = O(n2) intersections

• Sweep line algorithm (near optimal algorithm):– O(n log n + k) time and O(n) space– O(n) space

Sweep Line Algorithm

Avoid testing pairs of segments that are far apart.

Idea: imagine a vertical sweep line passes through the given set of line segments, from left to right.

Sweepline

Assumption on Non-degeneracy

No segment is vertical. // the sweep line always hits a segment at // a point.

If a segment is vertical, imagine we rotate it clockwise by a tiny angle.This means:

For each vertical segment, we will consider its lower endpoint before upper point.

Sweep Line Status

The set of segments intersecting the sweep line.

It changes as the sweep line moves, but not continuously.

Updates of status happen only at event points. left endpointsright endpointsintersections

event points

AG

C

T

Ordering Segments

A total order over the segments that intersect the current position of the sweep line: •Based on which parts of the segments we are currently interested in

A

B

C

D

E

B > C > D(A and E not inthe ordering)

C > D(B drops out ofthe ordering)

At an event point, the sequence of segments changes:

Update the status.

Detect the intersections.

D > C(C and D swaptheir positions)

Status Update (1)

A new segment L intersecting the sweep line

L

M

K

Check if L intersects with the segment above (K) and the segment below (M).

new event point

Intersection(s) are new event points.

Event point is the left endpoint of a segment.

N

K, M, N K, L, M, N

O

Status Update (2)

The two intersecting segments (L and M) change order.

L

M

K

Check intersection with new neighbors (M with O and L with N).

Intersection(s) are new event points.

Event point is an intersection.

N

O

O, L, M, N O, M, L, N

Status Update (3)

The two neighbors (O and L) become adjacent.

L

M

K

Check if they (O and L) intersect.

Intersection is new event point.

Event point is a lower endpoint of a segment.

N

O, M, L, N O, L, N

O

Data Structure for Event Queue

Data structure: balanced binary search tree (e.g., red-black tree).

Ordering of event points:

by x-coordinates

by y-coordinates in case of a tie in x-coordinates.

Supports the following operations on a segment s.

inserting an event

fetching the next event

Every event point p is stored with all segments starting at p.

// O(log m)

// O(log m)

m = #event points in the queue

Data Structure for Sweep-line Status

Describes the relationships among the segments intersected by the sweep line.

Use a balanced binary search tree T to support the following operations on a segment s.

Insert(T, s) Delete(T, s)Above(T, s) // segment immediately above s Below(T, s) // segment immediately below s

e.g, Red-black trees, splay trees (key comparisons replaced by cross-product comparisons).

O(log n) for each operation.

An Example

L

K

M

N

O

K

L O

N M

K

L

N

O

The bottom-up order of the segments correspond to the left-to-right order of the leaves in the tree T.

Each internal node stores the segment from the rightmost leaf in its left subtree.

The Algorithm

FindIntersections(S)Input: a set S of line segmentsOuput: all intersection points and for each intersection the segment containing it.1. Q // initialize an empty event queue2. Insert the segment endpoints into Q // store with every left endpoint // the corresponding segments3. T // initialize an empty status structure4. while Q 5. do extract the next event point p6. Q Q – {p}7. HandleEventPoint(p)

Handling Event Points

Status updates (1) – (3) presented earlier.

Degeneracy: several segments are involved in one event point (tricky).

A

B

C

C

A

D

G

E

H

l

D

D

AG E

HC

A

C

G

ET:

G C

HA

B

A

G

C

B

(a) Delete D, E, A, C (b) Insert B, A, C

p

Motivation

Transformation of a topographic map

into a perspective view

TerrainsGiven: A number of sample points p1..., pn

Required: A triangulation T of the points resulting in a “realistic” terrain.

"Flipping" of an edge:900

930

2050

900

930

20

50

Goal: Maximise the minimum angle in the triangulation

Triangulation of Planar Point Sets

Given: Set P of n points in the plane (not all collinear).

A triangulation T(P) of P is a planar subdivision of the convex hull of P into triangles with vertices from P.

T(P) is a maximal planar subdivision.

For a given point set there are only finitely many differenttriangulations.

Size of TriangulationsTheorem : Let P be a set of n points in the plane, not all collinear and let k denote the number of points in P that lie on the boundary of the convex hull for P. Then any trianglation of P has 2n-2-k triangles and 3n-3-k edges.

Proof :

Let T be triangulation of P, and let m denote the # of triangles of T. Each triangle has 3 edges, and the unbounded face has k edges. nf = # of faces of triangulation = m + 1every edge is incident to exactly 2 faces. Hence, # of edges ne = (3m +k)/2. Euler‘s formula : n - ne + nf = 2.Substituting values of ne and nf , we obtain:

m = 2n – 2 – k and ne = 3n – 3 – k .

Angle Vector

Let T(P) be a triangulation of P ( set of n points). Suppose T(P) has m triangles. Consider the 3m angles of triangles of T(P), sorted by increasing value.A(T) = { a1..., a3m } is called angle-vector of T.

Triangulations can be sorted in lexicographical order according to A(T).

A triangulation T(P) is called angle-optimal if A(T(P)) A(T´(P)) for all triangulations T´ of P.

Illegal Edge

a1

a2

a6a5

a3

a4 Pj

Pi a1‘

a2‘ a4‘

a3‘ a5‘

a6‘

Pk

Pj Pi

Edge flip

The edge pi pj is illegal if 6i1

min 6i1

min

‘i

Note: Let T be a triangulation with an illegal edge e. Let T´ be the triangulation obtained from T by flipping e. Then, A(T´) A(T) .

i

Consider a quadrilateral:

Legal Triangulation

P i

Pk

P j

P l

Definition : A triangulation T(P) is called a legal triangulation, if T(P) does not contain any illegal edges.

Test for illegality

Lemma :

Let edge pi pj be incident to triangles pi pj pk

and pi pj pl , and let C be the circle thru pi, pj and pk. The edge pi pj is illegal iff the point pl

lies in the interior of C. Furthermore, if thepoints pi, pj, pk, pl form a convex quadri-lateral and do not lie on a common circle, then exactly one of pi pj orpk pl is an illegal edge.

Test of IllegalityObservation:

pl lies inside the circle through pi, pj and pk iff pk lies inside thecircle through pi, pj, pl . When all four points lie on circle, bothpi pj and pk pl are legal.

P i

Pk

P j

P l

P i

Pk

P j

P l

Thales Theorem

a

b

p

q

r

s

pipj

pk

pl

asb aqb = apb arb

illegal

Lemma: Let C be the circle through the triangle pi, pj, pk and let the point pl be the fourth point of a quadrilateral.The edge pi pj is illegal iff pl lies in the interior of C.

Thales Theorem

pi pj

pl

pk

pi pj

pl

pk

Consider the quadrilateral with pl in the interior of the circle that goes through pi, pj, pk.

Claim: The minimum angle does not occur at pk!

(likewise: Minimum angle does not occur at pl )

Goal: Show that pi pj is illegal

pi pj

pk

pl

a1a2

a3

a4

pi pj

pk

pl

a1a2

a3

a4

a1‘a2‘

a3‘ a4‘

pi pj

pk

pl

a1‘a2‘

a3‘ a4‘

W.l.o.g.

a4 minimal

Thales Theorem: Proof

a1‘ > a4

a2‘ > a2

pi pj

pk

pl

a1a2

a3

a4

a1‘a2‘

a3‘ a4‘

Circle criterion violated illegal edge

Thales Theorem: Proof

a4‘ > a1

a3‘ > a3

Hence, min{ai‘} > min{ai}

Circle Criterion

Definition:A triangulation fulfills the circle criterion if and only if the circumcircle of each triangle of the triangulation does not contain any other point in its interior.

Theorems

Theorem:A triangulation T(P) of a set P of points does not contain an illegal edge if and only if nowhere the circle criterion is violated.

Theorem:Every triangulation T(P) of a set P of points can be finally transformed into an angle-optimal triangulation in a finite number of steps.

Definition of the Delaunay Triangulation

A triangulation T(P) is a Delaunay Triangulation of P, denoted as DT(P), if and only if the circumcircle of any triangle of T does not contain any other point of P in its interior (i.e. T fulfills the circle criterion).

Equivalent Characterisations of the Delaunay Triangulation

1. DT(P) is the straight-line-dual of Voronoi Diagram VD(P).

2. DT(P) is a triangulation of P such that all edges are legal (local angle-optimal).

3. DT(P) is a triangulation of P such that for each triangle the circle criterion is fulfilled.

4. DT(P) is global angle-optimal triangulation.

5. DT(P) is a triangulation of P such that for each edge pi pj there is a circle, on which pi and pj lie and which does not contain any other point from P.

Algorithm DT(P) (randomized, incremental)

Given: Point set P = {p1..., pn }

Initially: Compute triangle (x, y, z), which includes the points p1..., pn.

m

z (0,3m)

y (3m,0)

x (-3m,-3m)

Algorithm DT(P)m = max { |xi|,|yi| }

T = ((3m, 0), (3m, 3m), (0, 3m))

1. initialize DT(P) as T.

2. permutate the points in P randomly.

3. for r = 1 to n do

find the triangle in DT(P), which contains pr ;

insert new edges in DT(P) to pr ;

legalize new edges.

4. remove all edges, which are connected with x, y or z.

Inserting a Point

pi

pj

pr

pi

pj

pkpr

pl

pi

pj

pk

pr

2 cases : pr is inside a triangle pr is on an edge

Legalize (pr ,pi pj ,T):if pi pj is illegal then let pi pj pk be the triangle adjacent

to pr pi pj along pi pj.flip pi pj ,i.e. replace pi pj by pr pk

Legalize (pr , pi pk, T)Legalize (pr , pk pj, T)

Algorithm Delaunay Triangulation Input: A set of points P = {p1..., pn } in general position

Output: The Delaunay triangulation DT(P) of P

1. DT(P) = T = (x, y, z)

2. for r = 1 to n do

3. find a triangle pi pj pk T, that contains pr.

4. if pr lies in the interior of the triangle pi pj pk

5. then split pi pj pk 6. Legalize(pr, pi pj), Legalize(pr, pi pk),

Legalize(pr, pj pk)

7. if pr lies on an edge of pi pj pk (say pi pj)

8. then split pi pj pk and pi pj pl Legalize (pr, pi pl), Legalize (pr, pi pk), Legalize (pr, pj pl), Legalize (pr, pj pk)

9. Delete (x, y, z) with all incident edges to P

CorrectnessLemma :Every new edge created in the algorithm for constructing DT during the intersection of pr is an edge of the Delaunay graph of { p1,...,pr } .

pq is a Delaunay edge iff there is a (empty) circle, which contains only p and q on the circumference.

Proof idea : Shrink a circle which was empty before addition of pr !

pr

pr

Observation: After insertion of pr , every new edge produced by edge-flips is incident to pr!

Correctness of the algorithm: Consider newly produced edges:Correctness

pr

pi

pj

pk

Edge-flips produce only legal edges.

Before inserting pr , circle that goes through pi, pj, pk was empty!

Edge-flips produce edges that are always incident to pr !

Edge Flips

Data Structure for Point Location

t1

t2

t3

t2

t3

t4t5

t4

t6

t7

t1 t2

t4 t5

t3

t6 t7

t1 t2 t3

t1 t2 t3

t1 t2

t4 t5

t3

Split t1

flip pi pj

flip pi pk

pi

pj

pi pk

Analysis of the Algorithm

Lemma :

The expected number of triangles created by the incremental algorithm for constructing DT(P) is atmost 9n + 1.

Analysis of the RuntimeTheorem :

The Delaunay triangulation of a set of P of n points in theplane can be computed in O(n log n) expected time, usingO(n) expected storage.

Proof :

Running time without Point Location :Proportional to the number of created triangles = O(n).

Point Location :The time to locate the point pr in the current triangulationis linear in the number of nodes of D that we visit.

Relation to Euclidean MST

• Problem:– Find a spanning tree on a set P of N points in a plane such that

the total length of all the edges of the tree is minimized (Euclidean MST)

• Relation of Delaunay Triangulation– A Euclidean MST is a sub-graph of a Delaunay Triangulation– Blindly applying MST finding methods like Kruskal’s or Prim’s on

this problem gives O(m) where m = n(n-1)/2.– If one builds the DT(P) in O(nlogn) and then does Kruskal’s or

Prim’s, one get an O(nlogn) solution.

Geometry and Air Traffic Control

» Avoid collisions at all costs.» Start by finding the two aeroplanes closest to each other, at specific elevations.

» During peak times over 5000 planes, per hour, around Frankfurt airspace.

Geometry and Astronomy» Find the two nearest objects in space.

Search Quickly… and Why?» Find the two nearest objects in space.

Problem Description

• Input: A set S of n points in the plane.• Output: The pair of points with the shortest Euclidean distance

between them.

• Other applications in fundamental geometric primitive:– Graphics, computer vision, geographic information systems, molecular

modeling

– Special case of nearest neighbour, Euclidean MST, Voronoi

Closest Pair of Points

• Naïve method: Brute force.– Check all pair of points p and q with Θ(n2) comparisons.

• 1D version– Simple if all points are on a single line.

– Run time O(n log n)

Key to reducing quadratic runtime of the naïve method:

Simply don´t visit each point n times.

Algorithm: DAC – Divide

• Sort S in ascending order of x-coordinates.

• Find the median line L, so that we have ½ n points in SL and ½ n points SR.

SL SR

Algorithm: DAC – Conquer

• Recursively, find the closest pair in SL and SR.

SL SR

Algorithm: DAC – Merge (Strip-Margins)

• Assuming that distance < S = min(SL, SR).

• Observation: Only need to consider the points within S of line L.

SL SRS

S = min(12,21)

Handling Merge

• Sort points in 2S-strip in ascending y-coordinate order.

• (First observation) Only check those points within 11 positions in sorted list.

SL SRS

S = min(12,21)

Comparing EfficientlyDefinition: Let si be the point in the 2S-strip with the

ith smallest coordinate.

Claim: If |i - j| 12, then the distance between si and sj is at least S.

Proof:

• The points si and sj cannot lie in same ½ S-by- ½ S box.

• si and sj are at least 2 rows apart, and have distance 2(½ S).

Fact:

• The above still holds if we replace 12 with 7.SS

Closest Pair: DAC AlgorithmAlgorithm: Closest-Pair( S = {p1, …, pn} )

Sort all n points in ascending x-coordinate order

Compute the median line L, and divide S into SL and SR

Set SL Closest-Pair(SL)

Set SR Closest-Pair(SR)

Set S min(SL,SR)

Delete all points further than S from separation line L

Sort remaining points in ascending y-coordinate order

Scan points in y-order and compare distance between each point and next 7 neighbours. Update S if new shorter distance is found.

Return S

Closest Pair: DAC Analysis

Run time:T(N) 2T(N/2) + O(N log N) T(N) = O(N log2 N)

Can we achieve a better run time?Yes. Don’t sort points in strip from scratch each time.• Each recursive step returns two lists: all points sorted by y-coordinate,

and all points sorted by x-coordinate.• Sort by merging two pre-sorted list.

T(N) 2T(N/2) + O(N) T(N) = O(N log N)

Space:O(n)

Algorithm: Scan-Line

• Sort S in ascending order of x-coordinates.

• Place sorted points p1, …, pn in the event-queue.

• Initialise distance(p1, p2).

• Start scanning from p3.» Event-points?

» Status structure?

The Status-Structure

• Maintain the status-structure (-slab):

– Insert the new event-point pi.

– Remove all points that are greater than distance to the left of the current event-point pi.

– All points in the status structure are maintained in ascending y-coordinate order.

The Status-Structure

• Maintain the status-structure (-slab):

– Insert the new event-point pi.

– Remove all points that are greater than distance to the left of the current event-point pi.

– All points in the status structure are maintained in ascending y-coordinate order.

– Perform the ‘Efficient Comparison’, as described earlier, BUT only between the new event point pi and at most 7 other nearest points to pi.

– Update if a shorter distance is found.

Closest Pair: Scan-Line Analysis

Invariant:

Whenever an event-point pi has been handled, then is the minimum distance between any pair of points encountered from p1 up to pi.

Run time:

O(n log n)

Space:

O(n)

Related Problems

• Nearest neighbour query:– Given a set S of n points, and a query point q,

determine which s S is closest to q.

– Brute force: O(n)

– Goal: O(n log n) pre-processing, O(log n) per query.

• Voronoi region, VR(S): – A set of points closest to a given point.

• Voronoi diagram, VD(S):– Planar subdivision deliniating Voronoi regions.

• Delaunay Triangulation (the dual of VD(S)).

Acknowledgment

• Slides prepared mainly from the materials of Prof. Dr. Thomas Ottman

a tutorial on computational geometry

Technology

pairs p

edge p

set p of points

grahamscan511 p

grahamscan611 p

grahamscan711 p

grahamscan811 p

grahamscan911 p