242-535 ADA: 16. CG Topics 1

• Objectiveo an examination of four important CG topicso just a taster of a very large research area

Algorithm Design and Analysis

(ADA)242-535, Semester 1 2014-2015

16. Computational Geometry Topics

242-535 ADA: 16. CG Topics 2

1. Intersection of Multiple Line Segmentso the sweeping algorithm

2. Finding the Convex Hullo Graham Scan, Jarvis' March, QuickHull

3. Finding the Closest Pair of Pointso divide-and-conquer

4. The Art Gallery Problem

Overview

242-535 ADA: 16. CG Topics 3

1.1. Multiple Line Segments

1.2. A Brute-Force Algorithm

1.3. The Sweeping Algorithm

1.4. Implementing Sweeping

1. Intersection of Multiple Line Segments

1.1. Multiple Line Segments

Input: a set of n line segments in the plane. Output: all intersections, and for each intersection the involved segments. .

1.2. A Brute-Force Algorithm

Look at each pair of segments, and check if they intersect.If so, output the intersection.

n(n-1)/2 comparison are needed in the worst case, so the running time is O(n2)

• Most segments do not intersect, or if they do, only with a few other segments

Need a faster algorithm that deals with such situations!

But the lines are sparsly distributed in practice:

1.3. The Sweeping 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

also known as the "Bentley-Ottmann" Algorithmand the Sweep Line Algorithm

Non-Degeneracy Assumptions

No segment is vertical. // this means that the sweep line will always hit a segment at a point.

If an input segment is vertical, then it is rotated clockwise by a tiny angle.

Sweep Line StatusThe 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. endpoints

intersections

event points

AG

C

T

Ordering SegmentsA total order over the segments that intersect the current position of the sweep line:

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.

later

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 eventpoint

¨ 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

242-535 ADA: 16. CG Topics 13

The algorithm manages two kinds of data:• 1. The sweep line status gives the relationships

among the objects intersected by the sweep line.

• 2. The event point queue is a sequence of event points that defines the halting positions of the sweep line.

1.4. Implementing Sweeping

Event Point Queue Operations

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

Manages the 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 currentlybeing managed

242-535 ADA: 16. CG Topics 15

• The sweep line status (T) requires the following operations:• INSERT(T, s): insert segment s into T.• DELETE(T, s): delete segment s from T.• ABOVE(T, s): return the segment immediately above segment

s in T.• BELOW(T, s): return the segment immediately below

segment s in T.

• Use a balanced binary search tree for T (e.g. Red-black trees): O(log n) for each operation

Sweep Line Operations

242-535 ADA: 16. CG Topics 16

ANY-SEGMENTS-INTERSECT(S)1 T = Ø2 sort the endpoints of the segments in S from left to right,breaking ties by putting left endpoints before right endpointsand breaking further ties by putting points with lowery-coordinates first

3 for (each point p in the sorted list of endpoints) {4 if (p is the left endpoint of a segment s) {5 INSERT(T, s)6 if (ABOVE(T, s) exists and intersects s) or

(BELOW(T, s) exists and intersects s)7 return TRUE

}8 if (p is the right endpoint of a segment s) {9 if (both ABOVE(T, s) and BELOW(T, s) exist) and

(ABOVE(T, s) intersects BELOW(T, s))10 return TRUE11 DELETE(T, s)

}}

12 return FALSE

Segments intersect Pseudocode

242-535 ADA: 16. CG Topics 17

Execution Example

edb

the intersection of segments d and b is detected when segment c is deleted

242-535 ADA: 16. CG Topics 18

• If the segment set contains n segments, then ANY-SEGMENTS-INTERSECT runs in time O(n log n)o Line 1 takes O(1) time. o Line 2 takes O(n log n) time, using merge sort or

heapsort. o The for loop of lines 3–11 iterates at most once per

event point, and so with 2n event points, the loop iterates at most 2n times.

o Each iteration takes O(log n) time, since each tree operation takes O(log n) time and each intersection test takes O(1) time (by using the intersection function from part 15).

o Total cost = O(1) + O(n * (log n + 1)) = O(n log n)

Running Time

242-535 ADA: 16. CG Topics 19

2.1. Convex & Concave Sets2.2. The Convex Hull2.3. The Graham Scan2.4. Jarvis’ March2.5. QuickHull2.6. Lower Bound of O(n log n)

2. Finding the Convex Hull

2.1. Convex and Concave Sets

A planar region R is called convex if and only if for any pair of points p, q in R, the line segment pq lies completely in R.

Otherwise, it is called concave.

Convex

p

q

R1

p

q

R2

Concave

2.2. The Convex HullThe convex hull CH(P) of a set of points P is the smallest convex region that contains all of P.

Rubber band

When P is finite, its convex hull is the unique convex polygon whose vertices are from P and that contains all points of P.

The Convex Hull Problem

Input: a set P = { p , p , …, p } of points

Output: a list of vertices of CH(P) in counterclockwise order.

1 2 n

Example

8

pp

p

p

p

p

p

p

pp

9

210

5

76 1

3 4

CH(P) = [ p5, p9, p2, p8, p10, p7 ]

Not all the pointsin P are in CH(P)

Edges of a Convex Hull

For every edge both endpoints p, q P.

pq

All other points in P lie to the same side of the line passing through p and q

all points onthis side of theq – p edge

Floating Arithmetic is not Exact

p

r

qNearly colinear points p, q, r.

p to the left of qr. q to the left of rp. r to the left of qp.

All three accepted as edges!

The algorithm is not robust – it could fail due to small numerical error.

2.3. The Graham Scan

p 0

pp

pp

pp

p

pp

p

12

3

4

p5

6

7

8

9

10

11

The center point has the minimum y-coordinate

Labels are in the polar angle order.(What if two points have the same polar angle?)

How to break a tie?

sort by polar angle

handling degeneracies

242-535 ADA: 16. CG Topics 26

• Consider each of the points in the sorted sequence.

• For each point, is moving from the two previously considered points to this point a "left turn" or "right turn"?

• "Right turn": this means that the second-to-last point is not part of the convex hull and should be removed. o this process is continued for as long as the set of the last

three points is a "right turn"

• "Left turn": the algorithm moves to the next point in the sequence.

A Turning Algorithm

242-535 ADA: 16. CG Topics 27

Turning in Action

C is a left turn compared to A – B; add C

D is a right turn compared to B - C; remove C

D is a left turn compared to A - B; add D

X

Graham-Scan(P) let p0 be the point in P with minimum y-coordinate let p1 , p2 , …, pn-1 be the remaining points in P sorted in counterclockwise order by polar angle around p0 Stack s = new Stack(); s.push(p0) s.push(p1) s.push(p2) for i = 3 to n 1 while ( pi makes a nonleft turn from the line segment determined by s.top() and s.nextToTop() ) s.pop() s.push(pi) return s

The Graham Scan Algorithm

the convex hull pointsare stored on a stack

Stack Usage

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

2

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

3

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

4

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

4

p 5

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

4

p 6

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

4

p 6

p 7

p 8

p

pp

pp

pp

p

pp

p

0

12

3

4

p5

6

7

8

9

10

11

p

p

p

S

0

1

4

p 6

p 7

p

pp

p

p

pp

p

pp

p

0

12

3

4

p5

6

78

9

10

11

p

p

p

S

0

1

4

p 6

p 9

p10

p

pp

p

p

pp

p

pp

p

0

12

3

4

p5

6

78

9

10

11

p

p

p

S

0

1

4

p 6

p 9

p11

Finish

p

pp

p

p

pp

p

pp

p

0

12

3

4

p5

6

78

9

10

11

p

p

p

S

0

1

4

p 6

p 9

p11

Proof of CorrectnessEach point popped from stack S is not a vertex of CH(P).

pp

p

p

0

i

jk

p

p

p

p0

i

j

k

Two cases when pj is popped:

In neither case can pj become a vertex of CH(P).

Proof

pk is a rightturn comparedto pi - pj

pk is a 0 angleturn comparedto pi - pj

X X

The points on stack S always form the vertices of a convex polygon in counterclockwise order (an invariant).

• Popping a point from S preserves the invariant.

The regioncontaining pi The invariant still holds.

Proof • The claim holds at S initialization when p0, p1, p2 form a triangle (which is obviously convex)

• Consider a point pi being pushed onto S.

p0

pj

pi

Running Time

The running time of Graham’s Scan is O(n log n).

#operations time / operation total

Push n O(1) (n)

Pop n 2 O(1) O(n)

Sorting 1 O(n log n) O(n log n)

Why?

Findingp0

1 (n) (n)

2.4. Jarvis’ March A “package/gift wrapping” technique using two "chains"

242-535 ADA: 16. CG Topics 43

• We choose the first vertex as the lowest point p0, and start the right chain.o The next vertex, p1, has the smallest polar angle of any

point with respect to p0. o Then, p2 has the smallest polar angle with respect to p1. o The right chain goes as high as the highest point p3.

• Then,we return to p0 and build the left chain by finding smallest polar angles with respect to the negative x-axis.

The Operation of Jarvis’ March

Running Time of Jarvis’ March

Let h be the number of vertices of the convex hull.

• For each vertex, finding the point with the minimum Polar angle, that is, the next vertex, takes time O(n)

• The comparison between two polar angles can be done using the cross product.

Thus O(nh) time in total.

242-535 ADA: 16. CG Topics 45

• Concentrate on points close to hull boundary• Named for similarity to Quicksort

o O(n log n)

2.5. QuickHull

Set QuickHull(a, b, S) if S = 0 return {} else c = index of point with max distance from a-b A = points strictly right of (a, c) B = points strictly right of (c, b) return QuickHull(a, c, A) + {c} + QuickHull(c, b, B)

a

b

finds one of upper or lower hull

c

A

242-535 ADA: 16. CG Topics 46

• The worst-case time to find the convex hull of n points using a decision tree model is O(n log n)

• Proof based on sorting:o Given an unsorted list of n numbers: (x1,x2 ,…,

xn)o Form an unsorted set of points: (xi, xi

2) for each xi

o Convex hull of these points produces a sorted list!

• a parabola: so every point is on the convex hull

o Finding the convex hull of n points is therefore at least as hard as sorting n points, so worst-case time is in O(n log n)

2.6. Lower Bound of O(n log n)

Convex hullof red pts

242-535 ADA: 16. CG Topics 47

• There are a set of n points P = { p1,…pn }.

• Find a pair of points p, q such that |p – q| is the minimum of all |pi – pj|

• Easy to do in O(n2) timeo for all pi ≠ pj, compute ║pi - pj║ on all the pairs

and choose the minimum, which involves n(n-1)/2 comparisons

• We will aim for O(n log n) time

3. Finding the Closest Pair of Points

242-535 ADA: 16. CG Topics 48

• Divide:o Compute the median of the x-coordinateso Split the points into a left half PL and right half

PR, each of size n/2

• Conquer: compute the closest pairs for PL and PR

• Combine the results (the hard part)

Divide and Conquer

PL PR

median line

242-535 ADA: 16. CG Topics 49

• dL = closestDist(PL)dR = closestDist(PR)d = min( d1, d2 )

• Observe:o Need to check only pairs which

cross the dividing lineo Only interested in pairs within

distance < d of each other

• It's enough to look at only the points in the 2d wide strip around the median line

Combine

PL PR

median line

242-535 ADA: 16. CG Topics 50

• Sort all points in the strip by their y-coords, forming q1…qk, k ≤ n.

• Let yi be the y-coord of qi dmin = d for i = 1 to k { j = i - 1 while (yi - yj < d){ if(║qi-qj║< d) dmin =║qi-qj║ j = j-1 } }

• Report dmin (and the corresponding pair)

Scanning the Strip

242-535 ADA: 16. CG Topics 51

• Combine: O(n log n) because we sort the y-coords. But, we can:o Sort all points the y-coords at the beginning, outside of

the main loop since the Divide stage preserves the y-order of points

o Then this combine stage only takes O(n)

• We get T(n)=2T(n/2)+O(n), so T(n) is O(n log n)

Running Time

divide the data by half and calculate the closest pair on PL and PR

combine

4. The Art Gallery Problem

camera

How many cameras are needed to guard a gallery and where should they be placed?

Simple Polygon ModelModel the art gallery as a region bounded by some simple polygon (no self-crossing). Regions with holes are not allowed.

convex polygonone camera

an arbitrary n-gon (n vertices)

Bad news: finding the minimum number of cameras for a given polygon is NP-hard (exponential time).

TriangulationTo make things easier, we decompose a polygon into pieces that areeasy to guard. Draw diagonals between pairs of vertices.

diagonals

Triangulation: decomposition of a polygon into triangles by a maximal set of non-intersecting diagonals.

Guard the art galleryby placing a camera in every triangle …

No. of CamerasTheorem: Every simple polygon has a triangulation. Any triangulation of a simple polygon with n vertices consists of n – 2 triangles. n – 2 cameras can guard the simple polygon.

Note: a camera sitting on a diagonal guards two triangles. the no. of cameras can be reduced to roughly n/2

Note: a vertex is adjacent to many triangles so placing cameras at vertices can reduce the number even more…

3-Coloring of Vertices Idea: Select vertices, such that every triangle has at least oneof those vertices.

Assign each vertex a color: pink, green, or yellow.

Any two vertices connected by an edge or a diagonal must be assigned different colors.

Thus the vertices of every trianglewill be in three different colors.

Choose the smallest color set, and place cameras at all the vertices using that color n/3 cameras.

The Dual GraphA dual graph G has a node inside every triangle and an edge between every pair of nodes whose corresponding triangles share a diagonal.

The dual graph G is actually a tree (marked in red on this slide), and it is possible to build one using DFS.