Computational Geometry - uni-wuerzburg.de

1 - 1

Computational Geometry

Lecture #1

Convex Hullor

Mixing Things

Dr. Philipp Kindermann Winter Semester 2018

1 - 2

Computational Geometry

Lecture #1

Convex Hullor

Mixing Things

Dr. Philipp Kindermann Winter Semester 2018

2 - 1

Chair I

a) Efficient Algorithms

b) Knowledge-Based SystemsProf. Dietmar Seipel

c) Theoretical Comp. ScienceProf. Christian Glaßer

2 - 2

Chair I

a) Efficient Algorithms

b) Knowledge-Based SystemsProf. Dietmar Seipel

Alexander Wolff Philipp Kindermann

Thomas van Dijk

Steven Chaplick

c) Theoretical Comp. ScienceProf. Christian Glaßer

Professor PostDoc

PostDocMyroslav Kryven

Oksana Firman

Johannes ZinkAndre Loffler

PostDoc

3 - 1

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

3 - 2

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Wurzburg 2006 – 2011: Study

3 - 3

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Wurzburg 2006 – 2011: Study2011 – 2015: PhD

3 - 4

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Wurzburg

Hagen

2006 – 2011: Study2011 – 2015: PhD

2015 – 2018:PostDoc

3 - 5

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Wurzburg

Hagen

2006 – 2011: Study2011 – 2015: PhD

2015 – 2018:PostDoc

3 - 6

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Waterloo

3 - 7

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Waterloo

3 - 8

Philipp Kindermannfirstname.lastname@uni-wuerzburg.deEmail:

Office hours: Wed, 14–15h

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Philipp Kindermannfirstname.lastname@uni-wuerzburg.de

Office:

Email:

E14 @ Former Math Build. (M8)Office hours: Wed, 14–15h

3 - 10

Philipp Kindermannfirstname.lastname@uni-wuerzburg.de

Office:

Email:

E14 @ Former Math Build. (M8)Office hours: Wed, 14–15h

3 - 11

Philipp Kindermannfirstname.lastname@uni-wuerzburg.de

Office:

Email:

E14 @ Former Math Build. (M8)Office hours: Wed, 14–15h

Research areas:

– Comp. Geometry– Graph Algorithms– Graph Visualization

4 - 1

Computational GeometryLearning goals: At the end of this lecture you will be able to

4 - 2

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

At the end of this lecture you will be able to

4 - 3

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

At the end of this lecture you will be able to

– analyze new problems and find – with the concepts ofthe lecture – efficient solutions.

4 - 4

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

Requirements:

At the end of this lecture you will be able to

– analyze new problems and find – with the concepts ofthe lecture – efficient solutions.

– Big-Oh notation (Landau); e.g., O(n log n)

4 - 5

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

Requirements:

At the end of this lecture you will be able to

– analyze new problems and find – with the concepts ofthe lecture – efficient solutions.

– Some basic Algorithms & Data Structures– Big-Oh notation (Landau); e.g., O(n log n)

(Balanced) binary search tree, priority queue

4 - 6

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

Requirements:

At the end of this lecture you will be able to

– analyze new problems and find – with the concepts ofthe lecture – efficient solutions.

– Some basic Algorithms & Data Structures

– Some basic Algorithmic Graph Theory

– Big-Oh notation (Landau); e.g., O(n log n)

(Balanced) binary search tree, priority queue

Breadth-first search, Dijkstra’s algorithm

4 - 7

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

Requirements:

At the end of this lecture you will be able to

– analyze new problems and find – with the concepts ofthe lecture – efficient solutions.

– Some basic Algorithms & Data Structures

– Some basic Algorithmic Graph Theory

– Oral exam at the end of the semester

– Big-Oh notation (Landau); e.g., O(n log n)

(Balanced) binary search tree, priority queue

Breadth-first search, Dijkstra’s algorithm

4 - 8

Computational GeometryLearning goals:

– decide which algorithms help to solve a number offundamental geometric problems,

Requirements:

My vision: • “hands-on”• interactive

At the end of this lecture you will be able to

– analyze new problems and find – with the concepts ofthe lecture – efficient solutions.

– Some basic Algorithms & Data Structures

– Some basic Algorithmic Graph Theory

– Oral exam at the end of the semester

– Big-Oh notation (Landau); e.g., O(n log n)

(Balanced) binary search tree, priority queue

Breadth-first search, Dijkstra’s algorithm

5

Content1. Convex Hull in 2D2. Map Overlay3. Polygon Triangulation4. Linear Programming5. Orthogonal Range Queries6. Point Location7. Voronoi Diagram8. Delaunay Triangulation9. Convex Hull in 3D

10. Motion Planning11. Simplex Range Searching12. Arrangements and Duality13. Binary Space Partition14. Randomized Triangulation

6 - 1

LiteratureM. de Berg, O. Cheong, M. van Kreveld, M. Overmars:Computational Geometry: Algorithms & Applications.Springer, 3rd edition, 2008

Main resource for this course!Abbreviated as: Comp. Geom A&A

6 - 2

Literature

Rolf Klein:Algorithmische Geometrie: Grundlagen, Methoden,Anwendungen.Springer, 2nd edition, 2005

M. de Berg, O. Cheong, M. van Kreveld, M. Overmars:Computational Geometry: Algorithms & Applications.Springer, 3rd edition, 2008

Main resource for this course!Abbreviated as: Comp. Geom A&A

6 - 3

Literature

Rolf Klein:Algorithmische Geometrie: Grundlagen, Methoden,Anwendungen.Springer, 2nd edition, 2005

M. de Berg, O. Cheong, M. van Kreveld, M. Overmars:Computational Geometry: Algorithms & Applications.Springer, 3rd edition, 2008

Ketan Mulmuley:Computational Geometry: An Introduction ThroughRandomized Algorithms. Prentice Hall, 1st edition, 1993

Main resource for this course!Abbreviated as: Comp. Geom A&A

7

Week 1

Mixing Thingsor

Convex Hull

[Comp. Geom A&A : Chapter 1, (see also Ch. 11)]

8 - 1

Mixing Things

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

Given...

8 - 2

Mixing Things

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

?using s1, s2, s3

8 - 3

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

?using s1, s2, s3

8 - 4

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

8 - 5

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

8 - 6

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

Observe: Given a set S ⊂ R2 of substances, we can mix asubstance q ∈ R2 using the substances in S ⇔

8 - 7

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

Observe: Given a set S ⊂ R2 of substances, we can mix asubstance q ∈ R2 using the substances in S ⇔q ∈ CH(S).

8 - 8

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

Observe: Given a set S ⊂ R2 of substances, we can mix asubstance q ∈ R2 using the substances in S ⇔q ∈ CH(S).

s4

8 - 9

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

Observe: Given a set S ⊂ R2 of substances, we can mix asubstance q ∈ R2 using the substances in S ⇔q ∈ CH(S).

s4

s5

8 - 10

Mixing Things

.1 .2 .3 .4

subst. fract. A fract. Bs1 10 % 35 %s2 20 % 5 %s3 40 % 25 %

.4

.3

.2

.1

B

A

s1

s2

s3

Given...

subst. fract. A fract. Bq1 25 % 28 %q2 15 % 15 %

can we mix

q1

q2

?using s1, s2, s3

Observe: Given a set S ⊂ R2 of substances, we can mix asubstance q ∈ R2 using the substances in S ⇔q ∈ CH(S).

d

d

s4

s5

9 - 1

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

9 - 2

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach:

9 - 3

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope

9 - 4

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go

9 - 5

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go

9 - 6

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go– take area inside (and on) the rope

9 - 7

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go– take area inside (and on) the rope

9 - 8

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go– take area inside (and on) the rope

Math approach:

9 - 9

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go– take area inside (and on) the rope

Math approach: – define convex

9 - 10

Formally. . .

Given S ⊂ R2, how do we define the convex hull CH(S)?

Physics approach: – take (large enough) elastic rope– stretch and let go– take area inside (and on) the rope

Math approach: – define convex– define CH(S) =

⋂C⊇S : C convex

C

10 - 1

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

10 - 2

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

This set is HUGE!

10 - 3

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

This set is HUGE!

Maybe we can do with a little less?

10 - 4

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

This set is HUGE!

Maybe we can do with a little less?

Claim: CH(S) =

10 - 5

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

This set is HUGE!

Maybe we can do with a little less?

Claim: CH(S) =⋂

H⊇S :

H

H closed halfplane

10 - 6

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

This set is HUGE!

Maybe we can do with a little less?

Claim: CH(S) =⋂

H⊇S :

H =

H closed halfplane

10 - 7

Towards Computation

Problem with math approach:

CH(S) =def⋂

C⊇S : C convex

C

This set is HUGE!

Maybe we can do with a little less?

Claim: CH(S) =⋂

H⊇S :

H =⋂

H⊇S : H cl. halfplane,

H

|∂H ∩ S| ≥ 2H closed halfplane

11 - 1

Computer Science ApproachInput: set S of n points in the plane, that is, S ⊂ R2

11 - 2

Computer Science ApproachInput: set S of n points in the plane, that is, S ⊂ R2

11 - 3

Computer Science ApproachInput: set S of n points in the plane, that is, S ⊂ R2

11 - 4

Computer Science ApproachInput:

Output:

set S of n points in the plane, that is, S ⊂ R2

list of vertices of CH(S) in clockwise order

11 - 5

Computer Science ApproachInput:

Output:

set S of n points in the plane, that is, S ⊂ R2

list of vertices of CH(S) in clockwise order

Observation. (p, q) is an edge of CH(S) ⇔

p q

11 - 6

Computer Science ApproachInput:

Output:

set S of n points in the plane, that is, S ⊂ R2

list of vertices of CH(S) in clockwise order

Observation. (p, q) is an edge of CH(S) ⇔each point in S lies– strictly to the right of the directed line −→pq or– on the line segment pq

p q−→pq

12 - 1

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

12 - 2

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

// test: (p, q) edge of CH(S)?

12 - 3

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

// test: (p, q) edge of CH(S)?

12 - 4

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

12 - 5

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

Important:Test takes O(1) time!

12 - 6

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

Important:Test takes O(1) time!

12 - 7

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

Important:Test takes O(1) time!

12 - 8

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

Important:Test takes O(1) time!

12 - 9

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

Important:Test takes O(1) time!

12 - 10

Finally, an Algorithm

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

r strictly right of −→pqm∣∣∣∣∣∣

xr yr 1xp yp 1xq yq 1

∣∣∣∣∣∣ < 0

Important:Test takes O(1) time!

13 - 1

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

13 - 2

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

13 - 3

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

Θ(1)

13 - 4

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

Θ(1)

Θ(n)

13 - 5

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

Θ(1)

Θ(n)

(n2 − n)·

13 - 6

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

Θ(1)

Θ(n)

(n2 − n)·

Θ(n

3 )

13 - 7

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

Θ(1)

Θ(n)

(n2 − n)·

Θ(n

3 )O(n

2 )

13 - 8

Running Time Analysis

FirstConvexHull(S)E← ∅foreach (p, q) ∈ S× S with p 6= q do

valid← trueforeach r ∈ S do

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

if valid thenE← E ∪ {(p, q)}

from E construct sorted list L of vertices of CH(S)return L

Θ(1)

Θ(n)

(n2 − n)·

Θ(n

3 )O(n

2 )

Lemma. We can compute the convex hull of n pts in theplane in Θ(n3) time.

14 - 1

Discussion if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

14 - 2

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

14 - 3

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

14 - 4

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

Test may returnwrong answer(floating pt arithmetic!):

r right of −→pq :-(

14 - 5

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

E

Test may returnwrong answer(floating pt arithmetic!):

r right of −→pq :-(

p

r

q

14 - 6

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

E

Test may returnwrong answer(floating pt arithmetic!):

r right of −→pq :-(

p not right of −→rq

p

r

q

14 - 7

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

E

Test may returnwrong answer(floating pt arithmetic!):

r right of −→pq :-(

p not right of −→rq

q not right of −→pr

p

r

q

14 - 8

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

E

Test may returnwrong answer(floating pt arithmetic!):

r right of −→pq :-(

p not right of −→rq

q not right of −→pr

p

r

q

14 - 9

Discussion

r

p

q E

if not (r strictly right of −→pq or r ∈ pq) thenvalid← false

E

Test may returnwrong answer(floating pt arithmetic!):

r right of −→pq :-(

p not right of −→rq

q not right of −→pr

p

r

q

Observation. Algorithm FirstConvexHull is not robust.

15 - 1

New Ideas (Graham Scan)

15 - 2

New Ideas (Graham Scan)• split computation in two

15 - 3

New Ideas (Graham Scan)• split computation in two

upper convex hull

15 - 4

New Ideas (Graham Scan)• split computation in two

upper convex hull

lower convex hull

15 - 5

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hull

15 - 6

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hull• proceed incrementally

15 - 7

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hullUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

• proceed incrementally

15 - 8

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hullUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

• proceed incrementally

15 - 9

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hullUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

• proceed incrementally

15 - 10

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hullUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

• proceed incrementally

// compute upper convex hull of {p1, p2, . . . , pi}

15 - 11

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hullUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

• proceed incrementally

// compute upper convex hull of {p1, p2, . . . , pi}

15 - 12

New Ideas (Graham Scan)• split computation in two

• bring pts in lexicographic order

upper convex hull

lower convex hullUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

• proceed incrementally

// compute upper convex hull of {p1, p2, . . . , pi}

16 - 1

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

16 - 2

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

16 - 3

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

16 - 4

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

16 - 5

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

Amortized analysis:

16 - 6

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

Amortized analysis:– each pt p2, . . . , pn−1 pays 1e for its potential removal later on

16 - 7

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

Amortized analysis:– each pt p2, . . . , pn−1 pays 1e for its potential removal later on– this pays for the total effort of all executions of the while loop

16 - 8

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

Amortized analysis:– each pt p2, . . . , pn−1 pays 1e for its potential removal later on– this pays for the total effort of all executions of the while loop

16 - 9

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

Amortized analysis:– each pt p2, . . . , pn−1 pays 1e for its potential removal later on– this pays for the total effort of all executions of the while loop

O(n)

16 - 10

Running Time AnalysisUpperConvexHull(S: set of pts in the plane)〈p1, p2, . . . , pn〉 ← sort S lexicographicallyL← 〈p1, p2〉for i← 3 to n do

L.append(pi)while |L| > 2 and last 3 pts in L make no right turn do

remove second last pt from L

return L

O(n log n)

(n− 2)·

?

Amortized analysis:– each pt p2, . . . , pn−1 pays 1e for its potential removal later on– this pays for the total effort of all executions of the while loop

O(n)

Theorem. We can compute the convex hull of n pts in theplane in O(n log n) time – in a robust way.

17 - 1

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 2

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 3

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 4

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 5

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 6

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 7

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 8

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 9

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 10

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

17 - 11

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

Runtime?

17 - 12

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

Runtime? O(n · h)

17 - 13

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

. . . where h = |CH(S)| = size of the output

Runtime? O(n · h)

17 - 14

Output-Sensitive Algorithms• Jarvis’ gift-wrapping algorithm (aka Jarvis’ march)

• Chan’s exponential-search algorithm O(n log h)

. . . where h = |CH(S)| = size of the output

Runtime? O(n · h)

18 - 1

Chan’s Algorithm P1P2

P4

P3

18 - 2

Chan’s Algorithm P1P2

P4

P3

[in O(m log m) time]

18 - 3

Chan’s Algorithm P1P2

P4

P3

[in O(m log m) time]

p1

p0

18 - 4

Chan’s Algorithm P1P2

P4

P3

[in O(m log m) time]

p1

p0

18 - 5

Chan’s Algorithm P1P2

P4

P3

q1

q3q4

q2

[in O(m log m) time]

pk

pk−1

pk+1

p1

p0

18 - 6

Chan’s Algorithm P1P2

P4

P3

q1

q3q4

q2

[in O(m log m) time]

pk

pk−1

pk+1

p1

p0

18 - 7

Chan’s Algorithm P1P2

P4

P3

q1

q3q4

q2

[in O(m log m) time]

pk

pk−1

pk+1

p1

p0

19

Chan’s AlgorithmInitially, we assume that the value of h is known and make a parameter m = h. This assumption is not realistic, but weremove it later. The algorithm starts by arbitrarily partitioning P into at most 1 + n

m subsets Q with at most m pointseach. Then, it computes the convex hull of each subset Q using an O(n log n) algorithm – Graham’s scan. Note that, asthere are O(n/m) subsets of O(m) points each, this phase takes O(n/m) ·O(m log m) = O(n log m) time.

The second phase consists of executing the Jarvis’ march algorithm algorithm and using the precomputed convex hullsto speed up the execution. At each step in Jarvis’s march, we have a point pi in the convex hull, and need to find a pointpi+1 = f (pi , P) such that all other points of P are to the right of the line pi pi+1. If we know the convex hull of a set Qof m points, then we can compute f (pi , Q) in O(log m) time, by using binary search. We can compute f (pi , Q) for allthe O(n/m) subsets Q in O(n/m log m) time. Then, we can determine f (pi , P) using the same technique as normallyused in Jarvis’s march, but only considering the points that are f (pi , Q) for some subset Q. As Jarvis’s march repeatsthis process O(h) times, the second phase also takes O(n log m) time, and therefore O(n log h) time if m = h.By running the two phases described above, we can compute the convex hull of n points in O(n log h) time, assumingthat we know the value of h. If we make m < h, we can abort the execution after m + 1 steps, therefore spending onlyO(n log m) time (but not computing the convex hull). We can initially set m as a small constant (we use 2 for ouranalysis, but in practice numbers around 5 may work better), and increase the value of m until m > h, in which case weobtain the convex hull as a result.If we increase the value of m too slowly, we may need to repeat the steps mentioned before too many times, and theexecution time will be large. On the other hand, if we increase the value of m too quickly, we risk making m much largerthan h, also increasing the execution time. Similar to strategy used by Chazelle and Matousek’s algorithm, Chan’salgorithm squares the value of m at each iteration, and makes sure that m is never larger than n. In other words, at

iteration t (starting at 1), we have m = min(

n, 22t)

. The total running time of the algorithm is

dlog log he∑

t=1O(

n log(22t)

)= O(n)

dlog log he∑

t=1O(2t ) = O

(n · 21+dlog log he) = O(n log h).

To generalize this construction for the 3-dimensional case, an O(n log n) algorithm to compute the 3-dimensionalconvex hull should be used instead of Graham scan, and a 3-dimensional version of Jarvis’s march needs to be used.The time complexity remains O(n log h).

[Text copied on October 17, 2017 from:https://en.wikipedia.org/wiki/Chan’s algorithm]