cs 372: computational geometry lecture 13 arrangements and

CS 372: Computational GeometryLecture 13

Arrangements and Duality

Antoine Vigneron

King Abdullah University of Science and Technology

November 28, 2012

Antoine Vigneron (KAUST) CS 372 Lecture 13 November 28, 2012 1 / 41

1 Introduction

2 Point-line duality

3 Upper envelopes

4 Arrangements of lines

5 An application of arrangements and duality


Outline

We saw planar graph duality in Lecture 10.I In this lecture, a different type of duality: point-line duality.

We introduce line arrangements.

These two notions can often be combined to solve apparently difficultcomputational geometry problems.

References:

Textbook Chapter 8.

Dave Mount’s lecture notes, Lectures 8, 19 and 20.


http://0-www.springerlink.com.library.kaust.edu.sa/content/k18243

http://www.cs.umd.edu/~mount/754

Point-Line Duality

Point-Line Duality: A transformation that exchanges points and lines.

Motivation: For the same amount of work, twice as many results.

DualityTheorem 1 Theorem 1∗

involves points and lines involves lines and points

DualityProgram 1 Program 1∗

deals with points and lines deals with lines and points


Point-Line Duality: Example

(0,−1)

(0.5, 0)

p(2, 1)` : y = 2x − 1

notation:p = `∗


Dual Point

Let ` be a non-vertical line.I ` has equation ` : y = àx − `b.

F à is the slope of `.

I We associate the point `∗ = (à, `b) to `.I `∗ is called the dual of `.

We say that ` is a line in the primal plane.

Here, the primal plane is associated with a coordinate frame Oxy .

We say that `∗ is a point in the dual plane.

We will use a coordinate frame Uab for the dual plane.


Dual Point

O

` : y = àx − `b

1

à

U

`∗(à, `b)

(0,−`b)

Primal plane Dual plane

y b

ax


Dual Line

Let p(px , py ) be a point in the primal plane.

Its dual is a line p∗ in the dual plane with equation

p∗ : b = pxa− py .

O


y b

a

1

px

(0,−py )

Ux

p(px , py )

p∗ : b = pxa− py


Self Inverse

Property (Self inverse)

For any point p in the primal plane, (p∗)∗ = p.

Proof:

p(px , py ) in the primal plane.

p∗ : b = pxa− py .

(p∗)∗ has coordinates (px ,−(−py )).

Property (Self inverse)

For any line ` of the primal plane, (`∗)∗ = `.

Proof:

` : y = àx − `b.

`∗ (à, `b).

(`∗)∗ : y = àx − `b.


Point-Line Duality is Incidence Preserving


`∗

`

p∗

p


Point-Line Duality is Incidence Preserving

Property

p ∈ ` iff `∗ ∈ p∗.

Proof:

Assume p ∈ `.I It means py = àpx − `b.I So `b = pxà − py .I Thus (à, `b) ∈ p∗.

Now assume that `∗ ∈ p∗.I Then (p∗)∗ ∈ (`∗)∗.I By the self-inverse property, it yields p ∈ `.


Multiple Incidence

Corollary

p1, p2 and p3 are collinear iff p∗1 , p∗

2 and p∗3 intersect at a common point.


p1

p2

p3

p∗1

p∗2

p∗3


Order Reversing

Property (Order reversing)

p lies below ` iff p∗ is above `∗.

p

`

`∗

p∗



Example: Dual of a line segment

Let s = pq be a line segment in R2.

How to define its dual?

Its dual s∗ is the union of the duals of the points of s.

All the points in s are collinear, so all the lines in s∗ pass through onepoint.

So it is a double wedge. (See next slide.)

Property

A line ` intersect a segment s iff `∗ is in s∗.


Example: Dual of a line segment

q∗

s∗

s

`

`∗

p

q

p∗



Upper Envelope of Lines


Upper Envelope of Lines

Definition (Upper envelope)

The upper envelope of a set of lines is the set of the points that are aboveall lines.

How to compute the upper envelope of a set L of n lines?

Idea: use duality.

Lines configuration ⇒ points configuration.

We denote L∗ = `∗ | ` ∈ L.


Observation

Assume ` appears as a segment pq in the upper envelope.

` p q

Interpretation in the dual space?


Observation

p and q are on or above all the lines in L.

So p∗ and q∗ are on or below all the points in L∗.

`∗

L∗

p∗

q∗


Observation

So p∗ and q∗ are on the lower part of CH(L∗).

`∗ = p∗ ∩ q∗ is also on the lower part of CH(L∗).

We denote this lower part by LH(L∗).

`∗

CH(L∗)

p∗

q∗


Consequences

The lines that appear in the upper envelope of L correspond to the pointsthat appear in the lower hull of L∗. How to compute the upper envelope?

Compute the lower hull LH(L∗).

Traverse this chain from left to right, output the dual of the vertices.

This gives you a list of lines of L.

These are the lines that appear in the upper envelope.

They are in the same order as they appear in this upper envelope,from left to right.

I Why?


Consequences

We use the algorithm from Lecture 2.

It takes O(n log n) time.

So we can compute an upper envelope of lines in O(n log n) time.

So we can compute the intersection of n halfplanes in O(n log n)time.

I Split into two sets of halfplanes: those that go upward and those thatgo downward.

I The intersection of the upward halfplanes is an upper envelope of lines.I The other subset is a lower envelope. (Similar idea.)I Intersect these two chains. (Plane sweep for instance.)I Overall, it takes O(n log n) time.


Remarks

We have just seen that, in the plane, the following three problems areequivalent:

I Convex hull of a point set.I Upper (lower) envelope of lines.I Halfspace intersection.

In higher dimension, it is similar.I But the intersection of n half-spaces is a polytope that can have

Ω(nbd/2c) vertices.I Voronoi diagrams and Delaunay triangulations can be seen as upper

envelopes in one dimension higher.

These problems, that are all related, are fundamental problems incomputational geometry.

RIC works well for these problems.


Arrangements of Lines

L: Set of n lines

Arrangement A(L)

A Vertex

A cell

An edge


Definition

Definition (Arrangement of lines)

Let L be a set of n lines in R2. These lines subdivide R2 into severalregions, called cells. The edges of this subdivision are line segments orhalf-lines. The vertices are intersection points between two lines of L. Thissubdivision, with adjacency relation between vertices, edges and cells, iscalled the arrangement A(L) of L.


Bounding Box

We restrict our attention to a bounding box B that contains all thevertices of A(L).

B

Now all edges and faces are bounded.

How fast can we compute such a bounding box?


General Position Assumptions

No two lines are parallel.

No three lines intersect at one point.

- Three lines intersect at one point.

A degenerate case:- Two lines are parallel.


Combinatorial Complexity

Definition (Combinatorial complexity)

The combinatorial complexity of an arrangement A(L) is the total numberof vertices, edges, and faces in A(L).

This quantity is Θ(n2) for an arrangement of n lines.

More precisely, if L is in general position.I A(L) has

(n2

)vertices.

I A(L) has n2 edges.I A(L) has

(n2

)+ n + 1 faces.

Proof?


Zone

Definition (Zone)

The zone of a line ` /∈ L is the set of cells in A(L) that intersect `.

`

zone of `


Zone Theorem

Theorem (Zone Theorem)

The total number of edges of all the cells in the zone of a line ` /∈ L inA(L) is O(n).

In other words: the combinatorial complexity of the zone of a line islinear.

Why is it not obvious?

Some lines of L appear in several cells of the zone; See previous slide.


Proof

Definition

Left-bounding and right-bounding edges are edges that bound a cell of thezone of L from the left and from the right, respectively.

`

Left-boundingedge

left and right

Right-boundingedge


Proof

We insert lines of L from left to right along `.

Inserting a line increases by at most 3 the number of left boundingedges.

It follows that the zone has at most 3n left bounding edges.

So the zone of ` has at most 6n edges.

See picture next slide, and detailed proof in D. Mount’s notes.


Proof

new

`n

split

split

`

Inserting `n: Two left bounding edges are split and one is created.

Total: +3 left bounding edges.


Constructing an Arrangement

`


Constructing an arrangement

Incremental algorithm.

We insert the lines one by one and update the arrangement.

Arrangement maintained in a Doubly Connected Edge List.

Insertion of a new line `.

Find the leftmost point of ` in the bounding box and the cell thatcontains it.

It takes O(n) time.

Traverse the zone of ` from left to right and update the arrangementaccordingly.

By the Zone Theorem, can be done in O(n) time.

Overall, we compute a DCEL of A(L) in O(n2) time.


An application of arrangements and duality

Problem (Smallest triangle)

Given a set P of n points in R2, find the triangle with smallest area whosevertices are in P.

We assume general position: No three points are collinear.


Characterization

Let (a, b) ∈ P2

How can we find c ∈ P such that area of a, b, c is minimized?

Find the largest empty corridor along line ab.

bc

a

`c

c lies on its boundary.


Characterization

c lies on a line `c such that:

`c is parallel to ab,

and there is no other line with same slope between ab and `c .

What does it mean in the dual plane?

We call (ab)∗ the dual of line ab.

(`c)∗ is on c∗.

(`c)∗ and (ab)∗ have same abscissa.

No line p∗ where p ∈ P crosses the line segment with endpoints (ab)∗

and (`c)∗.


Characterization

b∗

(`c)∗

c∗

a∗

(ab)∗


Consequences

(`c)∗ is in the same cell of A(P∗) as (ab)∗.

(`c)∗ is vertically above or below (ab)∗.

Once A(P∗) is computed, only two candidates involving a and b.

We compute A(P∗) in O(n2) time.

For all cell of this arrangement, we compute by plane sweep the pointof the boundary that is vertically above or below every vertex of thecell.

It takes time linear in the number of edges of the cell as they aregiven in counterclockwise order.


Algorithm

For each triple a∗, b∗, c∗ we found, we can compute the area oftriangle a, b, c in O(1) time.

We maintain the minimum value found so far in O(1) time by trianglewe consider.

Overall, the running time of our algorithm is proportional to thecombinatorial complexity of A(P∗).

So it runs in O(n2) time.


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