Computational Geometry

Piyush Kumar(Lecture 4: Planar Graphs, Polygons and Art Gallery)

Welcome to CIS5930

Definitions

Planar Straight Line Graph No two edges intersect except at

endpoints

Courtesy Lovasz

Every planar graph can be drawnin the plane with straight edges

Fáry-Wagner

PSLG

A graph is called planar if it can be drawn in the plane in such a way that no two edges intersect except possibly at endpoints. 4K

The clique of size 4

PSLG : Definitions

• A plane graph cuts the plane into regions called faces.

4K

4 faces! 6 edges 4 vertices

v – e + f = ?

PSLGWhat about K3,3 ?

PSLG

Another example

v – e + f = ?

Euler 1752

For any connected planar graph G, vertices – edges + faces = 2

Let v = # of vertices e = # of edges f = # of faces

Why study PSLGs?

PSLG

Any planar graph can be triangulated Draw the planar graph with straight

edges For every face with more than 3

edgesoInsert new edges

PSLG

Any planar graph can be embedded on the sphere with ‘straight line’ edges Stereographic projection

oWrapping the plane on a sphere

Euler’s Formula: Proof

Sum of angles of a triangle?

What about a triangle on a sphere?

Courtesy Hopf

Formula for the spherical excess

Area of a 2-gon = ?Formula for spherical excess?

Euler’s Formula Proof

2v –f = 4; 3f = 2e; and hence v-e+f = 2Note that f = O(v) and e=O(v)

Food for thought

Prove that on a torus V – e + f = 0

In general for g handles V-e + f = 2-2g

Euler’s formula extensions

In 3D, a polyhedral subdivision can already have e in O(v2) Dehn-Sommerville relations relate the maximum number of edges vertices and faces of various dimensions.

Kuratowski’s Theorem [1930]

A graph is planar if and only if it contains no subgraph obtainable from K5 or K3,3 by replacing edges with paths.

Graph Coloring

A coloring of a graph is an assignment of colors to the vertices of the graph such that the endpoints of every edge has different colors.

Four-Color Theorem [1976]

The vertices of any planar graph can be 4-colored in such a way that no two adjacent vertices receive the same color.

Appel-Haken

Representing PSLGs

Doubly connected edge list Winged Edge Quad Edge Facet edgeSplit edge, Corner edge obj file format ;)

Obj file format

Drawback: How do you find which all vertices a vertex connects to? What if you wanted to jump from a face to an adjacent one?

Half Edges (DCEL)

Vertex List Knows its coordinates Knows one of its incident edges

Face List Pointer to one of its edges

Edges Split the edge into two records

DCEL: Edge record

Quad Edge

Considers the PSLG and its dual

Quad Edge

Edge Record Two vertices , two faces

Vertex A circular list of adjacent edges

Face A circular list of adjacent edges

Quad Edge example

Art Gallery

Visibility inside polygons

When is x visible to y if both x and y are inside a polygon P?

Art gallery problem

Art Gallery Problem Art gallery room is a polygon P. A guard can see all around (360

degrees) Place G(n) guards such that they cover

P. What is minimum G(n) that is

occasionally necessary and always sufficient?

Theorem

Every simple polygon admits a triangulation with n-2 triangles. Proof: Lemma: Every polygon has a diagonal For a triangle the theorem is true. Use MI now.

G(n) from previous theorem

Every n vertex polygon can be guarded using n-2 cameras. (Why?) What if we place the cameras on the diagonals? (n-3 diagonals / 2) What about vertices? If we place a guard on a vertex, it can cover

all the triangles incident on the vertex.

3-colorablity of polygon triangulations

If T be a triangulation of a polygon P, then the vertices of P can be 3-colored.

Why can we do 3-coloring?

Look at the dual graph of the polygon triangulation.

What is the maximum degree of the dual? Each edge of the dual is associated with a diagonal of the polygon. Removal of each edge of the dual splits the dual into two connected components

Why can we do 3-coloring

The dual is a free tree No cycles Every free tree has a leaf (An EAR)

Proof of 3-colorability Use MI For a triangle its trivial For a polygon P, color and remove an

ear, by induction, we are done!

The Art Gallery Theorem

Given a simple polygon with n vertices there exists a set of g(n) = floor(n/3) guards that can cover it. Proof: Use vertex guards: the minimum cardinality vertex set of same color in the 3-coloring of the polygon triangulation.

Art Gallery theorem

In the following example you would use red.

Art Gallery Theorem

Let r <= g <= b be the number of nodes colored with red, green and blue colors (w.l.o.g) Can r > n/3? If not, then r <= n/3 But r has to be an integer So r must be <= floor(n/3)For what kind of Polygon is r = floor(n/3)?

Polygon Triangulation

Triangulation of a polygon refers to the decomposition of P into triangles using non-intersecting diagonals such that no more diagonals can be further added. A maximal set of non-intersecting

diagonals.

Polygon Triangulation

Naïve algorithm? Find diagonal, add it, recurse T(n) = T(n-1) + O(n2)*n = O(n3)

Less Naïve? Find diagonal using existence

theorem T(n) = T(n-1) + O(n) = O(n2)

What about O(n) or O(nlogn)?

History

O(nlogn) Monotone pieces 1978 O(nlogn) D&C 1982 O(nlogn) Plane Sweep 1985 O(nlog*n) Randomized 1991 O(n) Polygon cutting, 1991 O(n) Randomized 2000 O(n) implementable with small constants? ??

What about point set triangulation?

A triangulation T of V is a set of triangles such that Each t in T is incident on 3 vertices of V

and does not contain any other vertex If t1, t2 in T => t1 does not intersect t2 T is a decomposition of the convex hull

of V.

Point Set Triangulation

Point Set triangulation

Can be built using one vertex at a time. Three cases New point outside current convex

hull New point inside triangle New point on an edge of the current

triangulation

Back to Polygon Triangulations

Courtesy Martin Held

Polygon Triangulation

Partition the polygon into monotone pieces O(nlogn) Triangulate them in O(n) time

Monotone polygons

A polygon is x-monotone if the intersection of P with a vertical line is connected ( a point, a line segment or O ).

Monotonicity

in x Implies you can sort all the x-coordinates of the vertices in O(n) implies that there are no “cusps”. A reflex vertex is a vertex with internal

angle > π A reflex vertex is a cusp if its neighbors

both lie to the left or to the right of the vertical line.

Monotone polygons => no cusps

Th: Any polygon that contains no cusps is monotone. Proof: We will prove that any polygon that is non-monotone has a cusp. Lets assume that P’ is a non-monotone polygon due to its upper chain. Let v1, v2…vk be its upper chain and vi be the first vertex such that v(i+1) lies left to vi.

Monotone polygons => no cusps

The vertex leftmost from vi…in the chain vi to vk….has to be a cusp.

Polygon Triangulation

First we partition the polygon into polygons without cusps. (Remove all cusps). O(nlogn) Then we triangulate all the monotone polygons. O(n)

Polygon Triangulation

Sweep from left to right. Remove leftward pointing cusps

Sweep right to left Remove rightward pointing cusps

Triangulate all monotone polygons

Regularization

Remove all cusps that point in the same direction

Regularization

Regularize from left to right removing all leftward pointing cuspsExactly in the same way, Regularize from right to left removing all rightward pointing cusps

Regularization

Once you find a leftward pointing cusp, how do you find a diagonal?

A

BX

V

W

Regularization

Once you find a leftward pointing cusp, how do you find a diagonal?

A

BX

V

W

Monotone Polygon Triangulation

Greedy Algorithm Sweep from left to right adding

diagonals whenever you can.

Greedy plane sweep

Sort vertices from left to right in O(n) Process vertices from left to right Stack keeps vertices that have been encountered but not yet used in a diagonal. When a vertex is encountered add as many diagonals as you can Each diagonal removes a triangle and a vertex

from the stack

Vertices on Stack: A Funnel

Courtesy Nancy Amato

•One boundary is chain of reflex vertices except for rightmost vertex (top of stack).•Other boundary is (partial) edge. (We have not seen bottom endpoint yet).

•Sweepline Invariant: Vertices on the stack conform to this funnel structure.

Vertices on Stack: A Funnel

12

3

5

9

11

10

8

6

4

12

7Split off triangle

Funnel

Partial Edge

Reflex chainLast encounteredvertex

What happens when we encounter a new vertex?

Case 1: Vj is on the side opposite the reflex chain on the stack Easy case: Can add diagonals from

Vj to all vertices on reflex chain except Vi1

2

3

5

9

11

10

8

6

4

12

7

Vj

Vk

Vi

Stack pop all push vj,vk

What happens when we encounter a new vertex?

Case 2: Vj is on the same side as the reflex chain on the stack

VjPopped and pushed

The only interesting case in this case, If Vj extends the reflex chain, we justneed to push it on the stack

Case 2

Can add diagonals from Vj to a consecutive set (possibly empty) of vertices in its same reflex chain (starting from the top of the stack/rightmost vertex.

Check each vertex in turn until finding one that does not make a diagonal

Homework

Textbook question 3.11

New Problem

Given a set of half-planes, compute their intersection in O(nlogn) time. How can we do it using what we

know?

Projective Geometry

Basic Elements: Points , Lines and Planes

Studies properties of geometric figures that remain unchanged under projection Lengths and ratios of lengths are NOT invariant under projectionAsserts parallel lines meets at infinity

Duality

The most remarkable concept in projective geometry The terms point and line are dual and can be interchanged in any valid statement to yield another valid statement

Example

A line contains at least 3 collinear points of the plane. A point contains at least 3 coincident lines of the plane.

Point LineJoin IntersectionCollinear Coincident

Duality in higher dimensions?

In space, the terms plane, line, and point are interchanged with point, line, and plane, respectively, to yield dual statements

Point Line Duality

Duality is usually denoted by a * in the superscript. A point and line are uniquely determined by two

parameters.

p = (a,b) is mapped to p*: y = ax– b l: y = ax - b is mapped to l* = (a, b)

Characteristics((p*)*) = p

p lies above l iff l* lies above p*p = (px, py) lies on l: y = ax-b iff l* lies on p*

Property : p lies above l iff l* lies above p*

p = (px, py)

(px,apx - b)

l: y = ax-b

p lies above lpy > apx - b

l* = (a,b)

p*: y = pxx - py

(a, pxa – py)

l* lies above p*b > pxa – py

iff py > pxa - b

Characteristics of the duality transform

Summary

Observations:

1. Point p on straight line l iff point l * on straight line p *

2. p above l iff l * above p *

New Problem

Given n lines, compute the lower envelope of these lines in O(nlogn) time.