AMS 345/CSE 355

Computational Geometry

Lecture: Polygons, Guarding

Joe Mitchell

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Meets: Friday 1:05 - 2:30 pm at CSE 2120

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Simple Polygons

Definition: A simple polygon P is the (closed) region bounded by a “simple closed polygonal curve”.

Simple Polygon

Definition in [O’Rourke]:

Simple Polygons

Alternate Definition: P is a simple polygon if it is a simply connected (i.e., no “holes”) subset of the plane whose boundary consists of a connected finite union of straight line segments.

Simple Polygons

Some definitions would allow this as a “degenerate” simple polygon

Definitions: Visibility, Diagonals

For p,q in P, p is visible to q if segment pq lies within (closed) P

p q

Definitions: Visibility, Diagonals

For p,q in P, p is visible to q if segment pq lies within (closed) P

p is clearly visible to q if p is visible to q AND the only points in common between pq and P are possibly p and q

p q

q’

p clearly sees q but does not clearly see q’ p sees q’

Definitions: Visibility, Diagonals

vivj is a diagonal if vi and vj are vertices that clearly see each other (versus: chord pq, with p and q on the boundary of P)

vi

vj p

q

pq is a chord (not a diagonal) vivj is a diagonal vkvm is not a diagonal

vk

vm

Diagonals

Triangulation

Definition: A partition of P into triangles by a maximal set of noncrossing diagonals.

Triangulation: Existence

Proof: Induction on n.

3D: Polyhedra

Polyhedra: Tetrahedralization

Polyhedron with No

Tetrahedralization

Combinatorics: Triangulations

Proof: induction on n. True for n=3 (trivially). Assume true for n=k (the Induction Hypothesis). Take an n-gon, P, with n=k+1. We know it has a diagonal (since n>3). The diagonal partitions P into two polygons, of sizes n1 and n2 , with n1 +n2 =n+2 (since the endpoints of the diagonal are counted twice) By IH, the two subpolygons have triangulation with n1-2 and n2-2 triangles; glue together along the diagonal to get a total of (n1-2)+(n2-2)=n-2 triangles in the overall triangulation. Similar for number of diagonals.

Ears

A diagonal of the form vi-1vi+1 is an ear diagonal; the triangle vi-1vivi+1 is an ear, and vi is the ear tip

Note that there are at most n ears (and that a convex polygon has exactly n ears)

vi-1 vi+1

vi

Ears

Proof(1): There are n edges of P and n-2 triangles in any triangulation. Imagine dropping the n edges into the n-2 “pigeonholes” corresponding to the triangles: Each edge appears on boundary of some triangle. By pigeonhole principle, at least 2 triangles get 2 edges “dropped in their box”. (2) Consider the planar dual (excluding the face at infinity) of a triangulation of P. Claim: The dual graph for a triangulated simple polygon is a TREE. Any tree of 2 or more nodes has at least 2 nodes of degree 1.

Ear-Clipping Triangulation

Ear-clipping applet

Input: Simple polygon P

vi-1 vi+1

pq is a diagonal, cutting off a single triangle (the “ear”)

Naive: O(n3) Smarter: Keep track of “ear tip status” of each vi (initialize: O(n2) ) Each ear clip requires O(1) ear tip tests ( @ O(n) per test ) Thus, O(n2) total, worst-case

vi

Number of Triangulations

The diagonals, together with the edges of convex polygon P, form the complete graph, Kn

Number of Triangulations:

Convex Polygons

The first Catalan numbers, for n=1,2,3,… are given by: 1, 2, 5, 14, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452,…

Number of Triangulations:

Convex Polygons

Number of Triangulations:

Convex Polygons

Number of Triangulations:

Simple Polygons

Question: Is it possible to find a simple polygon P having exactly k triangulations, for every positive integer k?

Exercise

Can you find a simple polygon that has exactly 3 different triangulations?

Number of Triangulations:

Simple Polygons

Number of Triangulations:

Simple Polygons

Example: Diagonals Set of all diagonals, together with the edges of P, forms the visibility graph of P

Forced diagonals: Must be in ANY triangulation of P: Any diagonal that is not crossed by other diagonals.

Number of Triangulations:

Simple Polygons

Example:

Total number of different triangulations:

1 * 2 * C5-2 = 1 * 2 * 5 = 10

1

2

5

More Examples: HW1

Counting Triangulations:

Example

Counting Triangulations:

Example

Forced diagonals: Those not crossed by any other diagonals

Step 1: Draw all diagonals

Counting Triangulations:

Example Step 2: Count number of triangulations in each separate subpolygon

3

C4 = 14

Overall: 4*3*14

4

By case analysis

Counting Triangulations:

Example How many triangulations of subpolygon?

0

1

2

3

4

5 Case analysis: (1) Use 03: one completion (since, once we use 03, diagonals 24, 14, and 14 are ruled out, leaving us with just 04 and 02; i.e., diagonal 03 splits polygon into two nonconvex quadrilaterals, each with a unique triangulation)

Counting Triangulations:

Example How many triangulations?

0

1

2

3

4

5 Case analysis: (1)Use 03: one completion (2) Do NOT Use 03: Then, 24 is forced, and there are 3 triangulations possible (using 15 and 14, or 04 and 14, or 04 and 02). Total for cases (1) and (2): 1+3=4

Counting Triangulations

Efficient algorithm?

Yes! (for simple polygons P, but not for polygons with holes)

Idea: use “dynamic programming” (recursive solution)

Input: Set S of n points

Input: Other shapes

3D: Surfaces and solids (tetrahedralization)

Triangulation

Triangulation applet for simple polygons

Planar Straight-Line Graph (PSLG)

Simple polygon

Polygon with holes

Triangulation Theory in 2D

Thm: A simple polygon has a triangulation.

• Lem: An n-gon with n4 has a diagonal.

Thm: Any triangulation of a simple n-gon has n-3 diagonals, n-2 triangles.

Thm: The “dual” graph is a tree.

Thm: An n-gon with n4 has 2 “ears”.

Thm: The triangulation graph can be 3-colored.

Proofs: Induction on n.

Also with holes

But, NOT true in 3D!

Triangulating a Polygon

Simple “ear-clipping” methods: O(n2 )

Cases with simple O(n) algorithms:

• Convex polygons (trivial!)

• Monotone polygons, monotone mountains

General case (even with holes!):

• Sweep algorithm to decompose into

monotone mountains

• O(n log n)

Best theoretical results:

• Simple polygons: O(n) [Chazelle’90]

• Polygons with h holes: O(n+h log1+ h), (n+h log h) [BC]

Good practical method: FIST [Held], based on clever methods of ear clipping (worst-case O(n2 ) )

Not practical!

fan

Tetrahedralizing Polyhedra

Tetrahedralizing Polyhedra

Today, 9/5/13

Review triangulation

Guarding problem

Art Gallery Theorem

Computing guard numbers by hand

Examples

Begin Convex Hulls

Guarding Polygons

V(p) = visibility polygon of p inside P = set of all points q that p sees in P

Guarding Polygons

Goal: Find a set of points (“guards”) within P so that their VP(p) sets cover P “Guard cover” “Point guards” versus “vertex guards” Regular visibility versus “clear visibility”

Determine a small set of guards to see all of a given

polygon P

5 guards suffice to cover P (what about 4 guards? 3?)

Computing min # of guards, g(P), for n-gon P is NP-hard Challenge/open: Compute g(P) approximately

Min-Guard Coverage Problem

Art Gallery Theorem

Answers a question of Victor Klee: How many guards are needed to see a simple n-gon?

Proofs: Chvatal (induction); Fisk (simple coloring argument)

g(P) = min number of guards for P G(n) = max of g(P), over all n-gons P What is G(n)? Answer: G(n) = floor(n/3)

The Combinatorics of Guarding

In fact, floor(n/3) vertex guards suffice

Chvatal Comb: Necessity of n/3

Guards in Some Cases

Shows that some n-gons require at least n/3 guards, since we can place “independent witness points”, wi , near each tip, and must have a separate guard in each of their visibility regions (triangles) Can extend to cases where n is not a multiple of 3, showing lower bound of floor(n/3). Thus: G(n) ≥ floor(n/3)

w1 w2

Fisk Proof: Floor(n/3) Guards

Suffice: G(n) ≤ floor(n/3)

1. Triangulate P (we know a triangulation exists) 2. 3-color the vertices (of triangulation graph) 3. Place guards at vertices in smallest color class (claim: every point of P is seen, since each

triangle has a guard at a corner, and that guard sees all of the (convex) triangle)

Vertex Guarding a Simple Polygon

Vertex guarding applet

11 yellow vertices 11 blue vertices 16 white vertices Place guards at yellow (or blue) vertices: at most n/3 vertex guards (here, n=38)

Computing g(P) by Inspection

By inspection, find a large set of “visibility independent witness points” within P

If we find w indep witness points, then we know that g(P)≥w

By inspection, find a small set of m guards that see all of P: g(P)≤m

If we are lucky, m=w; otherwise, more arguments are needed!

Lower Bound on g(P)

Fact: If we can place w visibility independent witness points, then g(P) ≥ w.

g(P) 4

Witness Number

Let w(P) = max # of independent witness points possible in a set of visibility independent witness points for P

Then, g(P) ≥ w(P)

Note: It is hard to find g(P); it is poly-time to find w(P)

Some polygons have g(P)=w(P); I call these perfect polygons – they are very special; most polygons P have a “gap”: g(P)>w(P)

Witness Number: Vertex

Guards We say that a set, W, of points inside P

are independent with respect to vertex guards if, for any two points of W, the set of vertices of P they see are disjoint

Let wV(P) = max # of witness points possible in a set of witness points for P that are indep wrt vertex guards

Then, gV(P) ≥ wV(P)

Note: It is hard to find gV(P); a recent (unpublished) algorithm computes wV(P) in polytime.

Useful Polygon Example

“Godfried’s Favorite Polygon”

g(P)=2, but w(P)=1

Useful Polygon Examples

“Godfried’s Favorite Polygon” variations

Examples

For each of these polygons P, find the point guard number, g(P), and the vertex guard number, gV(P). Also, find the witness numbers w(P) and wV(P)

Visibility Graphs

VG of P: (V,E), where E = set of all pairs of vertices of P that see each other, V = vertices

(clear-VG: demand clear visibility: i.e., edge set E is set of all diagonals)

Art Gallery Theorem: Orthogonal

(Rectilinear) Polygons

Polygons with Holes

Art Galley Theorem: floor( (n+h)/3 ) guards suffice and are sometimes necessary

(easy: floor( (n+2h)/3 ) suffice – do you see why?)

Exterior Guarding: Fortress

Problem

Edge Guards

Mirrored Galleries

Mobile Guards

Find shortest route (path or tour) for a mobile guard within P: Watchman route problem

Efficient algorithms for simple polygons P

NP-hard for polygons with holes (as hard as the TSP)

Mobile Robotic Guard Visit all visibility polygons

Watchman Route Problem

Subject to: stay inside polygonal domain P

Guarding Polyhedra

Note: Guards at vertices are NOT enough!

Scissors Congruence in 2D

Dissections

Scissors Congruence in 2D

Scissors Congruence in 2D

Scissors Congruence in 2D

Scissors Congruence: Rectangles

Scissors Congruence: Rectangles

Scissors Congruence: Polygons

Scissors Congruence: Polygons

Scissors Congruence: Polygons

Fair Partitions: Polygons

Scissors Congruence: 3D

Scissors Congruence: 3D