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Introduction Approximation Algorithm for Art Gallery Problem Terrain Guarding Problem General Terrain Guarding Problem

Approximation Algorithms for Art Gallery Problems

Subhas C. Nandy([email protected])

Advanced Computing and Microelectronics UnitIndian Statistical Institute

Kolkata 700108, India.


Outline

1 Introduction

2 Approximation Algorithm for Art Gallery Problem

3 Terrain Guarding Problem

3 General Terrain Guarding Problem


Introduction

Art Gallery Problem:

Given an art gallery (museum) P of a simple polygonal shape, findthe minimum number of guards along with their positions inside Psuch that each point inside P is visible to one of the guards.

Victor Klee proposed this problem in 1976.


Introduction

Chvtal’s Art Gallery Theorem

bn3c guards are always sufficient, andsometimes necessary to illuminate asimple polygon with n vertices.

An worst case example


Introduction

Status of the Problem

The decision version of the art gallery problem is NP-complete.

Our Objective

Designing an approximation algorithm for the art gallery problem.


Introduction

Variations of the Art Gallery Problem

Art gallery problem for rectilinearpolygons

Available Result: Upper bound bn4c

Art gallery problem with holes

Available Result: Upper bound bn+2h3 cfor a simple polygon with h holes and atotal of n vertices.

An example with 24vertices 3 holes and 9guards.


Vertex guard problem, where guards canonly be placed at the vertices of thepolygon.

Edge guard problem, where guards canpatrol along the edges of the polygon.

Objective: Choose minimum numberof edges of the polygon where from theentire polygon is weakly visible.


Approximation Algorithm for Vertex Guard Problem

An important note

Guarding the boundary of the polygon is not enough to guard thewhole polygon.

1

23

4

5 68

910

11

13

14

15

16

17

712

Vertices 7, 12 and 17 together can see the entire boundary of the

polygon but the shaded region is not visible from any of these vertices.


Approximation Algorithm

Definition

A simple polygon is called a fan if thereexists a vertex that is visible from allpoints in the interior of the polygon.

Main Idea of the vertex guard problem

The vertex guard problem can betreated as a polygon decompositionproblem in which each piece is a fan.



Three fans (vertices 1, 4 and 7) arenecessary to cover the polygon if onlyedge extensions are allowed.

1

2

3

4

5 6

7

8 9

1011

12

1314

1516

Whereas two fans (vertices 1 and 7)suffice if we allow the boundary ofconvex components to be bounded bysegments that contains any two verticesof the polygon.

1

2

3

4

5 6

7

8 9

1011

12

1314

1516



Arrangement AJoin every pair of vertices if the linesegment lies inside the polygon.

Extend them in one or both sidestill it (they) hit(s) the boundary ofthe polygon.

Observation:

Every cell in the arrangement A of these line segments insidethe polygon is convex,

Each cell in A is either totally visible or totally invisible from avertex of the polygon.

Each cell in A must lie in at least one of the fanscorresponding to the vertices of the polygon.


Set Cover Problem

Set Cover Problem

Given:

C = {c1, c2, . . . , cn} a set of elements, andS = {S1, S2, . . . ,Sk} be a set of subsets of C

Objective: Choose the minimum number of members in S thatcovers all the elements in C.


Formulation as a Set Cover Problem

Formulation of the Art Gallery Problem

C = The convex cells of the arrangement.S = Set of vertices of the polygon.

Each member Si ∈ S consists of those convex cells that lie in thefan of the corresponding vertex of P.

Choose the minimum number of vertices of P whose fanscover all the cells of the arrangement C.


Results

Approximation Factor: logn

using the existing O(log n) factor approximation algorithm for theset cover problem.

Time Complexity: O(n5)

Number of cells on the arrangement A = O(n4) (worst case)Computing each subset Si needs O(n) time.


Terrain Guarding Problem

Terrain

An x-monotone polychain.

A vertical line intersects the polychainin exactly one point.

The Problem

Find minimum number of points on the boundary of the terrainthat can see all the vertices of the terrain.

Status of the Problem - NP-hard

See J. King and E. Krohn, Terrain Guarding is NP-Hard, SODA-2009.


Rectilinear Terrain

A simpler version of the problem is therectilinear terrain,

Reflex Vertex

Reflex Vertex

(Right)Convex Vertex

(Right)Convex Vertex

(Left) Covex Vertex

Definitions

A terrain T is rectilinear terrain if each edge e ∈ E (T ) is eitherhorizontal or vertical.

A vertex vi of T is convex if the angle formed by the edges ei−1 andei above T is 90o .A vertex vi of T is reflex if the angle formed by the edges ei−1 andei above T is 270o .A convex vertex vi is called left convex if the edge ei−1 is vertical.

A convex vertex vi is called right convex if the edge ei is vertical.


Properties of the Rectilinear Terrain

Let a, b, c and d be four points on aterrain T , such that ax < bx < cx < dx .Now, if a can see c and b can see d ,then a can see d .

a

b

c

d

Let v be a right convex vertex on arectilinear terrain T , and p be a pointon T . If p can see v , then px ≤ vx . V

P



If a set G of points on a rectilinear terrain T guards a subsetV ′ ⊆ V` ∪ Vr , where V` (resp. Vr ) is a subset of left-convex(resp. right convex) vertices of T , then there exists a Ĝ ⊆ Vof reflex vertices, such that Ĝ guards V ′, and |Ĝ | ≤ |G |.

If G ⊆ V guards all the convex vertices of a rectilinear terrainT , then G guards all the vertices of T , and all the verticaledges of T .



Consider a situation where

u, v and w are three right-convexvertices such that ux < vx < wx .

There exists two vertices g1 and g2of the terrain T such that g1 cansee both u and v and g2 can seeboth u and w .

UV

W

g1

g2

then

∃ a vertex g that can see all three vertices u, v and w .Moreover, g is one among g1 and g2(the one that precedes the other in the sequence of verticesdefining T ).


Splitting the Problem

P1: Covering the right-convex vertices, and

P2: Covering the left-convex vertices.

Our Strategy

Compute optimum solution of both the problems P1 and P2.

Report both the solutions.

Result

A 2× OPT solution.


Solving Problem P1

Construct a graph G r

Its vertices are the the right-convex vertices Vr of T .Put an edge between a pair of vertices u, v ∈ Vr if and only ifthere exists a vertex g of the terrain T that can see both uand v .

Results

G r is a chordal graph.

Size of the minimum clique cover of G r is the minimumnumber of vertex guards for covering all the right-convexvertices in T .Minimum clique cover of G r can be computed in polynomialtime.


General Terrain Guarding Problem

Simpler Variation: Left Guarding

We are given two sets of points N and G . Each point in N is to beguarded using only guards from G which are strictly to its left.

Linear programming formulation: LP1

minimize∑

g∈G xgsubject to

∑g∈SL(p) xg ≥ 1 ∀p ∈ N

(* SL(p) is a subset of G where from p ∈ N can be guarded *)xg ∈ {0, 1} ∀g ∈ G

Important Note

In the defnition of left-guarding or right-guarding the guard doesnot see the point on which it lies


Optimal Solution for Left Guarding

Important Property

Any left-visibility matrix is totally balanced. In other words, there

is no submatrix of the form

(1 11 0

).

g1 g2 . . . gnn1n2...

nm

Resulta There is an efficient combinatorial algorithm for finding anoptimal integral solution to (LP1)

aA. Kolen.Location problems on trees and in the rectilinear plane. PhDthesis, Matematisch Centrum, Amsterdam, 1982.


Optimal Algorithm for Left Guarding

left-guarding (T , N, G )

• A← Φ• for p ∈ N do (* process members in N from left to right *)− if p is not yet seen by the existing members in A then− A← A ∪ {L(p)} (* append (left-most member) L(p) to A *)

• Return A

Result

The algorithm can be implemented in O(|N| log |G |) time using aprocedure similar to Graham’s scan


One Sided Guarding

Problem Statement

We are given a set of points N and two sets of guards GL and GR ,where each guard in GL (respectively, GR) can only guard points ofN strictly from its right (respectively, strictly from its left).

Right-Guarding Left-Guarding


One Sided Guarding

Integer programming formulation:LP2minimize

∑g∈GL xg ,L +

∑g∈GR xg ,R

subject to∑

g∈SL(p)∩GL xg ,L +∑

g∈SR(p)∩GR xg ,R ≥ 1, ∀p ∈ N

xg ,L ∈ {0, 1}, ∀g ∈ GL

xg ,R ∈ {0, 1}, ∀g ∈ GR


2-factor approximation Algorithm for One Sided Guarding

Step1 Find an optimal fractional solution x∗ to (LP2)

Step2 Guided by x∗, we divide the points into two sets:NL = {p ∈ N|

∑g∈SL(p)∩GL Xg ,L

∗ ≥ 12}, andNR = {p ∈ N|

∑g∈SR(p)∩GR Xg ,R

∗ ≥ 12}Step3 Optimally Solve the left-guarding problem for the pair

(NL, GL) and the right-guarding problem for the pair (NR , GR)

Step4 Final Solution is the combination of these two.

Result

We get a 2-factor approximation for one sided guarding problem.

Reason: Here, each point of the terrain T is either guarded by amember in GL or a member of GR or both.


4-factor approximation for General Terrain

Observation

• Let G ∗ be an optimum set of guards for a given instance T ,and G ∗L is the optimum solution that uses only left guards tocovers the entire NL. Then

w(G ∗L ) ≤ 2∑g∈GL

x∗g .

Reason: Put xg ,L = 2x∗g for guarding NL. It is a feasible solution

for guarding NL, and its cost is 2∑

g∈GL x∗g


4-factor approximation for General Terrain (contd.)

Similarly,

• If G ∗ be an optimum set of guards for a given instance T , andG ∗R is the optimum solution that uses only left guards tocovers the entire NR . Then

w(G ∗R) ≤ 2∑

g∈GR

x∗g .

• A feasible solution to the one-sided discrete version withGL = GR = G , also constitutes a feasible solution to thegeneral problem.

Result

So if we compute the one sided versions with GL = GR = G , thenthat would lead to a 4-factor approximation result.


Thanks

IntroductionApproximation Algorithm for Art Gallery ProblemTerrain Guarding ProblemGeneral Terrain Guarding Problem

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